Solution Manual for Concepts of Genetics 4th Edition by Robert Brooker.

Solution Manual for Concepts of Genetics 4th Edition By RobertBrooker

CONCEPTS OF GENETICS, 4/e ANSWERS TO PROBLEM SETS Chapters 1-24

CHAPTER 1 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 1. 1 Understanding our genes may help with diagnoses of inherited diseases. It may also lead to the development of drugs to combat diseases. Other answers are possible.

FIGURE 1. 2 Many ethical issues are associated with human cloning. Is it the wrong thing to do? Does it conflict an individual’s religious views? And so on.

FIGURE 1. 3 Because females mate only once, sorting out the male mosquitoes and releasing sterile males into the environment can limit mosquito reproduction.

FIGURE 1. 4 DNA is a macromolecule.

FIGURE 1. 5 DNA and proteins are found in chromosomes. A small amount of RNA may also be associated with chromosomes when transcription is occurring, and as discussed in Chapter 18, some non-coding RNAs may bind to chromosomes.

FIGURE 1.6 The information to make a polypeptide is stored in DNA. FIGURE 1. 7 The dark-colored butterfly has a more active pigment-producing enzyme.

FIGURE 1. 8 Genetic variation is the reason the frogs look different.

FIGURE 1. 9 These are examples of variation in chromosome number.

FIGURE 1. 10 If this girl had been given a standard diet, she would have developed the harmful symptoms of PKU, which include mental impairment and foul-smelling urine.

FIGURE 1. 11 A corn gamete contains 10 chromosomes. (The leaf cells are diploid.)

FIGURE 1. 12 The horse populations have become adapted to their environment, which has changed over the course of many years.

FIGURE 1.13 There are several possible examples of other model organisms, including rats and frogs.

End-of-chapter Questions: Conceptual Questions C1. A chromosome is a very long polymer of DNA. A gene is a specific sequence of DNA within that polymer; the sequence of bases creates a gene and distinguishes it from other genes. Genes are located in chromosomes, which are found within living cells.

C2. At the molecular level, a gene (a sequence of DNA) is first transcribed into RNA. The genetic code within the RNA is used to synthesize a protein with a particular amino acid sequence. This second process is called translation.

C3. A. Molecular level. This is a description of a how an allele affects protein function.

B. Cellular level. This is a description of how protein function affects cell structure.

C. Population level. This is a description of how the two alleles affect members of a population.

D. Organism level. This is a description of how the alleles affect the traits of an individual.

C4. Genetic variation is the occurrence of genetic differences within members of the same species or different species. Within any population, variation may occur in the genetic material. Variation may occur in particular genes, so some individuals carry one allele and other individuals carry a different allele. Examples include differences in coat color among mammals or flower color in plants. At the molecular level, this type of genetic variation is caused by changes in the DNA sequences of genes.

There may also be variation in chromosome structure and number.

C5. An extra chromosome (specifically an extra copy of chromosome 21) causes Down syndrome.

C6. You can pick almost any trait. For example, flower color in petunias would be an interesting choice.

Some petunias are red and others are purple. There must be different alleles in a flower color gene that affect this trait in petunias. In addition, the amount of sunlight, fertilizer, and water also affects the intensity of flower color.

C7. The term diploid means that a cell has two copies of each type of chromosome. In humans, nearly all of the cells are diploid except for gametes (i.e., sperm and egg cells). Gametes usually have only one set of chromosomes.

C8. A DNA sequence is a sequence of nucleotides. Each nucleotide may have one of four different bases (i.e., A, T, G, or C). When speaking of a DNA sequence, the focus is on the sequence of those bases.

C9. The genetic code is the way in which the sequence of bases in RNA is read to produce a sequence of amino acids within a protein.

C10. A. A gene is a segment of DNA. For most genes, the expression of the gene results in the production of a polypeptide, which is a unit of a protein. The functioning of proteins within living cells largely determines the traits of an organism.

B. A gene is a segment of DNA that usually encodes the information for the production of a specific polypeptide. Genes are found within chromosomes. Many genes are found within a single chromosome.

C. An allele is an alternative version of a particular gene. For example, suppose a plant has a flower color gene. One allele could produce a white flower, while a different allele could produce an orange flower. The white allele and orange allele are alleles of the flower color gene.

D. A DNA sequence is a sequence of bases, which are found within nucleotides. The information within a DNA sequence (which is transcribed into an RNA sequence) specifies the amino acid sequence within a polypeptide.

C11. The statement in part A is not correct. Individuals do not evolve. Populations evolve because certain individuals are more likely to survive and reproduce and pass their genes to succeeding generations.

C12. A. How genes and traits are transmitted from parents to offspring.

B. How the genetic material functions at the molecular and cellular levels.

C. Why genetic variation exists in populations, and how it changes over the course of many generations.

Application and Experimental Questions E1. There are many possible answers. Some common areas to discuss might involve the impact of genetics in the production of new medicines, the diagnosis of diseases, the production of new kinds of food, and the use of DNA fingerprinting to solve crimes.

E2. A genetic cross involves breeding two different individuals.

E3. This would be used to a great extent by molecular geneticists. The sequence of DNA is a molecular characteristic of DNA. In addition, the sequence of DNA is interesting to transmission and population geneticists as well.

E4. You would see 47 chromosomes instead of 46. There would be three copies of chromosome 21 instead of two copies.

E5. A. Transmission geneticists. Dog breeders are interested in how genetic crosses affect the traits of dogs.

B. Molecular geneticists. This is a good model organism to study genetics at the molecular level.

C. Both transmission geneticists and molecular geneticists. Fruit flies are easy to cross and study the transmission of genes and traits from parents to offspring. Molecular geneticists have also studied many genes in fruit flies to see how they function at the molecular level.

D. Population geneticists. Most wild animals and plants would be the subject of population geneticists. In the wild, you cannot make controlled crosses. But you can study genetic variation within populations and try to understand its relationship to the environment.

E. Transmission geneticists. Agricultural breeders are interested in how genetic crosses affect the outcome of traits.

E6. You need to follow the scientific method. You can take a look at an experiment in another chapter to see how the scientific method is followed.

CHAPTER 2 Note: the answers to Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 2. 1 Compartmentalization means that cells have membrane-bound compartments.

FIGURE 2. 2 The chromosomes would not be spread out very well, and would probably be overlapping. It would be difficult to see individual chromosomes.

FIGURE 2. 3 Homologs are similar in size, banding pattern, and carry the same types of genes. However, the alleles of a given gene may be different.

FIGURE 2. 4 FtsZ assembles into a ring at the future site of the septum and recruits to that site other proteins that produce a cell wall between the two daughter cells.

FIGURE 2. 5 In the G1 phase of the cell cycle, a cell may be preparing to divide. By comparison, the G 0 phase is a phase in which a cell is either not advancing through the cell cycle or has committed to never divide again.

FIGURE 2. 6 Homologs are genetically similar; one is inherited from the mother and the other from the father. By comparison, chromatids are the product of DNA replication. The chromatids within a pair of sister chromatids are genetically identical.

FIGURE 2 .7 One end of a kinetochore microtubule is attached to a kinetochore on a chromosome. The other end is within the centrosome.

FIGURE 2. 8 Anaphase FIGURE 2. 9 Ingression occurs because myosin motor proteins shorten the contractile ring, which is formed from actin proteins.

FIGURE 2. 10 The end result of crossing over is that homologous chromosomes have exchanged pieces.

FIGURE 2. 11 The cells at the end of meiosis are haploid, whereas the mother cell is diploid.

FIGURE 2. 12 In metaphase of mitosis, each pair of sister chromatids is attached to both poles, whereas in metaphase of meiosis I, each pair of sister chromatids is attached to just one pole.

FIGURE 2. 13 Polar bodies are small cells that are produced during oogenesis and then degenerate.

FIGURE 2. 14 All of the nuclei in the embryo sac are haploid. The central cell has two haploid nuclei, and all of the other cells, including the egg, have just one.

End-of-chapter Questions: Conceptual Questions C1. They are genetically identical, barring rare mutations, because they receive identical copies of the genetic material from the mother cell.

C2. A homolog is one of the members of a chromosome pair. Homologs are usually the same size and carry the same types and order of genes. They may differ in that the genes they carry may be different alleles.

C3. Sister chromatids are identical copies derived from the replication of a chromosome. They remain attached to each other at the centromere. They are genetically identical, barring rare mutations and crossing over with homologous chromosomes.

C4. Metaphase is the organization phase, and anaphase is the separation phase.

C5. G1, there should be six linear chromosomes. In G2, there should be 12 chromatids that are attached to each other in pairs of sister chromatids.

C6. In metaphase of meiosis I, each pair of chromatids is attached to only one pole via the kinetochore microtubules. In metaphase of mitosis, there are two attachments (i.e., to both poles). If the attachment is lost, a chromosome will not migrate to a pole and may not become enclosed in a nuclear membrane after telophase. If left out in the cytosol, it would eventually be degraded.

C7. A. During mitosis and meiosis II B. During meiosis I C. During mitosis, meiosis I, and meiosis II D. During mitosis and meiosis II C8. The reduction occurs because there is a single DNA replication event but two cell divisions.

Because of the nature of separation during anaphase of meiosis I, each cell receives one copy of each type of chromosome.

C9.

C10. It means that the maternally derived and paternally derived chromosomes are randomly aligned along the metaphase plate during metaphase of meiosis I.

C11. Mitosis—two diploid cells containing 10 chromosomes each (two complete sets). Meiosis—four haploid cells containing 5 chromosomes each (one complete set) C12. The number of different, random alignments equals 2 n, where n equals the number of chromosomes per set. In this case, there are three per set, so the possible number of arrangements equals 23, which is 8.

C13. (1/2)n = (1/2)4 = 1/16 or 6.25% C14. The probability would be much lower because pieces of maternal chromosomes would be incorporated into the paternal chromosomes. Therefore, a gamete would be unlikely to carry a chromosome that was completely paternally derived.

C15. Bacteria do not need to sort their chromosomes because they only have one type of chromosome.

Though not discussed in the text, the attachment of the two copies of the chromosomes to the cell membrane prior to cell division also helps to ensure that each daughter cell receives one copy.

C16. During interphase, the chromosomes are greatly extended. In this conformation, they might get tangled up with each other and not sort properly during meiosis and mitosis. The condensation process probably occurs so that the chromosomes easily align along the equatorial plate during metaphase without getting tangled up.

C17. To produce identical quadruplets, fertilization begins with one sperm and one egg cell. This fertilized egg then could divide twice by mitosis to produce four genetically identical cells. These four cells could then separate from each other to begin the lives of four distinct individuals. Another possibility is that mitosis could produce two cells that separate from each other. These two cells could then divide by mitosis to produce two pairs of cells, which also could separate to produce four individual cells.

C18. During prophase of meiosis II, your drawing should show four replicated chromosomes (i.e., four structures that look like Xs). Each chromosome is one homolog. During prophase of mitosis, there should be eight replicated chromosomes (i.e., eight Xs). During prophase of mitosis, there are pairs of homologs. The main difference is that prophase of meiosis II has a single copy of each of the four chromosomes, whereas prophase of mitosis has four pairs of homologs. At the end of meiosis I, each daughter cell has received only one copy of a homologous pair, not both. This is due to the alignment of homologs during metaphase of meiosis I and their separation during anaphase of meiosis I. C19. The products of meiosis have only one copy of each type of chromosome. For example, one human gamete may contain the paternally derived copy of chromosome 11, whereas a different gamete may contain the maternally derived copy of chromosome 11. These two homologs may carry different alleles of the same genes and therefore are not identical. In contrast, mitosis produces genetically identical daughter cells that have both copies of all the pairs of homologous chromosomes.

C20. DNA replication does not take place during interphase II. The chromosomes at the end of telophase of meiosis I have already replicated (i.e., they are found in pairs of sister chromatids). During meiosis II, the sister chromatids separate from each other, yielding individual chromosomes.

C21.

Prophase/Prometaphase Telophase Nuclear membrane dissociates. Nuclear membrane re-forms.

Mitotic spindle forms.

Mitotic spindle disassembles.

Chromosomes condense.

Chromosomes decondense.

Chromosomes attach to spindle. Chromosomes detach from the spindle.

C22. A. 20 B. 10 C. 30 D. 20 C23. The hybrid offspring would have 44 chromosomes (i.e., 25 + 19). The reason for infertility is because each chromosome does not have a homologous partner. Therefore, the chromosomes cannot properly pair during metaphase of meiosis I, and the gametes do not receive one copy of each homolog. Gametes will be missing certain chromosomes, which makes them infertile.

C24. Male gametes are usually small and mobile. Animal and some plant male gametes have flagella, which make them motile. The mobility of the male gamete makes it likely that it will come in contact with the female gamete. Female gametes are usually much larger and contain nutrients to help the growth of the embryo after fertilization occurs.

C25. To produce sperm, a spermatogonial cell first goes through mitosis to produce two cells. One of these remains a spermatogonial cell and the other advances through meiosis. In this way, the testes continue to maintain a population of spermatogonial cells.

C26. During oogenesis in humans, the cells are arrested in prophase of meiosis I for many years until selected primary oocytes advance through the rest of meiosis I and begin meiosis II. If fertilization occurs, meiosis II is completed.

C27. There is a 1/2 chance that the mother will transmit her abnormal chromosome and a 1/2 chance that the father will. You use the product rule to calculate the chances of both outcomes happening. So the answer is 1/2 × 1/2 = 1/4, or 25%. The probability that such a child will pass both chromosomes to an offspring is also 25% because that child had a 1/2 chance of passing either chromosome.

Application and Experimental Questions E1. A. G2 phase (it could not complete prophase) B. Metaphase (it could not enter anaphase) C. Telophase (it could not divide into two daughter cells) D. G2 phase (it could not enter prophase) E2. During interphase, the chromosomes are longer, thinner, and much harder to see. In metaphase, they are highly condensed, which makes them thicker and shorter.

E3. You could karyotype other members of the family and see if affected members always carry the abnormal chromosome.

Questions for Student Discussion/Collaboration 1. It’s not possible to give a direct answer, but the point is for students to be able to draw chromosomes in different configurations and understand the various phases. The chromosomes may or may not be: 1. In homologous pairs 2. Connected as sister chromatids 3. Associated in bivalents 4. Lined up in metaphase

5. Moving toward the poles.

2. A major advantage of sexual reproduction is that it fosters genetic diversity in future populations. A major disadvantage is that to reproduce, each individual must find a mate of the opposite sex.

CHAPTER 3 Note: the answers to Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 3. 2 The male gamete is found within pollen.

FIGURE 3. 3 The white flower is providing the sperm and the purple flower is providing the eggs.

FIGURE 3. 4 A true-breeding strain maintains the same trait over the course of many generations.

FIGURE 3. 6 Segregation means that the T and t alleles separate from each other so that a haploid cell receives one of them, but not both.

FIGURE 3. 7 In this hypothesis, two different genes are linked. The alleles of the same gene are not linked.

FIGURE 3. 9 Independent assortment allows for new combinations of alleles among different genes to be found in future generations of offspring.

FIGURE 3. 10 Such a parent could make two types of gametes, Ty and ty, in equal proportions.

FIGURE 3. 12 Homologous chromosomes separate at anaphase of meiosis I. FIGURE 3. 13 chromosomes could line up in four different ways.

FIGURE 3. 14 Horizontal lines connect two individuals that have offspring together, and they connect all of the offspring that produced by the same two parents.

End-of-chapter Questions: Conceptual Questions C1. Mendel’s work showed that genetic determinants are inherited in a dominant/recessive manner. This was readily apparent in many of his crosses. For example, when he crossed two true-breeding plants for a trait such as height (i.e., tall versus dwarf), all the F1 plants were tall. This is inconsistent with blending. Perhaps more striking was the result obtained in the F2 generation: 3/4 of the offspring were tall and 1/4 were short. In other words, the F2 generation displayed phenotypes that were like the parental generation. There did not appear to be a blending to create an intermediate phenotype.

Instead, the genetic determinants did not seem to change from one generation to the next.

C2. In plants, cross-fertilization occurs when the pollen and eggs come from different plants; in selffertilization they come from the same plant.

C3. The genotype is the type of genes that an individual inherits while the phenotype is the individual’s observable traits. Tall pea plants, red hair in humans, and vestigial wings in fruit flies are phenotypes. Homozygous, TT, in pea plants; a heterozygous carrier of the cystic fibrosis allele; and homozygotes for the cystic fibrosis allele are descriptions of genotypes. It is possible to have different genotypes and the same phenotype. For example, a pea plant that is TT or Tt would both have a tall phenotype.

C4. A true-breeding organism is homozygote that has two copies of the same allele.

C5. Conduct a cross in which the unknown individual is mated to an individual that carries only recessive alleles for the gene in question.

C6. Diploid organisms contain two copies of each type of gene. When they make gametes, only one copy of each gene is found in a gamete. Two alleles cannot stay together within the same gamete.

C7. B. This statement is not correct because these are alleles of different genes.

C8. Genotypes: 1 Tt : 1 tt Phenotypes: 1 tall : 1 dwarf C9. The recessive phenotype must be a homozygote. The dominant phenotype could be either homozygous or heterozygous.

C10. In this cross, c is the recessive allele for constricted pods; Y is the dominant allele for yellow color.

The cross is ccYy × CcYy. Follow the directions for setting up a Punnett square, as described in Section 3.3. The genotypic ratio is 2 CcYY : 4 CcYy : 2 Ccyy : 2 ccYY : 4 ccYy : 2 ccyy. This 2:4:2:2:4:2 ratio can be reduced to a 1:2:1:1:2:1 ratio.

The phenotypic ratio is 6 smooth pods, yellow seeds : 2 smooth pods, green seeds: 6 constricted pods, yellow seeds : 2 constricted pods, green seeds. This 6:2:6:2 ratio could be reduced to a 3:1:3:1 ratio.

C11. The genotypes are 1 YY : 2 Yy : 1 yy. The phenotypes are 3 yellow : 1 green.

C12. Offspring with a nonparental phenotype are consistent with the idea of independent assortment. If two different traits were always transmitted together as unit, it would not be possible to get nonparental phenotypic combinations. For example, if a true-breeding parent had two dominant traits and was crossed to a true-breeding parent having the two recessive traits, the F2 generation could not have offspring with one recessive and one dominant trait. However, because independent assortment can occur, it is possible for F2 offspring to have one dominant and one recessive trait.

C13. (a) It behaves like a recessive trait because unaffected parents sometimes produce affected offspring. In such cases, the unaffected parents are heterozygous carriers. (b) It behaves like a dominant trait. An affected offspring always has an affected parent. However, recessive inheritance cannot be ruled out.

C14. A. Barring a new mutation during gamete formation, the probability is 100%. They must be heterozygotes in order to produce a child with a recessive disorder.

B. Construct a Punnett square. There is a 50% chance of heterozygous offspring.

C. Use the product rule. The chance of being phenotypically unaffected is 0.75 (i.e., 75%), so the answer is 0.75 × 0.75 × 0.75 = 0.422, which is 42.2%.

D. Use the binomial expansion equation, where n = 3, x = 2, p = 0.75, q = 0.25. The answer is 0.422, or 42.2%.

C15. A. 100% because they are genetically identical.

B. Construct a Punnett square. You know the parents are heterozygotes because they produced a blue-eyed child. The fraternal twin is not genetically identical, but she or he has the same parents as its twin. The answer is 25%.

C. The probability that an offspring inherits the allele is 50% and the probability that this offspring will pass it on to his/her offspring is also 50%. You use the product rule: (0.5)(0.5) = 0.25, or 25%.

D. Barring a new mutation during gamete formation, the chance is 100% because they must be heterozygotes in order to produce a child with blue eyes.

C16. First construct a Punnett square. The chances are 75% of producing a solid pup and 25% of producing a spotted pup.

A. Use the binomial expansion equation, where n = 5, x = 4, p = 0.75, q = 0.25. The answer is 0.396, or 39.6%.

B. You can use the binomial expansion equation for each litter. For the first litter, n = 6, x = 4, p = 0.75, q = 0.25; for the second litter, n = 5, x = 5, p = 0.75, q = 0.25. Because the litters are in a specified order, you use the product rule and multiply the probability of the first litter times the probability of the second litter. The answer is 0.070, or 7.0%.

C. To calculate the probability of the first litter, you use the product rule and multiply the probability of the first pup (0.75) times the probability of the remaining four. You use the binomial expansion equation to calculate the probability of the remaining four, where n = 4, x = 3, p = 0.75, q = 0.25. The probability of the first litter is 0.316. To calculate the probability of the second litter, you use the product rule and multiply the probability of the first pup (0.25) times the probability of the second pup (0.25) times the probability of the remaining five. To calculate the probability of the remaining five, you use the binomial expansion equation, where n = 5, x = 4, p = 0.75, q = 0.25. The probability of the second litter is 0.025. To get the probability of these two litters occurring in this order, you use the product rule and multiply the probability of the first litter (0.316) times the probability of the second litter (0.025). The answer is 0.008, or 0.8%.

D. Because this is a specified order, you use the product rule and multiply the probability of the firstborn (0.75) times the probability of the second born (0.25) times the probability of the remaining four. You use the binomial expansion equation to calculate the probability of the remaining four pups, where n = 4, x = 2, p = 0.75, q = 0.25. The answer is 0.040, or 4.0%.

C17. If B is the black allele, and b is the white allele, the male is bb, the first female is probably BB, and the second female is Bb. You are uncertain of the genotype of the first female. She could be Bb, although it is unlikely because she didn’t produce any white pups out of a litter of eight.

C18. A. Use the product rule: (1/4)(1/4)=1/16 B. Use the binomial expansion equation: n = 4, p = ¼, q = ¾, x = 2 P = 0.21, or 21% C. Use the product rule: (1/4)(3/4)(3/4) = 0.14, or 14% C19. The parents must be heterozygotes, so the probability is 1/4.

C20. A. 1/4 B. 1, or 100%

C. (3/4)(3/4)(3/4) = 27/64 = 0.42, or 42% D. Use the binomial expansion equation, where n = 7, p = 3/4, q = 1/4, x = 3 P = 0.058, or 5.8% E. The probability that the first plant is tall is 3/4. To calculate the probability that among the next four, any two will be tall, you use the binomial expansion equation, where n = 4, p = 3/4, q = 1/4, x = 2.

The probability P equals 0.21.

To calculate the overall probability of these two outcomes: (3/4)(0.21) = 0.16, or 16% C21. A. T Y R, T y R, T Y r, T y r B. T Y r, t Y r C. T Y R, T Y r, T y R, T y r, t Y R, t Y r, t y R, t y r D. t Y r, t y r C22. The gamete violates the law of segregation because two copies of one gene are in the gamete. The two alleles for the A gene did not segregate from each other.

C23. It is recessive inheritance. The pedigree is shown here. Affected individuals are shown with filled (black) symbols.

The mode of inheritance appears to be recessive. Unaffected parents (who must be heterozygous) produce affected children.

C24. Based on this pedigree provided, the disease is likely to be a dominant trait because an affected child always has an affected parent. In fact, Marfan syndrome is a dominant disorder.

C25. A. 3/16 B. You can either use the binomial expansion equation where n = 3, x = 3, and p = 9/16 or you could use the product rule. The answer is 0.18.

C. Use the multinomial expansion equation, where n = 5, a = 2, b = 1, c = 2, p = 9/16, q = 3/16, and r = 1/16. The answer is 0.007.

D. Another way of looking at this is that the probability it will have round, yellow seeds is 9/16.

Therefore, the probability that it will not is 1 – 9/16 = 7/16.

C26. It is impossible for the F1 individuals to be true-breeding because they are all heterozygotes.

C27. This problem is a bit unwieldy, but you can solve it using the multiplication method.

For height, the ratio is 3 tall : 1 dwarf.

For seed texture, the ratio is 1 round : 1 wrinkled.

For seed color, they are all yellow.

For flower location, the ratio is 3 axial : 1 terminal.

Thus, the product is (3 tall + 1 dwarf)(1 round + 1 wrinkled)(1 yellow)(3 axial + 1 terminal) Multiplying this out, the answer is 9 tall, round, yellow, axial 9 tall, wrinkled, yellow, axial 3 tall, round, yellow, terminal 3 tall, wrinkled, yellow, terminal 3 dwarf, round, yellow, axial 3 dwarf, wrinkled, yellow, axial 1 dwarf, round, yellow, terminal 1 dwarf, wrinkled, yellow, terminal C28. 2 TY, tY, 2 Ty, ty, TTY, TTy, 2 TtY, 2 Tty This question is a bit tricky, but you get 2 TY and 2 Ty because either of the two T alleles could combine with Y or y. Also, you get 2 TtY and 2 Tty because either of the two T alleles could combine with t and then combine with Y or y.

C29. The drone is sB and the queen is SsBb. According to the laws of segregation and independent assortment, the male can make only sB gametes, while the queen can make SB, Sb, sB, and sb, in equal proportions. Therefore, male offspring will be SB, Sb, sB, and sb, and female offspring will be SsBB, SsBb, ssBB, and ssBb. The phenotypic ratios, assuming an equal number of males and females, will be: Males Females 1 normal wings/black eyes 2 normal wings, black eyes 1 normal wings/white eyes 2 short wings, black eyes 1 short wings/black eyes 1 short wings/white eyes C30. The genotype of the F1 plants is Tt Yy Rr. According to the laws of segregation and independent assortment, the alleles of each gene will segregate from each other, and the alleles of different genes will randomly assort into gametes. A Tt Yy Rr individual could make eight types of gametes: TYR, TyR, Tyr, TYr, tYR, tyR, tYr, and tyr, in equal proportions (i.e., 1/8 of the gametes will be of each type). To determine genotypes and phenotypes, you could make a large Punnett square that contains 64 boxes. You would need to line up the eight possible gametes across the top and along the side, and then fill in the 64 boxes. Alternatively, you could use the multiplication method or the forkedline method. The genotypes and phenotypes are as follows: 1 TT YY RR 2 TT Yy RR 2 TT YY Rr 2 Tt YY RR 4 TT Yy Rr

4 Tt Yy RR 4 Tt YY Rr 8 Tt Yy Rr = 27 tall, yellow, round 1 TT yy RR 2 Tt yy RR 2 TT yy Rr 4 Tt yy Rr = 9 tall, green, round 1 TT YY rr 2 TT Yy rr 2 Tt YY rr 4 Tt Yy rr = 9 tall, yellow, wrinkled 1 tt YY RR 2 tt Yy RR 2 tt YY Rr 4 tt Yy Rr = 9 dwarf, yellow, round 1 TT yy rr 2 Tt yy rr = 3 tall, green, wrinkled 1 tt yy RR 2 tt yy Rr = 3 dwarf, green, round 1 tt YY rr 2 tt Yy rr = 3 dwarf, yellow, wrinkled 1 tt yy rr = 1 dwarf, green, wrinkled C31. Construct a Punnett square to determine the probability of these three phenotypes. The probabilities are 9/16 for round, yellow; 3/16 for round, green; and 1/16 for wrinkled, green. Use the multinomial expansion, where n = 5, a = 2, b = 1, c = 2, p = 9/16, q = 3/16, r = 1/16. The answer is 0.007, or 0.7%, of the time.

C32. The woolly haired male is a heterozygote, because he has the trait and his mother did not. (He must have inherited the normal allele from his mother.) Therefore, he has a 50% chance of passing the woolly allele to his offspring; his offspring have a 50% of passing the allele to their offspring; and these grandchildren have a 50% chance of passing the allele to their offspring (the woolly haired man’s great-grandchildren). Because this is an ordered sequence of independent outcomes, you use the product rule: 0.5 × 0.5 × 0.5 = 0.125, or 12.5%. Because no other Scandinavians are on the island, the chance is 87.5% for the offspring being normal (because they could not inherit the woolly hair allele from anyone else). You use the binomial expansion equation to determine the likelihood that one out of eight great-grandchildren will have woolly hair, where n = 8, x = 1, p = 0.125, q = 0.875. The answer is 0.393, or 39.3%, of the time.

C33. A. Construct a Punnett square. Because it is a rare disease, you would assume that the mother is a heterozygote and the father is normal. The chances are 50% that the man in his thirties will have the allele.

B. Use the product rule: 0.5 (chance that the man has the allele) times 0.5 (chance that he will pass it to his offspring), which equals 0.25, or 25%.

C. You use the binomial expansion equation. From part B, you calculated that the probability of an affected child is 0.25. Therefore, the probability of an unaffected child is 0.75. For the binomial expansion equation, n = 3, x = 1, p = 0.25, q = 0.75. The answer is 0.422 or 42.2%.

C34. Use the product rule. If the woman is heterozygous, there is a 50% chance of having an affected offspring: (0.5)7 = 0.0078, or 0.78%, of the time. This is a pretty small probability. If the woman has an eighth child who is unaffected, however, she has to be a heterozygote, because it is a dominant trait. She would have to pass a normal allele to an unaffected offspring. The answer is 100%.

Application and Experimental Questions E1. Pea plants are relatively small and hardy. They produce both pollen and eggs within the same flower. Because a keel covers the flower, self-fertilization is quite easy. In addition, crossfertilization is possible by the simple manipulation of removing the anthers in an immature flower and later placing pollen from another plant. Finally, peas exist in several variants.

E2. The difference lies in where the pollen comes from. In self-fertilization, the pollen and eggs come from the same plant. In cross-fertilization, they come from different plants.

E3. Two generations would take two growing seasons. About 1 and 1/2 years.

E4. According to Mendel’s law of segregation, the genotypic ratio should be 1 homozygote dominant : 2 heterozygotes : 1 homozygote recessive. The data table considers only the plants with a dominant phenotype. The genotypic ratio should be 1 homozygote dominant : 2 heterozygotes. The homozygote dominants would be true-breeding while the heterozygotes would not be true-breeding.

This 1:2 ratio is very close to what Mendel observed.

E5. In a single-factor experiment, the experimenter is only concerned with the outcome of a single trait.

In a two-factor experiment, the experimenter follows the pattern of inheritance for two different traits.

E6. All three offspring had black fur. The ovaries from the albino female could only produce eggs with the dominant black allele (because they were obtained from a true-breeding black female). The actual phenotype of the albino mother does not matter. Therefore, all offspring were heterozygotes (Bb) with black coats.

E7. The data are consistent with two genes (let’s call them gene 22 and gene 24) that exist in two alleles each, a susceptible allele and a resistant allele. The observed data approximate a 9:3:3:1 ratio. This is the expected ratio if two genes are involved, and if resistance is dominant to susceptibility.

E8. If you construct a Punnett square according to Mendel’s laws, you expect a 9:3:3:1 ratio. Because a total of 556 offspring were observed, the expected number of offspring are 556 × 9/16 = 313 round, yellow 556 × 3/16 = 104 wrinkled, yellow 556 × 3/16 = 104 round, green 556 × 1/16 = 35 wrinkled, green If you plug the observed and expected values into the chi square equation, you get a value of 0.51.

With four categories, the degrees of freedom are n – 1, or 3. If you look up the value of 0.51 in the chi square table (see Table 3.2), you see that it falls between the P values of 0.80 and 0.95. This means that the probability is a 80% to 95% that a value equal to or greater than 0.51 is expected to occur due to random sampling error. Therefore, you accept the hypothesis. In other words, the results are consistent with the law of independent assortment.

E9. No, the law of independent assortment applies to transmission patterns of two or more genes. In a single-factor experiment, you are monitoring only the transmission pattern of a single gene.

E10. A. Let c+ represent straight wings and c represent curved wings, and e+ represents gray body and e represents ebony body: Parental cross: cc e+e+  c+c+ee.

F1 generation is heterozygous: c+c e+e An F1 offspring crossed to a fly with curved wings and ebony body is c+c e+e  ccee The F2 offspring would have this genotypic ratio: c+c e+e : c+cee : cc e+e : ccee B. The phenotypic ratio of the F2 flies is 1:1:1:1, as follows: straight wings, gray body : straight wings, ebony bodies : curved wings, gray bodies : curved wings, ebony bodies C. From part B, you expect 1/4 of each category. There are a total of 444 offspring. The expected number of each category is 1/4  444, which equals 111.  2  (114 111)  (105 111)  (111 111)  (114 111) 2

111

2

2

111

111

2

111

  0.49 With 3 degrees of freedom, a value of 0.49 or greater is likely to occur between 80% and 95% of the time. Therefore, you accept our hypothesis.

E11. You would expect a ratio of 3 normal : 1 long neck. In other words, there should be 1/3 as many long-necked mice as normal mice. If you multiply 522 times 1/3, the expected value is 174.

However, only 62 were observed. Therefore, it appears that 174 – 62, or 112, mice died during early embryonic development; 112 divided by 174 gives us the percentage that died, which equals 0.644, or 64.4%.

E12. Follow the basic chi square strategy. You expect a 3:1 ratio, or 3/4 of the dominant phenotype and 1/4 of the recessive phenotype.

The observed and expected values are as follows (rounded to the nearest whole number): 2

Observed*

Expected

5,474 1,850 6,022 2,001 705 224 882 299 428 152

5,493 1,831 6,017 2,006 697 232 886 295 435 145

(O  E)

2

E

0.066 0.197 0.004 0.012 0.092 0.276 0.018 0.054 0.113 0.338

651 207 787 277

644 215 798 266

0.076 0.298 0.152 0.455 2  = 2.15 *Due to rounding, the observed and expected values may not add up to precisely the same number.

Because n = 14, there are 13 degrees of freedom. If you look up this value in the chi square table, you have to look between 10 and 15 degrees of freedom. In either case, you expect a value of 2.15 or greater to occur more than 99% of the time. Therefore, you accept the hypothesis.

E13. This means that a deviation value of 1.005 or greater (between the observed and expected data) would occur 80% of the time. In other words, it is fairly likely to obtain this value due to random sampling error. Therefore, you accept our hypothesis.

E14. The dwarf parent with terminal flowers must be homozygous for both genes, because it is expressing these two recessive traits: ttaa, where t is the recessive dwarf allele, and a is the recessive allele for terminal flowers. The phenotype of the other parent is dominant for both traits.

Because this parent was able to produce dwarf offspring with axial flowers, it must have been heterozygous for both genes: TtAa. E15. Your hypothesis is that disease sensitivity and herbicide resistance are dominant traits and they are governed by two genes that assort independently. According to this hypothesis, the F2 generation should yield a ratio of 9 disease sensitive, herbicide resistant : 3 disease sensitive, herbicide sensitive : 3 disease resistant, herbicide resistant : 1 disease resistant, herbicide sensitive. Because there are a total of 288 offspring produced, the expected numbers would be 9/16  288 = 162 disease sensitive, herbicide resistant 3/16  288 = 54 disease sensitive, herbicide sensitive 3/16  288 = 54 disease resistant, herbicide resistant 1/16  288 = 18 disease resistant, herbicide sensitive  2  (157 162)  (57  54)  (54  54)  (20 18) 2

162

2

54

2

54

2

18

 2  0.54

If you look up this value in the chi square table under 3 degrees of freedom, the value lies between the 0.95 and 0.80 probability values. Therefore, you expect a value equal to or greater than 0.54, at least 80% of the time, due to random sampling error. Therefore, you accept the hypothesis.

E16. You need to make crosses to understand the pattern of inheritance of traits (determined by genes) from parents to offspring. And you need to microscopically examine cells to understand the pattern of transmission of chromosomes. The correlation between the pattern of transmission of chromosomes during meiosis, and Mendel’s laws of segregation and independent assortment, is what led to the chromosome theory of inheritance.

Questions for Student Discussion/Collaboration 1. The methods for making a Punnett square and using the multiplication method are described in the chapter. Presumably, the multiplication method (or the forked-line method) will work a lot faster.

2.

3.

If you construct a Punnett square, the following probabilities will be obtained: tall with axial flowers: 3/8 dwarf with terminal flowers: 1/8 The probability of being tall with axial flowers or dwarf with terminal flowers: 3/8 + 1/8 = 4/8 = 1/2 You use the product rule to calculate the probability of the ordered outcome of the first three offspring being tall/axial or dwarf/terminal, and fourth offspring being tall/axial: (1/2)(1/2)(1/2)(3/8) = 3/64 = 0.047 = 4.7% Ignore the other genes. The cross is Rr × RR. All of the offspring will have round seeds. The probability is 100%

CHAPTER 4 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 4. 1 In the X-Y system, the presence of the Y chromosome causes maleness, whereas in the X-0 system, it is the ratio between the number of X chromosomes and number of sets of autosomes that determines sex. A ratio of 0.5 results in a male and a ratio of 1.0 produces a female.

FIGURE 4. 2 A male bee is not produced by sexual reproduction because it is produced from an unfertilized egg.

Sexual reproduction involves the union of gametes.

FIGURE 4. 3 A higher average temperature would favor a high percentage of male alligators. Assuming that males are not limiting for alligator reproduction, fewer females would be expected to decrease the population size of alligators.

FIGURE 4. 4 The sporophytes are the opposite sexes in dioecious plants.

FIGURE 4. 5 The Barr body is more brightly stained because it is very compact.

FIGURE 4. 6 X-chromosome inactivation first occurs during embryonic development. It is then maintained into adulthood.

FIGURE 4. 7 Only the maintenance phase occurs in an adult female.

FIGURE 4. 9 In the F2 generation, only the males had white eyes.

FIGURE 4. 10 The key pedigree feature that points to X-linked inheritance is that only males are affected with the disorder. Also, carrier females often have affected brothers.

FIGURE 4. 11

The reason why the reciprocal cross yields a different result is because females carry two copies of an X-linked gene whereas males have only one.

End-of-chapter Questions: Conceptual Questions C1. In the X-Y system the male is heterogametic whereas in the Z-W system the female is heterogametic.

C2. A. Dark males and light females; reciprocal: all dark offspring B. All dark offspring; reciprocal: dark females and light males C. All dark offspring; reciprocal: dark females and light males D. All dark offspring; reciprocal: dark females and light males C3. A. The fly is a male because the ratio of X chromosomes to sets of autosomes is 1/2, or 0.5.

B. The fly is female because the ratio of X chromosomes to sets of autosomes is 1.0.

C. The fly is male because the ratio of X chromosomes to sets of autosomes is 0.5.

D. The fly is female because the ratio of X chromosomes to sets of autosomes is 1.0.

C4. A. Female; there are no Y chromosomes.

B. Female; there are no Y chromosomes.

C. Male; the Y chromosome determines maleness.

D. Male; the Y chromosome determines maleness.

C5. Dosage compensation refers to the phenomenon that the level of expression of genes on the sex chromosomes is similar in males and females, even though they have different numbers of sex chromosomes. In many species it seems necessary so that the balance of gene expression between the autosomes and sex chromosomes is similar between the two sexes.

C6. A Barr body is a mammalian X chromosome that is highly condensed. It is found in somatic cells with two or more X chromosomes. Most genes on the Barr body are inactive.

C7. In mammals, one of the X chromosomes is inactivated in females; in Drosophila, the level of transcription on the X chromosome in males is doubled; in C. elegans, the level of transcription of the X chromosome in hermaphrodites is decreased by 50% of that of males.

C8. X-chromosome inactivation in heterozygous females produces a mosaic pattern of gene expression.

During early embryonic development, some cells have the maternal X chromosome inactivated and other cells have the paternal X chromosome inactivated; these embryonic cells will divide and produce billions of cells. In the case of a female that is heterozygous for a gene that affects pigmentation of the fur, this produces a variegated pattern of coat color. Because it is a random process in any given animal, two female cats will vary as to where the orange and black patches occur. A variegated coat pattern could not occur in female marsupials due to X-chromosome inactivation because the paternal X chromosome is always inactivated in the somatic cells of females.

C9. X-chromosome inactivation begins with the counting of Xics. If there are two X chromosomes, in the process of initiation, one is targeted for inactivation. During embryogenesis, this inactivation begins at the Xic locus and spreads to both ends of the X chromosome until it becomes a highly condensed Barr body. The Xist gene, which is located in the Xic region, remains transcriptionally active on the inactivated X chromosome. It is thought to play an important role in X-chromosome inactivation by coating the inactive X chromosome. After X-chromosome inactivation is

established, it is maintained in the same X chromosome in somatic cells during subsequent cell divisions. In germ cells, however, the X chromosomes are not inactivated, so that an egg can transmit either copy of an active (noncondensed) X chromosome.

C10. The male is XXY. The person is male due to the presence of the Y chromosome. Because two X chromosomes are counted, one of the X chromosomes is inactivated to produce a Barr body.

C11. A. One B. Zero C. Two D. Zero C12. A. In females, one of the X chromosomes is inactivated. When the X chromosome that is inactivated carries the normal allele, only the defective color-blindness allele will be expressed.

Therefore, on average, about half of a female’s eye cells are expected to express the common (not color blind) allele. Depending on the relative amounts of cells expressing the common versus the color-blindness allele, the end result may be partial color blindness.

B. In this female, as a matter of chance, X-chromosome inactivation in the right eye always, or nearly always, inactivated the X chromosome carrying the normal allele. The opposite occurred in the left eye. In the left eye, the chromosome carrying the color-blindness allele was primarily inactivated.

C13. The offspring inherited XB from its mother and XO and Y from its father. It is an XXY animal, which is male (but somewhat feminized).

C14. The spreading phase is when the X chromosome is inactivated (i.e., condensed) in a wave that spreads outward from the X-inactivation center (Xic). The condensation spreads from Xic to the rest of the X chromosome. The Xist gene is transcribed from the inactivated X chromosome. It encodes an RNA that coats the X chromosome, which subsequently attracts proteins that are responsible for the compaction.

C15. The outcome is that all the daughters will be affected and all the sons will be unaffected. The ratio is 1:1.

C16. First set up the following Punnett square:

A. 1/4 B. (3/4)(3/4)(3/4)(3/4) = 81/256 C. 3/4

D. The probability of an affected offspring is ¼, and the probability of an unaffected offspring is 3/4. For this problem, you use the binomial expansion equation, where x = 2, n = 5, p = 1/4, and q = 3/4. The answer is 0.26, or 26%, of the time.

C17. 1 affected daughter : 1 unaffected daughter : 1 affected son : 1 unaffected son Application and Experimental Questions E1. The first type of observation was based on cytological studies. The presence of the Barr body in female cells was consistent with the idea that one of the X chromosomes was highly condensed. The second type of observation was based on the phenotypes of heterozygous females. A variegated phenotype that is found only in females is consistent with the idea that certain patches express one allele and other patches express the other allele. This variegated phenotype would occur only if the inactivation of one X chromosome happened at an early stage of embryonic development and was inherited permanently thereafter.

E2. In general, you cannot distinguish between autosomal and pseudoautosomal inheritance from a pedigree analysis. Mothers and fathers have an equal probability of passing the alleles to sons and daughters. However, if an offspring had a chromosomal abnormality, you might be able to tell. For example, in a family tree involving the Mic2 allele, an offspring that was X0 would have less of the gene product and an offspring that was XXX or XYY or XXY would have an extra amount of the gene products. This may lead you to suspect that the gene is located on the sex chromosomes.

E3. Perhaps the most convincing observation was that all of the white-eyed flies of the F2 generation were males. This suggests a link between sex determination and the inheritance of this trait.

Because sex determination in fruit flies is determined by the number of X chromosomes, this suggests a relationship between the inheritance of the X chromosome and the inheritance of this trait.

E4. Actually, his data are consistent with this hypothesis. To rule out a Y-linked allele, he could have crossed an F1 female with a red-eyed male rather than an F1 male. The same results would be obtained. Because the red-eyed male would not have a white allele, this would rule out Y-linkage.

E5. The basic strategy is to set up a pair of reciprocal crosses. The phenotype of sons is usually the easiest way to discern the two patterns. If it is Y linked, the trait will be passed only from father to son. If it is X linked, the trait will be passed from mother to son.

E6. To be a white-eyed female, a fly must inherit two X chromosomes and both must carry the whiteeye allele. This could occur only if both X chromosomes in the female stayed together and the male gamete contained a Y chromosome. The white-eyed females would be XXY. To produce a red-eyed male, a female gamete lacking any sex chromosomes could unite with a normal male gamete carrying the Xw+. This would produce an X0, red-eyed male.

E7. If you use the data from the F1 mating (i.e., F2 results), there were 3,470 red-eyed flies. You would expect a 3:1 ratio between red- and white-eyed flies. Therefore, assuming that all red-eyed offspring survived, there should have been about 1,157 (i.e., 3,470/3) white-eyed flies. However, there were only 782. If you divide 782 by 1,157, you get a value of 0.676, or a 67.6% survival rate.

E8. The rare female flies would be XXY. Both X chromosomes would carry the white-eye allele and the miniature allele. These female flies would have miniature wings because they would have inherited both X chromosomes from their mother.

E9. Originally, individuals who had abnormalities in their composition of sex chromosomes provided important information. In mammals, X0 individuals are females, while in flies, X0 individuals are males. In mammals, XXY individuals are males, while in flies, XXY individuals are females. These results indicate that the presence of the Y chromosome causes maleness in mammals, but it does not in flies. A further analysis of flies with abnormalities in the number of sets of autosomes revealed that it is the ratio between the number of X chromosomes and the number of sets of autosomes that determines sex in flies.

Questions for Student Discussion/Collaboration 1. The wide variety of sex determination mechanisms suggests that such mechanisms may have arisen independently on multiple occasions. It would seem that it may have arisen relatively recently after the major groups of animals diverged from each other.

2. One possibility is fertilization of an abnormal female gamete. The white-eyed male parent could make sperm carrying the X chromosome that could fertilize a female gamete without any sex chromosomes. This would produce an X0, white-eyed male. This is the most likely explanation.

Another possibility is that the flies in Morgan’s lab were not completely true-breeding. The allele creating the white phenotype may have occurred several generations earlier in a female fly.

Perhaps, among the many red-eyed sisters, one of them could have already been a heterozygote.

The rare sister could produce white-eyed male offspring. New mutations are a third, but unlikely possibility, because you already know that the mutation rate is very low. It took Morgan 2 years to get one white-eyed male.

CHAPTER 5 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 5. 1 Both of these colors are considered wild type because both are prevalent in natural populations.

FIGURE 5. 2 Yes. The PP homozygote probably makes twice the amount of protein that is needed for purple pigment formation.

FIGURE 5. 3 Individual III-2 shows incomplete penetrance.

FIGURE 5. 4 Genes and the environment determine an organism’s traits.

FIGURE 5. 5 50% of the functional protein is not enough to give a red color.

FIGURE 5. 6 It is often easier to observe incomplete dominance at the molecular or cellular level.

FIGURE 5. 7 In this case, the heterozygote is resistant to malaria.

FIGURE 5. 8 The scenario in part (a) explains the effects of the sickle-cell allele.

FIGURE 5. 9 The i allele is a loss-of-function allele.

FIGURE 5. 10 A heterozygous female does not have scurs.

FIGURE 5. 11 Certain traits are expressed only in males or females, possibly due to differences in the levels of sex hormones or other factors that differ between the sexes.

FIGURE 5. 12 The heterozygote has one normal copy of the gene, which allows for development to proceed in a way that is not too far from normal. Having two mutant copies of the gene probably adversely affects development to a degree that is incompatible with survival.

FIGURE 5. 14 Epistasis means that the alleles of one gene mask the phenotypic effects of the alleles of a different gene.

Complementation occurs when two strains exhibiting the same recessive trait produce offspring that show the dominant (wild-type) trait. This usually means that the alleles for the recessive trait are in two different genes.

FIGURE 5. 15 The two genes are redundant. Having one functional copy of either gene produces a triangular capsule.

If both genes are nonfunctional, an ovate capsule is produced.

End-of-chapter Questions: Conceptual Questions C1. Dominance occurs when one allele completely exerts its phenotypic effects over another allele.

Incomplete dominance is a situation in which two alleles in the heterozygote have an intermediate phenotype. In codominance, both alleles exert their effects independently in the heterozygote.

Overdominance is a case in which the heterozygote has a phenotype that is superior to either homozygote.

C2. Sex-influenced traits are affected by the sex of the individual even though the gene that governs the trait is autosomally inherited. Scurs in cattle is an example. The expression of a sex-limited trait is limited to one sex. For example, colorful plumage in certain species of birds is limited to the male sex. Sex-linked inheritance are traits whose genes are found on the sex chromosomes. Examples in humans include hemophilia and color blindness.

C3. The term gene interaction refers to the phenomenon that two or more different genes can have an impact on the same trait. This can occur, for example, if two genes encode enzymes in the same metabolic pathway. If the pathway is disrupted by a mutation in either of these two genes, the same net result may occur.

C4. If the functional allele is dominant, then one copy of the gene produces a sufficient amount of the protein. Having twice as much of this protein, as in the normal homozygote, does not alter the phenotype. If the allele is incompletely dominant, this means that one copy of the normal allele does not produce the same trait as the homozygote.

C5. Recessive alleles are often loss-of-function alleles. It would generally be more likely for a recessive allele to eliminate a trait or function rather than create one. Therefore, the peach carries the dominant allele, whereas the nectarine has a loss-of-function allele that prevents fuzz formation.

The recessive allele is in a gene that is necessary for fuzz formation.

C6. The ratio would be 1 normal : 2 star-eyed individuals.

C7. If individual 1 is ii, individual 2 could be IAi, IAIA, IBi, IBIB, or IAIB. If individual 1 is IAi or IAIA, individual 2 could be IBi, IBIB, or IAIB. If individual 1 is IBi or IBIB, individual 2 could be IAi, IAIA, or IAIB. Assuming individual 1 is the parent of individual 2: If individual 1 is ii, individual 2 could be IAi or IBi. If individual 1 is IAi, individual 2 could be IBi or IAIB. If individual 1 is IAIA, individual 2 could be IAIB. If individual 1 is IBi, individual 2 could be IAi or IAIB. If individual 1 is IBIB, individual 2 could be IAIB. C8. Types O and AB provide an unambiguous genotype. Type O can only be ii, and type AB can only be IAIB. It is possible for a couple to produce children with all four blood types. The couple would have to be IAi and IBi. If you construct a Punnett square, you will see that they can produce children with AB, A, B, and O blood types.

C9. The father could not be IAIB, IBIB, or IAIA. He is contributing the O allele to his offspring.

Genotypically, he could be IAi, IBi, or ii and have type A, B, or O blood, respectively.

C10. A. 1/4 B. 0 C. (1/4)(1/4)(1/4) = 1/64 D. Use the binomial expansion equation: n!

P px q ( n  x ) x!(n  x)! n  3, p  1/ 4, q  1/ 4, x  2 P  3/ 64  0.047, or 4.7% C11. Perhaps it should be called codominant at the “hair level” because one or the other allele is dominant with regard to a single hair. However, this is not the same as codominance in blood types, in which every cell can express both alleles.

C12. All of the F1 generation will be white because they have inherited the dominant white allele from their Leghorn parent. Construct a Punnett square. Let W and w represent one gene, where W is dominant and causes a white phenotype. Let A and a represent the second gene, where the recessive allele causes a white phenotype in the homozygous condition. The genotype of the F1 birds is WwAa. The phenotypic ratio of the F2 generation will be 13 white : 3 brown. The only brown birds will be 2 wwAa and 1 wwAA. C13. X-linked recessive (unaffected mothers transmit the trait to sons) C14. You know that the parents must be heterozygotes for both genes. The genotypic ratio of their offspring is 1 ScPScP : 2 ScPScA : 1 ScAScA The phenotypic ratio depends on sex: 1 ScPScP male with scurs : 1 ScPScP female with scurs : 2 ScPScA males with scurs : 2 ScPScA females without scurs : 1 ScAScA male without scurs : 1 ScAScA female without scurs.

A. 50% B. 1/8, or 12.5% C. (3/8)(3/8)(3/8)= 27/512 = 0.05, or 5% C15. A. Could be.

B. No, because an unaffected father has an affected daughter.

C. No, because two unaffected parents have affected children.

D. No, because an unaffected father has an affected daughter.

E. No, because both sexes exhibit the trait.

F. Could be.

C16. You would look at the pattern within families over the course of many generations. For a recessive trait, 25% of the offspring within a family are expected to be affected if both parents are unaffected carriers, and 50% of the offspring are expected to be affected if one parent is affected. You could look at many families and see if these 25% and 50% values are approximately true. Incomplete penetrance would not necessarily yield such numbers. Also, for very rare alleles, incomplete penetrance would probably have a much higher frequency of affected parents producing affected offspring. For rare recessive disorders, it is most likely that both parents are heterozygous carriers.

Finally, the most informative pedigrees would be those in which two affected parents produce children. If they can produce an unaffected offspring, this indicates incomplete penetrance. If all of their offspring were affected, this would be consistent with recessive inheritance.

C17. Molecular: The β-globin gene for HbA homozygotes encodes a β-globin polypeptide with a normal amino acid sequence compared to HbS homozygotes, whose β-globin genes encode a polypeptide that has an abnormal structure. The abnormal structure affects the ability of hemoglobin to carry oxygen.

Cellular: Under conditions of low oxygen, HbSHbS cells form a sickle shape compared to the normal biconcave disk shape of HbAHbA cells.

Organism: In HbSHbS individuals, the sickle shape decreases the life span of the red blood cell, which causes anemia. Also, the clogging of red blood cells in the capillaries causes tissue damage and painful crises. This does not occur in HbAHbA individuals.

C18. The probability of a heterozygote passing the allele to his/her offspring is 50%. The probability of an affected offspring expressing the trait is 80%. You use the product rule to determine the likelihood of these two independent outcomes.

(0.5)(0.8) = 0.4, or 40% of the time C19.

Note: III-1 could have either short or long index fingers because his mother is heterozygous.

C20. This pattern of inheritance is an example of incomplete dominance. The heterozygous horses are palominos. For example, if C represents chestnut and c represents cremello, the chestnut horses are CC, the cremello horses are cc, and the palominos are Cc. Application and Experimental Questions E1. The red and white seed packs should be from true-breeding (homozygous) strains. The pink pack should be seeds from a cross between white- and red-flowered plants.

E2. Mexican hairless dogs are heterozygous for a dominant allele that is lethal when homozygous. In a cross between two Mexican hairless dogs, you expect 1/4 to be normal, 1/2 to be hairless, and 1/4 to die.

E3. Two redundant genes are involved in feathering. The unfeathered Buff Rocks are homozygous recessive for the two genes. The Black Langhans are homozygous dominant for both genes. In the F2 generation (which is a double heterozygote crossed to another double heterozygote), 1 out of 16 offspring will be doubly homozygous for both recessive genes. All of the others will have at least one dominant allele for one of the two (redundant) genes.

E4. The first offspring must be homozygous for the horned allele. The father’s genotype is still ambiguous; he could be heterozygous or homozygous for the horned allele. The mother’s genotype must be heterozygous because her phenotype is polled (she cannot be homozygous for the horned allele) but she produced a horned daughter (who must have inherited a horned allele from its mother).

E5. The reason why all the puppies have black hair is because albino alleles are found in two different genes. If you let the letters A and B represent the two different pigmentation genes, then one of the dogs is AAbb and the other is aaBB. Their offspring are AaBb and therefore are not albinos because they have one dominant copy of each gene.

E6. It is a sex-limited trait, where W (white) is dominant but expressed only in females. In the cross of two yellow butterflies, the male is Ww but is still yellow because the white phenotype is limited to females. The female is ww and yellow. The offspring would be 50% Ww and 50% ww. All males would be yellow. Half of the females would be white (Ww) and half would be yellow (ww). Overall, this would yield 50% yellow males, 25% yellow females, and 25% white females.

E7. The sandy variation may be due to a homozygous recessive allele at one of two different genes in these two varieties of sandy pigs. Let’s call them genes A and B. One variety of sandy pig could be aaBB and the other AAbb. The F1 generation in this cross would be heterozygotes for both genes and are all red. This tells us that the A and B alleles are dominant. In the F2 generation, 6 out of 16 will be homozygous for either the aa or bb alleles and become sandy. One out of 16 will be doubly homozygous and be white. The remaining 9 will contain at least one dominant allele for both genes.

E8. The yellow squash has to be wwgg. It has to be ww because it is colored, and it has to be gg because it is yellow. Because the cross produced 50% white and 50% green offspring, the other parent (i.e., the white squash) must be WwGG. This cross would produce 50% offspring that are WwGg (white) and 50% that are wwGg (green).

E9. Let’s use the letters A and B for these two genes. Gene A exists in two alleles, which are called A and a. Gene B exists in two alleles, B and b. The uppercase alleles are dominant to the lowercase alleles. The true-breeding long-shaped squash is aabb and the true-breeding disk-shaped is AABB. The F1 offspring are AaBb. You can construct a Punnett square, with 16 boxes, to determine the outcome of self-fertilization of the F1 plants. To get the disk-shaped phenotype, an offspring must inherit at least one dominant allele from both genes.

1 AABB + 2 AaBB + 2 AABb + 4 AaBb = 9 disk-shaped offspring To get the round phenotype, an offspring must inherit at least one dominant allele for one of the two genes but must be homozygous recessive for only one of the two genes.

1 aaBB + 1 AAbb + 2 aaBb + 2 Aabb = 6 round-shaped offspring To get the long phenotype, an offspring must inherit all recessive alleles: 1 aabb = 1 long-shaped offspring E10. For this cross, you expect a 9:7 ratio of red to white flowers. In other words, 9/16 will be red and 7/16 will be white. Because there are a total of 345 plants, the expected values are 9/16  345 = 194 red 7/16  345 = 151 white

2  

(O  E)2 E (201 194)2 (144 151)2 2   194 151 2   0.58 With 1 degree of freedom, this chi square value is too small to reject your hypothesis. Therefore, you accept that it may be correct.

E11. The results obtained when crossing two F1 offspring appear to yield a 9:3:3:1 ratio, which would be expected if eye color is affected by two different genes that exist in dominant and recessive alleles.

Neither gene is X linked. Let pr+ represent the red allele of the first gene and pr the purple allele.

Let sep+ represent the red allele of the second gene and sep the sepia allele.

The first cross is: prpr sep+ sep+  pr+ pr+sep sep All of the F1 offspring would be pr+pr sep+sep. They have red eyes because they have a dominant red allele for each gene. When the F1 offspring are crossed to each other, the following results would be obtained:

In this case, one gene exists as the red (dominant) or purple (recessive) allele and the second gene exists as the red (dominant) or sepia (recessive) allele. If an offspring is homozygous for the purple allele, it will have purple eyes. Similarly, if an offspring is homozygous for the sepia allele, it will have sepia eyes. An offspring that is homozygous for both recessive alleles has purplish sepia eyes.

To have red eyes, it must have at least one copy of the dominant red allele for both genes. Based on an expected ratio of 9 red : 3 purple : 3 sepia : 1 purplish sepia, the observed and expected numbers of offspring are as follows: Observed Expected 146 purple eyes 148 purple eyes (791  3/16) 151 sepia eyes 148 sepia eyes (791  3/16) 50 purplish sepia eyes 49 purplish sepia eyes (791  1/16) 444 red eyes 445 red eyes (791  9/16) 791 total offspring If you plug the observed and expected values into the chi square formula, you obtain a chi square value of about 0.11. With 3 degrees of freedom, this is well within the expected range of values, so you cannot reject our hypothesis that purple and sepia alleles are in two different genes and that these recessive alleles are epistatic to each other.

Questions for Student Discussion/Collaboration 1. At the cellular or molecular level, the issue of dominance relies on the level of gene expression. The level of expression is quantitatively related to the genotype. For example, if an individual has one functional copy of a gene and second copy that is completely defective, it is common that only 50% of the normal amount of protein will be made. At the cellular level, this 50% level is less than the amount

2.

of the protein that would be made in a normal homozygote. At the organism level, however, 50% of the protein may be sufficient to give a normal phenotype.

Let's refer to the alleles as B dominant, b recessive and G dominant, g recessive.

The parental cross is BBGG × bbgg.

All of the F1 offspring are BbGg. Make a Punnett square for BbGg × BbGg. According to your Punnett square, the genotypes that are homozygous for the b allele and have at least one copy of the dominant G allele are gray. To explain this phenotype, you could hypothesize that the B allele encodes an enzyme that can make lots of pigment, whether or not the G allele is present.

Therefore, you get a black phenotype when one B allele is inherited. The G allele encodes a somewhat redundant enzyme, but maybe it does not function quite as well or its pigment product may not be as dark. Therefore, in the absence of a B allele, the G allele will give a gray pigment.

CHAPTER 6 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 6. 1 A nucleoid is not surrounded by a membrane, and a cell nucleus is.

FIGURE 6. 2 A reciprocal cross is a cross in which the sexes and phenotypes of the parents are reversed compared to a first cross.

FIGURE 6. 3 No. Once a patch of tissue is white, it has lost all of the normal chloroplasts, so it could not produce a patch of green tissue unless there was a new mutation.

FIGURE 6. 4 One mitochondrial nucleoid usually contains multiple copies of the mitochondrial chromosome.

FIGURE 6. 5 Mitochondria need rRNAs and tRNAs to synthesize polypeptides within the mitochondrial matrix.

FIGURE 6. 6 Chloroplasts and mitochondria have lost most of their genes during evolution. Many of these have been transferred to the cell nucleus.

FIGURE 6. 7 All of the offspring would be normal size because they would inherit a functional copy of the gene from their father.

FIGURE 6. 8 Erasure allows eggs to transmit unmethylated copies of the gene to the offspring.

FIGURE 6. 9 .

Maintenance methylation is automatic methylation that occurs when a methylated gene replicates and is transferred to daughter cells. It occurs in somatic cells. De novo methylation is the methylation of a gene that is not already methylated. It occurs in germ-line cells.

FIGURE 6. 10 The F2 offspring are all dextral because all of the F1 mothers are Dd, and the genotype of the mother determines the phenotype of the offspring.

FIGURE 6. 11 The egg cell will receive both D and d gene products.

End-of-chapter Questions: Conceptual Questions C1. Extranuclear inheritance is the transmission of genetic material (in eukaryotes) that is not located in the cell nucleus. Examples include mitochondria in animals, mitochondria in plants, and chloroplasts in plants.

C2. Extranuclear inheritance does not always occur via the female gamete. Sometimes it occurs via the male gamete. Even in species in which maternal inheritance is prevalent, paternal leakage may occur, that is, the paternal parent occasionally provides mitochondria (or chloroplasts) via the sperm. Maternal inheritance is the most common form of extranuclear inheritance because the female gamete is relatively large and more likely to contain cell organelles.

C3. Paternal leakage means that, in a small percentage of cases, an organelle is inherited from the paternal parent. If paternal leakage was 3%, then 3% of the time the offspring would inherit the organelle from the father. If the father was transmitting a dominant allele in the organellar genome (and the mother did not), then 3% of the offspring would exhibit the trait. Thus, among a total of 200 offspring, 6 would be expected to inherit paternal mitochondria.

C4. The mitochondrial and chloroplast genomes are composed of a circular chromosome found in one or more copies in a region of the organelle known as the nucleoid. The number of genes per chromosome varies from species to species. Chloroplast genomes tend to be larger than mitochondrial genomes. See Tables 6.1 and 6.3 for examples of the variation among mitochondrial and chloroplast genomes.

C5. Mitochondria and chloroplasts evolved from an endosymbiotic relationship in which bacteria took up residence within a primordial eukaryotic cell. Throughout evolution, there has been a movement of genes out of the organellar genomes and into the nuclear genome. The genomes of modern mitochondria and chloroplasts contain only a fraction of genes necessary for organellar structure and function. Nuclear genes encode most of the proteins that function within chloroplasts and mitochondria. Long ago, these genes were originally in the mitochondrial and chloroplasts genomes but have been subsequently transferred to the nuclear genome.

C6. A. Yes. B. Yes. C. No, it is determined by a gene in the chloroplast chromosome.

D. No, it is determined by a gene in the mitochondrial chromosome.

C7. Superficially, the tendency to develop this form of leukemia seems to be inherited from mother to offspring, much like the inheritance of mitochondria. To prove that it is not, you could separate newborn mice from their mothers and place them with mothers that do not carry AMLV. These offspring would not be expected to develop leukemia, even though their mother would.

C8. Biparental extranuclear inheritance might resemble Mendelian inheritance in that offspring could inherit alleles of a given gene from both parents. The patterns differ, however, when considered from the perspective of heterozygotes. For a Mendelian trait, the law of segregation tells us that a heterozygote passes one allele for a given gene to an offspring, but not both. In contrast, if a parent has a mixed population of mitochondria (e.g., some carrying a mutant gene and some carrying a functional gene), that parent could pass both types of genes to a single offspring, because more than one mitochondrion could be contained within a sperm or egg cell.

C9. Most of the genes originally within mitochondria and chloroplasts have been transferred to the nucleus. Therefore, mitochondria and chloroplasts have lost most of the genes that would be needed for them to survive as independent organisms.

C10. Genomic imprinting is a a modification that occurs to a nuclear gene or chromosome and alters gene expression, but is not permanent. Examples of imprinting include the Igf2 gene, and the pattern of X-chromosome inactivation in certain species.

C11. Erasure and reestablishment of the imprint occurs during gametogenesis. It is necessary to erase the imprint because each sex will transmit either inactive or active alleles of a gene. In somatic cells, the two alleles for a gene are imprinted according to the sex of the parent from which the allele was inherited.

C12. The process of de novo methylation will not occur in somatic cells but may occur in germ-line cells.

C13. A maternal effect gene is one in which the genotype of the mother determines the phenotype of the offspring. At the cellular level, this happens because maternal effect genes are expressed in diploid nurse cells and then the gene products are transported into the egg. These gene products play key roles in the early steps of embryonic development.

C14. A. Genotypes: All Nn Phenotypes: All with a small body B. Genotypes: All Nn Phenotypes: All with a normal body C. Genotypes: 1 NN : 2 Nn : 1 nn Phenotypes: All with a normal body, because the mother is heterozygous and N is dominant C15. The genotype of the mother must be bic– bic–. That is why it produces abnormal offspring. Because the mother is alive and able to produce offspring, its mother (the maternal grandmother) must have been bic+ bic– and passed the bic– allele to its daughter (the mother in this problem). The maternal grandfather also must have passed the bic– allele to its daughter. The maternal grandfather could be either bic+ bic– or bic– bic–.

C16. A. For an imprinted gene, you need to know whether the mutant Igf2 allele is inherited from the mother or the father.

B. For a maternal effect gene, you need to know the genotype of the mother.

C. For maternal inheritance, you need to know whether the mother has the trait, because the offspring will inherit her mitochondria.

C17. The mother must be heterozygous. She is phenotypically abnormal because her mother must have been homozygous for the abnormal recessive allele. However, because she produces all normal offspring, she must have inherited the normal dominant allele from her father. She produces all

normal offspring because this is a maternal effect gene, and the gene product of the normal dominant allele is transferred to the egg.

C18. The mother is hh. You know this because it is a maternal effect gene and all of its offspring have small heads. The offspring are all hh because their mother is hh and their father is hh.

C19. Maternal effect genes exert their effects because the gene products are transferred from nurse cells to eggs. The gene products, mRNA and proteins, do not last a very long time before they are eventually degraded. Therefore, they can only exert their effects during early stages of embryonic development.

Application and Experimental Questions E1. Based on these results, the pattern of inheritance may be cytoplasmic inheritance involving plastids.

In most cases, plastids follow a maternal inheritance pattern, but a low percentage of the time, paternal leakage does occur.

E2. Let’s first consider the genotypes of male A and male B. Male A must have two functional copies of the Igf2 gene. You know this because male A’s mother was Igf2 Igf2; the father of male A must have been a heterozygote, Igf2 Igf2―, because half of the litter that contained male A also contained dwarf offspring. But because male A was not dwarf, it must have inherited the functional allele from its father. Therefore, male A must be Igf2 Igf2. You cannot be completely sure of the genotype of male B. It must have inherited the functional Igf2 allele from its father because male B is normal size. You do not know the genotype of male B’s mother, but she could be either Igf2― Igf2― or Igf2 Igf2―. In either case, the mother of male B could pass the Igf2― allele to an offspring, but you do not know for sure if she did. So, male B could be either Igf2 Igf2― or Igf2 Igf2.

For the Igf2 gene, you know that the maternal allele is inactivated. Therefore, the genotypes and phenotypes of females A and B are irrelevant. The phenotype of the offspring is determined only by the allele that is inherited from the father. Because you know that male A has to be Igf2 Igf2, you know that it can produce only normal size offspring. Because both females A and B both produced dwarf offspring, male A cannot be the father. In contrast, male B could be either Igf2 Igf2 or Igf2 Igf2―. Because both females gave birth to dwarf babies (and because male A and male B were the only two male mice in the cage), you conclude that male B must be Igf2 Igf2― and is the father of both litters.

E3. The results of each succeeding generation depended on the genotypes of the mothers of the preceding generation. For example, if a mother was dd, the F1 offspring were all sinistral. The genotypes of the F2 mothers were 1 DD : 2 Dd : 1 dd. The DD and Dd mothers produced dextral offspring of the F3 generation and the dd mothers produced sinistral offspring. As expected, the ratio of dextral to sinistral was 3:1, which is derived from the genotypes of the F 2 mothers, which were 1 DD : 2 Dd : 1 dd.

E4. A haploid egg should express either the shorter or longer mRNA, but not both (because it has only one copy of the gene). The nurse cells, however, can express both types of mRNAs if the female is heterozygous. Therefore, if you begin with heterozygous females, you could separate the nurse cells from the eggs. You could then isolate mRNA from the nurse cells and (in a separate tube) isolate mRNA from eggs. As described in Chapter 20, the mRNA would then be run on a gel and subjected to Northern blotting, using a probe that is complementary to both types of mRNAs. According to our knowledge of maternal effect genes, you would expect the egg to contain both types of mRNAs,

because it receives them from the nurse cells. Both types of mRNAs would also be found in the nurse cells.

E5. The pattern of inheritance is consistent with imprinting. In every cross, the allele that is inherited from the father is expressed in the offspring, whereas the allele inherited from the mother is not.

E6. Mate the female to a dd male. If all of the offspring coil to the left, you know the female must be dd.

If they all coil to the right, she could be either DD or Dd. If the F1 offspring coil to the right, you can let them mate with each other to produce an F2 generation. If the original mother was Dd, then half of the F1 female offspring will be Dd and half will be dd. Therefore, half of the F2 snails will coil to the right and half to the left. In contrast, if the original mother was DD, all of the F1 female offspring will be Dd. In this case, all of the F2 snails will coil to the right.

