Genetic Analysis An Integrated Approach, 3rd EditionMark F Sanders Test Bank

Genetic Analysis An Integrated Approach, 3rd Edition BY Mark F. Sanders

Email: richard@qwconsultancy.com

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 1 The Molecular Basis of Heredity, Variation, and Evolution 1.1 Multiple-Choice Questions 1) Sexual reproduction uses ________ to generate ________ gametes, which join at fertilization.

A) meiosis; haploid B) mitosis; haploid C) meiosis; diploid D) mitosis; diploid E) mitosis; identical Answer: A Section: 1.1 Skill: Remembering/Understanding 2) When a diploid cell divides by mitosis, the result is ________.

A) identical haploid cells B) identical diploid cells C) unique diploid cells D) unique haploid cells E) a zygote Answer: B Section: 1.1 Skill: Remembering/Understanding 3) Modern genetics consists of three major branches. Which of these branches, also known as "transmission genetics," involves the study of the transmission of traits and characteristics in successive generations?

A) evolutionary B) Mendelian C) molecular D) population E) reproductive Answer: B Section: 1.1 Skill: Remembering/Understanding 4) You identify a new unicellular organism with multiple chromosomes organized by proteins within the cell's nucleus. Into which of the three domains of life might this organism fit?

A) Archaea B) Bacteria C) Eukarya D) Archaea or Bacteria E) Archaea or Eukarya Answer: C Section: 1.1 Skill: Applying/Analyzing 1 .

5) Watson and Crick used evidence from several studies to determine the structure of DNA. What conclusion were they able to draw from Rosalind Franklin's X-ray diffraction data, specifically?

A) DNA consists of four types of nucleotide bases: A, T, C, and G. B) DNA nucleotides form complementary base pairs.

C) Adenine pairs with thymine and cytosine pairs with guanine when they are on opposite DNA strands.

D) DNA is a duplex, with two strands forming a double helix.

E) The DNA strands are antiparallel, and the strands are held together by hydrogen bonds.

Answer: D Section: 1.2 Skill: Remembering/Understanding LO: S1.2 6) What kind of bond is formed between the 5′ phosphate group of one nucleotide and the 3′ hydroxyl (OH) group of the adjacent nucleotide?

A) ionic bond B) phosphodiester bond C) hydrogen bond D) disulfide bond E) hydroxyl bond Answer: B Section: 1.2 Skill: Remembering/Understanding LO: S1.2 7) What kind of bond is formed between complementary base pairs to join the two DNA strands into a double helix?

A) ionic bond B) phosphodiester bond C) hydrogen bond D) disulfide bond E) peptide bond Answer: C Section: 1.2 Skill: Remembering/Understanding LO: S1.2

2 .

8) Identify which of the following includes three possible components of a RNA nucleotide?

A) deoxyribose, guanine, phosphate group B) deoxyribose, uracil, phosphate group C) ribose, thymine, phosphate group D) deoxyribose, cytosine, phosphate group E) ribose, adenine, phosphate group Answer: E Section: 1.2 Skill: Remembering/Understanding LO: S1.2 9) What chemical group appears on the 5' carbon of a DNA nucleotide?

A) carboxyl group B) phosphate group C) amino group D) hydroxyl group E) nitrogenous base Answer: B Section: 1.2 Skill: Remembering/Understanding LO: S1.2 10) If a eukaryotic chromosome was composed of 20% adenine, how much cytosine should theoretically be present in that same chromosome?

A) 30% B) 20% C) 60% D) 40% E) 10% Answer: A Section: 1.2 Skill: Applying/Analyzing LO: G4, S1.2

3 .

11) Use the data in the following table to determine which nucleic acid sample can be ALL of the following 4 types: double-stranded DNA, single-stranded DNA, double-stranded RNA, or single-stranded RNA. Nucleic Acid Sample Sample 1 Sample 2 Sample 3 Sample 4 Sample 5

Data 25% of the bases are thymine 35% of the bases are adenine 25% of the bases are uracil 55% of the bases are cytosine 50% of the five-carbon sugars are deoxyribose

A) Sample 1 B) Sample 2 C) Sample 3 D) Sample 4 E) Sample 5 Answer: B Section: 1.2 Skill: Applying/Analyzing LO: G1, S1.2 12) What is the sequence and polarity of the DNA strand complementary to the strand 5' AAATGTCCATGC 3'?

A) 5' TTTACAGGTACG 3' B) 3' AAATGTCCATGC 5' C) 3' TTTACAGGTACG 5' D) 5' UUUACAGGUACG 3' E) 3' UUUACAGGUACG 5' Answer: C Section: 1.2 Skill: Applying/Analyzing LO: 1.3b 13) Messenger RNA (mRNA) is ________.

A) the major structural material making up ribosomes B) the molecule that carries the genetic information from DNA and is used as a template for protein synthesis C) the major structural component of chromosomes D) a molecule that incorporates a specific amino acid into the growing protein when it recognizes a specific group of three bases E) the monomer of polypeptides Answer: B Section: 1.3 Skill: Remembering/Understanding LO: S1.3a 4 .

14) What are the DNA regulatory sequences recognized by RNA polymerase called?

A) anticodons B) proteomes C) introns D) promoters E) termination sequences Answer: D Section: 1.3 Skill: Remembering/Understanding LO: S1.3a 15) What is the process of synthesizing proteins from mRNA sequences?

A) replication B) transcription C) translation D) transformation E) transduction Answer: C Section: 1.3 Skill: Remembering/Understanding LO: S1.3a 16) What is the process of synthesizing single-stranded RNA from template DNA? A) replication B) transcription C) translation D) transformation E) transduction Answer: B Section: 1.3 Skill: Remembering/Understanding LO: S1.3a 17) What kind of bond is formed between successive amino acids during translation?

A) ionic bond B) phosphodiester bond C) hydrogen bond D) disulfide bond E) peptide bond Answer: E Section: 1.3 Skill: Remembering/Understanding LO: S1.3a

5 .

18) Retroviruses carry their genetic information in the form of RNA, which is subsequently coded into DNA after the virus enters its host cell. What enzyme does the retrovirus use to produce this initial DNA? A) reverse transcriptase B) RNA polymerase C) DNA polymerase D) ribosomes E) reverse translationase Answer: A Section: 1.3 Skill: Remembering/Understanding LO: S1.3a 19) Only sixty-one of the sixty-four codons specify an amino acid. In what process do the other three codons function?

A) initiation of transcription B) initiation of translation C) initiation of replication D) termination of transcription E) termination of translation Answer: E Section: 1.3 Skill: Applying/Analyzing LO: S1.3a 20) The movement of DNA or RNA in gel electrophoresis is often a matter of molecular weight alone. Which of the following molecular parameters usually influence the movement of protein?

A) only weight B) only charge C) only shape D) only weight and shape E) weight, charge, or shape Answer: E Section: 1.4 Skill: Remembering/Understanding LO: S1.4a

6 .

21) Which of the following statements is NOT consistent with the DNA fragments shown in the gel?

A) Band 1 has a lower electrophoretic mobility than Band 2 B) Band 1 has a lower molecular mass than Band 2 C) Band 1 is closer to the origin of migration than Band 2 D) Band 1 must have a negative charge E) Band 1 must have been stained or hybridized by a molecular probe Answer: B Section: 1.4 Skill: Applying/Analyzing LO: G3, S1.4a 22) Hereditary anemia known as sickle cell disease (SCD) results from inheritance of a variant form of β-globin protein (βS), rather than the wild-type β-globin protein (βA). Which of the following did Linus Pauling find following gel electrophoresis of hemoglobin protein from individuals with the following three genotypes: βAβA, βAβS, or βSβS? A) the lane containing the hemoglobin from the heterozygote (βAβS) individual had two protein bands with differing electrophoretic mobility B) the lane containing the hemoglobin from the homozygous (βAβA) individual had two protein bands with differing electrophoretic mobility C) the lane containing the hemoglobin from the homozygous (βSβS) individual with SCD had two protein bands D) all three lanes had the same two protein bands with the same electrophoretic mobility E) all three lanes had just one protein band with the same electrophoretic mobility Answer: A Section: 1.4 Skill: Applying/Analyzing LO: G2, S1.4b

7 .

23) Hereditary anemia known as sickle cell disease (SCD) results from inheritance of a variant form of β-globin protein (βS), rather than the wild-type β-globin protein (βA). The βS protein does not migrate as far as the βA protein. Which of following does NOT explain why the gel electrophoresis lane containing the hemoglobin protein from the heterozygous (βAβS) individual has two protein bands?

A) The band closer to the origin of migration contained βS protein and the band farther from the origin of migration contained βA protein.

B) The βA protein has a higher electrophoretic mobility.

C) The βS protein has a lower electrophoretic mobility.

D) The protein bands migrated different distances based solely on differences in molecular weight.

E) The different electrophoretic mobility of the two proteins was a result of differences in their molecular weight, charge, and/ or shape.

Answer: D Section: 1.4 Skill: Evaluating/Creating LO: G3, S1.4b 24) You have digested a molecule of DNA and want to identify a specific fragment of interest.

The DNA is subjected to gel electrophoresis, but you get two bands that are very close in size.

What could you use to determine which band is the correct one?

A) eastern blot B) western blot C) northern blot D) southern blot E) stain with ethidium bromide Answer: D Section: 1.4 Skill: Applying/Analyzing LO: S1.4d

8 .

25) Which of the follow refers to all the RNA produced by transcription of DNA? A) proteome B) transcriptome C) genome D) translatome E) population genetics Answer: B Section: 1.4 Skill: Remembering/Understanding LO: S1.4e 26) Which evolutionary process describes the movement of members of a species from one population to another?

A) natural selection B) migration C) mutation D) random genetic drift E) population genetics Answer: B Section: 1.5 Skill: Remembering/Understanding LO: S1.5a 27) Which evolutionary process is most pronounced in small populations where statistical fluctuations in allele frequencies can be significant from one generation to the next?

A) natural selection B) migration C) mutation D) random genetic drift E) population genetics Answer: D Section: 1.5 Skill: Remembering/Understanding LO: S1.5a 28) Which evolutionary process involves the slow addition of allelic variation that increases the hereditary diversity of populations, ultimately leading to evolutionary change?

A) natural selection B) migration C) mutation D) random genetic drift E) population genetics Answer: C Section: 1.5 Skill: Remembering/Understanding LO: S1.5a 9 .

29) Which evolutionary process relies on the premise that individuals with the best adaptations are most successful at reproducing and leave more offspring than those with less adaptive forms?

A) natural selection B) migration C) mutation D) random genetic drift E) population genetics Answer: A Section: 1.5 Skill: Remembering/Understanding LO: S1.5a 30) Which term describes a set of organisms that descended from a single common ancestor and are more closely related to other members of the group than to organisms outside the group?

A) paraphyletic group B) phylogeny C) species D) monophyletic group E) parsimony Answer: D Section: 1.5 Skill: Remembering/Understanding LO: S1.5b 31) Morphological or molecular characters shared by members of a clade are called ________.

A) common ancestors B) paraphyletic groups C) homoplasmies D) monophyletic groups E) synaptomorphies Answer: E Section: 1.5 Skill: Remembering/Understanding LO: S1.5b

10 .

32) In the phylogenetic tree below, which feature distinguishes snakes and mammals from frogs and salamanders?

A) legs B) head C) amniotic egg D) milk E) notochord Answer: C Section: 1.5 Skill: Applying/Analyzing LO: S1.5a

11 .

1.2 Short Answer Questions 1) What are the three domains of life?

Answer: Bacteria, Archaea, and Eukarya Section: 1.1 Skill: Remembering/Understanding 2) With the assistance of William Bateson, Archibald Garrod produced the first documented example of a human hereditary disorder that shaped the study of biochemical pathways. Which disorder were they describing?

Answer: alkaptonuria Section: 1.1 Skill: Remembering/Understanding LO: G5, S1.1a 3) The physical units of heredity composed of defined DNA sequences that collectively control gene transcription and contain the information to produce RNA molecules or proteins are better known as what?

Answer: genes Section: 1.1 Skill: Remembering/Understanding LO: S1.1b 4) A complete set of chromosomes is transmitted to produce identical daughter cells in which cell division process?

Answer: mitosis Section: 1.1 Skill: Remembering/Understanding 5) The genotypes of humans are more than 99% similar. What is the term that describes the alternative forms of genes that contribute to human genetic variation?

Answer: alleles Section: 1.1 Skill: Applying/Analyzing LO: S1.1b 6) In eukaryotes, most of the cells' DNA is found in the form of chromosomes in the nucleus.

Which organelles contain their own genomes (descended from ancient endosymbiotic bacteria)?

Answer: mitochondria and chloroplasts Section: 1.1 Skill: Remembering/Understanding

12 .

7) During DNA replication, nascent DNA strands are synthesized in only one direction.

Nucleotides are added only to which end of the nascent strand?

Answer: the 3′ hydroxyl end Section: 1.2 Skill: Remembering/Understanding LO: S1.2 8) Messenger RNA codons pair with tRNA anticodons at which cell structure?

Answer: the ribosome Section: 1.3 Skill: Remembering/Understanding LO: S1.3a 9) Peptidyl transferase and other proteins power the continuous progression of the ribosome along mRNA and catalyze what type of bond formation in the growing polypeptide chain?

Answer: peptide bonds Section: 1.3 Skill: Remembering/Understanding LO: S1.3a 10) Before transferring DNA from a gel to the membrane in Southern blotting, the DNA must be denatured (usually by soaking the gel in NaOH). Why is this step necessary?

Answer: to make the DNA single stranded so that the molecular probe can bind via complementary base pairing to its target DNA Section: 1.4 Skill: Applying/Analyzing LO: S1.4c 11) What process proposed by Wallace and Darwin describes the higher rates of survival and reproduction of certain forms of a species over alternative forms?

Answer: natural selection Section: 1.5 Skill: Remembering/Understanding LO: S1.5a 12) As natural selection increases the frequency of one morphological form over another in the population, what changes at the genotypic level?

Answer: allele frequency Section: 1.5 Skill: Applying/Analyzing LO: G2, S1.5a

13 .

13) What type of diagram would you use to depict morphological or molecular similarities and differences that identify evolutionary relationships?

Answer: phylogenetic tree Section: 1.5 Skill: Remembering/Understanding 14) Both sugar gliders and flying squirrels have evolved characteristics that allow them to glide, despite being geographically separated. Similar traits that have independent origins arise as a result of what phenomenon?

Answer: convergent evolution Section: 1.5 Skill: Applying/Analyzing LO: S1.5a 15) Phylogenetic trees are constructed based on morphological characteristics, but molecular phylogenetic trees are constructed based on which feature?

Answer: nucleic or amino acid sequence Section: 1.5 Skill: Remembering/Understanding LO: S1.5b 1.3 Fill-in-the-Blank Questions 1) The work of Walter Sutton and Theodor Boveri suggested that the hereditary units, or genes, described by Mendel are located on ________.

Answer: chromosomes Section: 1.1 Skill: Remembering/Understanding LO: S1.1b 2) Genetic experiments have revealed the relationship between the observable traits of an organism, or ________, and the genetic constitution of an organism, or ________.

Answer: phenotype; genotype Section: 1.1 Skill: Remembering/Understanding LO: S1.1b 3) DNA replication is called ________ because the newly replicated DNA consists of a parental strand (from the original DNA) and a newly synthesized daughter strand.

Answer: semiconservative Section: 1.2 Skill: Remembering/Understanding LO: S1.2

14 .

4) The ________, first proposed by Francis Crick, summarizes the relationships between DNA, RNA, and protein.

Answer: central dogma of biology Section: 1.3 Skill: Remembering/Understanding LO: S1.3a 5) A general labeling compound called ________ attaches to all DNA or RNA in a gel by binding to the sugar-phosphate backbone, thus allowing researchers to visualize the nucleic acids when the gel is exposed to UV light.

Answer: ethidium bromide (EtBr) Section: 1.4 Skill: Remembering/Understanding LO: S1.4a

15 .

1.4 Essay Questions 1) DNA strands can be pulled apart by adding heat and "melting" the double-stranded DNA. The temperature required that melts a region of DNA changes based on the base-pair composition.

Based on the structure of the A-T and C-G bonds in the accompanying figure, which type of bond would require more energy (heat) to break? How might this help you predict which regions of the DNA helix may be the most stable and harder to break apart?

Answer: C-G bonds contain three hydrogen bonds, whereas A-T bonds have only two hydrogen bonds. The more hydrogen bonds in a particular region of DNA, the more energy required to break those bonds apart. Thus, regions of DNA with large numbers of C and G residues will be more heat resistant (and probably transcribed less often) than A-T rich regions.

Section: 1.2 Skill: Evaluating/Creating LO: G2, S1.2

16 .

2) The DNA sequence below encodes the first five amino acids of a large protein.

5' ATGTTAGGATATCAG 3' 3' TACAATCCTATAGTC 5' a. Identify the coding and template strands. b. Write the sequence and polarity of the mRNA transcript produced by this sequence. Which process in the central dogma of biology did you perform? Where does this process occur in eukaryotes? c. Write the amino acid sequence of the amino acids produced using the three-letter code for amino acids. (See genetic code table in text.) Which process in the central dogma of biology did you perform? Where does this process occur in eukaryotes?

Answer: a. The top strand is the coding strand. The bottom strand is the template. b. 5' AUGUUAGGAUAUCAG 3'. Transcription occurs in the nucleus in eukaryotes. c. Met-Leu-Gly-Tyr-Gln. Translation occurs on ribosomes.

Section: 1.3 Skill: Evaluating/Creating LO: G1, S1.3b 3) What are the three major types of RNA and their functions? What would happen to translation if each type of RNA were degraded?

Answer: 1. Messenger RNA (or mRNA) is transcribed from the DNA template and translated into proteins.

2. Ribosomal RNA (or rRNA) forms part of the ribosomes, the plentiful cellular structures where protein assembly takes place.

3. Transfer RNA (or tRNA) carries amino acids, the building blocks of proteins, to ribosomes.

If any of these types of RNA were degraded, then translation would not occur. Degrading mRNA would prevent translation of that particular gene. Degrading rRNA or tRNA would prevent translation of any mRNAs because the ribosome would not form properly, and the transfer RNA would not bring the correct amino acid to the growing polypeptide chain.

Section: 1.3 Skill: Evaluating/Creating LO: G2, S1.3a 4) Describe what is meant by adaptive and nonadaptive evolution. Which type of evolution might be represented by the differences in the shape of finch beaks on different islands with different food sources, and which type by the presence of both attached and detached earlobes in a given population?

Answer: Adaptive evolution implies that one form reproduces in greater numbers than others in a population because of being better adapted to the conditions driving natural selection. Finch beak shape is an example of adaptive evolution. Nonadaptive evolution describes the evolution of characteristics that are reproductively equivalent to other forms in the population.

Nonadaptive traits are neutral with respect to natural selection, conferring neither a selective advantage nor a selective disadvantage to their bearer (e.g., attached versus detached earlobes).

Section: 1.5 Skill: Evaluating/Creating LO: G8, S1.5a 17 .

5) Describe the evolutionary relationship of lancelets to tunicates and to hagfishes. Are lancelets more closely related to tunicates or to hagfishes, or are they equally related?

Answer: Lancelets are equally related to tunicates and to hagfishes. The most recent common ancestor of lancelets and tunicates is the common ancestor of chordates. The most recent common ancestor of lancelets and hagfishes is the same (the common ancestor of chordates).

Section: 1.5 Skill: Evaluating/Creating LO: G2, S1.5b

18 .

6) Based on molecular evidence, the ancestor of snakes had legs. How might you explain the loss of legs in modern snakes?

Answer: In a given environment, it was an advantage for the ancestors of modern snakes to be limbless. Due to natural selection, the legs became minimized over many generations to the point where they were eventually lost. So, just as traits can be gained by evolution, they can be lost if there is an evolutionary advantage to that change.

Section: 1.5 Skill: Evaluating/Creating LO: G2, S1.5b

19 .

7) You obtain the following sequence data from a group of related populations: Base #: 123 456 789 Ancestral sequence: AAT TCA GGA Descendant population #1: AAT TCA GGA Descendant population #2: AAT CCA GAA Descendant population #3: AAT CAA GGA Descendant population #4: AAT CAA GGG Construct a phylogenetic tree that fits the data and requires the least amount of genetic change, in other words, the most parsimonious outcome. Indicate which genetic changes occurred, if any, that were passed down to descendant populations.

Answer:

Section: 1.5 Skill: Evaluating/Creating LO: G3, S1.5c

20 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 2 Transmission Genetics 2.1 Multiple-Choice Questions 1) Genetic crosses in which F1 plants heterozygous for a given allele are crossed to generate a 3:1 phenotypic ratio in the F2 generation are known as ________.

A) replicate crosses B) reciprocal crosses C) test crosses D) monohybrid crosses E) dihybrid crosses Answer: D Section: 2.2 Skill: Remembering/Understanding LO: S2.2a, S2.2b 2) In peas, the smooth allele is dominant over the wrinkled allele. A plant with round peas was crossed to a plant with wrinkled peas and all of the resulting plants had smooth peas. What are the genotypes of the parents in this cross?

A) RR × rr B) RR × Rr C) Rr × rr D) Rr × Rr E) rr × rr Answer: A Section: 2.2 Skill: Applying/Analyzing LO: S2.2c 3) In peas, the yellow allele is dominant over the green allele. A plant with yellow peas was crossed to a plant with green peas. The resulting plants were 50% yellow and 50% green. What are the genotypes of the parents in this cross?

A) YY × yy B) YY × Yy C) Yy × yy D) Yy × Yy E) yy × yy Answer: C Section: 2.2 Skill: Applying/Analyzing LO: S2.2c

1 .

4) What genotypic ratio would you expect to observe among the progeny of a monohybrid cross?

A) 1:3 B) 9:3:3:1 C) 1:2:1 D) 1:3:2:1 E) 3:1 Answer: C Section: 2.2 Skill: Applying/Analyzing LO: S2.2b 5) You count 1000 F2 seeds from a monohybrid cross. How many seeds do you expect to display the dominant phenotype?

A) 1000 B) 750 C) 500 D) 250 E) 0 Answer: B Section: 2.2 Skill: Applying/Analyzing LO: S2.2b 6) Assuming independent assortment, what phenotypic ratio would you expect to see if an individual with the genotype RrGg is self-crossed?

A) 1:3 B) 9:3:3:1 C) 1:2:1 D) 1:3:2:1 E) 3:1 Answer: B Section: 2.3 Skill: Applying/Analyzing LO: S2.3a

2 .

7) In peas, axial (A) flower position is dominant to terminal (a), tall (L) is dominant to short (l), and yellow (Y) is dominant to green (y). If a plant that is heterozygous for all three traits is allowed to self-fertilize, how many of the offspring would show the dominant phenotype for all three traits?

A) 3/64 B) 9/64 C) 27/64 D) 32/64 E) 64/64 Answer: C Section: 2.3 Skill: Applying/Analyzing LO: S2.3d 8) In peas, axial (A) flower position is dominant to terminal (a), and tall (L) is dominant to short (l). If a plant that is heterozygous for both traits is allowed to self-fertilize, how many of the offspring would also be heterozygous for both traits?

A) 9/16 B) 1/4 C) 3/16 D) 1/8 E) 1/16 Answer: B Section: 2.3 Skill: Applying/Analyzing LO: S2.3b 9) What phenotypic ratio would you expect as a result of a test cross between two individuals where one that is homozygous recessive for alleles at two independent loci?

A) 3:1 B) 1:2:1 C) 1:1:1:1 D) 9:3:3:1 E) 9:4:2:1 Answer: C Section: 2.3 Skill: Applying/Analyzing LO: S2.3a

3 .

10) If a plant with purple, axial flowers and green, inflated pods is heterozygous for all four genes, how many different types of gametes can it produce?

A) 1 B) 4 C) 8 D) 9 E) 16 Answer: E Section: 2.3 Skill: Applying/Analyzing LO: S2.3e 11) In Guinea pigs, short hair (S) is dominant over long hair (s), rough coat (R) is dominant over smooth coat (r), and black hair (B) is dominant over white hair (b). Which of the following individuals could produce these (and only these) four possible gametes: SRb, Srb, sRb, srb.

A) SSRRbb B) ssRrBB C) SsRrbb D) SsRrBb E) SSRrBb Answer: C Section: 2.3 Skill: Applying/Analyzing LO: G3, S2.3e 12) The gene L determines hair length in rabbits. The gene B determines hair color. A rabbit with long, black hair is crossed to a rabbit with short, white hair. All the offspring have long, black hair. What are the genotypes of the parents?

A) LLBB × llbb B) LlBb × LlBb C) LlBb × llbb D) Llbb × llBb E) Impossible to determine from the information given Answer: A Section: 2.3 Skill: Applying/Analyzing LO: S2.3d

4 .

13) In rabbits, long hair and black fur are produced by the dominant alleles L and B, which assort independently. The genotype ll produces short hair and the genotype bb produces white fur. A cross between a male with short, black fur and a female with long, white fur produces four offspring with short, black fur, four offspring with long, white fur, four offspring with short, white fur, and four offspring with long, black fur. What are the genotypes of the parents?

A) llBB × LLbb B) LlBb × LlBb C) llBb × Llbb D) LLBB × llbb E) Impossible to determine from the information given.

Answer: C Section: 2.3 Skill: Applying/Analyzing LO: S2.3d 14) A couple has four children. What is the probability that they have four boys?

A) 1/2 B) 1/4 C) 1/8 D) 1/16 E) 1/32 Answer: D Section: 2.4 Skill: Applying/Analyzing LO: G4, S2.4 15) By convention, when an OBSERVED experimental outcome has a probability of occurrence of less than 5% (<0.05), the experimental results are considered to be ________.

A) within normal expected range B) statistically significant and different from the expected outcome C) not significant D) less than one standard deviation from the mean E) equal to the mean Answer: B Section: 2.5 Skill: Remembering/Understanding LO: G4, S2.5

5 .

16) The statistical interpretation of a chi-square value is determined by identifying the ________.

A) mean B) degrees of freedom C) average D) P value E) joint probability Answer: D Section: 2.5 Skill: Remembering/Understanding LO: G4, S2.5 17) The P value is a quantitative expression of the probability that the results of another experiment of the same size and structure will DEVIATE FROM EXPECTED RESULTS AS MUCH AS OR MORE THAN BY CHANCE. The greater the difference between observed and expected results of an experiment, ________.

A) the lower the χ2 value and the lower the P value B) the greater the χ2 value and the greater the P value C) the greater the χ2 value and the lower the P value D) the lower the χ2 value and the greater the P value E) the greater the χ2 value; but the P value is unaffected Answer: C Section: 2.5 Skill: Remembering/Understanding LO: G4, S2.5 18) Humans have a gene, T, that is involved in muscle formation of the tongue. Individuals homozygous for one allele can roll their tongues, while individuals homozygous for the other allele cannot. If both parents can roll their tongues, but their child cannot, what can be said about the mode of inheritance?

A) Tongue rolling is dominant.

B) Tongue rolling is recessive.

C) The parents were both homozygous, but the child was heterozygous.

D) Tongue rolling is dominant, and both parents were heterozygous (Tt).

E) Tongue rolling is recessive, and both parents were heterozygous (Tt).

Answer: D Section: 2.6 Skill: Applying/Analyzing LO: S2.6a, S2.6b

6 .

19) In mice, black coat color is dominant to white coat color. In the pedigree below, mice with a black coat are represented by darkened symbols, and those with white coats are shown as open symbols.

What is the genotype of III-1?

A) homozygous dominant or heterozygous B) heterozygous C) homozygous dominant D) heterozygous or homozygous recessive E) homozygous recessive Answer: B Section: 2.6 Skill: Applying/Analyzing LO: G3, S2.6b 20) In mice, black coat color is dominant to white coat color. In the pedigree below, mice with a black coat are represented by darkened symbols, and those with white coats are shown as open symbols.