Questions for Student Discussion/Collaboration 1. Having most genes in one location may facilitate coordinate gene regulation. In response to environmental changes, such as changes in extracellular nutrient levels, a cell may be better able to respond when most of the genes are in cell nucleus. For example, if a cell’s environment switched from having glucose (which is metabolized via glycolysis by enzymes in the cytosol) to citric acid (which is metabolized by enzymes of the citric acid cycle in the mitochondrion), it could more easily switch off the genes that encode glycolytic enzymes and switch on the genes the encode enzymes of the citric acid cycle.

2. Obviously, you cannot maintain a population of flies that are homozygous for a recessive lethal allele.

However, heterozygous females can produce viable offspring and these can be crossed to heterozygous or homozygous males to produce homozygous females. An experimenter would routinely have to make crosses and determine that the recessive allele was present in a population of flies by identifying homozygous females that were unable to produce any viable offspring. Maintaining a population of flies that carry a lethal recessive allele can be much easier if the recessive lethal allele is closely linked to a dominant allele that is not lethal, such as one affecting eye color or some other trait. If a fly exhibits the dominant trait, it is likely that it is also carrying the recessive lethal allele.

CHAPTER 7 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 7. 1 The offspring found in excess are those with purple flowers, long pollen and those with red flowers, round pollen.

FIGURE 7. 2 No, such a crossover would not change the arrangement of these alleles.

FIGURE 7. 3 A single crossover can produce offspring with these four phenotypic combinations: gray body, red eyes, miniature wings; gray body, white eyes, miniature wings; yellow body, red eyes, long wings; and yellow body, white eyes, long wings.

FIGURE 7. 4 When genes are relatively close together, a crossover is relatively unlikely to occur between them.

Therefore, the nonrecombinant offspring are more common.

FIGURE 7. 5 The reason is that the eye color and wing length genes are farther apart than the body color and eye color genes.

FIGURE 7. 6 Crossing over can change the combination kernel phenotypes and also it can change the morphologies of the chromosomes compared to the original chromosomes.

FIGURE 7. 7 Genetic maps are useful: (1) for understanding the complexity and genetic organization of a species and understanding the underlying basis of inherited traits; (2) in cloning genes; (3) in understanding evolution and the evolutionary relationships among different species; (4) in diagnosing and treating diseases, and predicting the likelihood of a couple having offspring with genetic diseases; and (5) in helping breeders of livestock and crops to improve strains.

FIGURE 7. 8 Crossing over occurred during oogenesis in the female parent of the recombinant offspring.

FIGURE 7. 9 The possible outcomes of multiple crossovers prevent the maximum percentage of recombinant offspring from exceeding 50%.

FIGURE 7. 10 Mitotic recombination occurs in somatic cells.

End-of-chapter Questions: Conceptual Questions C1. Genetic recombination is a term that refers to a new combination of alleles in an offspring. Crossing over is a physical event that involves the exchange of pieces of genetic material between two chromosomes.

C2. An independent assortment hypothesis is proposed because it enables you to calculate the expected values based on Mendel’s ratios. Using the observed and expected values, you can calculate whether or not the deviations between the observed and expected values are too large to occur as a matter of chance. If the deviations are very large, you reject the hypothesis of independent assortment.

C3. A. 1 and 4 B. 2 and 3 C. 2 and 4, or 1 and 3 C4. The single crossover produces ABC, Abc, aBC, and abc.

A. Between 2 and 3, between genes B and C B. Between 1 and 4, between genes A and B C. Between 1 and 4, between genes B and C D. Between 2 and 3, between genes A and B C5. There are 7 chromosomes per haploid genome. If you divide 20,000 by 7, there are about 2,857 genes per chromosome, on average.

C6. The likelihood of scoring a basket will be greater if the basket is larger. Similarly, the chances of a crossover initiating in a region between two genes is proportional to the size of the region between the two genes. There are a finite number (usually a few) of crossovers that occur between homologous chromosomes during meiosis, and the likelihood that a crossover will occur in a region between two genes depends on how big that region is.

C7. If the sweet pea has seven linkage groups, this means it has seven chromosomes per set. The sweet pea is diploid, meaning it has two sets of chromosomes. Therefore, the sweet pea has 14 chromosomes in leaf cells.

C8. The pedigree suggests a linkage between the dominant allele causing nail-patella syndrome and the IB allele of the ABO blood type gene. In every case, the individual who inherits the IB allele also inherits the disorder.

C9. The four phenotypic categories for the F2 offspring are: brown fur, short tails; brown fur, long tails; white fur, short tails; and white fur, long tails. The recombinants are brown fur, long tails, and white fur, short tails. The F2 offspring will occur in a 1:1:1:1 ratio if the two genes are not linked. In other words, there will be 25% of each of the four phenotypic categories. If the genes are linked, there will be a lower percentage of the recombinant offspring.

C10. Ass-1 43 Sdh-1 5 Hdc 9 Hao-1 6 Odc-2 8 Ada-1 The numbers in boldface are the distances between the genes.

C11. You use the product rule. The likelihood of a double crossover is 0.1  0.1, which equals 0.01, or 1%. The likelihood of a triple crossover is 0.1  0.1  0.1 = 0.001, or 0.1%.

C12. The inability to detect double crossovers causes the map distance to be underestimated. In other words, more crossovers occur in the region than you realize. When there is a double crossover, there are no recombinant offspring (in a two-factor cross). Therefore, the second crossover cancels out the effects of the first crossover.

C13. The map distance between genes A and B would appear to be greater than between C and D because of the higher rate of crossing over. You would obtain more recombinant offspring with regard to genes A and B, so the computed map distance would be higher, even though the physical distance between A and B, and between C and D, is identical.

C14. Mitotic recombination is crossing over between homologous chromosomes during mitosis in somatic cells. Mitotic recombination is one explanation for the blue patch. Following mitotic recombination, the two chromosomes carrying the b allele could segregate into the same cell and produce the blue color. Another explanation could be chromosome loss; the chromosome carrying the B allele could be lost during mitosis.

C15.

If the chromosomes (on the right side) labeled 2 and 4 move into one daughter cell, that will lead to a patch that is albino and has long fur. The other cell will receive chromosomes 1 and 3, which will produce a patch that has dark, short fur.

Application and Experimental Questions E1. If you hypothesize two genes independently assorting, the predicted ratio is 9:3:3:1. The total number of offspring is 427. The expected numbers of offspring are 9/16  427 = 240 purple flowers, long pollen 3/16  427 = 80 purple flowers, round pollen 3/16  427 = 80 red flowers, long pollen 1/16  427 = 27 red flowers, round pollen Plugging these values into our chi square formula, along with the observed numbers, you get: (296  240)2 (19  80)2 (27  80)2 (85  27)2 2     240 80 80 27 2   13.1  46.5  35.1  124.6

 2  219.3 Looking up this value in the chi square table under 1 degree of freedom, you find that such a large value is expected by chance less than 1% of the time. Therefore, you reject the hypothesis that the genes assort independently and conclude that the genes are linked.

E2. They could have used a strain with two abnormal chromosomes. In this case, the recombinant chromosomes would either look normal or have abnormalities at both ends.

E3. A gene on the Y chromosome in mammals would only be transmitted from father to son. It would be difficult to genetically map Y-linked genes because a normal male has only one copy of the Y chromosome, so you do not get any crossing over between two Y chromosomes. Occasionally, abnormal males (XYY) are born with two Y chromosomes. If such males were heterozygous for alleles of Y-linked genes, one could examine the normal male offspring of XYY fathers and determine if crossing over has occurred.

E4. The rationale behind a testcross is to determine if recombination has occurred during meiosis in the heterozygous parent. The other parent is usually homozygous recessive, so you cannot tell if crossing over has occurred in the recessive parent. It is easier to interpret the data if a testcross uses a completely homozygous recessive parent. However, in the other parent, it is not necessary for all of the dominant alleles to be on one chromosome and all of the recessive alleles on the other. The parental generation provides us with information concerning the original linkage pattern between the dominant and recessive alleles.

E5. The answer is explained in More Genetics TIPS #4. You cannot get more than 50% recombinant offspring because the pattern of multiple crossovers can yield an average maximum value of only 50%. When a testcross does yield a value of 50% recombinant offspring, it can mean two different things. Either the two genes are on different chromosomes or the two genes are on the same chromosome but at least 50 mu apart.

E6. The reason why the percentage of recombinant offspring is more accurate when the genes are close together is because fewer double crossovers occur. The inability to detect double crossovers causes the map distance to be underestimated. If two genes are very close together, very few double crossovers occur, so that underestimation due to double crossovers is minimized.

E7. If two genes are at least 50 mu apart, you would need to map genes between them to show that the two genes were actually in the same linkage group. For example, if gene A was 55 mu from gene B, there might be a third gene (e.g., gene C) that was 20 mu from A and 35 mu from B. These results would indicate that A and B are 55 mu apart, assuming two-factor testcrosses between genes A and B yielded 50% recombinant offspring.

E8. Morgan determined this by analyzing the data in gene pairs. This analysis revealed that there were fewer recombinants between certain gene pairs (e.g., body color and eye color) than between other gene pairs (e.g., eye color and wing length). From this comparison, he hypothesized that genes that are close together on the same chromosome will produce fewer recombinants than genes that are farther apart.

E9. A. Because they are 12 mu apart, you expect 12% (or 120) recombinant offspring. This would be approximately 60 Aabb and 60 aaBb plus 440 AaBb and 440 aabb.

B. You would expect 60 AaBb, 60 aabb, 440 Aabb, and 440 aaBb. E10. You consider the genes in pairs: There should be 10% offspring due to crossing over between genes A and B, and 5% due to crossing over between A and C. A. This genotype is due to a crossover between B and A. The nonrecombinants are Aa bb Cc and aa Bb cc. The 10% recombinants are Aa Bb Cc and aa bb cc. If you assume an equal number of both types of recombinants, 5% are Aa Bb Cc. B. This genotype is due to a crossover between A and C. The nonrecombinants are Aa bb Cc and aa Bb cc. The 5% recombinants are aa Bb Cc and Aa bb cc. If you assume an equal number of both types of recombinants, 2.5% are aa Bb Cc. C. This genotype is also due to a crossover between A and C. The nonrecombinants are Aa bb Cc and aa Bb cc. The 5% recombinants are aa Bb Cc and Aa bb cc. If you assume an equal number of both types of recombinants, 2.5% are Aa bb cc.

E11. Due to the large distance between the two genes, they will assort independently even though they are actually on the same chromosome. According to independent assortment, you expect 50%

nonrecombinant and 50% recombinant offspring. Therefore, this cross will produce 150 offspring in each of the four phenotypic categories.

E12. A. One basic strategy to solve this problem is to divide the data into gene pairs and determine the map distance between two genes.

184 tall, smooth 13 tall, peach 184 dwarf, peach 12 dwarf, smooth 13  12 Map distance   6.4 mu 184  13  184  12  153 tall, normal  44 tall, oblate 155 dwarf, oblate  41 dwarf, normal  44  41 Map distance   21.6 mu 153  44  155  41



163 smooth, normal 33 smooth, oblate 31 peach, normal 166 peach, oblate 33  31  16.3 mu 163  33  31  166 Use the two shortest distances to compute the map: tall, dwarf 6.4 smooth, peach 16.3 normal, oblate E13. A. If you hypothesize the two genes are independently assorting, the predicted ratio is 1:1:1:1.

There are a total of 390 offspring. The expected number of offspring in each category is about 98.

Plugging the figures into our chi square formula, Map distance 

 2  (117  98)  (115  98)  (78  98)  (80  98) 2

2

98 98   3.68  2.95  4.08  3.31

2

98

2

98

2

 2  14.02 Looking up this value in the chi square table under 1 degree of freedom, you reject the hypothesis, because the chi square value is above 7.815.

B. Map distance: Map distance =

78  80 117  115  78  80

 40.5 mu

Because the value is relatively close to 50 mu, it is probably a significant underestimate of the true distance between these two genes.

E14. The two nonrecombinant types are the homozygotes that cannot make either enzyme, and heterozygotes that can make both enzymes. The recombinants can make one enzyme but not both.

Because the two genes are 12 mu apart, 12% of the offspring would be recombinants and 88% would be nonrecombinant types. Because two nonrecombinant types are produced in equal numbers, you expect 44% of the mice to be unable to make either enzyme.

E15. The percentage of recombinants for the green, yellow and wide, narrow is 7%, or 0.07; there will be 3.5% of the green, narrow and 3.5% of the yellow, wide. The remaining 93% nonrecombinants will be 46.5% green, wide and 46.5% yellow, narrow. The third gene assorts independently. There will be 50% long and 50% short with respect to each of the other two genes. To calculate the number of offspring out of a total of 800, you multiply 800 by the percentages in each category.

(0.465 green, wide)(0.5 long)(800) = 186 green, wide, long (0.465 yellow, narrow)(0.5 long)(800) = 186 yellow, narrow, long (0.465 green, wide)(0.5 short)(800) = 186 green, wide, short (0.465 yellow, narrow)(0.5 short)(800) = 186 yellow, narrow, short (0.035 green, narrow)(0.5 long)(800) = 14 green, narrow, long (0.035 yellow, wide)(0.5 long)(800) = 14 yellow, wide, long (0.035 green, narrow)(0.5 short)(800) = 14 green, narrow, short (0.035 yellow, wide)(0.5 short)(800) = 14 yellow, wide, short E16. A. If you use B (bushy tail) and b (normal tail) for one gene, and Y (yellow) and y (white) for the second gene: Parental generation: BBYY  bbyy F1 generation: All BbYy (Note: if the two genes are linked, B would be linked to Y and b would be linked to y.)

Testcross: F1 BbYy  bbyy Nonrecombinant offspring from testcross: BbYy and bbyy BbYy males—bushy tails, yellow bbyy males—normal tails, white BbYy females—normal tails, yellow bbyy females—normal tails, white Recombinant offspring from testcross: Bbyy and bbYy Bbyy males—bushy tails, white bbYy males—normal tails, yellow Bbyy females—normal tails, white bbYy females—normal tails, yellow You cannot use the data on female offspring, because you cannot tell if females are recombinant or nonrecombinant. All females have normal tails. However, you can tell if male offspring are recombinant.

If you use the data on male offspring to conduct a chi-square analysis, you expect a 1:1:1:1 phenotypic ratio. Because there are 197 male offspring in total, you expect 1/4, or 49 (rounded to the nearest whole number), of each of the four possible phenotypes. Computing the chi square value:

(28  49)2 (72  49)2 (68  49)2 (29  49)2    49 49 49 49 2   9.0  10.8  7.4  8.2

2 

 2  35.4 If you look up the value of 35.4 in the chi square table, with 1 degree of freedom, the value lies far beyond the 0.01 probability level. Therefore, it is very unlikely to get such a large deviation if our hypothesis of independent assortment is correct. Therefore, you reject the hypothesis and conclude that the genes are linked.

B. Computing the map distance: 28  29  100  28.9 mu 28  72  68  29 E17. Let’s use the following symbols: G for green pods, g for yellow pods, S for green seedlings, s for bluish green seedlings, C for normal plants, c for creepers.

The parental cross is GG SS CC crossed to gg ss cc.

The F1 plants would all be Gg Ss Cc. If the genes are linked, the alleles G, S, and C would be linked on one chromosome, and the alleles g, s, and c would be linked on the homologous chromosome.

The testcross is F1 plants, which are Gg Ss Cc, crossed to gg ss cc.

To measure the distances between the genes, you can separate the data into gene pairs.

Pod color, seedling color 2,210 green pods, green seedlings—nonrecombinant 296 green pods, bluish green seedlings—recombinant 2,198 yellow pods, bluish green seedlings—nonrecombinant 293 yellow pods, green seedlings—recombinant Map distance =

296  293  100 = 11.8 mu 2, 210  296  2,198  293

Pod color, plant stature 2,340 green pods, normal—nonrecombinant 166 green pods, creeper—recombinant 2,323 yellow pods, creeper—nonrecombinant 168 yellow pods, normal—recombinant Map distance =

166  168

2,340  166  2,323  168

 100 = 6.7 mu

Seedling color, plant stature 2,070 green seedlings, normal—nonrecombinant 433 green seedlings, creeper—recombinant 2,056 bluish green seedlings, creeper—nonrecombinant 438 bluish green seedlings, normal—recombinant Map distance =

433  438

2, 070  433  2, 056  438

 100 = 17.4 mu

The order of the genes is seedling color, pod color, and plant stature (or you could say the opposite order). Pod color is in the middle. If you use the two shortest distances to construct our map: S 11.8 G 6.7 C E18. A. Parent Parent b 7 A 4 C  B7 a 4 c b 7 A 4 C B 7 a 4 c ↓ b 7 A 4 C Offspring B 7 a 4 c B. A heterozygous F2 offspring would have to inherit a chromosome carrying all of the dominant alleles. In the F1 parent (of the F2 offspring), a crossover in the 7 mu region between genes b and A (and between B and a) would yield a chromosome that was BAC and bac. If an F2 offspring inherited the BAC chromosome from its F1 parent and the bac chromosome from the homozygous parent, it would be heterozygous for all three genes.

C. From part B, a crossover between genes b and A (and between B and a) would yield BAC and bac chromosomes. If an offspring inherited the bac chromosome from its F1 parent and the bac chromosome from its homozygous parent, it would be homozygous for all three genes. The chances of a crossover in this region are 7%. However, half of the crossovers yield chromosomes that are BAC and the other half yield chromosomes that are bac. Therefore, the chances are 3.5% of getting homozygous F2 offspring.

E19. Yes. Begin with females that have one X chromosome that is XNl and the other X chromosome that is XnL. These females have to be mated to XNLY males because a living male cannot carry the n or l allele. In the absence of crossing over, a mating between X NlXnL females to XNLY males should not produce any surviving male offspring. However, during oogenesis in these heterozygous female mice, there could be a crossover in the region between the two genes, which would produce an X NL chromosome and an Xnl chromosome. Male offspring inheriting these recombinant chromosomes will be either XNLY or XnlY (whereas nonrecombinant males will be XnLY or XNlY). Only the male mice that inherit XNLY will live. The living males represent only half of the recombinant offspring.

(The other half are XnlY, which are born dead.)

To compute map distance: Map distance =

2 (Number of male living offspring) Number of males born dead+ Number of males born alive

Questions for Student Discussion/Collaboration 1. The basic strategy is first to determine the percentage of recombination between the different gene pairs. The percentage of recombinants between normal/droopy ears and normal/flaky tail is 6%.

There will be 3% normal ears/flaky tail and droopy ears/nonflaky tail. The remaining 94% will be 47% normal ears/nonflaky tail and 47% droopy ears/flaky tail. The other two genes will assort independently of these two linked genes. For each phenotypic category, there will be 50% long tails

2.

3.

versus 50% short tails, and 50% normal gait versus 50% jerker. You multiply these percentages together and with the total value of 400 to determine the relative numbers of each type of offspring: (0.47 normal ears/nonflaky)(0.5 long tail)(0.5 normal gait)(400) = 47 normal ears/nonflaky tail/normal gait/long tail (0.47 droopy ears/flaky)(0.5 long tail)(0.5 normal gait)(400) = 47 droopy ears/flaky tail/normal gait/long tail (0.47 normal ears/nonflaky)(0.5 long tail)(0.5 jerker)(400) = 47 normal ears/nonflaky tail/jerker/long tail (0.47 droopy ears/flaky)(0.5 long tail)(0.5 jerker)(400) = 47 droopy ears/flaky tail/jerker/long tail (0.47 normal ears/nonflaky)(0.5 short tail)(0.5 normal gait)(400) = 47 normal ears/nonflaky tail/normal gait/short tail (0.47 droopy ears/flaky)(0.5 short tail)(0.5 normal gait)(400) = 47 droopy ears/flaky tail/normal gait/short tail (0.47 normal ears/nonflaky)(0.5 short tail)(0.5 jerker)(400) = 47 normal ears/nonflaky tail/jerker/short tail (0.47 droopy ears/flaky)(0.5 short tail)(0.5 jerker)(400) = 47 droopy ears/flaky tail/jerker/short tail (0.03 normal ears/flaky)(0.5 long tail)(0.5 normal gait)(400) = 3 normal ears/flaky tail/normal gait/long tail (0.03 droopy ears/nonflaky)(0.5 long tail)(0.5 normal gait)(400) = 3 droopy ears/nonflaky tail/normal gait/long tail (0.03 normal ears/flaky)(0.5 long tail)(0.5 jerker)(400) = 3 normal ears/flaky tail/jerker/long tail (0.03 droopy ears/nonflaky)(0.5 long tail)(0.5 jerker)(400) = 3 droopy ears/nonflaky tail/jerker/long tail (0.03 normal ears/flaky)(0.5 short tail)(0.5 normal gait)(400) = 3 normal ears/flaky tail/normal gait/short tail (0.03 droopy ears/nonflaky)(0.5 short tail)(0.5 normal gait)(400) = 3 droopy ears/nonflaky tail/normal gait/short tail (0.03 normal ears/flaky)(0.5 short tail)(0.5 jerker)(400) = 3 normal ears/flaky tail/jerker/short tail (0.03 droopy ears/nonflaky)(0.5 short tail)(0.5 jerker)(400) = 3 droopy ears/nonflaky tail/jerker/short tail The X and Y chromosomes are not completely distinct linkage groups. You might describe them as overlapping linkage groups having some but not most of their genes in common.

The discussion could lead into many areas, such as bioethics, the probability of this happening by chance, and so forth.

CHAPTER 8 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 8. 1

The staining of chromosomes results in banding patterns that make it easier to distinguish chromosomes that are similar in size and have similar centromeric locations.

FIGURE 8. 2 Deletions and duplications alter the total amount of genetic material.

FIGURE 8. 3 If a chromosomal fragment does not contain a centromere, it will not segregate properly. If it remains outside the nucleus, it will be degraded.

FIGURE 8. 5 Nonallelic homologous recombination occurs when the pairing of homologous sites between two homologous chromosomes involves sites that are not alleles of the same gene. The result is that the chromosomes are misaligned.

FIGURE 8. 10 The chromosomes form an inversion loop so that the homologous regions can pair with each other. For the inverted and noninverted regions to pair, a loop must form.

FIGURE 8. 11 The mechanism shown in part (b) may occur if transposable elements are found in different chromosomes. These elements may promote the pairing between nonhomologous chromosomes and a subsequent crossover could occur.

FIGURE 8. 13 These chromosomes form a translocation cross because homologous regions are pairing with each other.

FIGURE 8. 14 aneuploid, monosomic, monosomy 3.

FIGURE 8. 15 The genes on chromosome 2 are present in single copies whereas the genes on the other chromosomes are present in two copies. The expression of genes on chromosome 2 would be less (perhaps 50%) relative to a normal individual. This creates an imbalance between genes on chromosome 2 and those on the other chromosomes.

FIGURE 8. 18 About 512.

FIGURE 8. 19 Polyploid plants are larger and more robust. They may have larger flowers and produce more fruit.

FIGURE 8. 20 During meiosis in a triploid individual, the homologs cannot pair properly. This results in highly aneuploid gametes, which are usually nonviable. Also, if aneuploid gametes participate in fertilization, the offspring are usually nonviable.

FIGURE 8. 21 Nondisjunction means that pairs of chromosomes are not separating from each other properly during meiosis or mitosis.

FIGURE 8. 23 In autopolyploidy, multiple sets of chromosomes come from the same species. In allopolyploidy, multiple sets of chromosomes come from at least two different species.

End-of-chapter Questions:

Conceptual Questions C1. Duplications and deficiencies cause a change in the total amount of genetic material: duplications involve a repeat of some genetic material; deficiencies, a shortage of some genetic material.

Inversions do not cause a change in the total amount of genetic material. Inversions involve a change in the direction of the genetic material along a single chromosome. Translocations occur when a segment of genetic material becomes attached to a different chromosome or to a different part of the same chromosome. Reciprocal translocations do not change the total amount of genetic material. However, unbalanced translocations do change the total amount.

C2. Small deletions and duplications are less likely to affect phenotype simply because they usually involve fewer genes. If a small deletion did have a phenotypic effect, you could conclude that a gene or genes in the region of the deletion are required to have a normal phenotype.

C3. It usually occurs when there is a misalignment during the crossing over of homologous chromosomes. One chromosome ends up with a deficiency, and the other has a duplication.

Crossing over of chromosomes with inversions can also cause gene duplications.

C4. A gene family is a group of genes that are derived from the process of gene duplications. They have similar sequences, but the sequences have some differences due to the accumulation of mutations over many generations. The members of a gene family usually encode proteins with similar but specialized functions. The specialization may occur in different cells or at different stages of development.

C5. You would expect α1 and α2 to be more similar, because they have diverged more recently.

Therefore, there has been less time for them to accumulate random mutations that would make their sequences different.

C6. It has a pericentric inversion.

C7. There are four products from meiosis. One would be a normal chromosome and one would contain the inversion. The other two chromosomes would have the following order of genes: A B C centromere D E F G H I J B A M L K J I H G F E D centromere C K L M C8. There are four possible products: One is a normal chromosome and one contains the inversion shown in the drawing in conceptual question C8. The other two chromosomes will be dicentric or acentric with the following order of genes: centromere centromere  

A

B C D EFGH I J D C B A Dicentric M L K JI H G F E K L M Acentric C9. Individuals who carry inversions and reciprocal translocations have the same amount of genetic material (i.e., the same number of genes) as do normal individuals. Therefore, they commonly have a normal phenotype. In some cases, however, the breakpoint in an inversion or translocation disrupts an important gene and thereby has a phenotypic consequence. In other cases, the chromosomal rearrangement may have a position effect that alters the expression of an important gene.

C10. In the absence of crossing over, alternate segregation will yield half of the cells with two normal chromosomes and half with a balanced translocation. With adjacent-1 segregation, all cells will be unbalanced. Two cells will be ABCDE + AIJKLM And the other two cells would have HBCDE + HIJKLM C11. A terminal piece of chromosome 11 broke off and attached to the short arm of chromosome 15. A crossover occurred between the long arm of chromosome 15 and the long arm of chromosome 18.

C12. One of the parents may carry a balanced translocation between chromosomes 5 and 7. The phenotypically abnormal offspring has inherited an unbalanced translocation due to the segregation of translocated chromosomes during meiosis.

C13. Adjacent-2 segregation is expected to be rare because the normal driving force for segregation is the segregation of centromeres. For example, the centromeres on chromosome 2 normally align during meiosis and segregate from each other whether or not the chromosome contains a translocation. On rare occasions, a misalignment of centromeres can lead to adjacent-2 segregation in which both centromeres from one chromosome travel to one pole and both centromeres from the other chromosome travel to the opposite pole.

C14. A deletion and an unbalanced translocation are more likely to have phenotypic effects because they create genetic imbalances. With a deletion, there are too few copies of several genes, and for an unbalanced translocation, there are too many.

C15. It is because the homologous chromosomes are trying to synapse with each other. As shown in the top of Figure 8.13, the formation of a translocation cross allows the homologous parts of the chromosomes to line up (i.e., synapse) with each other.

C16. The chromosome is due to a crossover within the inverted region. Your drawing should show the inversion loop. The crossover occurred between P and U. C17. A. 16 B. 9 C. 7 D. 12 E. 17 C18. This person has a total of 46 chromosomes. However, the person would be considered aneuploid rather than euploid, because one of the sets is missing a sex chromosome and one set has an extra copy of chromosome 21.

C19. It may be related to genetic balance. In aneuploidy, there is an imbalance in gene expression between the chromosomes found in their normal copy number versus those that are either too many or too few. In polyploidy, the balance in gene expression is still maintained.

C20. Imbalances due to aneuploidy, deletions, and duplications are related to the copy number of genes.

For many genes, the level of gene expression is directly related to the number of genes per cell. If there are too many copies, as in trisomy, or too few, as in monosomy, the level of gene expression will be too high or too low, respectively. It is difficult to say why deletions and monosomies are more detrimental, although one could speculate that having too little of a gene product causes more cellular problems than having too much of a gene product. In addition, monosomies may unmask rare recessive alleles that are detrimental.

C21. Trisomies 13, 18, and 21 survive because the chromosomes are small and contain fewer genes compared to the larger chromosomes. Individuals with abnormal numbers of X chromosomes can survive because the extra copies are converted to transcriptionally inactive Barr bodies. The other trisomies are lethal because they cause a great amount of imbalance between the level of gene expression on the normal diploid chromosomes relative to the chromosomes that are trisomic.

C22. One explanation is that one lizard is diploid and the other is a closely related tetraploid species.

Their offspring would be triploid, which would explain the sterility. Another possibility is that one of the lizards may carry a large inversion.

C23. Endopolyploidy means that a particular somatic tissue is polyploid even though the rest of the organism is not. The biological significance is not entirely understood, although it has been speculated that an increase in chromosome number in certain cells may enhance their ability to produce specific gene products that are needed in great abundance.

C24. Mosaicism is a condition in which an organism contains a subset of cells that are genetically different from those of the rest of the organism. It may result from mitotic nondisjunction in which sister chromatids separate improperly, so that one daughter cell has three copies of a chromosome while the other has only one, or in which the sister chromatids separate, but one chromosome does not attach to a spindle and is therefore lost.

C25. In certain types of cells, such as salivary cells, the homologous chromosomes pair with each other and then replicate approximately nine times to produce a polytene chromosome. The centromeres from each chromosome aggregate with each other at the chromocenter. This structure has six arms that arise from the single arm of two telocentric chromosomes (the X and 4) and two arms each from the metacentric chromosomes 2 and 3.

C26. Polyploid plants are often more robust than their diploid counterparts. Such plants may produce a greater yield of fruits and vegetables. In the field, they tend to be more resistant to harsh environmental conditions. When polyploid plants have an odd number of sets, they are typically seedless. This can be a desirable trait for certain fruit-producing crops such as bananas and watermelons.

C27. Aneuploid should not be used.

C28. Polyploid, triploid, and euploid should not be used.

C29. There are 11 chromosomes per set, so there are 11 possible trisomies: trisomy 1, trisomy 2, trisomy 3, trisomy 4, trisomy 5, trisomy 6, trisomy 7, trisomy 8, trisomy 9, trisomy 10, and trisomy 11.

C30. The odds of producing a euploid gamete are (1/2) n-1, which equals (1/2)5, or 1 in 32. The chance of producing an aneuploid gamete is therefore 31 out of 32 gametes. You use the product rule to

determine the chances of getting a euploid individual, because a euploid offspring is produced from two euploid gametes: 1/32  1/32 = 1/1,024. In other words, if this plant self-fertilized, only 1 in 1,024 offspring would be euploid. The euploid offspring could be diploid, triploid, or tetraploid.

C31. Nondisjunction is a mechanism whereby the chromosomes do not segregate equally into the two daughter cells. This can occur during meiosis to produce cells with altered numbers of chromosomes, or it can occur during mitosis to produce a genetic mosaic individual. A third way to alter chromosome number is by interspecies crosses that produce an alloploid.

C32. The nondisjunction usually occurs during meiosis I when the homologs synapse to form bivalents.

During meiosis II and mitosis, the homologs do not synapse. Instead, the chromosomes align randomly along the metaphase plate and then the centromeres separate.

C33. A genetic change occurred during early embryonic development to create the blue patch of tissue.

One possibility is a mitotic nondisjunction in which the two chromosomes carrying the b allele went to one cell and the two chromosomes carrying the B allele went to the other daughter cell. A second possibility is that the chromosome carrying the B allele was lost. A third possibility is that the B allele was deleted.

C34. In meiotic nondisjunction, the bivalents are not separating correctly during meiosis I. During mitotic nondisjunction, the sister chromatids are not separating properly.

C35. In general, Turner syndrome could be due to nondisjunction during either oogenesis or spermatogenesis. However, the Turner individual with color blindness is due to nondisjunction during spermatogenesis. The sperm lacked a sex chromosome, due to nondisjunction, and the egg carried an X chromosome with the recessive color blindness allele. This X chromosome had to be inherited from the mother, because the father was not color-blind. The mother must be heterozygous for the recessive color blind allele, and the father is hemizygous for the normal allele. Therefore, the mother must have transmitted a single X chromosome carrying the color-blind allele to her offspring, indicating that nondisjunction did not occur during oogenesis.

C36. Complete nondisjunction occurs during meiosis I, and that one nucleus receives all the chromosomes and the other nucleus does not get any. The nucleus with all of the chromosomes then proceeds through a normal meiosis II to produce two haploid sperm cells.

Application and Experimental Questions E1. Polytene chromosomes can be viewed in greater detail under the microscope because they are much larger. This makes it much easier to detect very small changes in chromosome structure. Polytene chromosomes are produced from the sequential replication and alignment of chromosomes. As an example, suppose a toothpick has fine lines written on it. You would probably have trouble seeing the lines. However, if you took 1,000 toothpicks with the same lines and stacked them up in a parallel manner, the lines would be much easier to see. Similarly, small changes in chromosome structure are hard to see in a single chromosome but much easier to detect in a polytene chromosome.

E2. Colchicine interferes with the mitotic spindle apparatus and thereby causes nondisjunction. At high concentrations, it can cause complete nondisjunction and produce polyploid cells.

E3 The male offspring is the result of nondisjunction during oogenesis. The female produced an egg without any sex chromosomes. The male parent transmitted a single X chromosome carrying the red allele. This produces an X0 male offspring

E4. A. The F1 offspring of this cross would probably be phenotypically normal, because they would carry the correct number of genes.

B. The F1 offspring would have lowered fertility because they would be inversion heterozygotes.

Because this is a large inversion, crossing over is fairly likely in the inverted region. When crossing over occurs, it will produce deletions and duplications that will probably be lethal in the resulting F 2 offspring.

E5. The turtles are two distinct species that appear phenotypically identical. The turtles with 48 chromosomes are polyploid relatives (i.e., tetraploids) of the species with 24 chromosomes. In animals, it is somewhat hard to imagine how this could occur because most animals cannot selffertilize, so there had to be two animals (i.e., one male and one female) that became tetraploids. It is easy to imagine how one animal could become a tetraploid; complete nondisjunction could occur during the first cell division of a fertilized egg, thereby creating a tetraploid cell that continued to develop into a tetraploid animal. This would have to happen independently (i.e., in two individuals of opposite sex) to create a tetraploid species. If you mated a tetraploid turtle with a diploid turtle, the offspring would be triploid and probably phenotypically normal. However, the triploid offspring would be sterile because they would make highly aneuploid gametes.

E6. First, you would cross the two strains together. It is difficult to predict the phenotype of the offspring. Nevertheless, you would keep crossing offspring to each other and backcrossing them to the parental strains until you obtained a great-tasting tomato strain that was resistant to heat and the viral pathogen. You could then make this strain tetraploid by treatment with colchicine. If you crossed the tetraploid strain with your great-tasting diploid strain that was resistant to heat and the viral pathogen, you may get a triploid that had these characteristics. This triploid would probably be seedless.

E7. A G band is a dark band on a chromosome that has been stained with Giemsa. The pattern of G bands on chromosomes can be used to distinguish chromosomes from each other even if their sizes and centromere locations are very similar. Banding patterns are also used to detect changes in chromosome structure.

E8. A polytene chromosome is formed when a chromosome replicates many times, and the chromatids lie side by side, as shown in Figure 8.18. The homologous chromosomes also lie side by side.

Therefore, if one chromosome carries a deletion, there will be a loop in its homolog. The loop is the segment that is not deleted from one of the two homologs.

E9. You could cross the two species together to first create an allodiploid. The allodiploid could be treated with colchicine to make a segment of the plant an allotetraploid. As described in Figure

8.24, a cutting of the allotetraploid segment of the plant could be rooted to create a new allotetraploid plant. Alternatively, you could first use colchicine to make two different tetraploids (autopolyploidy), which would make diploid gametes. These two autopolyploids could be crossed to each other to make an allotetraploid.

Questions for Student Discussion/Collaboration 1. The pairing would be a translocation cross with an inversion loop.

2. There are many possibilities. The fields of agriculture and botany books offer many examples. In the insect world, there are interesting examples of euploidy affecting gender determination. Among amphibians and reptiles, there are also several examples of closely related species that have euploid variation.

3. There are many examples: ion channels, motor proteins, transcription factors, etc. The importance usually lies in specificity. The members of the gene family are specifically used in a particular cell type, at a particular stage of development, or under a specific environmental condition.

4. 1. Polyploid plants are often more robust and disease resistant.

2. Allopolyploids may have useful combinations of traits.

3. Hybrids are often more vigorous; they can be generated from monoploids.

4. Strains with an odd number of chromosome sets (e.g., triploids) are usually seedless.

CHAPTER 9 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 9. 1 To grow, those colonies must have functional copies of all five genes. This could occur by the transfer of the met+ and bio+ genes to the met– bio– thr+ leu+ thi+ strain or the transfer of the thr+, leu+, and thi+ genes to the met+ bio+ thr– leu– thi– strain.

FIGURE 9. 2 Because bacteria are too large to pass through the filter, the U-tube apparatus can be used to determine if direct cell-to-cell contact is necessary for gene transfer to occur.

FIGURE 9. 3 It would be found in an F+ cell.

FIGURE 9. 4 Relaxase is a part of the relaxosome, which is needed for the cutting of the F factor and its transfer to the recipient cell. The coupling factor guides the DNA strand to the exporter, which transports it into the recipient cell.

FIGURE 9. 5 An Fʹ factor carries a portion of the bacterial chromosome, whereas an F factor does not.

FIGURE 9. 6 Because conjugation occurred for a longer period of time, pro+ was transferred in the conjugation experiment to the recipient cell at the bottom right.

FIGURE 9. 8 This type of map is based on the time of gene transfer in conjugation experiments, which is determined in minutes.

FIGURE 9. 9 The lacZ gene is closer to the origin of transfer; its transfer began at 16 minutes.

FIGURE 9. 10 The normal process is for bacteriophage DNA to be incorporated into a phage coat. In transduction, a segment of bacterial chromosomal DNA is incorporated into a phage coat.

FIGURE 9. 11 If the recipient cell did not have a lys– gene, the lys+ gene could be incorporated by nonhomologous recombination.

End-of-chapter Questions: Conceptual Questions C1. All of these processes are similar, in that a segment of genetic material has been transferred from one bacterial cell to another. The main difference is the underlying mechanism by which this transfer occurs. In conjugation, two living cells make direct contact with each other, and genetic material is transferred from one cell to another. In transduction, a virus infects a cell and makes new virus particles that contain some genetic material from the bacterium, and then these virus particles are transferred to a new bacterial cell. Finally, in transformation, segments of DNA are released from a dead bacterium and later taken up by a living cell.

C2. Conjugation is not a form of sexual reproduction, in which two distinct parents produce gametes that unite to form a new individual. However, conjugation is similar to sexual reproduction in that the genetic material from two cells is somewhat mixed. In conjugation, there is not the mixing of two genomes, one from each gamete. Instead, there is a transfer of genetic material from one cell to another. This transfer can alter the combination of genetic traits in the recipient cell.

C3. If neither cell has a selective growth advantage, you would expect that the F+ cells would eventually overrun the population. This is because a mating starts with an F+ and F– cell and ends with two F+ cells. Therefore, F+ cells can convert F– cells into F+ cells, but the opposite cannot occur.

C4. An F+ strain contains a separate, circular piece of DNA that has its own origin of transfer. An Hfr strain has its origin of transfer integrated into the bacterial chromosome. An F + strain can transfer only the DNA contained on the F factor. If given enough time, an Hfr strain can actually transfer the entire bacterial chromosome to the recipient cell.

C5. The role of the origin of transfer is to provide a starting site where two important events occur: the DNA is nicked, and one strand begins its transfer into a recipient cell. The direction of transfer in Hfr-mediated transfer will determine the order of transfer of the genes. For example, if the origin is between gene A and B, it could be oriented so that gene A will be transferred first. Alternatively, it could be oriented in the opposite direction so that gene B will be transferred first.

C6. Sex pili promote the binding of donor and recipient cells.

C7. Originally, a segment of the normal bacterial chromosome may have been accidentally excised to form a circle. F factors have their own origin of replication and may carry bacterial genes.

C8. Most of the P1 phages would contain P1 genetic material. However, an occasional phage would contain a piece of the bacterial chromosome.

C9. The steps that occur during transformation are described in Figure 9.12. A competent cell is able to take up DNA from the environment. Several factors affect competency including temperature, ionic conditions, and nutritional growth conditions.

C10. The transfer of conjugative plasmids such as F factor DNA does not require recombination.

C11. One advantage of the uptake of DNA is that it could be used as a nutritional source. The bacterial cells could use the nucleotides for their own growth and metabolism. The integration of DNA into the bacterial chromosome (i.e., transformation) also has the advantage of possibly giving the bacterium a new trait that might be beneficial to its survival. For example, if a bacterium that cannot metabolize lactose is transformed with lac+ genes, the transformed bacterium can survive in a medium containing lactose as the sole carbon source for growth.

C12. A. If it occurs as a single step, transformation is the most likely mechanism because conjugation does not usually occur between different species, particularly distantly related species, and different species are not usually infected by the same bacteriophages.

B. The genetic transfer could occur directly, but it may be more likely to have involved multiple steps.

C. The use of antibiotics selects for the survival of bacteria that have resistance genes. If a population of bacteria is exposed to an antibiotic, those carrying resistance genes will survive and their relative numbers will increase in subsequent generations.

Experimental Questions E1. Because no colonies were observed when 108 cells of the met– bio– thr+ leu+ thi+ or met+ bio+ thr– leu– thi– strains were plated alone.

E2. Mix the two strains together and then put some of them on plates containing streptomycin and some of them on plates without streptomycin. If mated colonies are present on both types of plates, then the thr+, leu+, and thi+ genes were transferred to the met+ bio+ thr– leu– thi– strain. If colonies are found only on the plates that lack streptomycin, then the met+ and bio+ genes were transferred to the met– bio– thr+ leu+ thi+ strain. This answer assumes a one-way transfer of genes from a donor to a recipient strain.

E3. The U-tube can distinguish between conjugation and transduction because of the pore size. Because the pores are too small for the passage of bacteria, this prevents direct contact between the two bacterial strains. However, viruses can pass through the pores and infect cells on either side of the U-tube.

It might be possible to use a U-tube to distinguish between transduction and transformation if a filter was used that had a pore size that was too small for viruses but large enough to allow the passage of DNA fragments. However, this might be difficult, because the difference in sizes between DNA fragments and phages are relatively small compared to the differences in sizes between bacteria and phages.

E4. An interrupted mating experiment is a procedure in which two bacterial strains are allowed to conjugate, and then the conjugation is interrupted at various time points. The interruption occurs by agitation of the solution in which the bacteria are found. This type of study is used to map the locations of genes. It is necessary to interrupt conjugation so that you can vary the time and obtain information about the order of transfer; which gene transferred first, second, and so on.

E5. The time of entry is the time it takes for a gene to be initially transferred from one bacterium to another. To determine this time, measurements are made at various lengths of time and then these data are extrapolated back to the x-axis.

E6. Conjugate unknown strains A and B to the F– strain in your lab that is resistant to streptomycin and cannot use lactose. Do this in two separate tubes (i.e., strain A plus your F– strain in one tube, and strain B plus your F– strain in the other tube). Plate the conjugated cells on growth media containing lactose plus streptomycin. If colonies grow, the unknown strain had to be strain A, the F+ strain that had lactose utilization genes on its F factor.

E7.

E8. A. The curve for the hisE intersects the x-axis at about 3 minutes and that for the pheA gene intersects it at about 24 minutes. These are the values for the times of entry. Therefore, the distance between these two genes is 21 minutes (i.e., 24 minus 3).

B. _hisE pabB pheA 4  17 E9. First, conjugate the streptomycin-resistant strain to the strain that has the genes that allow the cell to metabolize lactose. You can select for conjugated cells on growth media containing lactose and streptomycin. These cells will initially have an F factor with the streptomycin-resistant gene.

Expose these cells to acridine orange in a media that also contains streptomycin. This will select for the survival of rare cells in which the F factor has become integrated into the chromosome to become Hfr cells.

Questions for Student Discussion/Collaboration 1. There are many advantages. Bacteria are haploid and easy to grow. In addition, they grow fast and have a short generation time. The most common phenotypic traits that researchers study in bacteria are those that affect nutritional growth requirements, colony morphologies, and resistance to toxic molecules such as antibiotics.

2. Conjugation requires direct cell contact, which is mediated by proteins that are found in the same species. So, it is less likely to occur between different species unless they are very closely related

evolutionarily. Similarly, gene transfer via transduction involves bacteriophages that are usually species-specific. Gene transfer via transformation is most likely to occur between different species.

Gene transfer has several consequences, including antibiotic resistance and the ability to survive under new growth conditions.

CHAPTER 10 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 10. 1 Viruses vary in the hosts that they infect. They also vary in their genomes. Viral genomes vary with regard to size, RNA versus DNA, and single-stranded versus double-stranded. The structures of viruses also vary with regard to the complexity of their capsids and whether or not they have an envelope.

FIGURE 10. 3 A virus may remain latent following its integration into the genome of the host cell.

FIGURE 10. 4 The lytic cycle produces new phage particles.

FIGURE 10. 5 An emerging virus is a virus that has arisen recently and is more likely than previous strains to cause disease.

End-of-chapter Questions: Conceptual Questions C1. Viruses do not possess many of the characteristics that are associated with living organisms. By themselves, they do not carry out metabolism, reproduce, or utilize energy.

C2. All viruses have a nucleic acid genome and a capsid composed of protein. Some eukaryotic viruses are surrounded by an envelope that consists of a membrane with embedded proteins.

C3. All viral genomes are composed of nucleic acid and carry genes. They differ in that some genomes are DNA whereas others are RNA. Some are single stranded, and others are double stranded. Also, viral genomes differ greatly in their relative sizes.

C4. A viral envelope is a membrane with embedded proteins. It is made when the virus buds from the host cell, taking with it a portion of the host cell’s plasma membrane.

C5. With regard to viruses, a host cell is a cell that can be infected by a virus and allows the virus to replicate. Host range refers the types of cells and species that any particular virus may infect.

C6. The attachment step usually involves the binding of the virus to a specific protein on the surface of the host cell. Only certain cell types will make that protein.

C7. For phage  the virus attaches to the surface of the host cell and then injects its genetic material into the cytoplasm. The entire virus does not enter. By comparison, HIV binds to the surface of its host cell and the entire virus capsid is taken into the cell after the viral envelop fuses with the host cell’s plasma membrane.

C8. Reverse transcriptase is used to copy the viral RNA into DNA so it can be integrated into a chromosome of the host cell.

C9. Lytic bacteriophages encode an enzyme called lysozyme that digests the bacterial cell wall. This causes the bacteria to burst, thereby releasing the newly made phages.

C10. A temperate phage can follow either the lytic or lysogenic cycle, whereas a virulent phage can follow only the lytic cycle.

C11. A prophage or provirus refers to viral DNA that has integrated into a chromosome of a host cell.

An episome is segment of DNA that exists independently of the host cell’s chromosomes. All of them encode the information to make new viruses, and may remain latent for a long period of time.

Application and Experimental Questions E1. First, Adolf Mayer determined that the disease caused by TMV could be spread by spraying the sap from one plant onto another. By subjecting this sap to filtration, Ivanovski showed that the causative agent was too small to be a bacterium. Martinus Beijerinck ruled out that it is a toxin by showing that sap could continue to transmit the disease after many plant generations.

E2. Electron microscopy must be used to visualize a virus.

E3. A reconstituted virus is a virus that has been reassembled by using purified proteins and nucleic acid molecules.

E4. The two traits that Fraenkel-Conrat and Singer analyzed were the lesions on leaves that the viruses caused, and the amino acid composition of the viral coat proteins. When the nucleic acid of the reconstituted virus came from the wild-type strain, the lesions and the protein composition of newly made viruses were wild-type. Alternatively, if the RNA of the reconstituted virus came from the Holmes ribgrass strain, the lesions and protein composition were like those of the Holmes ribgrass strain.

E5. Drugs that affect spike glycoproteins can interfere with the ability of a virus to bind to its host cell.

This would prevent the virus from entering its host cell.

E6. The drugs prevent the assembly of new viruses. Therefore, if new viruses cannot be made, the virus cannot spread.

1.

2.

Questions for Student Discussion/Collaboration With regard to similarities, both viruses follow the same basic (five or six) steps. However, the details vary. Phage λ attaches to the bacterium via tail fibers, whereas HIV attaches via spike glycoproteins. Phage λ injects its DNA into the bacterial cytoplasm, whereas HIV enters via endocytosis. For integration to occur, the DNA of phage λ is directly inserted into the bacterial chromosome, whereas the HIV RNA genome is first reverse transcribed into DNA and then inserted. The synthesis and assembly processes have similarities in that the viral components are made. However, HIV makes RNA that assembles with proteins at the cell membrane and acquires an envelope. The release of the viruses is very different. Phage λ is released by lysing the bacterial cell, whereas HIV is released by budding from the plasma membrane of the host cell.

There are many possibilities. One difference is that bacterial diseases can be treated with antibiotics, whereas viral diseases cannot. You could try to isolate a sample (e.g., of blood) from an infected mouse and see if there are bacteria present. (Bacteria are visible via light microscopy, whereas viruses are not.) You could try to filter the infected blood through filters that allow the passage of viruses (which are smaller) but do not allow the passage of bacteria, and then determine if the filtrate is able to infect other mice. Other answers are possible.

CHAPTER 11 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 11. 1 In this experiment, the type R bacteria had taken up genetic material from the heat-killed type S bacteria, which converted the type R bacteria into type S. This enabled them to proliferate within the mouse and kill it.

FIGURE 11. 2 RNase or protease were added to the DNA extract to rule out the possibility that small amounts of contaminating RNA or protein were responsible for converting the type R bacteria into type S. FIGURE 11. 4 Ribose and uracil are not found in DNA. FIGURE 11. 7 Deoxyribose and phosphate form the backbone of a DNA strand.

FIGURE 11. 8 Modeling is useful because it shows how atoms and molecules can fit together in a three-dimensional structure of a complex molecule.

FIGURE 11. 11 Hydrogen bonding between base pairs and base stacking holds the DNA strands together.

FIGURE 11. 12 The major and minor grooves are the indentations where the atoms of bases are in contact with water in the cellular fluid. The major groove is wider than the minor groove.

FIGURE 11. 13 B DNA is a right-handed helix and the backbone is helical, whereas Z DNA is a left-handed helix and the backbone appears to zigzag slightly. Z DNA has the bases tilted relative to the central axis, whereas they are perpendicular to that axis in B DNA. The two forms also differ in the number of base pairs per turn.

FIGURE 11. 14 Covalent bonds hold nucleotides together in an RNA strand.

FIGURE 11. 15 A hydrogen bonds with U and G hydrogen bonds with C. End-of-chapter Questions: Conceptual Questions C1. The term genetic material refers to the actual substance that contains genetic information. It is usually DNA, but in some viruses it can be RNA. C2. The transformation process is described in Chapter 9.

1. A fragment of DNA binds to the cell surface.

2. It penetrates the cell membrane.

3. It enters the cytoplasm.

4. It recombines with the chromosome.

5. The genes within the DNA are expressed (i.e., transcription and translation).

6. The gene products create a capsule. That is, they are enzymes that synthesize a capsule using cellular molecules as building blocks.

C3. Transformation means changing from one form to another. In bacterial genetics, transformation involves the uptake of DNA into another bacterium. This may change the form (i.e., phenotype) of the bacterium. For example, transformation may change a rough bacterial strain into a smooth strain. The form or phenotype of the strain has been changed.

C4. The building blocks of a nucleotide are a sugar (ribose or deoxyribose), a nitrogenous base, and a phosphate group. In a nucleotide, the phosphate is already linked to the 5 position on the sugar.

When two nucleotides are linked together, a phosphate on one nucleotide forms a covalent bond with the 3 hyrdroxyl group on another nucleotide.

C5. The structures can be deduced from Figures 11.4 - 11.6. Guanine is the base by itself. Guanosine is the base attached to a ribose sugar. Deoxyguanosine triphosphate is a base attached to a deoxyribose sugar with three phosphates.

C6. The structure is a phosphate group connecting two sugars at the 3 and 5 positions, as shown in Figure 11.7.

C7. The bases conform to the AT/GC rule of complementarity. There are two hydrogen bonds between the A and T bases, and three hydrogen bonds between the G and C bases. The planar structures of the bases stack on top of each other within the helical structure to provide even more stability.

C8. 3′–CCGTAATGTGATCCGGA–5′ C9. The DNA sequence refers to the sequence of bases.

C10. A drawing of a DNA helix with 10 bp per turn looks like Figure 11.11. To make 15 bp per turn, you would have to add 5 more base pairs, but the helix should still make only one complete turn.

C11. B DNA is a right-handed helix and the backbone is helical, whereas Z DNA is a left-handed helix and the backbone appears to zigzag slightly. Z DNA has the bases tilted relative to the central axis, whereas they are perpendicular in B DNA. There are also minor differences in the number of base pairs per turn.

C12. The bases occupy the major and minor grooves. Phosphates and sugars are found in the backbone.

If a DNA-binding protein does not recognize a nucleotide sequence, it probably is not binding in the grooves, but instead is binding to the DNA backbone (i.e., sugar-phosphate sequence). DNAbinding proteins that recognize a base sequence must bind into the major or minor groove of the DNA, which is where the bases are accessible to a DNA-binding protein. Most DNA-binding proteins that recognize a base sequence fit into the major groove. By comparison, other DNAbinding proteins such as histones, which do not recognize a base sequence, bind to the DNA backbone.

C13. The sugar in DNA is deoxyribose; in RNA it is ribose. DNA contains the base thymine, while RNA has uracil. DNA is a double helical structure. RNA is single stranded, although parts of it may form double-stranded regions.

C14. The structure is shown in Figure 11.5. You begin numbering at the carbon that is to the right of the ring oxygen and continue to number the carbon atoms in a clockwise direction. Antiparallel means

that the backbones are running in the opposite direction. In one strand, the sugar carbons are oriented in a 3 to 5 direction, while in the other strand they are oriented in a 5 to 3 direction.

C15. Here is an example for an RNA molecule that could form a hairpin that contains 24 nucleotides in the stem and 16 nucleotides in the loop.

5–GAUCCCUAAACGGAUCCCAGGACUCCCACGUUUAGGGAUC–3 The complementary stem regions are underlined.

C16. The structures are similar in that RNA and DNA double helices are helical and antiparallel and base pairing is due to complementarity. The shapes of the helices are slightly different with regard to the number of base pairs per turn. Another difference is that RNA base pairing involves A with U, whereas DNA base pairing involves A with T. C17. The sequence in part A would be more difficult to separate because it has a higher percentage of GC base pairs compared to the one in part B. GC base pairs have three hydrogen bonds compared with AT base pairs, which have only two hydrogen bonds.

C18. Its base sequence.

C19. Complementarity is important in several ways. First, it is needed to copy genetic information. This occurs during replication, when new DNA strands are made, and during transcription, when RNA strands are made. Complementarity is also important during translation for codon/anticodon recognition. It also allows RNA molecules to form secondary structures and to recognize each other.

C20. G = 32%, C = 32%, A = 18%, T = 18% C21. The number of bases per turn is different in an RNA double helix and a DNA double helix. Also, protein binding may be affected by the structure of the sugar, which is ribose in RNA, but deoxyribose in DNA. C22. Lysines and arginines, and polar amino acids C23. This DNA molecule contains 280 bp. There are 10 base pairs per turn, so there are 28 complete turns.

C24. They always run parallel.

C25. A hydroxyl group is at the 3 end and a phosphate group is at the 5 end.

C26. The viral genetic material is probably double-stranded RNA because the amount of A nearly equals that of U and the amount of G nearly equals that of C. Therefore, this molecule could be double stranded and obey the AU/GC rule. However, it is possible that it is merely a coincidence that A happens to equal U and G happens to equal C, and the genetic material is really single stranded.

C27. Not necessarily. The AT/GC rule is required only of double-stranded DNA molecules.

C28. There are 108 base pairs in the chromosome. In a double helix, a single base pair has a length of about 0.34 nm, which equals 0.34  10−9 m. If you multiply the two values together: 108 (0.34  10−9) = 0.34  10−1 m, or 0.034 m, or 3.4 cm The answer is 3.4 cm, which equals 1.3 inches! That is enormously long considering that a typical human cell is only 10 to 100 µm in diameter. As described in Chapter 12, the DNA has to be greatly compacted to fit into a living cell.

C29. The first thing you need to do is to determine how many base pairs are in this DNA molecule. The linear length of 1 base pair is 0.34 nm, which equals 0.34  10−9 m. One centimeter equals 10−2 meters.

102

 2.9  107 bp

0.34  10-9 There are approximately 2.9  107 bp in this DNA molecule, which equals 5.8  107 nucleotides. If 15% are adenine, then 15% must also be thymine. This leaves 70% for cytosine and guanine.

Because cytosine and guanine bind to each other, there must be 35% cytosine and 35% guanine. If you multiply 5.8  107 times 0.35, you get: (5.8  107)(0.35) = 2.0  107 cytosines, or about 20 million cytosines C30. Yes, a stem-loop could form, as long as there are sequences that are complementary and antiparallel to each other. These would be similar to the complementary double-stranded regions observed in RNA molecules (e.g., see Figures 11.15 and 11.16).

C31. The methyl group is not attached to one of the atoms that hydrogen bonds with guanine, so methylation would not directly affect hydrogen bonding. It could indirectly affect hydrogen bonding if it perturbed the structure of DNA. Methylation may affect gene expression because it could alter the ability of proteins to recognize DNA sequences. For example, a protein might bind into the major groove by interacting with a sequence of bases that includes one or more cytosines. If the cytosines are methylated, this may prevent a protein from binding into the major groove properly.

Alternatively, methylation could enhance protein binding.

C32. Region 1 cannot form a stem-loop with region 2 and region 3 at the same time. Complementary regions of RNA form base pairs, not base triplets. The region 1/region 2 interaction would be slightly more stable than the region 1/region 3 interaction because it is one base pair longer, and it has a higher amount of GC base pairs. Remember that GC base pairs form three hydrogen bonds, whereas AU base pairs form two hydrogen bonds. Therefore, helices with a higher GC content are more stable.

Application and Experimental Questions E1. A trait of pneumococci is the ability to synthesize a capsule. There needs to be a blueprint for this ability. The blueprint for capsule formation was being transferred from the type S to the type R bacteria. (Note: At the molecular level, the blueprint is a group of genes that encode enzymes that can synthesize a capsule.)

E2. A. These are possible reasons why most of the cells were not transformed.

1. Most of the cells did not take up any of the type S DNA. 2. The type S DNA was usually degraded after it entered the type R bacteria.

3. The type S DNA was usually not expressed in the type R bacteria.

B. The antibody and centrifugation steps were used to remove the bacteria that had not been transformed. It enabled the researchers to determine the phenotype of the bacteria that had been transformed. If this step was omitted, there would have been so many colonies on the plate it would have been difficult to identify any transformed bacterial colonies, because they would have represented a very small proportion of the total number of bacterial colonies.

C. The researchers were trying to determine that it was really the DNA in their DNA extract that was the genetic material. It was possible that the extract was not entirely pure and could contain contaminating RNA or protein. However, treatment with RNase and protease did not prevent

transformation, indicating that RNA and protein were not the genetic material. In contrast, treatment with DNase blocked transformation, confirming that DNA is the genetic material.

E3. 1. Isolate and purify DNA from resistant bacteria.

2. In three separate tubes, add DNase, RNase, or protease.

3. Add sensitive bacteria to each tube. A small percentage may be transformed.

4. Plate on petri plates containing tetracycline.

Expected results: Tetracycline-resistant colonies should only grow when the DNA has been exposed to RNase and protease, but not to DNase.

E4. Here are five advantages: 1. You can create lots of different shapes.

2. You can move things around very quickly with a mouse.

3. You can use mathematical formulas to fit things together in a systematic way.

4. Computers process information very rapidly.

5. You can store the information you have obtained from model building in a computer file.

E5. Chargaff’s results would probably not be very convincing if done on a single species. The strength of his data was that all species appeared to conform to the AT/GC rule, suggesting that this is a consistent feature of DNA structure. In a single species, the observation that A = T and G = C could occur as a matter of chance.

Questions for Student Discussion/Collaboration 1. The key issue is that it needs to have some structural features (like being double-stranded and obeying the AT/GC rule) that allow it to be copied and also store information. However, the structure does not have to be made from nucleotides.

2. There are many possibilities. You could use a DNA-specific chemical and show that it causes heritable mutations. Perhaps you could inject an oocyte with a piece of DNA and produce a mouse with a new trait.

CHAPTER 12 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 12. 1 The sequences that comprise genes constitute most of a bacterial genome.

FIGURE 12. 2 Two FIGURE 12.5 Strand separation is needed for certain processes such as DNA replication and RNA transcription.

FIGURE 12. 6 Eukaryotic chromosomes have centromeres and telomeres, but bacterial chromosomes do not. Also, eukaryotic chromosomes typically have many more repetitive sequences.

FIGURE 12. 7

One reason for variation in genome size is that the number of genes among different eukaryotes varies.

A second reason is that the amount of repetitive sequences varies.

FIGURE 12.9 Retrotransposition always causes the TE to increase in number. Transposition by DNA transpososons, by itself, does not increase the TE number. However, such transposition can increase the number of a TE if it occurs around the time of DNA replication.

FIGURE 12.13 Reverse transcriptase uses RNA as a template to make a strand of DNA. FIGURE 12. 14 11 nm at its widest point.

FIGURE 12. 17 One way is for two CTCFs to bind to different CTC recognition sites and then to each other to form a loop. Another way is for SMC proteins to wrap around two different segments to form a loop.

FIGURE 12. 18 A chromosome territory is a discrete region in the cell nucleus that is occupied by a single chromosome.

FIGURE 12. 19 Active genes are found in less compacted regions of euchromatin.

FIGURE 12. 20 In the conversion of a region from a 300 nm 700 nm diameter, the loop domains become closely compacted and nucleosomes come closer together.

End-of-chapter Questions: Conceptual Questions C1. The bacterial nucleoid is a region in a bacterial cell that contains a compacted circular chromosome.

Unlike the eukaryotic nucleus, a nucleoid is not surrounded by a membrane.

C2. A bacterium with two nucleoids is similar to a diploid eukaryotic cell in that is has two copies of each gene. The bacterium is different, however, with regard to alleles. A eukaryotic cell can have two different alleles for the same gene. For example, a cell from a pea plant could be heterozygous, Tt, for the gene that affects height. By comparison, a bacterium with two nucleoids has two identical chromosomes. Therefore, such a bacterium is homozygous for its chromosomal genes. Note: As discussed in Chapter 9, a bacterium can contain another piece of DNA, called an F factor, that can carry a few genes. The alleles on an F factor can be different from the alleles on the bacterial chromosome.

C3. One mechanism is the formation of loop domains. Loops of DNA are anchored to DNA-binding proteins. Secondly, the DNA double helix is supercoiled to make it more compact, much like twisting a rubber band C4. DNA is a double helix. The helix is a coiled structure. Supercoiling involves additional coiling of a structure that is already a coil. Positive supercoiling is called overwinding because it adds additional twists in the same direction as the DNA double helix, that is, in a right-handed direction. Negative supercoiling is in the opposite direction. Z DNA is a left-handed helix. Positive supercoiling in Z DNA is in a left-handed direction, whereas negative supercoiling is in the right-handed direction (the opposite of the direction of positive and negative supercoiling in B DNA).

C5. These drugs would diminish the amount of negative supercoiling in DNA. Negative supercoiling is needed to compact the chromosomal DNA, and it also aids in strand separation. Bacteria might not be able to survive and/or transmit their chromosomes to daughter cells if their DNA was not compacted properly. Also, because negative supercoiling aids in strand separation, these drugs would make it more difficult for the DNA strands to separate. Therefore, the bacteria would have a difficult time transcribing their genes and replicating their DNA, because both processes require strand separation. As discussed in Chapter 13, DNA replication is needed to make new copies of the genetic material to transmit from mother to daughter cells. If DNA replication was inhibited, the bacteria could not grow and divide into new daughter cells. As discussed in Chapters 14–16, gene transcription is necessary for bacterial cells to make proteins. If gene transcription was inhibited, the bacteria could not make many proteins that are necessary for survival.

C6. A. The three twists would create either three fewer or three more turns for a total of seven or thirteen, respectively.

B. If the helix now has seven turns, it was left-handed. The three right-handed twists you made would cause three fewer turns in a left-handed helix. If the helix now has thirteen turns, it was righthanded. The three right-handed twists you made would add three more turns to a right-handed helix (compare Figures 12.4a and d).

C. The twisting would probably not make supercoils because the two strings are not tightly interacting with each other. It’s easy for the two strings to change the number of coils.

D. If the strings were glued together with rubber cement, the three additional twists would probably make supercoils. A glued pair of strings is more like the DNA double helix. In a double helix, the two strands are hydrogen bonding to each other. The hydrogen bonding is like the glue. Additional twists tend to create supercoils rather than alter the number of coils.

C7.

C8. Topoisomers are different with regard to the number of supercoils they contain. They are identical with regard to the number of base pairs in the double helix.

C9. The centromere is the attachment site for the kinetochore, which attaches to the spindle. If a chromosome is not attached to the spindle, it is free to “float around” within the cell, and it may not be near a pole when the nuclear membrane re-forms during telophase. If a chromosome is left outside of the nucleus, it is degraded during interphase. That is why the chromosome without a centromere may not be found in daughter cells.

C10. Centromeres are structures found in eukaryotic chromosomes that provide an attachment site for kinetochore proteins so the chromosomes are sorted (i.e., segregated) during mitosis and meiosis.

They are most important during M phase.

C11. Direct repeats occur because transposase or integrase produces staggered cuts in the two strands of chromosomal DNA. The transposable element is then inserted into this site, which temporarily

leaves two gaps. The gaps are filled in by DNA polymerase. Because this gap filling occurs at sites that are complementary, the two gaps end up with the exact same sequence, thereby creating direct repeats.

C12. Retrotransposons have the greatest potential for proliferation because the element is transcribed into RNA as an intermediate. Many copies of this RNA could be transcribed and then copied into DNA by reverse transcriptase. Theoretically, many copies of the element could be inserted into the genome in a single generation.

C13. Transposable elements are mutagens, because they alter (disrupt) the sequences of chromosomes and genes within chromosomes. They do this by inserting themselves into genes.

C14. A. LTR and non-LTR retrotransposons B. Insertion elements and simple transposons C. All of them are flanked by direct repeats.

D. Insertion elements and simple transposons C15. A DNA transposon is excised as a segment of DNA, and then it transposes to a new location via transposase. A retrotransposon is transcribed as RNA, and then reverse transcriptase makes a double-stranded copy of DNA. Integrase then inserts this DNA copy into the chromosome. All transposable elements have direct repeats. Transposons that move as DNA have inverted repeats, and they may encode transposase (as well as other genes). Retrotransposons do not have inverted repeats, but they may or may not have LTRs. Autonomous retrotransposons encode reverse transcriptase and integrase.

C16. An autonomous transposable element has the genes that are necessary for transposition. For example, a cut-and-paste TE that is autonomous has the transposase gene. A nonautonomous transposable element does not have all of the genes that are necessary for transposition. However, if a cell contains an autonomous element and a nonautonomous element of the same type, the nonautonomous element can move. For example, if a Drosophila cell contained two P elements, one autonomous and one nonautonomous, the transposase expressed from the autonomous P element could recognize the nonautonomous P element and catalyze its transposition.

C17. A deletion may occur when there is recombination between two different elements that align in the same direction. The region between the elements is deleted. An inversion can happen when recombination occurs between two different elements that are oriented in opposite directions. A translocation can result when recombination occurs between transposable elements that are located on different (nonhomologous) chromosomes. In other words, the TEs on different chromosomes align themselves and a crossover occurs. This will produce a translocation, as shown here:

C18. A nucleosome is composed of 146 bp of DNA wrapped 1.65 times around an octamer of histones.

In a 30-nm fiber, histone H1 helps to compact the nucleosomes. The zigzag model describes how this compaction may occur to form a 30-nm fiber. The arrangement of nucleosomes in the zigzag model appears somewhat like it’s zigzagging with a relatively straight linker region.

C19. Heterochromatin is more tightly packed than euchromatin. This is due to a greater compaction of the loop domains. Functionally, euchromatin can be transcribed into RNA, whereas heterochromatin is inactive.

C20. During interphase, the chromosomes are found within the cell nucleus. They are less tightly packed and are transcriptionally active. During M phase, the chromosomes become highly condensed, and the nuclear membrane is fragmented into vesicles. The chromosomes eventually become attached to the spindle apparatus via microtubules that are attached to the kinetochore, which is attached to the centromere.

C21. The main activities that can occur during interphase are transcription and DNA replication. For these activities to occur, the DNA must be in a relatively loose conformation. During M phase, there is relatively little genetic activity, although there is evidence that a few genes are transcribed.

However, during M phase most genes are transcriptionally inactive.

C22. There are 146 bp around the core histones. If the linker region is 54 bp long, you expect 200 bp of DNA (i.e., 146 + 54) for each nucleosome and linker region. If you divide 46,000 bp by 200 bp, you get 230. Because there are two molecules of H2A for each nucleosome, you expect there will be 460 molecules of H2A in a 46,000-bp sample of DNA. C23. The role of the core histones is to form the nucleosomes. In a nucleosome, the DNA is wrapped 1.65 times around the core histones. Histone H1 binds to the linker region. It may play a role in compacting the DNA into a 30-nm fiber.

C24. A. There are 108 bp in this human chromosome. In a double helix, a single nucleotide is about 0.34 nm long, which equals 0.34  10−3 µm. If you multiply the two values together: 108 (0.34  10−3) = 0.34  105 µm, or 34,000 µm B. The 30-nm fiber is about 49 times shorter than a linear double helix. (Note: The 11-nm fiber compacts the DNA about seven times. The 30-nm fiber compacts the DNA an additional seven

times compared to the 11 nm fiber. Therefore, compared to linear DNA, the 30-nm fiber is 7  7 = 49 times more compact.) If you divide 34,000 µm by 49 you get 694 µm.

C. This 30-nm fiber not fit inside the nucleus if it were stretched out in a linear manner because the diameter of the nucleus is much smaller than 694 µm. However, the 30-nm fiber is very thin and is compacted by many loop domains.

C25. During interphase, much of the chromosomal DNA is in the form of the 30 nm fiber, and some of it is more highly compacted heterochromatin. During metaphase, all of the DNA is highly compacted, as shown in Figure 12.20d. A high level of compaction prevents gene transcription and DNA replication from taking place. Therefore, these events occur during interphase.

Application and Experimental Questions E1. The second possibility, in which molecule A has 4 positive supercoils and molecule B has 1 negative supercoil fits these data. Molecule A would be more compacted because it has more supercoils. Also, molecule B would be more transcriptionally active, because it is more negatively supercoiled. The first possibility does not fit the data because both molecules have the same level of supercoiling so molecule A and molecule B would have the same level of compaction. The third possibility does not fit the data because molecule B would be more compact.

E2. Supercoiled DNA appears curled up into a relatively compact structure. You could add different purified topoisomerases and use microscopy to see how they affect the structure. For example, DNA gyrase relaxes positive supercoils, while topoisomerase I relaxes negative supercoils. If you added topoisomerase I to a DNA preparation and it became less compacted, then the DNA was negatively supercoiled.

E3. 1. The repeating nucleosome structure was revealed from DNase I digestion studies.

2. Purification studies showed that the biochemical composition is an octamer of histones.

3. More recently, crystallography has shown the precise structure of the nucleosome.

4. Microscopy has revealed information about the 30 nm fiber and the attachment of chromatin to the nuclear matrix.

In general, it is easier to understand the molecular structure of something when it forms a regular repeating pattern. The eukaryotic chromosome has a repeating pattern of nucleosomes. The bacterial chromosome seems to be more irregular in its biochemical composition.

E4. With a salt solution of moderate concentration, the nucleosome structure is still preserved so the same pattern of results would be observed. DNase I would cut the linker region and produce fragments of DNA that would be in multiples of 200 bp. However, with a highly concentrated salt solution, the core histones would be lost, and DNase I could cut anywhere. On the gel, you would see fragments of almost any size. Because there would be a continuum of fragments of many different sizes, the lane on the gel would probably look like a smear rather than having a few prominent bands of DNA. E5. You would get DNA fragments of about 446 to 496 bp (i.e., 146 bp plus 300 to 350).

E6. There are many possibilities. You could digest the DNA with DNase I and see if it that yields multiples of 200 bp or so. You could try to purify proteins from the sample and see if eukaryotic proteins or bacterial proteins are present.

E7. Histones are positively charged and DNA is negatively charged. They bind to each other by these ionic interactions. Salt is composed of positively charged ions and negatively charged ions. For

example, when dissolved in water, NaCl becomes individual ions of Na+ and Cl−. When chromatin is exposed to a salt such as NaCl, the positively charged Na+ ions could bind to the DNA and the negatively charged Cl− ions could bind to the histones. This would prevent the histones and DNA from binding to each other.

E8. A. Because the Alu sequence is interspersed throughout all of the chromosomes, many brightly colored spots would be seen along all chromosomes.

B. Only the centromeric region of the X chromosome would be brightly colored.

Questions for Student Discussion/Collaboration 1. You need to have the DNA compact to fit in the cell. Compaction allows a cell to store more information. Without this ability, organisms would not have been able to evolve to the high levels of complexity they have achieved. The downside is that cells need to copy and access this information.

Therefore, cells need to make the structure looser. To achieve both goals, the structure of chromosomes is extremely dynamic.

2. This is a matter of opinion. Having so much DNA that seems to have no obvious function seems like a waste of energy. Perhaps it has a function that we don’t know about yet. On the other hand, evolution does allow bad things to accumulate within genomes, such as alleles that cause diseases, etc. Perhaps this is just another example of the negative consequences of evolution.

3. The DNA structure is the same, except the degree of supercoiling. The protein compositions are very different. Bacteria compact their chromosomes by supercoiling and looping. Eukaryotes wrap their DNA around histones to form nucleosomes. The nucleosomes associate with one another to form loop domains, which become more closely associated during M phase.

CHAPTER 13 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 13.1 The two features that allow DNA to be replicated are its double-stranded structure and the base pairing between A and T and between G and C. FIGURE 13. 2 The semiconservative model is correct.

FIGURE 13. 5 The DnaA boxes are recognized by DnaA proteins, which bind to them and cause the DNA strands to separate at the AT-rich region.

FIGURE 13. 6 Two replication forks are formed at the origin of replication.

FIGURE 13.7 Primase is needed for DNA replication because DNA polymerase cannot initiate DNA replication on a bare template strand.

FIGURE 13. 8 The template strand is read in the 3ʹ to 5ʹ direction.

FIGURE 13. 10 The leading strand is made as one, long continuous strand in the same direction in which the replication fork is moving. The lagging strand is formed from Okazaki fragments produced in the direction away from the replication fork.

FIGURE 13. 13 The oxygen in the newly made ester bond comes from the sugar.

FIGURE 13. 16 Because eukaryotic chromosomes are so large, multiple origins of replication are needed so that the DNA can be replicated in a reasonable length of time.

FIGURE 13. 21 Six times (36 divided by 6).

FIGURE 13.22 An advantage of genetic recombination is that it may foster genetic diversity, which may produce organisms that have reproductive advantages over other organisms.

FIGURE 13.23 A heteroduplex region may be produced after branch migration because the two strands are not perfectly complementary.

FIGURE 13.24 A D loop is a structure formed during the recombination process when a DNA strand from one homolog invades the same region of the other homologous chromatid.

FIGURE 13.26 The region of the gene that contained the genetic variation that created the b allele was digested away following the double-strand break in that chromosome. The same region in the homologous chromosome, which carried the B allele, was then used as a template to make another copy of the B allele.

End-of-chapter Questions: Conceptual Questions C1. It is a double-stranded structure that follows the AT/GC rule.

C2. Bidirectional replication refers to the fact that DNA replication proceeds in both directions starting from one origin.

C3. Statement C is not true. A new strand is always made from a preexisting template strand. Therefore, a double helix always contains one strand that is older than the other.

C4. A. TTGGHTGUTGG HHUUTHUGHUU B. TTGGHTGUTGG HHUUTHUGHUU  TTGGHTGUTGG CCAAACACCAA AACCCACAACC HHUUTHUGHUU

 TTGGHTGUTGG TTGGGTGTTGG CCAAACACCAA CCAAACACCAA AACCCACAACC AACCCACAACC GGTTTGTGGTT HHUUTHUGHUU C5. No. In a conservative mechanism, one double helix would always be fully methylated so the cell would not have any way to delay the next round of DNA replication via a methylation mechanism.

C6. Given 4,600,000 bp of DNA and assuming that DNA replication is bidirectional at a rate of 750 nucleotides per second, the time required if there was just a single replication fork would be 4,600,000/750 = 6,133 seconds, or 102.2 minutes Because replication is bidirectional: 102.2/2 = 51.1 minutes Actually, this is an average value based on a variety of growth conditions. Under optimal growth conditions, replication can occur substantially faster.

With regard to base-pair mistakes, if you assume an error rate of one mistake per 100,000,000 nucleotides: 4,600,000  1,000 bacteria = 4,600,000,000 nucleotides of replicated DNA 4,600,000,000/100,000,000 = 46 mistakes This is pretty amazing. In this population, DNA polymerase would cause only 46 single mistakes in a total of 1,000 bacteria, each containing 4.6 million bp of DNA. C7. 5′—————————DNA polymerase——>3′ 3′———————————————— ————————————————————5′ C8. DNA polymerase would slide from right to left. The sequence of the new strand would be 3′–CTAGGGCTAGGCGTATGTAAATGGTCTAGTGGTGG–5′ C9. 1. DnaA boxes—Binding sites for the DnaA protein.

2. Methylation sites—Sites of adenine methylation that are important for regulating DNA replication.

3. AT-rich region—Site where the DNA initially separates to form an opening that is sometimes called a replication bubble.

C10. A. When looking at Figure 13.5, the first, second, and fourth DnaA boxes are running in the same direction, and the third and fifth are running in the opposite direction. Once you realize that, you can see that the sequences are very similar to each other.

B. After flipping the third and fifth boxes and then lining all 5 of them up, the consensus sequence is TGTGGATAA ACACCTATT C. The consensus sequence is nine nucleotides long. Because there are four kinds of nucleotides (i.e., A, T, G, and C), the chance of this sequence occurring by random chance is 4 −9, or once every 262,144 nucleotides. Because the E. coli chromosome is more than 10 times longer than this, it is fairly likely that this consensus sequence occurs elsewhere. The reason why there are not multiple origins, however, is because the origin has five copies of the consensus sequence very close together. The chance of having five copies of this consensus sequence occurring close together (as a matter of random chance) is very small.

C11. A. The right Okazaki fragment was made first. It is farthest away from the replication fork. The fork (not seen in this diagram) would be to the left of the three Okazaki fragments, and moving from right to left.

B. The RNA primer in the right Okazaki fragment would be removed first. DNA polymerase would begin by elongating the DNA strand of the middle Okazaki fragment and remove the right RNA primer with its 5 to 3 exonuclease activity. DNA polymerase I would use the 3 end of the DNA of the middle Okazaki fragment as a primer to synthesize DNA in the region where the right RNA primer is removed. If the middle fragment was not present, DNA polymerase could not fill in this DNA (because it needs a primer).

C. You need DNA ligase only at the right arrow. DNA polymerase I begins at the end of the left Okazaki fragment and synthesizes DNA to fill in the region as it removes the middle RNA primer.

At the left arrow, DNA polymerase I is simply extending the length of the left Okazaki fragment.

No ligase is needed here. When DNA polymerase I has extended the left Okazaki fragment through the entire region where the RNA primer has been removed, it hits the DNA of the middle Okazaki fragment. This occurs at the right arrow. At this point, the DNA of the middle Okazaki fragment has a 5 end that is a monophosphate. DNA ligase is needed to connect this monophosphate with the 3 end of the region where the middle RNA primer has been removed.

D. As mentioned in the answer to part C, the 5 end of the DNA in the middle Okazaki fragment is a monophosphate. It is a monophosphate because it was previously connected to the RNA primer by a phosphoester bond. At the location of the right arrow, there was only one phosphate connecting this deoxyribonucleotide to the last ribonucleotide in the RNA primer. For DNA polymerase to function, the energy to connect two nucleotides comes from the hydrolysis of the incoming triphosphate. In this location shown at the right arrow, however, the nucleotide is already present at the 5 end of the DNA, and it is a monophosphate. DNA ligase needs energy to connect this nucleotide with the left Okazaki fragment. It obtains energy from the hydrolysis of ATP or NAD+.

C12. 1. DnaA proteins recognize the origin of replication.

2. They promote the separation of the DNA strands.

3. They recruit helicase to the region.

C13. The picture would depict a ring of helicase proteins traveling along a DNA strand and separating the two helices, as shown in Figure 13.6.

C14. An Okazaki fragment is a short segment of newly made DNA that becomes part of the lagging strand. It is necessary to make these short fragments because in the lagging strand, the replication fork is exposing nucleotides in a 5 to 3 direction, but DNA polymerase is sliding along the template strand in a 3 to 5 direction, away from the replication fork. Therefore, the newly made lagging strand is synthesized in short pieces that are eventually attached to each other.

C15. The active site of DNA polymerase has the ability to recognize a mismatched nucleotide in the newly made strand and remove it by exonuclease cleavage. Proofreading occurs in a 3 to 5 direction. After the mistake is removed, DNA polymerase resumes DNA synthesis in the 5 to 3 direction.

C16. A processive enzyme is one that remains clamped to one of its substrates. DNA polymerase remains clamped to the template strand as it makes a new daughter strand. This processivity ensures a fast rate of DNA synthesis.

C17. The inability to synthesize DNA in the 3 to 5 direction and the need for a primer prevent replication at the 3′ end of the DNA strands. Telomerase is different than DNA polymerase in that it uses a short RNA sequence, which is part of its structure, as a template for DNA synthesis. Because

it uses this sequence many times in row, it produces a tandemly repeated sequence in the telomere at the 3′ ends of linear chromosomes.

C18. The opposite strand is made in the conventional way by DNA polymerase, using the strand made via telomerase as a template.

C19. Fifty, because two replication forks emanate from each origin of replication. DNA replication is bidirectional.

C20. The ends labeled B and C could not be replicated by DNA polymerase. DNA polymerase makes a strand in the 5 to 3 direction using a template strand that is running in the 3 to 5 direction. Also, DNA polymerase requires a primer. At the ends labeled B and C, there is no place (upstream) for a primer to be made.

C21. A. Both reverse transcriptase and telomerase use an RNA template to make a complementary strand of DNA. B. Because reverse transcriptase does not have a proofreading function, it is more likely for mistakes to occur. This creates many mutant strains of the virus. Some mutations might prevent the virus from proliferating. However, other mutations might prevent the host’s immune system from battling the virus. These kinds of mutations would enhance the proliferation of the virus.

C22. As shown in Figure 13.21, the first step of replication involves a binding of telomerase to the telomere. The 3′ overhang binds to the complementary RNA in telomerase. For this reason, a 3′ overhang is necessary for telomerase to replicate the telomere region.

C23. At the molecular level, SCE and recombination involving homologs are very similar. Identical (SCE) or similar (homologous recombination) segments of DNA line up and then cross over. Due to the molecular similarities of the two processes, you would expect that the same types of proteins would catalyze both events. However, at the genetic level, the events are different. SCE does not result in the recombination of alleles because the chromatids are genetically identical. Homologous recombination usually results in a new combination of alleles after a crossover has taken place.

C24. Branch migration will not create a heteroduplex during SCE because the sister chromatids are genetically identical. There should not be any mismatches between the complementary strands.

Gene conversion cannot take place because the sister chromatids carry alleles that are already identical to each other.

C25. The two molecular mechanisms that can explain the phenomenon of gene conversion are mismatch DNA repair and gap repair synthesis. Both mechanisms could occur in the double-strand break model.

C26. A recombinant chromosome is one that has been derived from a crossover and contains a combination of alleles that is different from those in the original chromosomes. A recombinant chromosome is a hybrid of the original chromosomes.

C27. It depends on which way the breaks occur in the DNA strands during the resolution phase. If the two breaks occur in the crossed DNA strands, nonrecombinant chromosomes result. If the two breaks occur in the uncrossed strands, the result is a pair of recombinant chromosomes.

C28. Gene conversion occurs when a pair of different alleles is converted to a pair of identical alleles.

For example, a pair of Bb alleles could be converted to BB or bb.

C29. The Holliday model proposes two breaks, one in each chromatid, and then both strands exchange a single strand of DNA. The double-strand break model proposes two breaks, both in the same

chromatid. As in the Holliday model, single-strand migration occurs between both homologs. The double-strand break model also proposes strand degradation and gap repair synthesis. Finally, the double-strand break model generates two Holliday junctions, not just one.

C30. Gene conversion is likely to take place near the breakpoint in a recombinant chromosome.

According to the double-strand break model, a gap may be created by the digestion of one DNA strand in the double helix. Gap repair synthesis may result in gene conversion. A second way that gene conversion can occur is by mismatch repair. A heteroduplex may be created after DNA strand migration and may be repaired in such a way as to result in gene conversion.

C31. No, it does not necessarily involve DNA mismatch repair; it could also involve gap repair synthesis. The double-strand break model involves a migration of DNA strands and the digestion of a gap. Therefore, in this gap region, only one chromatid is providing the DNA strands. As seen in Figure 13.26, the top chromosome is using the top DNA strand from the bottom chromosome in the gap region. The bottom chromosome is using the bottom DNA strand from the bottom chromosome. After DNA synthesis, both chromosomes may have the same allele.

Application and Experimental Questions E1. A. Four generations: 7/8 light, 1/8 half-heavy; five generations: 15/16 light, 1/16 half-heavy B. All of the DNA double helices would be 1/8 heavy.

C. The CsCl gradient separates molecules according to their densities. 14N-containing compounds have a lighter density compared to 15N-containing compounds. The bases of DNA contain nitrogen.

If the bases contain only 15N, the DNA will be heavy; it will sediment at a higher density. If the bases contain only 14N, the DNA will be light; it will sediment at a lower density. If the bases in one DNA strand contain 14N and the bases in the opposite strand contain 15N, the DNA will be halfheavy; it will sediment at an intermediate density.

E2. A. In this case, the researcher would probably still see a band of DNA, but only see a heavy band.

B. The researcher would probably not see a band because the DNA would not be released from the bacteria. The bacteria would sediment to the bottom of the tube.

C. The researcher would not see a band. UV light is needed to see the DNA, which absorbs light in the UV region.

E3. You might be able to determine the number of replication forks and their approximate locations. For chromosomes with a single origin, you could determine that replication is bidirectional. However, you would not get any molecular information about the DNA replication process.

E4. For a DNA strand to grow, a phosphoester bond is formed between the 3 –OH group on one nucleotide and the innermost 5 phosphate group on the incoming nucleotide (see Figure 13.13). If the –OH group is missing, such a bond cannot form.

E5. A. Heat is used to separate the DNA strands, so you do not need helicase.

B. Each primer must be a sequence that is complementary to one of the DNA strands. There are two types of primers, and each type binds to one of the two complementary strands.

C. A thermophilic DNA polymerase is used because DNA polymerases isolated from nonthermophilic species would be permanently inactivated during the heating phase of the PCR cycle. Remember that DNA polymerase is a protein, and most proteins are denatured by heating.

However, proteins from thermophilic organisms have evolved to withstand heat, which is how thermophilic organisms survive at high temperatures.

D. With each cycle, the amount of DNA is doubled. Because there are initially 10 copies of the DNA,   227 copies will occur after 27 cycles. 10  227 = 1.34  109 = 1.34 billion copies of DNA. As you can see, PCR can amplify the amount of DNA by a staggering amount!

E6. The more likely mechanism in this case is DNA gap repair synthesis. At any given base, mismatch repair has a 50% chance of causing gene conversion. The odds of converting all six bases to those found in the B allele would therefore be (1/2)6, or 1/64. Because this is a relatively small number, gap repair synthesis is perhaps more likely because it could remove a region that contains all six bases and use a DNA strand from the B allele as a template to make a new strand.

Questions for Student Discussion/Collaboration 1. Lots of possibilities. The main point is the idea of chemical recognition. There must be strict rules about how two structures recognize each other.

2. This is a matter of opinion. DNA replication needs to be fast so that life can occur on the time scale that we are accustomed to. That it is error free is very important. DNA stores information and that information needs to be maintained with high fidelity. Finally, the regulation of DNA replication during cell division is necessary to ensure the proper amount of DNA per cell.

CHAPTER 14 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 14.2 The mutation would not affect the length of the mRNA, because it would not terminate transcription.

However, the encoded polypeptide would be shorter.

FIGURE 14. 5 For a group of related sequences within DNA, the consensus sequence consists of the most common base found at each location within those sequences.

FIGURE 14. 6 Parts of sigma factor must fit into the major groove so that it can recognize a promoter and form hydrogen bonds with its bases.

FIGURE 14. 7 The -10 sequence is an AT rich region and thus has fewer hydrogen bonds compared to a region with a lot of G-C base pairs FIGURE 14. 10 Such a mutation would prevent -dependent termination of transcription.

FIGURE 14. 11 NusA helps RNA polymerase to pause, which facilitates transcriptional termination.

FIGURE 14. 13 The TATA box provides a precise starting point for the transcription of eukaryotic protein-coding genes.

FIGURE 14. 14 The phosphorylation of the CTD allows RNA polymerase to proceed to the elongation phase of transcription.

FIGURE 14. 17 An endonuclease can cleave within a DNA or RNA strand, whereas an exonuclease digests a strand, one nucleotide at a time, starting at one end.

FIGURE 14. 18 Splicing via a spliceosome is very common in eukaryotes.

FIGURE 14. 20 snRNPs are involved in recognizing the intron boundaries, cutting out the intron, and connecting the two adjacent exons together.

FIGURE 14.22 Exon 4 would be spliced out. Exon 3 would be retained within the mRNA. FIGURE 14. 23 The 7-methylguanosine cap is important for the proper splicing of pre-mRNA, the exit of mRNA from the nucleus, and the binding of mRNA to a ribosome.

FIGURE 14. 25 A functional consequence of RNA editing is that the amino acid sequence of an encoded polypeptide may be changed, which can affect its function.

End-of-chapter Questions: Conceptual Questions C1. According to the central dogma, the usual flow of genetic information is from DNA to RNA to proteins. This information is passed from cell to cell and transmitted from parent to offspring via DNA replication. Exceptions to the central dogma do occur. For example, some viruses can use RNA as a template to make DNA. C2. The release of  factor marks the end of the initiation stage of transcription.

C3. A consensus sequence is the most common nucleotide sequence that is found within a group of related sequences. An example is the –35 and –10 consensus sequences found in bacterial promoters. At –35, it is TTGACA, but it can differ by one or two nucleotides and still function efficiently as a promoter. In the consensus sequences within bacterial promoters, the –35 sequence is primarily for recognition by sigma factor. The –10 sequence, also known as the Pribnow box, is the site where the double-stranded DNA will begin to unwind to allow transcription to occur.

C4. GGCATTGTCA C5. Mutations that make a sequence more like the consensus sequence are likely to be up promoter mutations, whereas mutations that cause the promoter to deviate from the consensus sequence are likely to be down promoter mutations. Also, in the –10 sequence, AT pairs are favored over GC pairs, because the role of this region is to form the open complex. AT pairs are more easily separated, because they form only two hydrogen bonds compared to GC pairs, which form three hydrogen bonds.

A. Up promoter B. Down promoter C. Up promoter

C6. The most highly conserved positions are the first, second, and sixth. In general, when promoter sequences are conserved, they are more likely to be important for binding. That explains why changes do not occur at these positions. If a mutation altered such a position, the promoter would probably not work very well; such an unfavorable mutation would likely be eliminated by natural selection. By comparison, changes are tolerated occasionally at the fourth position and frequently at the third and fifth positions. The positions that tolerate changes are less important for binding by  factor.

C7. In Figure 14.6, each α helix appears to occupy a region that is approximately one-half of a complete turn of the double helix. Because a complete turn of a DNA double helix involves 10 bp, each α helix is bound to about 5 bp in the major groove of the DNA; 5 bp would span a linear distance of approximately 1.7 nm (as described in Chapter 11, Figure 11.14). If you divide 1.7 nm by 0.15 nm/amino acid, you obtain a value of 11.3 amino acids per helix. This would create about 3.15 turns of an α helix; there are three complete turns per helix, for a total of six complete turns for both helices 2 and 3.

C8. This mutation will not affect transcription. However, it will affect translation by preventing the initiation of polypeptide synthesis.

C9. RNA polymerase holoenzyme consists of sigma factor plus the core enzyme, which is composed of five subunits, α2ββ′ω. The role of sigma factor (σ) is to recognize the promoter sequence. The α subunits are necessary for the assembly of the core enzyme and for loose DNA binding. The β and β subunits catalyze the covalent linkages between adjacent ribonucleotides. The ω subunit is important for the proper assembly of the core enzyme.

C10.  factor can slide along the major groove of the DNA. This protein contains -helices that fit into the major groove of the DNA. In this way, it is able to recognize base sequences that are exposed in the groove. When  factor encounters a promoter sequence, hydrogen bonding between it and the bases promote a tight and specific interaction.

C11. The mutation would alter the bases so that hydrogen bonding with sigma factor would not occur or would not occur in the correct way. From Figure 14.6, if you look at the –35 sequence, you may expect that changing the first two nucleotides to anything besides T might inhibit transcription, changing the third base to an A (perhaps), changing the fourth base to a G or T, or changing the last base to anything besides an A. You would think these substitutions may have an effect, because they would deviate from known functional bacterial promoters.

C12. DNA-G/RNA-C DNA-C/RNA-G DNA-A/RNA-U DNA-T/RNA-A The template strand is 3–CCGTACGTAATGCCGTAGTGTGATCCCTAG–5 and the coding strand is 5–GGCATGCATTACGGCATCACACTAGGGATC–3. The promoter is to the left of where the template and coding strands are located (e.g., to the left of the 5 end of the coding strand).

C13. RNA polymerase slides down the DNA and forms an open complex as it goes. The open complex is a DNA bubble of about 17 bp. Within the open complex, the DNA strand running in the 3 to 5 direction is used as a template for RNA synthesis. This occurs as individual nucleotides hydrogen

bond to the DNA template strand according to the rules described in question C12. As the RNA polymerase slides along, the DNA behind the open complex rewinds back into a double helix.

C14. When transcriptional termination occurs, the hydrogen bonds are broken between the DNA and the part of the newly made RNA transcript that is located in the open complex.

C15. In ρ-dependent termination, the ρ protein binds to the rut site in the RNA transcript after the ρ site has been transcribed. Eventually, an RNA sequence forms a stem-loop structure that will cause RNA polymerase to pause in the transcription process. As it is pausing, the ρ protein, which functions as a helicase, will catch up to RNA polymerase and break the hydrogen bonds between the DNA and RNA within the open complex. When this occurs, the completed RNA strand is separated from the DNA along with RNA polymerase. In ρ-independent transcription, there is no ρ protein. The RNA forms a stem-loop complex that causes RNA polymerase to pause. However, when it pauses, a uracil-rich region is bound to the DNA template strand in the open complex.

Because this is holding on by fewer hydrogen bonds, it is rather unstable. Therefore, it tends to dissociate from the open complex and thereby end transcription.

C16. DNA helicase and ρ protein bind to a nucleic acid strand and travel in the 5 to 3 direction. When they encounter a double-stranded region, they break the hydrogen bonds between complementary strands. The ρ protein is different from DNA helicase in that it moves along an RNA strand, whereas DNA helicase moves along a DNA strand. The function of DNA helicase is to promote DNA replication, whereas the function of ρ protein is to promote transcriptional termination.

C17. RNA and DNA polymerase are similar in the following ways: 1. They both use a template strand.

2. They both synthesize in the 5 to 3 direction.

3. The chemistry of synthesis is very similar in that they use incoming nucleoside triphosphates and make a phosphoester bond between the previous nucleotide and the incoming nucleotide.

4. They are both processive enzymes that slide along a template strand of DNA. RNA and DNA polymerase are different in the following ways: 1. RNA polymerase makes RNA and DNA polymerase makes DNA. 2. RNA polymerase does not require a primer.

3. RNA polymerase does not have a proofreading function or exonuclease activity.

C18. A. Mutations that alter the uracil-rich region by introducing Gs and Cs, and mutations that prevent the formation of the stem-loop structure.

B. Mutations that alter the termination sequence, and mutations that alter the ρ recognition site.

C. Eventually, somewhere downstream from the gene, another transcriptional termination sequence would be found, and transcription would terminate there. This second termination sequence might be in a random location, or it might be at the end of an adjacent gene.

C19. A. Ribosomal RNA (5.8S, 18S, and 28S) B. All of the mRNA and certain snRNA genes C. All of the tRNA genes and the 5S rRNA genes C20. Core promoters in eukaryotic protein-coding genes are somewhat variable with regard to the pattern of sequence elements. In the case of protein-coding genes that are transcribed by RNA polymerase II, it is common to have a TATA box, which is about 25 bp upstream from a transcriptional start site. The TATA box is important in the identification of the transcriptional start

site and the assembly of RNA polymerase and various transcription factors. The transcriptional start site defines where transcription actually begins.

C21. A. RNA polymerase would not bind to the core promoter.

B. TFIID contains the TATA-binding protein. If it were missing, RNA polymerase would not bind to the core promoter.

C. The formation of the open complex would not take place.

C22. The two models are described in Figure 14.15. The allosteric model is more like -independent termination, and the torpedo model is more like -dependent termination. In the torpedo model, a protein knocks RNA polymerase off the DNA, much like the effects of  protein.

C23. TFIID and TFIIB would play equivalent roles to sigma factor. Sigma factor does two things: it recognizes the promoter (as does TFIID), and it recruits RNA polymerase to the promoter (as does TFIIB).

C24. Hydrogen bonding is usually the predominant type of interaction in an assembly-and-disassembly process. In addition, ionic bonding and hydrophobic interactions could occur. Covalent interactions would not occur. High temperature and high salt concentration tend to break hydrogen bonds and thus would inhibit assembly and stimulate disassembly.

C25. Only the first intron would be spliced out. The mature RNA would contain: exon 1–exon 2–intron 2–exon 3.

C26. In bacteria, the 5 end of the tRNA is cleaved by RNaseP. The 3 end is cleaved by a different endonuclease, and then a few nucleotides are digested away by an exonuclease that removes nucleotides until it reaches a CCA sequence.

C27. The spliceosome is composed of several subunits known as snRNPs (containing snRNA and proteins). The function of the spliceosome is to cut the RNA in two places, hold the ends together, and then catalyze covalent bond formation between the two ends. During this process, small snRNA could be involved in binding the pre-mRNA and/or it could be catalytic, like the RNA in RNase P. C28. A ribozyme is an RNA molecule with catalytic activity. Examples are RNaseP and self-splicing group I and II introns. The spliceosome also contains a ribozyme.

C29. A gene is colinear when the sequence of bases in the coding strand of the DNA (i.e., the DNA strand that is complementary to the template strand for RNA synthesis) corresponds to the sequence of bases in the mRNA. Most prokaryotic genes and many eukaryotic genes are colinear. Therefore, you can look at the gene sequence in the DNA and predict the amino acid sequence in the polypeptide. Many eukaryotic genes, however, are not colinear. They contain introns that are spliced out of the pre-mRNA. C30. Self-splicing means that an RNA molecule can splice itself without the aid of a protein. Group I and II introns are self-splicing, although proteins can enhance the rate of that splicing.

C31. In eukaryotes, pre-mRNA can be capped, tailed, spliced, edited, and then exported out of the nucleus.

C32. In alternative splicing, variation occurs in the pattern of splicing, and so the resulting mRNAs contain alternative combinations of exons. The biological significance is that two or more different proteins can be produced from a single gene. This is a more efficient use of the genetic material. In multicellular organisms, alternative splicing is often used in a cell-specific manner.

C33. This is called cleavage or processing rather than splicing because none of the mature rRNA fragments are ever ligated (i.e., connected) to each other after the cleavage of the primary rRNA transcript.

C34. As shown at the left side of Figure 14.18, guanosine, which binds to the guanosine-binding site, does not have a phosphate group attached to it. This guanosine is the nucleoside that winds up at the 5 end of the intron. Therefore, the intron does not have a phosphate group at its 5 end.

C35. A. U1 and U4/U6 B. U5 C. U2 C36. U5 C37. The 60-nucleotide sequence would be found in the closed loop. The closed loop is the region between the 5 splice site and the branch site.

Application and Experimental Questions E1. The location of the intron within the cDNA is shown below. cDNA: 5′-ATTGCATCCAGCGTATACTATCTCGGGCCCAATTAATGCCAGC GGCCAGACTATCACCCAACTCG...INTRON...GTTACCTACTAGTATATCCCATATACTAG CATATATTTTACCCATAATTTGTGTGTGGGTATACAGTATAATCATATA–3′ You can figure this out by finding where the sequence of the genomic DNA begins to differ from the sequence of the cDNA. The genomic DNA has the normal splice sites that are described in Figure 14.19.

Genomic DNA: 5′– ATTGCATCCAGCGTATACTATCTCGGGCCCAATTAATGCCAGCGGCCAGACTATCACCC AACTACTCGGCCCACCCCCCAGGTTTACACAGTCATACCATACATACAAAAATCGCAG TTACTTATCCCAAAAAAACCTAGATACCCCACATACTATTAACTCTTTCTTTCTAGTTAC CTACTAGTATATCCCATATACTAGCATATATTTTACCCATAATTTGTGTGTGGGTATACA GTATAATCATATA–3′ The splice donor and acceptor sites are underlined. The space indicates where the strands in the corresponding RNA would be cut. The branch site is also underlined. The bold A is the adenine that participates in the splicing reaction.

E2. The 1,100-nucleotide band would be observed for a normal individual (lane 1). A deletion that removed the –50 to –100 region would greatly diminish transcription, so the homozygote would produce hardly any of the transcript (just a faint amount, as shown in lane 2), and the heterozygote would produce roughly half as much of the 1,100-nucleotide transcript (lane 3) compared to a normal individual. A nonsense codon would not have an effect on transcription; it affects only translation. So, the individual with this mutation would produce a normal amount of the 1,100nucleotide transcript (lane 4). A mutation that removed AG sequence at the 3′ splice site would prevent splicing. Therefore, this individual would produce a 1,550-nucleotide transcript (actually, 1,547 nucleotides to be precise, 1,550 minus 3). The Northern blot is shown here:

E3. When the 900 bp fragment is mixed with TFIID (lane 1), its movement would be slowed down because TFIID would bind. When mixed with TFIIB (lane 2), its movement would not be slowed down because TFIIB cannot bind without TFIID. Compared to lane 1, the movement of 900 bp fragment would be slowed down even more when mixed with TFIID and TFIIB (lane 3), because both transcription factors could bind. Its movement would not be slowed down when mixed with TFIIB and RNA polymerase (lane 4) because you do not have TFIID, which is needed for the binding of TFIIB and RNA polymerase. Finally, when mixed with TFIID, TFIIB, and RNA polymerase/TFIIF, the movement of the 900 bp fragment would be slowed down a great deal, because all four factors could bind (lane 5).

E4. A. The movement would not be slowed down because ρ protein would not bind to the mRNA that is encoded by a gene that is terminated in a ρ-independent manner. The mRNA from such genes does not contain the sequence near the 3′ end that acts as a recognition site for the binding of ρ protein.

B. The movement would be slowed down because ρ-protein would bind to the mRNA. C. The movement would be slowed down because U1 would bind to the pre-mRNA. D. The movement would not be slowed down because U1 would not bind to mRNA that has already had its introns removed. U1 binds only to pre-mRNA. Questions for Student Discussion/Collaboration 1. It would seem like a sufficient reason. Natural selection tends to produce species that are very efficient with regard to their DNA storage abilities. Therefore, having a single gene that encodes several different polypeptides makes an efficient use of the genetic material. Other people may not see it that way. For example, in Chapter 12, we discussed how some organisms, like amphibians, have a lot of unused DNA. Therefore, it doesn’t seem like a small genome size is necessarily an important attribute.

2. RNA transcripts come in two basic types: those that function as RNA (e.g., tRNA, rRNA, etc.) versus those that are translated (i.e., mRNA). As described in this chapter, RNAs play a myriad of functional

roles. RNAs that form complexes with proteins have some interesting roles. In some cases, the role is to bind other types of RNA molecules. For example, rRNA in bacteria binds to mRNA. In other cases, RNA plays a catalytic role. An example is RNaseP. The structure and function of an RNA molecule may be enhanced by forming a complex with one or more proteins.

CHAPTER 15 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 15. 1 Alkaptonuria occurs when homogentisic acid oxidase is defective.

FIGURE 15. 2 The strain 2 mutants are unable to convert O-acetylhomoserine into cystathionine.

FIGURE 15. 3 The role of DNA in polypeptide synthesis is storage. It stores the information that specifies the amino acid sequence of a polypeptide.

FIGURE 15. 5 Concept check: Which two amino acids do you think are the least soluble in water?

Tryptophan and phenylalanine are the least soluble in water.

FIGURE 15. 6 Hydrogen bonding promotes the formation of secondary structures in proteins.

FIGURE 15. 8 Only one radiolabeled amino acid was found in each sample. If radioactivity was trapped on the filter, this meant that the RNA triplet specified that particular amino acid.

FIGURE 15. 10 A tRNA has an antiocodon that recognizes a codon in the mRNA. It also has a 3′ acceptor stem where the correct amino acid binds.

FIGURE 15. 11 A charged tRNA has an amino acid attached to it.

FIGURE 15. 12 The wobble rules allow for a smaller population of tRNAs to recognize all of the possible mRNA codons.

FIGURE 15. 14 A site in mRNA promotes the binding of the mRNA to the ribosome during initiation. The codons in mRNA specify the polypeptide sequence during elongation. The stop codon terminates translation.

FIGURE 15. 16 The Shine-Dalgarno sequence in bacterial mRNA is complementary to a region in the 16S rRNA within the small ribosomal subunit. These complementary sequences hydrogen bond with each other.

FIGURE 15. 17 Peptidyl transferase catalyzes the formation of peptide bond between the amino acid at the A site and last amino acid in the growing polypeptide.

FIGURE 15. 18

The structures of release factors, which are proteins, resemble the structures of tRNAs. End-of-chapter Questions: Conceptual Questions C1. The start codon begins at the fifth nucleotide. The amino acid sequence would be Met Gly Asn Lys Pro Gly Gln STOP. C2. When saying that the genetic code contains degeneracy, it means that more than one codon can specify the same amino acid. For example, GGG, GGC, GGA, and GGU all specify glycine. In general, the genetic code is nearly universal, because it is used in the same way by viruses, prokaryotes, fungi, plants, and animals. As shown in Table 15.2, there are a few exceptions, which occur primarily in protists and in yeast and mammalian mitochondria.

C3. A. True B. False C. False C4. A. The mutant tRNA would recognize glycine codons in the mRNA but would put in tryptophan where glycine was supposed to be in the polypeptide chain.

B. This mutation indicates that the aminoacyl-tRNA synthetase is primarily recognizing other regions of the tRNA molecule besides the anticodon region. In other words, tryptophanyl-tRNA synthetase (the aminoacyl-tRNA synthetase that attaches tryptophan) primarily recognizes other regions of the tRNATrp sequence (that is, other than the anticodon region). If aminoacyl-tRNA synthetases recognized only the anticodon region, you would expect glycyl-tRNA synthetase to recognize this mutant tRNA and attach glycine. That is not what happens.

C5. As shown in Figure 15.11, the energy comes from ATP. It is this energy input that explains the term “charged tRNA.” C6. A. Three tRNAs. There are six leucine codons: UUA, UUG, CUU, CUC, CUA, and CUG. The anticodon AAU recognizes UUA and UUG. Two other tRNAs are needed to efficiently recognize the other four leucine codons. These could be GAG and GAU or GAA and GAU. B. One tRNA. There is only one methionine codon, AUG, so only one tRNA with the anticodon UAC is needed.

C. Three tRNAs. There are six serine codons: AGU, AGC, UCU, UCC, UCA, and UCG. Only one tRNA is needed to recognize AGU and AGC. This tRNA could have the anticodon UCG or UCA. Two tRNAs are needed to efficiently recognize the other four tRNAs. These could be AGG and AGU or AGA and AGU. C7. There are four proline codons, four glycine codons, one methionine codon, and six serine codons.

You apply the product rule to solve this problem.

4  4  1  6 = 96 C8. 3–CUU–5 or 3–CUC–5 C9. The codon is 5–CCA–3, which specifies proline.

C10. It can recognize 5–GGU–3, 5–GGC–3, and 5–GGA–3. All of these specify glycine.

C11. An anticodon that was 3–UUG–5 would recognize the two codons. To recognize 5–AAA–3, it would have to be modified to 3–UUI–5.

C12. All tRNA molecules have a cloverleaf structure with three stem-loops. The second stem-loop contains the anticodon sequence that recognizes the codon sequence in mRNA. At the 3 end, there is an acceptor stem, with the sequence CCA, that serves as an attachment site for an amino acid.

Most tRNAs also have base modifications that occur within their nucleotide sequences.

C13. They are very far apart, at opposite ends of the molecule.

C14. The role of aminoacyl-tRNA synthetase is to specifically recognize tRNA molecules and attach the correct amino acid to them. This ability is sometimes described as the “second genetic code” because the specificity of the attachment is a critical step in deciphering the genetic code. For example, if a tRNA has a 3–GGG–5 anticodon, it will recognize a 5–CCC–3 codon, which should specify proline. The aminoacyl-tRNA synthetase known as prolyl-tRNA-synthetase recognizes this tRNA and attaches proline to the 3 end. Other aminoacyl-tRNA synthetases will not recognize this tRNA. C15. In the context of translation, an activated amino acid has had AMP attached to it. This provides necessary energy so that the amino acid can be attached to the correct tRNA. C16. Bases that have been chemically modified can occur at various locations throughout a tRNA molecule. The significance of all of these modifications is not entirely known. However, within the anticodon region, base modification alters base pairing to allow the anticodon to recognize two or more different bases within the codon.

C17. A formyl group is covalently attached to methionine after the methionine has been attached to the tRNA containing a UAC anticodon.

C18. No, it is not. According to the wobble rules, the 5 base in the anticodon of a tRNA can recognize two or more bases in the third (3) position of the mRNA. Therefore, any given cell type can synthesize fewer than 61 types of tRNAs. C19. Translation requires mRNA, tRNAs, ribosomes, proteins such as initiation, elongation, and termination factors, and many small molecules. ATP and GTP are small molecules that contain high-energy bonds. The mRNA, tRNAs, and proteins are macromolecules. The ribosomes are a large complex of macromolecules.

C20. The assembly process is very complex at the molecular level. In eukaryotes, 33 proteins and one rRNA assemble to form a 40S subunit, and 49 proteins and three rRNAs assemble to form a 60S subunit. This assembly occurs within the nucleolus.

C21. A protein subunit is a polypeptide. A ribosomal subunit is a complex that is composed of RNA and many proteins. A ribosomal subunit is a much larger structure compared to a protein subunit.

C22. A. Within the 30S subunit and at the interface between the two subunits B. Within the 50S subunit C. Within the 50S subunit D. Within the 30S subunit C23. Initiation: The mRNA, initiator tRNA, and initiation factors associate with the small ribosomal subunit; then the large subunit associates.

Elongation: The ribosome moves one codon at a time down the mRNA, adding one amino acid at a time to the growing polypeptide. Three sites on the ribosome, the A, P, and E sites, are important in this process. The A site is where the tRNA (except for the initiator tRNA) binds to the ribosome and recognizes the codon in the mRNA. The growing polypeptide is then transferred to the amino acid

attached to this tRNA. The ribosome then translocates so that this tRNA is now moved to the P site.

The empty tRNA that was in the P site is moved into the E site. This empty tRNA in the E site is then expelled and the next charged tRNA can bind to the A site.

Termination: A stop codon is reached and a termination factor binds to the A site. The hydrolysis of GTP initiates a series of events that leads to the disassembly of the ribosomal subunits and the release of the completed polypeptide.

C24. Most bacterial mRNAs contain a Shine-Dalgarno sequence, which is necessary for the binding of the mRNA to the small ribosomal subunit. This sequence, UUAGGAGGU, is complementary to a sequence in the 16S rRNA. Due to this complementarity, these sequences hydrogen bond to each other during the initiation stage of translation.

C25. A. The initiator tRNA would not bind to the 40S ribosomal subunit.

B. mRNA would not bind to the 40S ribosomal subunit, and/or the start codon would not be recognized because the secondary structure in the mRNA was not unwound.

C. The 60S ribosomal subunit would not assemble after the start codon had been identified.

C26. The ribosome binds at the 5 end of the mRNA and then scans in the 3 direction in search of an AUG start codon. If it finds one that satisfies Kozak’s rules, it will begin translation at that site.

Aside from an AUG start codon, two other important features are a guanosine at the +4 position and a purine at the –3 position.

C27. 1. GCCACCAUGG 2. GACGCCAUGG 3. GCCUCCAUGC 4. GCCAUCAAGG The last one does not have a start codon, so it would not work. The third one may be translated, but very poorly.

C28. The A (aminoacyl) site is the location where a tRNA carrying a single amino acid initially binds.

The only exception is the initiator tRNA, which binds to the P (peptidyl) site. The growing polypeptide is removed from the tRNA in the P site and transferred to the amino acid attached to the tRNA in the A site. The ribosome translocates in the 3′ direction, with the result that the two tRNAs in the P and A sites are moved to the E (exit) and P sites, and the uncharged tRNA in the E site is released.

C29. The amino acid sequence is methionine-tyrosine-tyrosine-glycine-alanine. Methionine is at the amino terminal end and alanine is at the carboxyl terminal end. The peptide bonds should be drawn as shown in Figure 15.4.

C30. The initiation phase involves the binding of the Shine-Dalgarno sequence to the rRNA in the 30S subunit. The elongation phase involves the binding of anticodons in tRNA to codons in mRNA. C31. The nucleolus is a region inside the eukaryotic nucleus where the assembly of ribosomal subunits occurs.

C32. A. The E site and the P site (Note: A tRNA without an amino acid attached is only briefly found in the P site, just before translocation occurs.)

B. The P site and the A site (Note: A tRNA with a polypeptide chain attached is only briefly found in the A site, just before translocation occurs.)

C. Usually the A site, except the initiator tRNA, which can be found in the P site.

C33. A polysome is an mRNA molecule with many ribosomes attached to it.

C34. The tRNAs bind to the mRNA because their anticodon and codon sequences are complementary.

When the ribosome translocates in the 5 to 3 direction, the tRNAs remain bound to their complementary codons, and the two tRNAs shift from the A site and P site to the P site and E site.

If the ribosome moved in the 3 direction, it would have to dislodge the tRNAs and drag them to a new position where they would not (necessarily) be complementary to the mRNA. C35. A. False.

B. True.

C36. 52 C37. This means that translation can begin before transcription of the mRNA is completed. This cannot occur in eukaryotic cells because transcription and translation occur in different cellular compartments. In eukaryotic cells, transcription occurs in the nucleus, whereas translation occurs in the cytosol.

Application and Experimental Questions E1. A codon contains three nucleotides. Because G and C are present at 50% each, if you multiply 0.5  0.5  0.5, you get a value of 0.125, or 12.5%, of each of the eight possible codons. If you look the codons up in Table 15.1, you would expect 25% glycine (GGG and GGC), 25% alanine (GCC and GCG), 25% proline (CCC and CCG), and 25% arginine (CGG and CGC).

E2. The initiation phase of translation is very different in bacteria and in eukaryotes, so mRNAs would not be translated very efficiently in either case. A bacterial mRNA would not be translated very efficiently in a eukaryotic translation system, because it lacks a cap structure attached to its 5 end.

A eukaryotic mRNA would not have a Shine-Dalgarno sequence near its 5 end, so it would not be translated very efficiently in a bacterial translation system.

E3. An understanding of the structure of ribosomes can be quite helpful in elucidating the process of translation. For example, the sequencing of bacterial 16S rRNA showed that a sequence within this rRNA was complementary to the Shine-Dalgarno sequence. This helped molecular biologists understand how mRNA initially binds to the small ribosomal subunit. Electron microscopy studies revealed the structure of ribosomal subunits and how the mRNA, ribosomal subunits, and tRNAs “fit together” during the translation process. Overall, the studies of molecular biologists provide important pieces to the puzzle of how a large macromolecular complex can catalyze the synthesis of a polypeptide with a defined amino acid sequence based upon an mRNA with a defined nucleotide sequence.

E4.

E5. The sample in lane 2 came from an individual who is homozygous for a mutation that introduces an early stop codon into the coding sequence. As seen in lane 2, the protein is shorter than normal. The sample in lane 3 came from an individual who was homozygous for a mutation that prevented the expression of this polypeptide. It could be a down promoter mutation (i.e., a mutation in the promoter that decreases the rate of transcription), or it could be a mutation in the coding sequence that causes the protein to be degraded very rapidly. The sample in lane 4 came from an individual who is homozygous for a mutation that changed one amino acid to another amino acid. This type of mutation, termed a missense mutation, may not be detectable on a gel. However, a single amino acid substitution within a polypeptide may block protein function, which would explain the albino phenotype.

E6. A. If codon usage was significantly different between kangaroo and yeast cells, this would inhibit the translation process. For example, if the preferred leucine codon in kangaroos was CUU, translation would probably be slow in a yeast translation system. The cell-free translation system from yeast cells would primarily contain leucine tRNAs with the anticodon sequence AAC, because this tRNALeu would match the preferred yeast leucine codon, which is UUG. In a yeast translation system, there probably would not be a large amount of tRNA with the anticodon sequence GAA, which would match the preferred leucine codon, CUU, of kangaroos. For this reason, kangaroo mRNA would not be translated very well in a yeast translation system, but it probably would be translated to some degree.

B. The advantage of codon bias is that a cell can rely on a smaller population of tRNA molecules to efficiently translate its proteins. A disadvantage is that mutations, which do not change the amino acid sequence but do change a codon (e.g., UUG to UUA), may inhibit the production of a polypeptide if a preferred codon is changed to a nonpreferred codon.

Questions for Student Discussion/Collaboration 1. In some ways, it seems like a waste of energy, but the process of translation is pretty amazing and complex. You are starting with a sequence of nucleotides, and using that sequence as a "director" to make a sequence of amino acids within a polypeptide. As a metaphor, it’s like taking a string of pearls and using them to copy them into a string of colored glass beads. To do this reproducibly the same way every time, you would need to examine each pearl and have some rules that would correlate each pearl with a particular colored bead. You would need tools to carefully examine each pearl, and then another set of tools to actually make the string of beads. When you think about it in this type of context, it makes sense that translation at the molecular level would involve a large amount of cellular proteins and other molecules.

2. There are similarities along several lines: 1. Both processes involve a lot of molecular recognition, either between two nucleic acid molecules or between proteins and nucleic acid molecules.

2. Biosynthesis occurs in both processes. Small building blocks are being connected, which requires an input of energy.

3. There are genetic signals that determine the beginning and ending of these processes.

There are also many differences: 1. Transcription produces an RNA molecule with a similar structure to the DNA, whereas translation produces a polypeptide with a structure that is very different from RNA.

3.

2. Depending on your point of view, it seems that translation is more biochemically complex, requiring more proteins and RNA molecules to accomplish the task.

The recognition between the Shine-Dalgarno sequence and 16S rRNA is an RNA/RNA recognition.

Likewise, the binding of codons and anticodons is RNA/RNA. The scanning of the start codon in eukaryotes is an RNA recognition process. Protein/protein recognition is also important in ribosome assembly. Protein/RNA recognition occurs during the termination process.

CHAPTER 16 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 16.1 Gene regulation enhances efficiency in terms of energy use. A cell does not waste energy making RNAs and proteins it does not need.

FIGURE 16. 2 Activators and repressors are regulatory proteins. Inducers, corepressors, and inhibitors are small effector molecules.

FIGURE 16. 3 The lacZ, lacY, and lacA genes are under the control of the lac promoter.

FIGURE 16. 5 This repressor protein is bound to the lac operon when allolactose is not bound to the repressor, which occurs when lactose is not available.

FIGURE 16. 8 The repressor protein allows the bacterial cell to avoid turning on the operon in the absence of lactose.

The activator protein allows the cell to choose between glucose and lactose.

FIGURE 16.9 The data for the cases in which O2 and O3 are deleted indicate that O1, by itself, is not very strong in repressing the lac operon.

FIGURE 16. 11 Tryptophan acts as a corepressor that causes trp repressor to bind to the trp operon and repress transcription.

FIGURE 16. 12 Hydrogen bonding between complementary regions of the RNA transcribed from the beginning of the trp operon causes the stem-loops to form.

FIGURE 16. 13 The presence of tryptophan causes the trpL gene to be translated to its stop codon. This blocks both regions 1 and 2, which allows a 3-4 stem-loop to form. The 3-4 stem-loop causes termination of transcription.

FIGURE 16. 14 The micF antisense RNA binds to the ompF mRNA and inhibits its translation.

FIGURE 16. 15 Feedback inhibition prevents a bacterium from over producing the product of a metabolic pathway.

FIGURE 16. 16 The RNA conformation with an antiterminator stem-loop favors transcription.

FIGURE 16 .17 The RNA conformation with the Shine-Dalgarno antisequestor favors translation.

End-of-chapter Questions: Conceptual Questions C1. A constitutive gene is unregulated, which means that its expression level is relatively constant. In contrast, the expression of a regulated gene varies under different conditions. In bacteria, the regulation of genes often occurs at the level of transcription by combinations of regulatory proteins and small effector molecules. In addition, gene expression can be regulated at the level of translation or the function of a protein can be regulated after translation is completed.

C2. Gene regulation greatly enhances the efficiency of cell growth. It takes a lot of energy to transcribe and translate genes. Therefore, a cell is much more efficient and better at competing in its environment if it expresses a gene only when the gene product is needed. For example, a bacterium will express only the genes that are necessary for lactose metabolism when the bacterium is exposed to lactose. When the environment is missing lactose, these genes are turned off. Similarly, when tryptophan levels are high within the cytoplasm, the genes that are required for tryptophan biosynthesis are repressed.

C3. In this case, an inhibitor molecule and an activator protein are involved. The binding of the inhibitor molecule to the activator protein would prevent it from binding to the DNA and thereby inhibit its ability to activate transcription.

C4. A. Regulatory protein B. Effector molecule C. DNA segment D. Effector molecule E. Regulatory protein F. DNA segment G. Effector molecule C5. B and C are correct. In both of these cases, the presence of the small effector molecule will turn off transcription. In contrast, the presence of an inducer turns on transcription.

C6. A mutation that has a cis-effect is within a genetic regulatory sequence, such as an operator site, that affects the binding of a genetic regulatory protein. A cis-effect mutation affects only the adjacent gene(s) that the genetic regulatory sequence controls. A mutation having a trans-effect is usually in a gene that encodes a genetic regulatory protein. A trans-effect mutation can be complemented in a merozygote experiment by the introduction of a normal gene that encodes the regulatory protein.

C7. The term enzyme adaptation means that a particular enzyme is made only when a cell is exposed to the substrate for that enzyme. It occurs because the gene that encodes the enzyme that is involved in the metabolism of the substrate is expressed only when the cells have been exposed to the substrate.

C8. A. No transcription would take place. The lac operon could not be expressed.

B. No regulation would take place. The operon would be continuously turned on.

C. The rest of the operon would function normally but no galactoside transacetylase would be produced.

C9. It would be impossible to turn the lac operon on even in the presence of lactose because the repressor protein would remain bound to the operator site.

C10. Diauxic growth refers to a phenomenon due to gene regulation in which a cell first uses up one type of sugar (e.g., glucose) before it begins to metabolize a second sugar (e.g., lactose). When a bacterial cell is exposed to both glucose and lactose, the uptake of glucose causes the cAMP levels in the cell to fall. When this occurs, the catabolite activator protein (CAP) is removed from the lac operon so it is not able to be activated by CAP. C11. A. Cis-effect. It would affect only the genes that are in the adjacent operon.

B. Trans-effect. This is a mutation that affects a protein that can move throughout the cell.

C. Cis-effect. It would affect only the genes that are in the adjacent operon.

C12. A mutation that prevented lac repressor from binding to the operator would make the lac operon constitutive only in the absence of glucose. However, this mutation would not be entirely constitutive because transcription would be inhibited in the presence of glucose. The disadvantage of constitutive expression of the lac operon is that the bacterial cell would waste a lot of energy transcribing the genes and translating the mRNAs when lactose was not present.

C13. Attenuation means that transcription is ended before it has reached the end of an operon. Because it causes an end to transcription, it is a form of transcriptional regulation even though the translation of the trpL region plays a key role in the attenuation mechanism.

C14. A. Attenuation will not occur because a 2–3 stem-loop will form.

B. Attenuation will occur because a 2–3 stem-loop cannot form, so a 3–4 stem-loop will form.

C. Attenuation will not occur because a 3–4 stem-loop cannot form.

D. Attenuation will not occur because a 3–4 stem-loop cannot form.

C15. A defective tryptophanyl-tRNA synthetase would make attenuation less likely. This is because the bacterial cell would have a lower amount of charged tRNATrp. Therefore, it would be more likely for the ribosome to stall at the tryptophan codons found within the trpL gene, even if the concentration of tryptophan amino acids in the cell was high. When the ribosome stalls at these tryptophan codons, this prevents attenuation.

C16. The addition of G and C bases into the U-rich sequence would prevent attenuation. The U-rich sequence promotes the dissociation of the mRNA from the DNA, when the terminator stem-loop forms. This causes RNA polymerase to dissociate from the DNA and thereby causes transcriptional termination. The UGGUUGUC sequence would probably not dissociate because of the G and C bases. Remember that G-C base pairs have three hydrogen bonds and are more stable than A-U base pairs, which have only two hydrogen bonds.

C17. If you look very carefully at the RNA sequence in Figure 16.12, you will notice that a UAA codon is found just past region 2. Therefore, in this mutant strain, the UGA stop codon at the end of region 1 could be read by the mutant tRNAGly and then the ribosome would stop at the UAA codon just past region 2. If the ribosome paused here, it would probably cover up a portion of region 3, and therefore the terminator 3–4 stem-loop would not form. According to this scenario, attenuation could not occur. However, you should also keep in mind the issue of timing. The ribosome would have to be really close to RNA polymerase to prevent attenuation in this nonsense suppressor strain.

It is possible that the 3–4 stem-loop might form before the ribosome reaches the UAA stop codon just past region 2. Therefore, attenuation might occur anyway because the 3–4 stem-loop might form before the ribosome reaches the UAA stop codon.

C18. It takes a lot of cellular energy to translate mRNAs into proteins. A cell wastes less energy if it prevents the initiation of translation rather than preventing a later stage such as elongation or termination.

C19. Antisense RNA is RNA that is complementary to a functional RNA such as mRNA. The binding of antisense RNA to mRNA inhibits translation.

C20. One possible mechanism is that histidine acts as a corepressor that shuts down the transcription of the histidine synthetase gene. A second mechanism is that histidine acts as an inhibitor via feedback inhibition. A third possibility is that histidine inhibits the ability of the mRNA encoding histidine synthetase to be translated. Perhaps it induces a gene that encodes an antisense RNA. If the amount of histidine synthetase protein is identical in the presence and absence of extracellular histidine, a feedback inhibition mechanism is favored, because this would affect only the activity of the histidine synthetase, not the amount of the enzyme. The other two mechanisms would diminish the amount of this enzyme.

C21. 1. lac operon: The binding of allolactose causes a conformational change in the repressor protein and removes it from the operator site and thereby allows transcription.

2. lac operon: The binding of cAMP to CAP allows it to bind to the promoter region and facilitate transcription.

3. trp operon: The binding of tryptophan to trp repressor causes it to bind to the operator site and inhibits transcription.

C22. The two proteins are similar in that both bind to a segment of DNA and repress transcription. They are different in three ways: (1) They recognize different effector molecules (i.e., lac repressor recognizes allolactose, and trp repressor recognizes tryptophan. (2) Allolactose causes lac repressor to release from its operator, whereas tryptophan causes trp repressor to bind to its operator. (3) The sequences of the operator sites that these two proteins recognize are different. Otherwise, lac repressor could bind to the trp operator, and trp repressor could bind to the lac operator.

C23. A. Antisense RNA or a translational repressor would shut down protein synthesis the fastest. A transcriptional repressor would also shut down the synthesis of mRNA, so it would eventually shut down protein synthesis once all of the preexisting mRNA had been degraded. Feedback inhibition would have no effect on protein synthesis.

B. Only a transcriptional repressor protein would shut down the synthesis of mRNA. C. Feedback inhibition is the fastest way to shut down the function of a protein. Antisense RNA and transcriptional repressors eventually prevent protein function once all of the pre-existing mRNA and proteins have been degraded.

Application and Experimental Questions E1. A. β-ONPG was used to measure the level of expression of the lac operon. More specifically, the cleavage of β-ONPG to produce a yellow compound requires the action of β-galactosidase. Thus the assay is indirectly measuring the amount of β-galactosidase protein. The reason why there was no yellow color in one of the tubes was because the repressor was preventing the expression of the lac operon, and this prevents the expression of β-galactosidase enzyme. Some other methods to measure the expression level of the lac operon could include the following: 1. Conduct a Northern blot to measure the amount of mRNA that is produced from the lac operon.

2. Conduct a Western blot using antibodies against one of the three proteins encoded by the lac operon.

3. Measure the uptake of radiolabeled lactose into the cells. This would measure the amount of the lactose permease that is expressed from the lac operon.

B. The merozygote has two copies of the lacZ gene, so it makes twice as much β-galactosidase.

E2. In samples loaded in lanes 1 and 4, the repressor should bind to the operator because no lactose is present. In the sample loaded into lane 4, the CAP protein could still bind cAMP because there is no glucose. However, there is no difference between lanes 1 and 4, so CAP does not appear to activate transcription when lac repressor is bound. If you compare samples loaded into lanes 2 and 3, lac repressor is not bound in either case, and CAP is not bound in the sample loaded into lane 3. There is less transcription in lane 3 compared to lane 2, but because there is some transcription observed in lane 3, you can conclude that the removal of CAP (because cAMP levels are low) is not entirely effective at preventing transcription. Overall, the results indicate that the binding of lac repressor is more effective at preventing transcription of the lac operon than is the removal of CAP. E3. In the normal strain, the lac operon would be fully induced, so maximal expression of the mRNA (lane 1) would occur. The same thing would happen in lane 2, because the lac repressor has been inactivated. In lane 3, you would not see any mRNA because the lac repressor does not bind to allolactose. Therefore, the lac repressor would remain bound to the lac operator even in the presence of lactose. In lane 4, the CAP would not activate transcription, although the repressor would not prevent transcription. You would probably see a little bit of transcription, but not as much as the maximal level seen in lane 1. In media without lactose, you would see a band only in lane 2. The only strain that would be induced would be the one that lacked a functional lac repressor.

E4. A. Yes, if the researcher did not sonicate the cells, then β-galactosidase will not be released from them, and not much yellow color will be observed. (Note: A little yellow color may be seen because some of the β-ONPG may be taken into the cells.)

B. No, yellow color should still be observed in the first two tubes even if the researcher forgot to add lactose because the unconjugated strain does not have a functional lac repressor.

C. Yes, if the researcher forgot to add β-ONPG, there will be yellow color because the cleavage of β-ONPG by β-galactosidase is what produces the color.

E5. The data indicate that two operators are needed because you need O1 plus either O2 or O3 to get a high level of repression. If all three operators were needed, the deletion of any one of them should have prevented high levels of repression. This was not observed.

E6. You could conjugate a strain that has an F factor carrying a normal lac operon and a normal lacI gene to the mutant strain. Because the mutation is in the operator site, expression of β-galactosidase would continue to occur, even in the absence of lactose.

E7. It appears to be a mutation in the operator site, so that a repressor protein cannot bind there and prevent lac operon expression in the absence of lactose.

Questions for Student Discussion/Collaboration 1. Transcriptional regulation is the most efficient form of regulation from the point of view of energy.

If a protein is not needed by a bacterium, turning off the gene (e.g., via a repressor protein) prevents the synthesis of the mRNA as well as the protein itself. Therefore, the bacterium does not waste

2.

energy making a protein it doesn’t need. On the other hand, transcriptional control is fairly slow because it takes time to transcribe a gene and synthesize a polypeptide. By comparison, the regulation of protein function via feedback inhibition or covalent modification is very fast.

Translational control, which involves the regulation of mRNA translation is in the middle. It is not as efficient as transcriptional control because mRNA is made, but it is faster than transcriptional control. However, it is not as fast as the regulation of protein function.

A DNA loop may inhibit transcription by preventing RNA polymerase from recognizing the promoter. Or, it may inhibit transcription by preventing the formation of the open complex.

Alternatively, a bend may enhance transcription by exposing the base sequence that sigma factor of RNA polymerase recognizes. The bend may expose the major groove in such a way that this base sequence is more accessible to binding by sigma factor and RNA polymerase.

CHAPTER 17 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 17. 1 Transcription regulation is the most energy-efficient way to regulate a gene, because a cell avoids wasting energy making RNAs or proteins that aren’t needed.

FIGURE 17. 3 If TFIID cannot bind to the TATA box, RNA polymerase will not be recruited to the core promoter, and therefore transcription will not begin.

FIGURE 17. 4 When an activator interacts with mediator, mediator phosphorylates the CTD, which causes RNA polymerase to proceed to the elongation phase of transcription.

FIGURE 17. 6 The glucocorticoid receptor binds only next to genes that have an adjacent GRE. FIGURE 17. 7 Eviction of histone octamers may allow certain proteins to access particular sites in the DNA and bind there, which could affect transcription.

FIGURE 17. 8 Histone modifications may directly affect the interaction between histones and the DNA, or they may affect the binding of other proteins to the chromatin.

FIGURE 17. 9 An NFR is needed at the core promoter so that transcription factors can recognize enhancers and so the preinitiation complex can form.

FIGURE 17. 10 Histone eviction or displacement is needed for elongation to occur because RNA polymerase cannot transcribe through nucleosomes. It needs to unwind the DNA for transcription to take place.

FIGURE 17. 12 In part (a), DNA methylation blocks an activator protein from binding to the DNA. This prevents transcriptional activation.

FIGURE 17. 13 De novo methylation occurs on unmethylated DNA, whereas maintenance methylation occurs on hemimethylated DNA. FIGURE 17 . 16 A reader domain recognizes a specific type of structure in a histone protein, such as an H3K9me3 modification, and binds to it.

FIGURE 17.18 Each nucleosome contains two copies of histone H3, so there can be a maximum of two H3K9me3 modifications in a single nucleosome.

FIGURE 17 . 20 The choice occurs only during embryonic development.

FIGURE 17. 21 A knotlike structure would prevent the binding of proteins, such as TFIID and RNA polymerase II, to the gene and thereby inhibit transcription.

FIGURE 17.23 At the level of DNA sequences, queens and worker bees are not different from each other. However, differences in gene expression due to epigenetics can explain the morphological differences between a queen and the worker bees.

FIGURE 17. 25 The mutation would cause the overproduction of ferritin, because ferritin synthesis would occur even if iron levels were low.

End-of-chapter Questions: Conceptual Questions C1. 1. Transcription. This includes regulatory transcription factors; DNA methylation (the attachment of methyl groups, which usually inhibits transcription); and changes in the arrangements and composition of histones and nucleosomes.

2. RNA level. This includes RNA processing, regulation of alternative splicing via SR proteins; RNA stability, regulation of RNA half-life and RNA translation via RNA binding proteins; regulation via miRNA and siRNA. 3. Protein level. This includes feedback inhibition, small molecules that modulate enzyme activity; and posttranslational modification-covalent changes to protein structure that affect protein activity.

C2. Regulatory elements are relatively short DNA sequences that are recognized by regulatory transcription factors. After the regulatory transcription factor has bound to a regulatory element, it will affect the rate of transcription, either increasing or decreasing it, depending on the action of the regulatory protein. Regulatory elements are typically located in the upstream region near the promoter, but they can be located almost anywhere (i.e., upstream and downstream) and even quite far from the promoter.

C3. Transcription factor modulation refers to different ways that the function of transcription factors can be regulated. The three general ways are the binding of an effector molecule, protein-protein interactions, and covalent modifications. These changes can enhance the function of a transcription factor or inhibit it.

C4. Transcriptional activation occurs when a regulatory transcription factor binds to a regulatory element and activates transcription. These transcriptional activator proteins may interact with TFIID and/or mediator to promote the assembly of RNA polymerase II and general transcription factors at the promoter region. They may also alter the structure of chromatin so that RNA polymerase and transcription factors are able to gain access to the promoter. Transcriptional inhibition occurs when a regulatory transcription factor inhibits transcription. Such a repressor also may interact with TFIID and/or mediator to inhibit RNA polymerase II. C5. A. True.

B. False.

C. True.

D. False, it causes up regulation.

C6. For the glucocorticoid receptor to bind to a GRE, a steroid hormone must first enter the cell. The hormone then binds to the glucocorticoid receptor, which releases HSP90. The release of HSP90 exposes a nuclear localization signal (NLS) within the receptor, which enables it to dimerize and then enter the nucleus. Once inside the nucleus, the dimer binds to a GRE, which activates transcription of the adjacent genes.

C7. 1. It could be in the DNA-binding domain, so that the receptor would not recognize a GRE. 2. It could be in the HSP90 domain, so that HSP90 would not be released when the hormone binds.

3. It could be in the dimerization domain, so that the receptor would not dimerize.

4. It could be in the nuclear localization domain, so that the receptor would not travel into the nucleus.

5. It could be in the domain that activates RNA polymerase, so that the receptor would not activate transcription, even though it could bind to GREs. C8. The glucocorticoid receptor binds to the glucocorticoid hormone with a certain affinity. The binding is a reversible process. Eventually, the glucocorticoid hormone will be degraded by the cell. Once the concentration of the hormone falls below the affinity of the hormone for the receptor, the receptor will no longer have the hormone bound to it. When the hormone is released, the glucocorticoid receptor will change its conformation, and it will no longer bind to the DNA. C9. The enhancer found in A would work, but the ones found in B and C would not. The sequence that is recognized by the transcriptional activator is 5′–GTAG–3′ in one strand and 3′–CATC–5′ in the opposite strand. This is the same arrangement found in A. In B and C, however, the arrangement is 5′–GATG–3′ and 3′–CATC–5′. In the arrangement found in B and C, the two middle bases (i.e., A and T) are not in the correct order.

C10. ATP-dependent chromatin remodeling complexes may change the positions of nucleosomes, evict histones, and/or replace histones with histone variants.

C11. A histone variant is a histone with an amino acid sequence that is slightly different from a core histone. Histone variants play specialized roles with regard to chromatin structure and function.

C12. The attraction between DNA and histones occurs because the histones are positively charged and the DNA is negatively charged. The covalent attachment of acetyl groups decreases the amount of positive charge on the histone proteins and thereby may decrease the binding of the DNA. In addition, histone acetylation may attract proteins to the region that loosen chromatin compaction.

C13. The histone code hypothesis is the idea that the patterns of covalent modifications of histones are recognized by proteins that affect chromatin structure and function. In this way, the modification of histones plays a role in gene regulation.

C14. A nucleosome-free region (NFR) is a location on a chromosome where nucleosomes are missing.

These regions are typically found at the beginnings and ends of genes. An NFR at the beginning of a gene is thought to be important for activation of the gene. An NFR at the end of a gene may be important for its proper termination.

C15. A potentially harmful consequence would be that transcription may be initiated at multiple points within a gene, thereby producing many nonfunctional transcripts. This would be waste of energy.