What is the probability that I-1, I-2, II-2, AND III-1 are all heterozygous?

A) 0 B) 1/4 C) 1/2 D) 2/3 E) 1 Answer: E Section: 2.6 Skill: Applying/Analyzing LO: G3, G4, S2.6b, S2.6c

7 .

21) In mice, black coat color is dominant to white coat color. In the pedigree below, mice with a black coat are represented by darkened symbols, and those with white coats are shown as open symbols.

What could you conclude regarding the genotype of mouse II-3 if a testcross resulted in 5 mice with black coat color and 6 mice with white coat color?

A) The genotype of II-3 could be homozygous recessive or heterozygous.

B) The genotype of II-3 could be homozygous dominant or heterozygous.

C) The genotype of II-3 must be homozygous dominant.

D) The genotype of II-3 must be heterozygous.

E) The genotype of II-3 must be homozygous recessive.

Answer: D Section: 2.6 Skill: Applying/Analyzing LO: G2, G3, G4, S2.2a, S2.6b 22) In the accompanying figure, the chance that individual III-5 is a heterozygous carrier is ________.

A) 0% B) 25% C) 50% D) 75% E) 100% Answer: E Section: 2.6 Skill: Applying/Analyzing LO: G3, G4, S2.6b, S2.6c 8 .

23) In the accompanying figure, the chance that individual IV-7 is a heterozygous carrier is ________.

A) 1/4 B) 1/3 C) 1/2 D) 2/3 E) 3/4 Answer: D Section: 2.6 Skill: Applying/Analyzing LO: G3, G4, S2.6b, S2.6c

9 .

24) In the accompanying figure, if individual IV-7 has three children with individual IV-2 what is the probability (to the nearest hundredth) that they would have exactly two affected offspring?

A) 0.02 B) 0.07 C) 0.17 D) 0.67 E) 0.44 Answer: B Section: 2.6 Skill: Applying/Analyzing LO: G3, G4, S2.6b, S2.6c 25) Huntington's disease is an autosomal dominant trait. Given the pedigree below, if individual IV-2 has three children with a normal man, what is probability that exactly two of the three children would have the disorder?

A) 0 B) 1/8 C) 1/2 D) 3/8 E) 7/8 Answer: D Section: 2.6 Skill: Applying/Analyzing LO: G3, G4, S2.4, S2.6b, S2.6c

10 .

26) Huntington's disease is an autosomal dominant trait. Given the pedigree below, if individual IV-4 has three children with a normal woman, what is probability that they would have at least one child with the disorder?

A) 0 B) 1/8 C) 1/2 D) 3/8 E) 7/8 Answer: E Section: 2.6 Skill: Applying/Analyzing LO: G2, G3, G4, S2.4, S2.6b, S2.6c 27) The genes responsible for some of the traits that Mendel observed have been recently identified and have helped in determining how molecular variation produces morphologic variation in pea plants. Allelic variation in the Sbe1 gene, which produces starch-branching enzyme 1, is responsible for which trait in peas?

A) smooth and wrinkled pea shape B) yellow and green pea color C) purple and white flowers D) tall and short plant height E) axial and terminal flower position Answer: A Section: 2.6 Skill: Remembering/Understanding LO: S2.1 28) In 1997, a gene called Le was discovered by two research groups led by David Martin and Diane Lester. Allelic variation in the Le gene, which controls elongation of the plant stem between branches, is responsible for which trait in peas?

A) inflated and constricted pod shape B) yellow and green pod color C) purple and white flowers D) tall and short plant height E) axial and terminal flower position Answer: D Section: 2.6 Skill: Remembering/Understanding LO: S2.1 11 .

2.2 Short Answer Questions 1) Mendel performed numerous controlled genetic crosses to obtain strains that consistently produced a single phenotype without variation. What are these strains that consistently produce the same phenotype called?

Answer: pure-breeding or true-breeding strains Section: 2.1 Skill: Remembering/Understanding LO: S2.1 2) In a test cross, a pure-breeding plant is crossed with a plant suspected to be heterozygous (Aa).

What is the genotype of the pure-breeding plant?

Answer: aa Section: 2.1 Skill: Applying/Analyzing LO: S2.1 3) Why did Mendel cut off the nascent anthers during the process of artificial cross-fertilization?

Answer: to prevent self-fertilization or to prevent uncontrolled crosses Section: 2.1 Skill: Remembering/Understanding LO: G1, S2.1 4) In some of Mendel's experiments, a cross in which one plant provides the pollen and another with the same genotype provides the egg is followed be another cross in which the first plant provides the egg while the second provides the pollen. During his experiments, these ________ crosses produced identical results.

Answer: reciprocal Section: 2.1 Skill: Remembering/Understanding LO: G1, S2.1 5) What simple type of cross that investigates the inheritance of only one trait could be used to illustrate Mendel's law of segregation?

Answer: monohybrid cross Section: 2.2 Skill: Remembering/Understanding LO: S2.2a 6) A cross between a short pea plant and a tall pea plant results in a 1:1 genotypic AND phenotypic ratio in the offspring. What are the genotypes of the parent plants?

Answer: Ss × ss (heterozygous × homozygous recessive) Section: 2.2 Skill: Applying/Analyzing LO: S2.2c

12 .

7) The law of independent assortment predicts that crossing of dihybrid F1 plants to one another would produce nine genotypes in a ________ ratio among F2 progeny.

Answer: 9:3:3:1 Section: 2.3 Skill: Remembering/Understanding LO: S2.3a 8) What is the probability of rolling one six-sided die and obtaining a 1 or a 2?

Answer: 1/6 + 1/6 = 2/6 = 1/3 Section: 2.4 Skill: Applying/Analyzing LO: S2.4 9) What is the probability of rolling one six-sided die and obtaining any number but 6?

Answer: 1 - 1/6 = 5/6 Section: 2.4 Skill: Applying/Analyzing LO: S2.4 10) What is the probability of rolling two six-sided dice and obtaining two 4's?

Answer: 1/6 × 1/6 = 1/36 Section: 2.4 Skill: Applying/Analyzing LO: S2.4 11) What is the probability of rolling two six-sided dice and obtaining at least one 3?

Answer: Probability of die 1 being a 3 and die 2 not: 1/6 × 5/6 = 5/36 Probability of die 2 being a 3 and die 1 not: 1/6 × 5/6 = 5/36 Probability of die 1 and 2 being a 3: 1/6 × 1/6 = 1/36 Probability of any of these possibilities = addition rule: 5/36 + 5/36 + 1/6 = 11/36 Section: 2.4 Skill: Applying/Analyzing LO: S2.4 12) What is the probability of rolling two six-sided dice and obtaining an odd number on at least one die?

Answer: 9/36 + 9/36 + 9/36 = 27/36 = 3/4 Probability of rolling odd number the first die only = 3/6 (odd) × 3/6 (even) = 9/36 Probability of rolling odd number the second die only = 3/6 (even) × 3/6 (odd) = 9/36 Probability of rolling odd number both dice = 3/6 (odd) × 3/6 (odd) = 9/36 Probability of any one of these three possible scenarios = addition rule Section: 2.4 Skill: Applying/Analyzing LO: S2.4

13 .

13) Geneticists must be able to compare the outcomes they obtain in their experiments to the outcomes that might be expected to occur. Which test would they use to confirm that the difference between observed and expected outcomes could be attributed to chance?

Answer: chi-square test Section: 2.5 Skill: Remembering/Understanding LO: G4, S2.5 14) If an affected individual is born to parents who are unaffected, what is the likely mode of inheritance?

Answer: autosomal recessive Section: 2.6 Skill: Remembering/Understanding LO: S2.6a 15) The pedigree suggests which mode of inheritance for an allele on chromosome 15?

Answer: autosomal recessive Section: 2.6 Skill: Applying/Analyzing LO: G3, S2.6a 2.3 Fill-in-the-Blank Questions 1) One key to Mendel's success was choosing to observe ________ traits, which exhibit one of two possible phenotypes.

Answer: dichotomous Section: 2.1 Skill: Remembering/Understanding LO: S2.1 2) A ratio of 9:3:3:1 is expected among the F2 progeny of a dihybrid cross as a result of ________ of alleles at two loci.

Answer: independent assortment Section: 2.3 Skill: Remembering/Understanding LO: S2.3a

14 .

3) In a cross between individuals who are both heterozygous (carriers) for a recessive disease, such as albinism, you would like to determine the risk of one or more children inheriting the recessive phenotype. ________ probability can be used to calculate the probability of a particular combination of events that each have two alternative outcomes?

Answer: Binomial Section: 2.4 Skill: Remembering/Understanding LO: S2.4 4) You have self-fertilized a plant with round seeds that is heterozygous, and you want to determine what proportion of the offspring will be dominant and true breeding. ________ probability can be used to calculate the probability of obtaining a particular outcome when specific information about that outcome modifies the probability calculation?

Answer: Conditional Section: 2.4 Skill: Remembering/Understanding LO: S2.4 5) The P value is dependent on the number of ________, which is equal to the number of independent variables in an experiment.

Answer: degrees of freedom (df) Section: 2.5 Skill: Remembering/Understanding LO: S2.5

15 .

2.4 Essay Questions 1) Describe the traits that make Pisum sativum an ideal organism for genetic studies.

Answer: There are many varieties of peas with distinct, heritable features in the form of dichotomous phenotypes that can be easily observed and quantified. In addition, mating of plants can be closely controlled. Since each pea plant has both sperm-producing (stamens) and eggproducing (carpels) organs, they can be self-crossed to generate true-breeding plants. After creating these true-breeding plants, Mendel could test for dominant or recessive inheritance patterns by cross-pollination (fertilization between different plants).

Section: 2.1 Skill: Evaluating/Creating LO: G2, G8, S2.1 2) Describe the blending theory of heredity and how Mendel's results help to reject this theory.

Answer: The blending theory viewed the traits of progeny as a mixture of the characteristics possessed by the two parental forms. Under this theory, progeny were believed to display characteristics that were approximately intermediate between those of the parents. Mendel reasoned that if the blending theory were true, he would see evidence of it in each trait. If no blending was seen in individual traits, the blending theory would be disproven. F1 experimental results reject the blending theory of heredity because all F1 progeny have the same phenotype (i.e., the dominant phenotype) that is indistinguishable from the phenotype of one of the purebreeding parents. This specifically contradicts the blending theory prediction that the F1 would display a mixture of the parental phenotypes. The persistence of the dominant phenotype and the reemergence of the recessive phenotype in the F2 also contradict the blending theory.

Section: 2.1 Skill: Evaluating/Creating LO: G2, G8, S2.1 3) What are Mendel's first and second laws of inheritance, and what do they state?

Answer: First Law: Law of Segregation—The two alleles for each trait will separate from one another during gamete formation, and each allele will have an equal probability (1/2) of inclusion in a gamete. Random union of gametes at fertilization will unite one gamete from each parent to produce progeny in ratios that are determined by chance.

Second Law: Law of Independent Assortment—During gamete formation, the segregation of alleles at one locus is independent of the segregation of alleles at another locus.

Section: 2.2 and 2.3 Skill: Remembering/Understanding LO: S2.2b, S2.3a

16 .

4) In Guinea pigs, short hair (S) is dominant over long hair (s), rough coat (R) is dominant over smooth coat (r), and black hair (B) is dominant over white hair (b). List all the different possible gametes that can be produced by each of the individuals below. a. SSRRbb b. ssRrBB c. SsRrbb d. SsRrBb Answer: a. SSRRbb: SRb b. ssRrBB: sRB, srB c. SsRrbb: SRb, Srb, sRb, srb d. SsRrBb: SRB, SRb, SrB, Srb, sRB, sRb, srB, sr Section: 2.3 Skill: Applying/Analyzing LO: S2.3e

17 .

5) A geneticist is investigating the inheritance of two autosomal recessive genes in mice, one for obesity (LEP) and another for autism (oprm1). The table below provides the number of offspring observed, for each phenotype, when dihybrid mice are crossed: Phenotype Observed(O) Expected (E) wild-type 154 obese 69 autistic 58 obese, autistic 23

O-E (O-E)2

(O-E)2/E

∑ (O-E)2/E= Fill in the table above and determine: a. the chi-square (χ²) value (to the nearest hundredth) for the chance hypothesis that the two genes assort independently b. the degrees of freedom c. whether or not you would reject your chance hypothesis that the two genes assort independently, based on your approximate P value, and the justification for why Answer: Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E wild-type 154 171 -17 289 1.690 obese 69 57 12 144 2.526 autistic 58 57 1 1 0.018 obese, autistic 23 19 4 16 0.842 ∑ (O-E)2/E= 5.08 a. 5.08 b. 3 c. You would fail to reject the chance hypothesis since the observed outcomes have a P value greater than 5% (> 0.05). Therefore, the chance hypothesis that the two genes assort independently cannot be rejected.

Section: 2.5 Skill: Applying/Analyzing LO: G2, G3, G4, G9, S2.3b, S2.3c, S2.5

18 .

6) A geneticist is investigating the inheritance of two autosomal recessive genes in mice, one for obesity (LEP) and another for autism (oprm1). The table below provides the number of offspring observed, for each phenotype, when dihybrid mice are testcrossed: Phenotype Observed(O) Expected (E) wild-type 88 obese 68 autistic 64 obese, autistic 80

O-E (O-E)2

(O-E)2/E

∑ (O-E)2/E= Fill in the table above and determine: a. the chi-square (χ²) value (to the nearest hundredth) for the chance hypothesis that the two genes assort independently b. the degrees of freedom c. whether or not you would reject your chance hypothesis that the two genes assort independently, based on your approximate P value, and the justification for why Answer: Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E wild-type 88 75 13 169 2.533 obese 68 75 -7 49 0.653 autistic 64 75 -11 121 1.613 obese, autistic 80 75 5 25 0.333 ∑ (O-E)2/E= 4.85 a. 4.85 b. 3 c. You would fail to reject the chance hypothesis since the observed outcomes have a P value greater than 5% (> 0.05). Therefore, the chance hypothesis that the two genes assort independently cannot be rejected.

Section: 2.5 Skill: Applying/Analyzing LO: G2, G3, G4, G9, S2.3a, S2.3c. S2.5

19 .

7) Tim and Mary are planning to start a familiy. While neither of them have cystic fibrosis, a rare autosomal recessive disease, they are concerned that their children could have the disease since Tim's father, John, AND Mary's brother, Ralph both have the disease. a. Draw a pedigree of the inheritance of the cystic fibrosis allele in this family. b. On your pedigree, indicate any individuals that must be carriers. c. What is the probability that Tim will be a carrier for the disease allele? d. What is the probability that Mary will be a carrier for the disease allele? e. What is the probability that Tim and Mary will have a child with the disease?

Answer: a.

b.

c. 1 d. 2/3 e. 1/6 Section: 2.6 Skill: Evaluating/Creating LO: G2, G3, G4, G5, S2.6b, S2.6c S2.6d

20 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 3 Cell Division and Chromosome Heredity 3.1 Multiple-Choice Questions 1) Mitosis is a process of cell division that results in ________.

A) two unique daughter cells B) four unique daughter cells C) two identical daughter cells D) three identical daughter cells E) four identical daughter cells Answer: C Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a 2) During which phase of the cell cycle does the cell actively transcribe and translate protein products necessary for normal cellular structure and function while not advancing the cell cycle?

A) Metaphase B) Gap 1 phase C) Gap 2 phase D) Gap 0 phase E) Interphase Answer: E Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a 3) During which phase of the cell cycle does the cell initiate replication of its nuclear DNA? A) Metaphase B) Gap 1 phase C) Gap 2 phase D) Gap 0 phase E) Synthesis phase Answer: E Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a

1 .

4) After which stage or phase of the cell cycle does cytokinesis occur?

A) M phase B) G1 C) G2 D) S E) G0 Answer: A Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a 5) Certain kinds of cells (e.g., some cells in the eyes and bones) mature and differentiate into a state in which they have a specialized function but do not divide or progress through the cell cycle. These cells are "stuck" in which stage?

A) M phase B) G1 C) G2 D) S E) G0 Answer: E Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a 6) During mitotic cell division, if chromosomal material is improperly divided between the two daughter cells, one cell could receive three copies of a chromosome, and the other cell could receive only one. This is likely due to a defect in which process?

A) cytokinesis B) karyokinesis C) crossing over D) synapsis E) homologous recombination Answer: B Section: 3.1 Skill: Applying/Analyzing LO: G1, S3.1b

2 .

7) A eukaryotic cell is diploid and contains 6 chromosomes during G1 phase. If the cell undergoes mitosis, how many daughter cells would be produced, and how many chromosomes would each one contain?

A) 2 daughter cells with 3 chromosomes each B) 2 daughter cells with 6 chromosomes each C) 2 daughter cells with 12 chromosomes each D) 4 daughter cells with 6 chromosomes each E) 4 daughter cells with 12 chromosomes each Answer: B Section: 3.1 Skill: Applying/Analyzing LO: G4, S3.1c 8) During prophase of mitosis, a diploid cell with 2 nuclear chromosomes has how many sister chromatids and how many contiguous DNA molecules?

A) 2 sister chromatids and 2 DNA molecules B) 2 sister chromatids and 4 DNA molecules C) 4 sister chromatids and 2 DNA molecules D) 4 sister chromatids and 4 DNA molecules E) 8 sister chromatids and 4 DNA molecules Answer: D Section: 3.1 Skill: Applying/Analyzing LO: G4, S3.1c 9) You discover a new species of snail, Biologica terificia, and find that its haploid number is 4 (n = 4). The somatic cells of Biologica terificia are diploid. How many chromosomes are in a somatic cell of the snail in G1 phase?

A) 4 B) 8 C) 16 D) too many to count Answer: B Section: 3.1 Skill: Applying/Analyzing LO: G1, S3.1c 10) What is the role of the centromere during cell division?

A) Hold together the DNA of a single chromatid.

B) Attach one sister chromatid to another sister chromatid.

C) Attach two chromosomes to each other.

D) Serve as the initiation site for DNA synthesis.

Answer: B Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1b 3 .

11) A single chromosome at metaphase contains how many DNA molecules?

A) one B) two C) four D) more than four E) impossible to determine Answer: B Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1b 12) Which organelle migrates during M phase to form the two opposite poles of the dividing cell and acts as the source of the spindle fiber microtubules?

A) centromere B) kinetochore C) centrosome D) chiasmata E) chromosome Answer: C Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1b 13) Which structure is responsible for chromosome movement during cell division?

A) synaptonemal complex B) kinetochore microtubules C) metaphase plate D) nuclear envelope E) cell membrane Answer: B Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1b 14) The separation of sister chromatids during anaphase I of mitosis is known as chromosome ________.

A) cytokinesis B) karyokinesis C) crossing over D) synapsis E) disjunction Answer: E Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1b

4 .

15) A dihybrid has the genotype AaBb. After a normal mitotic division into two daughters cells, one daughter cell will have the genotype ________ and the other will have the genotype ________.

A) Aa; Bb B) AaBb; AaBb C) AB; ab D) Ab; aB E) AABB; aabb Answer: B Section: 3.1 Skill: Applying/Analyzing LO: G1, S3.1d 16) Tumor suppressors are genes whose protein products regulate cell cycle checkpoints and ________.

A) promote advancement of the cell cycle B) phosphorylate proteins involved in the cell cycle C) block progression of the cell cycle D) prevent apoptosis in normal cells E) activate proto-oncogenes Answer: C Section: 3.1 Skill: Applying/Analyzing LO: G1, S3.1b 17) Vinblastine is a commonly used chemotherapy drug that acts by interfering with the assembly of microtubules. How does it likely target cancer cells?

A) by suppressing cyclins B) by inhibiting protein kinases C) by inhibiting DNA synthesis D) by denaturing myosin and preventing the cleavage furrow from forming E) by disrupting mitotic spindle formation Answer: E Section: 3.1 Skill: Applying/Analyzing LO: G2, S3.1b 18) Why might a parent cell undergo mitosis in a multi-cellular tissue?

A) Parent cell is damaged and needs to perform apoptosis.

B) Parent cell needs to out-compete other cells around it for resources.

C) Growth factor concentrations are low.

D) Cyclins are not phosphorylated.

E) Parent cell needs to replace nearby cells that have performed apoptosis.

Answer: E Section: 3.1 Skill: Synthesis/Evaluation LO: G2, S3.1b 5 .

19) If a cell begins to divide rapidly (as often seen in wound repair), what would you expect to happen to the timing of G1 phase?

A) It would require the same amount of time as any other cell.

B) It would require less time than other cells.

C) It would require more time than other cells.

D) The time it would require would depend on what other cells in the vicinity are doing.

E) It is not possible to say what would happen to the timing of G1 phase.

Answer: B Section: 3.1 Skill: Applying/Analyzing LO: G2, S3.1b 20) The five substages of M phase are designed to accomplish two main goals: equal partitioning between the two cells of chromosomal material and of cytoplasmic contents. What are these two processes called?

A) Nondisjunction and Centrosomes B) Depolymerization and Disjunction C) Kinetochore and Cohesion D) Karyokinesis and Cytokinesis E) There are not specific names for these process and are best described as in the question.

Answer: D Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1b 21) During prometaphase, microtubules grow from the ________ to attach to the ________.

A) Centrosomes; Kinetochores B) Nucleolus; Chromosomes C) Metaphase plate; Daughter chromosomes D) Spindle; Chromatids E) Centrioles; Organelles Answer: A Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1b

6 .

22) Why are chromosomes more easily visible under a microscope during Metaphase as compared to G2 phase?

A) They become stained by natural pigments in the cell.

B) There is twice as much DNA during Metaphase than during G2 phase.

C) Chromosomes are contained in the nucleus during G2 but not during metaphase.

D) Proteins that bind chromosomes during Metaphase are highly visible.

E) They are highly condensed in Metaphase.

Answer: E Section: 3.1 Skill: Applying/Analyzing LO: G2, S3.1b 23) When cyclin D1 is expressed in normal levels, it stimulates the cell cycle by promoting transition from one cell cycle phase to another. If overexpressed, it can promote cancer by allowing cells to overproliferate. The normal cyclin D1 gene regulates which general cell cycle process?

Answer: cell cycle checkpoints Section: 3.1 Skill: Applying/Analyzing LO: G2, S3.1b 24) During meiosis I, when does homologous chromosome pairing and recombination occur?

A) prophase I B) pro-metaphase I C) metaphase I D) anaphase I E) telophase I Answer: A Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a 25) The binding of nonsister chromatids by a synaptonemal complex draws the homologs into close contact so that what process can occur?

A) cytokinesis B) karyokinesis C) independent assortment D) synapsis E) disjunction Answer: D Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a

7 .

26) Contact points between nonsister chromatids that mark the locations of DNA-strand exchange are called ________.

A) synaptonemal complexes B) metaphase plates C) chiasmata D) kinetochores E) centrosomes Answer: C Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a 27) Genes A and B are on different chromosomes. A dihybrid with the genotype AaBb undergoes meiosis. Which of the following depicts a possible genotype in a gamete?

A) Aa B) Bb C) AB D) A E) B F) AaBb Answer: C Section: 3.2 Skill: Applying/Analyzing LO: G1, S3.2c 28) WT yeast (TRP+) can synthesize the amino acid tryptophan and can grow on media plates lacking tryptophan. In contrast, mutant alleles (trp−) block tryptophan synthesis and result in an inability of mutant strains to grow on media lacking tryptophan. A diploid yeast was made through the cross MATα ADE+ × MATα ade−. Meiosis was induced in the diploid strain to produce an ascus with four haploid spores, which were separated and grown on plates lacking tryptophan. How many spores are expected to grow into colonies on the plate?

A) 0 B) 1 C) 2 D) 3 E) 4 Answer: C Section: 3.2 Skill: Applying/Analyzing LO: G1, S3.2c

8 .

29) A dihybrid has the genotype AaBb. The diagram depicts the chromosome composition of different cells from the animal. Which of the cells could represent a mature gamete?

A) Cell G B) Cell H C) Cell K D) Cell M E) Cell P Answer: A Section: 3.2 Skill: Applying/Analyzing LO: G3, S3.2b

9 .

30) Which of the following represents the chromosomes in a cell at the beginning of meiosis phase 2?

A) Cell G B) Cell H C) Cell K D) Cell M E) Cell P Answer: B Section: 3.2 Skill: Applying/Analyzing LO: G3, S3.2b

10 .

31) Which of the following cells could be created from a mitotic division but would not be created during a normal meiosis?

A) Cell G B) Cell H C) Cell K D) Cell M E) Cell P Answer: C Section: 3.2 Skill: Applying/Analyzing LO: G3, S3.2b 32) Why might a cell undergo meiosis rather than mitosis?

A) It is more efficient to perform meiosis than mitosis.

B) Diploid gametes are effective for asexual reproduction.

C) Haploid gametes can be genetically diverse.

D) Cells that perform meiosis can out-compete cells that perform mitosis.

E) Meiosis protects against mutations during reproduction.

Answer: C Section: 3.2 Skill: Synthesis/Evaluation LO: G2, S3.2a 33) While female and male beetles both contain twenty diploid chromosomes, males have one smaller chromosome than in females. What can be inferred from this observation?

A) Males require fewer genes than females.

B) The number of autosomal chromosomes varies between gametes.

C) Crossing over occurs between homologous chromosomes.

D) Biological sex is determined by the presence of an X or Y chromosome.

E) Chromosomes assort independently to determine biological sex.

Answer: D Section: 3.3 Skill: Applying/Analyzing LO: G2, S3.3 11 .

34) Crossing white-eyed male Drosophila to red-eyed females produces all red-eyed F1 offspring, but the one half of the males in the F2 generation are white-eyed. What concept best explains this observation of eye color?

A) autosomal dominance B) random mutation C) crossing over D) independent assortment E) X-linked inheritance Answer: E Section: 3.3 Skill: Applying/Analyzing LO: G1, S3.3 35) Biological sex in some species, such as Drosophila, is determined by genes present on both the ________.

A) haploid and diploid chromosomes B) nuclear and mitochondrial DNA C) stem and gonad cells D) autosomal and sex chromosomes E) prophase and Telophase chromosomes Answer: D Section: 3.3 Skill: Knowledge/Comprehension LO: G1, S3.3 36) Chromosomal nondisjunction of the X chromosome in female gametes of humans means that live offspring can include ________.

A) females with XX or XXX, and males with XY or XXY B) females with XX, and males with XY C) females with XX, and males with XY or XO D) females with XX or XXX, and males with XY or XO E) females with XX or XO, and males with XY or OY Answer: A Section: 3.4 Skill: Applying/Analyzing LO: G1, S3.4a

12 .

37) The biological sex of birds and some other mammals can be explained by females containing ________ chromosomes and males containing ________ chromosomes.

A) XX; XY B) XY; XX C) ZW; ZZ D) Z; W E) X; Y Answer: D Section: 3.4 Skill: Knowledge/Comprehension LO: G1, S3.4a 38) If a trait is X-linked recessive, who would express the trait?

A) females homozygous for the dominant allele and males hemizygous for the recessive allele B) heterozygous females and males hemizygous for the dominant allele C) females homozygous for the recessive allele and males hemizygous for the recessive allele D) females homozygous for the recessive allele and males hemizygous for the dominant allele E) the same proportions of females and males Answer: C Section: 3.5 Skill: Applying/Analyzing LO: G1, S3.5e 39) If a trait is X-linked dominant, who would express the trait?

A) females homozygous for the dominant allele and males hemizygous for the recessive allele B) heterozygous females and males hemizygous for the dominant allele C) females homozygous for the recessive allele and males hemizygous for the recessive allele D) females homozygous for the recessive allele and males hemizygous for the dominant allele E) significantly more females than males Answer: B Section: 3.5 Skill: Applying/Analyzing LO: G1, S3.5e 40) Red-green color blindness is an X-linked recessive disorder. A woman with normal vision whose father was colorblind has children with a man with normal color vision. What is the probability that their first child will be colorblind?