C16. The attachment of the acetyl group (—COCH3) eliminates the positive charge on the lysine side chain, thereby disrupting the electrostatic attraction between the histone protein and the negatively charged DNA backbone. This makes it easier for the histones to be displaced from the DNA. C17. DNA methylation is the attachment of a methyl group to a base within the DNA. In many eukaryotic species, this occurs on cytosine at a CG sequence. After de novo methylation has occurred, it is passed from mother to daughter cell. Because DNA replication is semiconservative, the newly made DNA contains one strand that is methylated and one that is not. DNA methyltransferase recognizes this hemimethylated DNA and methylates the cytosine in the unmethylated DNA strand; this event is called maintenance methylation.

C18. Perhaps the DNA methyltransferase is responsible for methylating and inhibiting a gene that causes a cell to become a muscle cell. The methyltransferase is inactivated by the mutation.

C19. A CpG island is a stretch of 1,000 to 2,000 base pairs in length that contains a high number of CpG sites. CpG islands are often located near promoters. When the island is methylated, this inhibits transcription. This inhibition may be the result of the inability of the transcriptional activators to recognize the methylated promoter and/or the effects of methyl-CpG-binding proteins, which may promote a closed chromatin conformation.

C20. Epigenetics is the study of mechanisms that lead to changes in gene expression that can be passed from cell to cell and are reversible, but do not involve a change in the sequence of DNA. Not all epigenetic changes are passed from parent to offspring. For example, those that occur in somatic cells, such as lung cells, would not be passed to offspring.

C21. 1. DNA methylation-the attachment of methyl groups to cytosines in DNA. This often silences transcription.

2. Covalent histone modification-the covalent attachment of groups to the amino terminal tails of histones. This may silence or activate genes.

3. Chromatin remodeling-changes in the positions of nucleosomes. This may lead to a closed or open conformation for transcription.

4. Histone variants-replacement of standard histones for histone variants. This may silence or activate genes.

5. Feedback loop-the activation of a gene encoding a transcription factor. After the transcription factor is made, it continues to activate its own expression as well as the expression of other genes.

C22. Some key functions of heterochromatin include chromosome sorting during cell division, gene silencing, prevention of movement of transposable elements, and prevention of viral proliferation.

C23. Constitutive heterochromatin is located close to a centromere, called the pericentric region, and also at telomeres. Facultative heterochromatin may occur at multiple discrete sites that are located between the centromere and telomeres.

C24. A, B, and D would inhibit heterochromatin formation because DNA methylation, the methylation of histones (e.g., H3K9), and the removal acetyl groups all favor heterochromatin formation. In part C, a drug that prevents histone acetylation would favor heterochromatin formation.

C25. HP1 forms crossbridges that bring nucleosomes closer together. CTCFs and SMC proteins promote the formation of loop domains, which become closer together in heterochromatin. LADs bind to the nuclear lamina.

C26. During oogenesis, the ICR and DMR are not methylated. This allows CTCF to bind to these sites and to each other to form a loop. This loop prevents a nearby enhancer from activating the Igf2 gene. During sperm formation, methylation of these sites prevents CTCF binding and thereby prevents loop formation. The nearby enhancer activates the Igf2 gene. Therefore, the copy of the Igf2 gene inherited from the father via sperm is active.

C27. The Igf2 gene inherited from the mother would be active. For silencing to occur, a loop must form.

The loop could not form if the ICR was missing.

C28. See Figure 17.16. The choice of which X chromosome remains active occurs when the pluripotency factors and CTCFs, which were previously bound to both X chromosomes, shift entirely to one of the X chromosomes. Because the pluripotency factors stimulate the expression of the Tsix gene, the X chromosome to which they shift is chosen as the active X chromosome. By comparison, on the other X chromosome, the Tsix gene is not expressed, which permits the expression of the Xist gene.

C29. After the Xist RNA coats the X chromosome, it recruits proteins that silence genes, such as DNA methyltransferase, and proteins that make it more compact and form a Barr body. One gene on Xi that is not inactivated is the Xist gene.

C30. Epigenetic modifications lead to changes in gene expression that persist from early development to adulthood. These changes allow some genes to be expressed in certain cell types but not in others.

The trithorax and polycomb group complexes are involved in causing such epigenetic changes.

C31. The trithorax and polycomb group complexes are involved with gene activation and gene repression during development, respectively. They cause epigenetic changes that allow genes to either remain active or permanently repressed. Such changes occur during embryonic development and are maintained during subsequent stages.

C32. See Figure 17.17 C33. The consequences would be that many genes would be expressed in cell types where they should not be expressed. This would cause abnormalities in development and would likely be lethal.

C34. Dietary factors can affect the occurrence of epigenetic changes, particularly those that happen during early stages of development. For example, dietary factors that affect the level of DNA methylation appear to be responsible or partly responsible for altering coat color in mice and affecting female bee development. These dietary factors may affect the level of S-adenosyl methionine and/or the function of DNA methyltransferase and thereby influence whether particular genes are methylated or not. If genes are methylated, this typically inactivates them.

C35. The binding of IRP to the IRE inhibits the translation of ferritin mRNA and enhances the stability of the transferrin receptor mRNA. The increase in the stability of transferrin receptor mRNA increases the concentration of this mRNA and ultimately leads to more transferrin receptor proteins.

Conditions of low iron promote the binding of IRP to the IRE, leading to a decrease in ferritin proteins and an increase in transferrin receptor proteins. When the iron concentration is high, iron binds to IRP, causing it to be released from the IRE. This allows the ferritin mRNA to be translated and also causes a decrease in the stability of transferrin receptor mRNA. Under these conditions, more ferritin proteins are made, and less transferrin receptors are made.

C36. This person would be unable to make ferritin, because the IRP would always be bound to the IRE. The amount of transferrin receptor mRNA would be high, even in the presence of high amounts of iron, because the IRP would always remain bound to the IRE, stabilizing the transferrin receptor mRNA. Such a person would not have any problem taking up iron into his or her cells and, in fact, would take up a lot of iron via the transferrin receptor, even when the iron concentrations were high.

Therefore, this person would not need more iron in the diet. However, excess iron in the diet would be very toxic for two reasons. First, the person cannot make ferritin, which prevents the toxic buildup of iron in the cytosol. Second, when iron levels are high, the person would continue to synthesize the transferrin receptor, which functions in the uptake of iron.

C37. A cell may need to respond rather quickly to a toxic substance in order to avoid cell death.

Translational and posttranslational mechanisms are much faster than transcriptional activation, in which it is necessary to up-regulate the gene, synthesize the mRNA, and then translate the mRNA to make a functional protein.

Application and Experimental Questions E1. These results indicate that the fibroblasts undergo maintenance methylation because they can replicate and methylate DNA if it has already been methylated. However, the cells do not undergo de novo methylation, because if the donor DNA was unmethylated, the DNA in the daughter cells remains unmethylated.

E2. If the DNA fragment is 3,800 bp, it means that the site is unmethylated, because it has been cut by NotI (and NotI cannot cut methylated DNA). If the DNA fragment is 5,300 bp, the DNA is not cut

by NotI, so you assume that it is methylated. For the sample in lane 4, the gene is isolated from root tissue, and the DNA is not methylated because the sample runs at 3,800 bp. This suggests that gene T is expressed in root tissue. In the other samples, the DNA runs at 5,300 bp, indicating that the DNA is methylated. The pattern of methylation seen here is consistent with the known function of gene T. You would expect it to be expressed in root cells, because it functions in the uptake of phosphate from the soil. It would be silenced in the other parts of the plant via methylation.

E3. The results indicate that protein X binds to the DNA fragment and slows down its mobility (lanes 3 and 4). However, the hormone is not required for DNA binding. Because you already know that the hormone is needed for transcriptional activation, it must play some other role. Perhaps the hormone activates a signaling pathway that leads to the phosphorylation of the transcription factor, and phosphorylation is necessary for translational activation.

E4. The coat color relies on the expression of the Agouti gene. In this case, the Agouti gene is under the control of a promoter that is found in a transposable element, causing it to be overexpressed. When the promoter is not methylated and the gene is overexpressed, the coat color is yellow because the

Agouti gene product promotes the synthesis of yellow pigment. When the promoter is methylated, this inhibits the Agouti gene and coat color is darker because less yellow pigment is made.

E5. It would probably cause their coat color to be more yellow. When the promoter within the TE is not methylated, this results in the overexpression of the Agouti gene, which promotes the synthesis of yellow pigment. 5-azocytidine would prevent methylation of the promoter and thereby allow the Agouti gene to be overexpressed.

E6. The blot shown in (a) is correct for ferritin mRNA regulation. The presence and absence of iron does not affect the amount of mRNA; it affects the ability of the mRNA to be translated. The blot shown in (b) is correct for transferrin receptor mRNA. In the absence of iron, the IRP binds to the IRE in the mRNA and stabilizes it, so the mRNA level is high. In the presence of iron, the iron binds to IRP and IRP is released from the IRE, and the transferrin receptor mRNA is degraded.

Questions for Student Discussion/Collaboration 1. DNA methylation can regulate gene expression in a tissue-specific way because de novo methylation may occur only next to particular genes in particular tissues. This would tend to keep them permanently turned off during later stages of development. Genes may be methylated in the germ line to prevent their expression. However, if a gene was methylated in the germ line, it would have to be demethylated during gamete formation or at an early stage of embryonic development in the offspring so that it could be expressed in particular tissues in the offspring. Alternatively, genes that are regulated in a tissue-specific way may not be methylated in the germline, but instead are methylated during embryonic development in tissues where their expression should be silenced.

2. This list could a long one; it could include agents in cigarette smoke, pesticides, herbicides, heavy metals, etc.

CHAPTER 18 Note: The answers to the Comprehension questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 18.1 DNA, RNAs, proteins, and small molecules can bind to an ncRNA. FIGURE 18. 6 RISC binds to a specific mRNA because the small RNA within RISC is complementary to that mRNA. The bonding is hydrogen bonding between complementary bases.

FIGURE 18. 7 The hydrolysis of GTP promotes the release of SRP, which then allows translation to resume.

FIGURE 18. 8 The crRNA binds directly to one of the strands of the bacteriophage DNA strands via hydrogen bonding between complementary bases.

FIGURE 18 . 9 piRNAs and PIWI proteins cause TE RNAs to be degraded, and they prevent TEs from being transcribed.

End-of-chapter Questions: Conceptual Questions C1. ncRNAs can bind to DNA and RNA due to complementary base pairing. Stem-loops in ncRNAs can bind to the surface of proteins. Also, segments of ncRNAs can form pockets for the binding of small molecules.

C2. A. Scaffold and guide B. Ribozyme C. Guide D. Guide C3. First, HOTAIR acts a scaffold for the binding of PRC2 and LSD1. Second, HOTAIR acts a guide and binds to GA-rich sequences next to particular target genes. The histones associated with those genes are then covalently modified via the actions of PRC2 and LSD1.

C4. RNA interference refers to the phenomenon in which the presence of a double-stranded RNA molecule results in the silencing of a complementary mRNA. First, the double-stranded RNA is processed by dicer into small RNA fragments (miRNA or siRNA). The miRNA or siRNA associates with a complex called the RNA-induced silencing complex (RISC). RISC then binds to a complementary mRNA. This either leads to mRNA degradation or inhibits translation.

C5. First, the double-stranded miRNA could be produced by transcription of a gene, as in pri-miRNA. Second, it could come from a virus. Third, double-stranded RNA could be made experimentally.

C6. MicroRNAs (miRNAs) are ncRNAs that are transcribed from endogenous eukaryotic genes—genes that are normally found in the genome. They play key roles in regulating gene expression, particularly during embryonic development in animals and plants. Most commonly, a single type of miRNA inhibits the translation of several different mRNAs. By comparison, small-interfering RNAs (siRNAs) are ncRNAs that usually originate from exogenous sources, which means that they are not normally made by cells. They usually promote mRNA degradation. The sources of siRNAs can be viruses that infect a cell, or researchers can make siRNAs in order to study gene function experimentally.

C7. SRP in eukaryotes is composed of one ncRNA and six proteins. SRP recognizes a signal sequence near the beginning of a polypeptide and pauses translation. In eukaryotes, SRP then guides the polypeptide-ribosome complex to an SRP receptor in the ER membrane that is next to a channel.

SRP is then released and translation resumes. In this way, the protein is targeted to the ER. C8. If the SRP RNA did not stimulate the GTPase activities, SRP would remain bound to the polypeptide and translation would not resume.

C9. The role of tracrRNA is a guide. It binds to crRNA and guides it to Cas9. The role of crRNA is also that of a guide. It guides the crRNA-tracrRNA-Cas9 complex to the bacteriophage DNA. C10. tracrRNA is a guide, directing crRNA to Cas9.

C11. Cas1 and Cas2 are needed for the adaptation phase in which segments of bacteriophage DNA are incorporated into the Crispr gene. Cas9 is an endonuclease that cleaves the bacteriophage DNA during the interference phase.

C12. See Figure 18.7. The role of piRNAs is to guide the PIWI proteins to the TE RNA, where they either cause transcription to be inhibited or cause degradation of the TE RNA. C13. See Table 18.2.

C14. The miR-200 family plays an essential role in tumor suppression by inhibiting an event called the epithelial-mesenchymal transition (EMT), which is the initiating step of metastasis. During the EMT, cells lose their adhesion to neighboring cells. This loss of adhesion is associated with a decrease in expression of E-cadherin, which is a membrane protein that adheres adjacent cells to each other. When E-cadherin levels are low, cells can more easily move to new sites in the body. In an adult, when the miR-200 family of miRNAs is expressed at normal levels, these miRNAs inhibit the EMT. This inhibition maintains a normal level of E-caderin and thereby prevents metastasis.

However, if miR-200 miRNAs are expressed at low levels, EMT is not inhibited, and metastasis can occur.

Application and Experimental Questions E1. The mutation would prevent the targeting of trypsin to the ER membrane and thereby prevent it from being secreted. It would remain in the cytosol.

E2. The researchers were injecting pre-siRNA. In their experiment, it was a perfect match to the mex-3 RNA and caused its degradation.

E3. When they injected just antisense mex-3 RNA, they observed lower but detectable levels of mex-3 mRNA. However, the injection of both sense and antisense RNA, which would form a doublestranded structure, resulted in a complete loss of mex-3 mRNA in the cells. In this way, doublestranded RNA silenced the expression of the mex-3 gene.

E4. The CRISPR-Cas system is used to alter genes by causing a small deletion within a gene or producing a change in the sequence of a gene. It can be used to cause a small deletion within a gene or to produce a change in the sequence of gene. In this system, the tracrRNA and crRNA form one RNA called a single guide RNA (sgRNA). The sgRNA recognizes both the Cas9 protein and a target gene that a researcher wants to mutate. After the sgRNA-Cas9 complex binds to the target gene, Cas9 cleaves the gene. Following nonhomologous end joining, the result is often a deletion of a small segment of the target gene, or if donor DNA is added, a double crossover can introduce a point mutation into the target gene.

E5. The term anti-miRNA oligonucleotides is a broad term that refers to an oligonucleotide that inhibits miRNA function. Two types are locked nucleic acids and antagomirs. Locked nucleic acids contain a ribose sugar that has an extra bridge connecting the 2′ oxygen and 4′ carbon. The bridge locks the ribose in a conformation that causes it to bind more tightly to the complementary miRNAs. Antagomirs have one or more base modifications that may promote a stronger binding to complementary miRNAs. E6. miRNA replacement therapy is aimed at restoring the function of miRNAs that have been downregulated in cancer cells. One example of this treatment approach is the use of a viral vector to restore a particular miRNA. Another example is to increase the levels of miRNAs by targeting the miRNA-processing machinery via drugs such as enoxacin, which enhances the miRNA-processing machinery in certain cancer cells and thereby results in an increase of many miRNAs. A third example of miRNA replacement therapy is the use of DNA demethylating agents and histone deacetylase inhibitors. These compounds reverse the epigenetic silencing of miRNAs that behave as tumor suppressors.

Questions for Student Discussion/Collaboration 1. This is a matter of opinion. The ncRNAs that are needed for translation, like rRNAs and tRNAs, may have arisen during the RNA world because they may have preceded the formation of proteins in living cells. Other ncRNAs, like those that affect transcription, protein targeting, and genome defense, interact with DNA and/or proteins and probably arose after the modern DNA/RNA/protein world came into being.

2. The search will lead to many examples of roles played by ncRNAs that are not discussed in this text. It should be easy and interesting to make such a list.

CHAPTER 19 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 19.2 A position effect has occurred when the expression of a gene has been altered due to a change in its position along a chromosome.

FIGURE 19.3 In part (b), the DNA sequence of the eye color gene has not been changed. The change in its expression compared with part (a) is due to a position effect.

FIGURE 19.5 A trait due to a somatic mutation is not passed from parent to offspring.

FIGURE 19. 6 Because the colonies on the secondary plates appear in the same locations, the mutations occurred on the master plate before exposure to T1. Therefore, the mutations did not occur as a physiological adaptation to T1 exposure.

FIGURE 19. 7 The probability is 75% because 3 of the 4 nucleotides that may be added to the newly made strand opposite the apurinic site are not the correct one.

FIGURE 19. 8 The deamination of methylcytosine is more difficult to repair because thymine is a normal base found in DNA. This makes it difficult for DNA repair enzymes to distinguish between the correct and altered base.

FIGURE 19.10 A reactive oxygen species is a molecule that contains oxygen and may react with cellular molecules in a way that may cause harm.

FIGURE 19.13 The base analog 5-bromouracil causes a transition.

FIGURE 19.14 Thymine dimer formation is often the result of exposure to UV light. It most commonly occurs in skin cells.

FIGURE 19.15

Enzymes within the rat liver extract may convert nonmutagenic molecules into mutagenic agents.

Adding the extract allows researchers to identify molecules that may be mutagenic in people.

FIGURE 19.17 Cuts are made on both sides of the damaged region so that it can be removed by UvrD. FIGURE 19. 19 MutH distinguishes between the parental strand and the newly made daughter strand, which ensures that the daughter strand is repaired.

FIGURE 19.20 An advantage of NHEJ as a repair system is that it can occur at any stage of the cell cycle. A disadvantage is that it may be imprecise and result in a short deletion in the DNA. End-of-chapter Questions: Conceptual Questions C1. A. G→A, which is a transition.

B. T→G, which is a transversion.

C. A single-nucleotide deletion.

D. C→G, which is a transversion.

C2. It is a gene mutation, a point mutation, a base substitution, a transition mutation, a deleterious mutation, a mutant allele, a nonsense mutation, a conditional mutation, a temperature-sensitive mutation and a lethal allele.

C3. A suppressor mutation suppresses the phenotypic effects of some other mutation. Intragenic suppressors are within the same gene as the first mutation. Intergenic suppressors are in another gene.

C4. A. A nonsense mutation would probably inhibit protein function, particularly if did not occur near the end of the coding sequence.

B. A missense mutation may or may not affect protein function, depending on the nature of the amino acid substitution and on whether the substitution is in a critical region of the protein.

C. An up promoter mutation would increase the amount of functional protein.

D. Such a mutation may affect protein function if the alteration in splicing changes an exon in the mRNA in a way that results in a protein with an altered structure.

C5. The X rays did not produce a mutation, because a mutation is a heritable change in the genetic material. In this case, the X rays have killed the cell, so changes in DNA structure cannot be passed from cell to cell or from parent to offspring.

C6. A. Not appropriate, because the second mutation occurred at different codon B. Appropriate C. Not appropriate, because the second mutation affected the same gene as the first mutation D. Appropriate C7. A. Silent, because the same amino acid (glycine) is encoded by GGA and GGT. B. Missense, because a different amino acid is encoded by CGA compared to GGA. C. Missense, because a different amino acid is encoded by GTT compared to GAT. D. Frameshift, because an extra base is inserted into the sequence.

C8. The consensus sequences for many bacterial promoters are –35: 5′–TTGACA–3′ and –10: 5′– TATAAT–3′. Most mutations that alter the consensus sequence would be expected to decrease the

rate of transcription. For example, a mutation that changed the –35 region to 5′–GAGACA–3′ would decrease transcription.

The sequence 5′–TATAAT–3′ is recognized by the transcription factor TFIID. If this sequence was changed to 5′–TGTAAT–3′, TFIID would not recognize it very well and the adjacent gene would probably show a lower rate of transcription.

C9. One possibility is that a translocation may move a gene next to a heterochromatic region of another chromosome and thereby diminish its expression or it could be moved next to a euchromatic region and increase its expression. Another possibility is that the translocation breakpoint may move the gene next to a new promoter or regulatory sequences that may now influence the gene’s expression.

C10. Random mutations are more likely to be harmful than beneficial. The genes within each species have evolved to work properly. They have functional promoters, coding sequences, terminators, and so on, that allow them to be expressed. Mutations are more likely to disrupt these sequences. For example, mutations within the coding sequence may produce early stop codons, frameshift mutations, and missense mutations that result in a nonfunctional polypeptide. On rare occasions, however, mutations are beneficial; they may produce a gene that is expressed better than the original gene or produce a polypeptide that functions better.

C11. A. No, the position (i.e., chromosomal location) of a gene has not been altered.

B. Yes, the expression of a gene has been altered because it has been moved to a new chromosomal location.

C. Yes, the expression of a gene has been altered because it has been moved to a new chromosomal location.

C12. Yes, a person with cancer is a genetic mosaic. The cancerous tissue contains gene mutations that are not found in noncancerous cells of the body.

C13. If a mutation within the germ line is passed to an offspring, all the cells of the offspring’s body will carry the mutation. A somatic mutation affects only the somatic cell in which it originated and all of the daughter cells that the somatic cell produces. If a somatic mutation occurs early during embryonic development, it may affect a fairly large region of the organism. Because germ-line mutations affect the entire organism, they are potentially more harmful (or beneficial), but this is not always the case. Somatic mutations can cause quite harmful effects such as cancer.

C14. The drawing should show the attachment of a methyl or ethyl group to a base within the DNA. The presence of the alkyl group disrupts the proper base pairing between the alkylated base and the normal base in the opposite DNA strand.

C15. A thymine dimer can interfere with DNA replication because DNA polymerase cannot slide past the dimer and add bases to the newly growing strand. Alkylating mutagens such as nitrous acid will cause DNA replication to make mistakes in the base pairing. For example, an alkylated cytosine will base pair with adenine during DNA replication, thereby creating a mutation in the newly made strand. A third example is 5-bromouracil, which is a thymine analog. It may base pair with guanine instead of adenine during DNA replication.

C16. A. Nitrous acid causes mutations that change A to G or C to T, which are transition mutations.

B. 5-bromouracil causes mutations that change T to A, which are transitions.

C. Proflavin causes small additions or deletions, which may result in frameshift mutations.

C17. During TNRE, a trinucleotide repeat sequence gets longer. If someone was mildly affected with a TNRE disorder, he or she might be concerned that an expansion of the repeat might occur during

gamete formation, yielding offspring more severely affected with the disorder, a phenomenon called anticipation. This phenomenon may depend on the sex of the parent with the TNRE. C18. A spontaneous mutation originates within a living cell. It may be due to spontaneous changes in nucleotide structure, errors in DNA replication, or products of normal metabolism that may alter the structure of DNA. The causes of induced mutations originate from outside the cell. They may be physical agents, such as UV light or X rays, or chemicals that act as mutagens. Both spontaneous and induced mutations may cause a harmful phenotype such as a cancer. In many cases, induced mutations are avoidable if the individual can prevent exposure to the environmental agent that acts as a mutagen.

C19. According to the random mutation theory, spontaneous mutations can occur in any gene and do not involve exposure of the organism to a particular environment that selects for specific types of mutation. However, the structure of chromatin may cause certain regions of the DNA to be more susceptible to random mutations. For example, DNA in an open conformation may be more accessible to mutagens and more likely to incur mutations. Similarly, hot spots – certain regions of a gene that are more likely to mutate than other regions – can occur within a single gene. Also, another reason that some genes mutate at a higher rate is that some genes are larger than others, which provides a greater chance for mutation.

C20. Nitrous acid can change a cytosine to uracil. Nucleotide excision repair could remove the defect and replace it with the correct base.

C21. Excision repair systems could fix this damage. Also, homologous recombination repair could fix the damage.

C22. A. True B. False; the TNRE is not within the promoter, it is within the coding sequence.

C. True; CAG is a codon for glutamine.

D. False; CCG is a codon for proline.

C23. Anticipation means that the TNRE expands even further in future generations. Anticipation may depend on the sex of the parent with the TNRE. C24. The mutation rate is the number of new mutations in a gene per generation. There are 13 babies who did not have a parent with achondroplasia; thus, thirteen is the number of new mutations. To calculate the mutation rate as the number of new mutations per generation, you divide 13 by 2,844,000. The mutation rate is 4.6 × 10–6.

C25. A. TTGGHTGUTGG HHUUTHUGHUU  First round of replication TTGGHTGUTGG CCAAACACCAA AACCCACAACC HHUUTHUGHUU  Second round of replication B. TTGGHTGUTGG TTGGGTGTTGG CCAAACACCAA CCAAACACCAA AACCCACAACC AACCCACAACC GGTTTGTGGTT HHUUTHUGHUU C26. The effects of mutations are cumulative. If one mutation occurs in a cell, this mutation will be passed to the daughter cells. If a mutation occurs in the daughter cell, that cell will have two mutations. These two mutations will be passed to the next generation of daughter cells, and so forth.

The accumulation of many mutations eventually kills the cells. That is why mutagens are more

effective at killing dividing cells than nondividing cells. It is because the number of mutations accumulates to a lethal level.

There are two main side effects to these types of treatments. First, some normal (noncancerous) cells of the body, particularly skin cells and intestinal cells, are actively dividing. These cells are killed by chemotherapy and radiation therapy. Second, it is possible that the therapy may produce mutations that will cause noncancerous cells to become cancerous.

C27. The answer is 5-bromouracil. If this chemical is incorporated into DNA, it can change an A-T base pair into a C-G base pair. A lysine codon, AAG, could be changed into a glutamic acid codon, CAG, with this chemical. By comparison, UV light is expected to produce thymine dimers, which would not lead to a mutation creating a glutamic acid codon, and proflavin is expected to cause a frameshift mutation.

C28. A. Likely B. Not likely; the albino trait affects the entire individual.

C. Not likely, the early-producing trait affects the entire apple tree.

D. Likely C29. A. If UvrA was missing, the repair system would not be able to identify a damaged DNA segment.

B. If UvrC was missing, the repair system could identify the damaged segment, but it would be unable to make cuts in the damaged DNA strand.

C. If UvrD was missing, the repair system could identify the damaged segment and make cuts in the damaged strand, but it could not unwind the damaged and undamaged strands and thereby remove the damaged segment.

D. If DNA polymerase was missing, the repair system could identify the damaged segment, make cuts in the damaged strand, and unwind the damaged and undamaged strands to remove the damaged segment, but it could not synthesize a complementary DNA strand using the undamaged strand as a template.

C30. Mismatch repair is aimed at eliminating base pair mismatches that may have occurred during DNA replication. In this case, the wrong base is in the newly made strand. The binding of MutH, which occurs on a hemimethylated sequence, provides a means of distinguishing between the unmethylated and methylated strands. In other words, MutH binds to the hemimethylated DNA in a way that allows the mismatch repair system to distinguish which strand is methylated and which is not.

C31. The two main mechanisms by which cells can repair DSBs are homologous recombination (HR) and nonhomologous end joining (NHEJ).

C32. The underlying genetic defect that causes xeroderma pigmentosum is a defect in one of the genes that encode a polypeptide involved with nucleotide excision repair. Individuals with this disease are defective at repairing DNA abnormalities such as thymine dimers and abnormal bases. Therefore, they are very sensitive to environmental agents such as UV light, which is more likely to cause mutations in these people compared to unaffected individuals. For this reason, people with XP develop pigmentation abnormalities and premalignant lesions, and have a high predisposition to skin cancer.

C33. The mismatch repair system, which detects mistakes made by DNA polymerase, would not function properly. As described in Figure 19.20, the MutH protein recognizes the sequence GATC when it is hemimethylated. If an E. coli strain was missing the dam methylase, this sequence would

not be methylated, and MutH could not bind there. Therefore, mistakes, involving base pair mismatches caused by DNA polymerase, would not be repaired. This would increase the spontaneous mutation rate in this strain.

C34. Both types of repair systems recognize an abnormality in the DNA and excise a segment of the strand containing the abnormality. The normal strand is then used as a template to synthesize a complementary strand of DNA. The systems differ in the types of abnormalities they detect. The mismatch repair system detects base pair mismatches, while the nucleotide excision repair system recognizes thymine dimers, chemically modified bases, missing bases, and certain types of crosslinks. The mismatch repair system operates immediately after DNA replication, allowing it to distinguish between the daughter strand (which contains the wrong base) and the parental strand.

The nucleotide excision repair system can operate at any time in the cell cycle.

Application and Experimental Questions E1. When cells from a master plate were replica plated onto two plates containing selective media with the T1 phage, T1-resistant colonies were observed at the same locations on both plates. These results indicate that the mutations occurred randomly while on the master plate (in the absence of T1) rather than occurring as a result of exposure to T1. In other words, mutations are random events and selective conditions may promote the survival of mutant strains that occur randomly.

E2. To show that antibiotic resistance is due to random mutations, you could follow the same basic strategy as Lederberg and Lederberg did except the secondary plates would contain the antibiotic instead of T1 phage. If the antibiotic resistance arose as a result of random mutations in bacteria on the master plate, you would expect the antibiotic-resistant colonies to appear at the same locations on two different secondary plates.

E3. Haploid cells are more sensitive to mutation because they have only a single copy of each gene.

Therefore, recessive mutations that inhibit cell growth are easily detected. In diploid cells, it would take a mutation in both copies of the same gene to detect its phenotypic effects.

E4. You would expose the bacteria to the physical agent. Use of the rat liver extract is probably not necessary for two reasons. First, a physical mutagen is not something that a person would eat.

Therefore, the actions of digestion via the liver are probably irrelevant with regard to a physical mutagen. Second, the rat liver extract would not be expected to alter the properties of a physical mutagen.

E5. Absence of mutagen: 17/10,000,000 = 17 × 10–7 = 1.7 × 10–6 Presence of mutagen: 2,017/10,000,000 = 2.0 × 10–4 The mutagen increases the rate of mutation more than 100-fold.

E6. The results suggest that the strain is defective with regard to nucleotide excision repair system. If you compare the results for the normal and mutant strains that have been incubated for 2 hours at 37°C, much of the radioactivity in the normal strain has been transferred to the soluble fraction because it has been excised. For the mutant strain, however, less of the radioactivity has been transferred to the soluble fraction, suggesting that this strain is not as efficient at removing thymine dimers.

Questions for Student Discussion/Collaboration

1.

2.

3.

A strain that has a defect in any of the DNA repair systems discussed in this chapter would be a mutator strain because it would be defective at repairing mutations. Also, a strain that has a defect in the proofreading function of DNA polymerase would be a mutator strain. In addition, defects in the enzymes that promote recombination can have high rates of mutation. Strains that tend to produce mutagenic metabolic products could be mutator strains. Finally, strains that carry transposable elements can have high rate of mutation. Although it would be difficult to predict, research has shown that strains which lack a proofreading function tend to have the highest rates of spontaneous mutations.

The worst time to be exposed to mutagens is at very early stages of embryonic development. An early embryo is most sensitive to mutagens because any mutation will affect a large region of the body. Adults must also worry about mutagens for several reasons. Mutations in somatic cells can cause cancer, a topic that is discussed in Chapter 22. Also, adults should be careful to avoid mutagens that may affect the ovaries or testes because these mutations could be passed to offspring.

It’s a matter of opinion. Some ideas might be the following: A. Testing of mutagens would enable us to know what the mutagens are and thereby avoid them.

On the other hand, one might argue that there are so many now, that it’s difficult to avoid them anyway.

B. Investigating molecular effects may help us find a cure for diseases such as cancer, or help us to prevent mutations. On the other hand, it may not.

C. Similarly, investigating DNA repair mechanisms may lead to ways of preventing mutations. On the other hand, it may not.

D. Other places: educating the public about mutagens; tighter regulations of substances that contain mutagens; alternative methods of agriculture that may diminish the level of mutagens in food; and many others...

CHAPTER 20 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 20. 1 Eight FIGURE 20. 2 In this experiment, the selectable marker gene selects for the growth of bacteria that have taken up a plasmid; they can grow in the presence of ampicillin.

FIGURE 20. 3 The name reverse transcriptase refers to the idea that this enzyme catalyzes the opposite of transcription. It uses an RNA template to make DNA, whereas during transcription a DNA template is used to make RNA. FIGURE 20. 4 The advantage of a DNA library is that it lacks introns. This is useful if researchers want to focus their attention on the coding sequence of a gene or if they want to express the gene in cells that do not splice out introns.

FIGURE 20. 6 The products that predominate after four cycles is the type of DNA fragment containing only the region flanked by the primers. This occurs because this type of fragment is used in each cycle as a template to make another copy of itself.

FIGURE 20. 8 The probe must be cleaved to separate the quencher from the reporter.

FIGURE 20.10 The nucleus of the oocyte is removed so that the resulting organism contains (nuclear) genetic material only from the somatic cell.

FIGURE 20.11 No. Carbon Copy did not receive genetic material from a different species.

FIGURE 20.12 When stem cells divide, they produce one cell that remains a stem cell and another cell that differentiates. This pattern maintains a population of stem cells.

FIGURE 20.14 Hematopoietic stem cells are multipotent.

FIGURE 20. 17 Some examples are to study the core promoter of a gene; to study regulatory elements; to identify splice sites; and to study the importance of particular amino acids with regard to protein function.

FIGURE 20.18 In the bacterial defense system, tracrRNA and crRNA are separate molecules. In gene editing using CRIPSR-Cas technology, tracrRNA and crRNA are covalently linked to form a single guide RNA molecule (sgRNA).

FIGURE 20. 21 The secondary antibody is labeled or attached to an enzyme, which provides a way to detect the protein of interest.

End-of-chapter Questions: Conceptual Questions C1. First, cloned genes can be used for DNA sequencing. This has allowed researchers to understand genetics at the molecular level. Second, cloned genes can be mutated using site-directed mutagenesis or CRISPR to identify gene sequences such as promoters and regulatory elements and to see how specific mutations alter the structure and function of DNA. Third, cloned genes can be used as probes to identify similar genes or RNA. Fourth, cloned genes can be introduced into a different cell type or species.

C2. A restriction enzyme recognizes and binds to a specific DNA sequence and then cleaves a (covalent) phosphoester bond in each of two DNA strands.

C3. Here is an example: GGGCCCATATATATGGGCCC CCCGGGTATATATACCCGGG C4. cDNA is DNA that is made using RNA as the starting material. Unlike eukaryotic genomic DNA, it lacks introns.

C5. A. With regard to maternal effect genes, the phenotype would depend on the animal that donated the egg. It is the cytoplasm of the egg that accumulates the gene products of maternal effect genes. B. The extranuclear traits depend on the mitochondrial genome. Mitochondria are found in the egg and in the somatic cell. So, theoretically, both cells could contribute extranuclear traits. In reality, however, researchers have found that the mitochondria in Dolly were from the animal that donated the egg. It is not clear why she had no mitochondria from the mammary cell.

C. The cloned animal would be genetically identical to the animal that donated the nucleus with regard to traits that are determined by nuclear genes, which are expressed during the lifetime of the organism. The cloned animal would/could differ from the animal that donated the nucleus with regard to traits that are determined by maternal effect genes and extranuclear genes. Such an animal is not a true clone, but it is likely that it would greatly resemble the animal that donated the nucleus, because the vast majority of genes are found in the cell nucleus.

C6. A dideoxynucleotide is missing the OH group at the 3 position. When the 5 end of a dideoxynucleotide is added to a growing strand of DNA, another phosphoester bond cannot be formed at the 3 position. Therefore, the dideoxynucleotide prevents any further addition of nucleotides to the growing strand of DNA. Application and Experimental Questions E1. Sticky ends, which are complementary in their DNA sequence, promote the binding of DNA fragments to each other. This binding is due to hydrogen bonding between the sticky ends.

E2. Remember that A-T base pairs form two hydrogen bonds, while G-C base pairs form three hydrogen bonds. The order (from stickiest to least sticky) is as follows: BamHI = Pst I = Sau3A I > EcoRI > NaeI. E3. All vectors have the ability to replicate when introduced into a living cell. This ability is due to a DNA sequence known as an origin of replication, which determines the host cell specificity of a vector. Modern vectors also contain convenient restriction sites where geneticists can insert fragments of DNA, and they contain selectable markers, which are genes that confer some selectable advantage for the host cell that carries them. The most common selectable markers are antibiotic-resistance genes, which confer resistance to antibiotics that would normally inhibit the growth of the host cell.

E4. In conventional gene cloning, many copies are made because the vector replicates to a high copy number within the cell, and the cells divide to produce many more cells. In PCR, the replication of the DNA to produce many copies is facilitated by primers, deoxyribonucleoside triphosphates (dNTPs), and Taq polymerase.

E5. First, the chromosomal DNA that contains the gene of interest must be obtained from a cell (tissue) sample. A vector must also be obtained. The vector and chromosomal DNA are digested with a restriction enzyme. They are mixed together to allow the sticky ends of the DNA fragments to bind to each other, and DNA ligase is then added to promote covalent bonds, hopefully creating a recombinant vector. The DNA is introduced into a living cell. The vector will replicate and the cells will divide to produce a colony of cells that contain cloned DNA pieces.

E6. A recombinant vector is a vector that has a piece of foreign DNA inserted into it. The foreign DNA may be from the chromosomal DNA of some organism. To construct a recombinant vector, the

vector and the source of foreign DNA are digested with the same restriction enzyme. The complementary ends of the fragments are allowed to hydrogen bond to each other (i.e., sticky ends are allowed to bind), and then DNA ligase is added to create covalent phosphoester bonds. In some cases, a piece of the foreign DNA will become ligated to the vector, thereby creating a recombinant vector. In other cases, the two ends of the vector ligate back together, restoring the vector to its original structure.

As shown in Figure 20.2, the insertion of foreign DNA can be detected using X-Gal. The insertion of the foreign DNA causes the inactivation of the lacZ gene. The lacZ gene encodes the enzyme β-galactosidase, which converts the colorless compound X-Gal to a blue compound. If the lacZ gene is inactivated by the insertion of foreign DNA, the enzyme will not be produced, and the bacterial colonies will be white. If the vector has simply recircularized, and the lacZ gene remains intact, the enzyme will be produced, and the colonies will be blue.

E7. A DNA library is a collection of recombinant vectors that contain different pieces of DNA from a source of chromosomal DNA. Because it is a diverse collection of many different DNA pieces, the name library seems appropriate.

E8. It would be necessary to use cDNA so that the gene would not carry any introns. Bacterial cells do not contain spliceosomes (which are described in Chapter 14). To express a eukaryotic protein in bacteria, a researcher would clone cDNA into the bacteria, because the cDNA does not contain introns.

E9. A thermostable form of DNA polymerase (e.g., Taq polymerase) is used in PCR because each PCR cycle involves a heating step to denature the DNA. This heating step would inactivate most forms of DNA polymerase. However, Taq polymerase is thermostable and can remain functional after many cycles of heating and cooling. It is not necessary to use a thermostable form of DNA polymerase in the techniques of dideoxy sequencing or site-directed mutagenesis. In these methods, DNA polymerase can be added after the annealing step, and the sample can be incubated at a temperature that does not inactivate most forms of DNA polymerase.

E10. Initially, the mRNA would be mixed with reverse transcriptase and nucleotides to create a complementary strand of DNA. Reverse transcriptase also needs a primer. This could be a primer that is known to be complementary to the β-globin mRNA. Alternatively, mature mRNAs have a polyA tail, so you could add a primer that consists of many Ts, called a poly-dT primer. After the complementary DNA strand had been made, the sample would then be mixed with primers, Taq polymerase, and nucleotides and subjected to the standard PCR protocol. (Note: the PCR reaction would have two kinds of primers. One primer would be complementary to the 5 end of the mRNA and would be unique to the β-globin sequence. The other primer would be complementary to the 3 end. This second primer could be a poly-dT primer, or it could be a unique primer that would bind slightly upstream from the polyA-tail region.)

E11. The probe is a molecule whose fluorescence increases as the PCR product increases. One example is TaqMan. It has a fluorescent molecule and a quencher, which are connected by a primer that binds to the DNA that is amplified. When PCR is occurring, Taq polymerase digests the primer and thereby separates the fluorescent molecule from the quencher. Therefore, the fluorescence increases as the PCR product increases. Alternatively, other probes intercalate into the DNA, which causes

their fluorescence to increase. In this case, as the PCR product increases, the fluorescence also increases because the probes are intercalating into the newly made DNA. E12. The exponential phase of PCR is analyzed because it is during this phase that the amount of PCR product is proportional to the amount of the original DNA in the sample.

E13. A. Dolly’s chromosomes may seem so old because they were already old when they were in the nucleus that was incorporated into the enucleated egg. They had already become significantly shortened in the mammary cells. This shortening was not repaired by the egg.

B. Dolly’s age does not matter. Remember that shortening does not occur in germ cells. However, Dolly’s eggs are older than they seem by about 6 or 7 years, because Dolly’s germ-line cells received their chromosomes from a sheep that was 6 years old, and the cells were grown in culture for a few doublings before a mammary cell was fused with an enucleated egg. Therefore, the calculation would be: 6 or 7 years (the age of the mammary cells that produced Dolly’s germ-line cells) plus 8 years (the age of Molly), which equals 14 or 15 years. However, only half of Molly’s chromosomes would appear to be 14 or 15 years old. The other half of her chromosomes, which she inherited from her father, would appear to be 8 years old.

C. Chromosome shortening is a bit disturbing, because it suggests that aging has occurred in the somatic cell, and this aging is passed to the cloned organism. If cloning was done over the course of many generations, this may eventually have a major impact on the life span of the cloned organism.

It may die much earlier than a noncloned organism. However, chromosome shortening may not always occur. It did not seem to occur in mice that were cloned for six consecutive generations.

E14. Reproductive cloning is the cloning of entire multicellular organisms. With plants, this is easy.

Most species of plants can be cloned by asexual cuttings. In animals, cloning occurs naturally, as in identical twins. Identical twins are genetic replicas of each other because they begin from the same fertilized egg. (Note: Some somatic mutations could occur in identical twins that would make them slightly different.) Recently, as in the case of Dolly, reproductive cloning has been accomplished by fusing somatic cells with enucleated eggs. The advantage, from an agricultural point of view, is that reproductive cloning would allow farmers to choose the best animal in a herd and make many clones from it. Breeding would no longer be necessary. Also, breeding may be less reliable because the offspring inherit traits from both the mother and father.

E15. The A closest to the bottom of the gel is changed to a G. E16. AGGTCGGTTGCCATCGCAATAATTTCTGCCTGAACCCAATA E17. 5–CCCCCGATCGACATCATTA–3. The mutagenic base is underlined and bold.

E18. There are many different strategies one could follow. For example, you could mutate every other base and see what happens. It would be best to make very nonconservative mutations such as a purine for a pyrimidine or a pyrimidine for a purine. If the mutation prevents protein binding in the EMSA, then the mutation is probably within the response element. If the mutation has no effect on protein binding, it probably is outside the response element.

E19. You would conclude that it might be important. The only amino acid substitution that gave a substantial amount of functional activity was an aspartic acid. Glutamic acid and aspartic acid have very similar amino acid side chains (see Chapter 15); they both contain a carboxyl (COOH) group.

Based on these results, you would suspect that a carboxyl group at this location in the protein might be important for its function.

E20. The purpose of gel electrophoresis is to separate the many RNA molecules or proteins that were obtained from the sample being analyzed. This separation is based on molecular mass and allows you to identify the molecular mass of the RNA molecule or protein that is being recognized by the probe.

E21. The purpose of a Northern blotting experiment is to identify a specific RNA within a mixture of many RNA molecules, using a fragment of cloned DNA as a probe. It can tell you if a gene is transcribed in a particular cell or at a particular stage of development. It can also tell you if a premRNA is alternatively spliced into two or more mRNAs of different sizes.

E22. The Northern blot is shown here. The female mouse expresses the same total amount of this mRNA as the male does. In the heterozygous female, there is 50% of the 900 bp band and 50% of the 825 bp band.

E23. It appears that this mRNA is alternatively spliced to create a high molecular mass and a lower molecular mass product. Nerve cells produce a very large amount of the larger mRNA, whereas spleen cells produce a moderate amount of the smaller mRNA. Both types are produced in small amounts by the muscle cells. It appears that kidney cells do not transcribe this gene.

E24. Binding occurs due to the hydrogen bonding of complementary sequences. Due to the chemical properties of DNA and RNA strands, they form double-stranded regions when the base sequences are complementary.

E25. Lane 1 shows that β-globin is made in normal red blood cells. In red blood cells from a thalassemia patient (lane 2), however, very little is made. Perhaps this person is homozygous for a down promoter mutation, which diminishes the transcription of the gene. As shown in lanes 3 and 4, βglobin is not made in muscle cells.

E26. The Western blot is shown here. The sample in lane 2 came from a plant that was homozygous for a mutation that prevented the expression of this polypeptide. Therefore, no protein was observed in this lane. The sample in lane 4 came from a plant that is homozygous for a mutation that introduces an early stop codon into the coding sequence; thus, the polypeptide is shorter than normal (13.3 kDa). The sample in lane 3 was from a heterozygote that expresses about 50% of each type of polypeptide. Finally, the sample in lane 5 came from a plant that is homozygous for a mutation that changed one amino acid to another amino acid. This type of mutation, termed a missense mutation, may not be detectable on gel. However, a single amino acid substitution could affect polypeptide function.

E27. The products of protein-coding genes are proteins with a particular amino acid sequence.

Antibodies can specifically recognize proteins due to their amino acid sequence. Therefore, an antibody can detect whether or not a cell is making a particular type of protein.

E28. In this case, the transcription factor binds to the response element when the hormone is present.

Therefore, the hormone promotes the binding of the transcription factor to the DNA and thereby promotes transcriptional activation.

E29. The rationale behind an electrophoretic mobility shift assay (EMSA) is that a segment of DNA with a protein bound to it will migrate more slowly through a gel than will the same DNA without any bound protein. A shift in a DNA band to a higher molecular mass provides a way to identify DNA-binding proteins.

E30. TFIID can bind to this DNA fragment by itself, as seen in lane 2. However, TFIIB and RNA polymerase II cannot bind to the DNA by themselves (lanes 3 and 4)? As seen in lane 5, TFIIB can bind if TFIID is also present, because the band is higher than with TFIID alone (compare lanes 2 and 5). In contrast, RNA polymerase II cannot bind to the DNA when only TFIID is present. The band in lane 6 is at the same location in the gel as in lane 2, indicating that only TFIID is bound.

Finally, in lane 7, when all three components are present, the band is higher than when both TFIIB and TFIID are present (compare lanes 5 and 7). This result means that all three proteins are bound to the DNA. Taken together, the results indicate that TFIID can bind by itself, TFIIB needs TFIID to bind, and RNA polymerase II needs both proteins to bind to the DNA. Questions for Student Discussion/Collaboration  . To know what type of cell in which a gene is expressed. For example, in what cell type(s) is/are genes involved with cell contraction expressed? Proteins such as myosin and actin would be found in muscle cells.

2. To know what stage of development a gene is expressed. Different globin genes are expressed at different stages of development. . To know if alternative splicing is occurring. As discussed in Chapter 17, the gene encoding troponin produces a pre-mRNA that is alternatively spliced. . To know if environmental factors affect the level of gene expression. As discussed in Chapter 16, the genes of the lac operon are expressed when lactose is present. . To know if mutations affect the level of gene expression. Many human genetic diseases, PKU, cystic fibrosis, etc., involve mutations that alter expression levels.

2. Here are just a few possible research questions:

Does a particular amino acid within a protein sequence play a critical role in the protein's structure or function?

Does a DNA sequence function as a promoter?

Does a DNA sequence function as a regulatory site?

Does a DNA sequence function as a splicing junction?

Is a sequence important for correct translation?

Is a sequence important for RNA stability?

CHAPTER 21 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 21.1 A genetic map is a diagram that shows the order and relative locations of genes and other DNA sequences along a chromosome.

FIGURE 21.2 The probe binds to a specific site because the site has a complementary DNA sequence.

FIGURE 21.5 Microsatellites for which the number of repeat sequence varies are polymorphic.

FIGURE 21.9 Yes. The sequence is determined as the DNA is synthesized.

FIGURE 21.12 A key point is that mRNA is made only when a gene is expressed. In this experiment, mRNA is first isolated and then used to make cDNA, which is fluorescently labeled. The fluorescent spots indicate which genes have been transcribed to produce mRNA. FIGURE 21.13 The antibody recognizes and binds to the protein of interest that is covalently crosslinked to DNA. The antibody is also attached to a heavy bead, which makes it easy to separate it and the DNA-protein complex from the rest of the cellular components by centrifugation. The procedure provides an easy way to purify the protein of interest along with its attached DNA.

End-of-chapter Questions: Conceptual Questions C1. You would conclude that she has a deletion of the gene that the probe recognizes.

C2. A. Yes B. No; this is only one chromosome in the human genome.

C. Yes D. Yes C3. A. False, they do not have to carry genes.

B. True.

C. False, the marker may not carry a gene that affects phenotype.

D. True.

E. True.

Application and Experimental Questions E1. They are complementary to each other.

E2. Fluorescence in situ hybridization is a cytological method of mapping genes. A probe that is complementary to a chromosomal sequence is used to locate the gene microscopically within a mixture of many different chromosomes. Thus, this method can be used to cytologically map the location of a gene sequence. When more than one probe is used, the order of genes along a particular chromosome can be determined.

E3. Because normal cells contain two copies of chromosome 14, you would expect that a probe would bind to complementary DNA sequences on both of these chromosomes. If a probe recognized only one of two chromosomes, this means that one of the copies of chromosome 14 has been lost, or it has suffered a deletion in the region where the probe binds. With regard to cancer, the loss of this genetic material may be related to the uncontrollable cell growth.

E4. The term fixing refers to procedures that chemically freeze cells and prevent degradation. After fixation has occurred, the contents of the cells do not change their morphology. In a sense, they are frozen in place. In a FISH experiment, fixing keeps all of the chromosomes within a cell in the vicinity of each other; they cannot float around on the slide and get mixed up with chromosomes from other cells. A group of chromosomes observed in a FISH experiment comes from a single cell.

It is necessary to denature the chromosomal DNA so that the probe can bind to it. The probe is a segment of DNA that is complementary to the DNA of interest. The strands of chromosomal DNA must be separated (i.e., denatured) so that the probe can bind to complementary sequences.

E5. After the cells and chromosomes have been fixed to the slide, it is possible to add two or more different probes that recognize different sequences (i.e., different sites) within the genome. Each probe has a different fluorescence emission wavelength. Usually, a researcher will use computer imagery that recognizes the wavelength of each probe and then assigns that probe a bright color.

The color seen by the researcher is not the actual color emitted by the probe; it is a secondary color assigned by the computer. In a sense, the probes, with the aid of a computer, are “painting” the regions of the chromosomes that are recognized by a probe. An example of chromosome painting is shown in Figure 21.3. In this example, human chromosome 5 is painted with six different colors.

E6. If the sample was from an unaffected individual, two spots (one on each copy of chromosome 21) would be observed. Three spots would be observed if the sample was from a person with Down syndrome, because that person would have three copies of chromosome 21.

E7. Child 1 and child 3 belong to father 2; child 2 and child 4 belong to father 1; child 5 could belong to either father.

E8. The discrepancy arises because the rate of recombination between homologous chromosomes is different during oogenesis compared to spermatogenesis. Physical mapping measures the actual distance (in bp) between markers. The physical mapping of chromosomes from males and females reveals that they are the same lengths (excluding the Y chromosome). Therefore, the sizes of chromosomes in males and females are the same. The differences observed in linkage maps are due to differences in the rates of recombination during oogenesis versus spermatogenesis.

E9. A. One homolog contains the STS-1 that is 289 bp and STS-2 that is 422 bp, while the other homolog contains STS-1 that is 211 bp and STS-2 that is 115 bp. This is based on the observation that 28 of the sperm have either the 289 bp and 422 bp bands or the 211 bp and 115 bp bands.

B. There are two recombinant sperm; see lanes 12 and 18. Because there are two recombinant sperm out of a total of thirty: 2 Map distance =  100 30 = 6.7 mu C. In theory, this method could be used. However, there is not enough DNA in one sperm to carry out an RFLP analysis unless the DNA is amplified by PCR. E10. A sequence-tagged site (STS) is a segment of DNA, usually quite short (e.g., 100 to 400 bp in length), that is found at a unique site in a genome. STSs are identified using primers in a PCR reaction. STSs serve as molecular markers in genetic mapping studies. Sometimes the region within an STS may contain a microsatellite. A microsatellite is a short repetitive DNA segment that is variable in length. When a microsatellite is within an STS, the length of the STS varies among different individuals or even in the same individual if the individual is heterozygous for the STS. Such an STS is polymorphic. Polymorphic STSs can be used in linkage analysis, because their transmission can be followed in family pedigrees and through crosses of experimental organisms.

E11. You can calculate the probability that a base will not be sequenced using this approach with the following equation: P = e-m where, P is the probability that a base will be left unsequenced e is the base of the natural logarithm; e = 2.72 m is the number of bases sequenced divided by the total genome size In this case, m = 19 divided by 4.4, which equals 4.3, P = e-m = e-4.3 = 0.0136 = 1.36% This means that if you randomly sequence 19.0 Mb, you are likely to miss only 1.36% of the genome.

With a genome size of 4.4 Mb, you would miss about 59,840 base pairs out of approximately 4,400,000.

E12. The overall advantage of next-generation sequencing technologies is that a large amount of DNA can be sequenced in a short period of time, and at a lower cost. Also, subcloning fragments of DNA into vectors is no longer necessary. Further, some of the methods allow parallel analysis of an enormous number of samples simultaneously.

E13. Sequencing by synthesis is a form of next-generation DNA sequencing in which the base sequence is determined as the DNA strand is being made. Pyrosequencing is an example.

E14. First, researchers obtain a sample from the environment, which could be a soil sample, a water sample from the ocean, or a fecal sample from a person. After the sample is filtered, the cells within the sample are lysed and the DNA is extracted and purified. During this procedure, the purified DNA is sheared into fragments of different sizes. The DNA fragments are then randomly inserted

into cloning vectors and transformed into host cells. The result is a DNA library of thousands or tens of thousands of cells, each carrying a DNA fragment from the metagenome. All of these cells together constitute the metagenomic library. The members of the library are then subjected to shotgun DNA sequencing to identify the genes they may contain. (Note: The cloning step may be skipped if certain types of DNA sequencing methods, such as pyrosequencing, are used.)

E15. A. A DNA microarray is a small slide that is dotted with many different fragments of DNA. In some microarrays, DNA fragments, which were made synthetically (e.g., by PCR), are individually spotted onto the slide. The DNA fragments are typically 500 to 5,000 bp in length, and a few thousand to tens of thousands are spotted to make a single array. Alternatively, short oligonucleotides can be directly synthesized on the surface of the slide. In this process, the DNA sequence at a given spot is produced by selectively controlling the growth of the oligonucleotide using narrow beams of light. In this case, there can be hundreds of thousands of different spots on a single array.

B. In most cases, fluorescently labeled cDNA is hybridized to the microarray, though labeled genomic DNA or RNA could also be used.

C. After hybridization, the array is washed and placed in a laser scanner that scans each pixel (the smallest element in a visual image). After correction for local background, the final fluorescence intensity for each spot is obtained by averaging across the pixels in each spot. This results in a group of fluorescent spots at defined locations in the microarray.

E16. Complementary DNA (cDNA) is subjected to a next-generation DNA sequencing method.

Questions for Student Discussion/Collaboration 1. A molecular marker is a segment of DNA, not usually encoding a gene, that has a known location within a particular chromosome. It marks the location of a site along a chromosome. RFLPs and microsatellites are examples. It is easier to use these types of markers because they can be readily identified by molecular techniques such as restriction digestion analysis and PCR. The locations of functional genes are usually more difficult to use because this relies on conventional genetic mapping approaches whereby allelic differences in the gene are mapped by making crosses or following a pedigree. For monomorphic genes, this approach doesn't work.

2. This is a matter of opinion. Many people would say that the ability to identify many human genetic diseases is the most important goal. In addition, we will better understand how humans are constructed at the molecular level. As we gain a greater understanding of our genetic makeup, some people are worried that this may lead to greater discrimination. Insurance or health companies could refuse people who are known to carry genetic abnormalities. Similarly, employers could make their employment decisions based on the genetic makeup of potential employees rather than their past accomplishments.

At the family level, genetic information may affect how people choose mates, and whether or not they decide to have children.

CHAPTER 22 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 22.2 The key feature is that both parents of affected offspring are unaffected by the disease. The parents are heterozygous carriers.

FIGURE 22.3 The key feature is that affected offspring have an affected parent.

FIGURE 22.4 The key feature is that all of the affected individuals are males. Furthermore, these males all have mothers who were descendants of Queen Victoria.

FIGURE 22.5 A haplotype is the linkage of alleles or molecular markers on a single chromosome.

FIGURE 22.7 The original mutation in the HTT gene that resulted in the allele that causes Huntington disease occurred in a germ-line cell or gamete of the founder and that allele was closely linked to the G8-C marker.

Therefore, individuals in this pedigree that carry the G8-C marker are also likely to carry the disease-causing allele.

FIGURE 22.12 Such a mutation would keep the Ras protein in an active state and thereby promote cancerous growth.

The cell-signaling pathway would be turned on.

FIGURE 22.13 The bcr gene is expressed in white blood cells. This translocation causes the abl gene to be under the control of the bcr promoter. The abnormal expression of abl in white blood cells causes leukemia.

FIGURE 22.14 E2F will be active all of the time, and this will lead to uncontrolled cell division.

FIGURE 22.15 A checkpoint is a point in the cell cycle at which checkpoint proteins determine if the cell is in the proper condition to divide. If an abnormality such as DNA damage is detected, these proteins will halt the cell cycle.

FIGURE 22.18 An individual with a predisposition for developing familial breast cancer inherits only one copy of the mutant allele. Therefore, the disease shows a dominant pattern of inheritance. However, for the individual to actually get breast cancer, the other allele must become mutant in somatic cells.

End-of-chapter Questions: Conceptual Questions C1. Recessive X-linked traits are distinguished from the other two by their prevalence in males.

Dominant X-linked traits (which are exceedingly rare) are always passed from father to daughter and never from father to son. Autosomal recessive and dominant traits are distinguished primarily by the pattern of transmission from parents to offspring. A person with a dominant trait usually has an affected parent unless it is due to a new mutation or incomplete penetrance is observed. Also, two affected parents can have unaffected children, which is not expected to occur with recessive traits. In the case of rare recessive traits, affected children usually have unaffected parents.

C2. When a disease-causing allele affects a trait, it is causing a deviation from normality, but the gene involved is not usually the only gene that governs the trait. For example, an allele causing hemophilia prevents the normal blood-clotting pathway from operating correctly. Inheritance of hemophilia follows a simple Mendelian pattern because a single gene affects the phenotype. Even so, it is known that normal blood clotting is due to the actions of many genes.

C3. Locus heterogeneity refers to the phenomenon in which a particular type of disease may be caused by mutations in different genes. Hemophilia and thalassemia are two examples. Locus heterogeneity can confound pedigree analysis because alleles in different genes that cause the same disorder may be inherited in different ways. For example, there are autosomal and X-linked forms of hemophilia.

If a pedigree contained individuals with both types of alleles, the data would not be consistent with either an autosomal or X-linked pattern of inheritance.

C4. Changes in chromosome number, and unbalanced changes in chromosome structure, tend to affect phenotype because they create an imbalance of gene expression. For example, individuals with Down syndrome have three copies of chromosome 21 and therefore three copies of all the genes on that chromosome. The result is a relative overexpression of genes that are located on chromosome 21 compared to the genes on other chromosomes. Reciprocal translocations and inversions often lack phenotypic consequences because the total amount of genetic material is not altered, and the level of gene expression is not significantly changed.

C5. A. False. Dominant alleles require the inheritance of only one copy, and X-linked alleles require the inheritance of only one copy in males.

B. True.

C. True.

D. False. In many cases, it is difficult to assess the relative contributions of genetics and environment. In some cases, however, the environment seems most important. For example, with PKU, a low-phenylalanine diet will prevent harmful symptoms, even if an individual is homozygous for the mutant PKU alleles.

C6. Here are a few possible answers: Dwarfism occurs in people and dogs. Breeds like the dachshund and basset hound are examples of dwarfism in dogs. There are diabetic people and mice. There are forms of inherited obesity in people and mice. Hip dysplasia affects people and dogs.

C7. The age of onset of a genetic disease often coincides with a particular stage of development.

Throughout the life span of an individual, from embryonic through adult life, the pattern of gene expression changes. Some genes are turned on only during embryonic stages, some are turned on during fetal stages, others are turned on at birth, others at adolescence, etc. An individual who is carrying a mutant gene may not manifest any disorder until the time in life when the gene is supposed to be expressed. For example, many genetic diseases manifest themselves after birth. Prior to birth, an individual may develop properly and be born a healthy baby. After birth, the baby begins a new developmental phase, and at this point, new genes may be turned on. If one or both copies of a “newborn” gene (i.e., a gene turned on at birth) are mutant, this may lead to a serious disorder, with an age of onset that occurs during infancy.

An infectious disease would probably not have a particular age of onset, because people could be exposed to the infectious agent at any time in their lives. Therefore, most infectious diseases do not have a predictable age of onset, although some are more common in children or older people.

C8. A. Because a person must inherit two defective copies of the gene that encodes β- glucosidase and the gene is known to be on chromosome 1, inheritance of Gaucher disease is autosomal recessive.

Both parents must be heterozygous, because they had to have one affected parent (who transmitted the mutant allele to them) and their phenotypes are unaffected (so they must have received the normal allele from their other parent). Because both members of the couple are heterozygotes, there is a 1/4 chance that they will produce an affected child (a homozygote) with Gaucher disease. If you let G represent the nonmutant allele, and g the mutant allele, you obtain the following Punnett square:

B. From this Punnett square, you can see that there is also a 1/4 chance of producing a homozygote with both normal copies of the gene.

C. You need to apply the binomial expansion equation (see Chapter 3 for a description of this equation) with n = 5, x = 1, p = 0.25, q = 0.75. The answer is 0.396, or 39.6%.

C9. In this pedigree, every affected member has an affected parent (except I-1, whose parentage is unknown). Also, the disease is found in both males and females. Therefore, this pattern suggests an autosomal dominant mode of transmission. Recessive transmission cannot be ruled out, but it is unlikely because it is a rare disorder and it would be extremely unlikely that unrelated individuals in this pedigree (i.e., I-2, II-1, II-7, III-4, and III-5) would all be heterozygous carriers.

C10. The pattern of inheritance is autosomal recessive. All of the affected individuals do not have affected parents. Also, the disorder is found in both males and females. If the inheritance pattern were X-linked recessive, individual III-1 would have to have an affected father, which she does not.

C11. Because both members of this couple are phenotypically unaffected, and they have two sons with the disorder, the mother must be a heterozygous carrier, and the father has a normal allele (see the following Punnett square).

There is a 50% chance that their daughter is a carrier. If she is a carrier, there is a 50% chance that she will pass the mutant allele to her sons. So, if you multiply 0.5 times 0.5, this gives a value of 0.25, or 25%. This means that there is a 25% chance that she will have an affected son if she gives birth to a son. Another way to view this calculation, however, is to consider that this female may have sons or daughters. When she gives birth, there is a 25% chance of having a male child with Fabry disease if she is a carrier. If you do the calculation based on this perspective, the odds of this female having an affected son are 0.5 times 0.25, which equals 0.125, or 12.5%. In other words, there is a 12.5% chance that this female will have an affected son each time she has a child.

C12. The 13 babies have acquired a new mutation. In other words, during spermatogenesis or oogenesis or after the egg was fertilized, a new mutation occurred in the fibroblast growth factor gene. These 13 individuals have the same chances of passing the mutant allele to their offspring as the 18 individuals who inherited the mutant allele from a parent. The chance is 50%.

C13. Based on this pedigree, it appears to be an X-linked recessive trait (which it is known to be). All of the affected members are males who had unaffected parents. From this pedigree alone, you could not rule out autosomal recessive, but it is less likely because there are no affected females.

C14. Because this is inherited as a dominant trait, the mother must have two normal copies of the gene, and the father (who is affected) is most likely to be a heterozygote. (Note: the father could be a homozygote, but this is extremely unlikely because the dominant allele is very rare.) If you let M represent the mutant Marfan allele, and m the normal allele, you can construct the following Punnett square:

A. There is a 50% chance that this couple will have an affected child.

B. You need to use the product rule. The odds of having an unaffected child are 50%. So you multiply 0.5 × 0.5 × 0.5, which equals 0.125, or a 12.5% chance of having three unaffected offspring.

C15. A. The mode of transmission is autosomal recessive. All of the affected individuals do not have affected parents. Also, the disorder is found in both males and females. If it were X-linked recessive, individual III-1 would have to have an affected father, which she does not.

B. If the disorder is autosomal recessive, individuals II-1, II-2, II-6, and II-7 must be heterozygous carriers because they have affected offspring. Because individuals II-2 and II-6 are heterozygotes, one of their parents (I-1 or I-2) must also be a carrier. Because this is rare disease, you would assume that only one of them (I-1or I-2) would be a carrier. Therefore, there is a 50% chance that II-3, II-4, and II-5 are also heterozygotes.

C16. 1. The disease-causing allele had its origin in a single individual known as a founder, who lived many generations ago. Since that time, the allele has spread throughout portions of the human population.

2. When the disease-causing allele originated in the founder, it occurred in a region on a chromosome with a particular haplotype. The haplotype is not likely to have changed over the course of several generations if the disease-causing allele and markers in this region are very close together.

C17. An oncogene is abnormally activated to cause cancer, while a tumor-suppressor gene is inactivated to cause cancer. Ras and src are examples of oncogenes, and Rb and p53 are tumor-suppressor genes.

C18. A proto-oncogene is a normal cellular gene that typically plays a role in cell division. It can be altered by mutation to become an oncogene and thereby cause cancer. At the level of protein function, a proto-oncogene can become an oncogene by synthesizing too much of a protein or synthesizing the same amount of a protein that is abnormally active.

C19. The predisposition to develop cancer is inherited in a dominant fashion because the heterozygote has the higher predisposition. The mutant allele is actually recessive at the cellular level. But because we have so many cells in our bodies, it is relatively likely that a defective mutation will occur in the single normal gene and lead to a cancerous cell. Some heterozygous individuals may not develop the disease as a matter of chance. They may be lucky and not get a defective mutation in the normal gene or perhaps their immune system is better at destroying cancerous cells once they arise.

C20. Conversion of a proto-oncogene to an oncogene can occur by missense mutations, gene amplifications, chromosomal translocations, and viral integration. Examples are given in Table 22.6. The genetic changes are expected to increase the amount of the encoded protein or alter its function in a way that makes it more active.

C21. If an oncogene was inherited, it may cause uncontrollable cell growth at an early stage of development and thereby cause an embryo to develop improperly. This could lead to an early spontaneous abortion and thereby explain why we do not observe individuals with inherited oncogenes. Another possibility is that inherited oncogenes may adversely affect gamete survival, which would make it difficult for them to be passed from parent to offspring. A third possibility is

that oncogenes could affect the fertilized zygote in a way that would prevent the correct implantation in the uterus.

C22. A. No, because E2F is inhibited B. Yes, because E2F is not inhibited C. Yes, because E2F is not inhibited D. No, because there is no E2F C23. The role of p53 is to sense DNA damage and prevent damaged cells from proliferating. Perhaps, prior to birth, the fetus is in a protected environment, so that DNA damage may be minimal. In other words, the fetus may not really need p53. After birth, agents such as UV light may cause DNA damage. At this point, p53 is important. A p53 knockout is more sensitive to UV light because it cannot repair its DNA properly in response to this DNA-damaging agent, and it cannot kill cells that have become irreversibly damaged.

C24. A. True B. True C. False; most cancer cells are caused by mutations that result from environmental mutagens.

D. True C25. The p53 protein is a regulatory transcription factor; it binds to DNA and influences the transcription rate of nearby genes. This transcription factor (1) activates genes that promote DNA repair; (2) activates a protein that inhibits cyclin/CDK protein complexes that are required for cell division; and (3) activates genes that promote apoptosis. If a cell is exposed to DNA damage, it has a greater potential to become malignant. Therefore, an organism wants to avoid the proliferation of such a cell. When exposed to an agent that causes DNA damage, a cell will try to repair the damage.

However, if the damage is too extensive, the p53 protein will stop the cell from dividing and program it to die. This helps to prevent the proliferation of cancer cells in the body.

C26. Though they don’t change the DNA sequence, epigenetic modifications can affect gene expression.

Such changes could increase gene expression and thereby result in oncogenes or they could inhibit the expression of tumor suppressor genes. Either type of change could contribute to cancer. For example, DNA methylation of a tumor suppressor gene could promote cancer.

Application and Experimental Questions E1. A. False, this would suggest an infectious disease, because people living in the same area would be exposed to the same kinds of infectious agents. Relatives living together and apart would exhibit the same frequency for a genetic disease.

B. This could be true, because the individuals living in one area may be more genetically related.

On the other hand, a particular infectious agent may be found only in southern Spain, and this might explain the high frequency in this region.

C. A specific age of onset is consistent with a genetic disease.

D. A higher likelihood of developing a disease among monozygotic twins compared to dizygotic twins is consistent with a genetic basis for a disease because monozygotic twins are more genetically similar (in fact, identical) compared to dizygotic twins.

E2. Perhaps the least convincing observation is the higher incidence of a disease in particular populations. Because populations living in specific geographic locations are exposed to their own unique environment, it is difficult to distinguish genetic versus environmental causes for a particular

disease. The most convincing evidence might be the higher incidence of the disease in related individuals or the ability to correlate the disease with the presence of a mutant gene. Overall, however, the conclusion that a disease has a genetic component should be based on as many observations as possible.