A) 1/4 B) 1/2 C) 3/4 D) 0 E) 1 Answer: A Section: 3.5 Skill: Applying/Analyzing LO: G4, S3.5d 13 .

41) Which dosage compensation mechanism is employed by female placental mammals?

A) synteny B) Y-inactivation C) X-inactivation D) X nondisjunction E) X chromosome crossing over Answer: C Section: 3.6 Skill: Knowledge/Comprehension LO: G1, S3.6a 3.2 Short Answer Questions 1) How many CHROMOSOMES are found in a human somatic cell nucleus?

Answer: 46 Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a 2) How many PAIRS OF HOMOLOGOUS CHROMOSOMES are found in a human somatic cell nucleus?

Answer: 23 Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a 3) How many total CHROMATIDS are found in a human somatic cell nucleus during G2 phase?

Answer: 92 Section: 3.1 Skill: Applying/Analyzing LO: G1, S3.1a 4) What process is used to make mature germ-line cells, or GAMETES? Answer: meiosis Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a 5) What is the short segment of the cell cycle during which genetic material is partitioned equally to two daughter cells?

Answer: the M phase Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a

14 .

6) A cell in G1 phase has the option of entering which two phases?

Answer: S phase or G0 Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a 7) What separates in meiosis I to reduce the diploid number (2n) of chromosomes to the haploid number (n)?

Answer: homologous chromosomes Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a 8) At the end of meiosis II, what is produced?

Answer: four haploid gametes Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a 9) The gametes of the two sexes are often dramatically different in size and morphology. If both gametes have a haploid nucleus, then what do female gametes contain more of?

Answer: cytoplasm and organelles Section: 3.2 Skill: Applying/Analyzing LO: G1, S3.2a 10) The number of occurrences of which event correlates closely with the number of recombination nodules along each homologous chromosome arm?

Answer: crossover event Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a 11) Which protein assures that sister chromatids of each chromosome remain firmly joined and can resist the pull of the kinetochore microtubules?

Answer: cohesin Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a 12) Females have two copies of the X chromosome, so they can be homozygous or heterozygous for any genes found on the X chromosome. Males have only one copy of the X chromosome.

What term is used to describe this inheritance state?

Answer: hemizygous Section: 3.3 Skill: Knowledge/Comprehension LO: G1, S3.3 15 .

13) What is the mammalian transcription factor that elicits a cascade of gene transcription and developmental events that ultimately produce male internal and external structures?

Answer: SRY Section: 3.4 Skill: Knowledge/Comprehension LO: G1, S3.4a 14) Which term describes any mechanism that compensates for differences in the number of copies of genes due to the different chromosome constitutions of males and females?

Answer: dosage compensation Section: 3.6 Skill: Knowledge/Comprehension LO: G1, S3.6a 3.3 Fill-in-the-Blank Questions 1) Of the trillions of cells in the human body, most are ________, the cells that form organs and tissues.

Answer: somatic cells Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a 2) The number of chromosomes in ________ species varies and is identified non-specifically as 2n.

Answer: diploid Section: 3.1 Skill: Knowledge/Comprehension LO: G1, S3.1a 3) Gametes and some species are known as ________ because of the n number of chromosomes contained in their nuclei.

Answer: haploid Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a 4) In meiosis, interphase is followed by two successive rounds of cell division, meiosis I and II. Meiosis results in four haploid daughter cells because ________ does not occur between the end of telophase I and the start of prophase II. Answer: DNA replication or S phase Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a

16 .

5) Synapsis initiates formation of a protein bridge called the synaptonemal complex, a tri-layer protein structure that maintains synapsis by tightly binding nonsister chromatids of ________ to one another.

Answer: homologous chromosomes Section: 3.2 Skill: Knowledge/Comprehension LO: G1, S3.2a 6) An individual is heterozygous for a gene at a specific locus. ________ will have the same form of alleles at that locus after S phase.

Answer: Sister chromatids Section: 3.1 Skill: Applying/Analyzing LO: G1, S3.1a 3.4 Essay Questions 1) What is the difference between cytokinesis and karyokinesis? Which process needs to be more precise for accurate cell division?

Answer: Karyokinesis is the equal partitioning of the chromosomal material in the nucleus of the parental cell between the nuclei of the two daughter cells. This process requires first that each of the chromosomes in the nucleus be fully and accurately duplicated and then that the duplicate copies of each chromosome be separated so that one copy goes to the nucleus of one daughter cell and the second copy goes to the other daughter nucleus. Karyokinesis is followed by cytokinesis, the partitioning of the cytoplasmic contents of the parental cell into the daughter cells.

Cytokinesis does not demand the same degree of equivalency required in karyokinesis. The cytoplasm of the parental cell contains an abundance of the proteins and organelles that the daughter cells require in order to function, so the division of this material need not be equal.

Section: 3.1 Skill: Synthesis/Evaluation LO: G2, S3.1b 2) What are three important differences between mitosis and meiosis I? Answer: In meiosis I, but not in mitosis, the following three processes occur: 1. Homologous chromosome pairing 2. Crossing over between homologous chromosomes 3. Segregation (separation) of the homologous chromosomes, which reduces chromosomes to the haploid number Section: 3.2 Skill: Synthesis/Evaluation LO: G2, S3.2a

17 .

3) Describe how microtubules and cohesion work together to align the chromosomes along the metaphase plate during the process of sister chromatid cohesion.

Answer: Kinetochore microtubules are attached to the kinetochore at each centromere of sister chromatids. Because they are tethered to kinetochore microtubules from opposite centrosomes, the sister chromatids experience opposing forces that are critical to the positioning of chromosomes along an imaginary midline at the equator of the cell. This imaginary line is called the metaphase plate. The tension created by the pull of kinetochore microtubules is balanced by a companion process known as sister chromatid cohesion. Sister chromatid cohesion is produced by the protein cohesin that localizes between the sister chromatids and holds them together to resist the pull of kinetochore microtubules. As microtubules move chromosomes toward the midline of the cell, cohesin helps keep the sister chromatids together to ensure proper chromosome positioning and to prevent their premature separation.

Section: 3.2 Skill: Synthesis/Evaluation LO: G2, S3.2a 4) Which meiotic division(s) reduce(s) chromosome number (i.e., meiosis I, meiosis II, or both)?

Explain your answer.

Answer: Meiosis I reduces chromosome number when homologs separate. After meiosis I, each cell contains a haploid number of chromosomes. Meiosis II reduces DNA content, but does not reduce chromosome number.

Section: 3.2 Skill: Synthesis/Evaluation LO: G2, S3.2a 5) Describe how Mendel's two laws are illustrated by the movement of chromosomes in meiosis.

Answer: Mendel's Law of Segregation is illustrated in meiosis I when homologs separate.

Alternate alleles of a gene found on homologs segregate from each other and end up in separate gametes. Mendel's Law of Independent Assortment is illustrated by the behavior of two pairs of homologs during metaphase I. The alignment of one homologous pair is independent of the alignment of other pairs.

Section: 3.2 Skill: Applying/Analyzing LO: G2, S3.2a

18 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 4 Gene Interaction 4.1 Multiple-Choice Questions 1) Which mode of inheritance produces heterozygotes with phenotypes that differ from either homozygote but typically more closely resembles one homozygous phenotype than the other?

A) complete dominance B) incomplete dominance C) codominance D) epistasis E) incomplete penetrance Answer: B Section: 4.1 Skill: Remembering/Understanding LO: S4.1c 2) Which mode of inheritance results in the phenotype of a heterozygote being indistinguishable from that of an organism homozygous for the dominant allele?

A) complete dominance B) incomplete dominance C) codominance D) epistasis E) incomplete penetrance Answer: A Section: 4.1 Skill: Remembering/Understanding LO: S4.1c 3) Which mode of inheritance results in both alleles being detected equally in the heterozygous phenotype?

A) complete dominance B) incomplete dominance C) codominance D) epistasis E) incomplete penetrance Answer: C Section: 4.1 Skill: Remembering/Understanding LO: S4.1c

1 .

4) A mutation results in an enzyme that is partially active compared to the wild-type allele. This type of "leaky" mutation is classified as ________.

A) null/ amorphic B) hypomorphic C) hypermorphic D) neomorphic E) dominant negative Answer: B Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 5) A mutation resulting in an inactive gene product is classified as ________.

A) null/ amorphic B) hypomorphic C) hypermorphic D) neomorphic E) dominant negative Answer: A Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 6) Two proteins interact to form a multimeric complex. When one of the proteins is mutated, there is a substantial loss of functional activity in the multimeric protein. This type of mutation is classified as ________.

A) null/ amorphic B) hypomorphic C) hypermorphic D) neomorphic E) dominant negative Answer: E Section: 4.1 Skill: Remembering/Understanding LO: S4.1f

2 .

7) Many oncogenes result from mutations that cause excessive expression of a protein in cells where it is normally not expressed or is expressed at inappropriate times during development.

This type of mutation can be described as ________.

A) amorphic B) hypomorphic C) hypermorphic D) neomorphic E) dominant negative Answer: C Section: 4.1 Skill: Applying/Analyzing LO: S4.1f 8) A mutation results in a gene product with a novel function that is not normally found in wildtype organisms. This type of mutation is known as ________.

A) amorphic B) hypomorphic C) hypermorphic D) neomorphic E) dominant negative Answer: D Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 9) Which of the following is correct regarding individuals who are blood type AB? A) They express B-transferase that adds N-acetylgalactosamine to the H antigen.

B) They carry both the anti-A and anti-B antibodies.

C) They do not carry the A or B antigen.

D) They are the universal recipients.

E) Their blood cells clump when they receive blood from an individual with the genotype IAi. Answer: D Section: 4.1 Skill: Applying/Analyzing LO: G5, S4.1f

3 .

10) A man with blood type A (whose mother was blood type O) has children with a woman that has blood type AB. The man and the woman are also heterozygous for the H antigen. What is the probability that they will have a child with blood type A? A) 0 B) 1/8 C) 3/8 D) 1/2 E) 3/4 Answer: C Section: 4.1 Skill: Remembering/Understanding LO: G4, S4.1a 11) The parental cross between a rabbit with the Himalayan phenotype (genotype chch) and an albino rabbit (genotype cc) results in F1 rabbits that all have the Himalayan phenotype (genotype chc).

If the resulting F1 rabbits are crossed, what proportion of the F2 offspring will have the Himalayan phenotype and what causes the hypomorphic ch allele to be unstable and result in the distinctive fur pattern?

A) 1/4; temperature B) 3/4; temperature C) 1/2; complementation D) 3/4; complementation E) 1/2; temperature Answer: B Section: 4.1 Skill: Remembering/Understanding LO: G3, S4.1a

4 .

12) What type of allele is often detected as a distortion in segregation ratios, where one class of expected progeny is missing?

A) dominant negative allele B) temperature-sensitive allele C) lethal allele D) partially dominant allele E) incompletely penetrant allele Answer: C Section: 4.1 Skill: Remembering/Understanding LO: S4.1c 13) You discover a new allele of a gene important for tail formation in mice. WT mice have long tails, but mice heterozygous for the allele have short tails. When you cross two heterozygous mice together, you obtain a 2:1 ratio of short-tailed mice to long-tailed mice. None of the shorttailed progeny are homozygous. What type of allele results in short tails?

A) dominant negative allele B) temperature-sensitive allele C) lethal allele D) partially dominant allele E) incompletely penetrant allele Answer: C Section: 4.1 Skill: Applying/Analyzing LO: S4.1c 14) The amount of enzyme activity in a cell that is homozygous for a mutant allele is 400 units.

The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 300 units. What type of allele is the mutant allele?

A) null/ amorphic B) hypomorphic C) hypermorphic D) dominant negative E) neomorphic Answer: C Section: 4.1 Skill: Applying/Analyzing LO: S4.1f

5 .

15) The amount of enzyme activity in a cell that is homozygous for a mutant allele is 400 units.

The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 300 units. What is the dominance relationship between the WT and mutant allele for the phenotype of amount of enzyme per cell?

A) The WT allele is dominant.

B) The mutant allele is dominant.

C) The WT and mutant alleles are codominant.

D) The WT and mutant alleles show incomplete dominance.

E) It is impossible to determine from the information given.

Answer: D Section: 4.1 Skill: Applying/Analyzing LO: S4.1f 16) The amount of enzyme activity in a cell that is homozygous for a mutant allele is 0 units. The amount of enzyme activity in a cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 0 units. What type of allele is the mutant allele?

A) null/ amorphic B) hypomorphic C) hypermorphic D) dominant negative E) neomorphic Answer: D Section: 4.1 Skill: Applying/Analyzing LO: S4.1f 17) Which phenomenon explains differences in the inheritance patterns of the appearance of a chin beard between males and females of certain species of goats, even when their genotypes are the same?

A) sex-limited trait B) sex-influenced trait C) incomplete penetrance D) variable expressivity E) lethal allele Answer: B Section: 4.2 Skill: Applying/Analyzing LO: S4.1e

6 .

18) In certain goat breeds, appearance of a chin beard is a sex-influenced trait. Recall that bearding is inherited as an autosomal trait determined by two alleles, B1 and B2, and females must be homozygous for the bearded allele, B2, to have a beard. What genotypes must a bearded billy goat (male) and a beardless female goat have if they have a bearded female offspring?

A) The bearded billy goat must be heterozygous for the bearded allele, while the beardless female must be homozygous for the bearded allele.

B) Both the bearded billy goat and beardless female must be heterozygous for the bearded allele.

C) The bearded billy goat could be heterozygous or homozygous for the bearded allele, while the beardless female must be heterozygous for the bearded allele.

D) Both the bearded billy goat and beardless female must be homozygous for the bearded allele.

E) The bearded billy goat could be heterozygous or homozygous for the bearded allele, while the beardless female must be homozygous for the bearded allele.

Answer: C Section: 4.2 Skill: Applying/Analyzing LO: G2, G3, S4.1e 19) Brachydactyly type D is human autosomal dominant condition in which the thumbs are abnormally short and broad. In most cases, both thumbs are affected, but occasionally just one thumb is involved. The pedigree above shows a family in which brachydactlyly type D is segregating. Filled circles and squares represent females and males who have involvement of both thumbs. Half-filled in symbols represent family members with just one thumb affected?

There is evidence of variable expressivity and incomplete penetrance in this family. Which individual is most likely nonpenetrant for the trait?

A) III-11 B) IV-1 C) IV-5 D) III-10 E) II-4 Answer: D Section: 4.2 Skill: Applying/Analyzing LO: G3, S4.2b

7 .

20) Brachydactyly type D is human autosomal dominant condition in which the thumbs are abnormally short and broad. In most cases, both thumbs are affected, but occasionally just one thumb is involved. The pedigree above shows a family in which brachydactlyly type D is segregating. Filled circles and squares represent females and males who have involvement of both thumbs. Half-filled in symbols represent family members with just one thumb affected?

There is evidence of variable expressivity and incomplete penetrance in this family. Which individual is most likely nonpenetrant for the trait?

A) III-11 B) IV-1 C) IV-5 D) III-5 E) II-4 Answer: D Section: 4.2 Skill: Applying/Analyzing LO: G3, S4.2b 21) King George III of England and other members of the royal family were afflicted with a series of strange, seemingly unrelated symptoms including abdominal pain, rapid pulse, convulsions, and insanity. It has been determined that he likely suffered from porphyria, caused by a mutation in a single allele. What is the genetic term describing the alteration of multiple, distinct traits of an organism by a mutation in a single gene?

A) pleiotropy B) incomplete dominance C) codominance D) epistasis E) incomplete penetrance Answer: A Section: 4.2 Skill: Remembering/Understanding LO: S4.3a

8 .

22) Gene interactions in which an allele of one gene modifies or prevents expression of alleles of another gene is known as ________.

A) pleiotropy B) incomplete dominance C) codominance D) epistasis E) incomplete penetrance Answer: D Section: 4.3 Skill: Remembering/Understanding LO: S4.3a 23) Bateson and Punnett crossed two white-flowered lines and saw all purple flowers in the F1 generation. They concluded this was an example of complementary gene interactions because a cross of the F1 plants yielded what ratio in the F2 generation?

A) 8 purple to 8 white B) 16 purple to 0 white C) 0 purple to 16 white D) 7 purple to 9 white E) 9 purple to 7 white Answer: E Section: 4.3 Skill: Applying/Analyzing LO: G3, S4.3b 24) The 9:6:1 ratio seen in the dihybrid cross of summer squash indicates what genetic relationship between the two genes controlling fruit shape?

A) complementary gene interaction B) dominant gene interaction C) recessive epistasis D) dominant epistasis E) dominant suppression Answer: B Section: 4.3 Skill: Remembering/Understanding LO: S4.3d

9 .

25) In the biosynthetic pathway for conversion from homoserine to methionine, you identify a Neurospora crassa double mutant Met1/Met2. This mutant will grow only if which supplement(s) are added to the minimal media?

A) homoserine only B) cysteine only C) homocysteine only D) cystathionine and homocysteine E) methionine only Answer: E Section: 4.3 Skill: Evaluating/Creating LO: G3, G9, S4.3f

10 .

26) Which step is catalyzed by the enzyme responsible for the Met 2 mutant?

A) homoserine → cysteine B) cysteine → cystathionine C) cystathionine → homocysteine D) homocysteine → methionine E) methionine → homoserine Answer: C Section: 4.3 Skill: Evaluating/Creating LO: G3, G9, S4.3f 27) Wild-type bacteria can grow on minimal medium. Four mutants that cannot grow on minimal medium but can grow on minimal medium supplemented with the nutrient "H" are isolated. It is suspected that metabolites T, P, and A are in the biochemical pathway for synthesis of H, so each mutant is tested for the ability to grow on minimal medium supplemented with these metabolites: A. Mutant 1: can grow on minimal medium supplemented with T, but not P or A B. Mutant 2: is unable to grow on minimal medium supplemented with T, P, or A C. Mutant 3: is able to grow on minimal medium supplemented with A or T, but not P D. Mutant 4: can grow on minimal medium supplemented with T, P, or A. Which of the following is NOT consistent with this information?

A) Mutant 3 blocks the conversion of metabolite P → metabolite A B) Mutant 4 blocks a step before all the metabolites C) Mutant 1 will cause a build up of metabolite A D) Mutant 2 blocks a step right before metabolite H E) Mutant 3 will cause a build up of metabolite T Answer: E Section: 4.3 Skill: Evaluating/Creating LO: G1, G2, G3, S4.3g

11 .

28) Wild-type bacteria can grow on minimal medium. Four mutants that cannot grow on minimal medium but can grow on minimal medium supplemented with the nutrient "H" are isolated. It is suspected that metabolites T, P, and A are in the biochemical pathway for synthesis of H, so each mutant is tested for the ability to grow on minimal medium supplemented with these metabolites: A. Mutant 1: can grow on minimal medium supplemented with T, but not P or A B. Mutant 2: is unable to grow on minimal medium supplemented with T, P, or A C. Mutant 3: is able to grow on minimal medium supplemented with A or T, but not P D. Mutant 4: can grow on minimal medium supplemented with T, P, or A. Which of the following is NOT consistent with this information?

A) Mutant 3 blocks the conversion of metabolite P → metabolite A B) Mutant 4 blocks a step before all the metabolites C) Mutant 1 will cause a build up of metabolite A D) Mutant 2 blocks a step right before metabolite H E) Mutant 1 blocks the conversion of metabolite T → metabolite H Answer: E Section: 4.3 Skill: Evaluating/Creating LO: G1, G2, G3, S4.3g 29) Independent assortment predicts a 9:3:3:1 ratio with four different phenotypes in the F2 progeny. If the alleles are epistatic, what would you predict?

A) more than four phenotypes B) fewer than four phenotypes C) heterozygotes with a novel phenotype between the dominant and recessive homozygotes D) 1/3 of the progeny with the dominant phenotypes and 2/3 recessive E) no change in the 9:3:3:1 ratio Answer: B Section: 4.3 Skill: Applying/Analyzing LO: G2, S4.3b 30) Two pure-breeding mutant plants produce white flowers. When they are crossed, all of the progeny have wild-type purple flowers. What does this genetic complementation tell you?

A) The genes are part of two distinct biosynthetic pathways.

B) The two lines exhibit different mutations in the same gene.

C) More than one gene is involved in determining the phenotype.

D) The allele is pleiotropic.

E) The allele exhibits incomplete dominance.

Answer: C Section: 4.3 Skill: Applying/Analyzing LO: S4.3a

12 .

31) You are looking at the color of feathers in ducks and find that yellow ducks (Y) are dominant to green ducks (y). However, a second gene, H, controls whether the color will be expressed in the feathers. If the duck is hh, the duck will always be white, because the pigment does not go into feathers. What ratio of phenotypes would you expect following a dihybrid cross?

A) 9:6:1 B) 9:3:4 C) 13:3 D) 15:1 E) 12:3:1 Answer: B Section: 4.3 Skill: Applying/Analyzing LO: S4.3b 32) In sheep, coat color is influenced by two genes. Gene A influences pigment production, while gene B produces black or brown pigment. If two heterozygous white sheep resulted in 12 white sheep, 3 black sheep, and 1 brown sheep, which genotype(s) of the white sheep explain this data?

A) The white sheep could be A_B_ or aaB_.

B) The white sheep could be A_B_ or aabb.

C) The white sheep could be A_B_ or A_bb.

D) The white sheep must all be aabb.

E) The white sheep must all be A_B_.

Answer: C Section: 4.3 Skill: Applying/Analyzing LO: G2, G3, S4.3c 33) A certain species of morning glories produces flowers that are blue, red, or purple. Two pure-breeding purple lines are crossed and produce F1 progeny that all make blue flowers. The F1 are allowed to self and produce 320 F2 progeny with the following distribution: 185 blue, 115 purple, and 20 red.

Which the following is NOT consistent with this information?

A) Red-flowering plants are homozygous recessive for both genes.

B) The pure-breeding parental parents are homozygous recessive for mutations in two different genes.

C) Dominant gene interaction appears to result in a 9:6:1 ratio.

D) Blue-flowering plants are either A_bb or aaB_.

E) Analysis of the F1 and F2 progeny phenotypes suggests epistasis.

Answer: D Section: 4.3 Skill: Applying/Analyzing LO: G1, G2, G3, S4.3b, S4.3c, S4.3d

13 .

34) Deafness is caused by recessive mutations in any one of at least five genes. Two deaf individuals have nine children, all of whom have normal hearing. Which of the following can you conclude?

A) The parents have mutations in the same gene.

B) The parents have mutations in different genes.

C) The parents have the same mutated protein involved in inner ear development.

D) The mutations are codominant to the normal allele.

E) The mutations are incompletely dominant to the normal allele.

Answer: B Section: 4.4 Skill: Applying/Analyzing LO: S4.4 35) In yeast, there are three gene products required to synthesize the amino acid lysine. You cross two haploid lysine auxotrophs to form a diploid. The diploid fails to grow on plates lacking lysine. Which of the following can you definitively conclude?

A) The haploid strains have mutations in the same gene.

B) The haploid strains have mutations in different genes.

C) The haploid strains have identical mutations in the same genes.

D) The haploid strains have identical mutations in different genes.

E) The haploid strains must belong to the complementation group encoding the first enzyme in the biosynthetic pathway.

Answer: A Section: 4.4 Skill: Evaluating/Creating LO: G2, G9, S4.4

14 .

36) Complementation tests of distinct recessive mutants, 1 through 8, produce the data in the matrix below. A plus (+) indicates complementation, meaning the phenotype of the combined alleles is wild type, and a minus (-) indicates a failure to complement meaning that a mutant phenotype results. Assume that the missing mutant combinations would yield data consistent with the entries that are shown.

How many complementation groups are formed by these eight mutants?

A) 2 B) 3 C) 4 D) 5 E) 6 Answer: B Section: 4.4 Skill: Evaluating/Creating LO: G2, G3, G9, S4.4

15 .

4.2 Short Answer Questions 1) A metabolic reaction requires 40 units of enzymatic activity to proceed. If a dominant allele D can generate 40 units of enzyme and a mutant allele d" generates 20 units of enzyme, what can be said of the dominant wild-type allele?

Answer: D is haplosufficient.

Section: 4.1 Skill: Applying/Analyzing LO: S4.1f 2) A metabolic reaction requires 10 units of enzymatic activity to proceed. If a dominant allele D can generate 8 units of enzyme and a recessive allele d can generate 2 units of enzyme, what can be said of the dominant wild-type allele?

Answer: D is haploinsufficient.

Section: 4.1 Skill: Applying/Analyzing LO: S4.1f 3) Are loss-of-function mutations more likely to be dominant or recessive?

Answer: recessive Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 4) What are the three categories of loss-of-function mutations?

Answer: amorphic (null), hypomorphic (leaky), dominant negative Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 5) What are the two categories of gain-of-function mutations?

Answer: neomorphic and hypermorphic Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 6) You cross a pure-breeding white flower with a pure-breeding red flower, and the offspring are all pink. This is an example of what type of inheritance?

Answer: incomplete dominance Section: 4.1 Skill: Applying/Analyzing LO: S4.1c

16 .

7) You cross a pure-breeding white flower with a pure-breeding red flower, and the offspring are white with red spots. This is an example of what type of inheritance?

Answer: codominance Section: 4.1 Skill: Applying/Analyzing LO: S4.1c 8) Alleles of the Sbe1 gene, which controls pea shape, can be detected by DNA analysis. In heterozygous plants, two distinct bands of DNA are visible. What is the relationship between the alleles of the heterozygote at the molecular level?

Answer: codominant Section: 4.1 Skill: Remembering/Understanding LO: S4.1c 9) The four different human blood types are caused by how many different alleles? What are the alleles?

Answer: three: IA, IB, and i Section: 4.1 Skill: Remembering/Understanding LO: 4.1b 10) Which antigen, expressed on the surface of all red blood cells, is modified by A- or Btransferases?

Answer: H antigen Section: 4.1 Skill: Remembering/Understanding LO: G5 11) The allele responsible for the Siamese coat-color pattern produces an unstable tyrosinase enzyme. This type of gene product is an example of what type of allele?

Answer: temperature-sensitive Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 12) Most people with the dominant mutant polydactyly allele have extra digits but at least 25% have the normal number of digits. What is the genetic explanation for this observation?

Answer: incomplete penetrance Section: 4.2 Skill: Remembering/Understanding LO: S4.2a

17 .

13) People with the dominant mutant polydactyly allele can have extra digits on one or both of their hands. What is the genetic explanation for this observation?

Answer: variable expressivity Section: 4.2 Skill: Applying/Analyzing LO: S4.2a 14) To better understand which genes are involved in developmental pathways, geneticists use experimental analyses of mutant phenotypes. What is this analytic approach called?

Answer: genetic dissection Section: 4.3 Skill: Remembering/Understanding LO: G9 15) Mendel studied tall and short pure-breeding lines of pea plants. Inherited genetic variation would dictate that one line would produce tall plants and the other short plants, but the genes and phenotypes are influenced by what external factor?

Answer: environmental influence Section: 4.3 Skill: Remembering/Understanding LO: G5 4.3 Fill-in-the-Blank Questions 1) Amorphic, hypomorphic, and dominant negative mutations are ________ mutations, which decrease or eliminate gene activity.

Answer: Loss-of-function Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 2) Hypermorphic and neomorphic mutations are ________ mutations, which cause overexpression or result in new functions.

Answer: Gain-of-function Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 3) Most combinations of different ABO alleles result in complete dominance of one allele.

Which combination results in codominance?

Answer: IAIB Section: 4.1 Skill: Remembering/Understanding LO: S4.1c

18 .

4) If an organism with a particular genotype fails to produce the corresponding phenotype, the organism is said to be ________ for the trait.

Answer: nonpenetrant Section: 4.1 Skill: Remembering/Understanding LO: S4.1f 5) In Labrador retrievers, coat color is controlled by gene interaction in which homozygosity for a recessive allele can mask the phenotypic expression of a second gene. This genetic interaction is known as ________ and has a characteristic ________ ratio.

Answer: recessive epistasis; 9:3:4 Section: 4.3 Skill: Remembering/Understanding LO: S4.3d

19 .

4.4 Essay Questions 1) Proper cross-matching of blood type is essential for safe blood transfusion. What is the general rule for safe blood transfusion, and what are the consequences if a patient receives the incorrect blood type by accident? If a patient's blood type is unknown, what is the safest course of action?

Answer: The general rule for safe blood transfusion is that the recipient blood must not contain an antibody that reacts with an antigen in the donated blood. When such a reaction occurs, blood clots produced by clumping blood cells form at the site of transfusion. The clots block circulation, deprive tissues of oxygen, and can potentially cause life-threatening complications.

Thus, if a patient's blood type is not known, they are given type O-negative blood because it contains no antigens for the patient's own blood to react with.

Section: 4.1 Skill: Evaluating/Creating LO: G2, G5 2) You have crossed two Mexican hairless dogs, and the offspring are 1/3 hairy and 2/3 hairless.

Given this phenotypic ratio, draw the Punnett square for this cross. What are the genotypes of the P1 and F1 dogs in this cross? List the predicted genotype as well as the phenotype for each of the offspring. Which genotypes are hairless, and which are hairy? Can you design a genetic cross that would yield a true-breeding hairless line (where all offspring are hairless)?

Answer: The 1:2 ratio in the offspring provides strong evidence that this trait involves a lethal allele. Thus, the Punnett square for the cross would be

You cannot design a true-breeding cross, because the HH genotype is lethal. Thus, a maximum of 2/3 of all of the offspring from any cross are predicted to be hairless.

Section: 4.1 Skill: Evaluating/Creating LO: G1, S4.1b, S4.1c

20 .

3) Explain how the ch (Himalayan) allele and tyrosinase control the Siamese coat-color pattern.

What is the underlying reason that certain parts of a Siamese cat (e.g., tail and ears) are darker than the cat's trunk?

Answer: The tyrosinase enzyme produced by the hypomorphic ch (Himalayan) allele is unstable and is inactivated at a temperature very near the normal body temperature of most mammals.

Thus, ch is a TEMPERATURE-SENSITIVE allele. The parts of cats that are farthest away from the core of the body (the paws, ears, tail, and tip of the nose) at most times tend to be slightly cooler than the trunk. At these cooler extremities, the temperature-sensitive tyrosinase produced by the ch allele remains active, producing pigment in the hairs there. However, in the warmer central portion of the body, the slightly higher temperature is enough to cause the tyrosinase produced by the ch allele to denature, or unravel. This inactivates the enzyme and leads to an absence of pigment in the central portion of the body. Animals that are chch or chc have the Himalayan phenotype. The final allele in the series, c, is a null allele that does not produce functional tyrosinase.

Section: 4.1 Skill: Evaluating/Creating LO: G2, G8, S4.1f 4) Describe the difference between incomplete penetrance and variable expressivity. It is often difficult to pinpoint the cause of incomplete penetrance or variable expressivity. What possible interactions may be responsible?

Answer: In incomplete penetrance, a genotype is not expressed by every organism in which it is present. In variable expressivity, the organisms that share a genotype express the corresponding phenotype to different degrees. Both are explained by genetic or nongenetic interactions that modify or prevent the consistent expression of a genotype.

Three kinds of interactions may be responsible: (1) other genes interactin ways that modify the expression of the mutant allele, (2) environmental or developmental (i.e., nongenetic) factors interact with the mutant allele to modify its expression, and (3) some combination of other genes and environmental factors interact to modify expression of the mutation. In inbred laboratory strains of model genetic organisms, variation in genetic factors can be eliminated experimentally to allow separation of gene–gene and gene–environment variability, something that cannot be done in organisms such as humans.

Section: 4.2 Skill: Evaluating/Creating LO: G8, S4.2a

21 .

5) Phenylketonuria (PKU) is caused by the absence of the enzyme phenylalanine hydroxylase, which catalyzes the first step of the pathway that breaks down the amino acid phenylalanine, a common component of dietary protein. Explain how environmental intervention is commonly practiced to prevent the development of this human autosomal recessive condition.

Answer: PKU damages neurons because the body is unable to break down dietary phenylalanine, ultimately causing irreversible mental retardation and neuron death. There are two main ways in which the environment has been changed to affect the development of PKU. First, newborns are routinely screened for PKU so they can be placed on a phenylalanine-free diet and will avoid PKU complications. Second, individuals living with PKU are warned of foods containing phenylalanine by strict labeling of foods (e.g., foods and beverages containing the artificial sweetener aspartame). Thus, while they will always have the mutated allele, they are able to avoid developing the disease through strict control of diet.

Section: 4.3 Skill: Evaluating/Creating LO: G5, G7, G8

22 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 5 Genetic Linkage and Mapping in Eukaryotes 5.1 Multiple-Choice Questions 1) Syntenic genes can assort independently when ________.

A) they are very close together on a chromosome B) they are located on different chromosomes C) crossing over occurs rarely between the genes D) they are far apart on a chromosome and crossing over occurs frequently between the genes E) they are far apart on a chromosome and crossing over occurs very rarely between the genes Answer: D Section: 5.1 Skill: Remembering/Understanding LO: S5.1a 2) The alleles of linked genes tend to ________.

A) segregate together during gamete production B) assort independently C) be mutated more often than unlinked genes D) experience a higher rate of crossing over E) assort independently and show a higher rate of crossing over Answer: A Section: 5.1 Skill: Remembering/Understanding LO: S5.1a 3) Genetic linkage leads to the production of a significantly greater than expected number of gametes containing chromosomes with ________.

A) allele combinations that are different from parental combinations B) parental combinations of alleles C) mutant alleles D) dominant alleles E) recessive alleles Answer: B Section: 5.1 Skill: Remembering/Understanding LO: S5.1a

1 .

4) The syntenic genes B and T are linked. A cross between two parents, with genotypes BBtt and bbTT, produces F1 progeny with the BbTt genotype. What are the possible arrangements of alleles on the F1 progeny's chromosomes?

A) BT/bt B) Bt/bT C) Bb/Tt D) BT/Bt or bt/bT E) B/T or b/t Answer: B Section: 5.1 Skill: Applying/Analyzing LO: G3, S5.1b 5) In an individual that is dihybrid for completely linked genes, which alleles would you expect in the gametes?

A) one parental allele combination occurring more frequently than another B) two equally frequent gametes containing only parental allele combinations C) only recombinant gametes D) one parental allele combination occurring more frequently than another and no recombinant gametes E) two equally frequent gametes containing only parental allele combinations and no recombinant gametes Answer: E Section: 5.1 Skill: Applying/Analyzing LO: G2, S5.1b 6) What is a two-point test cross, involving a dihybrid F1 fly, used for when mapping genes?

A) to determine if the F1 fly is homozygous dominant or heterozygous B) to determine the allele combinations that the F1 fly can produce C) to determine the allele combinations that a homozygous recessive fly can create D) to determine the recombination frequency between two genes E) to determine which alleles are dominant and which are recessive Answer: D Section: 5.1 Skill: Remembering/Understanding LO: S5.1c

2 .

7) Genes A and B are 7.5 map units apart on one chromosome. Genes C and D are 8.0 map units apart on a second non-homologous chromosome. The recombination frequency between A and C is ________.

A) 0.851 B) 0.003 C) 0.006 D) 0.500 E) 0.155 Answer: D Section: 5.1 Skill: Applying/Analyzing LO: G3, S5.2a 8) In fruit flies, red eyes (pr+_) are dominant to purple eyes (prpr) and normal wings (vg+_) are dominant to vestigial wings (vgvg). The genes are located on the same chromosome. A pure breeding red-eyed fly with vestigial wings was crossed with a pure-breeding purple-eyed fly with normal wings. All of the F1 progeny had a WT phenotype. Which of the following represents the arrangement of alleles on the F1's chromosome?

A) prvg/pr+vg+ B) pr+vg/prvg+ C) prpr+/vgvg+ D) prvg/prvg E) pr+vg+/pr+vg+ Answer: B Section: 5.1 Skill: Applying/Analyzing LO: G3, S5.1b 9) In fruit flies, red eyes (pr+_) are dominant to purple eyes (prpr) and normal wings (vg+_) are dominant to vestigial wings (vgvg). The genes are located on the same chromosome. A purebreeding red-eyed fly with vestigial wings was crossed with a pure-breeding purple-eyed fly with normal wings. All of the F1 progeny had a WT phenotype. The recombination frequency between the two genes is 15%. If an F1 individual were test crossed, what percentage of the progeny would you expect to have the WT phenotype?

A) 7.5% B) 15% C) 50% D) 85% E) 92.5% Answer: A Section: 5.1 Skill: Applying/Analyzing LO: G2, G3, G4, S5.1d

3 .

10) What organism exhibits complete genetic linkage, meaning there is no recombination between homologous chromosomes?

A) Male Drosophila B) Female Drosophila C) Zea mays (corn) D) Male Homo sapiens E) Female Homo sapiens Answer: A Section: 5.1 Skill: Remembering/Understanding 11) William Bateson and Reginald Punnett were not able to detect genetic linkage in sweet peas.

Which of the following explains why Thomas Morgan was able to detect autosomal genetic linkage?

A) He sequenced the DNA and was able to determine the distance between the genes.

B) He performed a three-point test cross.

C) He crossed two heterozygous individuals and was able to determine offspring genotypes from their phenotypes.

D) He crossed a heterozygous individual with a homozygous recessive individual and was able to determine progeny genotypes from their phenotypes.

E) He performed Lod score analysis to determine the distance between the genes he was studying.

Answer: D Section: 5.1 Skill: Remembering/Understanding LO: G2, G9 12) In sweet peas, the genes for flower color and pollen grain shape are 11 cM apart. A purebreeding purple flowering plant with round pollen grains is crossed to a pure-breeding red flowering plant with long pollen grains. The resulting F1 offspring are all purple flowering plants with long pollen grains. What percent of offspring from a test cross analysis of the F1 individuals would you expect to be red flowering plants with long pollen grains?

A) 44.5 B) 11 C) 5.5 D) 89 E) 39 Answer: A Section: 5.2 Skill: Applying/Analyzing LO: G4, G9, S5.2a, 5.2b

4 .

13) In sweet peas, the genes for flower color and pollen grain shape are 11 cM apart. A purebreeding purple flowering plant with round pollen grains is crossed to a pure-breeding red flowering plant with long pollen grains. The resulting F1 offspring are all purple flowering plants with long pollen grains. What percent of offspring from a test cross analysis of the F1 individuals would you expect to be purple flowering plants with long pollen grains?

A) 44.5 B) 11 C) 5.5 D) 89 E) 39 Answer: C Section: 5.2 Skill: Applying/Analyzing LO: G4, G9, S5.2a, S5.2b 14) If you are given a recombination frequency of 34% between genes X and Y and 27% between X and Z, can you predict the order of the three genes?

A) Yes; the order is X-Z-Y. B) Yes; the order is X-Y-Z. C) Yes; the order is Z-X-Y. D) No; based on this data alone, the order could be Z-Y-X or X-Y-Z. E) No; based on this data alone, the order could be X-Z-Y or Z-X-Y. Answer: E Section: 5.2 Skill: Applying/Analyzing LO: G3, S5.2a 15) What type of test would you use to determine whether observed data constitute evidence of genetic linkage or are simply a case of chance variation from expected values?

A) test cross B) two-point test cross C) three-point test cross D) chi-square analysis E) recombination frequency (r) calculation Answer: D Section: 5.2 Skill: Remembering/Understanding LO: G1, S5.2c

5 .

16) You perform a test cross of the dihybrid AaBb and score the phenotypes of 1000 progeny.

Assuming independent assortment, how many of the progeny do you expect to display the dominant phenotype for both the A and B genes?

A) 100 B) 200 C) 250 D) 500 E) 750 Answer: C Section: 5.2 Skill: Applying/Analyzing LO: G3, G4, S5.2b 17) A geneticist is mapping the chromosomes of the newly captured gremlin. Stripe is heterozygous for three linked genes with alleles Ee, Hh, and Bb, that determine if gremlins are evil (E), have hair (H), and biting teeth (B). In order to determine if the three genes are linked, a standard testcross was done, and the 1000 offspring had the following genotypes: 48 36 400 4 426 46 38 2

ee Hh bb ee hh Bb ee Hh Bb Ee Hh Bb Ee hh bb Ee hh Bb Ee Hh bb ee hh bb

What is the recombination frequency between genes E and H? A) 8% B) 92% C) 7.4% D) 8.4% E) 40% Answer: A Section: 5.3 Skill: Applying/Analyzing LO: G3, G4, S5.3a

6 .

18) A geneticist is mapping the chromosomes of the newly captured gremlin. Stripe is heterozygous for three linked genes with alleles Bb, Ee, and Hh that determine if gremlins have biting teeth (B), are evil (E), and have hair (H). In order to determine if the three genes are linked, a standard testcross was done, and the 1000 offspring had the following genotypes: 48 36 400 4 426 46 38 2

bb ee Hh Bb ee hh Bb ee Hh Bb Ee Hh bb Ee hh Bb Ee hh bb Ee Hh bb ee hh

What is the recombination frequency between genes E and B? A) 9.4% B) 10.0% C) 9.0% D) 8.4% E) 42.6% Answer: B Section: 5.3 Skill: Applying/Analyzing LO: G3, G4, S5.3a 19) Incomplete genetic linkage of three genes in a trihybrid produces eight genetically different gamete genotypes. How many different gamete genotypes are produced in a four-gene cross with incomplete genetic linkage?

A) 4 B) 8 C) 16 D) 20 E) 24 Answer: C Section: 5.3 Skill: Applying/Analyzing LO: G2, G4

7 .

20) Assuming three genes are linked, how many recombinant genotypes would you expect and at what frequency?

A) two recombinant genotypes, more frequent than expected by independent assortment B) four recombinant genotypes, more frequent than expected by independent assortment C) six recombinant genotypes, more frequent than expected by independent assortment D) four recombinant genotypes, less frequent than expected by independent assortment E) six recombinant genotypes, less frequent than expected by independent assortment Answer: E Section: 5.3 Skill: Remembering/Understanding LO: G2, G3, S5.3d 21) For a given cross, the expected number of double recombinants is 18 and the observed number of double recombinants is 12. What is the coefficient of coincidence (COC)?

A) 0.33 B) 0.50 C) 0.30 D) 0.67 E) 1.50 Answer: D Section: 5.3 Skill: Applying/Analyzing LO: G4, S5.3c 22) For a given cross, the expected number of double recombinants is 18 and the observed number of double recombinants is 12. What is the interference (I)?

A) 0.33 B) 0.50 C) 0.30 D) 0.67 E) 1.50 Answer: A Section: 5.3 Skill: Applying/Analyzing LO: G4, S5.3c

8 .

23) Genes A and B are located 10 cM from each other on a chromosome. Gene C is located 25 cM from gene A and 15 cM from gene B. Assuming that I = 0, what is the probability that the trihybrid ABC/abc will produce an AbC gamete?

A) 76.5% B) 63.75% C) 25% D) 1.5% E) 3.75% Answer: D Section: 5.3 Skill: Applying/Analyzing LO: G4, S5.3d 24) The order of the genes on a plant chromosome is A, B, C, where A and B are located 10 cM apart and B and C are located 3 cM apart. What is the probability that the trihybrid ABC/abc will produce any kind of recombinant gamete?

A) 0.30% B) 1.27% C) 8.73% D) 87.30% E) 12.70% Answer: E Section: 5.3 Skill: Applying/Analyzing LO: G4, S5.3d 25) The order of genes on a plant chromosome is A, B, and D, where A and B have a recombination frequency of 0.2, and B and D have a recombination frequency of 0.3. A purebreeding plant that has the dominant A and D phenotypes with a recessive b phenotype is crossed to a pure-breeding plant that has the recessive a and d phenotypes with the dominant B phenotype. The resulting hybrid is crossed to a plant that has all three recessive phenotypes.

Under the assumption that the value for interference in this region is zero, what percentage of each progeny type would result from a single crossover between B and D? A) 24% ABd/ abd; 24% abD/ abd B) 7% ABD/ abd; 7% abd/ abd C) 12% Abd/ abd; 12% aBD/ abd D) 50% AbD/ abd; 50% aBd/ abd E) 9% ABD/ abd; 9% abd/ abd Answer: C Section: 5.3 Skill: Applying/Analyzing LO: G2, G3, G4, S5.3d

9 .

26) Which of the following is true regarding comparison of the physical maps and recombination maps?

A) There is a linear relationship between recombination frequency and physical distance if the genes are separated by more than 50 cM. B) Only with recent genome sequence analysis, which compares physical and recombination maps, has it been possible to determine that recombination hotspots are distributed evenly throughout the Drosophila genome.

C) Map distances calculated by recombination frequencies can underestimate the physical distance between two genes.

D) Two genes with a coldspot between them have larger recombination frequencies than predicted by their physical distances.

E) Two genes with a hotspot between them have smaller recombination frequencies than predicted by their physical distances.

Answer: C Section: 5.4 Skill: Remembering/Understanding LO: G2, G9, S5.4 27) Recombination frequencies between genes can be altered by all the following except?

A) temperature B) concentration of cofactors for DNA binding proteins C) being the homogametic rather than the heterogametic sex D) evolution E) addition of a chromosomal segment outside of the two genes being studied Answer: E Section: 5.4 Skill: Remembering/Understanding LO: G2, S5.4 28) For a given haplotype, the frequencies of alleles for gene Y are Y = 0.55 and Y' = 0.45, and the frequencies at gene Z are Z = 0.15 and Z' = 0.85. What is the predicted frequency of the YZ haplotype in a population in linkage equilibrium?

A) 0.2475 B) 0.1275 C) 0.9175 D) 0.3825 E) 0.4000 Answer: D Section: 5.5 Skill: Applying/Analyzing LO: G3, G4, S5.6b

10 .

29) After analysis of 100 pedigrees, the Zmax from the analysis of linkage between a disease gene D and the DNA marker P was found to be 3.5 at θ = 0.25. Which of the following is the best interpretation of these results?

A) No conclusion can be made regarding linkage of D and P. B) The evidence supports linkage of D and P at 25 cM. C) The evidence supports linkage of D and P at 2.5 cM. D) The evidence supports linkage of D and P at 0.25 cM. E) The evidence indicates that D and P are not linked.

Answer: B Section: 5.5 Skill: Applying/Analyzing LO: G3, G4, S5.5b 30) After analysis of 100 pedigrees, the Zmax from the analysis of linkage between a disease gene D and the DNA marker P was found to be -2.5 at θ = 0.01. Which of the following is the best interpretation of these results?

A) No conclusion can be made regarding linkage of D and P. B) The evidence supports linkage of D and P at 10 cM. C) The evidence supports linkage of D and P at 1.0 cM. D) The evidence supports linkage of D and P at 0.1 cM. E) The evidence indicates that D and P are not linked.

Answer: E Section: 5.5 Skill: Applying/Analyzing LO: G3, G4, S5.5b 31) The Zmax from the analysis of a disease gene N and the DNA marker H was 2.5 at θ = 0.1.

Which of the following is the best interpretation of this result?

A) N and H are not linked.

B) N and H are separated by 10 cM. C) N and H are separated by more than 10 cM. D) N and H are separated by less than 10 cM. E) No conclusion can be made regarding the linkage of N and H. Answer: E Section: 5.5 Skill: Applying/Analyzing LO: G3, G4, S5.5b

11 .

32) Genes associated by GWAS studies have all of the following characteristics except?

A) They have a specific group of alleles that co-occur with a specific trait.

B) They have a specific group of alleles that are usually identified by SNPs. C) They have a significant probability value statistic.

D) They are all located in the same chromosomal region.

E) They are most useful when they include both genes that contribute to disease and nearby genetic markers.

Answer: D Section: 5.5 Skill: Remembering/Understanding LO: S5.7b 33) Which of the following contributes to maintaining linkage disequilibrium in a population?

A) migration of individuals with new SNP haplotypes into the population B) mutation C) recombination D) natural selection E) having a very small distance between genes and their syntenic genetic markers Answer: C Section: 5.5 Skill: Remembering/Understanding LO: G2, S5.7c 5.2 Short Answer Questions 1) What is observed when syntenic genes are close enough to one another that they are unable to assort independently?

Answer: genetic linkage Section: 5.1 Skill: Remembering/Understanding LO: 5.1a 2) What is the term for chromosomes that do not reshuffle the alleles of linked genes?

Answer: parental chromosomes Section: 5.1 Skill: Remembering/Understanding LO: 5.1a 3) Incomplete genetic linkage is far more common than complete linkage. What is the term for gametes produced when recombination shuffles the alleles of linked genes?

Answer: nonparental or recombinant gametes Section: 5.1 Skill: Remembering/Understanding LO: 5.1a

12 .

4) What is the highest possible frequency of recombination between linked genes that can be generated by any type of crossover event?

Answer: 50% Section: 5.1 Skill: Remembering/Understanding LO: S5.1a 5) Two genes, A and X, exhibit incomplete linkage. The frequency of each parental gamete (AX and ax) is 45%. What is the approximate frequency of the Ax gamete?

Answer: 5% Section: 5.1 Skill: Applying/Analyzing LO: G3, G4, S5.2b 6) Construct a map of the chromosome, with the most accurate map distances, from the following recombination frequencies between individual pairs of genes: C – F 13%, V – F 20%, V – C 8%, S – F 4%, C – S 9%, Answer: (V)-----8 cM-----(C)------9 cM------(S)---4 cM---(F). Due to interference, the map distances calculated between genes that are far apart are less accurate than those for genes that are closer together. Therefore, smaller recombination frequencies must be used to find all of the distances for the most accurate map.

Section: 5.2 Skill: Applying/Analyzing LO: G3, S5.1d 7) What unit of physical distance between genes on a chromosome provides a convenient way to relate the recombination frequencies for linked genes with their positions and order along a chromosome?

Answer: map unit (m.u.) or centimorgan (cM) Section: 5.2 Skill: Remembering/Understanding 8) Genes A, B, and C lie at map positions 2.5, 7.5, and 17.5, respectively. In a three-point test cross, you observed 2 double crossovers in a total of 1000 progeny. What is the interference in this region?

Answer: 60% Section: 5.3 Skill: Applying/Analyzing LO: G2, G3, G4, S5.3d

13 .

9) A geneticist is mapping the chromosomes of the newly captured gremlin. Stripe is heterozygous for three linked genes with alleles Ee, Hh, and Bb, that determine if gremlins are evil (E), have hair (H), and biting teeth (B). In order to determine if the three genes are linked, a standard testcross was done, and the 1000 offspring had the following genotypes: 48 36 400 4 426 46 38 2

ee Hh bb ee hh Bb ee Hh Bb Ee Hh Bb Ee hh bb Ee hh Bb Ee Hh bb ee hh bb

Diagram the alleles on the homologous chromosomes of the heterozygous individuals with the correct gene order.

Answer: hEb HeB Section: 5.3 Skill: Applying/Analyzing LO: G3, G4, S5.3a 10) In most tests of genetic linkage, the number of double crossovers is LESS than the number expected due to what effect, which limits the number of crossovers that can occur in a short length of chromosome?

Answer: interference (I) Section: 5.3 Skill: Remembering/Understanding LO: S5.3c 11) When constructing a genetic map of Zea mays, Creighton and McClintock used genetic markers as well as structural differences in the homologous copies of chromosome 9 that can be seen under the microscope. What are these structural differences called?

Answer: cytological markers Section: 5.4 Skill: Remembering/Understanding 12) Recombination frequency differs between males and females. Which sex has a higher rate of recombination, heterogametic (males) or homogametic (females)?

Answer: homogametic (females) Section: 5.4 Skill: Remembering/Understanding LO: S5.4

14 .

13) If there is a maximum Lod score of 4.2 at θ = 0.31, what can you say about the linkage and the distance between two genes?

Answer: linked and 31 cM apart Section: 5.5 Skill: Applying/Analyzing LO: G3, G4, S5.5b 14) If too little time has passed for crossing over to randomize haplotypes or if natural selection favors certain haplotypes, what would you expect to see?

Answer: linkage disequilibrium Section: 5.5 Skill: Remembering/Understanding LO: S5.6a 5.3 Fill-in-the-Blank Questions 1) A chromosome with a different combination of alleles than parental that is created by crossing over between homologous chromosomes is termed ________.

Answer: recombinant chromosome Section: 5.1 Skill: Remembering/Understanding 2) Alleles of linked genes usually segregate together during meiosis. When they don't, it is because ________ has occurred between them.

Answer: crossing over Section: 5.1 Skill: Remembering/Understanding LO: S5.1a 3) 5% recombination is equal to ________ map unit(s) (m.u.) or centimorgan(s) (cM) of distance between linked genes.

Answer: 5 Section: 5.2 Skill: Applying/Analyzing LO: G4, S5.2b 4) The specific array of alleles making up a set of linked genes on a single chromosome is called a ________.

Answer: haplotype Section: 5.5 Skill: Remembering/Understanding

15 .

5) Instead of creating a gene map along a chromosome, ________ is a method of analysis that detects and locates the genes that, as a group, influence form or appearance.

Answer: genome-wide association studies (GWAS) Section: 5.5 Skill: Remembering/Understanding LO: S5.7a

16 .

5.4 Essay Questions 1) What is the relationship between linked genes and syntenic genes? Are syntenic genes always linked?Are linked genes always syntenic? Describe what is meant by each term.

Answer: Linked genes are always inherited together with no independent assortment. Syntenic genes are genes found on the same chromosome. Linked genes are always found on the same chromosome, but they are unable to sort independently because there is no crossing over observed between these two genes. Syntenic genes can be unlinked, and their alleles will assort independently if they are far enough apart on the chromosome for crossing over to generate independent assortment of the alleles.

Section: 5.1 Skill: Evaluating/Creating LO: G8, S5.1a

17 .

2) Refer to this map to answer the questions:

a) b) c) d) e) f) g) h) i) j)

What is the distance between gene E and gene F? Assuming I = 0, what is the probability of no crossovers between gene D and gene E? Assuming I = 0, what is the probability of no crossovers between gene E and gene F? Considering both gene pairs, what is the proportion of nonrecombinant gametes?

What is the predicted frequency of each parental gamete (DEF/def )?

What is the recombination frequency between gene D and gene E? What is the recombination frequency between gene E and gene F? What is the frequency of two single recombinant gametes between genes D and E? What is the frequency of two single recombinant gametes between genes E and F? What is the frequency of each of the double-recombinant gametes, DeF and dEf?

k) Fill in the blanks below to prove that the sum of the frequencies of the eight predicted gamete genotypes is equal to 1.0: ________ + ________ + ________ + ________ + ________ + ________ + ________ + ________ = 1 DEF def Def dEF DEf deF DeF dEf Answer: a) 12 m.u. b) 92% c) 88% d) (0.92)(0.88) = 0.8096 e) (0.5)(0.8096) = 0.4048 f) 8% g) 12% h) (0.08)(0.88)(0.5) = 0.0352 i) (0.12)(0.92)(0.5) = 0.0552 j) (0.08)(0.12)(0.5) = 0.0048 k) 0.4048 + 0.0048 + 0.0352 + 0.0352 + 0.0552 + 0.0552 + 0.0048 + 0.0048 = 1 Section: 5.3 Skill: Evaluating/Creating LO: G3, G4, S5.3d

18 .

3) A geneticist is mapping the chromosomes of a new fish species. These fish have genes for the dominant traits non-bioluminescent (B), ventral eyes (D), and short spikes (L) with corresponding recessive traits bioluminescent (b), dorsal eyes (d), and long spikes (l). He testcrosses one of these fish who is dominant at all three loci to determine if the three genes are linked resulting in the data shown below: 345 b D L 355 B d l 42 B D l 38 b d L 105 b D l 95 B d L 9 BDL 11 b d l 1000 a) Why are these data consistent with genetic linkage among the three genes? b) Perform a chi-square test to determine if these data show significant deviation from the expected phenotype distribution. c) Determine the recombination frequencies between each pair of genes. d) Determine the coefficient of coincidence (COC) and interference (I) values for this data set.