E3. The term genetic testing refers to the use of laboratory tests to determine if an individual is a carrier or affected by a genetic disease. Testing at the protein level means that the amount or activity of the protein is assayed. Testing at the DNA level means that the researcher tries to detect the mutant allele at the molecular or chromosomal level. Examples of approaches are described in Table 22.5.

E4. You would probably conclude that the disease does not have a genetic component. If it were rooted primarily in genetics, it would be likely to be observed in the Central American population. Of course, there is a chance that very few or none of the people who migrated to Central America were carriers of the mutant gene, but that is somewhat unlikely for a large migrating population. By comparison, one might suspect that an environmental agent that is present in South America but not present in Central America may underlie the disease. Researchers could try to search for this environmental agent (e.g., a pathogenic organism).

E5. A. As seen in lane 3, the α-galactosidase A polypeptide is shorter in cells obtained from Pete. This indicates that Pete’s disease is caused by a mutation that either is a deletion in the gene or introduces an early stop codon. In Jerry’s case (lane 6), there does not appear to be any of the αgalactosidase A polypeptide in his cells. This could be due to a deletion that removes the entire gene, a promoter mutation that prevents the expression of the gene, a mutation that prevents translation (e.g., a mutation in the start codon), or a mutation that results in a polypeptide that is very unstable and rapidly degraded.

B. Amy appears to have two normal copies of the α-galactosidase A gene. She will not pass a mutant allele to her offspring. Nan is a heterozygote. She has a 50% chance of passing the mutant allele. Half of her sons would be affected with the disease. Likewise, Aileen also appears to be a heterozygote because the amount of α-galactosidase A polypeptide seems to be about 50% of normal. She also would have a 50% chance of passing the mutant allele to her offspring; half of her sons would be affected.

E6. Males I-1, II-1, II-4, II-6, III-3, III-8, and IV-5 have a normal copy of the gene. Males II-3, III-2, and IV-4 are hemizygous for an inactive mutant allele. Females III-4, III-6, IV-1, IV-2, and IV-3 have two normal copies of the gene, whereas females I-2, II-2, II-5, III-1, III-5, and III-7 are heterozygous carriers of a mutant allele.

E7. Most inherited forms of cancer are inherited in a dominant manner. This can often be revealed by a pedigree analysis, because affected individuals are much more likely to have affected offspring.

Because cancer may be caused by the sequential accumulation of mutations, the correlation between affected parents and affected offspring may be relatively low because the offspring may not accumulate the other postzygotic mutations that are necessary for cancer to occur.

E8. One possible category of drugs would be GDP analogues (i.e., compounds whose structure resemble that of GDP). Perhaps you could find a GDP analogue that binds to the Ras protein and locks it in the inactive conformation.

One way to test the efficacy of such a drug would be to incubate the drug with a type of cancer cell that is known to have an overactive Ras protein, and then plate the cells on solid media. If the drug locked the Ras protein in the inactive conformation, it should inhibit malignant growth of cells.

There are possible side effects of such drugs. First, they might block the growth of normal cells, because Ras protein plays a role in normal cell proliferation. Second (as you might know from a cell biology course), there are many GTP/GDP-binding proteins in cells, and the drugs could somehow inhibit cell growth and function by interacting with these proteins.

E9. Molecular profiling is used to identify gene expression patterns in tumors. Sometimes tumors that look morphologically similar have very different expression patterns. Molecular profiling makes it possible to distinguish such tumors. This may be useful in deciding on treatment options and in the prognosis of tumor progression.

Questions for Student Discussion/Collaboration 1. The overall benefit is that genetic screening may lead to better treatments of disease. For example, if an individual discovered that they were carrying an allele that caused a higher predisposition to develop cancer, they could be more careful about monitoring their health in this regard. Or if a person discovered that they were the carrier of a hypercholesterolemia allele, they could try to eat a very lowfat diet. On the negative side, people may be discriminated against by their employers or by insurance companies because of the alleles that they carry.

2. There isn’t a clearly correct answer to this question, but it should stimulate interesting discussion.

CHAPTER 23 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 23.1 A local population is a group of individuals that are more likely to interbreed with each other compared with individuals in more distant populations.

FIGURE 23.2 Polymorphisms are very common in nearly all natural populations.

FIGURE 23.6 The word directional means that selection is favoring a particular phenotype. Selection is moving the population toward prevalence of that phenotype.

FIGURE 23.8 No, directional selection is not promoting genetic diversity because it is eliminating individuals that are sensitive to DDT. FIGURE 23. 9 The HbS allele is an advantage in the heterozygous condition because it confers resistance to malaria.

The heterozygote advantage outweighs the homozygote disadvantage.

FIGURE 23.10 With negative frequency-dependent selection, the rarer phenotype has a higher fitness, which improves its reproductive success.

FIGURE 23.11

Yes, disruptive selection fosters polymorphism. The fitness values of the resulting phenotypes depend on the environment. Some phenotypes are the fittest in one environment, whereas other phenotypes are the fittest in another environment.

FIGURE 23. 13 Stabilizing selection decreases genetic diversity because it eliminates individuals that carry alleles that promote more extreme phenotypes.

FIGURE 23.14 Genetic drift tends to have a greater effect in small populations, where it can lead to the rapid loss or fixation of an allele.

FIGURE 23.15 At the bottleneck, genetic diversity may be reduced because the population consists of fewer individuals.

Following the bottleneck, genetic drift may promote the loss of certain alleles and the fixation of other alleles, thereby lowering genetic diversity even more.

FIGURE 23.16 Inbreeding increases the likelihood of homozygosity, and therefore tends to increase the likelihood that an individual will exhibit a recessive trait. Homozygosity becomes more likely because an individual can inherit both copies of the same allele from a common ancestor.

FIGURE 23. 18 DNA fingerprinting is used for identification and for relationship testing.

End-of-chapter Questions: Conceptual Questions C1. A gene pool is all of the genes present in a particular population. Each type of gene within a gene pool may exist in one or more alleles. The prevalence of an allele within the gene pool is described by its allele frequency. If a gene is monomorphic, the allele frequency is close to 100%. If it is polymorphic, each allele has a frequency that is between 1 and 99%. The sum of all the allele frequencies for a particular gene will add up to 100%.

C2. A population is a group of interbreeding individuals. An example is a squirrel population in a forested area. Over the course of many generations, several things could happen to this population.

A forest fire, for example, could dramatically decrease the number of individuals and thereby create a bottleneck. This would decrease the genetic diversity of the population. A new predator may enter the region, and natural selection may then select for the survival of squirrels that are best able to evade the predator. Another possibility is that a group of squirrels within the population may migrate to a new region and found a new squirrel population.

C3. The term genetic polymorphism refers to the phenomenon in which a gene is found in two or more alleles within a population. Mutation is the ultimate source of genetic variation.

C4. A. Phenotype frequency and genotype frequency B. Genotype frequency C. Allele frequency C5. When a trait is polymorphic, this means that different individuals show phenotypic variation with regard to the trait. For example, petunias can have red or white flowers. Flower color is polymorphic in petunias. When a gene is polymorphic, it exists in two or more alleles. At the molecular level, alleles of a given gene have different DNA sequences. These differences could be

very slight (e.g., a single-base change) or they could involve significant additions or deletions.

Different alleles may cause differences in phenotype. For example, in Mendel’s pea plants, the T allele resulted in tall plants, while the t allele caused dwarf plants. However, alleles do not always cause differences in phenotype. A single-base substitution in a gene may not affect the amino acid sequence of the encoded polypeptide (e.g., the base change may affect the wobble base) or a singlebase change might alter the amino acid sequence, but not in a way that alters the protein’s function.

Therefore, gene polymorphism does not always result in phenotypic polymorphism.

C6. A. The genotype frequency for the homozygotes for the mutant allele is 1/2,500, or 0.004. This value equals q2. The allele frequency is the square root of this value, which equals 0.02. The frequency of the corresponding non-disease-causing allele equals 1 – 0.02 = 0.98.

B. The genotype frequency for homozygous affected individuals is 0.004; for the unaffected homozygous individuals, it is (098)2 = 0.96; and for the heterozygous individuals, it is 2(0.98)(0.02), which equals 0.039.

C. If a person is known to be a heterozygous carrier, the chances that this particular person will happen to choose another as a mate is equal to the frequency of heterozygous carriers in the population, which is 0.039, or 3.9%. The chances that two heterozygous carriers will choose each other as mates equals 0.039 × 0.039 = 0.0015, or 0.15%.

C7. For two alleles, the heterozygote is at a maximum when they are 0.5 each. For three alleles, the two heterozygotes are at a maximum when each allele is 0.33.

C8. Applying the Hardy-Weinberg equation gives these genotype frequencies: BB = (0.67)2 = 0.45, or 45% Bb = 2(0.67)(0.33) = 0.44, or 44% bb = (0.33)2 = 0.11, or 11% The actual data show a higher percentage of homozygotes (compare 45% with 50% and 11% with 13%) and a lower percentage of heterozygotes (compare 44% with 37%) than expected. Therefore, these data are consistent with inbreeding, which increases the percentage of homozygotes and decreases the percentage of heterozygotes.

C9. Because this is a recessive trait, only the homozygotes for the rolling allele will be able to roll their tongues. If p equals the rolling allele and q equals the nonrolling allele, the Hardy-Weinberg equation predicts that the frequency of homozygotes who can roll their tongues would be p2. In this case, p2= (0.6)2 = 0.36, or 36%.

C10. Migration, genetic drift, and natural selection are the main mechanisms that alter allele frequencies within a population. Natural selection acts to eliminate harmful alleles and promote beneficial alleles. Genetic drift involves random changes in allele frequencies that may eventually lead to elimination or fixation of alleles. It is thought to be important in the establishment of neutral alleles in a population. Migration is important because it introduces new alleles into neighboring populations.

C11. Genetic drift alters allele frequencies without any regard to whether the changes are beneficial or not. An example would be changes that alter the sequences of introns. Natural selection is an adaptive process that tends to eliminate harmful alleles from a population and increase the frequency of beneficial alleles. An example would be the long snout of the anteater, which enables it to feed on ants. At the molecular level, a neutral mutation does not affect gene expression; for example, it could alter the wobble base without affecting the amino acid sequence of a polypeptide.

In contrast, adpative mutations may alter the coding sequence of a gene and change the structure and function of the protein in a way that is beneficial. For example, the sickle-cell allele alters the structure of hemoglobin, and in the heterozygous condition, this decreases the proliferation of the malaria pathogen.

C12. Darwinian fitness is the relative likelihood that an individual with a particular genotype will survive and contribute to the gene pool of the next generation as compared to other genotypes. The genotype with the highest reproductive ability is given a value of 1.0. Characteristics that promote survival, ability to attract a mate, or enhanced fertility are expected to increase Darwinian fitness.

Examples are the thick fur of a polar bear, which helps it to survive in a cold climate; the bright plumage of male birds, which helps them to attract a mate; and the high number of gametes released by certain species of fish, which enhances their fertility.

C13. The intuitive meaning of the mean fitness of a population is the relative likelihood that members of a population will reproduce. If the mean fitness is high, it is likely that an average member will survive and produce offspring. Natural selection increases the mean fitness of a population.

C14. All of these patterns of natural selection favor one or more phenotypes because such phenotypes have a reproductive advantage. However, the patterns differ with regard to whether a single phenotype or multiple phenotypes are favored and whether the phenotype that is favored is in the middle of the phenotypic range or at one or both extremes. Directional selection favors an extreme phenotype. Over time, natural selection is expected to favor the fixation of alleles that result in such a phenotype. Disruptive selection favors two or more phenotypes. It will lead to a population with a balanced polymorphism for the trait. Balancing selection may occur because of heterozygote advantage or negative-frequency dependent selection. These promote a stable polymorphism in a population. Stabilizing selection favors individuals with intermediate phenotypes. It tends to decrease genetic diversity because alleles for extreme phenotypes are eliminated.

C15. A. Disruptive. There are multiple environments that favor different phenotypes.

B. Directional. The thicker the fur, the more likely that survival will occur.

C. Stabilizing. Low birth weight is selected against because it results in low survival. Also, very high birth weight is selected against because it could cause problems in delivery, which also could decrease the survival rate.

D. Directional. Sturdy stems and leaves will promote survival in windy climates.

C16. In genetic drift, allele frequencies are drifting. Genetic drift is an appropriate term because the word drift implies a random process. Nevertheless, drift can be directional. A boat may drift from one side of a lake to another. It would not drift in a straight path, but the drifting process will alter its location. Similarly, allele frequencies can “drift” up or down and eventually lead to the elimination or fixation of particular alleles within a population.

C17. Genetic drift is due to sampling error, and the degree of sampling error depends on the population size. In small populations, the relative proportion of sampling error is much larger. If genetic drift is moving an allele toward fixation, it will take longer in a large population because the degree of sampling error is much smaller.

C18. A. Probability of fixation for allele a = 1/2N = 1/2(4) = 1/8, or 0.125 B. t = 4N = 4(4) = 16

C. The calculations for parts A and B assume a constant population size. If the population grows after it has been founded by the four individuals, the probability of fixation will be lower and the time to reach fixation will be longer.

C19. During the bottleneck effect, allele frequencies are dramatically altered due to genetic drift. In extreme cases, some alleles are lost, while others may become fixed at 100%. The overall effect is to decrease genetic diversity within the population. This may make it more difficult for the species to respond in a positive way to changes in the environment. Species that are approaching extinction also face a bottleneck as their numbers decrease. The loss of genetic diversity may make it even more difficult for the species to rebound.

C20. A. True B. True C. False; it causes allele loss or fixation, which results in less genetic diversity.

D. True C21. When two populations intermix, both populations tend to have more genetic variation because each population introduces new alleles into the other population. In addition, the two populations tend to have similar allele frequencies, particularly when a large proportion migrates.

C22. A. Migration will increase the genetic diversity in both populations. A random mutation could occur in one population to create a new allele. This new allele could be introduced into the other population via migration.

B. The allele frequencies between the two populations will tend to be similar to each other, due to the intermixing of their alleles.

C. The rate of genetic drift depends on population size. When two populations intermix, this has the effect of increasing the overall population size. In a sense, the two smaller populations behave like one big population. Therefore, the effects of genetic drift are lessened when the individuals in two populations can migrate. The net effect is that allele loss and allele fixation are less likely to occur due to genetic drift.

C23. In the absence of other factors, inbreeding does not alter allele frequencies, because it does not favor the transmission of one allele over another. It simply increases the likelihood of homozygosity (which is a genotype frequency). In human populations, inbreeding increases the frequency of individuals who are homozygous for rare recessive alleles that cause human disease.

C24. A. Yes. B. The common ancestors are I-1 and I-2.

C. F = Σ(1/2)n (1 + FA) F = (1/2)9 + (1/2)9 F = 1/512 + 1/512 = 2/512 = 0.0039 D. Not that you can tell from this pedigree C25. A. The inbreeding coefficient is calculated using the formula F = Σ (1/2)n (1 + FA) In this case, there is one common ancestor, I-2. Because you have no prior history on I-2, you assume she is not inbred, which makes FA = 0. The inbreeding loop for IV-3 contains five people, III-4, II-2, I-2, II-5, and III-5. Therefore, n = 5.

F = (1/2)5(1 + 0) = 1/32 = 0.031 B. Based on the data shown in this pedigree, individual IV-4 is not inbred.

C26. A. The inbreeding coefficient is calculated using the formula F = Σ (1/2)n (1 + FA) In this case, there are two common ancestors, I-1 and I-2. Because you have no prior history on I-1 or I-2, you assume they are not inbred, which makes FA = 0. The two inbreeding loops for IV-2 contain five people: III-4, II-2, I-1, II-6, III-5; and III-4, II-2, I-2, II-6, III-5. Therefore, n = 5 for both loops.

F = (1/2)5 (1 + 0) + (1/2)5 (1 + 0) F = 0.031 + 0.031 = 0.062 B. Based on the data shown in the pedigree, individual III-4 is not inbred.

C27. A. Random mutation is the source of genetic variation that may lead to antibiotic resistance. A random mutation may create an antibiotic-resistance allele. This could occur in different ways. Two possibilities are: 1. Many antibiotics exert their effects by binding to an essential cellular protein within the microorganism and inhibiting its function. A random mutation could occur in the gene that encodes such an essential cellular protein; this could alter the structure of the protein in a way that would prevent the antibiotic from binding to the protein or inhibiting its function.

2. As another possibility, microorganisms, which are killed by antibiotics, possess many enzymes, which degrade related compounds. A random mutation could occur in a gene that encodes a degradative enzyme so that the enzyme now recognizes the antibiotic and degrades it.

B. When random mutations occur, they may be lost due to genetic drift. This is particularly likely when the frequency of the mutation is very low in a large population. Alternatively (and much less likely), a random mutation that confers antibiotic resistance could become fixed in a population.

C. If a random mutation occurs that confers antibiotic resistance, and if the mutation is not lost by genetic drift, natural selection will favor the growth of microorganisms that carry the antibioticresistance allele if the organisms are exposed to the antibiotic. Therefore, if antibiotics are widely used, this will kill microorganisms that are sensitive and favor the proliferation of ones that happen to carry antibiotic-resistance alleles.

C28. You can use the following equation to calculate the change in allele frequency after any number of generations: (1  u)t 

pt

po (1  10 )  0.5 / 0.6  0.833 4 t

(0.9999)t  0.833 t  1,827 generations

Application and Experimental Questions E1. A. At the DNA level, a clear-cut way to determine genetic variation is to clone and sequence genes.

If the same gene is cloned from two different individuals and the sequences are different, this shows that there is genetic variation. In addition, several other methods can be used to detect genetic

variation. For example, a comparison of Southern blots using samples from different individuals might reveal that a gene exists in different sizes or that it contains different restriction sites.

B. At the RNA level, a Northern blot may reveal genetic variation. If the RNA encoded by two different alleles has a different size, this can be detected in a Northern blot.

C. At the protein level, gel electrophoresis may reveal genetic variation. This method is described in the Appendix. Another approach is to study the function of an enzyme using a biochemical assay of its activity.

E2. In this case: (p + q + r + s)2 = 1 p2 + q2 + r2 + s2 + 2pq + 2qr +2qs +2rp + 2rs + 2sp = 1 Let p = C, q = cch, r = ch, and s = c.

A. The genotype frequency of albino rabbits is s2: s2 = (0.05) = 0.0025, or 0.25% B. Himalayan is dominant to albino but recessive to full and chinchilla. Therefore, genotypes for Himalayan rabbits are represented by r2 and by 2rs: r2 + 2rs = (0.44)2 + 2(0.44)(0.05) = 0.24, or 24% Among 1,000 rabbits, about 240 would have a Himalayan coat color.

C. Chinchilla is dominant to Himalayan and albino but recessive to full coat color. Therefore, heterozygotes with chinchilla coat color are represented by 2qr and by 2qs.

2qr + 2qs = 2(0.17)(0.44) + 2(0.17)(0.05) = 0.17, or 17% Among 1,000 rabbits, about 170 would be heterozygotes with chinchilla fur.

E3. A. Let W represent the white fat allele and w represent the yellow fat allele. Assuming HardyWeinberg equilibrium, you can let p2 represent the genotype frequency of WW animals, and then Ww would be 2pq and ww would be q2. The only genotype frequency you know is that of the ww animals.

76 ww  q2  5, 468 2 q = 0.014 q = 0.12, which is the allele frequency of w p=1–q p = 0.88, which is the allele frequency of W B. The heterozygous carriers are represented by 2pq. If you use the values of p and q, which were calculated in part A: 2pq = 2(0.88)(012) = 0.21 Approximately 21% of the animals would be heterozygotes with white fat.

If you multiply 0.21 times the total number of animals in the herd: 0.21  5,468 = 1,148 animals E4. A. Inuit M = 0.913 N = 0.087 Navajo M = 0.917 N = 0.083 Finns M = 0.673 N = 0.327 Russians M = 0.619 N = 0.381 Aborigines M = 0.176 N = 0.824

B. To determine if any of these populations are in equilibrium, you can use the Hardy-Weinberg equation and calculate the expected number of individuals with each genotype. For example: Inuit MM = (0.913)2 = 83.3 MN = 2 (0.913)(0.087) = 15.9 NN = (0.087)2 = 0.76 In general, these values agree pretty well with an equilibrium. The same is true for the other four populations.

C. Based on similar allele frequencies, the Inuit and Navajo seem to have interbred as have the Finns and Russians.

E5. The first thing you need to do is to determine the allele frequencies. Let’s let p represent i, q represent IA, and r represent IB. p2 is the genotype frequency of ii q2 is the genotype frequency of IAIA r2 is the genotype frequency of IBIB 2pq is the genotype frequency of IAi 2pr is the genotype frequency of IBi 2qr is the genotype frequency of IAIB 721 721  932  235  112 p2 = 0.36 p = 0.6 Next, you can calculate the allele frequency of IA. Keep in mind that two genotypes (IAIA and IAi) result in type A blood.

932 q2  2 pq  721  932  235  12 p2 

q2 + 2(0.6)q = 0.47 By solving this quadratic equation, you get q = 0.31 Now you can solve for r, p+q+r=1 0.6 + 0.31 + r = 1 r = 0.09 Based on these allele frequencies, you can compare the observed and expected values. To determine the expected values, you multiply the genotype frequencies times 2,000, which was the total number of individuals in this population. p2 is the genotype frequency of ii = (0.6)2(2,000) = 720 q2 is the genotype frequency of IAIA= (0.31)2(2,000) = 192 r2 is the genotype frequency of IBIB = (0.09)2(2,000) = 16 2pq is the genotype frequency of IAi = 2(0.31)(0.6)(2,000) = 744

2pr is the genotype frequency of IBi = 2(0.09)(0.6)(2,000) = 216 2qr is the genotype frequency of IAIB = 2(0.31)(0.09)(2,000) = 111 Type O Type A Type B Type AB

Expected Numbers 720 192 + 744 = 936 16 + 216 = 232 111

Observed Numbers 721 932 235 112

The observed and expected values agree quite well. Therefore, it does appear that this population is in Hardy-Weinberg equilibrium.

E6. A. ∆pC = m(pD – pR) With regard to the sickle cell allele: ∆pC = (550/10,550)(0.1 – 0.01) = 0.0047 pC = pR + ∆pC = 0.01 + 0.0047 = 0.0147 B. You need to calculate the genotypes separately: For the 550 migrating individuals, HbAHbA = (0.9)2 = 0.81 You expect (0.81)550 = 445.5.

HbAHbS = 2(0.9)(0.1) = 0.18 You expect (0.18)550 = 99 HbSHbS = (0.1)2 = 0.01 You expect (0.01)550 = 5.5.

For the original recipient population, HbAHbA = (0.99)2 = 0.98 You expect 9,801 individuals to have this genotype.

HbAHbS = 2(0.99)(0.01) = 0.0198 You expect 198 with this genotype.

HbSHbS = (0.01) 2 = 0.0001 You expect 1 with this genotype.

For the conglomerate population, (445.5 + 9801)/10,550 = 0.971 HbAHbA homozygotes (99 + 198)/10,550 = 0.028 heterozygotes (5.5 + 1)/10,550 = 0.00062 HbSHbS homozygotes C. In the next generation, the allele frequencies in the conglomerate population (calculated in part A), should yield the expected genotype frequencies according to Hardy-Weinberg equilibrium.

Allele frequency of HbS = 0.0147, so HbA = 0.985 HbAHbA = (0.985)2 = 0.97 HbAHbS = 2(0.985)(0.0147) = 0.029 HbSHbS = (0.0147)2 = 0.0002 E7. Let’s assume that the relative fitness values are 1.0 for the dominant homozygote and the heterozygote and 0 for the recessive homozygote. The first thing you need to do is to calculate the mean fitness of the population. _ 2 2 p wAA + 2pqwAa + q waa = w _ 2 2 (0.78) + 2(0.78)(0.22) + (0.22)(0) = w _

w = 0.95 The allele frequency (p) in the next generation equals

p= p= q=

p2wAA + pqwAa _ _ w w 0.82 1 - p = 0.18

You would follow the same general strategy for the second and third generations as well.

For the second generation, the mean fitness of the population now equals 0.97. Using the preceding equation, the allele frequency of A in the second generation approximately equals 0.85. The frequency of the recessive allele in the second generation would equal about 0.15 and the mean fitness would now be approximately 0.98. The allele frequency of A in the third generation would be approximately 0.87. The frequency of the recessive allele would be about 0.13.

E8. A. Probability of fixation = 1/2N(assuming equal numbers of males and females contributing to the next generation) Probability of fixation = 1/2(2,000,000) = 1 in 4,000,000 chance B. _ t = 4N where t = the average number of generations to achieve fixation N = the number of individuals in a population, assuming that males and females contribute equally to each succeeding generation _ t = 4(2 million) = 8 million generations C. If the blue allele conferred a slight survival advantage, the value calculated in part A would be slightly larger; there would be a higher chance of allele fixation. The value calculated in part B would be smaller; it would take a shorter period of time to reach fixation.

E9. The selection coefficients are sww = 1 – 0.19 = 0.81 sWW = 1 – 0.37 = 0.63 If the rats are not exposed to warfarin, the equilibrium will no longer exist, and natural selection will tend to eliminate the warfarin-resistance allele because the homozygotes are vitamin K deficient.

E10. You could mark snail shells with a dye and release equal numbers of dark-shelled and light-shelled snails into dimly lit forested regions and sunny fields. At a later time, recapture the snails and count them. It would be important to have a method of unbiased recapture because it would be easier to find the light-shelled snails in a forest and the dark-shelled snails in a field. Perhaps you could bait

the region with something that the snails like to eat and only collect snails that are at the bait. In addition to this type of experiment, you could also observe predation as it occurs.

E11. Male 2 is the potential father, because he contains the bands that are found in the offspring but are not found in the mother. To calculate the probability, you would have to know the probability of having each of the types of bands that match. In this case, for example, male 2 and the offspring have four bands in common. As a simple calculation, you could eliminate the four bands that the offspring shares with the mother. If the probability of having each paternal band is 1/4, the chances that this person is not the father are (1/4)4.

E12. DNA fingerprinting is a method of identification based on the properties of DNA. Minisatellites and microsatellite sequences vary in size in natural populations. This variation can be seen when DNA fragments are subjected to gel electrophoresis. Within a population, any two individuals (except for identical twins) will display a different pattern of DNA fragments; that pattern is called their DNA fingerprint.

E13. This percentage is not too high. Based on their genetic relationship, you expect that a father and daughter must share at least 50% of the same bands in a DNA fingerprint. However, the value can be higher than that because the mother and father may have some bands in common, even though they are not genetically related. For example, at one site in the genome, the father may be heterozygous for a 4,100 bp and 5,200 bp minisatellite, and the mother may also be heterozygous in this same region and have 4,100 bp and 4,700 bp minisatellites. The father could pass the 5,200 bp band to his daughter, and the mother could pass the 4,100 bp band. Thus, the daughter would inherit the 4,100 bp and 5,200 bp bands. This would be a perfect match to both of the father’s bands, even though the father transmitted only the 5,200 bp band to his daughter. The 4,100 bp band matches because the father and mother happened to have a minisatellite in common. Therefore, the 50% estimate of matching bands in a DNA fingerprint based on genetic relationships is a minimum estimate. The value can be higher than that.

E14. The minimum percentage of matching bands is based on the genetic relationships.

A. 50% B. 50% (on average, but it could be less or more) C. 25% (on average, but it could be less or more) D. 25% (on average, but it could be less or more) Questions for Student Discussion/Collaboration 1. Positive assortative mating: In natural populations, individuals with similar markings may tend to recognize each other as being members of the same species. In human populations, people tend to choose mates who are similar with regard to size, race, etc. Inbreeding in agriculture to produce larger tomatoes, greater beef yield, etc.

Negative assortative mating: This is less common although it does occur. In certain plant species, there are alleles that prevent selfhybridization. Therefore, outcrossing is favored which is somewhat like disassortative mating. Some people tend to prefer mates who are very dissimilar in their characteristics.

2. Mutation is responsible for creating new alleles but the rate of new mutations is so low that it cannot explain allele frequencies in this range. Let's call the two alleles B and b and assume that B was the

3.

original allele and b is more recent allele that arose as a result of mutation. Three scenarios to explain the allele frequencies are as follows: 1. The b allele arosee via a neutral mutation and reached its present frequency by genetic drift. It hasn't yet been eliminated or fixed.

2. The b allele arose via a beneficial mutation, and its frequency is increasing due to natural selection.

However, there hasn't been enough time to reach fixation.

3. The Bb heterozygote has a selective advantage leading to a balanced polymorphism.

It's a difficult question to answer. Over the short term, the elimination of harmful alleles is probably more significant at keeping the population healthy because the frequency of new mutations that are deleterious is much higher than the frequency of new mutations that are beneficial. Over the long term, however, every species needs variation that promotes its adaptation to its environment. In this regard, the selection of beneficial alleles is much more important.

In human populations, the selection of beneficial alleles may be less important because humans are becoming better at controlling their environment and enhancing their reproductive capabilities. On the other hand, humans are also getting better at curing genetic diseases so perhaps the elimination of harmful diseases may not be so critical either. Overall, it may be true that natural selection does not play as important of a role in human populations as it once did.

CHAPTER 24 Note: the answers to the Comprehension Questions are at the end of the chapter.

Concept Check Questions (in figure legends) FIGURE 24.1 In most populations (like the one shown), height exhibits a continuum.

FIGURE 24.2 About 95.4% are within 2 standard deviations, which means that 4.6% are outside of this range. Half of them, or 2.3%, fall more than two standard deviations above the mean.

FIGURE 24.3 When alleles are additive, these means they contribute in an incremental way to the outcome of a trait.

Having three red alleles will make the hull color of a wheat plant darker red than it will be with two red alleles.

FIGURE 24.4 Increases in gene number and in environmental variation tend to cause greater overlaps between different phenotypes due to different genotypes.

FIGURE 24.5 These two strains differ with regard to a quantitative trait and in their molecular markers.

FIGURE 24.9 Both natural selection and selective breeding affect allele frequencies as a result of differences in reproductive success. In the case of natural selection, reproductive success is determined by environmental conditions. For selective breeding, reproductive success is determined by the people who choose the parents for breeding.

FIGURE 24 .10

This could be a long list. Examples are weight, disease resistance, food quality, resistance to harsh environmental conditions, etc.

FIGURE 24.11 A selection limit may be reached because a population has become monomorphic for all of the desirable alleles. A second reason is because the desired effects of artificial selection are balanced by the negative effects on fitness.

End-of-chapter Questions: Conceptual Questions C1. Quantitative traits are described numerically. Examples include height, weight, speed, and metabolic rate. In a population, a trait may be given a mean value, and the degree of variation may be described by the variance and standard deviation.

C2. Quantitative traits often exhibit a continuum of phenotype because they are usually influenced by multiple genes that exist as multiple alleles. A large amount of environmental variation will also increase the overlap between genotypes and phenotypes for polygenic traits.

C3. A normal distribution varies in a symmetrical way around a mean value. When graphed, it exhibits a bell-shaped curve. It is common for quantitative traits in a population to have a normal distribution around a mean value. If a distribution is normal, the standard deviation can predict the percentage of individuals that will fall within certain limits. For example, approximately 68% of all individuals in a population will be within one standard deviation unit (plus or minus) from the mean.

C4. A discontinuous trait is one that falls into discrete categories. Examples include brown eyes versus blue eyes in humans and purple versus white flowers in pea plants. A continuous trait is one that does not fall into discrete categories. Examples include height in humans and fruit weight in tomatoes. Most quantitative traits are continuous; the trait exhibits a range of values. The reason why quantitative traits are continuous is because they are usually polygenic and greatly influenced by the environment. As shown in Figure 24.4b, this tends to create overlaps between phenotypes due to different genotypes.

C5. With regard to genetics, a frequency distribution is a graph that compares the number of individuals and their phenotypes for a given trait. This is a common way to depict the distribution of a quantitative trait in a population. To make a frequency distribution for a continuous quantitative trait, it is necessary to divide individuals into arbitrary groups. For example, if the trait is weight, individuals could be divided into groups that are in multiples of 10 pounds. Such a grouping would begin with individuals in the 0–10 pound range, 10–20 pound, 20–30 pound, etc. To make a frequency distribution, a population of individuals would be chosen, and then the number of individuals in each grouping would be determined. The groupings (weight categories) would be plotted on the x-axis, and the number of individuals in each group would be plotted on the y-axis.

C6. To rank in the top 2.5% requires a weight about two standard deviation units above the mean. If you take the square root of the variance, the standard deviation is 22 pounds. To rank in the top 2.5%, an animal has to be at least 44 pounds heavier than the mean, that is have a weight of 606 pounds. To rank in the bottom 0.13%, an animal has to be three standard deviations lighter than the mean weight, or at least 66 pounds lighter than the mean and thus weigh 496 pounds.

C7. A. The differences in variance could be explained by variation in the way the potatoes are affected by environmental variation, or they could be explained by genetic differences. The strain with a higher variance may have more genetic variation.

B. On the one hand, the strain with a higher variance may be better because the farmer may want to select for individuals that produce larger potatoes. On the other hand, if the farmer wants the size of his/her potatoes to be uniform, the population with a smaller variance would be better.

C. You would use the strain with a higher variance, hoping that a significant proportion of the variance was due to genetic variation.

C8. There is a positive correlation, but it could have occurred as a matter of chance alone. According to Table 24.2, this value could have occurred due to random sampling error. Further experimentation would be needed to determine if there is a significant correlation. If N = 500, the correlation is statistically significant, and you would conclude that the correlation did not occur as a matter of random chance. However, you could not claim cause and effect.

C9. A negative correlation means that two variables are related to each other in opposite ways. For example, a high amount of snowfall in the winter might be negatively correlated with the survival of wild species.

C10. When a correlation coefficient is statistically significant, it means that the association is likely to have occurred for reasons other than random sampling error. Statistical significance may indicate cause and effect but not necessarily. For example, large parents may have large offspring due to genetics (cause and effect). However, such a correlation may be related to the sharing of similar environments rather than being due to cause and effect.

C11. Polygenic inheritance refers to the transmission of traits that are governed by two or more different genes. Such traits are difficult to study because there may be multiple genes with multiple alleles, and environmental effects may be significant. Therefore, the discrete genotypes do not fall within discrete phenotypic categories. Instead, there tends to be overlap, which confounds a genetic analysis. Molecular approaches, such as QTL mapping, can greatly help the situation.

C12. A quantitative trait locus is a site on a chromosome that contains one or more genes that affect a quantitative trait. It is possible for a QTL to contain one gene, or it may contain two or more closely linked genes. QTL mapping, which analyzes linkage to known molecular markers, is commonly used to determine the locations of QTLs. C13. A. The probability of inheriting two light alleles from a parent would be 0.5  0.5 = 0.25. The probability of inheriting two light alleles from both parents would be 0.25  0.25 = 0.0625, or 6.25%.

B. The probability of inheriting three light alleles from a parent would be 0.5  0.5  0.5 = 0.125.

The probability of inheriting three light alleles from both parents would be 0.125  0.125 = 0.0156, or 1.56%.

C. The probability of inherited four light alleles from a parent would be 0.5  0.5  0.5  0.5 = 0.0625. The probability of inheriting four light alleles from both parents would be 0.0625  0.0625 = 0.0039, or 0.39% C14. If the broad sense heritability equals 1.0, it means that all of the variation in the population is due to genetic variation rather than environmental variation. It does not mean that the environment is

unimportant in the outcome of the trait. Under another set of environmental conditions, the trait may have turned out quite differently.

C15. The primary advantage of selective breeding in agriculture is that it improves the magnitude of quantitative traits. For example, selective breeding can increase fruit size, fruit number, etc. A disadvantage is that selective breeding will diminish the overall genetic diversity in the population, and this may result accidentally in the prevalence of traits that are not desirable. For example, as a tomato breeder selects for bigger and juicier tomatoes, he or she may unwittingly obtain a variety that has bigger and juicier tomatoes, but the plants may be more susceptible to pathogens compared to the original (nonselected) parental strain.

Undesirable traits, like pathogen sensitivity, are often inherited in a recessive manner. These are due to rare recessive alleles that happen to become monomorphic as a random result of selective breeding. The creation of hybrids may overcome the adverse effects of such rare recessive alleles.

Two different varieties that have been independently subjected to selective breeding are not likely to be monomorphic for the same recessive alleles. Therefore, when two inbred varieties are crossed, the hybrids become heterozygous and carry one copy of the normal allele at each gene and one copy of a detrimental recessive allele. If the allele is recessive, it will not have an adverse effect on the phenotype of the organism.

C16. When a species is subjected to selective breeding, the breeder is focusing on improving one particular trait. In this case, rose breeders focused on the size and quality of the flowers. Because a breeder usually selects a small number of individuals (e.g., the ones with best flowers) as the breeding stock for the next generation, this may lead to a decrease in the allelic diversity of other genes. For example, several genes affect flower fragrance. In an unselected population, these genes may exist as “fragrant” alleles and “nonfragrant” alleles. After many generations of breeding for large flowers, the fragrant alleles may be lost from the population, just as a matter of random chance. This is a common problem of selective breeding: Selecting for an improvement in one trait may inadvertently diminish the quality of an unselected trait.

Others have suggested that the lack of fragrance may be related to flower structure and function.

Perhaps the amount of energy that a flower uses to make beautiful petals somehow diminishes its capacity to make fragrance.

C17. When you examine a trait in a population, it often varies among different members. Sometimes the variation is large, and sometimes it is small. The variation can be caused by genetic differences among the members of the population, and it can be caused by the fact that each member of the population is exposed to a slightly or greatly different environment. The proportion of the variation that is caused by genetic differences among the members of the population is the heritability of the trait. It is applicable only to a particular population raised in a particular environment because each population has its own unique amount of genetic variation (based on the genetic heritage of its members), and the members of each population are exposed to their own unique environmental conditions.

C18. Broad-sense heritability takes into account all genetic factors that affect the phenotypic variation in a trait. Narrow-sense heritability considers only alleles that behave in an additive fashion. In many cases, the alleles affecting quantitative traits appear to behave additively. More importantly, if a breeder assumes that the heritability of a trait is due to the additive effects of alleles, it is possible to

predict the outcome of selective breeding. This estimated narrow-sense heritability is termed the realized heritability.

C19. A. False; the environment has very little impact on the amount of variation in the trait. However, the environment always contributes greatly to the outcome of every trait. You could not have a living organism without an environment.

B. True C. Probably true, but you cannot say for sure. Quantitative traits are usually polygenic.

D. False; you cannot say anything about the heritability of egg weight in Montana chickens. It depends on the amount of genetic variation in the population and on the type of environment in which they are raised.

C20. A. Due to good nutrition, individuals in the commune would tend to grow taller, increasing the mean height of that population. If the genetic variation affecting height was similar to the general population, the average member of the commune population would be taller than the average member of the general population.

B. If the environment is fairly homogeneous, then heritability values tend to be higher because the environment contributes less to the amount of variation in the trait. Therefore, in the commune, the heritability might be higher, because the members uniformly practice good nutrition. On the other hand, because the commune population is smaller than the general population, the amount of genetic variation might be less, which would make the heritability lower. However, because the problem states that the commune population is large, you would probably assume that the amount of genetic variation is similar to that in the general population. Overall, the best guess would be that the heritability in the commune population is higher because of the uniform nutrition standards.

C. As stated in part B, the amount of variation would probably be similar, because the commune population is large. As a general answer, larger populations tend to have more genetic variation.

Therefore, the general population probably has a bit more variation.

C21. In some cases, the alleles that are being selected will become monomorphic in the population, and further selection will have no effect. Alternatively, a second possible reason for a selection limit is natural selection. Some genes that respond to artificial selection have a negative impact on the overall fitness of a population. A selection limit is reached in which the desired effects of artificial selection are balanced by the negative effects on fitness.

C22. A natural population of animals is more likely to have a higher genetic diversity than a domesticated population. This is because domesticated populations have been subjected to many generations of selective breeding, which decreases genetic diversity. Therefore, VG is likely to be higher for the natural population. The other issue is the environment. It is difficult to say which group would have a more homogeneous environment. In general, natural populations tend to have a more heterogeneous environment, but not always. If the environment is more heterogeneous, this tends to cause more phenotypic variation, which makes VE higher.

Heritability = VG / VT = VG /(VG + VE) When VG is high, this increases heritability. When VE is high, this decreases heritability. In the natural wolf population, you would expect VG to be high. In addition, you would guess that VE might be high as well (but that is less certain). Nevertheless, if this were the case, the heritability of size for the wolf population might be similar to that for the domestic population, because the high

VG of the wolf population is balanced by its high VE. On the other hand, if VE is not that high in the wolf population, or if it is fairly high in the domestic population, then the wolf population would have a higher heritability for size.

Application and Experimental Questions E1. A. You first need to calculate the standard deviations for height and weight, and then the covariance for both traits: Height: variance = 140.02; standard deviation = 11.8 Weight: variance = 121.43; standard deviation = 11.02 Covariance = 123.3 CoV( X ,Y ) r ( X ,Y )  SDX SDY r(X,Y) = 123.3 (11.8)(11.02) r(X,Y) = 0.948 B. With 8 degrees of freedom, this value is statistically significant. This means the association between these two variables occurs more frequently than would be expected by random sampling error. It does not necessarily imply cause and effect.

E2. To calculate the mean, you add the values together and divide by the total number of flies: 1.9  2(2.4)  2.(2.1)  3(2.0)  2(2.2) 1.7 1.8  2(2.3)  1.6 Mean  15 Mean  2.1 The variance is the sum of the squared deviations from the mean divided by N – 1. The mean value of 2.1 must be subtracted from each value, and then the square is taken. These 15 values are added together and then divided by 14 (which is N – 1): 0.85 Variance= 14  0.061 The standard deviation is the square root of the variance: Standard deviation = 0.25 E3. The heritability reflects the amount of genetic variation that influences a trait. In the strain with seven QTLs, there are at least seven different genes that exist in two or more alleles that are influencing the outcome of the trait. In the other strain, there are the same types of genes, but three of them are monomorphic and therefore do not contribute to the variation in the outcome of the trait.

E4. These results suggest there are (at least) three different genes that influence the size of pigs. This is a minimum estimate because a QTL may have two or more closely linked genes. Also, it is possible that the large and small strains have the same RFLP band that is associated with one or more of the genes that affect size.

E5. Let’s assume there is an extensive molecular marker map for the rice genome. You would begin with two strains of rice, one with a high yield and one with a low yield, that greatly differ with regard to the molecular markers that they carry. You would make a cross between these two strains to get F1 hybrids. You would then backcross the F1 hybrids to both of the parental strains and then examine hundreds of offspring with regard to their rice yields and molecular markers. In this case, our expected results would be that six different markers in the high-producing strain would be correlated with offspring that produce higher yields. You might get fewer than six bands if some of these genes are closely linked and associate with the same marker. You also might get fewer than six if the two parental strains have the same marker that is associated with one or more of the genes that affect yield.

E6. A. If you assume that the highly inbred strain has zero genetic variance: VG (for the wild strain) = 3.2 g2 – 2.2 g2 = 1.0 g2 B. h 2 = 1.0 g2/3.2 g2 = 0.31 C. ItBis the same as h 2, so it also equals 0.31.

B

E7.

R

h 2 N

S R  XO  X S  XP  X In this problem: X equals 1.01 (as given in the problem) XO equals 1.11 (by calculating the mean for the parents) XP equals 1.09 (by calculating the mean for the offspring)

R = 1.09 – 1.01 = 0.08 S = 1.11 – 1.01 = 0.10 hN2 = 0.08/0.10 = 0.8 (which is a pretty high heritability value) E8. A.    

h  Xo  X N XP  X 0.21  (26.5 g - 25 g)/(27 g - 25 g) 2

XO  25 g = 2 g (0.21) XO  25.42g

B.

0.21  (26.5 g  25 g)/( X p  25 g) ( X p  25 g)(0.21)=1.5 g X p  32.14 g parents

However, because this value is so far from the mean, there may not be parents of that weight in the population of mice that you have available.

E9. The identical and fraternal twins, which probably share very similar environments, but who differ in the amount of genetic material they share, are a strong argument against an environmental bias. The differences in the observed correlations (0.49 versus 0.99) are consistent with the differences in the expected correlations (0.5 versus 1.0).

E10. A. After six or seven generations, the selective breeding seems to have reached a plateau. This suggests that the tomato plants have become monomorphic for the alleles that affect tomato weight.

B. Because Martin’s first generation has a mean tomato weight of 1.7 pounds, which is heavier than either Mary’s or Hector’s tomatoes, these results suggest that heterozygosity may increase tomato weight. This partially explains why Martin has obtained tomatoes that are heavier than 1.5 pounds.

However, this is not the whole story; it does not explain why Martin obtained tomatoes that weigh 2 pounds. Even though Mary’s and Hector’s tomatoes were selected for heavier weight, they may not have all of the heavy alleles for each gene that controls weight. For example, let’s suppose there are 20 genes that affect weight, with each gene existing in a light and heavy allele. During the early stages of selective breeding, when Mary and Hector picked their 10 plants as seed producers for the next generation, as a matter of random chance, some of these plants may have been homozygous for the light alleles at a few of the 20 genes that control weight. Therefore, just as a matter of chance, they probably lost a few of the heavy alleles that affect weight. So, after 12 generations of breeding, they have predominantly heavy alleles but also have light alleles for some of the genes. If you represent each heavy allele with a capital letter, and each light allele with a lowercase letter, Mary’s and Hector’s strains could have the following genotypes: Mary’s strain: AA BB cc DD EE FF gg hh II JJ KK LL mm NN OO PP QQ RR ss TT Hector’s strain: AA bb CC DD EE ff GG HH II jj kk LL MM NN oo PP QQ RR SS TT As seen here, Mary’s strain is homozygous for the heavy allele at 15 of the genes but carries the light allele at the other 5. Similarly, Hector’s strain is homozygous for the heavy allele at 15 genes and carries the light allele at the other 5. It is important to note, however, that the light alleles in Mary’s and Hector’s strains are not in the same genes. Therefore, when Martin crosses them together, he will initially get Martin’s F1 offspring: AA Bb Cc DD EE Ff Gg Hh II Jj Kk LL Mm NN Oo PP QQ RR Ss TT If the alleles are additive and contribute equally to the trait, you would expect about the same mean weight (1.5 pounds), because this hybrid has a total of 10 light alleles. However, genes (which were homozygous recessive in Mary’s and Hector’s strains) are heterozygous in the F 1 offspring, and this may make the plants healthier and contribute to a higher weight. If Martin’s F 1 strain is subjected to selective breeding, the 10 genes that are heterozygous in the F1 offspring may eventually become homozygous for the heavy alleles. This would explain why Martin’s tomatoes achieved a weight of 2.0 pounds after five generations of selective breeding.

E11. hN2 = robs/rexp

The value for rexp comes from the known genetic relationships: Mother/daughter robs = 0.36 rexp = 0.5 hN2 = 0.72 Mother/granddaughter robs = 0.17 rexp = 0.25 h 2= 0.68 N

Sister/sister Twin sisters (fraternal)

robs = 0.39 robs = 0.40

rexp = 0.5 rexp = 0.5

hN2= 0.78 h 2= 0.80

Twin sisters (identical)

robs = 0.77

rexp = 1.0

hN2= 0.77 0.75

N

The average heritability is 0.75.

E12. A. X X hN 2  O XP  X 269  254 hN2   0.56 281  254 B. 275  254 0.56  XP  254 XP  291.5 lb E13. These data suggest there might be a genetic component to blood pressure, because the relatives of people with high blood pressure also seem to have high blood pressure themselves. Of course, more extensive studies would need to be conducted to determine the role of environment. To calculate heritability, the first thing to do is to calculate the correlation coefficient between relatives to see if it is statistically significant. If it is, then you could follow the approach described in the experiment of Figure 24.8. You would determine the correlation coefficients between genetically related individuals as a way to determine the heritability for the trait. In this approach, heritability equals robs/rexp. It would be important to include genetically related pairs that were raised apart (e.g., uncles and nieces) to see if they had a similar heritability value compared to genetically related pairs raised in the same environment (e.g., brothers and sisters). If their values were similar, this would give you some confidence that the heritability value is due to genetics and not due to fact that relatives often share similar environments.

Questions for Student Discussion/Collaboration 1. Heritability is an important phenomenon in agriculture because a population with high heritability can be subjected to selective breeding as a way to improve quantitative traits. People misunderstand heritability because they think that it is the proportion of a trait that is governed by genetics. It is the proportion of the variation in a particular population that is governed by genetics. You can never tell the relative impact of genetics and environment on the final outcome of a trait.

2. Most traits depend on the influence of many genes. Also, genetic variation is a common phenomenon in most populations. Therefore, most individuals have a variety of alleles that contribute to a given trait. For quantitative traits that involve the size of the whole body or the size of body parts, some alleles promote a bigger size, and other alleles promote a smaller size. If a

population contains many different genes and alleles that govern a quantitative trait such as size, most individuals will have an intermediate phenotype because they will have inherited some alleles for larger size and some alleles for smaller size. Fewer individuals will inherit a predominance of just one type of allele. An example of a quantitative trait that does not fit a normal distribution is snail shell pigmentation. The dark-shelled snails and light-shelled snails are favored rather than the intermediate colors because they are less susceptible to predation in certain environments.

Test Bank for Concepts of Genetics 4th Edition By Robert Brooker

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Student name: 1)

The basic unit of heredity is the

A) B) C) D)

2)

.

individual gene macromolecule trait

A variation of a gene is called a(n)

.

A) species B) morph C) genome D) allele E) proteome

3)

Which of the following acts to accelerate chemical reactions in a cell?

A) B) C) D)

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nucleic acids lipids carbohydrates enzymes

2

4)

The building blocks of DNA are

.

A) amino acids B) carbohydrates C) enzymes D) nucleotides E) lipids

5)

A cellular structure that contains genetic information is called a

A) B) C) D)

.

nucleotide genetic code chromosome nucleic acid

6) If a carbohydrate is going to be broken down for energy, which of the following molecules would be directly involved in the breakdown?

A) enzymes B) nucleotides C) microtubules D) lipids E) chromosomes

7)

RNA is formed by the process of

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A) transcription B) translation C) both transcriptionand translation

8)

A characteristic that an organism displays is called

.

A) a gene B) a chromosome C) DNA D) geneexpression E) a trait

9) If a geneticist is studying the prevalence of a trait in a species, they are at the level of study.

A) B) C) D)

population organismal cellular molecular

10) The study of the processes of transcription and translation is at the biological organization.

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level of

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A) B) C) D)

11)

Genetic variation is ultimately based upon which of the following?

A) B) C) D)

12)

population organismal cellular molecular

morphological differences variations in nucleotide sequence of the DNA carbohydrate content of the cell translation

A species that contains two copies of each chromosome is called

.

A) a geneticmutation B) a morph C) haploid D) diploid E) alleles

13)

A diploid cell within an organism's body that is not a reproductive cell is _

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A) a gamete B) a somatic cell C) an allele D) rare E) a sperm cell

14) In many organisms, one set of chromosomes comes from the maternal parent, while the other set comes from the paternal parent. Similar chromosomes in these sets are said to be .

A) morphs B) alleles C) haploid D) homologs E) physiologicaltraits

15)

In humans, gametes are different than other cells of the body in that they are

A) B) C) D)

16)

.

diploid haploid geneticmutations morphs

What is natural selection?

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A) When nongenetically based traits are passed from one generation to the next B) A process that allows traits to remain the same over many generations C) A process in whichenvironmental constraints enable some phenotypes to be more successful than others D) When one phenotype is as successful as all other phenotypes

17) is the use of the information in gene sequences to synthesize functional proteins that affect cellular characteristics.

A) B) C) D)

18)

Loss-of-functionmutation Geneexpression The human genomeproject Proteomics

The differences in inherited traits among individuals in a population are called

A) B) C) D)

.

speciesvariation genetic mutations geneticvariation naturalselection

19) Three populations of an organism, each with drastically different external markings, but still members of the same species, would be called .

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A) homologs B) mutants C) communities D) alleles E) morphs

20) Which one of the following is NOT one of the general classes of macromolecules that are necessary for cellular function?

A) nucleic acids B) proteins C) ions D) carbohydrates E) lipids

21)

The changes in the genetic makeup of a population over time is called

.

A) homologousrecombination B) model organismsstudies C) geneticcrosses D) biologicalevolution E) hypothesistesting

22)

Which of the following could be used to study the effects of drugs on gene expression?

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A) B) C) D)

23)

Which of the following uses a genetic cross to determine patterns of inheritance?

A) B) C) D)

24)

population genetics transmission genetics molecular genetics quantitative genetics

population genetics transmission genetics molecular genetics evolutionary genetics

The traits of an individual organism can be influenced by both genes and the

A) B) C) D)

.

genome environment population size genetic variation within a population

25) Most cellular characteristics,such as structure and function, are the result ofthe synthesis and activity of different _ .

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A) B) C) D)

26)

DNA carbohydrates lipids proteins

Genetics is an experimental, as opposed to theoretical, science because

A) B) C) scientists D)

.

hypotheses are tested byperformingexperiments hypotheses are tested by reviewing the literature to see what others have found no hypotheses are accepted or rejected unless they are voted on by a council of it does not rely on observations but only hypothesis testing experiments

27) Performing a mating of two plants, one with a known genotype and the other with an unknown genotype, to determine the genotype of the individual with the unknown genotype would be an example what type of science?

A) B) C) D)

discovery-based science hypothesis testing unethical experimentation an impossible experiment

28) What is the first step that both scientists and students perform to answer questions in genetics?

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A) B) C) D)

Gathering background information Reaching a conclusion Analyzing data Performing an experiment

29) Mice have 20 chromosomes in their sperm cells. How many chromosomes does a somatic cell from a mouse contain?

A) B) C) D)

20 40 10 80

30) Chronic myelogenous leukemia cells are characterized by the so-called Philadelphia chromosome, which contains part of chromosome 22 fused with chromosome 9. The Philadelphia chromosome is the result of a translocation, in which two chromosomes exchange material. The genetic variation found in chronic myelogenous leukemia is therefore due to

A) gene mutations.

B) major alterations in the structure of a chromosome.

C) variation in the total number of chromosomes.

31) You work in a lab. You engineer a mutant mouse that doesn't synthesize a protein important for breakdown of the sugar galactose and study the results. What type of geneticist are you?

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A) B) C) D)

Transmission geneticist Molecular geneticist Population geneticist Proteome geneticist

32) Ball pythons of the same species can look very different from one another. In fact, there are at least 26 types of ball pythons, characterized by their color, eyes, and markings. The blueeyed leucistic ball python has blue eyes and white scales. The bumblebee ball python has black and yellow scales. The coral glow ball python has purple and orange scales. What term best describes these different types of pythons?

A) B) C) D)

Morphs Alleles Homologs Model organisms

33) You are a geneticist trying to understand why Tay-Sachs disease is more prevalent in Ashkenazi Jews compared to other groups of people. What level of biological organization are you studying?

A) B) C) D)

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Molecular level Cellular level Organism level Population level

12

Answer Key Test name: Chapter 01 Test Bank

1) B 2) D 3) D 4) D 5) C 6) A 7) A 8) E 9) A 10) D 11) B 12) D Version 1

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13) B 14) D 15) B 16) C 17) B 18) C 19) E 20) C 21) D 22) C 23) B 24) B 25) D 26) A

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27) B 28) A 29) B 30) B 31) B 32) A 33) D

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Student name: 1)

Select traits associated with prokaryotic cells. Check all that apply.

A) B) C) D)

2)

Genetic information is contained within a nucleoid region.

Genetic material is organized as a single circular chromosome.

They have a cell wall surrounding their plasma membrane.

They have membrane-bound organelles in their cytoplasm.

Select the cells that are eukaryotic. Check all that apply.

A) bacteria B) fungi C) protists D) plants E) animals

3)

Select the phases that are part of interphase. Check all that apply.

A) B) C) D)

4)

G1 phase G2 phase S phase Metaphase

Select the characteristics that are true of restriction points. Check all that apply.

A) An example is the restriction point that occurs in the G1 phase.

B) In many cases, molecular changes must be present at this point for the cell to continue through the cell cycle.

C) Cells passing this point are committed to the next stage of the cell cycle.

D) Cells passing this point can reverse to an earlier phase of the cell cycle.

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5)

Cytokinesis in animals occurs through the formation of a forms.

A) B) C) D)

6)

7)

.

structures thatcontain the genetic material membrane-boundcompartments of eukaryotic cells the region thatcontains the DNA in prokaryotic cells the outer, rigidcovering of a prokaryotic cell

A cytogeneticist would primarily do which of the following?

A) B) C) D)

8)

cleavage furrow; cell plate cell plate; cleavage furrow cleavage furrow; kinetochore kinetochore; cell plate

Organelles are

A) B) C) D)

, whereas in plants a

study the distribution of traits in a population study the evolutionary changes in a specific trait examine chromosomes using a karyotype determine the genetic sequence of a specific gene

A karyotype is a(n)

A) B) C) D)

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organelle ofeukaryotic cells stage of prophaseI in meiosis division of thecytoplasmic material following mitosis photographicrepresentation of the chromosomes of a cell

2

9) During sexual reproduction, each parent contributes one set of chromosomes. Similar chromosomes from each parent are called .

A) B) C) D)

10)

Which of the following would contain genetic material that is 100% identical?

A) B) C) D)

11)

homologous chromosomes sister chromatids X and Y chromosomes All of these choicesare identical.

The location of a gene on a chromosome is called its

A) B) C) D)

12)

karyotypes sisterchromatids homologs sexchromosomes

karyotype allele locus homolog

Cell division in prokaryotic cells is called .

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, while in eukaryotic cells it is called

3

A) B) C) D)

13)

binary fission; binary fission binary fission; mitosis mitosis; mitosis mitosis; binary fission

The process of binary fission is primarily used for asexual reproduction in

.

A) prokaryotes B) eukaryotes

14)

During this phase of the cell cycle, the sister chromatids are formed in eukaryotes.

A) G 1 phase B) G 2 phase C) S phase D) Prophase E) Cytokinesis

15)

Select the phase when chromosomes start to condense.

A) Metaphase B) Prometaphase C) Telophase D) Anaphase E) Prophase

16) cell.

Select the phase when sister chromatids separate and move towards opposite poles of the

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A) Metaphase B) Prometaphase C) Telophase D) Anaphase E) Prophase

17)

Select the phase during which the centrosomes move to opposite poles of the cell.

A) Metaphase B) Prometaphase C) Telophase D) Anaphase E) Prophase

18)

Select the phase when the chromosomes line up in the center of the cell.

A) Metaphase B) Prometaphase C) Telophase D) Anaphase E) Prophase

19)

Select the phase when the nuclear membrane starts to disassociate.

A) Metaphase B) Prometaphase C) Telophase D) Anaphase E) Prophase

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20)

Select the phase when the nuclear membrane re-forms around the chromosomes.

A) Metaphase B) Prometaphase C) Telophase D) Anaphase E) Prophase

21)

Select the phase when the microtubules of the mitotic spindle attach to the kinetochores.

A) Metaphase B) Prometaphase C) Telophase D) Anaphase E) Prophase

22) Select the phase when the separated sister chromatids are considered independent chromosomes.

A) Metaphase B) Prometaphase C) Telophase D) Anaphase E) Prophase

23)

Which of the following indicates the correct order of these events?

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A) B) C) D)

24)

In animals, somatic cells are A) B) C) D)

25)

.

diploid; diploid diploid; haploid haploid; diploid haploid; haploid

isogamous heterogamous diploid haploid

The general purpose of the synaptonemal complex is to

A) B) C) D)

27)

and gametes are

If the gametes of an organism are different morphologically, they are said to be .

A) B) C) D)

26)

Anaphase -Telophase - Prophase - Prometaphase - Metaphase Telophase -Prometaphase - Prophase - Metaphase - Anaphase Metaphase -Prometaphase - Prophase - Anaphase - Telophase Prophase -Prometaphase - Metaphase - Anaphase - Telophase

.

provide a linkbetween homologous chromosomes in meiosis enable thereformation of the cell wall during cytokinesis separate thesister chromatids during anaphase independentlyassort the chromosomes during metaphase of meiosis

Which of the following occurs during leptotene of prophase I?

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A) B) C) D)

28)

A bivalent contains how many sister chromatids?

A) B) C) D)

29)

The homologous chromosomes recognize one another by synapsis.

Crossing over occurs.

The replicated chromosomes condense.

The synaptonemal complex dissociates.

2 4 8 depends on the cell

When does crossing over occur?

A) During diplotene B) Prior to leptotene, when synapsis is complete C) Prior to pachytene, when synapsis is complete D) Prior to pachytene, before synapsis is complete E) Prior to leptotene, before synapsis is complete

30)

The bivalent structure forms during which of the following?

A) leptotene B) zygotene C) pachytene D) diplotene E) diakinesis

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31)

Which of the following represents the correct order of events during prophase I?

A) B) C) D)

32)

Pachytene - diplotene - diakinesis - leptotene - zygotene Leptotene - zygotene - pachytene - diplotene - diakinesis Zygotene - leptotene - pachytene - diakinesis - diplotene Diplotene - pachytene - leptotene - diakinesis - zygotene

The physical structure that is formed when two chromatids cross over is called a(n) .

A) B) C) D)

synaptonemal complex bivalent karyotype chiasma

33) If an organism has five pairs of chromosomes, how many chromosomal combinations are possible at metaphase I of meiosis?

A) 5 2 B) 10 5 C) 5 10 D) 2 5

34)

The end result of meiosis in animals is

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A) B) C) D)

35)

two diploidcells two haploidcells four diploidcells four haploidcells

The process of meiosis II is similar to that of

.

A) mitosis B) binaryfission C) meiosis I

36)

Oogenesis is a gametogenic process in which cells undergo .

to produce

A) binary fission; sperm cells B) mitosis; egg cells C) meiosis; egg cells D) meiosis; sperm cells E) mitosis; sperm cells

37) the

In plants, the haploid generation is called the .

A) B) C) D)

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and the diploid generation is called

sporophyte; spermatogenesis gametophyte; sporophyte sporophyte; gametophyte oogenesis; gametophyte

10

38)

In plants, spore production occurs by

A) spermatogenesis B) meiosis C) mitosis D) binaryfission E) oogenesis

39)

A pollen grain in a plant represents the A) B) C) D)

40)

.

male gametophyte female gametophyte male sporophyte female sporophyte

Which type of microtubule is paired to its correct function?

A) polar microtubules - attach to the kinetochore B) aster microtubules - position the spindle apparatus C) kinetochore microtubules - separate the poles

41) During sexual reproduction, gametes are made that contain material as a somatic cell in the organism.

A) B) C) D)

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amount of genetic

twice the half the the same a quarter of the

11

42)

Genes are physically located within A) B) C) D)

.

chromosomes centrosomes kinetochores microtubules

A diploid cell in G 1contains 5 pairs of chromosomes. During mitosis, how many 43) chromosomes will be lined up on the metaphase plate in this cell?

A) B) C) D)

5 10 15 20

44) A primary spermatocyte contains 8 total chromosomes. What will be the composition of the chromosomes in one of the secondary spermatocytes that results from this primary spermatocyte?

A) B) C) D)

8 unduplicated chromosomes 4 chromosomes, each containing 2 sister chromatids joined at the centromere 8 chromosomes, each containing 2 sister chromatids joined at the centromere 4 unduplicated chromosomes

45) A cell from an organism that has a diploid number of 6 chromosomes has 3 chromosomes lined up in the center the cell, each chromosome containing two joined sister chromatids. What phase of the cell cycle is this cell in?

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A) B) C) D)

metaphase I of meiosis metaphase II of meiosis metaphase of mitosis telophase I of meiosis

46) A species has a diploid number of 36. If meiosis didn't occur, and gametes were produced by mitosis, each gamete would contain chromosomes, and a fertilized egg would contain chromosome.

A) 36; 72 B) 36; 36 C) 18; 36 D) 72; 144

47) What is the arrangement of chromosomes in an oocyte in a female mammal at birth if the species has a diploid number of 20? Hint: Recall which phase of the cell cycle oocytes are arrested at in a female animal prior to birth.

A) B) C) D)

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20 chromosomes, each with two sister chromatids 10 chromosomes, each with two sister chromatids 20 unduplicated chromosomes 10 unduplicated chromosomes

13

Answer Key Test name: Chapter 02 Test Bank 1) [A, B, C] 2) [B, C, D, E] 3) [A, B, C] 4) [A, B, C] 5) A 6) B 7) C 8) D 9) C 10) B 11) C 12) B 13) A 14) C 15) E 16) D 17) B 18) A 19) E 20) C 21) B 22) D 23) D 24) B 25) B 26) A Version 1

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27) C 28) B 29) C 30) B 31) B 32) D 33) D 34) D 35) A 36) C 37) B 38) B 39) A 40) B 41) B 42) A 43) B 44) B 45) B 46) A 47) A

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Student name:_ 1)

Which of the following traits were studied by Mendel? Check all that apply.

A) B) C) D) E)

2)

flower color seed color pod color pollen color plant height

Choose all statements that constitute key tenets of the chromosome theory of inheritance.

A) Chromosomes are copies of the genetic material that are used to make proteins and are then destroyed.

B) Chromosomes contain the genetic material.

C) Chromosomes are contributed by the maternal parent only.

D) Chromosomes are replicated and transmitted during inheritance and during development.

E) Most cells are diploid, with two copies of each chromosome.

F) Gametes contain two copies of each chromosome so there is a backup copy if one is damaged.

G) During gamete formation, chromosomes segregate independently.

H) Each parent contributes one complete set of chromosomes.

I) Certain chromosomes are linked together, so that genes that work together are transmitted together.

3) The use of crosses between different strains of pea plants was not the first attempt to explainheredity. The theory of pangenesis was first proposed by .

A) B) C) D) E)

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Aristotle Galen Mendel Hippocrates None of these choicesare correct.

1

4) Mendel's data from single-factor crosses did not support a blending mechanism of inheritance. Which of the following is correct regarding the blending theory of inheritance?

A) B) C) D) E)

5)

It theorizedthat hereditary traits blended from one generation to the next.

It was possible for the blending to change the trait from one generation to the next.

It was supported by early research of tobacco plants by Joseph Kölreuter.

It was the prevailing theory of inheritance prior to Mendel.

All of these choicesare correct.

Mendel's work was rediscovered in 1900 by which of the following individual(s)?

A) B) C) D)

Carl Correns Erich von Tschermak Hugo de Vries All of these choicesare correct.

6) Which of the following characteristics made the pea plant Pisum sativum an ideal organism for Mendel's studies?

A) B) C) D)

It has the ability to self-fertilize.

It was possible to cross-fertilize one plant with another.

It has easily identifiable traits.

All of these choicesare correct.

7) The anthers represent the portion of the plant.

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portion of the plant; the ovules represent the

2

A) B) C) D)

female; male male; female female; female male; male

8) When studying a genetic cross, the second generation following the initial cross is identified by which of the following?

A) B) C) D) E)

P generation F 1 generation F 2 generation F 3 generation P 3 generation

9) A true-breeding line of green pod pea plants is crossed with a true-breeding line of yellow pod plants. All of their offspring have green pods. From this information, it can be stated that the green color is to the yellow color.

A) B) C) D) E)

10)

recessive dominant subservient blended None of these choicesare correct.

What theory did Mendel's work with monohybrid crosses support?

A) B) C) D)

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blending theory of inheritance particulate theory of inheritance chromosomal theory of inheritance pangenesis

3

11)

What resulted from Mendel's work with single-factor crosses?

A) B) C) D)

law of segregation law of independent assortment theory of natural selection law of biological evolution

12) When Mendel crossed two plants that were heterozygous for a single trait, what was the phenotypic ratio of their offspring?

A) 1:2:1 B) 9:3:3:1 C) 3:1 D) 7:4 E) Varied depending on the trait

13) When Mendel crossed two plants that were heterozygous for a single trait, what was the genotypic ratio of their offspring?

A) 1:2:1 B) 9:3:3:1 C) 3:1 D) 1:1 E) Varied depending on the trait

14)

An individual who has two identical alleles for a trait is said to be

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.

4

A) B) C) D)

15)

The genetic composition of an individual is called its

A) B) C) D) E)

16)

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.

phenotype genotype dominance genes None of these choicesare correct.

An individual who has two different alleles for a trait is called

A) B) C) D) E)

.

phenotype genotype hybrid dominance None of these choicesare correct.

The observable characteristics of an organism are called its

A) B) C) D) E)

17)

homozygous heterozygous isozygous a variant

.

haploid homozygous heterozygous isozygous true-breeding

5

18)

In a Punnett square diagram, the outside of the box represents the

A) B) C) D)

19)

.

diploidoffspring haploidoffspring diploidgametes haploidgametes

Mendel's work with two-factor crosses led directly to which of the following?

A) B) C) D) E)

chromosomal theory of inheritance particulate theory of inheritance law of segregation law of independent assortment theory of biological evolution

20) In a two-factor cross using Mendelian inheritance, if both parents are heterozygous for both traits, what will be the phenotypic ratio of their offspring?

A) 3:1 B) 1:2:1 C) 1:1 D) 9:3:3:1

21) If a Punnett square is used to visualize a three-factor cross between two parents that are both heterozygous for all three genes involved, how many boxes would be inside of the square?

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A) B) C) D) E)

22)

In humans, patterns of inheritance are often studied using which of the following?

A) B) C) D)

23)

3 8 48 64 can't be determined

two-factor crosses production of true-breeding lines pedigree analysis self-fertilization

The chance that a future event will occur is called

A) B) C) D) E)

.

probability goodness offit degrees offreedom randomselection All of these choicesare correct.

24) A coin is flipped 100 times, with a result of 53 heads and 47 tails. The deviation between the observed numbers and the expected 50-50 results is called .

A) B) C) D) E)

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probability degrees offreedom goodness offit random samplingerror standarderror

7

25) Which of the following would be used to determine the probability of three independent events in order?