Answer: a) The data show that the predominant progeny classes are the parental types (nonbioluminescent, ventral eyes, and short spikes as well as the reciprocal class bioluminescent, dorsal eyes, and long spikes). In addition, all the recombinant progeny classes are rare, which is consistent with genetic linkage between each pair of genes. b) Under the assumption that all three genes assort independently the expected distribution is eight classes, each present at a frequency of 1/8. For 1000 progeny, 1/8 is 125. The chi-square calculation is: + +

+ +

+ +

+ = 1148

The chi-square value is 1,148. For seven degrees of freedom, this corresponds to a P value less than 0.01, which indicates that the difference between the expected and observed results is highly statistically significant. c) The genes for bioluminescence and eye position are 10 m.u. apart. The genes for bioluminescence and spike length are 22 m.u. apart. d) COC= 0.909; I = 0.091 Section: 5.3 Skill: Evaluating/Creating LO: G2, G3, G4, G9, S5.3a, S5.3c, S5.3e

19 .

4) Compare and contrast the following three DNA genetic markers: variable number tandem repeats (VNTRs), single nucleotide polymorphisms (SNPs), and restriction fragment length polymorphisms (RFLPs).

Answer: All three of these genetic markers are typically in noncoding regions of the genome.

VNTRs consist of 3-20 bp sequences of DNA that are repeated end-to-end in a chromosome region. SNPs are single base pair substitutions in DNA sequences. RFLPs are changes in restriction recognition sites in DNA that lead to differences in restriction fragments following digestion with a restriction enzyme. SNPs are much more commonly used today.

Section: 5.5 Skill: Knowledge/Comprehension LO: G7, G8 5) The pedigrees below are for a dominant disease where individuals with the dominant disease allele (D) are affected. Individuals that were alive were also screened for a polymorphic DNA marker that has six alleles identified as P1, P2, P3, P4, P5, and P6. a) In Family B, explain why the P1 marker cannot be assumed to be transmitted with the dominant allele given that there is no haplotype data for the first generation. b) In Family A, how does the information regarding the haplotype of individuals in the first generation allow you to determine the allelic phase of the disease?

"Use Figure 5.15 in Chapter 5, page 169 from main title."

Answer: a) In Family B, three of the affected individuals in generation III (III-1, III-3, and III-4) received the P1 allele along with the disease allele, whereas the other affected individual (III-6) received the P2 allele along with the disease allele. This shows that the disease allele can be associated with both the P1 and P2 alleles. However, definitive identification of the allelic phase is not possible with the information provided in the pedigree for Family B. It should not be assumed that the disease allele is associated with the P1 variant, even though (1) it is more prevalent and (2) individual III-5's haplotype likely resulted from a recombination event. b) The P1 allele is carried with the disease allele because the affected maternal parent (I-2) transmits her P1D allele to her affected offspring (II-1). Furthermore, most of the offspring of individual II-1 that are affected also carry the P1 variant. This combined knowledge provided in the pedigree for Family A, supports the conclusion that P1 marker allele is transmitted with the disease allele.

Section: 5.5 Skill: Synthesis/Evaluation LO: G2, G3, G5

20 .

6) How is Newton Morton's statistical method (Lod score analysis) helpful for calculating the overall probability of genetic linkage when allelic phase is unknown?

Answer: Morton's method determines whether genetic linkage exists between genes for which allelic phase is unknown by comparing the likelihood of obtaining the genotypes and phenotypes observed in a pedigree if two genes are linked versus the likelihood of getting the same pedigree structure if the genes assort independently. This approach assesses the probability of genetic linkage between two genes at a time and compares the probability of genetic linkage to the probability of independent assortment of the genes.

Section: 5.5 Skill: Remembering/Understanding LO: G1, G9 7) What is the Lod score?

Answer: The ratio of the two likelihoods described in answer 4 above gives the "odds" of genetic linkage, and the Logarithm of the odds ratio generates the Lod score, a statistical value representing the probability of genetic linkage between two genes. A Lod score is a statistic that can attain significance favoring genetic linkage if the probability of genetic linkage is sufficiently greater than the probability of independent assortment, or significance against genetic linkage if the probability of independent assortment is sufficiently greater than the linkage probability. Lod scores can be interpreted for individual families, or they can be added together for as many families as are analyzed.

Section: 5.5 Skill: Remembering/Understanding LO: G1, G8 8) What would a Lod score of 3.2 tell you about genetic linkage? What information does the θ value tell you?

Answer: Scores of greater than 3.0 provide evidence of genetic linkage. θ values that correspond to a significant Lod score indicate an approximate distance (in cM) between linked genes.

Section: 5.5 Skill: Remembering/Understanding LO: G3, S5.5b

21 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 6 Genetic Analysis and Mapping in Bacteria and Bacteriophages 6.1 Multiple-Choice Questions 1) Which of the following is false regarding plasmids?

A) They are small double stranded circular DNA molecules.

B) They contain essential genes for metabolism.

C) They can have genes that promote their own transfer from one bacterium to another.

D) They can carry genes for antibiotic resistance.

E) They generally replicate autonomously.

Answer: B Section: 6.1 Skill: Remembering/Understanding LO: S6.1 2) Which of the following is true of a bacterial strain that is his+ lac+ gal- ampR. A) It is auxotrophic for histidine.

B) It will grow on ampicillin.

C) It is unable to synthesize histidine.

D) It can use galactose as an energy source.

E) It is unable to break down lactose.

Answer: B Section: 6.1 Skill: Remembering/Understanding LO: S6.2c 3) A lab technician has four strains of bacteria. Although he forgot to label them, he knows that they include one F+ strain, one Hfr strain, one F' strain, and one F- strain. He observed the following results after plating these strains on different media. Which of the following plates can be used to identify the F- strain? strain F+ Hfr F’ F-

his + + + -

glu + + + -

cys + + + -

leu + + +

str + +

A) complete media B) minimal media with histidine, glutamate, and cysteine C) minimal media with histidine, glutamate, cysteine, and streptomycin D) minimal media with streptomycin E) minimal media with leucine and streptomycin Answer: C Section: 6.1 Skill: Applying/Analyzing LO: G2, G3, S6.2c 1 .

4) Which of the following is incorrect about genetic analysis and mapping in bacteria and bacteriophages?

A) Conjugation requires cell-to-cell contact.

B) Generalized transduction uses lytic phage and specialized transduction uses a lysogenic phage.

C) Transformation requires a vector to carry donor DNA fragments to the recipient cell.

D) Conjugation, transformation, and transduction all require recombination.

E) Conjugation, transformation, and transduction can all be used to map the order of genes on a bacterial chromosome.

Answer: C Section: 6.2 Skill: Applying/Analyzing LO: G7 5) Why does conjugation with an Hfr donor result in a much higher rate of gene transfer than conjugation with an F+ donor?

A) An F+ donor makes fewer pili.

B) An F+ donor's T strand is slow to reach the pilus.

C) The Hfr donor transfers genes from the chromosome and not the plasmid.

D) The Hfr donor segment must undergo recombination in the recipient.

E) The F+ donor segment must undergo recombination in the recipient.

Answer: C Section: 6.2 Skill: Applying/Analyzing LO: S6.1 6) Which of the following is NOT a step involved in rolling circle replication during conjugation?

A) Replication begins at the oriT after relaxosome binding.

B) Relaxase facilitates passage of the T strand into the recipient cell.

C) A protein complex known as the relaxosome cuts one strand of the F+ plasmid.

D) A double stranded plasmid is transferred to the F- cell through the conjugation pilus.

E) Rolling circle replication occurs in the donor.

Answer: D Section: 6.2 Skill: Remembering/Understanding LO: S6.1

2 .

7) When F+ conjugation occurs successfully, only one copy of the F plasmid single-stranded DNA is transferred. What else must minimally occur?

A) circularization of the recipient's new DNA B) circularization plus replication of the new recipient DNA C) stopping the rolling circle replication after one copy length, circularization, and replication D) continuation of rolling circle replication in the donor cell and replication in the recipient cell E) rolling circle replication in the recipient cell Answer: C Section: 6.2 Skill: Applying/Analyzing LO: S6.1 8) Which of the following is an accurate statement regarding F factor states?

A) An F+ cell can convert exconjugants to a donor state, as well as change their genotype.

B) An Hfr cell can convert exconjugants to a donor state, as well as change their genotype.

C) An F' cell can convert exconjugants to a donor state, as well as change their genotype.

D) F+ and Hfr cells cannot convert exconjugants to a donor state.

E) F+ and F' cells cannot transfer donor genes to exconjugants.

Answer: C Section: 6.2 Skill: Applying/Analyzing LO: G2, S6.1, S6.3 9) In the flow diagram shown below, which of the following is an inaccurate relationship between bacterial strains in the various F factor states?

A) Link 1 represents transfer of an entire F plasmid.

B) Link 2 represents integration of the F plasmid into the host chromosome.

C) Link 3 represents precise excision of the F plasmid from the chromosome.

D) Link 4 represents excision of the F plasmid plus some host DNA from the chromosome.

E) All four links represent accurate relationships.

Answer: E Section: 6.2 Skill: Applying/Analyzing LO: G2, S6.1, S6.3

3 .

10) To map genes of a bacterial strain, conjugation must be interrupted at given times. Suppose you have Hfr cells of genotype a+b+c+d+e+strR and F- cells of genotype a-b-c-d-e-strS and you combine these two cultures in liquid medium in four blenders at time 0. After intervals of 3, 6, 9, and 12 minutes, you turn on successive blenders. What role does the blender play in the experiment?

A) serves as a culture container only B) acts as a mechanism to bring cells into contact C) speeds up molecular movement of DNA D) shears the sex pili of conjugants E) hastens conjugation events Answer: D Section: 6.2 Skill: Applying/Analyzing LO: G2, G9, S6.1 11) To map genes of a bacterial strain, conjugation must be interrupted at given times. Suppose you have Hfr cells of genotype a+b+c+d+e+strR and F- cells of genotype a-b-c-d-e-strS and you combine these two cultures in liquid medium in four blenders at time 0. After intervals of 3, 6, 9, and 12 minutes, you turn on successive blenders. The resulting cultures were then plated on medium containing streptomycin. Why? A) to eliminate non-conjugating Hfr cells B) to eliminate all non-conjugated cells C) to eliminate non-conjugated F- cells D) to selectively eliminate cells that have taken in F+ genes Answer: C Section: 6.2 Skill: Applying/Analyzing LO: G2, S6.2c, S6.3

4 .

12) To map genes of a bacterial strain, conjugation must be interrupted at given times. Suppose you have Hfr cells of genotype a+b+c+d+e+strR and F- cells of genotype a-b-c-d-e-strS and you combine these two cultures in liquid medium in four blenders at time 0. After intervals of 3, 6, 9, and 12 minutes, you turn on successive blenders. The graph below shows the results of plating the cultures from each blender. What is the correct order of the genes on this chromosome?

A) a b c d e B) a c e b d C) b e c a (d is unknown) D) d a c e b E) b e c a d Answer: B Section: 6.2 Skill: Applying/Analyzing LO: G2, G3, S6.2a 13) Which of the following allows the compilation of time-of-entry maps?

A) the use of many replicates of each experiment B) the use of appropriate controls repeatedly C) the fact that gene entry can proceed in either direction D) the observation that Hfr genes recombine so efficiently E) the recognition of phenotype interaction Answer: C Section: 6.2 Skill: Applying/Analyzing LO: G2 5 .

14) In five Hfr strains, each of which was used to build a time-of-entry map, the genes entered the recipient cells as follows: Strain 1: S L A C T F Strain 2: N P F T C A Strain 3: T F P N U Y Strain 4: S H Y U N P Strain 5: U N P F T C Which of the following represents a correct compilation of these results A) S L A C T F P N H C U B) N P F T S L A C H U T C) T C A L S P N U Y H D) U N P F T C A L S T F E) S L A C T F P N U Y H Answer: E Section: 6.2 Skill: Applying/Analyzing LO: G3, S6.2a 15) An F′ donor includes F DNA plus a segment of bacterial chromosome DNA. If conjugation is interrupted before the entire F′ chromosome transfers, what could be the expected consequences?

A) integration of some but not all F′ genes into the recipient's chromosome B) a smaller F′ plasmid than had been present in the donor C) failure of the F′ strand to circularize in the recipient D) change of the recipient from F- to F+ E) formation of a recipient that is a stable partial diploid Answer: C Section: 6.2 Skill: Applying/Analyzing LO: G2, S6.1

6 .

16) Jacob and colleagues developed a time-of-entry mapping technique. They used the genes for threonine and leucine synthesis (thr and leu), which are near the oriT site, as selected markers.

Additionally, they used an E. coli F- strain P678 that has dysfunctional variants for all of the genes studied. They also used an HfrH strain that carried a functional allele for all genes studied.

Which of the following statements is FALSE regarding the time-of-entry profile they acquired?

A) Had the experiment continued, a time of entry map could be viewed for the entire E. coli chromosome.

B) The azi gene is closer to the thr and leu genes than the ton, lac, and gal genes.

C) The ton gene is found between the azi and lac genes.

D) The lac+ allele appears at about 16 or 17 minutes.

E) At 40 minutes, nearly 20% of exconjugants screened for the selected markers have the gal+ allele.

Answer: A Section: 6.2 Skill: Applying/Analyzing LO: G2, G3, G9, S6.2a

7 .

17) Which of the following is NOT a step in transformation?

A) Donor cell DNA binds to a receptor site on the recipient cell.

B) The donor cell lyses, releasing pieces of its chromosome into the environment.

C) Both strands of the donor cell DNA are degraded.

D) DNA fragments pair with homologous regions of the recipient cell chromosome.

E) DNA recombines with the recipient cell chromosome.

Answer: C Section: 6.3 Skill: Remembering/Understanding LO: G7 18) A competent bacterial strain with genes a, b, and c is transformed by a donor bacterial fragment. Cotransformation frequencies for each gene pair are as follows: a and b 0.04% a and c 0.02% b and c 0.0064% Which conclusion can be definitively made from this data?

A) Genes b and c are farthest apart.

B) Gene order is a, b, then c.

C) Gene a is closer to c than to b.

D) Gene b is closer to c than to a.

E) Cotransformation of b and c is so frequent, they must be one gene.

Answer: A Section: 6.3 Skill: Applying/Analyzing LO: G2, G3, S6.4 19) Which of the following statements about the T4 lytic life cycle is false?

A) The host DNA breaks down.

B) Under the direction of phage genes, transcription and translation produce new phage components.

C) Progeny phage particles are released by lysis from host bacteria.

D) The phage DNA is injected into the host.

E) The phage DNA is integrated into the host chromosome.

Answer: E Section: 6.4 Skill: Remembering/Understanding LO: S6.5a

8 .

20) In generalized transduction, a phage introduces a segment of donor DNA into the recipient cell. This is followed by recombination of the donor fragment with the recipient chromosome.

Which of the following must occur?

A) circularization of the donor fragment before recombination B) a pair (or even number) of crossovers between the donor segment and the recipient C) degradation of one of the two strands of phage genome D) replication of the donor segment before recombination E) a single crossover between the donor segment and the recipient chromosome Answer: B Section: 6.4 Skill: Applying/Analyzing LO: G9 21) In a cotransduction experiment the alleles a+, b+, and c+ were studied. The allele a+ alone was transduced into an auxotroph 1750 times, b+ alone 1700 times, and c+ alone 1725 times.

Alleles a+ and b+ were both found 11 times, b+ and c+ 117 times, and a+ and c+ 15 times. What can you definitely conclude?

A) The genes are all located extremely close to each other.

B) The gene order is a, b, c.

C) Genes a and b are more closely linked than a and c or b and c.

D) Genes b and c are found together only if double crossovers have occurred.

E) Genes b and c are more closely linked than a and b or a and c.

Answer: E Section: 6.4 Skill: Applying/Analyzing LO: G3, S6.5b 22) Which of the following statements regarding cotransduction mapping experiments is false?

A) A high co-transduction frequency between two genes is evidence of linkage.

B) A low co-transduction frequency between two genes is evidence of linkage.

C) A selected marker screen identifies transductants for a donor allele of interest.

D) An unselected marker screen identifies transductants for an additional donor allele of interest.

E) Selection identifies transductants that have acquired an even number of crossovers.

Answer: B Section: 6.4 Skill: Applying/Analyzing LO: G9, S6.5b

9 .

23) A scientist is setting up a co-transduction experiment to determine the order of five genes in an operon. What gene should be used as the selected marker and what type of media should be used to select for it?

A) a nearby essential metabolic gene upstream of the operon; complete media B) a nearby essential metabolic gene upstream of the operon; minimal media supplemented with the corresponding essential metabolite C) a nearby essential metabolic gene downstream of the operon; complete media D) the first gene in the operon; complete media E) the first gene in the operon; minimal media supplemented with the corresponding essential metabolite Answer: B Section: 6.4 Skill: Remembering/Understanding LO: G1, G2, S6.5c 24) A competent bacterial strain that has the genotype a- b- c+ is transduced with a P1 phage carrying a fragment of donor chromosome with the a+ b+ c- alleles. The following number of transductants were found after screening for the selectable marker a+.

Transductant Genotype a+ b- c+ a+ b+ c+ a+ b+ ca+ b- c-

Number 160 360 460 20

Which of the following conclusions is false?

A) The transduction frequency between a and b is 82%.

B) The transduction frequency between a and c is 48%.

C) Gene a and gene b are closer together than gene a and gene c.

D) The gene order is acb.

E) The selected marker screen will inhibit growth of transductants with the genotype a- b+ c-.

Answer: D Section: 6.4 Skill: Applying/Analyzing LO: G2, G3, G4, G9, S6.5d

10 .

25) Which methods did Seymour Benzer use to study 20,000 mutants of the rII region of the T4 bacteriophage?

A) genetic complementation analysis only B) intragenic recombination analysis only C) deletion-mapping analysis only D) intragenic recombination analysis and deletion-mapping analysis only E) genetic complementation analysis, intragenic recombination analysis, and deletion-mapping analysis Answer: E Section: 6.5 Skill: Remembering/Understanding LO: G9 26) Seymour Benzer found which of the following while studying 20,000 mutants of the rII region of the T4 bacteriophage?

A) Simultaneous infection of rII phage mutants resulted in T4 plaques when the mutations were in the same rII gene.

B) Intragenic recombination analysis could determine the distance between rIIA and rIIB. C) Intragenic recombination between mutants with partial deletions and point mutations could be used to map the position of individual rII mutations.

D) Deletion-mapping analysis found that a point mutation and a deletion mutation whose locations overlapped resulted in wild-type intragenic recombinants.

E) The size of the plaques on E. coli lawns determined the distance between rIIA and rIIB. Answer: C Section: 6.5 Skill: Applying/Analyzing LO: G1, G2, G9, S6.6a 27) Benzer's analysis of phage genomes included deletion mapping. Since recombination could not occur in the area of a deletion, an infection by two phage strains, one of which has a deletion and the other a point mutation, that results in no wild-type recombinants being produced indicates ________.

A) that the deletion includes the position of the point mutation in the same gene B) that the deletion includes the position of the point mutation in a different gene C) that the point mutation is dominant D) that the point mutation is recessive E) the distance between the two types of mutation Answer: A Section: 6.5 Skill: Applying/Analyzing LO: G2, S6.6a

11 .

28) A scientist is studying a gene (grn) carried by a bacteriophage that allows a host bacterium to produce a green metabolic product before lysis. The scientist has four bacteriophage mutants for this gene. Mutant 1 has a nonrevertible mutation in grn while Mutants 2, 3, and 4 each have a revertible mutation in grn. After plating pairs of mutants and subsequent coinfection of host cells, the scientist observed the following: I. Mutants 1 and 2: no green product II. Mutants 1 and 3: very rare green product III. Mutants 1 and 4: rare green product Which of the following is NOT consistent with these observations?

A) A partial deletion in Mutant 1 overlaps a base-pair substitution in Mutant 2.

B) Mutant 2 has a base-pair substitution at a different site than Mutant 3.

C) Mutant 3 has a base-pair substitution at a different site than Mutant 4.

D) Mutant 1's deletion is closer to the base-pair substitution in Mutant 3 than it is to the base-pair substitution in Mutant 4.

E) A green product should be observed if coinfection occurs with mutants 3 and 4.

Answer: D Section: 6.5 Skill: Applying/Analyzing LO: G2, G3, G9, S6.6a 29) In his analysis of rII in phage, Benzer assayed complementation. Suppose you try your hand at this with several mutants and get the following results (+ = complementation): 42 and 62 + 42 and 83 – 62 and 63 – 62 and 74 + 42 and 74 – 42 and 75 + 62 and 75 – What can you conclude?

A) Mutations 42 and 75 are in the same gene.

B) Mutations 62 and 74 are in the same gene.

C) Mutations 62, 63, and 74 are in the same gene.

D) Mutations 42 and 74 are in the same gene.

E) Mutation 63 is not in the same gene as either 62 or 42.

Answer: D Section: 6.5 Skill: Applying/Analyzing LO: G3, S6.6b

12 .

30) Which of the following regarding lateral gene transfer is false?

A) It may occur by conjugation, transformation, or transduction.

B) It facilitates rapid genome evolution.

C) It can convert benign bacteria into pathogens that cause disease.

D) It allows an organism to rapidly acquire whole segments of chromosomes containing multiple genes.

E) It increases the efficiency of antibiotics against normally sensitive strains.

Answer: E Section: 6.6 Skill: Remembering/Understanding LO: G5, S6.7 6.2 Short Answer Questions 1) One kind of plasmid in bacteria is called an R plasmid. What kinds of genes are typically present on such plasmids?

Answer: antibiotic-resistant genes, heavy metal resistance Section: 6.1 Skill: Remembering/Understanding 2) What is the difference between F+ and F- bacteria?

Answer: F+ bacteria have an F plasmid.

Section: 6.2 Skill: Remembering/Understanding LO: S6.1 3) In a bacterial cross of an Hfr donor a+b+ with a recipient F- a-b-, what would be a possible recombinant?

Answer: a+b-, a-b+, or a+b+ Section: 6.2 Skill: Applying/Analyzing LO: S6.1 4) All three types of bacterial recombination share one characteristic. What is it?

Answer: one-way transfer of gene(s), or transfer from donor to recipient Section: 6.2 Skill: Remembering/Understanding LO: G7 5) What is the role of the relaxosome in conjugation?

Answer: binds to F origin, or cleaves T strand, or binds to 5′ end to begin transfer Section: 6.2 Skill: Remembering/Understanding LO: S6.1

13 .

6) In conjugation, as the 5′ end of the T strand begins to move across the pilus, what form of replication occurs in the donor cell?

Answer: rolling circle (replication) Section: 6.2 Skill: Remembering/Understanding LO: S6.1 7) In Hfr transfer, what happens to linear DNA that does not recombine with the host chromosome?

Answer: (enzymatic) degradation Section: 6.2 Skill: Remembering/Understanding LO: S6.1 8) In an E. coli conjugation experiment, the donor cell is found to be thr+, leu+, his+, and the Fcell is thr-, leu-, and his-. When the mating is interrupted after 10 minutes and the recipients are plated, they are found to be leu+ and his+ but thr-. What can you conclude?

Answer: leu and his are close together.

Section: 6.2 Skill: Applying/Analyzing LO: S6.2a 9) In a conjugation experiment, the Hfr donor is thr+, leu+, his+, strR. The recipient is thr-, leu-, his-, strS. Why would you use medium containing streptomycin to analyze your results?

Answer: to identify strR recipients Section: 6.2 Skill: Applying/Analyzing LO: G9, 6.2c 10) If conjugation occurs with an F′ donor, how will the recipient be characterized genetically?

Answer: partially diploid Section: 6.2 Skill: Remembering/Understanding LO: S6.1 11) What characterizes a bacterial cell that can undergo transformation?

Answer: competence, or ability to take up double-stranded DNA Section: 6.3 Skill: Remembering/Understanding 12) What distinguishes generalized versus specialized transduction?

Answer: generalized uses lytic phage; specialized uses lysogenic phage Section: 6.4 Skill: Remembering/Understanding LO: S6.5a

14 .

13) Name the two cycles in which bacteriophages act when they infect bacteria.

Answer: lysis and lysogeny Section: 6.4 Skill: Remembering/Understanding LO: S6.5a 14) Seymour Benzer's fine structure studies of genomes used mutations found in specific regions of what kind of genome?

Answer: bacteriophage (phage) Section: 6.5 Skill: Remembering/Understanding LO: S6.6a 15) What is the difference between the revertible and nonrevertible rII mutants that Benzer generated?

Answer: revertible mutants were caused by DNA base-sequence substitutions (point mutations) and nonrevertible mutants were caused by partial deletion mutations Section: 6.5 Skill: Knowledge/Comprehension LO: S6.6a 6.3 Fill-in-the-Blank Questions 1) Bacteria that can grow on minimal medium are called ________ while bacteria that can require one or more essential compounds for growth are called ________.

Answer: prototrophs; auxotrophs Section: 6.1 Skill: Remembering/Understanding LO: S6.2c 2) Some bacteriophages can be either lytic or lysogenic. These are called ________ phages.

Answer: temperate Section: 6.4 Skill: Remembering/Understanding LO: S6.5a 3) Once a phage genome has become part of the host chromosome, it is called a ________.

Answer: prophage Section: 6.4 Skill: Remembering/Understanding LO: S6.5a

15 .

6.4 Essay Questions 1) How are selective media used in studying bacterial recombination?

Answer: Selective media permit certain bacteria to grow while suppressing the growth of other bacteria. During bacterial recombination, you can accurately assess transfer of genes using selective media in two ways.

First, you can use selective media to identify bacteria that have acquired a resistance gene and are recombinant. For example, E. coli can be transformed with a plasmid containing your gene of interest and a selectable marker, often an antibiotic resistance gene. You can add antibiotic to a bacterial culture and only the bacteria that have taken up the plasmid will be resistant. This will allow for identification of sequences that have been transferred between bacteria.

Second, you can use selective media to identify bacteria that have a mutation in a metabolic pathway by looking for mutants that are unable to complete normal metabolic processes. For instance, let's say you are studying a metabolic pathway and wish to identify bacteria that have mutations in this pathway. You can grow the bacteria in media with and without the final product of the pathway. If the pathway is mutated and one of the enzymes is absent, the bacteria must be supplemented with certain nutrients to be able to grow. This will allow you to identify transformants, or mutants of interest.

Section: 6.1 Skill: Applying/Analyzing LO: S6.2c, S6.5c, G2, G8

16 .

2) An interrupted mating study is carried out on the Hfr strains 1, 2, and 3. A small sample of each mixture is collected every minute for 20 minutes after conjugation is established. Results for each of the three Hfr strains are shown below. The total duration of conjugation (in minutes) is given for each transferred gene.

Hfr strain 1 oriT red

blu

grn

ylw

Duration (min)

10

16

19

oriT red

cyn

org

vlt

0

6

11

20

oriT ylw

ind

mrn vlt

0

8

12

Hfr strain 2 Duration (min) Hfr strain 3 Duration (min)

0

3

2

1

14

What are the interrupted mating results you would expect for a new Hfr strain if after 20 minutes of conjugation the genes transferred in the following order: oriT ind ylw grn blu and the first gene, ind, appeared after 2 minutes.

17 .

Answer: The difference between the time of entry of two adjacent genes is the map distance, in minutes, between the genes.

The time of entry for ylw (9 min), grn (12 min), and blu (18 min), is determined by the genetic map above.

Hfr strain 4 oriT ind

ylw

grn

blu

Duration (min)

9

12

18

0

2

Section: 6.2 Skill: Applying/Analyzing LO: G2, G3, G9, S6.2b

18 .

3) Conjugation, transduction, and transformation all occur in nature. Which do you expect would more likely account for the acquisition of the genes for resistance to an antibiotic? Why? Which do you think would best account for the acquisition of genes from one bacterial species by another of a different genus? Why? Answer: Through lateral gene transfer, genes located on a circular strand of DNA called an R plasmid contain antibiotic-resistant genes. Through conjugation, an antibiotic-resistant bacterium can transfer the R plasmid to a non-resistant bacterium. Because this R plasmid provides a selective growth advantage to some bacteria, those bacteria may "collect" many copies of various antibiotic resistance genes from other bacteria.

Transfer of genes between bacterial species likely involves transformation, the process by which extracellular fragments of DNA released when a donor bacterial cell lyses can be absorbed across the cell membrane of a competent recipient cell as transforming DNA. Specialized transduction involves bacteriophages, which can transfer host DNA from one host to the next through aberrant excision of a lysogenic prophage. However, it is more likely that gene transfer between species occurred through transformation rather than transduction.

Section: 6.4 Skill: Applying/Analyzing LO: G2, G5, G8 4) Explain when, why, and how recombination in bacteria is like crossing over in diploid cells.

Answer: Both bacterial recombination and crossing over during meiosis require several factors.

First, both processes require nicking of DNA. In bacteria, this is when the DNA is nicked in preparation for the transfer of single-stranded DNA via conjugation. In eukaryotes, one strand of the DNA is nicked during the formation of chiasmata. This allows for strand invasion and crossing over between homologous chromosomes.

Second, both processes use similar enzymes to mediate recombination.

Third, both processes require pairing of homologous sequences. In bacteria, when the chromosome DNA from the donor cell is transferred to a recipient bacterium, the homologous parts of the donor and recipient DNA molecules can undergo recombination that leads to a change in the genotype of the recipient cell. In eukaryotes, proper pairing of homologous sequences is necessary for even exchange of chromosomes during crossing over. Uneven pairing of homologous sequences can result in a region of chromosome being deleted or duplicated in the resulting gametes.

Section: 6.4 Skill: Evaluating/Creating LO: G2, G8

19 .

5) Which, if any, characteristics of bacterial gene transfer is/are useful in analyzing eukaryotic gene transfer?

Answer: Lateral gene transfer is the process by which genes are transferred between species, as opposed to vertical gene transfer in which genes are inherited from ancestors. This phenomenon is common in prokaryotes, and can occur by conjugation, transformation, or transduction. There is evidence of natural lateral gene transfer in eukaryotes, but it is far more relevant in terms of artificial gene transfer and genetic engineering. Genes from one organism can be successfully expressed in a different organism; for example, expression of GFP, luciferase, and various reporter genes. If you want to introduce a gene of interest into a eukaryotic cell, you will most often use transformation. Cells are incubated with the DNA of interest, and then subjected to artificial selection to select for cells that have taken up the DNA from the environment. This same process occurs naturally in bacteria and in single-celled eukaryotes, but can be a valuable asset for genetic engineering.

More recently, specially engineered viruses have also been used to introduce DNA into a eukaryotic cell as a strategy for creating transgenic cell lines or for gene therapy. This process is similar to the mechanism by which bacteriophages transfer DNA between bacteria through the process of transduction.

Thus, studying the mechanisms of lateral gene transfer is both helpful and relevant to understanding the processes by which eukaryotes acquire new genes naturally or through intentional genetic engineering.

Section: 6.6 Skill: Evaluating/Creating LO: G2, G8, S6.7

20 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 7 DNA Structure and Replication 7.1 Multiple-Choice Questions 1) The process by which Pneumococcus transfers DNA between living type RII and heat-killed type SIII cells is known as ________.

A) transformation B) replication C) transduction D) ligation E) conjugation Answer: A Section: 7.1 Skill: Knowledge/Comprehension LO: G1, S7.1 2) Avery, Macleod, and McCarty expanded on Griffith's experiment to prove that DNA is the hereditary molecule required for transformation. What treatment of the heat-killed SIII bacteria extract resulted in the mouse LIVING? A) Destruction of type SIII lipids and polysaccharides B) Destruction of type SIII DNA with DNase C) Destruction of type SIII RNA with RNase D) Destruction of type SIII proteins with protease E) Use of the control group (null treatment), in which all components are intact Answer: B Section: 7.1 Skill: Knowledge/Comprehension LO: G1, S7.1 3) In the Hershey-Chase experiment, bacteriophages were produced in either 32P-containing or 35S-containing medium. Where were these isotopes eventually detected when the radioactivelylabeled bacteriophages were introduced to a fresh bacterial culture?

A) Both 32P and 35S were associated with the bacterial cells.

B) The 32P was associated with the culture medium and 35S was associated with the phage particles.

C) The 32P was associated with the bacterial cells and 35S was associated with the phage particles.

D) The 32P was associated with the culture medium and 35S was associated with the bacterial cells.

E) Both 32P and 35S were associated with the phage particles.

Answer: C Section: 7.1 Skill: Knowledge/Comprehension LO: G1, S7.1 1 .

4) Which of the following are classified as pyrimidines?

A) adenine and thymine B) guanine and cytosine C) adenine and guanine D) thymine and cytosine E) adenine and uracil Answer: D Section: 7.2 Skill: Knowledge/Comprehension LO: G1, S7.2 5) What type of bond is formed between the hydroxyl group of one nucleotide and the phosphate group of an adjacent nucleotide, forming the sugar-phosphate backbone of DNA? A) glycosidic bond B) phosphodiester bond C) ester linkage D) hydrogen bond E) ionic bond Answer: B Section: 7.2 Skill: Knowledge/Comprehension LO: G1, S7.2 6) What is the difference between a nucleotide and a nucleoside?

A) Nucleotides are found in DNA, while nucleosides are found in RNA. B) Nucleotides are involved in eukaryotic DNA replication, while nucleosides are used in bacterial DNA replication.

C) Nucleosides contain only deoxyribose sugars.

D) A nucleoside with a phosphate ester linked to the sugar is a nucleotide.

E) Nucleosides are purines, while nucleotides are pyrimidines.

Answer: D Section: 7.2 Skill: Knowledge/Comprehension LO: G1, S7.2

2 .

7) What is the difference between a ribonucleic acid and a deoxyribonucleic acid?

A) Deoxyribonucleic acids have a 2'H instead of the 2'OH found in ribonucleic acids.

B) Deoxyribonucleic acids are used to build nucleic acid strands, while ribonucleic acids are used exclusively in enzymatic reactions.

C) Ribonucleic acids have a 3'OH, while deoxyribonucleic acids have a 3'H. D) Ribonucleic acid include A, T, G, and C, while deoxyribonucleic acids include A, U, G, and C. E) Deoxyribonucleic acids have fewer oxygen groups and so are found only in monophosphate form in the cell, while ribonucleic acids have more oxygen groups and are found only in triphosphate forms.

Answer: A Section: 7.2 Skill: Knowledge/Comprehension LO: G1, S7.2 8) What types of bonds are formed between complementary DNA bases?

A) glycosidic bonds B) phosphodiester bonds C) covalent bonds D) hydrogen bonds E) ionic bonds Answer: D Section: 7.2 Skill: Knowledge/Comprehension LO: G1, S7.2 9) If there is 24% guanine in a DNA molecule, then there is ________ cytosine.

A) 24% B) 26% C) 48% D) 52% E) Not possible to determine Answer: A Section: 7.2 Skill: Applying/Analyzing LO: G4, S7.2b 10) In a single strand of DNA, what fraction of the nucleotides in a molecule are cytosine and thymine (% C and % T added together)?

A) 25% B) 50% C) 75% D) 100% E) It depends on the DNA sequence Answer: E Section: 7.2 Skill: Applying/Analyzing LO: G4, S7.2b 3 .

11) If complementary DNA strands were arranged in a parallel manner, what would you expect to see?

A) Complementary nucleotides would be attracted to each other, forming ionic bonds that would make the helix stable but not uniform in width.

B) Some regions of the two strands may form atypical hydrogen bonds, but the overall structure of the two DNA strands would not be stable.

C) Complementary nucleotides would line up normally; but fewer hydrogen bonds would form, so the strands could be more easily pulled apart.

D) The phosphodiester backbones would be too close and repel one another.

E) There would be no discernable difference between DNA strands aligned in a parallel versus an antiparallel manner.

Answer: B Section: 7.2 Skill: Applying/Analyzing LO: G2, S7.2 12) In their famous experiment, which of the following would Meselson and Stahl have observed after one cycle of replication in 14N medium if DNA replication were CONSERVATIVE? A) DNA molecules containing one strand of 15N-DNA and one strand of 14N-DNA B) DNA molecules containing two 15N-DNA strands only C) DNA molecules containing two 14N-DNA strands only D) An equal number of DNA molecules containing two 15N-DNA strands and DNA molecules containing two 14N-DNA strands E) A mix of DNA molecules corresponding to A and B Answer: D Section: 7.3 Skill: Applying/Analyzing LO: G1, S7.3 13) A portion of one strand of DNA has the sequence 5′ AATGGCTTA 3′. If this strand is used as a template for DNA replication, which of the following correctly depicts the sequence of the newly synthesized strand in the direction in which it will be synthesized?

A) 3′ TTACCGAAT 5′ B) 5′ TAAGCCATT 3′ C) 3′ AATGGCTTA 5′ D) 5′ AATGGCTTA 3′ E) 5′ TTACCGAAT 3′ Answer: A Section: 7.3 Skill: Applying/Analyzing LO: G1, S7.3b

4 .

14) Based on the following replication bubble, which of these statements is true?

A) X and Y are leading strands, W and Z are lagging strands B) X and Z are leading strands, W and Y are lagging strands C) X and W are leading strands, Y and Z are lagging strands D) W and Z are leading strands, X and Y are lagging strands E) W and Y are leading strands, X and Z are lagging strands Answer: E Section: 7.4 Skill: Applying/Analyzing LO: G1, S7.4 15) The following represents a DNA strand in the process of replication. The bottom sequence is that of the DNA strand with polarity indicated and the top sequence represents the RNA primer.

GGGGCCUUU 5′ AAATCCCCGGAAACTAAAC 3′ Which of the following will be the first DNA nucleotide added to the primer?

A) A B) G C) C D) T E) U Answer: A Section: 7.4 Skill: Applying/Analyzing LO: G1, S7.3b

5 .

16) What is one difference between DNA replication in bacteria versus eukaryotes?

A) Eukaryotic chromosomes are replicated bi-directionally, while bacterial chromosomes are replicated in one direction.

B) Eukaryotic chromosomes have many origins of replication and replicate bi-directionally, while bacteria have only one origin of replication and replicate uni-directionally.

C) Bacterial chromosomes are replicated bi-directionally, while eukaryotic chromosomes are replicated in one direction.

D) Eukaryotic chromosomes have many origins of replication, while bacteria have only one origin of replication.

E) The process is identical in bacterial and eukaryotic DNA replication.

Answer: D Section: 7.3 Skill: Applying/Analyzing LO: G1, S7.3 17) What is the DNA replication fork?

A) It is the binding site on the chromosome for DNA polymerase.

B) It is the site where the DNA helix opens to two single DNA strands.

C) It is the DNA bound inside the DNA polymerase active site.

D) It is the site where the leading and lagging strands meet on chromosome.

E) It is the DNA strand that wraps around DNA polymerase and serves as the template for the lagging strand.

Answer: B Section: 7.3 Skill: Knowledge/Comprehension LO: G1, S7.3

6 .

18) Which arrow(s) point(s) to the location of helicase in the diagram shown?

A) A B) B C) C D) A and E E) B and D Answer: E Section: 7.4 Skill: Applying/Analyzing LO: G1, S7.4 19) Why would DNA synthesis occur in both directions from an origin of replication?

A) Leading strands extend from one side of the origin while lagging strand extends from the other side of the origin.

B) DNA is a double helix and DNA synthesis is 5' to 3'.

C) DNA direction synthesis is random and sometimes extends in one direction from an origin of replication and sometimes in the other direction.

D) Replication-associated proteins bind at two replication forks.

E) It allows DNA strands to overlap when reaching another origin of replication.

Answer: D Section: 7.4 Skill: Synthesis/Evaluation LO: G2, S7.4

7 .

20) Why do origins of replication in various bacteria have conserved DNA consensus sequences?

A) Consensus sequences are found throughout bacterial genomes and so would be found at the origin as well.

B) The consensus sequence allows the origin to wrap into a circle and recruit replication initiation proteins.

C) It is more efficient to use the same sequence multiple times in different genomes than to change it.

D) Fewer mutations in the DNA occur at the very beginning of replication at the origin, hence why they are conserved as consensus sequences.

E) Origins need to be recognized by replication initiation proteins to open up their AT-rich regions.

Answer: E Section: 7.4 Skill: Synthesis/Evaluation LO: G2, S7.4a 21) If Single-Stranded Binding protein (SSB) is not present during DNA replication, what would you expect to see?

A) The replisome complex would not assemble on the oriC region.

B) The DNA cannot bend, so hydrogen bonds in the 13-mer region of oriC remain intact.

C) Helicase activity is inhibited, so DNA strands cannot be separated.

D) SSB carries the helicase protein to the open region of DNA, so hydrolysis and strand separation will not occur.

E) SSB prevents reannealing of the separated strands, so strands would quickly reanneal and DNA replication cannot proceed.

Answer: E Section: 7.4 Skill: Applying/Analyzing LO: G2, S7.4b 22) DNA helicase inhibitors are well studied as potential drug targets. What would you expect to see if DNA helicase activity is inhibited?

A) The replisome complex would not assemble on the oriC region.

B) The DNA cannot bend, so hydrogen bonds in the 13-mer region of oriC remain intact.

C) Helicase catalyzes ATP hydrolysis and DNA strands separation, so the helix cannot be unwound and strands will not separate.

D) Helicase carries the SSB protein to the open region of DNA, so hydrolysis and strand separation will not occur.

E) Helicase prevents reannealing of the separated strands, so strands would quickly reanneal and DNA replication cannot proceed.

Answer: C Section: 7.4 Skill: Applying/Analyzing LO: G2, S7.4b

8 .

23) What is required for DNA polymerase to initiate DNA strand synthesis?

A) a short DNA primer synthesized by the enzyme primase B) a short RNA primer synthesized by the enzyme primase C) ATP and a short DNA primer synthesized by the enzyme topoisomerase D) ATP and a short RNA primer synthesized by the enzyme topoisomerase E) DNA polymerase initiates DNA strand synthesis without requiring any additional enzymes.

Answer: B Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4 24) Why are leading and lagging strand primers removed rather than joined with Okazaki fragments?

A) The primers do not hydrogen bond to the template strand correctly.

B) They have to be removed for leading and lagging strand synthesis to begin.

C) Primers are too short to efficiently bind to Okazaki fragments.

D) They contain nucleotides with 2'OH groups, and are targeted for excision by DNA Polymerase.

E) They lack the 3' OH that would permit them to be covalently linked to the Okazaki fragments.

Answer: D Section: 7.4 Skill: Synthesis/Evaluation LO: G2, S7.4 25) You identify a cell in which DNA polymerase III is functional, but it seems to exhibit extremely low processivity. This is likely a defect in what structure?

A) the t proteins B) the pol III holoenzyme C) the clamp loader D) the topoisomerase enzyme E) the sliding clamp Answer: E Section: 7.4 Skill: Applying/Analyzing LO: G2, S7.4b 26) The extraordinary accuracy of the DNA polymerase III enzyme lies in its ability to "proofread" newly synthesized DNA, a function of the enzyme's ________.

A) 3′-to-5′ exonuclease activity B) 5′-to-3′ exonuclease activity C) 3′-to-5′ polymerase activity D) 5′-to-3′ polymerase activity E) 3′-to-5′ helicase activity Answer: A Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4 9 .

27) What is the purpose of the 3'-to-5' exonuclease activity of DNA Polymerase?

A) Some polymerases do not have exonuclease activity.

B) Remove mismatched nucleotides in the newly synthesized strand.

C) Degrade excessive nucleotides on Okazaki fragments.

D) Remove mismatched nucleotides in the template strand.

E) Removing mismatched nucleotides at the polymerase active site increases the rate of synthesis in the 5' to 3' direction.

Answer: B Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4 28) Where would you expect to find telomerase activity?

A) At the end of a chromosome in a normal healthy eukaryotic body (somatic) cell.

B) At the end of a chromosome in a cancerous eukaryotic body cell.

C) On the leading strand of DNA in a normal bacterial cell.

D) On the lagging strand of DNA in a normal bacterial cell.

E) At the centromere of a chromosome in a healthy eukaryotic reproductive cell.

Answer: B Section: 7.4 Skill: Applying/Analyzing LO: G2, S7.4c 29) Why are telomeres problematic for eukaryotic chromosome replication?

A) They are highly repetitive and thus hard to replicate correctly.

B) Telomerase is more error-prone than the normal DNA Polymerase.

C) The T loop blocks formation of primers on the lagging strand.

D) Removal of the lagging strand primer leaves a gap in the one of the strand's DNA sequences.

E) Maintaining very long telomeres promotes cancer cell formation.

Answer: D Section: 7.4 Skill: Synthesis/Evaluation LO: G2, S7.4c 30) Which of the following would you find in a Sanger sequencing reaction but not in a polymerase chain reaction?

A) DNA template B) DNA polymerase C) dNTPs D) ddNTPs E) DNA primer Answer: D Section: 7.5 Skill: Synthesis/Evaluation LO: G2, S7.5

10 .

31) Which functional groups have been altered in a ddNTP compared to a dNTP? A) The ddNTPs have a 2′ OH and a 3′ H, while dNTPs have a 2′ H and a 3′ OH. B) The ddNTPs have a 2′ H and a 3′ H, while dNTPs have a 2′ H and a 3′ OH. C) The ddNTPs have a 2′ OH and a 3′ OH, while dNTPs have a 2′ H and a 3′ H. D) The ddNTPs have a 2′ H and a 3′ H, while dNTPs have a 2′ OH and a 3′ OH. E) The ddNTPs have a 2′ H and a 3′ OH, while dNTPs have a 2′ H and a 3′ H. Answer: B Section: 7.5 Skill: Knowledge/Comprehension LO: G1, S7.5 32) Which of the following temperature cycles would you expect to see in a standard polymerase chain reaction (from denaturation to annealing to extension)?

A) 95° → 72° → 55° B) 45° → 72° → 95° C) 95° → 55° → 72° D) 95° → 72° → 95° E) 75° → 95° → 45° Answer: C Section: 7.5 Skill: Knowledge/Comprehension LO: G1, S7.5a 33) Why are next generation DNA sequencing technologies known as sequencing-by-synthesis?

A) The sequencing occurs during S phase of the cell cycle.

B) RNA molecules are synthesized off of the DNA templates and the incorporated nucleotides are then determined.

C) Numerous synthesized fragments of DNA are sequenced to determine which nucleotides were incorporated.

D) Incorporated nucleotides are determined while they are being added to a growing DNA strand.

E) The complete DNA strands are synthesized then sequenced.

Answer: D Section: 7.5 Skill: Synthesis/Evaluation LO: G2, S7.5d

11 .

7.2 Short Answer Questions 1) Based on Griffith's results, what would you expect if you injected both heat-killed type RII and living type SIII? Answer: The mouse lives.

Section: 7.1 Skill: Applying/Analyzing LO: G1, S7.1 2) What are the three parts of a DNA nucleotide?

Answer: deoxyribose, base, phosphate Section: 7.1 Skill: Knowledge/Comprehension LO: G1, S7.2 3) What are the regions where DNA-binding proteins can make direct contact with exposed nucleotides?

Answer: major and minor grooves Section: 7.2 Skill: Knowledge/Comprehension LO: G1, S7.2 4) What is removed from dNTPs to provide the energy required for DNA polymerase to catalyze DNA strand elongation?

Answer: two phosphates (pyrophosphate group) Section: 7.2 Skill: Knowledge/Comprehension LO: G1, S7.2 5) What proteins bind the DNA sequences found at the origins of replication?

Answer: binding replication enzymes Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4 6) During DNA replication in E. coli, how does the cell prevent the DNA strands from following their natural tendency to seek maximum stability and reanneal?

Answer: SSB prevents reannealing.

Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4a 7) What is the overall direction of DNA strand elongation?

Answer: 5′ to 3′ Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4 12 .

8) What structure in E. coli delivers primase and accessory proteins to the oriC site to synthesize the primer needed for DNA replication?

Answer: primosome Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4a 9) What is the term for the daughter strand synthesized continuously during DNA replication?

Answer: leading strand Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4 10) What did Okazaki's discovery of the short segments of DNA reveal about DNA replication on the lagging strand?

Answer: It is discontinuous.

Section: 7.4 Skill: Applying/Analyzing LO: G1, S7.4 11) Which enzyme, a combination of several proteins and a molecule of RNA, acts as the template for the telomere repeat DNA sequence?

Answer: telomerase Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4c 12) How might you choose a region of DNA for a PCR primer so as to increase the temperature necessary for primer annealing (to minimize nonspecific PCR products)?

Answer: look for higher GC concentrations Section: 7.5 Skill: Applying/Analyzing LO: G2, S7.5a 13) What characteristic of the polymerase isolated from Thermus aquaticus makes it unique and highly useful for maintaining the enzyme efficiency during PCR? Answer: heat stability Section: 7.5 Skill: Knowledge/Comprehension LO: G1, S7.5a 14) What are the three steps of each PCR cycle?

Answer: denaturation, annealing, extension Section: 7.5 Skill: Knowledge/Comprehension LO: G1, S7.5a 13 .

7.3 Fill-in-the-Blank Questions 1) The large, complex aggregation of proteins and enzymes that assembles at the replication fork is known as the ________.

Answer: replisome Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4 2) Comparison of conserved sequences among related species usually leads to the identification of ________, which illustrate the nucleotides most often found at each position of DNA in the conserved region.

Answer: consensus sequences Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4a 3) DNA polymerase requires a primer sequence to provide a ________ functional group to which the new DNA nucleotide can be added.

Answer: 3′ OH Section: 7.4 Skill: Knowledge/Comprehension 4) The major and minor grooves of DNA are features of the helix that can be attributed to ________, which involves tight packing of DNA bases in a duplex.

Answer: base stacking Section: 7.2 Skill: Knowledge/Comprehension LO: G1, S7.4 5) ________ is the term for a multiprotein complex in which a core enzyme is associated with additional protein components leading to its function.

Answer: Holoenzyme Section: 7.4 Skill: Knowledge/Comprehension LO: G1, S7.4 6) ________ must be added to a PCR to direct the polymerase where to begin synthesis.

Answer: PCR primers Section: 7.5 Skill: Knowledge/Comprehension LO: G1, S7.5

14 .

7.4 Essay Questions 1) Describe the DNA helix structure proposed by Watson and Crick. What is the diameter of the helix and the distance between bases? What types of bonds are involved in stabilizing the DNA backbone, and what types of bonds are found between complementary base pairs? Why would a sequence containing high concentrations of G and C require more energy to break the bonds?

Answer: The DNA model contains two DNA polymer strands in a right-handed twist around a common axis. The strands are orientated in an antiparallel direction from 5′ to 3′ and 3′' to 5′.

The backbone consists of alternating sugar-phosphate groups stabilized by phosphodiester bonds, and it is located on the outside of the helix. Complementary base pairs are stacked in the center of the helix and are stabilized by hydrogen bonds. There are three hydrogen bonds between G and C, but only two between A and T. Thus, GC-rich regions require higher temperatures to break the bonds. The base pairs, which are perpendicular to the axis, are stacked with 10 base pairs per turn and a helical repeat of 34 Å. The overall diameter of the helix is 10 angstroms.

Section: 7.2 Skill: Synthesis/Evaluation LO: G2, S7.2 2) Briefly describe the Meselson and Stahl experiment that indicated that DNA replication is semiconservative. How would their results have differed if DNA replication were actually conservative in nature?

Answer: Meselson and Stahl grew bacteria in the presence of 15N and then replaced the 15N media with the "lighter" 14N media. DNA was extracted at various intervals to collect several successive generations. Using high-speed CsCl density-gradient ultracentrifugation, they observed that after the first generation, the "heavy" DNA (15N/15N) was not present, and the DNA was intermediate between light and heavy (14N/15N). After the second generation, there were two bands with equal amounts of intermediate and light bands (14N/15N and 14N/14N).

These results validated the semiconservative model, in which one strand of the parent DNA is present in each daughter strand of replicated DNA. If the conservative model were correct, they would have expected to observe DNA molecules with two distinct densities after generation 1 (15N/15N and 14N/14N). Their first-generation results excluded this model.

Section: 7.2 Skill: Synthesis/Evaluation LO: G2, S7.2

15 .

3) Describe the function of the oriC. What do you predict will happen to the cell if 100 bp between the 13-mer and 9-mer sequence is deleted as indicated in the figure?

Answer: A consensus sequence is the portion of a conserved region that carries out the essential function that is the focus of natural selection. The 13-mer and 9-mer consensus sequences that are part of oriC have been maintained by natural selection because they have essential functional roles in replication initiation, as we explain in the following section. Beyond the presence of the consensus sequences themselves, natural selection also acts to maintain specific spacing between the consensus sequences. Spacing is important because DNA-binding proteins must assemble at sites where consensus sequences are located. Different proteins are attracted to different consensus sequences, and each protein must have the physical space to bind to DNA and to interact with the other proteins bound to the consensus sequence region. The spacing between consensus sequence elements is also important to allow room on the DNA molecule for protein binding and interaction between bound proteins. If you delete this region, replication will not proceed because there is not enough space for replisome assembly.

Section: 7.4 Skill: Synthesis/Evaluation LO: G2, S7.4a 4) Many antibacterial agents and chemotherapy agents act as topoisomerase inhibitors. Why would inhibition of topoisomerase cause cell death in bacterial and eukaryotic cells?

Answer: During DNA replication, as DNA strands unwind it creates torsional stress that accumulates as the unwound region gets larger. Without free ends, covalently closed circular DNA readily accumulates superhelical twists (supercoiled DNA). The accumulating stress could break the DNA template at random locations, potentially leading to a breakdown of DNA replication and, ultimately, to cell death. Topoisomerases prevent this lethal event by catalyzing a controlled cleavage and rejoining of DNA, thus enabling over-twisted strands to unwind.

Topoisomerase inhibitors block the ligation step of the cell cycle, generating single- and doublestranded breaks that harm the integrity of the genome and ultimately result in apoptosis of the cancer cells or bacteria.

Section: 7.4 Skill: Synthesis/Evaluation LO: G2, S7.4b

16 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 8 Molecular Biology and Transcription and RNA Processing 8.1 Multiple-Choice Questions 1) What are two distinguishing features of RNA? A) RNA has a ribose sugar and uracil nitrogenous base.

B) RNA contains a methylated form of thymine.

C) RNA forms a double helix of reverse complementary strands.

D) RNA contains a pyrophosphate group bound to the ribose.

E) RNA contains phophodiester bonds as part of its sugar backbone.

Answer: A Section: 8.1 Skill: Knowledge/Comprehension LO: G1, S8.1 2) Which type of research technique was used to track newly synthesized RNA within a eukaryotic cell?

A) in situ hybridization B) band shift assay C) pulse-chase D) Southern blotting E) DNA footprint protection assay Answer: C Section: 8.1 Skill: Knowledge/Comprehension LO: G1, S8.1 3) Prokaryotes and eukaryotes produce which of the following types of RNA? A) mRNA, tRNA, rRNA B) mRNA, gRNA, siRNA C) rRNA, siRNA, snRNA D) siRNA, tRNA, miRNA E) miRNA, rRNA, snRNA Answer: A Section: 8.1 Skill: Knowledge/Comprehension LO: G1, S8.1

1 .

4) You wish to create a mutation that prevents access of RNA polymerase to the gene. Which region of a gene would you mutate?