A) B) C) D) E)

sum rule product rule chi square test binomial expansion random sampling error

26) A couple would like to know what the probability is that out of five children, three will be girls. This is solved using which of the following?

A) B) C) D) E)

sum rule product rule chi square test binomial expansion random sampling error

27) Recall that in pea plants, purple flower color is dominant and white is recessive. If two heterozygous plants are crossed, what is the probability that the first two offspring will have purple flowers?

A) 1/2 B) 1/4 C) 6/4 D) 9/16 E) 1/16

28) In a genetic cross, there are n classes of data. What would the degrees of freedom be for a chi square test on this data?

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A) B) C) D) E)

n n+1 n-1 2 n+1 x( n) where x equals the number of individuals in the cross

29) The likelihood that the variation of observed data from expectations is due to random chance is called the .

A) B) C) D)

30)

P value goodness offit degrees offreedom empiricalapproach

In the biological sciences, the hypothesis is usually rejected if the P value is A) B) C) D) E)

.

greater than1 less than0.30 less than0.95 less than0.05 less than 1

31)

Mendel'sdata and the study of chromosomes and meiosis did not support the idea of , which is the belief that seeds are produced by all parts of the body and transmitted to the next generation.

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A) B) C) D) E)

32) a

the chromosome theory of inheritance pangenesis the blending theory of inheritance the law of segregation the law of independent assortment

If two individuals with different distinct characteristics are mated, their offspring is called .

A) B) C) D) E)

strain true-breedingline gamete cross hybrid

33) If over several generations a character does not vary in a group of organisms, that group can be called a .

A) B) C) D) E)

heterozygote hybrid true-breedingline variant cross-fertilizedline

34) A cross in which a researcher investigates the patterns of inheritance of a single trait is called a .

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A) B) C) D) E)

35)

monohybrid cross multi-factor cross two-factorcross cross-fertilization self-fertilization

A specific version of a gene is called a(n) A) B) C) D) E)

trait character gamete allele variant

36) In a genetic cross, the not found in the parental lines.

A) B) C) D)

37)

represent offspring with genetic combinations that were

P generation nonrecombinants parentals nonparentals

The study of family trees in humans is called a

A) B) C) D) E)

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.

analysis.

pedigree monohybrid factorial statistical probability

11

38) Statistical analysis determines the expected from the original hypothesis.

A) B) C) D) E)

between observed data and what was

testcross degrees offreedom P values completehypothesis goodness offit

39) Mendel's law of independent assortment states that during the formation of haploid cells.

A) B) C) D)

randomly assort their

genes; alleles alleles; versions cells; mitochondria genes; nucleotides

40) Which situation below would violate the law of independent assortment, as understood

by Mendel?

A) A plant breeder is able to create a true-breeding line that combines the alleles for white flowers and wrinkled peas.

B) A mutant gene can prevent pod formation altogether, thus preventing expression of the round or wrinkled pea alleles.

C) A gene for petal color does not affect the expression of a gene for petal number.

D) A gene for leaf shape is linked to a gene for petal number nearby on the same chromosome, so their alleles usually segregate together.

E) Combining certain alleles of different genes results in a plant that grows poorly.

41)

What aspect of meiosis best explains Mendel's law of segregation?

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A) B) C) D) E)

42)

selective removal of genetic information to create haploid gametes separation of homologous chromosomes during Meiosis I random alignment of different bivalents during Meiosis I transmission of genes together when they lie nearby on the same chromosome separation of identical alleles duringMeiosis II

What aspect of meiosis best explains Mendel's law of independent assortment?

A) B) C) D) E)

selective removal of genetic information to create haploid gametes separation of homologous chromosomes during Meiosis I random alignment of different bivalents during Meiosis I transmission of genes together when they lie nearby on the same chromosome separation of identical alleles duringMeiosis II

43) In a certain family pedigree, you observe that 7 of a couple's 12 children display a disease trait. In the subsequent generation, some of the children of affected individuals have the disease, but none of the children of two unaffected parents do. You conclude that the disease is caused by a allele.

A) B) C) D)

recessive dominant redundant wild-type

44) To determine the genotype of a pea plant with purple flowers it was crossed with a pea plant with white flowers. Half of the offspring exhibited the dominant purple flower while the other half had white flowers. What was the genotype of the parental pea plant with the purple flowers?

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A) B) C) D)

heterozygous homozygous dominant homozygous recessive can’t be determined

45) Consider the pedigree shown. What can best explain the expression of the trait in individual IV-1?

©McGraw-Hill Education A) B) C) D) E)

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The trait is recessive, and both II-2 and III-3 were heterozygous carriers.

The trait is dominant, and both III-3 and III-4 were heterozygous.

The trait is dominant, and IV-1 received the allele from III-4.

The trait is recessive but only expressed in males.

The trait is recessive, and both II-1 and II-2 were heterozygous carriers.

14

46) Consider a tree in which in thick trunk ( T) is dominant to a skinny trunk ( t), back bark ( B) is dominant to brown bark ( b) and round leaves ( R) are dominant to oval leaves ( r).

You cross a tree that is heterozygous for all three traits to a tree that is homozygous recessive for all three traits. Using the forked line method or the multiplication method, calculate what fraction of the offspring will have a thick trunk, black bark, and round leaves.

A) 1/4 B) 1/2 C) 1/8 D) 1/16

47) In a certain species of plants, red flowers ( R) are dominant to pink flowers ( r) and round leaves ( L) are dominant to oval leaves ( l). A plant with red flowers and round leaves is crossed to a plant with pink flowers and oval leaves. In the F 1 generation, all of the plants have red flowers, but half of the plants have round leaves and half have oval leaves. What can we conclude about the parental plant with red flowers and round leaves?

A) B) C) D) E)

The genotype of the plant with red flowers and round leaves is RRll. The genotype of the plant with red flowers and round leaves is RRLL. The genotype of the plant with red flowers and round leaves is RrLl. The genotype of the plant with red flowers and round leaves is RRLl. The genotype of the plant with red flowers and round leaves is RrLL.

48) Consider a bird in which a high-pitched song ( H) is dominant to a low-pitched song ( h). You cross two birds that are heterozygous for the gene for song pitch. Predict the genotypes and phenotypes of the offspring.

A) B) C) D)

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1/4 HH: 1/2 Hh: 1/4 hh;3/4 high-pitched song: 1/4 low-pitched song 1/2 HH: 1/2hh;1/2high-pitched song: 1/2low-pitched song 1/4 HH: 1/2 Hh: 1/4 hh;1/2 high-pitched song: 1/2low-pitched song All Hh; Allhigh-pitched song

15

49) Mendel's work on inheritance had an immediate influence on the scientific community and theories of inheritance. ⊚ ⊚

true false

50) Differences in plant characteristics such as flower color or height are called variants or traits. ⊚ true ⊚ false

51)

The chi square test is used to prove that a hypothesis is correct. ⊚ true ⊚ false

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Answer Key Test name: Chapter 03 Test Bank 1) [A, B, C, E] 2) [B, D, E, G, H] 3) D 4) E 5) D 6) D 7) B 8) C 9) B 10) B 11) A 12) C 13) A 14) A 15) B 16) A 17) C 18) D 19) D 20) D 21) D 22) C 23) A 24) D 25) B 26) D Version 1

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27) D 28) C 29) A 30) D 31) B 32) E 33) C 34) A 35) D 36) D 37) A 38) E 39) A 40) D 41) B 42) C 43) B 44) A 45) A 46) C 47) D 48) A 49) FALSE 50) TRUE 51) FALSE

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Student name:_ 1) In which of the following organisms is the level of X chromosome expression altered in males?

A) B) C) D)

2)

mammals Drosophila C. elegans humans

Who originally identified a highly condensed structure in the interphase of nuclei?

A) Lyon B) Barr andBertram C) Ohno

3)

The product of the Xist gene is a(n)

A) B) C) D)

4)

.

protein RNA small molecule DNA

What gene is most responsible for X-inactivation?

A) B) C) D)

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Xic Xist TsiX Xce

1

5)

How many Barr bodies would an individual with a XXY genotype possess?

A) 0 B) 1 C) 2

6)

The coat color of calico cats is a result of

A) B) C) D)

7)

maternalinheritance X-inactivation imprinting extra nuclearinheritance

The Lyon hypothesis attempts to explain the molecular mechanism of

A) B) C) D)

8)

.

.

X-inactivation genomicimprinting maternalinheritance extra nuclearinheritance

Dosage compensation offsets the problems associated with differences in the number of chromosomes in many species.

A) B) C) D)

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sex autosome somatic nuclear

2

9) The genetic control of X-inactivation is governed by a short region on the X chromosome called the .

A) B) C) D)

10)

Y-inactivationcenter (Yic) condensationcenter X-inactivationcenter (Xic) X-initiationcenter

In humans, which sex is considered to be the heterogametic sex?

A) male B) female

11)

In a Z-W system, which is considered to be the homogametic sex?

A) males B) females

12) A male that is produced from an unfertilized haploid egg is an example of what type of sex determination system?

A) B) C) D)

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X-Y Z-W X-O haplo-diploid

3

13) The first person to experimentally determine that a trait is located on a specific chromosome was .

A) B) C) D)

14)

If a gene is located on the X chromosome, but not the Y, it is said to be an example of .

A) B) C) D) E)

15)

Mendel Boveri and Sutton Weissman Morgan

autosomalinheritance sex-linkage reciprocalcross pseudoautosomalinheritance holandric

A testcross

A) is used todetermine if an organism that is displaying a recessive trait is heterozygousor homozygous for that trait.

B) results in the F2generation having a phenotypic ratio of 1/2 dominant to 1/2 recessive.

C) is used when anindividual with a dominant phenotype and unknown genotype is crossed to anindividual with a recessive phenotype.

D) results inoffspring that always display the recessive phenotype.

16)

Sex determination in humans and Drosophila is similar in the fact that

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4

A) the ratio of Xchromosomes to sets of autosomes determines maleness or femaleness in both.

B) males in both willalways have one Y chromosome, whereas females will not have a Y chromosome.

C) females lack X chromosomes.

D) males have one X chromosome and females have two X chromosomes.

17)

An example of a mechanism in which sex is determined by environmental factors is

A) females are produced from fertilized eggs, whereas males are produced fromunfertilized eggs.

B) when the male cares for the eggs, all of the offspring will be male, whereasif the female cares for the eggs, all of the offspring will be female.

C) fertilized eggsincubated at one temperature produce males, whereas eggs incubated at adifferenttemperature produce females.

D) males will have a Y chromosome, whereas females will not have a Ychromosome.

18) A human individual has a karyotype that is XY but is phenotypically female. An explanation for this could be A) a deletion of the SRY gene on the Y chromosome, thus resulting in an female phenotype.

B) a translocation of the SRY gene to one of the X chromosomes has occurred, thus resulting in a female phenotype.

C) the SRY gene has been amplified 10 fold on the Y chromosome.

D) sex is determinedby genes located on autosomes, not sex chromosomes.

19) Brown spotting of the teeth in humans is caused by a dominant X-linked gene.If a man with normal teeth marries a woman with brown teeth who had a father with normal teeth, then of their daughters will have brown teeth.

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A) 100% B) 50% C) 25% D) 0%

20) Red-green colorblindness is a X-linked recessive trait in humans. If a woman who is a carrier for red-green colorblindness marries a normal male, what percent of their sons will be colorblind?

A) 100% B) 50% C) 25% D) 0%

21)

Pseudoautosomal inheritance occurs when

A) a gene on anautosome is silenced by inactivation.

B) a gene on a Y chromosome has the same inheritance pattern as a gene on an X chromosome.

C) a gene on anautosome has the same inheritance pattern as a gene on a sex chromosome.

D) a gene on a sexchromosome has the same inheritance pattern as a gene on an autosome.

22) X-linked hypophosphatemia (XLH) is an X-linked dominant disease that results in weak or soft bones. A girl with XLH has a dad with XLH. Her mother does not have XLH. What is the probability that her sisters will have XLH? What about her brothers?

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A) B) C) D) E)

All of her sisters and none of her brothers will have XLH. All of her brothers and none of her sisters will have XLH. 50% of her brothers and sisters will have XLH. None of her brothers or sisters will have XLH. All of her brothers and sisters will have XLH.

23) In certain birds, feather color is Z-linked, and green coloring is dominant to purple coloring. A purple feathered male mates with a green feathered female. Predict the phenotypic ratios in their offspring.

A) Half of the offspring will be green females and the other half will be purple males.

B) Half of the offspring will be purple females and the other half will be green males.

C) 1/4 of the progeny will be green females, 1/4 green males, 1/4 purple females, and 1/4 purple males.

D) All of the progeny will be green.

24) Glucose-6-phosphate dehydrogenase deficiency (G6PDD) is an X-linked recessive disease. What is the probability that the brother of a girl with G6PDD will also have G6PDD if their mother is unaffected.

A) 0% B) 25% C) 50% D) 100%

25) Flowering plants will produce flowers that are dioecious, meaning that the male and female parts are contained within the same flower. ⊚ ⊚

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true false

7

26)

With rare exceptions, a calico cat is likely to be female. ⊚ true ⊚ false

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Answer Key Test name: Chapter 04 Test Bank 1) B 2) B 3) B 4) B 5) B 6) B 7) A 8) A 9) C 10) A 11) A 12) D 13) D 14) B 15) C 16) D 17) C 18) A 19) B 20) B 21) D 22) A 23) B 24) C 25) FALSE 26) TRUE Version 1

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Student name:_ 1) Genes that are not required for survival, but are likely to be beneficial to the organism, are called .

A) B) C) D) E)

2)

Alleles that cause death in a certain temperature range are examples of A) B) C) D) E)

3)

essential genes lethal alleles semilethalalleles nonessentialgenes conditional lethalalleles

.

essential genes lethal alleles semilethalalleles nonessentialgenes conditional lethalalleles

The alleles that cause Huntington disease in humans are an example of

A) B) C) D) E)

.

essentialgenes lethal alleles semilethalalleles nonessentialgenes sex linked allele

4) A heterozygote possesses a phenotype that is intermediate between the homozygous dominant and homozygous recessive phenotypes. This is most likely an example of .

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A) B) C) D)

lethal alleles incompletedominance gene dosage sex-influencedinheritance

5) In four-o'clock plants, red flower color is dominant to white flower color. However, heterozygous plants have a pink color. If a pink-flowered plant is crossed with a white-flowered plant, what will be the phenotypic ratios of their offspring?

A) B) C) D) E)

¼ red, ½ pink, ¼ white all pink all white ½ pink, ½ white ½ red, ½ pink

6) In rabbits, full coat color (C ) is the dominant trait. A second allele, chinchilla (c ch ), is recessive to full coat color. Himalayan coat color (ch ) is recessive to chinchilla and full coat colors, and albino (c ) is recessive to all coat colors. If two chinchilla rabbits mate, what coat color is not possible in their offspring?

A) B) C) D) E)

full coat color chinchilla coat color himalayan coat color albino coat color All coat colors are possible.

7) In human blood groups, the fact that an individual can have an AB blood type is an example of .

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2

A) B) C) D) E)

incompletedominance incompletepenetrance sex-influencedtrait temperature-sensitive conditional allele codominance

8) An individual with type A blood and an individual with type B blood mate and have offspring. What blood type is not possible in their offspring?

A) B) C) D) E)

type O blood type A blood type B blood type AB blood All blood types are possible.

9) At the molecular level, type A and type B blood differ in which of the following characteristics?

A) B) C) D)

the antigens present on the surface of the red blood cells the type of sugar found in each type the antibodies that are generated against the other type of blood All of these choicesare correct.

10) For a certain trait, a heterozygous individual has a selective advantage over a homozygous dominant or homozygous recessive individual. This is called .

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3

A) B) C) D) E)

11)

codominance incompletedominance overdominance incompletepenetrance multiple allelesystems

Sickle-cell disease in humans is an example of

A) B) C) D)

.

codominance incompletepenetrance heterozygote advantage sex-limited resistance

12) At the molecular level, which of the following best explains heterozygous advantage and overdominance?

A) A heterozygous individual can produce more varieties of homodimer proteins.

B) The alleles produce two different proteins with slightly different functions.

C) The proteins produced by the alleles may provide a broader range of environmental tolerance, such as temperature ranges.

D) Infectious organisms may recognize only a specific functional protein.

E) All of the answers are possibilities.

13)

Polydactyly in humans is an example of

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4

A) B) C) D) E)

14)

If a geneticist describes a trait as being 70% penetrant, what would that mean?

A) B) C) D)

15)

The expression of the trait varies by individual.

It is lethal in 30% of the individuals who have the trait.

Only 70% of the individuals who have a particular genotypeexpress the trait.

The trait is present in 70% of the population.

The phenylketonuria phenotype in humans is an example of A) B) C) D) E)

16)

sex-limited inheritance incompletedominance incompletepenetrance codominance gene dosage

.

incompletepenetrance codominance anenvironmental-influenced trait incompletedominance All of these choicesare correct.

If an allele is dominant in one sex and recessive in another, it is an example of .

A) B) C) D)

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sex-limitedinheritance sex-influencedinheritance incompletedominance simple Mendelianinheritance

5

17)

A paralog

A) B) C) D) E)

18)

is found forevery gene in mammals is only found onthe X but not the Y chromosome can explain thelack of phenotype for a gene knockout cannot bemutated has the same DNAsequence as the original duplicated gene

The interaction of two genes to produce a phenotype was first described by

A) B) C) D) E)

19)

.

Morgan and Bridges Mendel Darwin Bateson and Punnett None of these choicesare correct.

Epistasis is

A) B) C) D) E)

.

.

another term for overdominance when one gene canmask the expression of a second gene a trait that isonly expressed in one sex of the species when two dominantalleles can be expressed in the same individual None of these choicesare correct.

20) In a dihybrid cross of two heterozygous individuals, you expect a 9:3:3:1 phenotypic ratio in the offspring, but observe a ratio of 9:7. What is the most likely explanation?

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A) B) C) D) E)

21)

codominance a sex-limited trait simple Mendelian inheritance incomplete penetrance epistatic interactions of the two genes

Which of the following is not correct concerning epistatic interactions?

A) They are often associated with enzymatic pathways.

B) They produce variations in the expected 9:3:3:1 ratio of a dihybrid cross.

C) They can result when a gene at one locus masks the expression of a gene at a different locus.

D) They always result in a 9:7 ratio of a dihybrid cross.

E) They are due to gene interactions.

22) When wild-type offspring are produced from a cross between parents that both display the same recessive phenotype, this illustrates the genetic phenomenon of .

A) B) C) D) E)

23)

simple recessivealleles incompletepenetrance complementation gene dosage a spontaneousmutation

The multiple effects of a single gene on the phenotype of an organism is called

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.

7

A) B) C) D) E)

epistasis penetrance expressivity overdominance pleiotropy

24) A allele typically encodes a protein that is made in the proper amount and functions normally.

A) B) C) D) E)

25)

loss-of-function mutant wild-type gain-of-function lethal

In overdominance, the

A) B) C) D) E)

genotype is beneficial over the

genotypes.

heterozygous; homozygous homozygous; heterozygous homozygous dominant; homozygous recessive homozygous recessive; homozygous dominant incomplete dominant; codominant

26) In cattle, the scurs trait follows a sex-influenced pattern of inheritance.A heterozygous male has the phenotype and a heterozygous female has a phenotype.

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A) B) C) D)

27) a(n)

scurs; scurs hornless; hornless hornless; scurs scurs; hornless

If a combination of two or more genes is required to produce a specific trait, it is called .

A) B) C) D) E)

overdominance simple Mendelian trait sex-influencedtrait sex-linkedtrait geneinteraction

28) Recall that scurs in cattle that exhibit sex-influenced inheritance. There are two alleles P Sc , which represents scurs being presentand ScA,which indicates scurs being absent. ScPis dominant in males and recessive in females. ScAis dominant in females and recessive in males.

From a mating between a heterozygous male with scurs and a heterozygous female without scurs, what proportion of male offspring will lack scurs (be hornless)?

A) 0% B) 25% C) 50% D) 75% E) 100%

29) An experiment in which plants with the same genotype are raised under conditions that differ in soil quality could evaluate .

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A) B) C) D)

the norm of reaction overdominance genetic polymorphisms gene interactions

30) ©McGraw-Hill Education Epistasis occurs when the alleles of one gene mask the phenotypic effects of the alleles of another gene at a different locus. This phenomenon can be observed in the sweet pea, Lathyrus odoratus: ● C (one purple-color-producing) allele is dominant to c (white) ● P (another purple-color-producing) allele is dominant to p (white) ● cc or pp masks the P or C alleles, producing white color Determine the genotypic and phenotypic ratios of a cross between a purple and white sweet pea above.

A) 1 CcPP(purple) : 1 CCPp (purple) : 1 ccPP (white) : 1 ccPp(purple) B) 1 CCPP(purple) : 2 CCPp (purple) : 2 CcPP (purple) : 1 CCpp(white) : 4 CcPp (purple) : 2 Ccpp (white) : 1 ccPP(white) : 2 ccPp (white) : 1 ccpp (white) C) 1 CcPP (purple) : 1 CcPp (purple) : 1 ccPP (white) : 1 ccPp (white) D) 9 CCPP(purple) : 7 CCPp (purple) E) 9 CCPP (purple) : 3 CcPp (purple) : 3 ccPP (purple) : 1 ccpp (white)

31) You cross two dogs with an Irish spotting phenotype. Some of the progeny have the extreme-white phenotype. What were the genotypes of the parents?

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A) B) C) D)

sIsw and swsw sIsIand swsw sIswand sI sw S+sI and sIsw

32) Hairlessness in dogs is inherited from a single dominant allele. Inheriting two dominant alleles is lethal for an embryo. Penelope, a hairless dog, mates with Arnold, a dog with hair. The probability Penelope will have hairless puppies is

A) B) C) D)

3/4 2/3 1/4 1/2

33) Several members of a family have polydactyly, and they differ in their numbers of extra fingers and toes. Yet, certain relatives who should have extra digits don't. An aunt has two children, a son with two extra fingers and a daughter with an extra toe. Also her father has an extra digit on each hand and foot. Polydactyly in this family is

A) B) C) D)

variably expressive and incompletely penetrant. invariably expressive and completely penetrant. pleiotropic. incompletely dominant.

34) A man who has normal hearing and a recessive deafness allele on chromosome 17 marries a woman who also has normal hearing and also has a recessive deafness allele, but this allele is on chromosome 3. The probability that their children will be deaf due to either of these mutated genes is closest to

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A) 100%.

B) 50%.

C) 25%.

D) 0%.

35) New research indicates that male-pattern baldness is a sex-linked trait in humans because of mutations of an androgen receptor gene on the X chromosome. ⊚ true ⊚ false

36)

In an epistatic interaction, the genes must be located on the same chromosome. ⊚ ⊚

37)

Typically, a recessive allele increases the expression of a functional protein. ⊚ ⊚

38)

true false

Incomplete dominance is an example of blending of phenotypes, not genotypes. ⊚ ⊚

39)

true false

true false

Human blood groups are determined by antigens on the surface of red blood cells. ⊚ ⊚

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true false

12

40) Incomplete penetrance indicates that individuals who possess a dominant trait always express the trait. ⊚ ⊚

41)

true false

Variable expressivity means that the phenotype of a trait can vary between individuals. ⊚ ⊚

true false

42) Paralogs are often functionally redundant and can compensate for the loss of either of the paralogs in a set. ⊚ ⊚

true false

43) Because loss-of-function alleles typically do not make a functional gene product, they usually follow the recessive allele inheritance pattern. ⊚ true ⊚ false

44) Because most lethal alleles are inheritedin a recessive manner, a cross between two heterozygous individuals could result in a 2:1 genotypic ratio in their surviving offspring. ⊚ true ⊚ false

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Answer Key Test name: Chapter 05 Test Bank 1) D 2) E 3) B 4) B 5) D 6) A 7) E 8) E 9) D 10) C 11) C 12) E 13) C 14) C 15) C 16) B 17) C 18) D 19) B 20) E 21) D 22) C 23) E 24) C 25) A 26) D Version 1

14

27) E 28) B 29) A 30) C 31) C 32) D 33) A 34) D 35) TRUE 36) FALSE 37) FALSE 38) TRUE 39) TRUE 40) FALSE 41) TRUE 42) TRUE 43) TRUE 44) TRUE

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Student name:_ 1)

Where does erasure and imprinting occur?

A) B) C) D)

2)

Which of the following is primarily responsible for the maternal effect?

A) B) C) D)

3)

sperm cells oocytes nurse cells placenta

What controls the coiling of shells in snail offspring?

A) B) C) D)

4)

nurse cells sperm cells oocytes somatic cells

mother'sphenotype father'sphenotype mother'sgenotype father'sgenotype

What is the molecular mechanism for imprinting a gene?

A) B) C) D)

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acetylation nitration phosphorylation methylation

1

5)

What disease is associated with imprinting?

A) Angelman Syndrome B) LHON C) Alzheimer's Disease

6)

Where is extranuclear DNA located in mammalian cells?

A) B) C) D)

7)

What type of inheritance is observed with extranuclear DNA? A) B) C) D)

8)

endoplasmic reticulum mitochondria ribosome plasma membrane

Mendelian inheritance Sex-linked inheritance Paternal inheritance Cytoplasmic inheritance

What is a disease associated with extranuclear inheritance?

A) B) C) D)

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Angelman syndrome Tay-Sachs disease LHON Muscular dystrophy

2

9) from

Mitochondria are thought to have originated from .

A) B) C) D)

10)

Schimper and Wallin Haig and Schimper Wallin and Correns Haig and Correns

Monoallelic expression is associated with which of the following?

A) B) C) D)

12)

cyanobacteria; tobacco mosaic virus purple bacteria; single-celled fungi purple bacteria; cyanobacteria single-celled fungi; tobacco mosaic virus

Who is largely responsible for proposing the endosymbiosis theory?

A) B) C) D)

11)

, and chloroplasts likely arose

X-inactivation genomic imprinting maternal inheritance extranuclear inheritance

Which of the following is true regarding mt genomes?

A) B) C) D)

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They are approximately 17,000 bp in length in humans.

They are linear.

A single copy of the mt genome is found within the mitochondrial nucleus.

They have all the genes necessary for mitochondrial function.

3

13) The inheritance of leaf pigmentation in the four-o'clock plant Mirabilis jalapa is an example of .

A) B) C) D)

14)

maternaleffect maternalinheritance epigenetic inheritance imprinting

In maternal effect, the

A) B) C) D) E)

of the mother determines the

of the offspring.

chloroplast;mitochondria phenotype;genotype genotype;phenotype phenotype; sex methylation;inheritance

15) What is the type of inheritance that is based on a reversible modification of genomic DNA that results in the alleles of only one parent being transcriptionally active?

A) B) C) D) E)

maternal epigenetic cytoplasmic extranuclear mitochondrial

16) The molecular explanation for imprinting in mammals involves differential various DNA regions.

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of

4

A) B) C) D) E)

phosphorylation mutations endosymbiosis methylation heteroplasmy

17) The inheritance patterns of genetic material that is not contained in the nucleus of the cell is called .

A) B) C) D)

18)

Heteroplasmy is associated with inheritance patterns involving

A) B) C) D) E)

19)

extranuclearinheritance cytoplasmicinheritance maternaleffect both extranuclearinheritance and cytoplasmic inheritance

.

ribosomes maternaleffect genomicimprinting nuclear genes chloroplasts

Contribution of mitochondria by the sperm cell to the zygote is called A) B) C) D) E)

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heteroplasmy. paternal leakage. genomic imprinting. paternal effect. endosymbiosis.

5

20)

Diseases associated with mitochondria mutations in humans are usually A) B) C) D) E)

21)

The symbiotic relationship where one organism lives inside another species is called A) B) C) D) E)

22)

endosymbiosis. genomic imprinting. cytoplasmic inheritance. heteroplasmy. exosymbiosis.

An example of genomic imprinting is

A) B) C) D) E)

23)

paternally inherited. imprinted. fatal. very common. degenerative.

Leber hereditary optic neuropathy. neurogenic muscleweakness. snail coiling inwater snails.

Igf2expression in mice. bicoid gene in Drosophila.

A nurse cell is a cell that

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A) B) C) D) E)

24)

In genomic imprinting, erasure of DNA methylation occurs in the A) B) C) D)

25)

lacks chloroplasts and mitochondria. destroys defective oocytes during embryogenesis. surrounds an oocyte during oogenesis. will fix a mutation in another cell. can repair its own DNA mutations.

mitochondria. chloroplasts. somatic cells of the embryo. cells destined to become gametes.

How is a chloroplast genome similar to a prokaryotic genome?

A) B) C) D)

They are both circular.

They both code for all the proteins they need for functioning.

They are both linear.

There is only one chromosome per cell or chloroplast.

26) Which mating of four-o'clock plants would produce progeny that were either green, variegated, or white if the color variation is due to a mutation in a chloroplast gene?

A) B) C) D)

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green female x variegated male variegated female x green male white female x white male white female x green male

7

27) R and r represent dominant and recessive alleles, respectively, for a gene that is imprinted. Imprinting of the allele occurs in the male and the allele is silenced when it is imprinted. Which mating would result in a ratio of 1:1 of dominant and recessive phenotypes in the offspring?

A) B) C) D)

rr female x Rr male RR female x rr male Rr female x Rr male rr female x RR male

28) G and g are dominant and recessive alleles, respectively, for a gene. If a mating of a gg female with a Gg male resulted in offspring that all have the recessive phenotype, this would most likely be an example of A) B) C) D)

a recessive lethal gene. a maternal effect gene. environmental influence on phenotype. imprinting which results in silencing of the maternal alleles.

29) What would be the genotype of the mother of a female snail that has a a shell with a left hand twist?

Dd genotype and

A) dd B) DD C) Dd D) Cannot tell since the shell twist phenotype of any offspring depends on the genotype of the father as well as the mother.

30) What is the genotype of the father for a snail that has the genotype Dd and has a left hand twist shell?

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A) B) C) D)

dd Dd DD It cannot be determined.

31) If a snail that has a right hand twist and is DD is mated to a snail that has a left hand twist and is dd what is the expected ratio of progeny?

A) B) C) D)

3 left hand twist to 1 right hand twist 3 right handtwist to 1 left hand twist It depends on which parent snail is female and which is male.

1 left handtwist to 1 right hand twist

32) You are a geneticist studying a newly discovered genetic disorder. The disorder follows a pattern of maternal inheritance, and both male and female children are affected. Strangely, within a family, all the children are affected but to very different degrees. What could explain this variation?

A) The disorder is caused by a mitochondrial mutation and the variation is due to heteroplasmy.

B) The disorder is caused by a mitochondrial mutation and the variation is due to imprinting.

C) The disorder is caused by a nuclear mutation and the variation is due to maternal effect.

D) The disorder is caused by a nuclear mutation and the variation is due to Xchromosome inactivation.

33)

Both parents usually imprint the same gene. ⊚ ⊚

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true false 9

34) mtDNA contains all of the genes necessary for the complete function of mitochondrial metabolism. ⊚ ⊚

true false

35) The symbiotic relationship of purple bacteria and cyanobacteria with eukaryotic cells was beneficial to both cell types. ⊚ ⊚

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true false

10

Answer Key Test name: Chapter 06 Test Bank 1) [B, C] 2) C 3) C 4) D 5) A 6) B 7) D 8) C 9) C 10) A 11) B 12) A 13) B 14) C 15) B 16) D 17) D 18) E 19) B 20) E 21) A 22) D 23) C 24) D 25) A 26) B Version 1

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27) C 28) B 29) A 30) D 31) C 32) A 33) FALSE 34) FALSE 35) TRUE

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Student name:_ 1)

Which of the following defines gene linkage?

A) Two or more genes that are physically connected on a chromosome.

B) Genes that are transmitted to the next generation as a group.

C) The process by which genetic information is exchanged between homologous chromosomes.

D) All of these choices are correct.

E) Both two or more genes that are physically connected on a chromosome and genes that are transmitted to the next generation as a group.

2) Assume that genes C and D are located on the same chromosome. On one chromosome alleles C and D are found, while the homolog contains alleles c and d. Which of the following would be an example of a recombination event?

A) alleles C and D together on one chromosome B) alleles c and d together on one chromosome C) alleles C and d together on one chromosome D) alleles c and D together on one chromosome E) both alleles C and d together on one chromosome and alleles c and D together on one chromosome

3) The first observational evidence that genes may be inherited together rather than by simple Mendelian inheritance was provided by .

A) B) C) D)

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Mendel Morgan and Bridges Bateson and Punnett Boveri and Sutton

1

4) Experimental evidence that crossing over occurs between the X chromosomes of female Drosophila was provided by .

A) B) C) D)

Morgan Punnett Darwin Bateson

5) Which of the following statistical tests is used to determine if two genes are linked or assorting independently?

A) B) C) D)

sum rule binomial expansion product rule chi square test

6) In a chi square test to determine if two genes are linked or assorting independently, what is the default (null) hypothesis that is tested?

A) B) C) D)

The genes are linked to one another.

The genes are assorting independently.

The genes are located on the sex chromosomes.

No crossing over occurs.

7) The visual proof that chromosomes exchange pieces of information during crossing over was provided by .

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2

A) B) C) D)

8)

Bateson and Punnett Morgan and Bridges Creighton and McClintock Watson and Crick

Twin spotting provides evidence of what genetic event?

A) B) C) D)

meiotic recombination mitotic recombination linkage mutation

9) An organism that contains patches of tissue that vary for a specific characteristic, such as a pigment, is an example of .

A) B) C) D)

linkage meiotic recombination mitotic recombination translocations

10) Which of the following are necessary characteristics of an organism in order to easily construct a genetic linkage map?

A) B) C) D)

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short generation times produces large numbers of offspring easily crossed All of these choices are correct.

3

11)

Crossing over is more likely to occur between genes that are

on a chromosome.

A) close together B) far apart C) not

12) A testcross is always performed between the individual that is heterozygous for the genes to be mapped and an individual who is .

A) B) C) D)

heterozygous forthe genes homozygousdominant for the genes homozygousrecessive for the genes lacking thegenes

13) While mapping two genes in Drosophila, you observe 30 recombinants among 200 total offspring. What is the distance between these genes?

A) B) C) D)

14)

30 map units 6.67 map units 200 map units 15 map units

A map distance of 23.6 between two genes indicates which of the following?

A) B) C) D)

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The genes are 23.6 millimeters apart.

There are 23.6 other genes between the two genes of interest.

23.6% of the offspring exhibit recombination between the two genes.

23.6% of the offspring do not survive.

4

15)

The individual who is credited with discovering genetic linkage in Drosophila is A) B) C) D)

.

Thomas Hunt Morgan Gregor Mendel Alfred Sturtevant Barbara McClintock

16) In a mapping experiment with three genes, which phenotype should occur most often in the F 2 offspring?

A) B) C) D)

parental phenotypes phenotypes of individuals with single crossover events phenotypes of individuals with double crossover events All of these choicesshould be equal in the F2 generation.

17) The middle gene of a three gene mapping experiment can be determined by examining the genotypes of which of the following?

A) B) C) D)

offspring that resemble the parents offspring that exhibit a single crossover event offspring that exhibit double crossover events None of these choices are correct.

18) Which of the following is not one of the properties of gene linkage that Morgan obtained from his experiments?

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5

A) Genes that are on the same chromosome may be inherited together.

B) Crossing over exchanges pieces of chromosomes and creates new allele combinations.

C) The likelihood of crossing over occurring between two genes is dependent on the distance of the genes from one another.

D) Genes that are on the same chromosome are always transmitted together as a unit.

19) In humans, there are linkage group.

A) B) C) D)

autosomal linkage groups, plus an X and Y chromosome

23 46 22 92

20) Another name for a chromosome is a inherited together.

A) B) C) D)

21)

, since it contains genes that are often

linkage group crossing overgroup geneticrecombinant bivalent

Two genes that are located on the same chromosome are said to be A) B) C) D)

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.

linked recombinant parental-like nonparental-like

6

22) Creighton and McClintock worked with as their model system to show that homologous chromosomes physically exchange genetic information during crossing over.

A) B) C) D)

fruit flies peas corn tobacco

23) In a mapping cross, you determine that the recombination frequency between theQ and P is 12% and betweenloci Q and L is 15%. If locus Q is in between loci P and L, then the recombination frequency between P and L should be approximately .

A) B) C) D)

3% 27% 50% 75%

24) The diploid garden pea plant has 14 chromosomes. The haploid fungus Neurospora crassa has 7 chromosomes. Neither organism has separate male and female individuals. Which of the following correctly identifies the number of linkage groups in each organism?

A) B) C) D)

Garden pea has 14 linkage groups, and Neurospora has 7.

Garden pea has 7 linkage groups, and Neurospora has 7.

Garden pea has 8 linkage groups, and Neurospora has 8.

Gardent pea has 15 linkage groups, and Neurospora has 8.

25) Suppose the genes for eye color and wing shape are both on the X chromosome (they are linked) and are known to be 68 map units apart. In a mating between a X w+m+X wm female and a X wmY male that produces 1000 progeny, how many recombinant progeny would you expect to observe?

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A) B) C) D)

500 680 340 320

A Xw+m+Xwm female was mated to a XwmY male. Out of 1000 total F1 progeny there were 26) 230 flies that were otherwise wild type with white eyes and 240 flies that were otherwise wild type with miniature wings. What is the distance between the genes for eye color and wing length?

A) B) C) D)

27)

The process of recombination may rarely occur during mitosis. ⊚ ⊚

28)

true false

A genetic linkage map indicates the precise distance between two genes of interest. ⊚ ⊚

29)

47 map units 53 map units 23 map units 24 map units

true false

The rearrangement of alleles by the process of crossing over is called genetic linkage. ⊚ ⊚

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true false

8

30) Map distance is the number of recombinant offspring divided by the total number of nonrecombinant offspring. ⊚ ⊚

true false

31) Following crossing over, chromosomes with genetic combinations that resemble the parents are called nonrecombinant. ⊚ true ⊚ false

32)

A map unit or centiMorgan is equal to a 10% recombination frequency. ⊚ true ⊚ false

33) Map distances above 50 are considered unreliable due to the occurrence of doublecrossovers between the genes. ⊚ ⊚

34)

true false

The locus is the physical place of a gene on a chromosome. ⊚ ⊚

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true false

9

Answer Key Test name: Chapter 07 Test Bank 1) E 2) E 3) C 4) A 5) D 6) B 7) C 8) B 9) C 10) D 11) B 12) C 13) D 14) C 15) A 16) A 17) C 18) D 19) C 20) A 21) A 22) C 23) B 24) B 25) A 26) A Version 1

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27) TRUE 28) FALSE 29) FALSE 30) FALSE 31) TRUE 32) FALSE 33) TRUE 34) TRUE

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Student name:_ 1)

Examples of euploidy include A) B) C) D) E)

2)

Select the pairs that correctly match an aneuploid condition to its genotype.

A) B) C) D) E)

3)

Edward syndrome - trisomy 18 Down syndrome - trisomy 21 Jacobs syndrome - XYY Turner syndrome - XXY Klinefelter - XO

Variations in chromosome structure are important for which of the following reason(s)?

A) B) C) D)

4)

diploidy. triploidy. tetraploidy. polyploidy. aneuploidy.

evolution of new species may result in the production of abnormal offspring may alter the phenotype of an organism All of these choices are correct.

Chromosomes may be identified based on which of the following characteristics?

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A) B) C) D)

5)

Which of the following would have the shortest p arm of the chromosome?

A) B) C) D)

6)

acrocentric metacentric telocentric submetacentric

A loss of an internal portion of a chromosome is called a(n) A) B) C) D)

7)

location of the centromere banding patterns size of the chromosome All of these choices are correct.

.

reciprocaltranslocation terminaldeficiency interstitialdeletion geneduplication

What process is the cause of human genetic diseases such a cri-du-chatsyndrome?

A) B) C) D)

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translocation duplication deletion inversion

2

8) Given the following sequence of genes on a chromosome, determine what change in chromosome structure occurred. The asterisk (*) indicates the centromere.

Before: A B C D * E F G H After: A C D * E F G H A) B) C) D)

terminal deletion interstitial deletion inversion gene duplication

9) Given the following sequence of genes on a chromosome, determine what change in chromosome structure occurred. The asterisk (*) indicates the centromere.

Before: A B C D * E F G H After: A B C D * E F E F G H A) B) C) D)

10)

What causes gene duplications?

A) B) C) D) E)

11)

terminal deletion interstitial deletion inversion gene duplication

the crossing over of misaligned chromosomes deletion of important genetic information reciprocal translocations position effect None of these choices are correct.

The production of gene families, such as the globin genes, is the result of

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.

3

A) B) C) D) E)

inversions deficiencies geneduplications simpletranslocations None of these choices are correct.

12) Given the following sequence of genes on a chromosome, determine what change in chromosome structure occurred. The asterisk (*) indicates the centromere.

Before: A B C D * E F G H After: A B G F E * D C H A) B) C) D) E)

reciprocal translocation pericentric inversion paracentric inversion gene duplication None of the answers are correct.

13) Inversions detectable by light microscopy are contained within what percent of the human population?

A) B) C) D)

14)

less than 1% approximately 2% approximately 5% greater than 10%

Inversion loops can occur in which of the following?

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4

A) B) C) D) E)

15)

paracentric inversions pericentric inversions gene duplications reciprocal translocations both paracentric inversions and pericentric inversions

An inversion heterozygote contains which of the following?

A) B) C) D)

two homologous chromosomes with inversions two normal chromosomes one normal chromosome and one chromosome with an inversion None of these choices are correct.

16) Which of the following generally has the least effect on the phenotype of the individual who carries it?

A) B) C) D)

17)

Robertsonian translocation unbalanced translocation balanced translocation All of these choicesare equally detrimental to the phenotype.

Robertsonian translocations usually occur between what types of chromosomes?

A) B) C) D)

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metacentric acrocentric telocentric submetacentric

5

18)

A translocation cross may occur in an individual which has which of the following?

A) B) C) D)

19)

Which of the following expressions denotes a trisomic organism?

A) B) C) D)

20)

.

aneuploidy allopolyploidy autopolyploidy translocations

What are Klinefelter and Turner syndromes examples of?

A) B) C) D)

22)

3n 2n-1 2n+1 2n+2

Edward and Patau syndromes are examples of A) B) C) D)

21)

reciprocal translocation unbalanced translocation simple translocations All of these choices are correct.

sex chromosome aneuploidy autosomal aneuploidy reciprocal translocations paracentric inversions

Which human cells exhibit endopolyploidy?

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6

A) B) C) D) E)

23)

The polytene chromosomes of Drosophila are an example of

A) B) C) D) E)

24)

.

aneuploidy polyploidy translocations inversionloops None of these choices are correct.

Which type of plant would usually be a seedless variety?

A) B) C) D)

25)

sex cells nerve cells all somatic cells liver cells red blood cells

aneuploid diploid triploid tetraploid

Which correctly describes mitotic nondisjunction?

A) B) C) D)

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It is a relatively rare event.

It results in the loss of chromosomes in some cells.

It usually does not affect all of the cells of the organism.

All of these choices are correct.

7

26)

The failure of chromosomes to separate during anaphase is called

A) B) C) D)

.

synapsis maternaleffect epistasis nondisjunction

27) Which of the following describes an organism with two complete sets of chromosomes from two different species?

A) B) C) D)

28)

tetraploid aneuploid allodiploid allotetraploid

The term

A) B) C) D)

refers to genetic differences between members of the same species.

chromosomalvariation geneticvariation chromosomaldeletions polyploidy

29) The short arm of a chromosome is denoted by the letter letter .

A) B) C) D)

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and the long arm by the

p; q s; l q; p c; d

8

30) A chromosome.

A) B) C) D)

31)

translocation occurs when a piece of one chromosome is attached to another

simple complex reciprocal balanced

is a drug that is used to experimentally produce polyploidy in organisms.

A) B) C) D)

Penicillin Colchicine Polymosca Karyocine

32) Genetic abnormalities that occur after fertilization may result in some tissues of the body differing in their genetic composition. This is called .

A) B) C) D)

mosaicism polyploidy polypheny genetic variation

33) If a disease is associated with copy number variation, that suggests that it is caused by the presence of

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9

A) B) C) D)

34)

mRNA transcripts missing the translation STOP sequence. a much longer version of a gene. a duplicated or deleted copy of a gene. extra telomeres.

Describe endopolyploidy, and how it might benefit an organism.

A) Portions of the organism have extra copies of all chromosomes, for producing additional gene products.

B) The entire organism has extra copies of certain chromosomes, for producing additional gene products.

C) The entire organism has extra copies of all chromosomes, to act as backup in case of DNA damage.

D) Portions of the organism have extra copies of certain genes, for the functions of specialized cell types.

35)

Polytene chromosomes are A) chromosomes that lack telomeres.

B) chromosomes with high copy number variation.

C) backup copies of chromosomes that are kept segregated away from the functional

copies.

D) bundles of hundreds of copies of the same chromosome.

36) A phenotypically normal person is having fertility issues.They consult a geneticist.Their karyotype shows they have 45 chromosomes. Which of the following is the most likely cause of their chromosome abnormality?

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A) B) C) D)

Inversion Partial deletion Balanced translocation Robertsoniantranslocation

37) In mammals, the majority of cardiomyocytes (cardiac muscle cells) undergo repeated DNA replication without cytokinesis as a part of normal development and in response to injury or heart attack. Many cardiomyocytes are tetraploid, and some cells have even higher DNA content. This variation in DNA content is an example of .

A) B) C) D)

alloploidy aneuploidy copy number variation endopolyploidy

38) Which of the following would produce the most euploid progeny, sperm cells that result from a nondisjunction following meiosis I or meiosis II? A) B) C) D)

Sperm that are produced following a non-disjunction event in meiosis II. Sperm that are produced following a non-disjunction event in meiosis I. Neither will produce any euploid progeny.

All progeny will be euploid with either non-disjunction event.

39) Trisomy 8 usually leads to the early miscarriage of a fetus. However, adult individuals have been found with cells that have three copies of chromosome 8 in them. How can this be?

A) B) C) D)

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The trisomic 8 adults likely have a mosaic region with trisomy 8.

The trisomic cells underwent complete nondisjunction.

The trisomic cells underwent a meiotic nondisjunction.

This individual must be triploid.

11

40) In meiotic nondisjunction, meiotic products can be n+1, n-1, or n depending on when nondisjunction occurs. If non disjunction occurs in meiosis I, what is the outcome?

A) B) C) D)

Two trisomic and two monosomic products One trisomic and three monosomic products One trisomic, one monosomic, and two normal products None of these choices are correct.

41) A diploid organism has a total of 36 chromosomes. Assuming all possible chromosome combinations are viable, if a mutant tetraploid version of this organism was created how many chromosomes would it have? If a mutant version of the diploidorganism was monosomic for chromosome 9 how many chromosomes would it have?

A) 144; 35 B) 72; 35 C) 144; 37 D) 72; 37

42) Hemophilia A is an X-linked recessive blood clotting disorder. A normal man and a woman with hemophilia A have a child with Turner Syndrome (XO). This child does not have hemophilia. In whom did nondisjunction occur? In meiosis I or meiosis II? A) Woman; meiosis I B) Woman; meiosis II C) Man; meiosis I D) Man; meiosis II E) Woman; there is not enough information to tell if the nondisjunction happened in meiosis I or II

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43) The salivary amylase gene (AMY1) is present in two diploid copies in chimpanzees.

Humans are known to have between 6-15 copies, an adaptation that is thought to be related to the high-starch diet of humans. This difference among humans is an example of .

A) B) C) D)

G banding copy number variation a simple translocation a reciprocal translocation

44) The common goldfish Carassius auratus has 100 chromosomes and is tetraploid. The goldfish therefore has sets of chromosomes containing chromosomes each.

A) B) C) D)

4; 25 2; 50 4; 100 1; 100

45) The African clawed frog ( Xenopus laevis) is an allotetraploid, likely as a result of an interspecies mating long ago, followed by a duplication of the entire genome. Xenopus laevis is fertile and has a normal life cycle. In contrast, mules, the allodiploid offspring of a male donkey and a female horse, are generally sterile. Why can Xenopus reproduce and mules cannot?

A) Frogs are less sensitive to multiple copies of the genome than mammals.

B) In allotetraploid organisms each chromosome has a chromosome to pair up with in meiosis, whereas in allodiploid organisms they do not.

C) The two frogs that interbred to form the Xenopus laevis must have been more closely genetically related than the donkey and horse are.

D) The mule must have a genetic mutation that prevents it from reproducing.

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46) A wholphin is a rare hybrid animal born from mating a female bottlenose dolphin with a male false killer whale. Wholphins are diploid. Interestingly, wholphins are fertile. What can you conclude from the fact that wholphins are fertile?

A) B) C) D)

Wholphins are allotetraploid.

Dolphins and false killer whales likely have the same number of chromosomes.

Dolphins and false killer whales are actually the same species.

One of the parents must have been aneuploid.

47) The effects of a deficiency on an organism are dependent on the importance of the missing genetic material. ⊚ true ⊚ false

48) The majority of autosomal aneuploid conditions are lethal, but sex chromosome aneuploids are usually not lethal. ⊚ true ⊚ false

49)

Variations in a gene are called allelic variations. ⊚ ⊚

50)

true false

Inversions represent a change in the total genetic material of a chromosome. ⊚ true ⊚ false

51) Polyploid plants tend to produce less fruits and flowers, be smaller in size, and cannot adapt as well to extreme environmental conditions as normal varieties.

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⊚ ⊚

true false

52) Due to the formation of an inversion loop, sections of DNA may either be duplicated or deleted. ⊚ true ⊚ false

53) Since no genetic material is lost or gained by an inversion, it will only influence phenotype if the breaks are within a gene or in important regulatory regions for a gene. ⊚ true ⊚ false

54)

Homologous genes within a species are called paralogs. ⊚ ⊚

55)

true false

The ends of chromosomes have areas of repeated DNA called centromeres. ⊚ ⊚

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true false

15

Answer Key Test name: Chapter 08 Test Bank 1) [A, B, C, D] 2) [A, B, C] 3) D 4) D 5) C 6) C 7) C 8) B 9) D 10) A 11) C 12) B 13) B 14) E 15) C 16) C 17) B 18) A 19) C 20) A 21) A 22) D 23) B 24) C 25) D 26) D Version 1

16

27) D 28) B 29) A 30) A 31) B 32) A 33) C 34) A 35) D 36) D 37) D 38) A 39) A 40) A 41) B 42) E 43) B 44) A 45) B 46) B 47) TRUE 48) TRUE 49) TRUE 50) FALSE 51) FALSE 52) TRUE 53) TRUE 54) TRUE 55) FALSE

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Student name:_ 1)

DNA may be transferred between bacteria via A) B) C) D)

2)

conjugation. transformation. transduction. fission.

Which correctly describe the process of conjugation?

A) The cells must be in physical contact.

B) It requires bacteriophage as an intermediary.

C) It involves a sex pilus.

D) Only competent cells can participate.

E) Donor strains can convert non-donor strains into donor strains.

F) F- cells act as donors.

3)

Which correctly describes F factors?

A) B) C) D) E)

"F factor" is short for "fertility factor".

If present, that cell can act as a donor.

It consists of a single gene that is present on the chromosome.

All natural isolates of E. coli have it.

After it is transferred by conjugation, the recipient cell becomes F+.

4) During conjugation, one gene (A) is found to transfer to the recipient bacteria 26 minutes following the start of conjugation, while a second gene (M) is found to transfer 37 minutes following the start of conjugation. A third gene (T) transfers 45 minutes following the start of conjugation. Based on this information, which of the following is true?

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A) B) C) D)

5)

Which of the following conditions must be correct for transformation to occur?

A) B) C) D) E)

6)

Genes A and M have a genetic distance of 11 minutes.

Genes A and T have a genetic distance of 19 minutes.

Genes M and T have a genetic distance of 18 minutes.

The order of the genes is A T M.

temperature ionic conditions nutrient balance F factor must be present bacteriophage strain

Horizontal gene transfer can include A) when a gene is acquired from another species.

B) when a gene is acquired directly from the parent.

C) when a gene is acquired by conjugation.

D) when a gene is acquired by transformation.

E) when a gene is acquired by transduction.

F) when a gene is lost from the genome.

7) You are in a lab and are planning to grow bacteria with the genotype met- bio- thr- leuthi+on agar plates. What amino acids or vitamins must you include on the plates in order for this bacteria to grow? Check all that apply.

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A) B) C) D) E)

Methionine Biotin Threonine Leucine Thiamine

8) Among the plasmid types listed here, which two have the most direct relevance to human health?

A) B) C) D) E)

9)

F factors R factors Degradative plasmids Col-plasmids Virulence plasmids

Which of the following is NOT true regarding bacterial genetics?

A) Bacteria are usually haploid organisms.

B) Bacteria primarily reproduce asexually.

C) The patterns of inheritance in bacteria are studied using the same techniques as eukaryotic organisms.

D) DNA is transferred between individuals by either conjugation, transduction, or transformation.

10)

Which process uses a bacteriophage as an intermediary for the genetic information?

A) B) C) D)

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transformation transduction conjugation fission

3

11) When a bacteria assimilates genetic material (DNA) directly from the environment, this is called A) B) C) D)

12)

transformation. transduction. conjugation. fission.

DNA transfer via a physical connection between two bacteria is known as A) B) C) D)

transformation. transduction. conjugation. fission.

13) This form of DNA transfer was first observed by Lederberg and Tatum (1946) in Escherichia coli.

A) B) C) D)

14)

transformation transduction conjugation fission

This form of DNA transfer uses a sex pilus to transfer the genetic material.

A) B) C) D)

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transformation transduction conjugation fission

4

15)

Bacterial cells that can participate in A) B) C) D)

16)

are known as competent cells.

transformation transduction conjugation fission

What best describes a histidine, methionine auxotroph?

A) B) C) D) E)

It can synthesize only histidine and methionine.

It can use only histidine and methionine as sources of nutrition.

It cannot synthesize histidine or methionine It is killed by histidine and methionine.

It lacks proteins with histidine or methionine.

17) After two cells make contact via the conjugation bridge, the protein complex that initiates DNA transfer is called the A) B) C) D)

origin of transfer.

T DNA. relaxosome. nucleoprotein complex.

18) After conjugation, the enzyme that rejoins the ends of the linear DNA into a circular plasmid is called .

A) B) C) D)

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relaxosome relaxase F factor nucleoprotein

5

19) The site on a plasmid that is cut, producing a single strand of DNA, to allow conjugation to occur is called the .

A) B) C) D)

20)

How do researchers interrupt the process of conjugation?

A) B) C) D)

21)

using tweezers using a microscope using a blender using a special gene

In experiments, interrupted mating may be used to A) B) C) D)

22)

origin oftransfer relaxase nucleoprotein T DNA

inhibit transduction. generate loss of function alleles. determine the relative position of genes. alter the genotype of the donor cell.

The units of genetic distance derived from bacterial conjugation studies are called A) B) C) D)

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seconds centimorgans recombinationunits minutes

6

23)

Who was the first to discover the process of natural transformation?

A) B) C) D)

24)

Morgan Sturtevant Griffith Avery

What do competence factors do?

A) B) C) D)

allow the formation of a sex pilus promote bacteriophage binding for transduction allow DNA in the environment to bind to the cell surface inhibit genetic transfer between donor and recipient cells

25) A structure called a(n) conjugation.

A) B) C) D)

26)

allows the transfer of genetic information during

F factor plasmid sex pilus transformationbridge

A cell with an F factor integrated into the bacterial chromosome is called

.

A) Hfr B) F+ C) a conjugative plasmid

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27)

Which does not require any recombination with the bacterial chromosome?

A) conjugation B) transduction C) transformation

28)

Loss of function mutations are easier to study in _

organisms, such as bacteria.

A) haploid B) diploid C) triploid

29)

Small circular pieces of DNA, separate from the bacterial chromosome, are called .

A) heteroduplexes B) plasmids C) relaxosomes

30) What is the most important result of horizontal gene transfer as it relates to human health?

A) It spreads antibiotic resistance in pathogenic bacteria.

B) It can explain the occurrence of nonheritable genetic diseases.

C) It can cause the disappearance of genes from the human genome, leading to disease.

D) It has been used successfully to treat dozens of diseases, by replacing damaged DNA in the cells of the patient.

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31) DNA transfer by transduction depends on what occasional mistake that can occur during a phage lytic cycle?

A) B) C) D)

32)

A piece of host cell DNA is packaged into a phage.

A host cell protein is incorporated into a phage.

The host cell undergoes lysis before any new phages are produced.

The phage and bacterial genome fragments recombine.

How does a transducing phage directly contribute to DNA transfer between bacteria?

A) By injecting a fragment of the bacterial genome from a previous host into another bacterial cell B) By stimulating the formation of a sex pilus C) By injecting phage DNA into a living bacterial cell D) By incorporating DNA fragments from two different bacterial cells so that recombination can occur within the phage

33) In the experiment performed by Lederberg and Tatum bacteria of different genotypes had to be mixed together to obtain a genetic transfer. However, Davis performed the experiment in a slightly different manner by separating cells with a filter incapable of allowing bacteria to pass.

What possibility did the experiment by Davis exclude that could have occurred by the protocol used by Lederberg and Tatum?

A) Transformation B) Conjugation C) Transduction

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34) What is the approximate distance in minutes between the argR and pyrB genes on the E. coli chromosome?

A) B) C) D)

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16 minutes 96 minutes 26 minutes 70 minutes

10

35) If the origin of transfer was located directly to the right of the galE gene on this map of the E. coli chromosome, and gene transfer proceeded counterclockwise, how many minutes after the galE gene was transferred would the dnaB gene be transferred in an interrupted mating experiment?

A) B) C) D)

25 minutes 75 minutes 107 minutes 16 minutes

36) During a phage life cycle sometimes a phage is packaged carrying a portion of the host cell chromosome, which results in transduction. During conjugation, which plasmid can most directly be compared to the phage that carries a fragment of the host cell chromosome?

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A) B) C) D)

F factor F ' factor R factor Col-plasmid

37) During transformation, if the extracellular endonuclease was not functioning properly, at which stage would transformation be halted?

A) B) C) D)

During heteroduplex repair During DNA transport into the cell During homologous recombination After DNA fragment binding to a cell surface receptor on the bacterium

38) You are performing transduction experiments using recipient bacteria that are auxotrophic for histidine, leucine, and tryptophan. You notice that following transduction, recombinant bacteria that are his+ are usually also trp+but that they are less likely to also be leu+. What can you conclude about this transduction experiment?

A) In the original phage genome, the genes for his+ and trp+ are close together.

B) In the donor bacterial genome, the genes for his+ and trp + are close together.

C) When these phages were made they took up two separate pieces of DNA, one containing his + and one containing trp+.

D) The donor bacterial genome was leu-.

39) In conjugation, the strand of DNA that is transferred from the donor cell to the recipient cell is single stranded. ⊚ true ⊚ false

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40) The process of conjugation may introduce new genes into the recipient strain, altering its genotype. ⊚ ⊚

41)

true false

The direct movement of DNA between two bacteria is called genetic transfer. ⊚ ⊚

true false

42) During transformation, homologous recombination may occur if the sequence of the introduced DNA has no similarity to the host DNA. ⊚ ⊚

true false

43) During homologous recombination, areas of mismatched DNA form structures called a heteroduplex. ⊚ ⊚

44)

true false

A minute is the basic unit of map distance in bacteria. ⊚ ⊚

Version 1

true false

13

45) An episome is a segment of bacterial DNA that can exist as a plasmid or be integrated into the bacterial chromosome. ⊚ ⊚

46)

true false

Competent cells are those used in studies of conjugation. ⊚ ⊚

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true false

14

Answer Key Test name: Chapter 09 Test Bank 1) [A, B, C] 2) [A, C, E] 3) [A, B, E] 4) [A, B] 5) [A, B, C] 6) [A, C, D, E] 7) [A, B, C, D] 8) [B, E] 9) C 10) B 11) A 12) C 13) C 14) C 15) A 16) C 17) C 18) B 19) A 20) C 21) C 22) D 23) C 24) C 25) C 26) A Version 1

15

27) A 28) A 29) B 30) A 31) A 32) A 33) A 34) C 35) A 36) B 37) D 38) B 39) TRUE 40) TRUE 41) TRUE 42) FALSE 43) TRUE 44) TRUE 45) TRUE 46) FALSE

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Student name:__ 1) What was the outcome of the experiment that reconstituted two different strains of tobacco mosaic viruses that had different infection characteristics?

A) The differences in infection characteristics depended only on the RNA sequence of the virus strain.

B) The differences in infection characteristics depended onlythe protein coat.

C) The differences in infection characteristics depended on an RNA and protein coat interaction.

2)

Which virus uses reverse transcriptase in its reproductive cycle?

A) B) C) D)

3)

Which is a property of emerging viruses?

A) B) C) D)

4)

HIV Phage lambda Both of the viruses use reverse transcriptase Neither of the viruses use reverse transcriptase

May lead to a significant loss of human life Recent origin More likely to cause infection compared to other strains All of these

What is true in bothphageλ and HIV infections?

A) There can be a long latent period where no sign of infection, such as cell death, is detectable.

B) Cell lysis always occurs immediately after virus introduction.

C) Assembled viral particles are always found within the cell.

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5) A bacteriophage that is physically integrated into the host chromosome is called a ___ __.

A) B) C) D)

heteroduplex envelope phage prophage plaque

6) A _______ uses the lytic cycle to begin the immediate manufacture of new phages in the host cells.

A) B) C) D)

virulent phage temperatephage bacteria eukaryoticcell

7) A _______ frequently integrates into the host genome, forming prophages and utilizing a lysogenic cycle to manufacture new phages.

A) B) C) D)

8)

virulent phage temperatephage bacteria eukaryoticcell

What is the genetic material in the parvovirus?

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A) B) C) D)

9)

single-stranded DNA double-stranded DNA single-stranded RNA double-stranded RNA

What is the genetic material in the T4 bacteriophage?

A) B) C) D)

double-stranded DNA single-stranded DNA single-stranded RNA double-stranded RNA

10) If protein was the genetic material of TMV rather than RNA, what would be the result of the Fraenkel-Conrat/Singer experiments when using HR proteins and wild-type RNA? A) B) C) D)

HR type lesions, histidine and methionine in the capsid protein HR type lesions, no histidine and methionine in the capsid protein Wild type lesions, histidine and methionine in the capsid protein Wild type lesions, no histidine and methionine in the capsid protein

11) Azidothymidine (AZT) is an HIV drug. AZT is a reverse transcriptase inhibitor. That means AZT is only effective during the _____ _____ stage of the viral life cycle.

A) B) C) D)

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entry integration viral assembly release

3

12) Vaccines have been very useful in preventing viral diseases.In fact, they have lead to the eradication of smallpox from the world's human population.Most vaccines result in the host producing antibodies (proteins that bind to specific amino acid sequences) that bind to specific viral proteins. What might be a reason why a vaccine for HIV has been so difficult to develop?

A) B) C) D)

The low mutation rate of HIV The fact that HIV genome is found as both DNA and RNA in a cell The high mutation rate of HIV The HIV virion is coated with cellular proteins

13) You perform an experiment like that performed by Fraenkel-Conrat and Singer on TMV. However, in the first step when the viruses are reconstituted, you decide to add RNase to your HR RNA and wild-type proteins. What will be the content of the newly made viral proteins after the reconstituted virus is placed on leaves?

A) The proteins will contain histidine and methionine.

B) The proteins will contain histidine and methionine, but not as much as HR proteins do.

C) The proteins will not contain histidine and methionine.

D) No new viral proteins will be made.

14) The HIV drug fostemsavir binds to the glycoprotein gp120, which is a viral envelope protein. Based on what you know about the lifecycle of HIV, with what step do you predict fostemsavir interferes?

A) attachment B) entry C) integration D) synthesis of viral components E) viral assembly F) release

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15) Most biologists consider viruses to be living organisms, which are classified with bacteria. ⊚ ⊚

16)

true false

The genetic material of HIV is single-stranded RNA. ⊚ true ⊚ false

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Answer Key Test name: Chapter 10 Test Bank 1) A 2) A 3) D 4) A 5) C 6) A 7) B 8) A 9) A 10) A 11) B 12) C 13) D 14) A 15) FALSE 16) TRUE

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Student name:_ 1)

Select the statements that are true about the Hershey-Chase experiments.

A) B) C) D)

2)

All nucleotides contain (check all that apply) A) B) C) D)

3)

a phosphate group. a five carbon sugar. one of five nitrogenous bases. a six-carbon sugar.

What features of DNA were suggested by Rosalind Franklin's X-ray diffraction data?

A) B) C) D)

4)

They radioactively labeled the protein coat of the bacteriophage with 35S. They radioactively labeled the DNA of the bacteriophage with 32P. The results of the experiment indicated only the presence of 35S in the cell.

The results suggested that DNA was the genetic material.

It had a helical structure.

The helix had more than one strand.

The helix contained about 10 bases per turn.

The helices were antiparallel.

Select the ways in which an RNA strand differs from a DNA strand.

A) B) C) D)

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It contains ribose instead of deoxyribose.

It runs 6′ to 3′ instead of 5′ to 3′.

It is usually much shorter than chromosomalDNA. It contains uracil instead of thymine.

1

5)

Select the criteria that all genetic material must meet.

A) B) C) D)

6)

It must contain the information necessary to construct the entire organism.

It must have no variation within a species.

It must be passed from parent to offspring.

It must be able to be copied.

Select the researchers that are correctly matched with their discovery.

A) Pauling provided information on secondary structure in biological molecules.

B) Franklin suggested that DNA was a helix with more than one strand and that there were about 10 bases per turn of the DNA. C) Watson and Crick collected a large amount of X-ray data on the structure of DNA. D) Chargaff demonstrated that the adenine and thymine bases and the cytosine and guanine bases interacted in some manner.

7)

Frederick Griffith is responsible for discovering what process?

A) B) C) D)

replication transmission transformation transduction

8) The fact that the type R and S strains of Streptococcus pneumoniae that Griffith worked with possessed small differences in capsule structure satisfies which of the following criteria for genetic material?

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A) B) C) D)

9)

The individuals who determined that Griffith's transforming principle was DNA were .

A) B) C) D)

10)

Hershey and Chase Avery, Macleod, andMcCarty Watson and Crick Creighton and McClintock

The building blocks of DNA are called

A) B) C) D)

12)

Hershey and Chase Avery, Macleod, andMcCarty Watson and Crick Creighton and McClintock

These individuals determined that DNA was the genetic material in T2 phage.

A) B) C) D)

11)

transmission replication information variation

.

amino acids codons nucleotides alleles

DNA differs from RNA in which way?

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A) B) C) D)

13)

Select the nitrogenous base found only in RNA. A) B) C) D) E)

14)

the structure of the nucleotide the five-carbon sugar it uses the size of the phosphate groups the number of bases attached to the sugar

cytosine guanine thymidine adenine uracil

The backbone of the DNA molecule is formed by

A) B) C) D)

.

peptide bonds ribose sugars nitrogenousbases phosphodiesterbonds

15) How is the directionality of a DNA moleculedescribed based on the orientation of sugar molecules within the strand?

A) B) C) D) E)

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left to right top to bottom 5′ to 3′ A to T or G to C All DNA molecules are different in their directionality.

4

16) The individual(s) who used ball-and-stick models to identify the three-dimensional structure of proteins was .

A) B) C) D) E)

Franklin Watson and Crick Hershey and Chase Pauling Chargaff

17) The researcher(s) who initially used X-ray diffraction to gather information on the DNA molecule was .

A) B) C) D) E)

Franklin Watson and Crick Hershey and Chase Pauling Chargaff

18) According to Chargaff's rule, if the DNA of a species contains 20% adenine, what percent of guanine will it contain?

A) B) C) D)

19) were

20% 30% 50% 75%

The first group of researchers to correctly identify the double-helix structure of DNA .

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5

A) B) C) D)

20) a(n)

McClintock and Franklin Hershey and Chase Pauling and Avery Watson and Crick

In a double-helix DNA strand, the adenine on one strand forms a hydrogen bond with on the other strand.

A) B) C) D)

adenine guanine thymine cytosine

21) How many bases are necessary to complete one complete twist (360 degrees) of a DNA double helix?

A) 5 B) 10 C) 100 D) 1000

22)

The fact that the two DNA strands are arranged in opposite orientations gives DNA its characteristics.

A) B) C) D)

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antiparallel complementary redundant water-soluble

6

23)

One strand of DNA is 5′ - AGGCCTTA - 3′. What is the opposite strand?

A) B) C) D)

24)

What form of DNA is most common in living organisms?

A) B) C) D)

25)

5′ - AGGCCTTA - 3′ 5′ - TCCGGAAT - 3′ 3′ - AGGCCTTA - 5′ 3′ - TCCGGAAT - 5′

A DNA B DNA Z DNA R DNA

Which of the following is a left-handed DNA molecule?

A) B) C) D)

A DNA B DNA Z DNA R DNA

26) Avery, Macleod, and McCarty used the enzyme cell extracts.

to remove the proteins from the

A) protease B) DNase C) RNase

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27) The experiments.

A) B) C) D)

28)

RNA DNA proteins carbohydrates

The purine bases are

A) B) C) D)

29)

was/were labeled using the radioisotope 32P in the Hershey-Chase

.

cytosine, thymine,and uracil adenine andguanine adenine andthymine cytosine andguanine

The pyrimidine bases are

A) B) C) D)

.

cytosine, thymine,and uracil adenine andguanine adenine andthymine cytosine andguanine

30) The idea that the adenine and thymine bases of the DNA interact in some manner was first proposed by .

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A) B) C) D)

Watson andCrick Franklin Pauling Chargaff

31) Adenine and thymine form guanine form bonds.

A) B) C) D)

32)

hydrogen bonds between them, while cytosine and

2; 3 3; 4 3; 2 4; 3

The term nucleic acids is used to describe RNA and DNA because A) they are acidic molecules and have a net negative charge at neutral pH. B) they are acidic molecules and have a net positive charge at neutral pH. C) they are acidic molecules and have no charge at neutral pH.

33) Imagine Rosalind Franklin had performed X-ray diffraction on double-stranded RNA as well as double-stranded DNA. What would be a key difference that she would have observed between these two molecules?

A) DNA but not RNA forms a helical structure.

B) RNA forms a left-handed helix while DNA forms a right-handed helix.

C) A DNA helix contains about 10 bp per turn, whereas an RNA helix contains 11 to 12 bp per turn.

D) DNA double helices are parallel whereas RNA double helices are antiparallel.

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34) Choose the RNA molecule that would be complementary to this DNA molecule: 5'TTAACCGG-3'.

A) B) C) D)

5'-UUAACCGG-3' 3'-AATTGGCC-5' 3'-AAUUGGCC-5' 5'-AAUUGGCC-3'

35) How many total hydrogen bonds would be found holding together the base pairs in a double stranded DNA molecule containing the following nucleotides: 5'-TAGCCCTT-3' 3'-ATCGGGAA-5' A) B) C) D)

36)

8 16 20 24

Which RNA structural pattern does NOTcontain single-stranded regions?

A) B) C) D)

Bulge loop Internal loop Multibranched junction Stem-loop

37) Select the feature that is incorrect about this DNA double strand: 5'-ATGCCCGGAAC-3' 5'-TACGGGCCTTG-3'

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A) B) C) D)

Chargaff's rules are not observed.

The nucleotides are not base paired correctly.

The strands are not anti-parallel.

U should be used instead of T.

38) In the Hershey-Chase experiments, the protein coat of the bacteriophage was labeled with 32 the P radioisotope. ⊚ ⊚

39)

true false

A nucleoside consists of only a five-carbon sugar and a phosphate group. ⊚ ⊚

true false

40) A purine on one strand of the DNA is always paired with a pyrimidine on the other strand. ⊚ ⊚

true false

41) The helical shape of the DNA contains major and minor grooves, which assist in the regulation of gene expression. ⊚ ⊚

true false

42) The two DNA strands of a double helix can be described as antiparallel but uncomplementary.

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⊚ ⊚

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true false

12

Answer Key Test name: Chapter 11 Test Bank 1) [A, B, D] 2) [A, B, C] 3) [A, B, C] 4) [A, C, D] 5) [A, C, D] 6) [A, B, D] 7) C 8) D 9) B 10) A 11) C 12) B 13) E 14) D 15) C 16) D 17) A 18) B 19) D 20) C 21) B 22) A 23) D 24) B 25) C 26) A Version 1

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27) B 28) B 29) A 30) D 31) A 32) A 33) C 34) C 35) C 36) C 37) C 38) FALSE 39) FALSE 40) TRUE 41) TRUE 42) FALSE

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Student name:__ 1)

What feature(s) may be found in insertion elements?

A) B) C) D) E)

inverted repeats reverse transcriptase gene antibiotic resistance gene transposase gene integrase gene

2) You wish to repeat the Noll experiment. However, after you run your gel for an appropriate amount of time, you are dismayed to find that all of the DNA is present at the very top of the agarose gel. What mistakes might have been made in your experiment? (Check all that apply.)

A) B) C) D)

3)

How many origins of replication are there in a bacterial chromosome?

A) B) C) D)

4)

You forgot to add DNase I to your DNA. You didn't sufficiently isolate your DNA from the attached proteins.

You forgot to stain the gel with ethidium bromide.

You added too much DNase I to your DNA.

0 1 2 More than two

Where is the bacterial chromosome located?

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A) B) C) D)

5)

What is a mechanism of condensation shared by both prokaryotes and eukaryotes?

A) B) C) D)

6)

nucleosomes loop domains 30 nm fiber None of these choices are correct.

Topoisomerase I does which of the following?

A) B) C) D)

7)

nucleus nucleolus nucleoid nuclear envelope

relax negative supercoils relieve positive supercoils introduce negative supercoils More than one of these choices are correct.

Quinolones kill bacteria by what mechanism?

A) B) C) D)

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They inhibitDNA gyrase.

They inhibitDNA compaction.

They inhibitDNA replication.

All of these choices are correct.

2

8) What is one reason why an amphibian, such as a salamander, might have a larger genome than a mammal, such as a chimpanzee?

A) B) C) D)

9)

Where do kinetochores attach to chromosomes?

A) B) C) D)

10)

telomeres specific genes on the chromosome centromeres They do not attach to DNA.

Which of the following is found at the end of a eukaryotic chromosome?

A) B) C) D)

11)

The amphibian is more complex than the mammal.

The amphibian has more structural genes than the mammal.

The amphibian has more repetitive sequences than the mammal.

The amphibian has more introns than the mammal.

telomeres centromeres kinetochores nucleoids

An Alu sequence is an example of what?

A) moderately repetitive element B) Unique sequences that are less frequent than moderately repetitive genes, but more frequent than other unique sequences.

C) highly repetitive DNA D) unique sequence DNA

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12) The majority of the nonrepetitive genes in an organism are found in which of the following?

A) B) C) D)

13)

unique sequences moderately repetitive sequences highly repetitive sequences None of these choices are correct.

Unique sequences make up approximately what percent of the human genome?

A) 5% B) 25% C) 40% D) 80%

14)

How many types of histone proteins are there?

A) B) C) D)

15)

4 5 7 8

What types of amino acids are most responsible for the binding of DNA to histones?

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A) B) C) D)

16)

hydrophobic amino acids polar amino acids positively charged amino acids negatively charged amino acids

About how many bases of DNA wrap around a histone complex?

A) <50 B) 146 C) 200 D) >1,000

17) Areas of the chromosome that are highly condensed and transcriptionally inactive are called ev: 02_04_2014_QC_44451 A) B) C) D)

18)

What is the firstlevel of chromosome compaction among the following choices?

A) B) C) D)

19)

euchromatin. superchromatin. heterochromatin. exons.

Loop domains 30 nm fibers Closer association of loop domains Nucleosomes

What is the highest level of chromosome condensation among the following choices?

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A) B) C) D)

20)

Loop domain 30 nm fibers Double helix Nucleosome

What is a nucleosome composed of?

A) It is composed of 2 copies each of H2A, H2B, H3, and H4 and 1 copy of H1.

B) DNA wrapped twice around a histone octet and includes approximately 200 bases that are not associated with the histones.

C) Two copies each of H2A, H2B, H3, and H4.

D) Two turns of DNA wrapped around an octet core of histone proteins.

21) The DNA of a bacterial cell must be compacted about ________fold to fit within the confines of the cell.

A) 10 B) 100 C) 150 D) 1000 E) 1,000,000

22)

The zigzag modelisassociated with the ________ level of DNA organization.

A) B) C) D) E)

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histone 11nm fiber beads-on-a-string 30 nm fiber scaffoldprotein

6

23)

A chromosome territory is defined as A) B) C) D)

a region in the nucleus that is occupied by a single chromosome. a region on a chromosome that is devoid of nucleosomes. a region on a chromosome that has no protein encoding genes. a region in the nucleus that may have several chromosomes.

24) The experiment by Noll, 1974, supported the observation that histones were regularly spaced on chromosomes because A) digestion with increased amounts of DNase I resulted in one or a few bands.

B) limited digestion with DNase I resulted in DNA fragments that differed in size by approximately 200 bp.

C) digestion with low or high amounts of DNase I resulted in a fragment that was found at the bottom of the gel.

D) a digestion performed with a moderate amount of DNase I gave results that were very similar to results obtained with higher amounts of DNase I.

25)

The correct order of compaction from least compacted to most compacted would be A) B) C) D)

26)

naked DNA, 30 nm fiber, nucleosome,loop, metaphase chromosome. naked DNA, loop, 30 nm fiber, nucleosome, metaphase chromosome. naked DNA, nucleosome, 30 nm fiber,loop, metaphase chromosome. naked DNA, metaphase chromosome,loop, 30 nm fiber, nucleosome.

The individual who first proposed the existence of transposable elements was ____.

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A) B) C) D) E)

27)

Transposons that use an RNA intermediate for transposition are called ____ _.

A) B) C) D)

28)

R elements RNAi transposons retrotransposons reverse transposons

Which of the following enzymes is required for simple transposition?

A) B) C) D)

29)

Sturtevant Morgan McClintock Franklin Watson

Integrase Jumpase Splicase Transposase

Long-terminal repeats are unique to which group of transposons?

A) B) C) D)

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integrase simple transposons LTR retrotransposons non-LTR retrotransposons

8

30) Which of the following is technically NOT part of the genetic information of a transposable element?

A) B) C) D)

31)

Which enzyme is unique to the retrotransposons?

A) B) C) D) E)

32)

insertion elements inverted repeats direct repeats transposase

reverse transcriptase reverse resolvase transposase integrase reverse integrase

Which statement is true concerning chromosome compaction during cell division?

A) B) C) D)

Loop domains are not anchored to the scaffolding proteins.

Overall chromosomal length is shortened.

Compaction results in increased transcriptional rates.

The 30 nm fibers are altered to 60 nm fibers.

33) A transposon is inserted into the sequence 5' GACTC 3'. Following simple transposition into this site, what sequences flank the transposon?

A) B) C) D)

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5' GACTC 3'; 5' GACTC 3' 5' GACTC 3'; 5' CTCAG 3' 5' CTCAG 3'; 5' GACTC 3' 5' CTCAG 3'; 5' CTCAG 3'

9

34) You are studying a protein in the frog Xenopus laevis. This organism has a high percentage of transposable elements (TEs) in its genome. After several months of experiments, you start experiencing strange results. It appears that your protein is no longer being made! You look at the chromosomes from the cells where the protein is not being produced, and they all appear normal. You also sequence the coding region of your gene and it has a normal sequence.

What is the most likely explanation for your missing protein?

A) B) C) D)

A chromosomal breakage has occurred due to excision of a TE. A mutation has occurred due to incorrect excision of a TE from your gene.

A TE has inserted in the regulatory sequences for your gene, altering gene regulation.

Exon shuffling has occurred, and there is a new exon in your gene.

35) A double stranded stretch of eukaryotic DNA is 250,000 bp long. If the average linker is about 54 base pairs, how many nucleosomes would be present on this piece of DNA?

A) 1250 B) 4630 C) 1712 D) 200

36) A double stranded stretch of eukaryotic DNA is 4,000 bp long. If the average linker is about 54 base pairs, how many molecules of histone H3 would be present on this piece of DNA?

A) 20 B) 40 C) 74 D) 148

37)

The formation of what structure would be impaired if a cell was lacking CTCF proteins?

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A) B) C) D)

Double helix Nucleosome 30-nm fiber Loop domains

38) During a retrotransposition event, 4 DNA copies of a retrotransposon are created. At the end of this event, how many total copies of the retrotransposon will be found in the genome?

A) 1 B) 4 C) 5 D) 8 E) 10

39) Consider a simple transposition event in a region of the genome where DNA replication is not actively occurring. At the end of the event how many copies of this transposon will be present in the genome?

A) B) C) D) present.

1 2 4 There is not enough information given to determine how many transposons will be

40) Human chromosome 3 is almost 200 million base pairs in length. How many origins of replication do you predict will be found on human chromosome 3?

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A) 2,000 B) 1 C) 2 D) 200,000 E) 20,000

41) You are performing an experiment to identify chromosome territories in a species with 6 pairs of homologous chromosomes, plus two sex chromosomes. You use 7 different colored fluorescent molecules(one for each chromosome and one for the sex chromosome) and link each one to a probe that is highly conserved at the centromere of each chromosome in this species.

What will be the result of your experiment?

A) The chromosomes will all be marked but you won't be able to tell which chromosome is which.

B) You will be able to identify the location and identity of each chromosome territory.

C) The chromosomes will all be marked with the same color.

D) None of the probes will bind to the chromosomes.

42) One would expect heterochromatic regions of DNA to be more compacted than euchromatic regions. ⊚ ⊚

true false

43) Overwinding of the DNA decreases the number of turns in the double helix, and thus results in supercoils in the DNA. ⊚ true ⊚ false

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44) To date, there are no indications that transposable elements offer a selective advantage to a species. ⊚ true ⊚ false

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Answer Key Test name: Chapter 12 Test Bank 1) [A, D] 2) [A, B] 3) B 4) C 5) B 6) A 7) A 8) C 9) C 10) A 11) C 12) A 13) C 14) B 15) C 16) B 17) C 18) D 19) A 20) D 21) D 22) D 23) A 24) B 25) C 26) C Version 1

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27) C 28) D 29) C 30) C 31) A 32) B 33) A 34) C 35) A 36) B 37) D 38) C 39) A 40) A 41) A 42) TRUE 43) FALSE 44) FALSE

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Student name:_ 1)

Select the statements that are true regarding the initiation of DNA replication in bacteria.

A) B) C) D)

2)

It involves a region of the DNA called oriC. Proteins bind to the DNA to begin separation of the strands.

The strands are initially separated at GC-rich regions of DNA. Following initial separation, DNA helicase enzymes continue to separate the strands.

Select the reasons that DNA synthesis occurs with a high degree of fidelity.

A) The hydrogen bonding between G and C or A and T is much more stable than between mismatched pairs.

B) The active site of DNA polymerase preferentially catalyzes the attachment of nucleotides when the correct bases are located in opposite strands.

C) DNA polymerase enzymatically removes mismatched nucleotides.

D) DNA polymerase has the ability to change the structure of the base in order to form the correct bond.

3)

Select the limitations of the DNA polymerases used for DNA replication.

A) DNA polymerase can attach new nucleotides only in the 5′ to 3′ direction.

B) DNA polymerases must begin synthesis using a primer or existing DNA strand.

C) DNA polymerases must have a template strand to copy from.

D) DNA polymerases can attach new nucleotides in the 3′ to 5′ direction only on the lagging strand.

4) Which of the following may account for the process of gene conversion? (Check all that apply)

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A) B) C) D) E)

5)

DNA mismatch repair DNA gap repair DNA proofreading Telomerase errors Flap endonuclease

The purpose of DNA replication is to produce

A) B) C) D)

.

two daughter strands two parental strands two template strands four daughter strands

6) In the attachment of a nucleotide to a DNA strand, the cleavage of the dNTP into deoxyribonucleoside monophosphate and pyrophosphate is .

A) exergonic B) endergonic C) energy neutral

7) You are studying DNA replication in human cells. You discover a mutant cell line that can carry out DNA replication, but the final product contains both RNA and DNA fragments.

What enzyme is likely to be missing in this cell line?

A) B) C) D)

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flap endonuclease DNA ligase DNA polymerase I DNA helicase

2

8) Which describes the mechanism of DNA replication in which both parental strands remain together following replication?

A) dispersive B) semiconservative C) conservative

9) Which describes the mechanism of DNA replication in which the newly made doublestranded DNA contains one parental strand and one daughter strand?

A) dispersive B) semiconservative C) conservative

10) The first generation of replication in the Meselson and Stahl experiment disproved which theory of replication?

A) B) C) D)

semiconservative conservative dispersive None—it took more than one generation to disprove the theory.

11) You have isolated what appears to be alien DNA. While studying its replication, you performed the exact experiment Meselson and Stahl did. After three generations, the DNA is subjected to a CsCl gradient and only one band appears. What type of replication does this DNA undergo?

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A) semiconservative B) conservative C) dispersive

12)

How many origins of replication does a bacterial chromosome contain?

A) B) C) D)

13)

DNA helicase enzymes move in what direction along the DNA during DNA replication?

A) B) C) D)

14)

5′ to 3′ 3′ to 5′ They travel in both directions.

They remain stationary.

What enzyme relaxes supercoiling ahead of the replication fork?

A) B) C) D) E)

15)

0 1 10 Depends on the size of the DNA

DNA ligase DNA primase topoisomerase DNA polymerase I DNA polymerase III

Which enzyme manufactures a 10-12 base segment of RNA?

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A) B) C) D) E)

16)

Which enzyme fills in small regions of DNA where the RNA primers were located?

A) B) C) D) E)

17)

DNA ligase DNA primase topoisomerase DNA polymerase I DNA polymerase III

Which enzyme is responsible for the majority of DNA synthesis during replication?

A) B) C) D) E)

18)

DNA ligase DNA primase topoisomerase DNA polymerase I DNA polymerase III

DNA ligase DNA primase topoisomerase DNA polymerase I DNA polymerase III

Which enzyme attaches adjacent Okazaki fragments, forming a continuous DNA strand?

A) B) C) D) E)

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DNA ligase DNA primase topoisomerase DNA polymerase I DNA polymerase III

5

19)

Which enzyme synthesizes the lagging strand of the DNA? A) B) C) D) E)

20)

How many DNA polymerases are found in prokaryotes?

A) B) C) D)

21)

DNA ligase DNA primase Topoisomerase DNA polymerase I DNA polymerase III

5 7 9 12

DNA polymerases add new nucleotides in what direction?

A) 5′ to 3′ B) 3′ to 5′ C) both directions

22) You have discovered a strain of E. coli that grows very slowly—the generation time is nearly 12 hours compared to the normal 20-30 minutes. Upon further investigation, you find a mutation in a gene for one of the subunits of DNA polymerase III. What subunit is this mutation most likely to be in?

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A) B) C) D)

23)

What is the purpose of Okazaki fragments?

A) B) C) D)

24)

To assist in the formation of the replication fork To allow DNA to be synthesized in the 5' to 3' direction on the leading strand To allow DNA to be synthesized in the 5′ to 3′ direction on the lagging strand To reform the double helix following replication

Which of the following is an example of a processive enzyme?

A) B) C) D)

25)

α β γ δ

DNA polymerase I DNA polymerase III DNA ligase Okazaki fragments

What functions are accomplished by the primosome?

A) tracking along DNA B) tracking along DNA, separating double-stranded DNA C) tracking along DNA, separating double-stranded DNA, synthesizing RNA primers D) tracking along DNA, separating double-stranded DNA, synthesizing RNA primers, adding nucleotides

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26)

The proofreading of the DNA occurs in the

.

A) 5′ to 3′ direction B) 3′ to 5′ direction C) both directions

27)

In yeast, which of the following is similar to the oriC region of prokaryotes?

A) B) C) D)

centromeres ARS elements promoters telomeres

28) Based on their known functions, which of the enzymes in the following pairs play a similar role in prokaryotes and eukaryotes?

A) DNA pol I : DNA pol α B) DNA pol II : DNA pol β C) DNA pol III : DNA pol γ

29) DNA polymerase III carries out leading and lagging strand synthesis in prokaryotes. How many enzymes accomplish this task in eukaryotes (not counting primer creation)?

A) B) C) D)

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one two three four

8

30)

DNA polymerases are unable to replicate what areas of the chromosome?

A) B) C) D)

31)

What process doesthe Holliday model help explain?

A) B) C) D)

32)

centromeres 3′ ends origins of replication 5′ ends

recombination between homologous chromosomes recombination between sister chromatids the creation of shortenedchromosomes replication initiation

Which recombination model involves formation of a D-loop?

A) double-stranded break model B) Holliday model C) single-stranded break model

33) In E. coli, which protein(s) are necessary for the cell to recognize double-stranded breaks and conduct recombination during meiosis?

A) B) C) D)

34)

RecG RecA RecBCD RuvABC

Which of the following best describes the function of the RecA protein in E. coli?

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A) B) C) D)

Promotes strand invasion and D-loop formation in the double-stranded break model Recognizes double-stranded breaks in the DNA Involved in the initial formation of Holliday junctions Resolves Holliday junctions into separate chromosomes

35) If eukaryotic cells evolved such that lagging strand DNA synthesis could occur continuously (without the use of Okazaki fragments), what enzyme would likely no longer be needed for DNA replication?

A) B) C) D)

Flap endonuclease Helicase Primase Topoisomerase

36) You are in the lab trying to synthesize DNA in vitro. You are upset because the lab seems to be out of dNTPs (deoxynucleosidetriphosphates) but you find a tube of dNMPs(deoxynucleoside monophosphates) in the freezer. You add this to your replication reaction instead of dNTPs. Will your reaction work? Why or why not?

A) No, cleavage of dNTP drives the formation of the covalent bond between the nucleoside monophosphate and the growing DNA strand.

B) Yes, only the nucleoside monophosphate ultimately is incorporated into the growing DNA strand.

C) No, all three phosphates on the nucleoside triphosphateare incorporated into the growing DNA strand.

D) Yes, the formation of the covalent bond between the nucleoside monophosphate and the growing DNA strand is energetically favorable.

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37) If the piece of partially double-stranded DNA below was added to a reaction mixture that contained DNA pol III and all the reaction components necessary to synthesize DNA, what would be the next base added?

5' AGCATGACGATCCAT 3' 3' ACTGCTAG 5' A) B) C) D)

38)

A G C T

Which answer would best represent the composition of double stranded DNA?

A) B) C) D)

20% A20% G30% C30% T 20% A20% G20% C40% T 30% A15% G15% C30% T 15% A25% G15% C35% T

39) Imagine you are able to watch DNA replication under a microscope in a mutant strain of bacteria. The parental strands appear to unwind and separate, but then they immediately join back together before any new DNA can be synthesized. What protein is missing or defective in this bacterial strain?

A) B) C) D) E)

Helicase Single-strand binding proteins Primase DNA polymerase Ligase

40) DNA polymerase III has the ability to begin synthesis of the new daughter strands immediately following the formation of the replication fork.

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⊚ ⊚

true false

41) The Meselson-Stahl experiments used 35S radioisotopes to determine the mechanism of DNA replication. ⊚ ⊚

42)

The Meselson-Stahl experiments supported the model of dispersive DNA replication. ⊚ ⊚

43)

true false

true false

The origin of replication in bacteria is called oriC. ⊚ ⊚

true false

44) Replication usually begins in GC-rich regions due to the presence of only 2 hydrogen bonds between the bases. ⊚ true ⊚ false

45)

The movement of the replication fork in bacterial replication is unidirectional. ⊚ ⊚

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true false

12

46) After the action of the helicase, single-stranded binding proteins keep the parental DNA strands from reforming a double helix. ⊚ true ⊚ false

47) The synthesis of the daughter strand of DNA occurs away from the replication fork in the leading strand. ⊚ ⊚

48)

true false

DNA polymerase III has an error in replication once every 100 million nucleotides. ⊚ ⊚

true false

49) The rate at which DNA polymerase III synthesizes a new strand of DNA is 300 nucleotides per minute. ⊚ true ⊚ false

50)

A primosome consists of a polymerase and a single-stranded binding protein. ⊚ ⊚

true false

51) The ability of the DNA polymerase to remove mismatched bases using exonuclease cleavage is called proofreading. ⊚ ⊚

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true false

13

52) The problem of replicating the 3′ ends of chromosomes is solved by using the enzyme telomerase. ⊚ ⊚

true false

53) Recombination between sister chromatids produces new combinations of alleles that may be beneficial to the species. ⊚ true ⊚ false

54) A heteroduplex is a region within a DNA double helix that contains base mismatches that can lead to gene conversion. ⊚ ⊚

55)

true false

Homologous recombination that leads to genetic diversity occurs during meiosis II. ⊚ ⊚

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true false

14

Answer Key Test name: Chapter 13 Test Bank 1) [A, B, D] 2) [A, B, C] 3) [A, B, C] 4) [A, B] 5) A 6) A 7) A 8) C 9) B 10) B 11) C 12) B 13) A 14) C 15) B 16) D 17) E 18) A 19) E 20) A 21) A 22) B 23) C 24) B 25) C 26) B Version 1

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27) B 28) B 29) B 30) B 31) A 32) A 33) C 34) A 35) A 36) A 37) D 38) C 39) B 40) FALSE 41) FALSE 42) FALSE 43) TRUE 44) FALSE 45) FALSE 46) TRUE 47) FALSE 48) TRUE 49) FALSE 50) FALSE 51) TRUE 52) TRUE 53) FALSE 54) TRUE 55) FALSE

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Student name:_ 1)

In transcription, the closed complex consists of A) B) C) D)

2)

Select the eukaryotic regulatory elements.

A) B) C) D)

3)

enhancers silencers cis-acting elements Pribnow box

Select the components of the eukaryotic preinitiation complex.

A) B) C) D) E)

4)

RNA polymerase. transcription factors. double-stranded DNA. single-stranded DNA.

TFIID DNA TFIIB RNA polymerase II mediator

Select the stages of transcription.

A) B) C) D) E)

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initiation elongation termination translation replication

1

5) You are studying transcription in two different unknown systems. Select the experimental questions you could test that would allow you to definitively determine if the system is eukaryotic or prokaryotic.

A) B) C) D) E)

Does transcription in the system require a promoter?

Does RNA capping occur?

Does RNA splicing occur, and if so is it self-splicing?

Does RNA tailing occur?

How many RNA polymerases are present?

6) What is the name of the site in eukaryotic DNA where general transcription factors associate?

A) B) C) D)

7)

promoter terminator silencer enhancer

Following transcription, the RNA has a complementary sequence to A) B) C) D)

regulatory sequences. termination sequences. the coding strand of DNA. the template strand of DNA.

8) Which of the following forms of RNA encodes the sequence of amino acids for a functional protein?

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A) B) C) D) E)

9)

tRNA snRNA mRNA rRNA scRNA

In prokaryotes, a holoenzyme is formed when what is added to the core enzyme?

A) B) C) D) E)

Start codons b b' σ a

10) In eukaryotic organisms, a mutation in which of the following would result in a failure of the RNA polymerase to maintain an open complex?

A) B) C) D) E)

TFIIB TFIID TFIIE TFIIH TFIIF

11) In eukaryotic organisms, what protein allows RNA polymerase II to interact with transcription factors bound to enhancers or silencers?

A) B) C) D)

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mediator TFIIB TFIID sigma factors

3

12) In eukaryotic organisms, a functional eukaryotic protein is typically made up of the information contained within what region of pre-mRNA? A) B) C) D)

13)

In eukaryotic organisms, what RNA modification is important for mRNA stability?

A) B) C) D) E)

14)

exons introns enhancers promoters

alternative splicing RNA editing base modification 3' polyA tailing processing

In eukaryotic organisms, what process is important for the initiation of translation?

A) B) C) D)

alternative splicing RNA editing 5' capping base modification

15) In eukaryotic organisms, the processing of the 45S rRNA into 5.8S, 18S, and 28S rRNA occurs where?

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A) B) C) D) E)

16)

in the cytoplasm in the nucleolus in the endoplasmic reticulum in the Golgi body throughout the cell

RNA polymerase I in eukaryotes produces which type of RNA?

A) B) C) D)

snRNA tRNA 5S rRNA 18S rRNA

17) The consensus sequences in a prokaryotic promoter are 5'TTGACA3' and 5'TATAAT3' at -35 and -10, respectively. What sequence would be most easily recognized by σ factor?

A) B) C) D)

5'TTGAAA3' 5'TAAATT3' 5'TATATA3' 5'TTCAGA3'

18) Which subunit of the prokaryotic RNA polymerase carries out catalytic synthesis of RNA? A) B) C) D)

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alpha omega beta sigma

5

19)

What region in eukaryotic genes contains the majority of regulatory elements?

A) 0 to 50 B) −50 to 0 C) −50 to −100 D) −100 to −150

20)

How many general transcription factors are needed for eukaryotic translation?

A) B) C) D)

21)

In eukaryotic organisms, what general transcription factor is a helicase?

A) B) C) D)

22)

3 4 5 6

TFIID TFIIH TFIIF TFIIB

Most eukaryotic genes are colinear.

A) True B) False C) It is unknown at this time.

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23)

What enables the splicing of group I and II introns?

A) B) C) D)

24)

spliceosomes ribozymes snRNA poly-A tail

Both the rho-dependent and rho-independent mechanisms of termination

A) B) C) D)

end transcription immediately after the stop codon. involve a sequence rich with A-U base pairs. require a helicase to separate the DNA-RNA complex. involve the formation of a stem-loop structure.

25) What is the purpose of phosphorylating the carboxy terminal domain (CTD) of RNA Polymerase II? A) B) C) D)

26)

convert from the initiation to the elongation stage to aid in transcriptional initiation phosphorylate transcription factors allow transcription factors to bind the TATA box

In prokaryotic organisms, what structural motif is used by sigma factor to bind to DNA? A) B) C) D)

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zinc-finger helix-turn-helix leucine zipper homeodomain

7

27)

In eukaryotic organisms, alternative splicing allows an organism to

.

A) produce several different proteins from a single gene.

B) produce only one protein from a gene.

C) produce nonfunctional variants of proteins from a gene

28) If a eukaryotic organism is missing the U5 snRNP, it will have trouble performing which step of splicing?

A) B) C) D)

29)

Binding of snRNPs to the 5' splice site Binding of snRNPs to the branch site Formation of the intron loop Exon degradation

Select the statement that is true regarding tRNA processing in E. coli.

A) B) C) D)

RNaseP acts as an endonuclease.

The recognition sequence of RNaseD is 3'CCA5'.

RNaseD acts as an endonuclease.

The action of RNaseP produces the correct 3' end of the mature tRNA.

30) In trying to distinguish between the two models of eukaryotic transcriptional termination, you identify a drug that specifically blocks exonuclease activity. When you add it at the appropriate time to an in vitro system that carries out all the stages of eukaryotic transcription, you find that transcription proceeds normally, but transcription doesn't terminate. This experiment provides support for which model of eukaryotic transcriptional termination?

A) B) C) D)

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Allosteric model Torpedo model Rho-dependent termination Rho-independent termination

8

31)

Which modifications are typically found on eukaryotic mRNAs?

A) B) C) D)

Capping only Addition of poly A only Capping and poly-A addition Eukaryotic mRNAs are not modified before translation

32) A gene that undergoes rho-dependent termination has a mutation in the rut site. How will this influence transcriptional activity?

A) Transcription will terminate normally as Rho termination relies on the rut site to enhance termination; it is not essential for termination.

B) Transcription will stop immediately because RNA polymerase will sense there is a mutation.

C) Transcription will continue until another termination sequence is reached, either Rho-dependent or independent.

D) There will be no transcription initiation.

33) This pre-mRNA contains 8 exons and 7 introns. The splice sites surrounding exon 4 are not readily recognized by the splicesome. In addition, a splicing repressor prevents the recognition of the 3' splice site in exon 7. What exons will be present in the mature mRNA?

© McGraw-Hill Education

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A) 1, 2, 3, 5, 6, 8 B) 1, 2, 3, 4, 5, 6, 7, 8 C) 1, 2, 3, 4, 5, 6, 8 D) 1, 2, 3, 5, 6, 7, 8 E) 1, 2, 3, 5, 6

34) You are working in a lab studying mRNAs from many different species. You need to distinguish between mRNAs in a mixed sample. You begin by having all of the mRNAs sequenced, including the mitochondrial mRNAs. You notice a high number of C to U conversions in the mitochondrial mRNAs from a certain portion of the sample. You conclude that this portion of the sample is from .

A) B) C) D) E)

mammals plants slime mold Drosophila Trypanosomes

35) A mutation occurs in one of the general transcription factors necessary for eukaryotic transcription. A pre-initiation complex forms, but the open complex does not form. Further, TFIIB, TFIIE, and TFIIH are not released from the DNA. The mutation is most likely in which general transcription factor?

A) B) C) D) E)

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TFIIB TFIID TFIIE TFIIF TFIIH

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36) You have an RNA molecule that contains introns. You wish to carry out an in vitro splicing reaction with this RNA and cellular extracts. You want to determine if splicing is occuring via self-splicing or through the spliceosome. What enzyme could you add to the reaction to make this determination?

A) B) C) D)

37)

A ribozyme is a catalytic protein that assists in the processing of tRNA. ⊚ ⊚

38)

Protease DNase RNase Lipase

true false

The unit of transcription at the molecular level is the gene. ⊚ ⊚

true false

39) The RNA polymerase forms a closed complex during the process of elongation during transcription. ⊚ ⊚

40)

true false

During translation, the mRNA is read in groups of nucleotides called quadrons. ⊚ ⊚

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true false

11

41) The Pribnow box contains a CACAAC consensus sequence located in the −100 region of the promoter. ⊚ ⊚

42)

true false

Sigma factor belongs to a general group of proteins called the transcription factors. ⊚ ⊚

true false

43) The formation of an open complex usually forms in GC-rich regions of DNA due to the decreased number of hydrogen bonds. ⊚ true ⊚ false

44)

Basal transcription in eukaryotes involves the action of only the core promoter. ⊚ ⊚

true false

45) The process of RNA editing can alter the base sequence of the mRNA following transcription. ⊚ ⊚

46)

true false

The term ribozyme is given to catalytic RNA molecules.

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⊚ ⊚

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true false

13

Answer Key Test name: Chapter 14 Test Bank 1) [A, B, C] 2) [A, B, C] 3) [A, B, C, D] 4) [A, B, C] 5) [B, C, E] 6) A 7) D 8) C 9) D 10) C 11) A 12) A 13) D 14) C 15) B 16) D 17) A 18) C 19) C 20) C 21) B 22) B 23) B 24) D 25) A 26) B Version 1

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27) A 28) C 29) A 30) B 31) C 32) C 33) A 34) B 35) E 36) A 37) FALSE 38) TRUE 39) FALSE 40) FALSE 41) FALSE 42) TRUE 43) FALSE 44) TRUE 45) TRUE 46) TRUE

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Student name: 1)

Which two things act together to terminate translation? (Choose two.)

A) Rho factor B) release factor C) Met codon D) filled P site E) stop codon F) nucleotide exchange factors

2) Based on our modern understanding, what revisions are necessary to the original onegene / one-enzyme hypothesis?

A) Alternative splicing allows one gene to encode multiple polypeptides.

B) Some genes encode RNAs, not enzymes.

C) Some genes encode lipids, not enzymes.

D) Some genes encode polypeptides that are not enzymes, such as structural proteins.

E) Proteins as functional units may be composed of several polypeptides, so genes may encode just one polypeptide within a larger protein.

F) Enzymes are not encoded by genes after all — they are constructed through separate biochemical processes.

3)

Which of the following are true regarding the experiments of Beadle and Tatum?

A) B) C) D)

They were able to create loss-of-function mutations in any enzyme they chose.

They used the mold Neurospora as a model organism.

They isolated strains that could not grow on minimal media.

They established the link between a gene and an enzyme or protein.

4) Mutations in the enzyme homogentisic acid oxidase are associated with alkaptonuria.

Select all characteristics of homogentisic acid oxidase.

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A) Is encoded by a structural gene B) Assists other polypeptides with folding into the correct secondary structure C) Has primary, secondary, and tertiary structure D) Has an amino-terminal and a carboxyl-terminal end E) Has primary structure but not secondary structure F) Catalyzes a reaction in a metabolic pathway

5) Taking into account your knowledge of wobble rules, select all anticodons that can pair with the codon 5'-UAC-3'. (Check all that apply.)

A) 5'-AUG-3' B) 5'-IUA-3' C) 5'-AUA-3' D) 5'-GUA-3' E) 5'-GUI-3'

6)

Which of the following is NOT a stop codon?

Copyright © McGraw-Hill Education

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A) B) C) D)

7)

UGA UUA UAG UAA

Francis Crick proposed the adaptor hypothesis, and later the adaptor was discovered to be .

A) siRNA B) amino acid C) tRNA D) mRNA E) micro RNA

8)

Which of the following codons is NOT degenerate to CUU?

Copyright © McGraw-Hill Education

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A) CUC B) UUG C) CUA D) AUU E) UUA

9)

The primary structure of a protein refers to A) B) C) D)

10)

What level of protein structure is exemplified by α-helices and β-sheets?

A) B) C) D)

11)

regular repeating shapes, such as beta-sheets. the three-dimensional shape of the protein. the linear sequence of the amino acids. the interaction of two or more peptide chains.

primary structure secondary structure tertiary structure quaternary structure

Which molecule contains an anticodon?

A) DNA B) mRNA C) rRNA D) tRNA E) snRNA

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12) What is the least number of tRNAs that can be used to recognize all of the codons of isoleucine, and what is/are the anticodon(s)? (In the answer choices, I = inosine.)

Copyright © McGraw-Hill Education A) B) C) D)

13)

one—5′IAU3′ one—5′UAU3′ two—5′GAU3′ and 5′UAU3′ two—5′AUA3′ and 5′UAI3′

What does eIF5 act to recruit during the initiation stage?

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A) the mRNA B) the polypeptide C) the large ribosomal subunit D) termination GTPases E) the 7-methylguanosine cap

14)

A tRNA that has an amino acid attached is called

.

A) rRNA B) degenerate tRNA C) coding tRNA D) charged tRNA E) None of the answers are correct.

15) Kozak's rules for translation are similar to those for transcription in prokaryotes in that they both contain consensus sequences. In prokaryotes, the promoter consensus sequences are at −35 and −10. Where is the consensus sequence for translational initiation according to Kozak's rules?

A) −6 to +4 B) −12 to −2 C) +1 to +10 D) −35 to −10

16)

Which of the following would be considered an activated amino acid?

A) B) C) D)

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Amino acid · ATP · aminoacyl-tRNA synthetase Amino acid · aminoacyl-tRNA synthetase Amino acid · AMP · PPi Amino acid · AMP

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17)

What site on the ribosome is primarily responsible for holding the growing polypeptide?

A) A B) E C) P

18)

What site does the initiator tRNA bind to on the ribosome?

A) A B) E C) P

19)

What is responsible for binding the initiator tRNA in prokaryotes and eukaryotes?

A) B) C) D)

IF2 / eIF2 EF-Tu / eEF1α RF3 / eRF3 EF-G / eEF2

20) If translation begins at the first start codon and ends at the first in frame stop codon, how many amino acids are encoded by the following mRNA? 5′GCCACCAUGGGCCAAUUACGAAGGUUUUGCUGA3′ A) B) C) D)

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8 9 10 11

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21)

Where are the ribosomal subunits assembled?

A) B) C) D)

22)

nucleus nucleoid nucleolus nuclear envelope

A tRNA's anticodon is 5′GGC3′. What amino acid is attached to it?

Copyright © McGraw-Hill Education A) B) C) D)

glycine proline alanine arginine

23) Which component of the ribosome is part of the peptidyl transferase in the 50S subunit and acts as a ribozyme to catalyze peptide bond formation?

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A) 16S rRNA B) P site C) Shine-Dalgarno sequence D) snRNA E) aminoacyl-tRNA synthetase F) 23S rRNA

24) Which component of the ribosome is able to detect when the wrong tRNA has inserted into the A site?

A) 16S rRNA B) P site C) Shine-Dalgarno sequence D) snRNA E) aminoacyl-tRNA synthetase F) 23S rRNA

25)

The peptidyltransferase complex is a component of which of the following?

A) DNA B) tRNA C) The ribosome D) The growing polypeptide E) mRNA

26)

The C-terminus of a polypeptide always contains

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A) a stop codon.

B) a carboxyl group.

C) an amino group.

D) carbon dioxide.

E) a stop amino acid.

F) Methionine.

27)

RF1 and RF2 are active during

A) B) C) D)

.

initiation elongation termination cotranslational import

28) In bacteria, the sequence that facilitates the binding of the mRNA to the ribosome is called the A) B) C) D)

Kozak sequence.

Shine-Dalgarno sequence. start sequence. rRNA sequence.

29) The scientist(s) studied patients who had defects in their ability to metabolize certain compounds, particularly the inherited disease alkaptonuria. He/They proposed that a relationship exists between the inheritance of the trait and the inheritance of a defective enzyme.

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A) Garrod B) Nirenberg and Matthaei C) Khorana and colleagues D) Nirenberg and Leder E) Beadle and Tatum

30) The scientist(s) studied Neurospora mutants that were altered in their nutritional requirements and hypothesized that one gene encodes one enzyme.

A) Garrod B) Nirenberg and Matthaei C) Khorana and colleagues D) Nirenberg and Leder E) Beadle and Tatum

31) The scientist(s) created a cell-free translation system. An mRNA template was added to a number of tubes, each containing amino acids with a different one radiolabeled.

Radiation was measured from the precipitated protein, and this corresponded to the mRNA template that had been added. This was the first experiment that demonstrated the ability to synthesize polypeptides from synthetic mRNA, and revealed a relationship between the two.

A) Garrod B) Nirenberg and Matthaei C) Khorana and colleagues D) Nirenberg and Leder E) Beadle and Tatum

32) The scientist(s) developed a novel method to synthesize RNA. Short RNAs of 2 to 4 nucleotides were linked together to create a copolymer. The resulting peptide created after translation revealed the 3-letter code for the amino acids.

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A) Garrod B) Nirenberg and Matthaei C) Khorana and colleagues D) Nirenberg and Leder E) Beadle and Tatum

33) The scientist(s) developed the triplet-binding assay. Samples containing ribosomes and a particular RNA triplet were exposed to charged tRNAs with different radiolabeled amino acids. Only the amino acid whose tRNA matched the triplet would bind the ribosome. The triplet-binding assay established relationships between particular triplet sequences and specific amino acids.

A) Garrod B) Nirenberg and Matthaei C) Khorana and colleagues D) Nirenberg and Leder E) Beadle and Tatum

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34) You are studying the DNA of a person who you know has two defective copies of the gene that encodes phenylalanine hydroxylase. You are surprised to find that this person also carries two defective copies of the gene for homogentisic acid oxidase. What disease symptoms will this person exhibit? (Assume pathway intermediates are not available from sources outside the phenylalanine breakdown pathway.)

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A) This person will exhibit symptoms of alkaptonuria.

B) This person will exhibit symptoms of tyrosinosis.

C) This person will exhibit symptoms of phenylketonuria.

D) This person will exhibit symptoms of alkaptonuria and phenylketonuria.

E) This person will be phenotypically normal.

35)

Copyright © McGraw-Hill Education You perform a cell-free translation experiment like Nirenberg and Matthaei. You start with 60% C and 40% A. What relative amount of radiolabeled proline do you expect in the translated polypeptides?

A) 36% B) 22% C) 14% D) 28%

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36) What is the minimal number of tRNAs that can be used to recognize all of the codons for threonine?

Copyright © McGraw-Hill Education A) B) C) D)

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1 2 3 4

16

37) Suppose you performed a triplet binding assay with the triplet 5′-UAA-3′. Among those listed below, whatradiolabeled amino acids do you expect to find bound to the filter?

Copyright © McGraw-Hill Education A) B) C) D)

38) than

Tyrosine Glycine Methionine None

A protein called actin consists of 376 amino acids. The mRNA for actin must be longer

A) B) C) D)

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125 nucleotides.

376 nucleotides.

1128 nucleotides.

3384 nucleotides.

17

39) You perform a cell free translation experiment like Nirenberg and Matthaei, but you forget to write down what nucleotides you added to make the mRNA. You precipitate the translated polypeptides and measure the relative amount of radiolabeled amino acids incorporated into them. You get 25% proline, 25% threonine, 12.5% glutamine, 12.5% lysine, 12.5% asparagine, and 12.5% histidine. What nucleotides and in what % did you add to make the mRNA? A) Equal amounts of A, C, U, and G B) 50% C and 50% A C) 70% C and 30% A D) 50% C and 50% U E) 50% A and 50% G

40) The genetic code — the correspondence between codons and amino acids — is unique for each species of organism. ⊚ true ⊚ false

41)

All mature polypeptides contain a methionine at the N-terminus. ⊚ ⊚

42)

true false

61 different tRNAs are required for translation in vivo. ⊚ true ⊚ false

43) The first person to suggest that a connection exists between the function of a gene and the production of an enzyme was Archibald Garrod. ⊚ ⊚

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true false

18

44)

The one-gene, one-enzyme hypothesis was first proposed by Beadle and Tatum. ⊚ ⊚

45)

true false

The wobble base is the second base of a codon. ⊚ true ⊚ false

46) Proteins that are composed of two or more polypeptides are said to have quaternary structure. ⊚ true ⊚ false

47)

The anticodon on the mRNA recognizes the codon on the tRNA. ⊚ ⊚

48)

true false

The enzyme that attaches free amino acids to the tRNA molecule is called tRNA attaches. ⊚ ⊚

true false

49) During elongation, the polypeptide is removed from the tRNA in the P site and transferred to the amino acid in the A site.

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⊚ ⊚

50)

true false

Transcription and translation may occur simultaneously in prokaryotic cells. ⊚ ⊚

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true false

20

Answer Key Test name: Chapter 15 Test Bank 1) [B, E] 2) [A, B, D, E] 3) [B, C, D] 4) [A, C, D, F] 5) [B, C, D] 6) B 7) C 8) D 9) C 10) B 11) D 12) A 13) C 14) D 15) A 16) D 17) C 18) C 19) A 20) A 21) C 22) C 23) F 24) A 25) C 26) B Version 1

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27) C 28) B 29) A 30) E 31) B 32) C 33) D 34) C 35) A 36) B 37) D 38) C 39) B 40) FALSE 41) FALSE 42) FALSE 43) TRUE 44) TRUE 45) FALSE 46) TRUE 47) FALSE 48) FALSE 49) TRUE 50) TRUE

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Student name:_ 1) A gene is inducible and under negative control. Which of the following pairs will allow expression of this gene?

A) B) C) D)

2)

Regulation of gene expression may occur at which of the following levels?

A) B) C) D)

3)

transcription translation posttranslation All of the answers are correct.

How many promoters are in an operon?

A) B) C) D)

4)

activator + repressor activator + inhibitor repressor + inducer repressor + corepressor

1 2 3 It depends on how many genes there are in the operon.

Which of the following is found in an operon?

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A) B) C) D) E)

5)

promoter terminator two or more genes operator All of the answers are correct.

Allosteric regulation is accomplished by A) B) C) D)

a small molecule that fits into an enzyme's active site. a large protein that blocks an enzyme's active site. a small molecule that fits into a site on the enzyme that is not the active site. a small molecule that covalently modifies a site on the enzyme that is not the active

site.

6)

What is the gene responsible for attenuation in the trp operon?

A) B) C) D)

7)

trpL trpR trpD trpC

What stem-loop conformations favor attenuation in the trp operon?

A) 1-2 B) 1-2 and 2-3 C) 2-3 D) 3-4

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8) If CAP could not bind to its CAP site, then what would be the result? Assume lactose is present in each scenario.

A) B) C) D)

Transcription would be difficult to repress in the presence of glucose.

Transcription would be difficult to activate in the presence of glucose.

Transcription would be difficult to activate in the absence of glucose.

Transcription would be difficult to repress in the absence of glucose.

9) In Jacob, Monod, and Pardee's experiment, how many functional copies of lacI were there in the merozygote?

A) B) C) D)

0 1 2 3

10) In the Jacob Monod merozygote experiment, what was indicated by the presence of a yellow color when b-ONPG was added?

A) B) C) D)

11)

Expression of the lac operon is constitutive whether lacI is functional or not.

Beta-galactosidase was present.

The lac operon was completely turned off.

The researcher added too much b-ONPG.

CAP affects which operon(s)?

A) B) C) D)

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the lac operon the trp operon both the lac and the trp operon neither the lac nor the trp operon

3

12) What would be the result if the U-rich sequence after the fourth stem loop in the trp operon was replaced by a UG-rich sequence?

A) B) C) D)

13)

Which of the following is not an example of translational regulation in prokaryotes?

A) B) C) D)

14)

Attenuation would occur if tryptophan was high.

Attenuation would occur if tryptophan is low.

Attenuation would not occur if tryptophan was high.

None of the answers are correct.

Sterically blocking the ribosome Phosphorylation of an enzyme Binding of antisense RNA to mRNA Altering the structure of the mRNA

Enzymes involved in metabolism are most likely regulated via A) B) C) D)

.

feedback inhibition acetylation methylation None of the answers are correct.

15) If a bacterium is placed in an environment that contains both glucose and lactose, the regulation of the lac operon will allow which nutrient to be processed first?

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A) B) C) D)

16)

The regulation of protein function, not gene expression is called

A) B) C) D)

17)

regulation.

posttranslational transcriptional translational posttranscriptional

Translational regulatory proteins recognize specific areas of what molecule?

A) B) C) D) E)

18)

glucose lactose Both will be processed equally.

Neither will be processed in this environment.

tRNA Ribosome rRNA mRNA None of the answers are correct.

Antisense RNA does which of the following?

A) B) C) D)

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inhibits the formation of the open complex in transcription occupies the A and P sites of the ribosome binds to a complementary RNA and prevents its translation prevents the correct folding of a newly formed peptide

5

19)

Riboswitches have been shown to affectregulation of A) B) C) D)

.

transcription and translation only transcription onlytranslation only RNA splicing

20) A deletion in an operon removes the terminator. How will that affect the transcript that is produced from the operon?

A) B) C) D) E)

The transcript will be produced and normal in length The transcript will be produced, but shorter than normal The transcript will be produced, but longer than normal The transcript will be produced, but will contain a deletion The transcript will not be produced

21) A deletion in an operon removes the promoter. How will that affect the transcript that is produced from the operon?

A) B) C) D) E)

The transcript will be produced and normal in length The transcript will be produced, but shorter than normal The transcript will be produced, but longer than normal The transcript will produced, but will contain a deletion The transcript will not be produced

22) In Jacob, Monod, and Pardee's experiment, what would have been the conclusion if all four tubes produced a yellow color when β-ONPG was added?

A) B) C) D)

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Expression of the lac operon is constitutive whether lacI is functional or not LacI provides the binding site for the repressor LacI encodes a diffusible repressor The researcher added too much β-ONPG

6

23) A mutation in the lacI gene prevents the gene product from binding allolactose. What will the expression level of the operon be in the absence of lactose?

A) B) C) D)

No transcription Positive regulation Constitutively active Transcription will occur only in the presence of glucose

24) In a particular E. coli strain, a mutation in the thiMD operon results in improper formation of the stem loop secondary structure making it impossible to bind TPP. There are two enzymes encoded by the thiMD operon. How many of the enzymes encoded in the thiMD operon are translated?

A) B) C) D)

Two Zero Three One

25) If the Trp codons in the trpL gene were mutated to encode another amino acid, what would the result be?

A) B) C) D)

The trp operon would never be transcribed.

The trp operon would always be transcribed.

The trp operon would only be transcribed when tryptophan in the cell was low.

The trp operon would only be transcribed when tryptophan in the cell was high.

26) A riboswitch is an RNA molecule that can exist in two different conformations.

Conversion from one to another is due to the binding of a small molecule to the riboswitch. ⊚ true ⊚ false Version 1

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27)

In the lac operon, the operator is an example of a trans-effect genetic regulation. ⊚ true ⊚ false

28)

Constitutive genes are those that have constant levels of expression. ⊚ ⊚

29)

Negative transcriptional regulation is conducted by activator proteins. ⊚ ⊚

30)

true false

true false

Repressor proteins are responsible for negative transcriptional regulation. ⊚ ⊚

true false

31) The term enzyme adaptationis used to describe an enzyme that appears in a living cell following exposure to a specific substrate. ⊚ ⊚

32)

true false

In the lac operon, the operator site is recognized by an activator protein. ⊚ true ⊚ false

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33)

cAMP is a small effector molecule. ⊚ true ⊚ false

34) The form of regulation that involves a physical change in the shape of an enzyme is called allosteric regulation. ⊚ ⊚

true false

35)

In the trp operon, tryptophan is a corepressor. ⊚ true ⊚ false

36)

In the E. coli riboswitch, mRNA gets translated when TPP is low. ⊚ true ⊚ false

37)

Because it stabilizes mRNA,antisense RNA enhances translation. ⊚ true ⊚ false

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Answer Key Test name: Chapter 16 Test Bank 1) C 2) D 3) A 4) E 5) C 6) A 7) D 8) C 9) B 10) B 11) A 12) C 13) B 14) A 15) A 16) A 17) D 18) C 19) A 20) C 21) E 22) A 23) A 24) A 25) A 26) TRUE Version 1

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27) FALSE 28) TRUE 29) FALSE 30) TRUE 31) TRUE 32) FALSE 33) TRUE 34) TRUE 35) TRUE 36) TRUE 37) FALSE

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Student name:_ 1) Feeding a specific diet to female mice may result in an increased proportion of pseudoagouti offspring if

A) the diet is low in folic acid and vitamin B 12.

B) the offspring carry the Avy allele.

C) the transposon at the Avy locus in the progeny is hypermethylated.

2)

Select all examples that are associated with expression of a single allele of a gene.

A) B) C) D)

3)

X-chromosome inactivation in female mammals Genomic imprinting Epigenetic modifications during cell differentiation Development of queen bees determined by exposure to royal jelly

How might heterochromatin silence the expression of genes? (Check all that apply.)

A) B) C) D)

Limiting the access of activators to the DNA Limiting the access of general transcription factors to the DNA Limiting the access of coactivators to the DNA Limiting the access of repressors to the DNA

4) Select the statements that are true about constitutive and facultative heterochromatin.

Check all that apply.

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A) The formation of facultative heterochromatin is reversible, whereas constitutive heterochromatin is stable.

B) Facultative heterochromatin typically activates genes, whereas constitutive heterochromatin silences genes.

C) Facultative heterochromatin may be found at many discrete sites between the centromere and telomere, whereas constitutive heterochromatin is usually located near the centromere and at the telomeres.

D) Both constitutive and facultative heterochromatin contain repeat sequences.

5)

Where is the IRE located in the regulation of ferritin translation?

A) B) C) D)

5′ end of DNA 5′ end of mRNA 3′ end of DNA 3′ end of mRNA

6) What effect will a mutation in IRPthat prevents it from binding to iron have on an individual?

A) B) C) D)

Ferritin will not be made, so iron intake must be maximized.

There will be excess ferritin, so iron intake must be lowered.

Transferrin will not be made, so iron intake must be maximized.

There will be excess transferrin, so iron intake must be lowered.

7) What general transcription factor is most often affected by regulatory transcription factors?

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A) B) C) D)

TFIIB TFIID TFIIE TFIIF

8) Regulatory transcription factors may influence gene expression in which of the following ways?

A) recruiting proteins to the promoter that enhance chromatin compaction B) by directly interacting with TFIII to inhibit its binding to the core promoter C) influencing the ability of the RNA polymerase to perform elongation D) directly recruiting RNA polymerase to the promoter without interacting with any other proteins

9)

Regulatory transcription factors may be regulated by A) B) C) D)

10)

phosphorylation. binding of carbohydrates. lipid modifications. sulfation.

Which of the following is incorrect regarding the glucocorticoid hormones?

A) B) C) D)

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They interact with receptors located in the plasma membrane of the cell.

After interacting with the receptor, they release HSP90.

The receptors form a homodimer that travels to the nucleus.

The homodimer interacts with GRE, activating transcription.

3

11)

Acetylation of histones results in A) B) C) D)

12)

CpG islands are associated with which of the following?

A) B) C) D)

13)

nucleosome location DNA methylation steroid hormone activity termination of translation

Genomic imprinting is a result of A) B) C) D)

14)

formation of an open chromatin structure. removal of histones from the histone octamer. formation of a closed chromatin structure. termination of transcription.

nucleosome location. histone activation.

DNA methylation. serine to leucine changes in the genetic code.

What is a major difference between the general and regulatory transcription factors?

A) General transcription factors only regulate the housekeeping genes while regulatory transcription factors control all of the other genes.

B) General transcription factors are essential for any transcription for all genes while regulatory transcription factors regulate transcription of specific genes.

C) General transcription factors act as heterodimers while regulatory transcription factors act as monomers.

D) General transcription factors regulate basal genes while regulatory transcription factors control all of the other genes.

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15)

Combinatorial factors include activators and repressors as well as

A) small effector molecules, proteins that alter the composition of nucleosomes, and DNA methylation.

B) small effector molecules, proteins that alter the composition of nucleosomes, DNA methylation, and basal transcription factors.

C) small effector molecules, proteins that alter the composition of nucleosomes, but not DNA methylation.

D) proteins that alter the composition of nucleosomes and DNA methylation, but not small effector molecules.

16)

Chromatin remodeling complexes alter nucleosomes, this occurs when

A) remodeling proteins bind to TFIID to promote or inhibit nucleosome formation.

B) complexes of ATP-dependent protein degrade histones and release the DNA from nucleosomes.

C) complexes of ATP-dependent proteins may reposition, evict, or change nucleosome composition.

D) remodeling proteins increase the supercoiling of the DNA, which can inhibit nucleosome formation.

17)

What are histone variants?

A) histone proteins that have slightly different amino acid sequences and have specialized functions B) histone proteins that have slightly different amino acid sequences but are found in nucleosomes throughout the chromosomes C) histone proteins that have been modified by acetylation D) histone proteins that have been modified by acetylation and phosphorylation

18)

What are some of the steps in eukaryotic transcriptional control?

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A) A transcriptional repressor recruits nucleosome remodeling proteins that alter nucleosome positioning, modification, and composition. This allows for the formation of a closed chromatin complex, and binding of general transcription factors and RNA polymerase II. Histone cores remain on the gene as transcript elongation occurs.

B) A transcriptional repressor recruits general transcription factors and RNA polymerase II without affecting nucleosome positioning.

C) A transcriptional activator recruits nucleosome remodeling proteins that alter nucleosome positioning, modification and composition. This allows for the formation of an open chromatin complex, and binding of general transcription factors and RNA polymerase II. Finally, as transcription elongation occurs, histone cores are evicted from the DNA to allow the RNA polymerase to transcribe the gene.

D) A transcriptional activator recruits general transcription factors and RNA polymerase II. Finally, as transcription elongation occurs, histone cores remain with the DNA.

19)

One definition of epigenetics would be

A) the study of irreversible changes in DNA sequence.

B) the study of reversible mechanisms that lead to changes in gene expression that involve changes in DNA sequence.

C) the study of irreversible mechanisms that lead to changes in gene expression that do not involve changes in DNA sequence.

D) the study of reversible mechanisms that lead to changes in gene expression that do not involve changes in DNA sequence.

20)

What are some types of the molecular changes that underlie epigenetic gene regulation?

A) B) C) D)

21)

depurination of cytosine and thymidine residues DNA methylation and chromatin remodeling DNA methylation and depurination of cytosine residues depurination and chromatin remodeling.

One difference between developmental and environmental epigenetic changes is that

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A) environmental epigenetic control is the result of developmental processes interacting with the environment.

B) developmental epigenetic control is a normal process that occurs in all offspring.

C) environmental epigenetic control is a normal process that occurs in all offspring.

D) the differences between the two lie in the mechanisms that each uses.

22)

What gene(s) is/are encoded in the Xic?

A) B) C) D)

23)

What genes appear to be controlled by the Xic?

A) B) C) D)

24)

Xce Xist Tsix Both the Tsix and Xist are genes are in the Xic region

Xist and Tsix Xist TsiX Xic

Expression of

A) B) C) D) E)

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would inhibit X-chromosome inactivation.

Xic Xist Tsix All of the answers are correct.

None of the answers are correct.

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25)

What gene is most responsible for X-chromosome inactivation?

A) B) C) D)

Xic Xist Tsix Xce

26) In imprinting of the Igf2 gene in mice, are the ICR and DMR methylatedon the maternal or paternal chromosome? Which chromosome expresses Igf2?

A) B) C) D)

27)

What is the molecular mechanism for imprinting a gene?

A) B) C) D)

28)

Paternal, paternal Paternal, maternal Maternal, paternal Maternal, maternal

Acetylation Nitration Phosphorylation Methylation

Most genes imprinted by methylation are silenced. What is one exception to this rule?

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A) B) C) D)

29)

Epigenetic inheritance may occur at which of the following stages?

A) B) C) D)

30)

Oogenesis Spermatogenesis Embryogenesis All of the answers are correct.

The differentially methylated region (DMR) is associated with which of the following?

A) B) C) D) E)

31)

H19 Igf2 Xist Agouti

X-chromosome inactivation Genomic imprinting Polycomb group (PcG) proteins mRNA stability All of the answers are correct.

Which of the following is an example of epigenetic inheritance?

A) B) C) D)

Expression of the Igf2 gene based on methylation of the ICR and DMR regions Inheritance of flower color as studied by Mendel Control of mRNA stability by an RNA-binding protein Activation of transcription by a general transcription factor

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32)

Which of the following are molecular mechanisms used in epigenetic gene regulation?

A) B) C) D)

33)

DNA methylation Covalent histone modification Chromatin remodeling All of the answers are correct.

Which of the following statements are correct?

A) Changes in gene expression based on environmental conditions are not considered normal while developmental changes are.

B) Environmental epigenetic changes can vary due to the exposure of the organism to different environmental conditions, while those programmed during development are the result of stimuli generated by the organism itself.

C) Environmental epigenetic gene regulation only occurs in reptiles and insects, while developmental epigenetic regulation occurs in all animals.

D) Environmental epigenetic gene regulation is typically reversible while developmental epigenetic gene regulation is typically not reversible.

34)

Which of the following is part of the process of X-chromosome inactivation?

A) Expression of Xist from both chromosomes at the start of the process B) Binding of multiple Xist transcripts to Xic on the X chromosome that will be inactivated C) Compaction of the active X chromosome into a Barr body D) Binding of Tsix transcripts to the X chromosome to be inactivated after the Xist transcripts binds to Xic

35) What is the difference between the trithorax (TrxG) and polycomb (PcG) complexes of proteins?

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A) TrxG proteins methylate lysine 27 on histone H3 while PcG complexes methylate lysine 4 on H3 B) Methylation of lysine at position 4 on H3 results in silencing of gene expression C) TrxG complex proteins activate gene expression by methylating lysine 4 on H3, while PcG complex proteins repress gene expression by methylating lysine 27 on H3.

D) Methylation of H3 on lysine 27 can result in activation of transcription

36) In cancer cells, one allele of a tumor suppressor gene called p53 is frequently mutated so that the protein is inactive, not produced, or deleted. The other allele usually has a normal sequence and the promoter remains intact, but the gene is not expressed. Sequencing with sodium bisulfite modification of DNA can be used to detect which cytosines are methylated. If cancer cell DNA is sequenced, which of the following results would be expected?

A) Cytosines in or near the promoter region will be methylated B) Cytosines in or near the promoter will not be methylated C) Cytosines in the coding region will have an increased methylation D) There will be no differences in the methylation pattern of the promoter of p53 from a cancer cell and a normal cell

37) Gene methylation can be detected through the use of restriction endonucleases. Usually these are used in pairs where one enzyme will digest only unmethylated DNA in its recognition sequence while the other is insensitive to methylation. Which of the following statements is correct?

A) The enzyme that is insensitive to methylation serves as a control to make sure the inability of the other enzyme to digest DNA is not due to a mutation.

B) These are used in pairs because experiments have to be replicated and this one way of performing a replication of the experiment.

C) The experimental design is flawed because there should be a third enzyme that would serve as a positive control D) This experimental design is able to identify the number ofmethylated C residues in a particular region of DNA.

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38) A common way of studying methylation in cells is to sequence DNA samples before and after modification with sodium bisulfite. The sodium bisulfite deaminates Cytosine residues, generating Uracil residues, therefore resulting in a change in the sequence as compared to the nonmodified DNA. Sodium bisulfite does not react with 5-methylcytosine so there will be no change in the sequence of those modified bases. When tumors are sequenced to study methylation patterns and epigenetic control, what would be the best control for sequencing using this technique?

A) Noncancerous tissue DNA from a different individual but from the same organ B) Liver DNA from the same individual that the cancer sample is from C) DNA from noncancerous tissue from the same individual in the same organ as the cancer D) A combination of DNA from liver, brain and muscle from the same individual that has the cancer

39) A queen bee is larger than the female worker bees in a hive. The queen develops functional ovaries that allow her to produce up to 2000 eggs a day. What determines which bees become queen bees?

A) B) C) D)

A single dominant allele of the Queen gene The larvae raised on a diet of royal jelly The larvae raised on a diet of pollen and nectar The egg that hatches first

40) Regions of chromatin that are more compact are called found in the nucleus.

A) B) C) D)

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; they are usually

heterochromatin; at the periphery euchromatin; centrally heterochromatin; centrally euchromatin; at the periphery

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41) Following mitosis, the two daughter cells usually have pattern of facultative heterochromatin and pattern of constitutive heterochromatin as was present in the mother cell.

A) B) C) D)

the same; the same the same; a different a different; the same a different; a different

42) Imagine all the histone H3 molecules in the cell have a mutation that changes their lysine-9 residue to an alanine. Assuming this cell could still live, what would be the result?

A) B) C) D)

The ability of the cell to form heterochromatin would be hindered.

The ability of the cell to form euchromatin would be hindered.

There would little effect because lysine and alanine are similar amino acids.

The ability of the cell to imprint genes would be hindered.

43) Which DNA sequence would also work in place ofthe following bi-directional enhancer?

5' GTTC 3' 3' CAAG 5' A) B) C) D)

5' GAAC 3'3' CTTG 5' 5' CTTG 3'3' GAAC 5' 5' CAAG 3'3' GTTC 5' 5' CTTC 3'3' GAAG 5'

44) What would be the result of a mutation in HSP90that prevented its binding to the glucocorticoid receptor?

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A) B) C) D)

Glucocorticoid receptor could not form a dimer.

The nuclear localization signal would no longer function.

Expression of the regulated genes would become constitutive.

The hormone would not be able to bind to the glucocorticoid receptor.

45) A particular cell contains all of the standard histones but lacks several histone variants.

Which of the following MAY be true of this cell?

A) B) C) D) E)

The cell will be completely unable to remodel its chromatin.

The cell will have an enhanced ability to chemically modify its histones.

The cell will be unable to express any of its genes at all.

The cell will express different sets of genes than other cells in the same organism.

The cell will not be able to divide.

46) A particular gene has a mutation in its NFR that causes it to bind abnormally tightly to histones. What effect would you expect this mutation has on the expression of this gene?

A) The expression of the gene would be abnormally high.

B) The effect on gene expression is impossible to predict.

C) The gene would encode a different protein instead.

D) The expression of the gene would be abnormally low or absent.

E) The promoter of the gene will be constantly available to transcription factors and RNA polymerase.

47) The Avyallele of the Agoutigene involves the insertion of a transposable element upstream from the normal Agoutipromoter. The transposable element carries a promoter that causes the overexpression of the Agoutigene. Mice carrying this allele tend to have coat colors that are more yellow than mice that don’t have this transposable element. If pregnant female mice are fed a diet that contains chemicals that inhibit DNA methylation, how would you expect that this diet would affect the coat color of offspring carrying the A vyallele?

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A) Their fur would be more yellow because theAgoutigene would tend to be over expressed.

B) Their fur would be more yellow because the Agoutigene would tend to be under expressed.

C) Their fur would be less yellow (darker brown) because theAgoutigene would tend to be over expressed.

D) Their fur would be less yellow (darker brown) because theAgoutigene would tend to be under expressed.

48) The Avyallele of the Agoutigene involves the insertion of a transposable element upstream from the normal Agoutipromoter. The transposable element carries a promoter that causes the overexpression of the Agoutigene. Mice carrying this allele tend to have coat colors that are more yellow than mice that don’t have this transposable element. If pregnant female mice are fed a diet that contains chemicals that increase DNA methylation, how would you expect that this diet would affect the coat color of offspring carrying the A vyallele?

A) Their fur would be less yellow (darker brown) because theAgoutigene would tend to be under expressed.

B) Their fur would be more yellow because the Agoutigene would tend to be over expressed.

C) Their fur would be more yellow because the Agoutigene would tend to be under expressed.

D) Their fur would be less yellow (darker brown) because theAgoutigene would tend to be over expressed.

49) In humans, a recent study has shown that maternal anxiety during pregnancy can diminish the methylation of the ICR located between the H19 and Igf2 genes of the embryo and fetus. How would this affect the expression of the maternal and paternal alleles of the Igf2 gene after the baby is born?

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A) It would not affect the expression of the paternal allele, but the maternal allele would be overexpressed.

B) It would not affect the expression of the maternal allele, but the paternal allele would be overexpressed.

C) It would not affect the expression of the paternal allele, but the maternal allele would be inhibited.

D) It would not affect the expression of the maternal allele, but the paternal allele would be inhibited.

50) Activator proteins bind to silencer sequences and repressor proteins bind to enhancer sequences. ⊚ ⊚

true false

51) A repressor protein would enhance the ability of TFIID to bind to the TATA box of the promoter. ⊚ ⊚

true false

52) Steroid hormones are an example of an effectorwhich regulates regulatory transcription factor activity. ⊚ ⊚

true false

53) DNA that contains actively transcribed genes would most likely contain chromatin in the closed configuration. ⊚ ⊚ Version 1

true false 16

54) Nucleosome location may be changed by a process called ATP-dependent chromatin remodeling. ⊚ ⊚

55)

true false

DNA methylation usually activates gene expression. ⊚ ⊚

true false

56)

The CpG islands upstream of housekeeping genes are unmethylated. ⊚ true ⊚ false

57)

Both parents usually imprint the same gene. ⊚ ⊚

58)

true false

Data suggests that the royal jelly fed to bee larvae causes inhibition of DNA methylation. ⊚ ⊚

true false

59) Vernalizationin Arabidopsis involves epigenetic changes ascold temperatures induce histone modifications that influence expression of genesfor germination or flowering in the spring when temperatures increase.

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⊚ ⊚

60)

Covalent histone modification is sometimes involved in cell differentiation. ⊚ ⊚

61)

true false

true false

Expression of a gene can be impaired if its promoter ishypermethylated. ⊚ ⊚

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true false

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Answer Key Test name: Chapter 17 Test Bank 1) [B, C] 2) [A, B] 3) [A, B, C] 4) [A, C, D] 5) B 6) D 7) B 8) A 9) A 10) A 11) A 12) B 13) C 14) B 15) A 16) C 17) A 18) C 19) D 20) B 21) B 22) D 23) A 24) C 25) B 26) A Version 1

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27) D 28) B 29) D 30) B 31) A 32) D 33) B 34) B 35) C 36) A 37) A 38) C 39) B 40) A 41) A 42) A 43) A 44) C 45) D 46) D 47) A 48) A 49) D 50) FALSE 51) FALSE 52) TRUE 53) FALSE 54) TRUE 55) FALSE 56) TRUE Version 1

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57) FALSE 58) TRUE 59) TRUE 60) TRUE 61) TRUE

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Student name:_ 1)

Select the four most common types of molecules to which ncRNAs bind.

A) Lipids B) Ions C) RNA D) Proteins E) Small molecules F) Water G) DNA

2)

Select functions that ncRNAs can perform. Choose all that apply.