A) coding region B) start codon C) promoter sequence D) terminator sequence E) stop codon Answer: C Section: 8.2 Skill: Applying/Analyzing LO: G1, S8.2a 5) A gene has acquired a mutation in which the protein product has 50 additional amino acids at the end. Which region of the gene was likely mutated?

A) coding region B) start codon C) promoter sequence D) terminator sequence E) stop colon Answer: E Section: 8.2 Skill: Applying/Analyzing LO: G2, S8.2a 6) What is the role of a promoter region of a gene?

A) Recruit RNA Polymerase to the transcriptional start site.

B) Recruit transcription factors that form the initiation complex.

C) Serve as the original region of transcription of a gene.

D) Protect the gene from mutations in intergenic regions.

E) Recruit rho protein to assist in transcription.

Answer: B Section: 8.2 Skill: Knowledge/Comprehension LO: G1, S8.2a 7) Which region(s) of a gene are NOT found within the mRNA transcript?

A) promoter region B) termination region C) stop codon D) promoter and stop codon E) promoter and termination region Answer: E Section: 8.2 Skill: Knowledge/Comprehension LO: G1, S8.2a

2 .

8) What is the consensus sequence for the Pribnow box from these sequences?

A) TTGATA B) TATGAT C) TATTAT D) ACCA E) TAGTAT Answer: B Section: 8.2 Skill: Applying/Analyzing LO: G3, S8.2b 9) What is the -35 consensus sequence for the following sequences?

A) TTGATA B) TATGAT C) TATTAT D) ACAA E) TAGTAT Answer: A Section: 8.2 Skill: Applying/Analyzing LO: G3, S8.2b

3 .

10) In a given bacterium, transcription of housekeeping genes is normal, but genes involved in nitrogen metabolism, stress, and chemotaxis are disrupted. Which sigma subunit is INTACT? A) σ28 B) σ32 C) σ35 D) σ54 E) σ70 Answer: E Section: 8.2 Skill: Applying/Analyzing LO: G1, S8.2a 11) What is the significance of the open complex when the RNA Polymerase binds the DNA? A) The RNA Polymerase binds the single-stranded template strand in its active site.

B) The growing RNA molecule can now fit inside the active site of the RNA Polymerase.

C) It permits transcription factors to bind to the RNA Polymerase.

D) The RNA Polymerase binds the single-stranded coding strand in its active site.

E) It assists with propagation of the RNA Polymerase along the DNA helix.

Answer: A Section: 8.2 Skill: Applying/Analyzing LO: G2, S8.2a 12) Why does rho-dependent transcriptional termination in bacteria require the rho protein?

A) RNA Polymerase stalls at various sites in the gene and rho helps push RNA Polymerase to the end of the gene.

B) The rho protein assists in formation of the termination stem-loop that pauses the RNA Polymerase.

C) The rho protein helps unwind the stem-loop structure after the RNA has been released by RNA Polymerase.

D) RNA Polymerase stalls on the termination stem-loop and rho is needed to displace the RNA Polymerase.

E) The stem-loop is insufficiently stable to displace the RNA Polymerase by itself and needs rho protein to assist.

Answer: D Section: 8.2 Skill: Applying/Analyzing LO: G2, S8.2a

4 .

13) You want to design a drug that prevents transcription of eukaryotic mRNAs but does not affect transcription of other RNAs. What enzyme would you target?

A) RNA polymerase I B) RNA polymerase II C) RNA polymerase III D) ribozyme E) methyl transferase Answer: B Section: 8.3 Skill: Applying/Analyzing LO: G1, S8.3 14) What is the type of each eukaryotic protein that primarily transcribes mRNA, tRNA, and rRNA, respectively?

A) RNA Polymerase I, II, III B) RNA Polymerase II, I, III C) RNA Polymerase I, III, II D) RNA Polymerase II, III, I E) RNA Polymerase III, II, I Answer: B Section: 8.3 Skill: Knowledge/Comprehension LO: G1, S8.3 15) What must eukaryotes do to initiate transcription of a gene?

A) Open the DNA template and then bind RNA Polymerase at the transcriptional initiation site.

B) Recruit the transcription factors and RNA Polymerase that compose the pre-initiation complex.

C) Recruit general transcription factors to produce an open complex and then recruit RNA Polymerase.

D) Bind RNA Polymerase to displace histone proteins that binding DNA in the promoter region.

E) Bind transcription factors from enhancer sequences to the RNA Polymerase.

Answer: B Section: 8.3 Skill: Applying/Analyzing LO: G2, S8.3c

5 .

16) Which assay allows you to identify the exact location of the protein-binding sequence within a promoter?

A) western/immuno blotting B) in situ hybridization C) pulse-chase assay D) Southern blotting E) DNA footprint protection assay Answer: E Section: 8.3 Skill: Applying/Analyzing LO: G1, S8.3b

6 .

17) Following a DNA footprint protection assay, which banding pattern would you expect to see if you identify a potential promoter region?

A)

B)

C)

D)

E)

7 .

Answer: C Section: 8.3 Skill: Applying/Analyzing LO: G3, S8.3b 18) RNA polymerase I transcribes which tandem-repeat genes in nucleoli?

A) mRNA B) tRNA C) rRNA D) siRNA E) all types of RNA Answer: C Section: 8.3 Skill: Knowledge/Comprehension LO: G1, S8.3 19) Which of the following is part of a DNA molecule?

A) sigma B) RNA polymerase C) transcription factor D) promoter E) activator Answer: D Section: 8.2 Skill: Knowledge/Comprehension LO: G1, S8.2a 20) Which of the following statements accurately describes tRNA? A) All organisms produce tRNAs corresponding to the 61 amino-acid coding codons.

B) tRNAs are a variety of lengths and fold into a variety of shapes.

C) Amino acids are bound to the 5' end of the tRNA. D) Post-transcriptional modifications of tRNAs are not necessary for their function.

E) Wobble in the anticodon allows a single tRNA to bind to multiple codons.

Answer: E Section: 8.4 Skill: Knowledge/Comprehension LO: G1, S8.4

8 .

21) Which of the following are present in your liver cells?

A) enhancers of genes expressed in the liver B) enhancers of genes expressed in the kidney C) promoters of genes expressed in the liver D) promoters of genes expressed in the kidney E) All enhancers and promoters are present in liver cells Answer: E Section: 8.3 Skill: Applying/Analyzing LO: G1, S8.3a 22) If transcription of this gene occurs from left to right in the accompanying diagram, which DNA strand is the CODING (non-template) strand?

A) impossible to determine B) B C) C D) D E) E Answer: D Section: 8.3 Skill: Applying/Analyzing LO: G3, S8.3e

9 .

23) Using the accompanying diagram, which of the following corresponds to the 3' untranslated region?

A) A B) B C) C D) F E) G Answer: C Section: 8.3 Skill: Applying/Analyzing LO: G3, S8.3d 24) Which enzyme is required to initiate 5′ capping of eukaryotic mRNA transcripts by removing the terminal phosphate group?

A) ribozyme B) guanylyl transferase C) methyl transferase D) adenylyl cyclase E) phosphodiesterase Answer: B Section: 8.4 Skill: Knowledge/Comprehension LO: G1, S8.4a

10 .

25) What defines the end of a eukaryotic gene?

A) A 3' UTR of at least 25 nucleotides recruits an RNase that cleaves the pre-mRNA. B) There is no clearly defined end to eukaryotic genes, unlike for bacterial genes.

C) A stem-loop structure in the transcriptional terminator region stalls the RNA polymerase.

D) Presence of a stop codon leads to RNA polymerase stalling and ceasing transcription.

E) Presence of a polyadenylation signal sequence leads to cleavage of the pre-mRNA. Answer: E Section: 8.4 Skill: Applying/Analyzing LO: G2, S8.4a 26) A cell has a defect in polyadenylation of mRNA. The RNA transcripts encoding which type of protein would NOT be affected by this defect because they are not polyadenylated?

A) histone proteins B) transcription factors C) transmembrane proteins D) DNA binding proteins E) SR proteins Answer: A Section: 8.4 Skill: Applying/Analyzing LO: G1, S8.4a 27) What is the general name for the components of the spliceosome, which removes introns from mRNAs? A) lariat intronic nucleolar proteins B) branch point adenine recognition proteins C) small nuclear ribonucleoproteins D) microRNA activators E) small interfering RNA enhancers Answer: C Section: 8.4 Skill: Knowledge/Comprehension LO: G1, S8.4a 28) What is the purpose of alternative splicing in eukaryotic cells?

A) Increase the number of genes that do not have to contain introns.

B) Improve the efficiency of transcription and translation.

C) Produce multiple types of tRNAs that can bind to different codons.

D) Regulate the quantity of any single protein being produced in the cell.

E) Produce multiple polypeptide sequences from a single primary transcript.

Answer: E Section: 8.4 Skill: Applying/Analyzing LO: G1, S8.4c

11 .

29) The rat α-tropomyosin (α-Tm) gene produces nine different mature mRNA proteins from a single gene using which three "alternative" mechanisms?

A) splicing, start codon, stop codon B) splicing, promoters, and polyadenylation C) promoters, start codon, poly(A) signal sequences D) 5' cap, branch points, polyadenylation E) 5' cap, self-splicing, polyadenylation Answer: B Section: 8.4 Skill: Applying/Analyzing LO: G1, S8.4c 30) What are catalytically active RNAs that can activate processes such as self-splicing?

A) ribozymes B) ribosomes C) rnzymes D) snRNAs E) pre-mRNAs Answer: A Section: 8.4 Skill: Knowledge/Comprehension LO: G1, S8.4 31) In humans, the 30S pre-RNA transcript yields three rRNA segments following transcription and RNA cleavage. Which RNA transcripts are generated by the 30S pre-RNA transcript in E. coli?

A) two rRNAs (15S each) B) three rRNAs (5S, 16S, and 23S) C) two tRNA transcripts D) three rRNAs (5.8S, 18S, and 28S) and two tRNA transcripts E) three rRNAs (5S, 16S, and 23S) and two tRNA transcripts Answer: E Section: 8.4 Skill: Knowledge/Comprehension LO: G1, S8.4 32) One type of RNA editing involves inserting uracils into edited mRNA with the assistance of which type of specialized RNA? A) small nucleolar RNA B) post-transcriptional editor RNA C) transposon RNA D) guide RNA E) telomerase RNA Answer: D Section: 8.4 Skill: Knowledge/Comprehension LO: G1, S8.4 12 .

8.2 Short Answer Questions 1) What are the two chemical structures found in RNA but not DNA? Answer: ribose sugar, uracil base Section: 8.1 Skill: Knowledge/Comprehension LO: G1, S8.1 2) Which three types of RNAs are found in both prokaryotes and eukaryotes?

Answer: mRNA, tRNA, and rRNA Section: 8.1 Skill: Applying/Analyzing LO: G1, S8.1 3) miRNA regulates protein production through which process?

Answer: RNA interference Section: 8.1 Skill: Knowledge/Comprehension LO: G1, S8.1 4) Which type of RNA interacts with nuclear proteins to form a ribonucleoprotein complex responsible for intron removal?

Answer: snRNA Section: 8.1 Skill: Knowledge/Comprehension LO: G1, S8.1 5) You wish to prevent transcription of all three types of RNA in bacteria. How many different types of RNA polymerase would you need to inhibit?

Answer: one Section: 8.2 Skill: Applying/Analyzing LO: G1, S8.2a 6) What are the mechanisms for transcription termination in bacteria?

Answer: intrinsic and rho-dependent Section: 8.2 Skill: Knowledge/Comprehension LO: G1, S8.2a 7) Transcription of inverted repeats produces an mRNA with complementary segments that fold to form what type of secondary structure?

Answer: stem-loop (hairpin) Section: 8.2 Skill: Knowledge/Comprehension LO: G1, S8.2a 13 .

8) Eukaryotes have how many different RNA polymerases?

Answer: 3 Section: 8.3 Skill: Knowledge/Comprehension LO: G1, S8.3 9) A plant cell exhibits defects in transcription of transfer RNA genes. The gene encoding which polymerase is likely to be mutated in this cell?

Answer: RNA polymerase III Section: 8.3 Skill: Knowledge/Comprehension LO: G1, S8.3 10) Prokaryotes have a -10 and -35 consensus sequence in their promoter. What are the three eukaryotic promoter sequence elements (or "boxes")?

Answer: GC-rich, CAAT, and TATA boxes Section: 8.3 Skill: Knowledge/Comprehension LO: G1, S8.3c 11) What proteins aid in the recognition of the promoter sequence and binding of RNA polymerase II in eukaryotes?

Answer: transcription factors (TFs) Section: 8.3 Skill: Knowledge/Comprehension LO: G1, S8.3 12) What proteins are bound to enhancers, forming a protein "bridge" that bends the DNA over the promoter to initiate transcription?

Answer: activator proteins Section: 8.3 Skill: Knowledge/Comprehension LO: G1, S8.3a 13) What proteins bind to silencer sequences, forming a protein "bridge" that bends the DNA over the promoter and prevents transcription?

Answer: repressor proteins Section: 8.3 Skill: Knowledge/Comprehension LO: G1, S8.3a

14 .

14) Processing of the 30S pre-RNA transcript in humans produces which three rRNAs after enzymatic cleavage?

Answer: 5.8S, 18S, 28S Section: 8.4 Skill: Knowledge/Comprehension LO: G1, S8.4 8.3 Fill-in-the-Blank Questions 1) RNA polymerase is called a ________, meaning it is an intact complex with full enzymatic capacity.

Answer: holoenzyme Section: 8.2 Skill: Knowledge/Comprehension LO: G1, S8.2a 2) In intrinsic termination, inverted repeat DNA sequences followed immediately by a string of ________ produce an mRNA stem-loop, followed by a string of ________.

Answer: adenines; uracils Section: 8.2 Skill: Knowledge/Comprehension LO: G1, S8.2a 3) Polyadenylation begins with the binding of ________ near a six-nucleotide mRNA sequence, AAUAAA, downstream of the stop codon.

Answer: cleavage and polyadenylation specificity factor (CPSF) Section: 8.4 Skill: Knowledge/Comprehension LO: G1, S8.4 4) Intron self-splicing occurs when two ________ reactions excise the intron and allow exons to ligate.

Answer: transesterification Section: 8.4 Skill: Knowledge/Comprehension LO: G1, S8.4

15 .

8.4 Essay Questions 1) In the following prokaryotic DNA sequence, label the highlighted regions.

Using the prokaryotic DNA sequence above, describe the relationship between the highlighted regions and how mutations in each region may affect gene expression.

Answer: The -10 (Pribnow) and -35 consensus sequences are important because of their nucleotide content, their location relative to one another, and their location relative to the start of transcription. Mutation in a consensus sequence is likely to alter how efficiently a protein binds to the promoter, and to decrease the amount of gene transcription. In contrast, mutations between consensus sequences are unlikely to alter gene transcription, because the sequences in these intervening regions do not bind tightly to RNA polymerase. The nucleotides between -10 and -35 are important as spacers between the consensus elements, but their specific sequences are not critical.

Section: 8.2 Skill: Synthesis/Evaluation LO: G2, S8.2b 2) Describe how band/mobility shift assays and DNA footprint protection assays can be used to identify promoter regions in DNA. Answer: A band/mobility shift assay uses two identical samples of DNA fragments that contain suspected promoter consensus sequences. The control sample has no transcriptional proteins added, while the experimental sample includes transcriptional proteins.

DNA footprint protection also begins with two identical samples of DNA fragments containing suspected promoter consensus sequences. All fragments are end-labeled with radioactive or fluorescent phosphorus, and once again the experimental DNA is mixed with transcriptional proteins, but the control sample is not. In this experiment, both samples are exposed to DNase I that randomly cuts DNA that is not protected by protein. The samples are run in separate lanes of an electrophoresis gel, and each end-labeled fragment produced is identified by autoradiography or other detection method.

Section: 8.3 Skill: Synthesis/Evaluation LO: G2, S8.3b

16 .

3) Describe the mechanism(s) used by bacteria and eukaryotes for transcription termination.

Answer: Bacterial transcription termination can be intrinsic or rho-dependent, and the type of termination is dependent on the presence of certain DNA sequences. In intrinsic termination, inverted repeat DNA sequences followed immediately by a string of adenines produce an mRNA stem-loop followed by a poly-U string. In rho-dependent termination, a rho protein binds to mRNA and catalyzes the release of the transcript from RNA polymerase.

Each of the eukaryotic RNA polymerases uses a different mechanism to terminate transcription.

Transcription by RNA polymerase III is terminated in a mechanism that resembles E. coli intrinsic termination. The RNA pol III transcribes a terminator sequence that creates a string of uracils in the transcript. The RNA pol III terminator sequence does not contain an inverted repeat, however, so there is no stem-loop structure near the 3′ end of RNA. Transcription by RNA pol I is terminated at a 17-bp consensus sequence that binds transcriptionterminating factor I (TTFI). A large rRNA precursor transcript is cleaved about 18 nucleotides upstream of the TTFI binding site, so the consensus sequence does not appear in the mature transcript.

Termination of transcription by RNA pol II is not completely clear, but it is known that the 3′ end of mRNA is not generated by transcriptional termination. The 3′ end of the pre-mRNA is created by enzymatic action that removes a segment from the 3′ end of the transcript and replaces it with a string of adenine nucleotides, the poly-A tail. This step of pre-mRNA processing is thought to be associated with subsequent termination of transcription.

Section: 8.3 Skill: Synthesis/Evaluation LO: G2, S8.3c 4) Describe the three main differences between bacterial and eukaryotic mRNA transcripts.

Answer: • Eukaryotic mRNA transcripts are more stable than bacterial transcripts.

• Eukaryotes exhibit a separation, in time and in location, between transcription and translation. Bacteria lack a membrane-bound nucleus, which leads to coupling of transcription and translation.

• Eukaryotic genes have introns, which also leads to alternative splicing of mRNA transcripts.

Bacteria do not have introns, and thus do not have intron splicing mechanisms.

Section: 8.4 Skill: Synthesis/Evaluation LO: G2, S8.4a

17 .

5) What are the three types of posttranscriptional processing, and what are the consequences of preventing each of these modifications in terms of gene expression?

Answer: (a) 5′ capping: the addition of a modified nucleotide to the 5′ end of mRNA. The 5′ cap has several functions, including (1) protecting mRNA from rapid degradation, (2) facilitating mRNA transport across the nuclear membrane, (3) facilitating subsequent intron splicing, and (4) enhancing translation efficiency by orienting the ribosome on mRNA. (b) 3′ polyadenylation: cleavage at the 3′ end of mRNA and addition of a tail of multiple adenines to form the poly-A tail. The 3′ poly-A tail has several functions, including (1) facilitating transport of mature mRNA across the nuclear membrane, (2) protecting mRNA from degradation, and (3) enhancing translation by enabling ribosomal recognition of messenger RNA. (c) intron splicing: RNA splicing to remove introns and ligate exons. Intron splicing requires exquisite precision to remove all intron nucleotides accurately without intruding on the exons and without leaving behind additional nucleotides, so that the mRNA sequence encoded by the ligated exons will completely and faithfully direct synthesis of the correct polypeptide.

Disrupting any one of the transcriptional processes would ultimately disrupt normal gene expression by resulting in mRNA degradation, reduced translation, or improper splicing and, thus, production of the wrong gene product.

Section: 8.4 Skill: Synthesis/Evaluation LO: G2, S8.4a

18 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 9 The Molecular Biology of Translation 9.1 Multiple-Choice Questions 1) What features of proteins does two-dimensional gel electrophoresis exploit in order to separate proteins?

A) shape and size B) charge and shape C) size and charge D) pH and polarity E) charge and pH Answer: C Section: 9.1 Skill: Knowledge/Comprehension LO: G1, S9.1 2) The Shine-Dalgarno sequence in bacteria ________.

A) is a purine-rich consensus sequence found in the 5′ UTR of the mRNA B) is a pyrimidine-rich consensus sequence found in the 3′ UTR of the mRNA C) is a purine-rich consensus sequence found in the 16S rRNA subunit D) is a region of the tRNA molecule involved in formation of charged tRNAs E) is a consensus sequence involved in the termination of translation Answer: A Section: 9.2 Skill: Knowledge/Comprehension LO: G1, S9.2b 3) During translation initiation in bacteria, the amino acid on the initiator tRNA is ________.

A) methionine (Met) B) N-formylmethionine (fMet) C) acetylated D) IF-1 E) added using ATP as the energy source Answer: B Section: 9.2 Skill: Knowledge/Comprehension LO: G1, S9.2b

1 .

4) Identification of ribosomal proteins involves two-dimensional gel electrophoresis, which separates the proteins on the basis of ________.

A) mass B) folded shape C) charge D) mass and charge E) mass, charge, and folded shape Answer: D Section: 9.2 Skill: Applying/Analyzing LO: G1, S9.2 5) How does the eukaryotic initiation complex locate the correct start codon?

A) The true start codon is the first ATG encountered downstream of the Shine-Dalgarno sequence.

B) The initiation complex moves the small ribosomal subunit through the 5′ UTR, scanning for the start AUG. C) The pre-initiation complex moves the ribosome through the 3′ UTR, scanning for the Kozak sequence.

D) The correct start codon is the first ATG encountered downstream of the Kozak sequence.

E) The correct start codon is the formyl-ATG, which will encode for fMet in the protein.

Answer: B Section: 9.2 Skill: Applying/Analyzing LO: G1, S9.2a 6) What is the cellular location of the stages of translation in bacteria and eukaryotes?

A) cytosol for bacteria and cytosol, mitochondrion, and plastid for eukaryotes B) cytosol for bacteria and nucleus for eukaryotes C) membrane for bacteria and cytosol and rough ER for eukaryotes D) nucleoid for bacteria and rough ER for eukaryotes E) cytosol for bacteria and eukaryotes Answer: A Section: 9.2 Skill: Applying/Analyzing LO: G1, S9.2a

2 .

7) How does the eukaryotic ribosomal small subunit recognize the start codon on the mRNA? A) It binds an Met-tRNA to the first AUG codon it encounters.

B) It performs an ATP hydrolysis within the small subunit once it encounters a Met-tRNA already bound to the AUG. C) It binds an Met-tRNA to the first AUG codon after the Kozak sequence.

D) It wraps the mRNA strand to bring initiation enhancer proteins into the vicinity of the start codon.

E) It undergoes a conformational charge that recruits other proteins when it hydrogens bonds to the correct tri-nucleotide sequence.

Answer: C Section: 9.2 Skill: Applying/Analyzing LO: G1, S9.2 8) A tRNA in the P site of the ribosome will enter the ________ site after translocation of the ribosome.

A) E B) A C) 3′ D) 5′ E) initiation Answer: A Section: 9.2 Skill: Knowledge/Comprehension LO: G1, S9.2 9) A portion of an mRNA attached to a ribosome reads: 5′ GACCAUUUUUGA 3′ If a tRNA with a Phenylalanine amino acid attached is in the P site of the ribosome, an empty tRNA will be present in the E site that delivered which amino acid?

A) aspartic acid B) proline C) serine D) histidine E) tyrosine Answer: D Section: 9.2 Skill: Applying/Analyzing LO: G1, S9.2

3 .

10) A portion of mRNA attached to a ribosome reads: 5′ GACCAUUUUUGA 3′ In the polypeptide produced, what amino acid will be attached to the amino group of the histidine encoded by this mRNA? A) aspartic acid B) proline C) serine D) phenylalanine E) tyrosine Answer: A Section: 9.2 Skill: Applying/Analyzing LO: G1, S9.2 11) What would you expect to find bound to the stop codon at the A site?

A) a charged tRNA with the anticodon TAG B) a charged tRNA with the anticodon ATC C) an uncharged tRNA D) a translation release factor E) Nothing binds to a stop codon, which is why the peptide is released.

Answer: D Section: 9.2 Skill: Knowledge/Comprehension LO: G1, S9.2 12) What is necessary for a eukaryotic RNA to be recognized and bound by the small subunit of the ribosome?

A) Formation of the pre-initiation complex before ribosome binding.

B) Formation of the initiation complex before ribosome binding.

C) Sufficiently large 5' UTR for ribosome scanning.

D) Presence of 5' methyl-G cap on the mRNA. E) Presence of an AUG start codon near the 5' end of the mRNA. Answer: D Section: 9.3 Skill: Knowledge/Comprehension LO: G1, S9.3

4 .

13) Elongation factors translocate the ribosome in the 3′ direction by a distance of ________.

A) one codon B) one nucleotide C) two nucleotides D) two codons E) three codons Answer: A Section: 9.3 Skill: Knowledge/Comprehension LO: G1, S9.3 14) A polycistronic mRNA contains multiple ________.

A) promoters B) Shine-Dalgarno sequences C) Kozak sequences D) mRNAs E) polypeptide-encoding sequences Answer: E Section: 9.3 Skill: Knowledge/Comprehension LO: G1, S9.3 15) Why are eukaryotic mRNAs not polycistronic, unlike some bacterial transcripts?

A) Eukaryote's genetic code is non-overlapping, and so coding sequences cannot overlap on the same mRNA. B) Eukaryotes have more complex translational machinery than bacteria that is also less efficient in initiating translation.

C) Eukaryotic mRNAs are generally shorter than bacteria mRNAs, and so do not contain sufficient information to encode additional polypeptides.

D) Bacteria couple their translation with transcription, while eukaryotes do not.

E) The eukaryotic ribosome must bind to the 5' end of the mRNA and scan, while the bacterial ribosome can bind to a Shine-Delgarno sequence anywhere in the mRNA. Answer: E Section: 9.3 Skill: Synthesis/Evaluation LO: G2, S9.3 16) What does it mean for two codons to be synonymous?

A) They share two of the same nucleotides in their codon sequence.

B) They share one of the same nucleotides in their codon sequence.

C) They encode the same amino acid.

D) They are adjacent on the mRNA. E) They occur in equal abundance in an mRNA sequence.

Answer: C Section: 9.4 Skill: Applying/Analyzing LO: G1, S9.4 5 .

17) What result would you expect if a mutation eliminates one of the four arms of a tRNA? A) The tRNA will fit into the A site but will not release the peptide at the P site.

B) The tRNA will not be recognized by tRNA synthetase and cannot be charged.

C) The tRNA will be charged with the wrong amino acid.

D) The tRNA will not be able to undergo traditional complementary base pairing.

E) There will be no effect on function, so long as the anticodon region is intact.

Answer: B Section: 9.4 Skill: Applying/Analyzing LO: G1, S9.4a 18) How many different aminoacyl-tRNA synthetases can be found in a given organism's cells?

A) 16 B) 20 C) At least 20, or more depending on the organism D) 61 E) The number varies greatly depending on the organisms type.

Answer: C Section: 9.4 Skill: Applying/Analyzing LO: G1, S9.4b 19) If a tRNA anticodon were mutated such that it no longer performed wobble, what would be the effect on encoded proteins?

A) The ribosome would be unable to translate proteins.

B) Many proteins would be truncated.

C) Many proteins would have several mutated amino acids throughout their sequence.

D) The rate of protein synthesis would be slowed.

E) A different amino acid would consistently replace the amino acid whose tRNA was mutated.

Answer: B Section: 9.4 Skill: Synthesis/Evaluation LO: G2, S9.4b 20) A mutagen has introduced a frame-shift mutation by adding one nucleotide base. Which of the following could be a reversion mutation for this particular mutant?

A) deleting 2 bases only B) adding 1 base only C) deleting 1 base or adding 1 base D) deleting 1 base or adding 3 bases E) deleting 1 base or adding 2 bases Answer: E Section: 9.5 Skill: Applying/Analyzing LO: G1, S9.4

6 .

21) Which of these choices represents one possible corresponding mRNA sequence that can be transcribed from the following DNA template?