A) Scaffold B) Guide C) Ribozyme D) Blocker E) Decoy F) Generation of cellular energy G) Alteration of protein function or stability

3) Select the ways in which modifications to the chromatin brought about by HOTAIR and associated proteinsare proposed to inhibit transcription.

A) Direct inhibition of RNA polymerase binding B) Formation of a double-stranded RNA structure between HOTAIR and the mRNA preventing translation C) Attraction of other chromatin-modifying enzymes D) Recruitment of repressors to the DNA

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4) Select possible outcomes after RISC binds to an mRNA during RNA interference.

Choose all that apply.

A) B) C) D) E)

5)

Inhibition of translation without mRNA degradation Expulsion of the RISC-mRNA complex from the cell Cell death mRNA degradation Movement of theRISC-mRNA complex to a P-body

In order of their occurrence what are the steps in RNA interference?

A) Processing of RNA transcript by DGCR8, transport to cytoplasm, processing by Dicer, binding to RISC, binding to target mRNA B) Processing by Dicer, processing of RNA transcript by DGCR8, transport to cytoplasm, binding to RISC, binding to target mRNA C) Processing of RNA transcript by DGCR8, processing by Dicer, transport to cytoplasm, binding to RISC, binding to target mRNA D) Processing of RNA transcript by DGCR8, binding to RISC, transport to cytoplasm, processing by Dicer, binding to target mRNA

6) In an in vitro translation assay radioactive amino acids can be incorporated into proteins and the proteins separated from the unincorporated amino acids (higher counts per minute indicates more protein synthesis). Given the following data what are the conclusions that you would come to?

RNA added to reaction

Amount of radioactivity in protein in counts per minute

Insulin mRNA alone

10,000

Insulin mRNA with antisense insulin RNA

3,000

Insulin mRNA with antisense fibronectin RNA

11,000

No mRNA

150

Insulin mRNA with double-stranded insulin RNA

500

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A) Double-stranded RNA is more efficient at inhibiting translation than antisense RNA B) Antisense RNA is more efficient at inhibiting translation than double-stranded RNA C) No conclusion about the efficiency of translation inhibition can be made as the positive and negative controls did not perform appropriately D) The experiment did not work since the antisense fibronectin did not have an effect on amino acid incorporation.

7) You are working in a lab that studies ncRNAs. You develop an assay to isolate ncRNAs and any associated molecules. You must then identify the molecules you have isolated. You find a molecule that is associated with an ncRNA that is not changed by protease, RNase, or DNase.

The associated molecule is most likely

A) B) C) D)

a small molecule. a protein. an mRNA. DNA.

8) A bacterial ncRNA called OxySregulates translation by binding to the ShineDalgarnosequence, which preventsribosome binding. What is the function of this ncRNA?

A) B) C) D)

9)

Blocker Decoy Ribozyme Scaffold

Ribozymes are RNA molecules with what type of activity?

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A) B) C) D)

10)

Catalytic Binding Protease Decoy

An RNA molecule with catalytic activity is called a(n)

A) B) C) D)

ribozyme. ncRNA. snRNA miRNA.

11) Select the ncRNA that facilitates the binding of telomerase to the telomere and acts as a template for DNA replication.

A) B) C) D)

TERC snRNA SRP RNA XistRNA

12) SmY RNAs are a family of ncRNAsin nematodes that are thought to be involved in a special type of RNA splicing. Based on this function, SmY RNAsaremost likely members of which category of ncRNAs?

A) B) C) D)

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snRNA tRNA snoRNA siRNA

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13)

The ncRNAHOTAIR recruits what type of proteins to target genes?

A) B) C) D)

Histone-modifyingcomplexes Transcription factors RNA polymerase Histone acetyltransferase

14) With which injection did Mello and Fire observe the lowest amount of mex-3 probe staining?

A) Injected with double-stranded RNA B) Injected with mex-3 antisense RNA C) Uninjected

15) Consider the two key roles that have been identified for SRP RNA. You have isolated a mutant SRP RNA. This RNA is able to perform its role as a scaffold for SRP proteins, but has lost its other function. How will this affect translation of SRP-containing proteins?

A) B) C) D)

16)

Partially translated proteins will be stuck in the cytosol.

Partially translated proteins will be stuck in the ER lumen.

Partially translated proteins will be stuck in the nucleus.

Translation will proceed normally.

What is the make up of the bacterial and eukaryotic SRP, respectively?

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A) B) C) D)

Bacterial: 1 ncRNA, 1 protein; Eukaryotic: 1 ncRNA, 6 proteins Bacterial: 1 ncRNA, 1 protein; Eukaryotic: 6ncRNAs, 1 protein Bacterial: 1 ncRNA, 6 proteins; Eukaryotic: 1 ncRNA, 6 proteins Bacterial: 6ncRNAs, 1 protein; Eukaryotic: 1 ncRNA, 1 protein

17) Into which gene of the CRISPR-Cas system are fragments of bacteriophage DNA inserted?

A) B) C) D) E)

18)

Crispr tracr Cas9 Cas1 Cas2

The CRISPR-Cas system in bacteria is most similar to what system in vertebrates?

A) B) C) D)

Immune system Circulatory system Nervous system Digestive system

19) Based on what you know about cartilage-hair hypoplasia, which strategy involving ncRNAsdo you anticipate would be useful in treating this disease?

A) B) C) D)

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A viral vector to restore a particular ncRNA Anti-ncRNA oligonucleotides Targeting of the ncRNA-processing machinery Use of DNA demethylating agents and histone deacetylase inhibitors

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20) You work in a lab that is studying PIWI proteins. You have isolated piRISCs and piRITSs from a cell. What would be one way to identify which complexes are piRISCs and which are piRITSs?

A) B) C) D)

21)

One of the proteins in the piRITS should have a nuclear localization signal.

One of the proteins in the piRITS should have nuclease activity.

One of the proteins in the piRITS should have acetylation activity.

One of the proteins in the piRITS should function as a transcription factor.

piRNA sequences are complimentary to

A) B) C) D)

mRNA from a transposable element. another RNA within the piRISC. the Crispr gene. bacteriophage DNA.

22) Abnormal levels of miRNAs are associated with nearly all forms of what type of human disease?

A) B) C) D)

23)

Neurological disorders Cancer Cardiovascular disease Gastrointestinal disease

What humandisease is associated with mutations in Drosha?

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A) B) C) D)

Familial amyotrophic lateral sclerosis Alzheimer disease Heart arrhythmias Lung cancer

24) What is the name of the type of chemical modification on anti-miRNA oligonucleotides used to inhibit miRNA function that involves ribose sugars with an extra bridge connecting the 2′ oxygen and the 4′ carbon?

A) B) C) D)

25)

Locked nucleic acids Antagomirs Enoxacin RNaseMRP

What effect would disabling exportin 5 have on RNA interference?

A) B) C) D)

In the absence of exportin 5 activity, no RNA interference could occur.

RNA interference would be unaffected by the lack of exportin 5 function. siRNAs could still carry out RNA interference, but miRNAs would not be made. miRNAs could still carry out RNA interference, but siRNAs would not be made.

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Answer Key Test name: Chapter 18 Test Bank 1) [C, D, E, G] 2) [A, B, C, D, E, G] 3) [A, C] 4) [A, D, E] 5) A 6) A 7) A 8) A 9) A 10) A 11) A 12) A 13) A 14) A 15) A 16) A 17) A 18) A 19) A 20) A 21) A 22) A 23) A 24) A 25) C

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Student name: TRUE/FALSE - Write 'T' if the statement is true and 'F' if the statement is false.

1) Reverse mutations involve changing a wild-type allele to a mutant allele. ⊚ true ⊚ false

2) The Lederbergs developed theprocess of replica plating to test whether advantageous mutations arose in response to the selective agent or were naturally present in the population before selection. ⊚ ⊚

3)

The mutation rate would be the same for all genes in a given culture. ⊚ ⊚

4)

true false

Silent mutations are possible due to the degenerate nature of the genetic code. ⊚ ⊚

7)

true false

A heritable change in the genetic material is called a mutation. ⊚ ⊚

6)

true false

The Ames test may be used to determine if an agent is a mutagen. ⊚ ⊚

5)

true false

true false

Ethyl methanesulfonate (EMS) is an example of a deaminating agent.

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⊚ ⊚

8)

true false

The most common genotype in a population is called the mutant genotype. ⊚ ⊚

true false

9) Breakpoints in chromosomes can lead to mutant phenotypes when they occur in the middle of a gene. ⊚ ⊚

true false

10) If an organism has a mutation in a gene that causes it to be sterile, a mutation in a second gene that restoresfertility is a suppressor mutation. ⊚ true ⊚ false

11) A mutation in a promoter region that causes the promoter sequence to more closely resemble the consensus sequence is called an up promoter mutation and decreases transcription. ⊚ true ⊚ false

12) TNRE repeats frequently result in the addition of extra histidine amino acids to the protein. ⊚ true ⊚ false

13)

Homologous recombination repair requires the presence of a sister chromatid. ⊚ true ⊚ false

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14)

Somatic cells are those that give rise to sperm and egg cells. ⊚ ⊚

15)

true false

An individual that is a genetic mosaic would be the result of a germ cell mutation. ⊚ ⊚

true false

16) When guanine is oxidized into 8-oxoguanine, it will base pairwith adenine during DNA replication, causing a transversion mutation. ⊚ true ⊚ false

17)

Mutagens change the structure of DNA, which may lead to mutations. ⊚ true ⊚ false

CHECK ALL THE APPLY. Choose all options that best completes the statement or answers the question.

18) Which of the following are causesof spontaneous mutations?

A) depurination B) deamination C) UV light D) tautomeric shifts E) nitrogenous base oxidation F) base analogs

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19) Consider the following DNA sequence, which codes for the first portionof a long proteinbeginning at the ATG (AUG in mRNA) start codon.

5' ATG CCC CGC AGTAGG GGG TGG AGA3' A mutation occurs changing this sequence to: 5'ATG CCC CGC AGTAGG GAG TGG AGA3' What type of mutation is this? (Check all that apply.)

A) Point mutation B) Transition C) Transversion D) Silent mutation E) Missense mutation F) Nonsense mutation G) Frameshift mutation

MULTIPLE CHOICE - Choose the one alternative that best completes the statement or answers the question.

20) In the following sequence of DNA, the italicized base has been mutated. What type of mutation is this?

5'- G A T C T C C G A A T T - 3'original strand 5'- G A T C T C C C A A T T - 3'mutated strand A) transition B) transversion C) neither

21) The wild-type eye color of Drosophila is red. A single-base mutation occurs that produces a white eye color. Which of the following is correct regarding this mutation?

A) B) C) D)

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It is an example of a mutation that alters protein function.

Individuals with white eyes are called reversions.

It would be an example of a silent mutation.

All of the answers are correct.

4

22) If a point mutation increases the number of offspring an organism can produce, it would be called a mutation.

A) neutral B) lethal C) beneficial D) conditional E) deleterious

23)

Mutations that change the configuration of a protein at a specific temperature are called mutations.

A) B) C) D)

24)

Anticipation is associated with which of the following?

A) B) C) D)

25)

nonsense mutations up-promoter mutations intergenic suppressors TRNE mutations

Most trinucleotide repeat expansion repeats involve expansion of which of the following?

A) B) C) D)

26)

neutral beneficial deleterious conditional

GAA CAG ATG any codon containing three of the same bases

Translocations and inversions may cause which of the following?

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A) B) C) D)

27)

TRNE anticipation position effect genome mutations

Which of the following is not an example of a cause of a spontaneous mutation?

A) DNA replication errors B) tautomeric shifts C) aberrant recombination D) UV light E) transposable elements

28) A translocation that moves a gene from an area of euchromatin to heterochromatin would typically cause in the expression of the gene.

A) a reduction B) an increase C) no change

29) The results of the replica plating experiments by the Lederbergs supported which of the following theories?

A) B) C) D)

30)

random mutation theory physiological adaptation theory both theories neither theory

Which of the following is correct regarding the rate of mutation?

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A) B) C) D)

Rates of mutation per cell generation typically range from 10−5 to 10−9.

Mutation rates are consistent across species.

Mutation rates are not influenced by environmental conditions.

Mutation rates are constant.

31) After screening a colony of bacteria for a given gene, you discover 100 mutant colonies out of 3 million total colonies. What is the mutation rate for this gene in the population?

A) 1.0 x 10 5 B) 1.0 x 10 -5 C) 3.0 x 10 5 D) 3.3 x 10 -5

32)

The conversion of cytosine to uracil in DNA is an example of A) B) C) D)

33)

depurination tautomeric shifts deamination None of the answers are correct.

The complete loss of either a guanine or adenine from DNA is an example of

A) B) C) D)

34)

.

.

depurination tautomeric shifts deamination None of the answers are correct.

What is an example of a suppressor mutation?

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A) an intragenic mutation that restores protein structure B) an intergenic mutation that increases the activity of a protein performing the same function as the mutated protein C) an intergenic mutation that alters a transcription factor to activate the expression of a compensating protein D) All of the answers are possible suppressor mutations.

35)

A temporary change in the structure of a nitrogenous base is called

A) B) C) D)

36)

.

depurination a tautomeric shift deamination None of the answers are correct.

Which environmental agent shown can induce mutations?

A) B) C) D)

UV radiation x-rays gamma rays All of the answers are correct.

37) Which of the following integrate into the double-helix of DNA, interfering with DNA replication?

A) B) C) D)

38)

EMS nitrous acid 5BU acridine dyes

Which of the following is an example of a base analog?

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A) EMS B) nitrous acid C) 5BU D) nitrogen mustards E) acridine dyes

39)

What process repairs damage from UV radiation?

A) homologous recombination repair B) direct repair C) base excision repair D) mismatch repair E) nucleotide excision repair F) nonhomologous end joining (NHEJ)

40)

Which DNA repair process utilizes MutL, MutH, and MutS proteins in E. coli?

A) homologous recombination repair B) direct repair C) base excision repair D) mismatch repair E) nucleotide excision repair F) nonhomologous end joining (NHEJ)

41)

Which DNA repair process often leads to deletion in chromosome sequence?

A) homologous recombination repair B) direct repair C) base excision repair D) mismatch repair E) nucleotide excision repair F) nonhomologous end joining (NHEJ)

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42)

Which DNA repair process identifies daughter strands by methylation?

A) homologous recombination repair B) direct repair C) base excision repair D) mismatch repair E) nucleotide excision repair F) nonhomologous end joining (NHEJ)

43)

How does position effect influence gene expression?

A) The movement of thegenetic material on the chromosome by inversions or translocations may place acoding sequence near a new regulatory region, thus activating the expression ofthe gene.

B) The movement of the gene may place it into a region that is highly condensed.

C) The movement of a gene may remove it from its normal promoter, thussilencing the gene.

D) All of the answers are correct.

44) In the nucleotide excision repair system, which of the following proteins is responsible for recognizing a thymine dimer to be repaired?

A) B) C) D)

UvrA/UvrB UvrC UvrD None of the answers are correct.

45) Which process is used to replicateDNA that containsdistortions due to unrepaired DNA damage?

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A) B) C) D)

46)

Translesion synthesis Homologous recombination repair Nonhomologous end joining Nucleotide excision repair

What would be a set of anticipated results from a "Lederberg" experiment?

A) Total number of colonies on a plate: 1500 Total number of resistant colonies on replica plate with T1: 15 B) Total number of colonies on a plate: 1500Total number of resistant colonies on replica plate with T1: 1500 C) Total number of colonies on a plate: 500Total number of resistant colonies on replica plate with T1: 1500 D) Total number of colonies on a plate: 1500Total number of resistant colonies on replica plate with T1: 0

47) There is a spectrum of syndromes in humans known as Xeroderma pigmentosum. The individuals that have XD most commonly have mutations in their nucleotide excision DNA repair mechanisms that make them particularly susceptible to environmental mutagens such as UV light. Individuals have to be careful with how much sunlight they are exposed to since they have an extremely elevated chance for developing skin cancer. There have been several cell lines that have been established from XD patients that can be studied in tissue culture. In an experiment, several different cell lines of unknown origin were tested for their ability to undergo unscheduled DNA synthesis (UDS), an assay for DNA repair. In this essay, the amount of radioactive nucleotides that are incorporated into DNA after the cell sustains a mutagenic event are measured. The amount of radioactivity incorporated is measured by the number of counts per minute (CPM). Below is a table from such an experiment. Which cell line is most likely from XD patient(s)?

Cell line

CPM in UDS assay

CPM if no mutagenic event

A

2,500

2,750

B

300,000

5,000

C

45,000

25,000

D

275,000

50,000

Wild-type control

750,000

75,000

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A) B) C) D)

A B C D

48) Beechdrops is a parasitic plant that cannot perform photosynthesis but relies on its host the Beech tree. However, beechdrops still retains many if not all of the genes for photosynthesis.

Snapdragons and gladiolas are common garden flowers that rely on their ability to perform photosynthesis. If you were to compare the gene sequences for these three plants for ribulose1,5-bisphosphate carboxylase/oxygenase (RuBisCO) a protein necessary for photosynthesis what would you predict?

A) The differences between gladiolas and snapdragons would most likely be silent mutations while those in beechdrops may be silent or missense B) Since these three plants are not highly related the sequences for RuBisCO would be very different between them C) The differences between gladiolas and snapdragons would most likely be in the second nucleotide of codons while beechdrops would have a higher number of mutations in the third nucleotide of the different codons D) The differences between gladiolas and snapdragons would most likely be missense mutations while those in beechdrops may be silent or missense

49)

Which types of mutations are least likely to be subjected to natural selection?

A) B) C) D)

Silent Missense Nonsense Insertion

50) Consider the following DNA sequence, which codes the first portionof a long proteinbeginning at the ATG (AUG in mRNA) start codon.

5' ATG CCC CGC AGTAGG GGG TGG AGA 3' Which of the mutatedsequences listed is most likely to be a deleterious mutation?

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A) 5'ATG CCC CGC AGTAGG GAG TGG AGA3' B) 5'ATG CCC CGC AGTAGG GGG TGAAGA3' C) 5'ATG CCGCGC AGTAGG GGG TGGAGA3'

51) You are a doctor tasked with diagnosing a patient that comes to you with mental impairment. Upon performing genetic tests, you find he has approximately 300 repeats of GCC in one of his genes. His father and grandfather had similar mental impairment but were never diagnosed with a condition prior to their death. His grandmother, mother, and sisters are unaffected. What disease do you predict that this patient has?

A) SBMA B) HD C) SCA1 D) FRAXA E) FRAXE

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Answer Key Test name: Chapter 19 Test Bank 1) FALSE 2) TRUE 3) FALSE 4) TRUE 5) TRUE 6) TRUE 7) FALSE 8) FALSE 9) TRUE 10) TRUE 11) FALSE 12) FALSE 13) TRUE 14) FALSE 15) FALSE 16) TRUE 17) TRUE 18) [A, B, D, E] 19) [A, B, E] 20) B 21) A 22) C 23) D 24) D 25) B 26) C Version 1

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27) D 28) A 29) A 30) A 31) D 32) C 33) A 34) D 35) B 36) D 37) D 38) C 39) E 40) D 41) F 42) D 43) D 44) A 45) A 46) A 47) A 48) A 49) A 50) B 51) E

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Student name:_ 1)

Select items that describe a stage of a PCR reaction. Choose all that apply.

A) DNA is heated to a temperature that causes the strands to separate.

B) Primers bind to complementary sequences on the template DNA. C) DNA gyrase enzyme relieves the supercoiling ahead of the replication fork.

D) The DNA polymerase synthesizes a new strand of DNA that is complementary to the template DNA.

2)

Which of the following cells are pluripotent? (Check all that apply) A) neuronal cells B) embryonic stem cells C) epithelial cells D) red blood cells E) embryonic germ cells F) adult stem cells

3) Small circular pieces of bacterial DNA that are used as vectors in cloning experiments are called .

A) B) C) D)

4) a(n)

phages viruses plasmids chromosomes

A plasmid that contains separate origins of replication for two different species is called .

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A) B) C) D)

R factor expression vector viral vector shuttle vector

5) Site-directed mutagenesis allows researchers to produce a mutation at a cloned DNA segment.

within a

A) specific site B) random site C) semilocalized site that can later be identified by sequencing

6) Select the technique that can be used to see how a specific mutation affects the expression of a gene and the function of a protein.

A) B) C) D)

7)

What is the origin of restriction endonucleases?

A) B) C) D)

8)

site-directed mutagenesis Western blotting real-time PCR DNA sequencing

They are part of DNA repair mechanisms in eukaryotic cells.

They are a defense mechanism against viruses in bacteria.

They are replication enzymes of yeast.

They are transposable elements of Drosophila.

Which of the following is an example of a palindromic DNA sequence?

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A) B) C) D)

9)

DNA ligase is needed in a cloning experiment for what reason?

A) B) C) D)

10)

to proofread the vector for mistakes to covalently link the sugar-phosphate backbones of the two DNA fragments to allow the vector to enter into the cell to act as a selectable marker in the identification of cells that took up the vector

What do competent cells do?

A) B) C) D)

11)

5′ - ATCGAC - 3′3′ - TAGCAG - 5′ 5′ - ATCATC - 3′3′ - ATCATC - 5′ 5′ - GCCGCC - 3′3′ - CGGCGG - 5′ 5′ - CTGCAG - 3′3′ - GACGTC - 5′

resist transfection by a viral vector rejoin DNA strands without the aid of DNA ligase utilize transposons as a vector take up DNA from the external environment

Which of the following would contain both vector DNA and chromosomal DNA?

A) B) C) D)

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recircularized vector hybrid vector both vectors neither vector

3

12) In a cloning experiment with chromosomal DNA, you use a vector that contains a lacZ gene as a selectable marker. If the competent cells are grown on X-Gal and IPTG, which colonies would contain cloned chromosomal DNA? A) B) C) D)

13)

cDNA is made using what as the starting material?

A) B) C) D)

14)

the white colonies the blue colonies half the total colonies none of the colonies

plasmid vectors viral DNA chromosomal DNA RNA

What is the purpose of RNaseH in the making of cDNA?

A) B) C) D)

It copies the RNA into a complementary DNA strand.

It reforms the sugar-phosphate backbone.

It generates short RNAs that are used as primers.

It proofreads the cDNA for errors.

15) Which of the following individual(s) developed the process of the polymerase chain reaction?

A) B) C) D)

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Watson and Crick McClintock Mullis Pauling

4

16)

What occurs during the annealing stage of a PCR reaction?

A) B) C) D)

The DNA strands separate The DNA polymerase copies the template DNA. The primers bind to the template DNA. The reaction stops.

17) In the DNA sequence 5′ GCGCAAATTTCCCTATACCC 3′, you wish to use sitedirected mutagenesis to change the T at position 8 to a C. Select the primer that is likely to give you the best results.

A) 5′ GCGCAAACTTCCCTATACCC 3′ B) 5′ CGCGTTTGAAGGATATGGG 3′ C) 5′ GCGCAGGCCTCCCTATACCC 3′

18)

What is the purpose of RT(reverse transcriptase)-PCR? A) B) C) D)

19)

to conduct a PCR reaction without the use of primers to follow the amount of a specific PCR product in real-time to quantify RNA in a cell to nonspecifically amplify DNA

What is a DNA library?

A) B) C) D)

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a collection of vectors, each containing a fragment of the genome or DNA of interest a complete sequence of the genetic material in an organism an electronic file that may be used for additional genetic analysis a collection of mRNA from a given cell

5

20)

A cDNA library differs from a genomic library in which way?

A) B) C) D)

It consists solely of RNA molecules.

It contains many copies of the gene of interest.

It contains only coding sequences, not introns.

It is typically 10-100 times the size of a DNA library.

21) Which procedure is used to identify a specific RNA within a mixture of many RNA molecules?

A) B) C) D)

Western blotting Northern blotting Southern blotting Eastern blotting

22) Which procedure is used to identify a specific protein within a mixture of many different protein molecules?

A) B) C) D)

23)

Western blotting Northern blotting Southern blotting Eastern blotting

Antibodies can be used as probes for which of the following techniques?

A) B) C) D)

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Western blotting Northern blotting Southern blotting Eastern blotting

6

24)

Which technique is best suited to identify DNA-protein interactions?

A) B) C) D)

25)

Western blotting PCR electrophoretic mobility shift assay restriction enzyme analysis

Which technique uses dideoxynucleotides?

A) B) C) D)

RT-PCR DNA sequencing restriction digest gene cloning

26) The use of fluorescently-labeled dideoxynucleotides is a characteristic of what type of sequencing?

A) automated DNA sequencing B) base-specific DNA cleavage

27)

Detection of a product in real-time PCR depends on A) B) C) D)

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the separation of the reporter from the quencher. the reporter and quencher remaining in close proximity to each other. the reporter binding to the quencher.

Taq polymerase destroying the reporter.

7

28) In real-time PCR, the initial template concentration is calculated using the of PCR.

phase

A) exponential B) linear C) plateau

29) You are performing a mobility shift assay with a protein complex that is composed of proteins A, B, C, & D, each of which can bind to the DNA of interest. You run the gel with the lanes as follows: Lane 1: Protein A + DNA Lane 2: Proteins A + B + DNA Lane 3: Proteins A + B + C + DNA Lane 4: Proteins A + B + C + D + DNA In which lane will the DNA run closest to the top of the gel?

A) B) C) D)

Lane 4 Lane 3 Lane 2 Lane 1

30) Although Dolly was only three years old, her chromosomes had the length of a 9- to 10year-old sheep. Which of the following best describes why this occurred?

A) The cell lines that created Dolly were aged prematurely in the lab and thus became longer.

B) Nonhomologous recombination occurred, which caused Dolly's chromosomes to shorten.

C) A mutation enhanced the rate of aging of Dolly's chromosomes.

D) Dolly was not a clone.

E) The telomeres of the somatic cells that Dolly originated from were already shortening.

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31) Which of the following terms describes a cell that can form any other cell of the organism?

A) B) C) D)

32)

Hematopoietic stem cells are A) B) C) D)

33)

pluripotent totipotent unipotent All of the answers are correct.

.

pluripotent totipotent unipotent multipotent

Bone marrow transplants typically use what type of cells?

A) B) C) D)

embryonic stem cells embryonic germ cells hematopoietic stem cells None of the answers are correct.

34) The origins of which of the following cell types creates the least amount of ethical debate?

A) B) C) D)

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embryonic germ cells hematopoietic stem cells embryonic stem cells All of the answers are equal.

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35)

Which of the following is an example of a pluripotent cell?

A) B) C) D) E)

36)

embryonic stemcells red blood cells fetal heartcells umbilical cordblood nerve cells

What is a property of stem cells?

A) B) C) D)

They can differentiate into one or more specialized cell types.

They can form any tissue in the body.

They are only capable of one to two cell divisions.

They will only form hematopoietic cells.

37) A scientist does not have an antibody to detect the protein of interest, but wants to know if the gene is expressed at low levels. What technique could the researcher use to detect, but not quantify the expression of the gene of interest?

A) B) C) D)

Western Blotting Real-time PCR Cloning Reverse transcriptase PCR

38) A researcher is interested in whether or not a transcription factor binds to a specific site on a strand of DNA. However, the samples accidentally contain SDS. What will the electrophoretic mobility shift assay results show?

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A) The protein will bind the DNA and retard the movement of the DNA through the gel.

B) The protein will not bind to the DNA and the movement of the DNA through the gel will not be affected.

C) The protein will bind the DNA and the movement of theDNA through the gel will not be affected.

D) The protein will not bind the DNA and the movement ofthe DNA through the gel will be affected.

39) You are using CRISPR-Cas technology to introduce a single nucleotide change in a gene of interest in living cells. You are designing your experiment. In which component of the reaction will you engineer the single nucleotide change?

A) B) C) D) E)

Donor DNA Spacer RNA Repeat RNA Linker RNA tracrRNA

40) You are sequencing the following DNA molecule 3'- GACTACCGAAATTAT -5'.

Assume the annealing primer binds immediately prior to this sequence. What nucleotide will be found closest to the bottom of the sequencing gel?

A) B) C) D)

G A C T

41) You are carrying out a sequencing reaction. You remember to add ddNTPs, but forget to add dNTPs. What will the outcome of your sequencing reaction be?

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A) Sequencing will proceed normally.

B) No sequencing will take place at all.

C) Sequencing will begin, but will not be able to proceed past the first nucleotide because the ddNTPs cannot be used for chain extension.

D) Four fragments will be made corresponding to the first A, T, C, and G in the sequence, but after that no more fragments will be made.

42) Prior to the discovery of Taq polymerase, PCR could be carried out in a more laborious manner by adding a non-thermostable polymerase to each round of the reaction. In order for this reaction to work, when should the non-thermostable polymerase be added?

A) Before denaturation B) After primer annealing C) After primer extension

43) The restriction enzyme BamHI recognizes the sequence: 5'-G/GATCC-3' 3'-CCTAG/G-5' It cuts between the G and G as noted by the slashes. Following the digestion of chromosomal DNA and vector DNA with BamHI, let's suppose that one fragment of chromosomal DNA gets inserted into a vector. No DNA ligase has been added. If both of the sticky ends holding the chromosomal DNA fragment are binding to the sticky ends of the vector, what is holding the two pieces of DNA together?

A) B) C) D)

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10 hydrogen bonds 16 hydrogen bonds 20 hydrogen bonds 26 hydrogen bonds

12

44) A primer used in dideoxy DNA sequencing is 20 nucleotides long and has the following sequence: 5'-GGATCCATGACTAGTCCGAC-3' A segment of DNA is cloned into a vector and then the vector DNA is denatured and subjected to dideoxy DNA sequencing. Below is the DNA sequence from a region of the vector. The primer-annealing site is shown in bold and underlined.

3'-CCCGATCGGCCTAGGTACTGATCAGGCTGAATGACT-5' Based on the sequence above, what would be the sizes of the bands (i.e. the number of nucleotides in each band) containing dideoxyT? A) 1, 2, and 5 B) 21, 22, and 25 C) 3 and 7 D) 23 and 27

45)

Restriction endonucleases recognize specific DNA sequences. ⊚ ⊚

true false

46) Since cDNA is made from RNA, it lacks introns and contains only the coding sequence for the functional protein. ⊚ ⊚

true false

47) The process of ligating DNA sequences into a vector and introducing the construct into a cell to create many copies is referred to as gene cloning. ⊚ ⊚

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true false

13

48)

A vector is a small segment of DNA that a gene of interest is inserted into for cloning. ⊚ ⊚

true false

49) Molecular biologists use restriction enzymesto amplify a specific section ofDNA within the genome. ⊚ ⊚

50)

true false

The first organism ever cloned by researchers was Dolly the sheep in 1997. ⊚ true ⊚ false

51) The primary interest of genetic research into stem cells is the desire to clone a human being. ⊚ true ⊚ false

52)

Unipotent cells may differentiate into all other cell types of the body. ⊚ ⊚

53)

true false

An organism that can be regenerated from somatic cells is called multipotent. ⊚ true ⊚ false

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Answer Key Test name: Chapter 20 Test Bank 1) [A, B, D] 2) [B, E] 3) C 4) D 5) A 6) A 7) B 8) D 9) B 10) D 11) B 12) A 13) D 14) C 15) C 16) C 17) A 18) C 19) A 20) C 21) B 22) A 23) A 24) C 25) B 26) A Version 1

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27) A 28) A 29) A 30) E 31) B 32) D 33) C 34) B 35) A 36) A 37) D 38) B 39) A 40) C 41) C 42) B 43) C 44) B 45) TRUE 46) TRUE 47) TRUE 48) TRUE 49) FALSE 50) FALSE 51) FALSE 52) FALSE 53) FALSE

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Student name:_ 1)

Which might be used as a marker for linkage mapping? (Check all that apply.)

A) AFLP B) RFLP C) microsatellite D) FISH E) PCR F) SNP G) STS

2)

Which researcher(s) are using a metagenomic approach?

A) A medical researcher studies the DNA of mixed bacterial populations from the intestines of patients with different diseases.

B) A plant biologist sequences the genomes of broccoli, cauliflower, and other Brassica species to determine which genes are responsible for their different features.

C) A developmental geneticist clones the mutated gene from a Drosophila strain with disrupted wing development.

D) An environmental researcher sequences DNA from collections of microbes living in polluted coastal waters, which cannot be cultured in the lab.

E) A pharmaceutical developer screens through DNA found in soil samples for novel antibiotics.

3)

For what situation(s) would a metagenomics strategy be appropriate?

A) to study organisms that cannot be cultivated in the lab B) to thoroughly study the genome of a new model organism C) to ensure that an entire genome is sequenced D) to study multiple species that interact in the environment E) to determine whether a patient carries a disease-causing allele F) to do a broad survey of genes from different species in a certain habitat

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4)

Select all advantages of RNA-Seq as compared to DNA microarrays.

A) B) C) D) E)

Better detection of transcripts present at low levels Identifies boundaries of introns and exons to determine splice variants More accurately quantifies the amount of each transcript Does not require isolation of RNA from the cell Does not require cDNA synthesis

5) A researcher isolates mRNA from healthy and diseased liver biopsies. Each mRNA sample is converted into cDNA and subjected to next-generation sequencing. Select all statements that describe this experiment and the possible results.

A) The researcher can identify transcripts that are more abundant in healthy liver than in diseased liver, which are candidate genes for further study.

B) The researcher can identify transcripts that are less abundant in healthy liver than in diseased liver, which are candidate genes for further study.

C) This is an RNA-Seq experiment.

D) This is a DNA microarray analysis.

E) Some genes' transcriptsmay be present in both samples but have different splice forms in healthy and diseased liver.

F) The intron sequences will be more abundant in the diseased liver samples.

G) The researcher should have combined the healthy and diseased mRNA samples before making cDNA, so that only one sample would need to be sequenced.

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6) A woman has a child and wants to determine who the father of the child is. There are four possible fathers. This group of individuals is analyzed with regard to three different STSs: STS-1 is 215 bp and 199 bp, STS-2 is 144 bp and 122 bp, and STS-3 is 101 bp and 85 bp. The child (C), mother (M), and 4 potential fathers (D1-D4) show the following results. Select the individual(s) who are heterozygous for STS-3. (Check all that apply.)

A) C B) M C) D1 D) D2 E) D3 F) D4

7) Which of the following procedures best determines the relative order, but not precise location, of a series of genes on a chromosome?

A) B) C) D)

physical mapping cytogenetic mapping linkage mapping shotgun sequencing

8) uses the unique banding patterns of a chromosome to determine the general location of a gene.

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A) B) C) D)

9)

What does physical mapping provide that the other types of genetic mapping do not?

A) B) C) D)

10)

cytogenetic mapping physical mapping linkage mapping sequencing

A site in the genome that varies between members of the population is said to be .

A) B) C) D)

12)

recombination distances between genes number of nucleotide base pairs between genes relationship of the gene to specific chromosomal banding patterns frequencies of alleles in a population

What is in situ hybridization used for?

A) B) C) D)

11)

Physical mapping Cytogenetic mapping Linkage mapping Shotgun sequencing

monomorphic polymorphic trimorphic neomorphic

If all members of a population have the identical DNA segment, it is said to be

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.

4

A) B) C) D)

13)

monomorphic polymorphic trimorphic neomorphic

The most common microsatellite in humans is

.

A) (AT) 100 B) (CA) n C) (TG) n D) (CAT) n E) AUG

14) The method of sequencing in which DNA fragments to be sequenced are randomly generated from larger DNA fragments is

A) B) C) D)

15)

cytogeneticmapping. shotgun sequencing. southernblotting. dideoxyblotting.

The Human Genome Project was a

A) B) C) D)

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-year effort.

5 13 3 50

5

16)

The human genome consists of approximately

base pairs of DNA.

A) 100,000 B) 1 million C) 1 billion D) 3 billion E) 2 trillion

17) Pyrosequencing is an example of a high-throughput, next-generation sequencing technology. What is a crucial detail of the procedure?

A) B) C) D)

including the enzyme luciferase mixing the solution of four nucleotides together in even amounts removing nucleotides from the target DNA one at a time using large wells to accommodate multiple beads

18) After a metagenomics researcher collects a mixed sample of DNA, what next step allows this DNA to be studied?

A) The DNA fragments are each ligated into a cloning vector to create a DNA library.

B) The pieces of DNA are ligated together into one complete "genome" so they can be sequenced.

C) The species responsible for each DNA are identified and individually cultured.

D) DNA is reconstructed into chromosomes by using the human genome as a scaffold.

19)

Gene chips are being used for which of the following purposes?

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A) B) C) D) E)

to examine patterns of gene expression under various environmental conditions to establish levels of genetic variation to determine gene expression specific to certain cell types to identify species of bacteria All of these choices are possible.

20) In the screening of a DNA microarray, detection of a fluorescent signal from one of the genes indicates which of the following?

A) B) C) D)

The labeled cDNAhybridized to that particular gene.

The labeled cDNAdid not hybridize to that particular gene.

The experiment was not conducted correctly.

None of these choices are correct.

21) make it possible for researchers to study how an entire genome responds to an environmental stimuli.

A) B) C) D)

22)

Gene knockouts PCR reactions DNA microarrays Shotgun sequences

To identify novel DNA binding sites for a protein, it would be best to use A) B) C) D)

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.

pyrosequencing microsatellites chromatin immunoprecipitation ChIP-chip

7

23)

Gene knockouts in mice can be useful because

A) they can be used to develop models of human disease.

B) they identify DNA binding proteins C) they limit mRNA production both from the knockout as well as the normal allele in diploids.

D) only genes that are actively transcribed in the embryo are susceptible to being knocked out.

24)

In shotgun sequencing, how is a complete sequence of a genome reconstructed?

A) The entire genome is directly sequenced chromosome by chromosome.

B) The genome is fragmented and sequences of each fragment are arranged so that overlapping regions are identified.

C) The sequences of a large number of individuals is obtained and “averaged” to obtain the complete sequence.

D) The position of each marker and gene are first mapped relative to regions for which the genomic sequence is not known.

25) The experimental process of determining the relative location of genes or other segments of DNA along individual chromosomes is called

A) B) C) D)

mapping. metagenomics.

DNA sequencing. proteomics.

26) An application of FISH technique, where multiple DNA probes are labeled with different fluorescent molecules that bind to a chromosome, is called . Using this technique, researchers can determine the sites of the DNA segments along the chromosome.

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A) B) C) D)

27)

chromosome painting Giemsa staining pyrosequencing physical mapping

A complex collection of genes from a specific environmental sample is a A) B) C) D)

.

microgenome metagenome linkage map microarray

28) Which of the following samples would NOT be suitable for a metagenomics-based approach?

A) Water samples from different bodies of water with different temperatures, salinities, and acidities B) A soil sample from a farm where a new planting schedule has been implemented C) A water sample from an industrial site that has been contaminated with pollutants D) A fecal sample from a patient with irritable bowel syndrome E) A homogeneous population of skin cancer cells collected from a patient biopsy

29) You are sequencing the genome of the yeast S. cerevisiae, with a genome size of 12,100,100 bp, using shotgun sequencing. Your sequencer malfunctions when only 24,200,000 bp in total have been sequenced. How much of the genome should have been sequenced at this point?

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A) 100% B) 0% C) 13.5% D) 86.5% E) 0.135% F) 0.865%

30) A woman has a child and wants to determine who the father of the child is. There are four possible fathers. This group of individuals is analyzed with regard to three different STSs: STS-1 is 215 bp and 199 bp, STS-2 is 144 bp and 122 bp, and STS-3 is 101 bp and 85 bp. The child (C), mother (M), and 4 potential fathers (D1-D4) show the following results. Which individual is the father of the child?

A) B) C) D)

D1 D2 D3 D4

31) You are carrying out a chromosome painting experiment. You have single-stranded DNAs that bind to 6 different regions of the X chromosome. You inadvertently label the DNA probes all with the same colored fluorescent molecule, which happens to be red in this experiment. What will be the result of your experiment?

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A) B) sequence.

C) D) probes.

The DNA probes will be unable to bind.

The probes will bind, but you will not be able to tell which spot corresponds to which The entire X chromosome will be labeled red along its entire length.

The probes will bind normally, and you will still be able to distinguish between the

32) Imagine that when shotgun sequencing was initially used to sequence the H. influenzae genome, the sonicator didn't work. How would this affect the sequencing experiment?

A) Because the DNA fragments would be longer than what is usually carried by a plasmid DNA, no sequence would be obtained.

B) Because the DNA fragments would be longer than what is usually carried by a plasmid DNA, there would likely be sequences missing in the sequencing results due to difficulties with DNA replication.

C) Because the DNA fragments would be longer than what is usually carried by a plasmid DNA, sequencing would occur more quickly than usual.

D) Because the DNA fragments would be shorter than what is usually carried by a plasmid DNA, there would likely be sequences missing in the sequencing results due to difficulties with DNA replication.

E) Because the DNA fragments would be shorter than what is usually carried by a plasmid DNA, no sequence would be obtained.

F) Because the DNA fragments would be shorter than what is usually carried by a plasmid DNA, sequencing would occur more quickly than usual.

33)

One map unit of distance always corresponds to 106 base pairs. ⊚ true ⊚ false

34) A microsatellite marker may be used to predict the likelihood that an individual carries a trait for a specific disease.

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11

⊚ ⊚

true false

35) Part of the Human Genome Project involved sequencing the genomes of model organisms, including bacteria and a plant. ⊚ true ⊚ false

36)

RFLP stands for restriction fragment linked position. ⊚ ⊚

37)

In a FISH procedure, the probe being used is labeled with a fluorescent tag. ⊚ ⊚

38)

true false

true false

cDNA uses chromosomal DNA as its template. ⊚ ⊚

true false

39) A small glass chip that contains thousands of microscopic pieces of DNA is called a DNA microarray. ⊚ ⊚

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true false

12

40) to it.

The termgene knockout refers to a genealteration that inhibitsthe genes that are adjacent ⊚ ⊚

true false

41) Shotgun sequencing is precise enough to achieve a complete genomic sequence without the need for other mapping methods. ⊚ true ⊚ false

42) In sequencing by synthesis (SBS) methods, the sequence of a sample is directly determined as nucleotides are incorporated into a DNA strand by DNA polymerase. ⊚ true ⊚ false

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Answer Key Test name: Chapter 21 Test Bank 1) [A, B, C, F, G] 2) [A, D, E] 3) [A, D, F] 4) [A, B, C] 5) [A, B, C, E] 6) [A, E, F] 7) C 8) B 9) B 10) A 11) B 12) A 13) B 14) B 15) B 16) D 17) A 18) A 19) E 20) A 21) C 22) D 23) A 24) B 25) A 26) A Version 1

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27) B 28) E 29) D 30) C 31) B 32) B 33) FALSE 34) TRUE 35) TRUE 36) FALSE 37) TRUE 38) FALSE 39) TRUE 40) FALSE 41) FALSE 42) TRUE

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Student name:_ 1)

What can cause epigenetic changes? (Select all that apply.)

A) B) C) D)

2)

DNA methylation Histone acetylation Histone methylation Histone phosphorylation

What are the potential advantages for personalized medicine? (Select all that apply.)

A) More effective dosing B) Knowing which drugs will be the most effective C) Being able to classify cancers on the way the cells look under a microscope as opposed to their genetic makeup D) Treatment of patients based on their specific physiology

3) This pedigree involves a human disease that is inherited in a simple Mendelian manner.

Which pattern(s) of inheritance is consistent with this pedigree? Check all that apply.

A) B) C) D)

Autosomal recessive Autosomal dominant X-linked recessive X-linked dominant

4) Select the genetic changes that are associated with the development of colorectal cancer.

(Check all that apply.)

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A) Loss of APC B) Activation of ras C) Loss of myc D) Loss of DCC E) Activation of NF1 F) Loss of p53

5)

Select the viruses that cause cancer in humans. Choose all that apply.

A) B) C) D) E)

6)

Rous sarcoma virus Hepatitis B virus Papillomavirus Hardy-Zuckerman-4 feline sarcoma virus Herpes virus

For a single-gene disorder, the concordance among monozygotic twins should be A) 0%.

B) 25%.

C) 50%.

D) 75%.

E) 100%.

7)

For a recessive single-gene disorder, the concordance among dizygotic twins should be A) 0%.

B) 25%.

C) 50%.

D) 75%.

E) 100%.

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8) Which of the following would indicate that a disease has a genetic, rather than environmental, cause?

A) The disease has a specific age of onset.

B) The disease is cured by taking an antibiotic.

C) The disease occurs at the same rate in individuals that are exposed to a similar environment.

D) The disease occurs as frequently between close relatives as it does in unrelated individuals.

9) You are analyzing a human pedigree for a new disease and discover that the disease occurs with the same frequency in both sexes and that 25% of the children from two unaffected heterozygous parents have the disease. This disease is displaying inheritance.

A) B) C) D)

autosomalrecessive autosomaldominant sex-linkedrecessive sex-linkeddominant

10) Which of the following is typically associated with an autosomal recessive pattern of inheritance?

A) B) affected C) D)

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an affected offspring has one or more affected parents an affected individual with one affected parent will have 50% of its offspring two affected heterozygous individuals will have 25% of their offspring unaffected an affected offspring has two unaffected parents

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11) In the analysis of a family, you notice that males are more likely to contract a certain disease, and the daughters of affected males produce 50% of their sons affected with the disease.

This disease is displaying which of the following patterns of inheritance?

A) B) C) D)

X-linked recessive autosomal recessive autosomal dominant None of these choices are correct.

12) Which technique can be used at the earliest stage in a pregnancy to provide a sample for fetal genetic testing?

A) amniocentesis B) chorionic villus sampling C) Both chorionic villus sampling and amniocentesis are performed at the same stage in a pregnancy.

13)

The term that indicates that cancer has begun to migrate to other parts of the body is .

A) B) C) D) E)

malignant benign metastatic invasive clonal

14) A gene that promotes the development of cancer when it sustains a gain-of-function mutation is called a

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A) B) C) D)

15)

An oncogene form of Ras would have which of the following characteristics?

A) B) C) D)

16)

decreased GTPase activity an increased number of copies in the cell decreased exchange of GDP for GTP increased GTPase activity

What are the general ways that tumor suppressors and oncogenes act?

A) B) C) D)

17)

housekeeping gene. tumor suppressor gene. proto-oncogene caretaker gene.

Tumor suppressors inhibit the cell cycle and oncogenes stimulate the cell cycle.

Oncogenes inhibit the cell cycle and tumor suppressors stimulate the cell cycle.

Both tumor suppressors and oncogenes stimulate the cell cycle.

Both tumor suppressors and oncogenes inhibit the cell cycle.

Caspases are active during which of the following?

A) B) C) D)

the normal cell cycle apoptosis viral integration DNA replication

18) The process of cells for genetic problems.

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removes amniotic fluid from a pregnant mother to examine fetal

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A) B) C) D)

19)

amniocentesis chorionic villussampling preimplantationgenetic diagnosis in vitrofertilization

Locus heterogeneity can cause difficulties in pedigree analysis since A) B) C) D)

the inheritance of only one gene is being analyzed. the inheritance of more than one gene is being analyzed. the cause of the disease may not have a genetic origin. the cause of the disease may have environmental as well as genetic origins.

20) The term that refers to the linkage of alleles or molecular markers along a single chromosome is A) B) C) D)

21)

loss of variation. association group. recombination group. haplotype.

What is the difference between genetic testing and genetic screening?

A) Genetic testing is concerned more with individuals and genetic screening is concerned with populations.

B) Genetic screening is concerned more with individuals and genetic testing is concerned with populations.

C) They use different tests and techniques to detect genetic mutations.

D) Genetic testing is concerned more with nonlethal syndromes and diseases, and genetic screening is concerned with lethal diseases and syndromes.

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22) The term for the use of information about a person's genotype and other clinical data in order to select a medication is referred to as A) B) C) D)

23)

best medical practices. molecular profiling. personalized medicine. minimalized drug therapy.

An example of personalized medicine would be A) B) C) D)

treatment of an infection based on the drug resistance of the bacterium. treatment of diabetics with synthetic instead of natural insulin. determining the CYP2C9 alleles present to help set the patient's coumarin dose. performing an appendectomy on a patient with appendicitis.

24) What kind of mutation would be expected both within oncogenes and tumor suppressor genes that would result in gain-of-function or loss-of-function, respectively?

A) B) C) D)

25)

Missense Amplifications Silent Neutral

Epigenetics may play a role in cancer by

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A) B) C) D)

causing a change in gene expression patterns. directly inducing inversion mutations. directly resulting in gene deletions. causing translocations.

26) Gene X can exist as a normal allele and also as a dominant mutant allele that causes a disease. You purify protein X from a homozygous normal individual and also from an individual that is homozygous for the mutant allele. You have an assay to determine the function of protein X. When you assay 1 mg of the purified protein from a normal homozygote, the activity is 100%.

When you mix together 0.5 mg of the normal protein and 0.5 mg of the mutant protein, the activity is 13%. Based on this experiment, what type of dominant mutant is this?

A) Gain of function B) Dominant negative C) Haploinsufficiency

27) How would a mutation that increases the rate of GTP hydrolysis by the Ras protein affect the EGF signaling pathway and how would it affect cell growth?

A) The signaling pathway would stay turned on and cell growth would be stimulated.

B) The signaling pathway would stay turned on and cell growth would be inhibited.

C) The signaling pathway would be turned off faster and cell growth would be stimulated.

D) The signaling pathway would be turned off faster and cell growth would be inhibited.

28)

The general function of the p53 protein could be described as

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A) a negative regulator of cell division.

B) a positive regulator of cell division.

C) maintaining genome integrity.

29)

If the individuals in the picture above had a child, that child could have which haplotype listed (assuming no crossing over)?

A) B) C) D)

1A2A3A4A; 1B2B3A4B 1A2B3B4C; 1C2C3C4A 1A2B3C4A; 1C2C3B4C 1A2A3A4A; 1C2C3C4A

30) Tamoxifen is a medication used for treating cancer. Many women were found to have resistance to this drug, and research showed that women with a mutation in their CYP2D6 gene could not break down Tamoxifen. Therefore Tamoxifen was not a useful treatment for their cancer. Women can now be screened for this mutation, allowing clinicians to use a therapy that is effective for each woman. This type of treatment is an example of .

A) B) C) D)

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personalized medicine haplotype association studies loss of heterozygosity karyotyping

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31) Fourcancer patients are treated with an experimental drug.Patients #2 and #4 have an extremely serious adverse reaction to the drug.Another patient (#5) requests to enter the trial and you do not wish to treat this patient if there is any indication that the new patientwill have the same side effect.The haplotypes for the region where the allele that controls this sensitivity has been mapped and the haplotypes for this region for these patients aregiven below.Should you enter patient #5 into the trial?

1: 1A 2B 3C 4B/1A 2B 3A 4C 2: 1A 2C 3B 4B/1B 2A 3B 4A 3: 1B 2A 3A 4B/1C 2B 3A 4A 4: 1B 2C 3B 4B/1A 2B 3A 4C 5: 1A 2B 3A 4C/1B 2A 3B 4A A) No, # 5 shares the 1A haplotype with #2 and #4.

B) Yes, #5 shares no haplotype combinations with #2 and #4.

C) No, # 5 shares the 3B haplotype with #2 and #4.

D) Yes, since there are no shared haplotypes between #2 and #4 these haplotypes are not correlating with the drug sensitivity.

32)

Genome wide association studies are used

A) B) C) D)

.

to identify the mutation that causes a disease to identify single nucleotide polymorphisms that are near disease causing genes to find the total variation that occurs in human genomes to describe the number of alleles for a specific gene

33) You are working in a lab where you are studying a disease that isknown to be caused by a single nucleotide change, although the effect this change ultimately has on the protein's structure/function is unknown. You have DNA samples from multiple patients that you suspect of having this disease. What is the most efficient way to test the samples for the relevant mutation?

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A) B) C) D)

Western blotting DNA sequencing Karyotyping Amniocentesis

34) In the United States, most newborns undergo a test for primary congenital hypothyroidism, a condition where people are unable to produce enough thyroid hormone. On a nationwide basis, this is an example of _.

A) B) C) D)

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genetic testing genetic screening amniocentesis preimplantation genetic diagnosis

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Answer Key Test name: Chapter 22 Test Bank 1) [A, B, C, D] 2) [A, B, D] 3) [A, B] 4) [A, B, D, F] 5) [B, C, E] 6) E 7) B 8) A 9) A 10) D 11) A 12) B 13) C 14) C 15) A 16) A 17) B 18) A 19) B 20) D 21) A 22) C 23) C 24) A 25) A 26) B Version 1

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27) D 28) C 29) D 30) A 31) B 32) B 33) B 34) B

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Student name:_ 1) The term for a group of individuals from the same species that can interbreed with one another is A) B) C) D) E)

species. population. race. community. kingdom.

2) If a gene has more than one allele and each allele has a frequency that is less than 99%, then the gene is considered to be A) B) C) D)

monomorphic. polymorphic. biallelic. polyallelic.

3) In a given population of Drosophila, curly wings (c ) is recessive to the wild-type condition of straight wings (c+ ). You isolate a population of 35 curly winged flies, 70 flies that are heterozygous for straight wings, and 45 that are homozygous for straight wings. What is the total number of curly or straight wing alleles in this population?

A) 2 B) 150 C) 230 D) 300 E) None of thesechoices are correct

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4) In a given population of Drosophila, curly wings (c ) is recessive to the wild-type condition of straight wings (c+ ). You isolate a population of 35 curly winged flies, 70 flies that are heterozygous for straight wings, and 45 that are homozygous for straight wings. What is the frequency of alleles in this population?

A) 35% c; 45% c+ B) 46.7% c; 53.3% c+ C) 50% c; 50% c+ D) 55% c; 45% c+ E) None of these choices are correct.

5)

The formula p

A) B) C) D) E)

6)

+ 2 pq + q

2

= 1 is associated with which of the following?

calculations of heterozygosity Hardy-Weinberg equilibrium calculations of recombination frequencies degrees of freedom None of these choices are correct.

In the equation p2 + 2pq + q2 = 1, what does the term 2pq represent?

A) B) C) D) E)

7)

2

the genotypic frequency of homozygous recessive individuals the genotypic frequency of homozygous dominant individuals the genotypic frequency of heterozygous individuals the sum of the phenotypic frequencies in the population None of these choices are correct.

Which of the following is NOT an assumption of the Hardy-Weinberg equation?

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A) B) C) D) E)

8)

There is no migration into or out of the population.

Individuals of the population mate randomly.

The population size is very large.

Selection favorsthe dominant allele.

There is no mutation occurring in the population.

The term for the mating for two genetically unrelated individuals is A) B) C) D)

outbreeding. inbreeding. disassortative mating. assertive mating.

9) The probability that two alleles will be identical for a given gene in an individual because the alleles are from the same ancestor is estimated by A) B) C) D)

10)

the calculation of average heterozygosity. the Hardy-Weinberg equation. the inbreeding coefficient. the chi-square test.

Genetic drift has the greatest influence on allele frequencies for which population size?

A) B) C) D)

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small large Genetic drift has the same impact on both large and small populations.

Genetic drift has no effect on the allele frequencies.

3

11) A natural disaster that resulted in the loss of genetic diversity in a population would be an example of a A) B) C) D)

founder effect. genetic drift. bottleneck effect. natural selection.

12) Migration of a random few individuals from one population to a new area to establish a new population is an example of A) B) C) D)

bottleneck effect. mutation. founder effect. selection.

13) Given the allele frequency for a gene in population A is 0.3 and the frequency for the same allele in population B is 0.5 and that 25 individuals from A migrate to and mate randomly with the 475 individuals in population B, what is the new allele frequency (PC) in the conglomerate?

A) 0.052 B) 0.2 C) 0.8 D) 0.49

14) Which of the following types of selection favors one extreme of a phenotypic distribution?

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A) B) C) D)

disruptive selection stabilizing selection directional selection balancing selection

15) Which of the following types of selection creates two phenotypic classes from a single original distribution?

A) B) C) D)

16)

disruptive selection stabilizing selection directional selection balancing selection

Natural selection occurs when

A) individuals whose phenotypes are best suited forenvironmental conditions have higher reproductive success and contribute more alleles to the next generation.

B) humans breed certain individuals with the most desirable phenotypes.

C) individuals whose phenotypes are best suited forenvironmental conditions have lower reproductive success but live longer.

D) the genotype of an individual does not determine whether or not is has reproductive success.

17)

A SNP would best be described as A) B) C) D) E)

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a gene that comes in multiple different alleles. a balanced polymorphism system. a single nucleotide difference between two DNA sequences. the least frequent type of mutation.

None of these choices are correct.

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18)

Which type of selection would lead to two distinct phenotypes?

A) B) C) D) E)

stabilizing selection phenotype selection fitness selection disruptive selection directional selection

19) The prevalence of the allele for sickle cell anemia in some populations is an example of which of the following?

A) B) C) D) E)

20)

Which of the following is NOT an assumption of the Hardy-Weinberg equation?

A) B) C) D) E)

21)

heterogeneous environments balancing selection inverted selection non-Darwinian selection nonrandom mating

The population is large.

There is no migration into or out of the population.

There is no selection against a given genotype.

There is no mutation in the gene being studied.

There is nonrandom mating.

Microevolution is defined as

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A) B) C) D)

changes in the gene pool from one generation to the next. changes in gene flow from one generation to the next. the ability of different genotypes to succeed in a particular environment. morphological changes that occur from one generation to the next.

22) The mechanism that results in a change in allele frequenciesdue to random processes is known as A) B) C) D)

genetic drift. natural selection. migration. nonrandom mating.

23) Calculate the inbreeding coefficient of an individual given an n of 4 (excluding the inbred offspring) with one common ancestor. The inbreeding of the common ancestor is unknown.

A) 25% B) 3.125% C) 6.25% D) 50%

24)

How does inbreeding affect the Hardy-Weinberg equilibrium?

A) B) C) D)

25)

It makes it more likely for the population to remain inHardy-Weinberg equilibrium.

It has no effect on the Hardy-Weinberg equilibrium.

It results in a higher level of either dominant or recessive homozygotes.

It results in a higher level of heterozygotes.

Horizontal gene transfer occurs

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A) B) C) D)

within and between species of eukaryotes and prokaryotes. between prokaryotic species only. between eukaryotic species only. only from prokaryotes to eukaryotes.

26) If in a population the rate of mutation that converts the A allele to the a allele is 10-6 and the current frequency of the A allele is 0.75 and the a allele is 0.25, then the frequency of the A and a alleles in the next generation will be A) A: 0.75a: 0.25 B) A: 0.74999925a: 0.25000075 C) A: 0.75000075a: 0.24999925 D) A: 0.74a: 0.26

27)

Repetitive sequences are useful for DNA fingerprinting becausethey A) B) C) D)

are not inherited and so are unique to every individual. are unique to every individual and change from generation to generation. are inherited and can show significant variability between individuals. never undergo mutation.

28) If the allele frequency of the dominant allele is 0.4, what value is used for the term p2 in the equation p2+ 2pq + q2= 1?

A) 0.4 B) 0.2 C) 0.16 D) 16

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29) If two individuals of a population, who vary in their phenotype, preferentially mate, it is called .

A) B) C) D)

Outbreeding Inbreeding Negative assortative mating Positive assortative mating

30) You are preparing to perform DNA fingerprinting by PCR for the first time. Select the reagent that would result in complications in the interpretation of your results.

A) B) C) D)

31)

Which term describes all the alleles of every gene in the population?

A) B) C) D)

32)

PCR primers that anneal to the repetitive region of the microsatellites PCR primers that anneal to regions flanking the microsatellites Human DNA Taq polymerase

conglomerate gene pool polymorphisms genotype frequency

Select the definition of genetic drift.

A) B) C) D)

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Changes in allele frequencies in a population due to random fluctuations Changes in allele frequencies in a population due to migration Changes in genotype frequencies in a population due to new mutations Changes in genome size in a species due to random mutations

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33) Consider a hypothetical gene B for which there are two alleles in the population. For every ten BB individuals that survive, six Bb individuals survive, and one bb individual survives.

What are the relative fitness values for each genotype?

A) wBB = 1.0; wBb = 0.6 ; wbb = 0.1 B) wBB = 100; wBb = 0.6 ; wbb = 0.01 C) wBB = 0.59; wBb = 0.35 ; wbb = 1.0 D) wB= 26; wb = 7

34) Given a population of 1,000 individuals what is the average number of generations it will take to fix a new mutation assuming no selection for or against the new mutation?

A) 100,000 B) 1,000 C) 4,000 D) 25,000

35) What is the probability of fixation of a gene in a population of 30 individuals (assume no selection for or against a mutation)?

A) 3.3% B) 3% C) 1.67% D) 0.167%

36) What is the frequency of an allele in a conglomerate population if 50 individuals from a population where the allele has a frequency of 0.6 migrates into a population of 450 individuals where the allele frequency is 0.3?

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A) 0.3 B) 0.33 C) 0.9 D) 0.18

37)

By what process do new alleles arise?

A) B) C) D) E)

38)

Natural selection Mutation Nonrandom mating Genetic drift Migration

Most natural populations are in Hardy-Weinberg equilibrium. ⊚ true ⊚ false

39) A population with an allele for a gene that constitutes 99% of the alleles in the population means that the population is called monomorphic for that gene. ⊚ true ⊚ false

40)

At equilibrium the allelic and genotypic frequencies of a population change over time. ⊚ true ⊚ false

41)

The probability that a gene will be altered by a mutation is called the mutation rate.

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⊚ ⊚

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true false

12

Answer Key Test name: Chapter 23 Test Bank 1) B 2) B 3) D 4) B 5) B 6) C 7) D 8) A 9) C 10) A 11) C 12) C 13) D 14) C 15) A 16) A 17) C 18) D 19) B 20) E 21) A 22) A 23) C 24) C 25) A 26) B Version 1

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27) C 28) C 29) C 30) A 31) B 32) A 33) A 34) C 35) C 36) B 37) B 38) FALSE 39) TRUE 40) FALSE 41) TRUE

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Student name:_ 1)

How do quantitative traits arise? (Check all that apply.)

A) the trait.

B) C) D)

2)

The trait is controlled only be a single gene and there is no environmental effect on The trait is polygenic.

The trait is controlled by a single gene and the environment.

The population is inbred with no environmental effect on the trait.

Which of the following is incorrect concerning quantitative traits?

A) B) C) D)

Individuals fall into distinct classes for comparison.

The phenotypic variation for the trait is continuous.

The frequency distribution follows a bell-shaped curve.

All of these choices are correct.

3) Which of the following best describes the degrees of freedom for the calculation of variance?

A) (n + 1) B) 2(n + 1) C) (n - 1)/n D) (n - 1)

4)

The square root of the variance is called the

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1

A) B) C) D)

mean standarddeviation standarderror covariance

5) When analyzing two variables, the strength of the association between the variables is called the .

A) B) C) D)

6)

covariance standarddeviation correlationcoefficient variance

What must be true for a correlation coefficient to be calculated between two traits?

A) B) C) D)

The values for the traits must have been obtained from unbiased sampling.

The relationship between the two traits must be linear.

The traits must follow a normal distribution.

All of these choices are correct.

7) In calculations of the degrees of freedom for a correlation coefficient, what value does the degrees of freedom take?

A) (n - 1) B) (n - 2) C) (n - 1)/2 D) 2(n - 1)

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8) After calculating a correlation coefficient, you discover that the value is not significant at the 5% level. What can be done that may result in a significant difference?

A) increase the sample size B) decrease the sample size C) Nothing, the value will remain the same regardless of sample size.

9)

Quantitative traits usually have which of the following characteristics?

A) B) C) D)

environmental influence polygenic basis continuous distribution All of these choices are correct.

10) QTL mapping determines the relationship between genes for quantitative traits and which of the following?

A) B) C) D)

11)

the centromere transposon sites structural genes molecular markers such as RFLPs

What must be true regarding the strains of organisms used to construct a QTL map?

A) they must differ for the quantitative trait B) they must differ for the molecular markers C) Both of these choices must be true.

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In QTL mapping, the parental strains are crossed and the F 1 generation backcrossed to 12) which of the following?

A) B) C) D)

13)

the F2 generation the parental generation a strain that is homozygous recessive for all markers a wild-type organism

A heritability value of 0.997 indicates which of the following?

A) B) C) D) E)

The majority of the phenotypic variation has a genetic basis.

The majority of the phenotypic variation has an environmental basis.

The trait is polygenic.

There is a significant difference between the two strains.

None of these choices are correct.

14) For a quantitative trait, the total variance of the population is due to which of the following?

A) B) C) D)

15)

The amount of variance due to genetic factors only The amount of variance due to environmental factors only The difference between the variance attributed to genetic and environmental factors The sum of the genetic and environmental variances

Inbreeding reduces which of the following variances to near zero?

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A) B) C) D)

16)

genetic variance environmental variance both variances neither variance

Heritability may be calculated as which of the following?

A) B) C) D) E)

the total variance minus environmental variance total variance minus genetic variance genetic variance divided by environmental variance genetic variance divided by total variance None of these choices are correct.

17) After several generations of artificial selective breeding, a plateau is reached where artificial selection is no longer effective. This is called the .

A) B) C) D)

realized limit realizedheritability selectioncoefficient selection limit

18) Which of the following steps is NOT used in conducting a backcross in order to map a QTL?

A) Two inbred strains that differ in a quantitative trait are crossed.

B) The F1 offspring are crossed with each other.

C) The F2 offspring contain different combinations of the parental chromosomes.

D) By using genetic markers, it is determined which parent each chromosome came from and thus genes are identified.

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19) If the difference between the mean of the parents and the mean of the starting population is 1.0 and the difference between the mean of the offspring and the mean of the starting population is 0.45 then the realized heritability is

A) B) C) D)

1.45 2.22 0.45 0.55

20) Calculate the mean plant height, the variance and the standard deviation for 10 plants that have the following heights: 5 cm, 6 cm, 6 cm, 7 cm, 7 cm, 7 cm, 8 cm, 9 cm, 9 cm, and 10 cm.

A) B) C) D)

mean: 7.4 cm variance: 2.5 cm2 standard deviation: 1.6 cm mean: 7 cm variance: 3.5 cm2standard deviation: 1.9 cm mean: 7.4 cm variance: 2.5 cm2 standard deviation: 1.9 cm mean: 7.5 cm variance: 3.5 cm2standard deviation: 1.6 cm

21) Calculate the narrow sense heritability for a grandparent-grandchild relationship when the observed correlation coefficient is 0.2.

A) 0.8 B) 0.4 C) 1.25 D) 2.5

22) Calculate the narrow-sense heritability of height between aunts and nieces if the observed phenotypic correlation coefficient is 0.1.

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A) 0.4 B) 2.5 C) 4 D) 0.25

23) A selective breeding program is trying to develop an agricultural plant with decreased stem length to aid in mechanical harvesting. The mean stem height of the population is 36 cm while the mean stem height of the parents is 30 cm and the mean stem height of the offspring is 33 cm. What is the realized heritability for stem height in this crop?

A) 0.5 B) 1 C) .33 D) 0

24) The heights of mothers and their female children are measured (in inches) with the following results Height of mother

Height of offspring at maturity

64

72

56

63

63

62

67

61

71

74

70

69

78

70

65

70

24.1)

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What is the mean height for the mothers and daughters?

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A) B) C) D)

24.2) A) B) C) D)

24.3) A) B) C) D)

Mothers: 66.75 inDaughters: 67.63 in Mothers: 65.75 inDaughters: 67.63 in Mothers: 66.75 inDaughters: 66.52 in Mothers: 65.67 inDaughters: 66.45 in

What is the variance for the mother's and daughter's heights?

Mothers: 42.21 in2Daughters: 24.27 in2 Mothers: 41.32 in2Daughters: 23.71 in2 Mothers: 43.78 in2Daughters: 24.27 in2 Mothers: 42.21 in2Daughters 23.78 in2

What is the standard deviation for the mother's and daughter's heights?

Mothers: 6.50 inDaughters: 4.93 in Mothers: 1781.68 inDaughters: 589.03 in Mothers: 42.21 inDaughters: 24.27 in Mothers: 5.50 inDaughters: 5.93 in

25) Heritability is the amount of genetic variation in a particular population and a particular environment. ⊚ ⊚

true false

26) Heritability values can be compared between populations if there is variation in the environmental conditions. ⊚ ⊚

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true false

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27)

A trait that varies measurably in a species is called a discontinuous trait. ⊚ true ⊚ false

28)

Traits that fall into discrete categories are called continuous traits. ⊚ ⊚

true false

29) Since quantitative traits are frequently associated with a number of genes and the environment, it is difficult to assign accurate phenotypic classes for a Punnett square diagram. ⊚ ⊚

true false

30) The sum of all of the values for a trait, divided by the total number of individuals, is called the standard deviation. ⊚ ⊚

true false

31) The variance is the sum of the standard deviation from the mean divided by the degrees of freedom. ⊚ true ⊚ false

32)

QTL stands for quantitative trait locus and may contain more than one gene.

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⊚ ⊚

true false

33) The amount of phenotypic variation in a group that is due to genetic variation is called heritability. ⊚ ⊚

true false

34)

A population that is homozygous for a given allele is called monomorphic. ⊚ true ⊚ false

35)

Many organisms used in modern agriculture are a result of artificial selection. ⊚ ⊚

36)

The process of mating between genetically unrelated individuals is called inbreeding. ⊚ ⊚

37)

true false

true false

Realized heritability may be determined from artificial selection experiments. ⊚ true ⊚ false

38) If a completely inbred population shows phenotypic variance, then that variance is caused by environmental effects.

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⊚ ⊚

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true false

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Answer Key Test name: Chapter 24 Test Bank 1) [B, C] 2) A 3) D 4) B 5) C 6) D 7) B 8) A 9) D 10) D 11) C 12) B 13) A 14) D 15) A 16) D 17) D 18) B 19) C 20) A 21) A 22) A 23) A 24) Section Break 24.1) A 24.2) A Version 1

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24.3) A 25) TRUE 26) FALSE 27) FALSE 28) FALSE 29) TRUE 30) FALSE 31) TRUE 32) TRUE 33) TRUE 34) TRUE 35) TRUE 36) FALSE 37) TRUE 38) TRUE

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