5′ - CTGTATCCTAGCACCCAAATCGCATTAGGAC - 3′ A) 5′ - ATG CGA TTT GGG TGC TAG - 3′ B) 5′ - AUG CGA UUU GGG UGC UAG - 3′ C) 5′ - AUG CGA UUU GGG UGC - 3′ D) 5′ - CTA GCA CCC AAA TCG CAT TAG - 3′ E) 3′ - GGA CAU AGG UAC GUG GGU UUA GCG UAA UCC UG - 5′ Answer: B Section: 9.4 Skill: Applying/Analyzing LO: G1, S9.4c 22) Given the following mRNA sequence, what is the amino acid sequence for the corresponding polypeptide?

5′ - AUG CGA UUU GGG UGC UAG - 3′ A) 5′-Met-Arg-Phe-Gly-Stop-3′ B) C–Met-Arg-Leu-Glu–N C) N–Met-Arg-Phe-Gly-Stop–C D) N–Met-Asp-Phe-Gly-Trp–C E) N–Arg-Phe-Gly-Stop–C Answer: C Section: 9.4 Skill: Applying/Analyzing LO: G1, S9.4c 23) Given the following mRNA sequence, which of the following mRNAs would encode a protein with a different sequence of amino acids?

5′ - AUG CAG UUA GCG UGC UAG - 3′ A) 5′ - AUG CAA UUA GCG UGC UAG - 3′ B) 5′ - AUG CAG UUA GCA UGC UAG - 3′ C) 5′ - AUG CAG UUG GCG UGC UAG - 3′ D) 5′ - AUG CAA UUA GCG UGU UAG - 3′ E) 5′ - AUG CAC UUA GCA UGC UAG - 3′ Answer: E Section: 9.4 Skill: Applying/Analyzing LO: G1, S9.4c

7 .

24) Which mRNA below would code for a premature stop codon from the following amino acid sequence?

N—Met-Gln-Leu-Arg-Cys—C A) 5′ - AUG CAG AUA GCG UGC UAG - 3′ B) 5′ - AUG AAG UUA GCG UGC UAG - 3′ C) 5′ - AUG CAG UAA GCG UGC UAG - 3′ D) 5′ - AUG CAG UUA UUG UGC UAG - 3′ E) 5′ - AUG CAG UUA GCG UGC AAG - 3′ Answer: C Section: 9.4 Skill: Applying/Analyzing LO: G1, S9.4c 25) How might a single base INSERTION into the second codon of the coding sequence of a gene affect the amino acid sequence of a protein encoded by the gene?

A) A single amino acid could change.

B) The amino acid sequence would be changed.

C) The mutation may have no effect on amino acid sequence.

D) A single extra amino acid would be present in the protein.

E) All of the above are possible outcomes.

Answer: B Section: 9.5 Skill: Applying/Analyzing LO: G1, S9.5b 26) You have identified a bacterial protein that has retained the starting fMet in its protein sequence. Which of the following is likely true of this protein?

A) It is likely nonfunctional, since bacteria use posttranslational cleavage of fMet to make functional proteins.

B) It will show improper protein sorting and will likely remain in the ER. C) It will be not be able to be chemically modified, so it will be sent to the Golgi for secretion.

D) It will be a functional bacterial protein, since all functional proteins must begin with fMet. E) It will likely form a disulfide bond with a second peptide chain, forming a protein complex.

Answer: A Section: 9.5 Skill: Applying/Analyzing LO: G1, S9.5

8 .

27) How is a tRNA able to recognize its proper mRNA codon?

A) The amino acid on the tRNA recognizes the mRNA codon through hydrogen and ionic bonding.

B) Appropriate shape of the tRNA allows it to fit onto the extended mRNA strand.

C) Complementary hydrogen bonding between tRNA and mRNA promote binding.

D) Ionic bonds between the tRNA and the active site of the ribosomal subunits promote binding to the mRNA. E) Hydrogen bonding between the ribosomal subunits and the mRNA creates the proper active site conformation to allow tRNA binding.

Answer: C Section: 9.5 Skill: Applying/Analyzing LO: G1, S9.5 28) If the first nucleotide in a codon is mutated to a different nucleotide, what would be the effect on the encoded protein?

A) No effect as the problem nucleotide would be corrected by RNA editing mechanisms.

B) A missense mutation from one encoded amino acid changing to another.

C) A frameshift mutation for all of the subsequent amino acids in the protein.

D) A silent mutation and no change in the encoded amino acid.

E) It depends on what the changed nucleotide is.

Answer: E Section: 9.5 Skill: Applying/Analyzing LO: G1, S9.5 29) In the unlikely event that a tRNA has been charged with the wrong amino acid, which highfidelity enzyme most likely caused the incorrect charging?

A) peptidyl transferase B) DNA polymerase I C) DNA polymerase III D) aminoacyl synthetase E) aminoacyl peptidase Answer: D Section: 9.5 Skill: Applying/Analyzing LO: G1, S9.5

9 .

30) If the genetic code were overlapping, how many complete codons would the following sequence encode before encountering a stop codon?

5′ - AUGCGAUUAAAGUGC - 3′ A) 4 B) 6 C) 8 D) 10 E) 13 Answer: C Section: 9.5 Skill: Applying/Analyzing LO: G1, S9.5a 9.2 Short Answer Questions 1) Ribosomal subunits are measured in which units, which describe the speed of sedimentation of a substance during centrifugation?

Answer: Svedberg units (S) Section: 9.1 Skill: Knowledge/Comprehension LO: G1, S9.1 2) What molecule provides the energy to form charged tRNAs? Answer: GTP Section: 9.2 Skill: Knowledge/Comprehension LO: G1, S9.2 3) The preinitiation complex forms in bacteria when the start codon is recognized through binding of 16srRNA and what region of mRNA? Answer: Shine-Dalgarno sequence Section: 9.2 Skill: Knowledge/Comprehension LO: G1, S9.2b 4) What are the three phases of translation?

Answer: initiation, elongation, termination Section: 9.2 Skill: Knowledge/Comprehension LO: G1, S9.2

10 .

5) During elongation, the charged tRNA is recruited to which location on the ribosome?

Answer: A site Section: 9.2 Skill: Knowledge/Comprehension LO: G1, S9.2 6) If you are designing an antibiotic that inhibits peptide bond formation, what enzyme would you target?

Answer: peptidyl transferase Section: 9.2 Skill: Applying/Analyzing LO: G1, S9.2 7) Binding of what protein initiates translation-termination events that result in polypeptide release and dissociation of ribosomal subunits?

Answer: release factor (RF) Section: 9.3 Skill: Knowledge/Comprehension LO: G1, S9.3 8) In a polyribosome, which end of the mRNA would have the shortest polypeptides?

Answer: 5′ end Section: 9.3 Skill: Applying/Analyzing LO: G1, S9.3 9) Thanks to flexible base pairing, the wobble nucleotides in anticodons can be one or more of the standard RNA nucleotides or which modified nucleotide?

Answer: inosine (I) Section: 9.4 Skill: Knowledge/Comprehension LO: G1, S9.4b 10) Eighteen of the amino acids have two or more synonymous codons. Which two amino acids are the exceptions?

Answer: Met and Trp Section: 9.4 Skill: Applying/Analyzing LO: G1, S9.4 11) Sidney Brenner demonstrated that the triplet code is, in fact, nonoverlapping. Assuming a nonoverlapping code, how many COMPLETE codons would the following sequence encode before encountering a stop codon? 5′ - AUGCGAUUAUAGUGC - 3′ Answer: 3 (AUG CGA UUA UAG) Section: 9.5 Skill: Applying/Analyzing LO: G1, S9.5a 11 .

12) Many antibiotics cause mispairing between codons and anticodons. What fatal effect would this mispairing have on the cell?

Answer: defective protein synthesis Section: 9.6 Skill: Applying/Analyzing LO: G1, S9.6c 13) In the Golgi complex, what type of protein modification will signal the polypeptide destination by determining the receptor to which the peptide will bind?

Answer: glycosylation Section: 9.6 Skill: Applying/Analyzing LO: G1, S9.6a 14) In the endoplasmic reticulum, misfolded proteins are identified and bound by what molecules?

Answer: chaperones Section: 9.6 Skill: Knowledge/Comprehension LO: G1, S9.6b 9.3 Fill-in-the-Blank Questions 1) ________ help control ribosome formation and binding of the initiator tRNA. Answer: Initiation factor proteins Section: 9.1 Skill: Knowledge/Comprehension LO: G1, S9.1 2) In eukaryotes, the initiation factor proteins eIF1A and eIF3 join with ________ to form the preinitiation complex.

Answer: charged tRNAMet Section: 9.2 Skill: Knowledge/Comprehension LO: G1, S9.2 3) Bacteria group their genes such that they share a single promoter and the mRNA transcript synthesizes several different polypeptides. Collectively, these are referred to as ________ mRNAs, which are part of the operon system.

Answer: polycistronic Section: 9.3 Skill: Knowledge/Comprehension LO: G1, S9.3

12 .

4) Using mathematical reasoning, a triplet genetic code gives a possible ________ different codons, while a doublet genetic code would yield ________ different codons. Thus, the triplet code accounts for 20 known amino acids and indicates redundancy in the genetic code.

Answer: 64; 16 (4n, where n = number of nucleotides per codon) Section: 9.4 Skill: Knowledge/Comprehension LO: G4, S9.4 5) A signal sequence at the N terminus of newly synthesized proteins will direct the protein to the ________ for protein packaging.

Answer: endoplasmic reticulum (ER) Section: 9.6 Skill: Knowledge/Comprehension LO: G1, S9.6b 9.4 Essay Questions 1) Describe the technique of two-dimensional gel electrophoresis. What are the two dimensions referring to? How is this technique different from conventional gel electrophoresis, and when would it be used?

Answer: In two-dimensional gel electrophoresis, proteins are separated in the first dimension by a version of gel electrophoresis known as isoelectric focusing. In this procedure, proteins are separated exclusively by their charge. A protein's pH environment affects its charge, and every protein has a pH–called the isoelectric point–at which it has neutral charge and cannot move in an electrical field. In isoelectric focusing, proteins migrate through a pH gradient to their isoelectric point, where they stop.

Once isoelectric focusing is complete, protein separation takes place in the second dimension that uses sodium dodecyl sulfate (SDS) gel electrophoresis. SDS is a strong anionic detergent that denatures proteins by disrupting the interactions that keep them folded. Denatured proteins migrate through the gel at a rate determined by their mass; that is, the rate is determined by the number of amino acids they contain. In the SDS gel dimension of two-dimensional gel electrophoresis, each protein has a unique starting point corresponding to its isoelectric point.

Proteins with large mass (more amino acids) migrate a short distance in the second dimension, whereas proteins with small mass (fewer amino acids) migrate a greater distance.

One-dimensional gel electrophoresis uses a buffered solution to maintain constant pH throughout the gel, while isoelectric focusing gels contain a pH gradient. Therefore, two-dimensional gel electrophoresis separates proteins based on charge and mass.

Section: 9.2 Skill: Synthesis/Evaluation LO: G2, S9.2

13 .

2) Starting translation at the authentic (correct) start codon is essential for translation of the correct polypeptide. Errant translation starting at the wrong codon, or even at the wrong nucleotide of the start codon, may produce an abnormal polypeptide and result in a nonfunctional protein. Compare and contrast the mechanisms used by bacteria and eukaryotes to identify the authentic start codon during translation initiation.

Answer: In bacteria (specifically E. coli), six components are involved in translation initiation: (1) mRNA, (2) the small ribosomal subunit, (3) the large ribosomal subunit, (4) the initiator tRNA, (5) three essential initiation factor proteins, and (6) GTP. The preinitiation complex forms when the authentic start codon sequence is identified by base pairing that occurs between the 16S rRNA in the 30S ribosome and a short mRNA sequence located a few nucleotides upstream of the start codon in the 5′ UTR of mRNA. The complex knows where to assemble on the mRNA because of the Shine-Dalgarno sequence, a purine-rich sequence of about six nucleotides located three to nine nucleotides upstream of the start codon. A complementary pyrimidine-rich segment containing the sequence UCCUCC is found near the 3′ end of 16S rRNA, and it pairs with the Shine-Dalgarno sequence to position the mRNA on the 30S subunit such that it recognizes the correct start codon.

In eukaryotes, the preinitiation complex is recruited to the 5′-cap region of mRNA, located at the end of the 5′ UTR. Once the eukaryotic initiation complex is formed, it embarks on a process called scanning, in which it uses ATP hydrolysis to move the small ribosomal subunit through the 5′ UTR in search of the start codon. About 90% of eukaryotic mRNAs use the first AUG encountered by the initiation complex as the start codon. The initiation complex is able to accurately locate the authentic start codon because the codon is embedded in a consensus sequence, the Kozak sequence, which reads 5′ - ACCAUGG - 3′.

Section: 9.2 Skill: Synthesis/Evaluation LO: G2, S9.2a 3) Prokaryotes and eukaryotes show differences in posttranscriptional modifications of mRNA. What are these differences, and why do eukaryotes show more modifications than prokaryotes?

Answer: In bacteria, the coupling of transcription and translation allows ribosomes to engage in translation of the 5′ region of mRNAs whose 3′ end is still under construction by RNA polymerase. Translation of the mRNA transcripts begins before transcription is complete, since there are no membrane-bound organelles to provide physical separation of the two processes. In eukaryotes, however, transcription and translation are uncoupled. Transcription takes place in the nucleus, where pre-mRNA is processed to form mature mRNA. Translation occurs in the cytoplasm after release of mature mRNA. Because the processes are physically (and temporally) separated, the mRNA receives a 5′ cap and 3′ tail to make the transcript stable and to help the ribosome locate the transcript when it exits the nucleus.

Section: 9.3 Skill: Synthesis/Evaluation LO: G2, S9.3

14 .

4) Polypeptides must be sorted after translation. Does protein sorting occur in both prokaryotes and eukaryotes? Describe the process of protein sorting and explain how the signal sequence is involved. What would you expect to see if the signal sequence has been mutated or deleted?

Answer: Protein sorting is a process that uses signal sequences of amino acids at the N-terminal end of a polypeptide to sort proteins and direct them to their cellular destinations.

Protein sorting is needed in bacterial cells because of the many proteins specifically destined for the cell membrane. In eukaryotes, however, protein sorting is far more complex than in bacteria; proteins are dispatched to particular cellular organelles, such as the chloroplast, mitochondrion, lysosome, and nucleus, and certain proteins are secreted from the cells. If the signal sequence is mutated, proteins may not be secreted or processed properly and will begin to accumulate within the cell. Ultimately, this will result in cell death and impaired protein secretion.

Section: 9.6 Skill: Synthesis/Evaluation LO: G2, S9.6b 5) Describe what is meant by a "conformational disease" and give an example. Why do you think these diseases exhibit late age of onset?

Answer: Diseases resulting from the death of cells due to accumulation of misfolded proteins are classified as conformational diseases. Several human protein conformational diseases are known, and they are often neurodegenerative disorders and dementias. For example, Alzheimer disease is produced by the accumulation of misfolded β-amyloid protein that leads to the destruction of certain brain cells. A familial form of Parkinson disease produced by the accumulation of misfolded α-synuclein protein also results in the destruction of certain brain cells. Similarly, Huntington disease is produced by the destruction of neurons due to the accumulation of misfolded huntington protein. Each of these diseases has a late age of onset that is due to gradual cell death caused by the presence of abnormal aggregates of misfolded protein.

These diseases are neuronal because that is where the genes encoding the proteins are expressed.

Section: 9.6 Skill: Synthesis/Evaluation LO: G2, S9.6a

15 .

6) Compare and contrast the mechanism of action of two commonly used antibiotics. Include details about which steps of translation are inhibited, and explain how the antibiotic accomplishes this inhibition.

Answer: Antibiotics target different aspects of microbe biology to kill microorganisms. Familiar antibiotics such as tetracycline, streptomycin, and chloramphenicol target different stages of microbial translation, as do less familiar antibiotics such as erythromycin, puromycin, and cycloheximide. Each antibiotic contains a different active compound that takes advantage of unique features of bacterial translation to disrupt the production of bacterial proteins while not interfering with the translation of proteins in our cells. That is how the antibiotic selectively kills the bacteria and does not harm our cells.

Answers may vary, but some possible answers are listed below.

Section: 9.6 Skill: Synthesis/Evaluation LO: G2, S9.6c

16 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 10 Eukaryotic Chromosome Abnormalities and Molecular Organization 10.1 Multiple-Choice Questions 1) Which type of chromosome has no p arms?

A) metacentric B) submetacentric C) acrocentric D) subacrocentric E) telocentric Answer: E Section: 10.1 Skill: Remembering/Understanding LO: S10.1a 2) What technique would you use to detect a target sequence in an intact chromosome using a labeled molecular probe?

A) polymerase chain reaction (PCR) B) fluorescent in situ hybridization (FISH) C) western/immuno blotting D) DNA sequencing E) karyotyping Answer: B Section: 10.1 Skill: Remembering/Understanding LO: S10.1d 3) Which of the following is FALSE regarding organization in an interphase nucleus?

A) Each chromosome is localized to a specific region of the nucleus.

B) Each chromosome may assume a position that allows for a characteristic pattern of gene transcription.

C) Each chromosome occupies exactly the same region in all nuclei within an organism.

D) Interchromosomal domains act as channels for the movement of proteins, RNAs, and enzymes.

E) The largest and most gene-rich chromosomes are located near the center of the nucleus.

Answer: C Section: 10.1 Skill: Applying/Analyzing LO: S10.1e

1 .

4) Which of the following is true of the gametes of a human female who has nondisjunction of her X chromosomes in meiosis I? A) All the gametes contain 23 chromosomes.

B) All the gametes contain 22 chromosomes.

C) All the gametes contain 24 chromosomes.

D) The gametes contain 22 or 24 chromosomes.

E) The gametes contain 22, 23, or 24 chromosomes.

Answer: D Section: 10.2 Skill: Applying/Analyzing LO: G7, S10.2a 5) If nondisjunction occurs in meiosis I of a mother, what proportion of her gametes could have been fertilized by a normal sperm to produce a child with Turner syndrome?

A) 0 B) 1/4 C) 1/2 D) 3/4 E) 1 Answer: C Section: 10.2 Skill: Applying/Analyzing LO: G2, G7, S10.2a 6) A karyotype shows that a child has Klinefelter syndrome (47,XXY). If the child is also colorblind (due to a recessive X-linked allele), despite his parents having normal color vision, in which parent and stage of meiosis did nondisjunction occur?

A) mother in meiosis I B) mother in meiosis II C) father in meiosis I D) father in meiosis II E) mother in meiosis I or II Answer: B Section: 10.2 Skill: Evaluating/Creating LO: G2, G5, G7, S10.2b

2 .

7) When nondisjunction occurs early in embryogenesis rather than gametogenesis, what would you expect in the resulting karyotype?

A) monosomy B) trisomy C) mosaicism D) uniparental disomy E) random X-inactivation Answer: C Section: 10.2 Skill: Remembering/Understanding LO: S10.2d 8) Which of the following is FALSE of Down syndrome?

A) Studying individuals with partial trisomy of chromosome 21 identified the Down syndrome critical region (DSCR) with a small number of genes responsible for the principal symptoms.

B) There are homologs in model organisms, such as mice and Drosophila, for genes in the Down syndrome critical region (DSCR).

C) Molecular genetic analysis of the chromosomes in infants with trisomy 21 determined that the majority of cases were attributable to maternal nondisjunction in meiosis I. D) Down syndrome is the most common triploidy (3n) in which individuals have three sets of chromosomes.

E) There is a link between the increased risk of Down syndrome and increased maternal age at conception.

Answer: D Section: 10.2 Skill: Applying/Analyzing LO: S10.2c 9) Prader-Willi and Angelman syndromes are caused by which type of chromosomal mutations, both in connection with chromosome 15?

A) monosomy B) trisomy C) mosaicism D) uniparental disomy E) random X-inactivation Answer: D Section: 10.2 Skill: Remembering/Understanding LO: S10.2d

3 .

10) In an allopolyploid organism, what is true regarding the fertility of interspecies hybrids?

A) Interspecies hybrids are fertile due to nonhomology of chromosomes.

B) Chromosome doubling by nondisjunction in gametocytes can lead to homologous chromosome pairing, disjunction, and fertile hybrids.

C) Interspecies hybrids will be fertile so long as there is an odd number of chromosomes in the offspring.

D) Mitotic nondisjunction results in haploid cells.

E) Meiotic nondisjunction produces three haploid gametes and one diploid gamete.

Answer: B Section: 10.3 Skill: Applying/Analyzing LO: S10.3a 11) Species A has 2n=16 chromosomes and species B has 2n=14 chromosomes. Offspring that are diploid for either 22 or 23 chromosomes are most likely?

A) autotriploids of species A B) autotetraploids of species B C) allotriploids D) allotetraploids E) allohexaploids Answer: C Section: 10.3 Skill: Applying/Analyzing LO: G4, S10.3b 12) Donkeys have a somatic chromosome number of 62. Hybrids produced by crossing donkeys and zebras are called zonkeys, are sterile, and have many characteristics intermediate between the two parental species. If a zonkey has 52 chromosomes, what is the somatic chromosome number of zebras?

A) 62 B) 42 C) 52 D) 21 E) 31 Answer: B Section: 10.3 Skill: Applying/Analyzing LO: G2, G4, G7, S10.2e

4 .

13) Which type of chromosome deletion is caused by two concurrent chromosome breaks (rather than a single break)?

A) acentric deletion B) microdeletion C) terminal deletion D) partial deletion E) interstitial deletion Answer: E Section: 13.3 Skill: Remembering/Understanding LO: S10.5d 14) What structure will form during synapsis in Prophase I if there is a partial chromosome deletion or duplication?

A) tetravalent complex B) nucleosome C) inversion loop D) unpaired loop E) solenoid Answer: D Section: 10.4 Skill: Remembering/Understanding LO: S10.5d 15) Which of the following is true regarding heterozygous carriers of large chromosome inversions or translocations?

A) They should be completely fertile because no genes were deleted.

B) They commonly exhibit phenotypic abnormalities.

C) They typically have a reduction in the number of viable gametes.

D) Inversion heterozygotes will only produce viable gametes if they have a paracentric inversion, but not a pericentric inversion.

E) Translocation heterozygotes will produce viable gametes regardless of the pattern of segregation of the involved chromosomes.

Answer: C Section: 10.5 Skill: Applying/Analyzing LO: G2, G5, S10.5d

5 .

16) A region of a chromosome spanning the centromere is broken and reattached in the reverse direction. This is an example of which type of chromosomal defect?

A) paracentric inversion B) pericentric inversion C) pericentric translocation D) dicentric inversion E) dicentric translocation Answer: B Section: 10.5 Skill: Applying/Analyzing LO: S10.5c 17) A chromosome has broken, and a piece of one chromosome is translocated to a nonhomologous chromosome. This is an example of what type of chromosomal alteration?

A) paracentric inversion B) dicentric bridge C) unbalanced translocation D) Robertsonian translocation E) inversion loop Answer: C Section: 10.5 Skill: Applying/Analyzing LO: S10.5c 18) A chromosome contains the following gene order: ABCD•EFGH Which of the following rearrangements represents a pericentric inversion?

A) A B C D • E F G H B) A B C D • E H G F C) A F G H B C D • E D) A C B D • E F G H E) A F E • D C B G H Answer: E Section: 10.5 Skill: Applying/Analyzing LO: S10.5a

6 .

19) A chromosome contains the following gene order: ABCD•EFGH Which of the following rearrangements represents a paracentric inversion?

A) A F E • D C B G H B) A B C E • D F G H C) A C B D • E F G H D) A B C D • E F G H E) A F G H B C D • E Answer: C Section: 10.5 Skill: Applying/Analyzing LO: S10.5a 20) Given a chromosome with the gene order, A B C D • E F G H, and an inverted chromosome with this gene order, A F E • D C B G H, which of the following recombinant chromosomes would result from a crossover between C and D? A) A B C C B G H and HGFE•DD•HEFA B) A B C C B G H and HGFE•DCBGH C) A B C D • E F A and H•GFEDCBG•H D) A B C D • E F A and HGFE•DCBGH E) A B C D E F A and H•GFEDCBG•H Answer: D Section: 10.5 Skill: Evaluating/Creating LO: G2, G3, G7, S10.5a 21) Which of the following is FALSE regarding individuals who are heterozygous for a large inversion?

A) Viable gametes will contain either the normal-order chromosome or the inversion-order chromosome.

B) Crossovers between homologous chromosomes that occur outside the inverted region produce nonviable gametes.

C) There is a higher probability of a crossover that would result in reduced fertility.

D) Crossover in the inversion loop of a paracentric inversion heterozygote results in viable and nonviable gametes.

E) Crossover in the inversion loop of a pericentric inversion heterozygote results in gametes with recombinant chromosomes that contain deletions and duplications.

Answer: B Section: 10.5 Skill: Applying/Analyzing LO: G2, S10.5a, S10.5d

7 .

22) A diploid organism with a genome size of n = 23 experienced a Robertsonian translocation.

How many chromosomes would you expect to see in the karyotype of a somatic cell in an affected individual?

A) 21 B) 25 C) 42 D) 44 E) 46 Answer: D Section: 10.5 Skill: Remembering/Understanding LO: S10.5d 23) In balanced translocation heterozygotes, viable gametes and progeny are produced by which pattern(s) of chromosome segregation?

A) adjacent-1 segregation B) adjacent-2 segregation C) alternate segregation D) both adjacent-1 and alternate segregation E) all three patterns of segregation (adjacent-1, adjacent-2, and alternate) Answer: C Section: 10.5 Skill: Remembering/Understanding LO: S10.5b 24) What is meant by the "beads on a string" model of chromatin?

A) The beads are the histones, and the string is the transcriptionally active DNA loops.

B) The beads are the nucleosomes, and the string is the linker DNA. C) The beads are molecules of DNA polymerase that attach to the DNA string.

D) The beads are ribosomes, and the string is the mRNA that has been transcribed from active chromatin.

E) The beads are the heterochromatic regions that are tightly compacted, and the strings are euchromatic regions that are being actively transcribed.

Answer: B Section: 10.6 Skill: Remembering/Understanding LO: S10.6b

8 .

25) Which histone protein is not part of the core nucleosome structure?

A) H1 B) H2A C) H2B D) H3 E) H4 Answer: A Section: 10.6 Skill: Remembering/Understanding LO: S10.6a 26) Which of the following correctly lists the levels of compaction in eukaryotes from naked DNA to the most compact?

A) solenoid, nucleosome, looped chromatin (300-nm fiber), metaphase chromosome B) solenoid, metaphase chromosome, nucleosome, looped chromatin (300-nm fiber) C) metaphase chromosome, nucleosome, solenoid, looped chromatin (300-nm fiber) D) nucleosome, solenoid, looped chromatin (300-nm fiber), metaphase chromosome E) nucleosome, looped chromatin (300-nm fiber), metaphase chromosome, solenoid Answer: D Section: 10.6 Skill: Remembering/Understanding LO: S10.6b 27) A cell can form 10 nm chromatin fibers, but not 30 nm fibers. Which molecule has likely been removed or mutated in this cell?

A) H1 B) H2A C) H2B D) H3 E) H4 Answer: A Section: 10.6 Skill: Applying/Analyzing LO: S10.6a 28) Active transcription occurs where with respect to matrix attachment regions (MARs)?

A) within euchromatin regions a large distance from MARs B) within euchromatin regions near MARs C) within heterochromatin regions a large distance from MARs D) within centromeric heterochromatin near MARs E) within the telomeres near MARs Answer: A Section: 10.6 Skill: Applying/Analyzing LO: S10.1c

9 .

29) Which of the following best describes the histones associated with eukaryotic DNA? A) The amino acid sequences of H4 vary extensively.

B) Thousands of base pairs of DNA often wrap around each nucleosome core particle more than a dozen times.

C) Along a 10,000 base pair segment of DNA, you expect to find twice as many molecules of histone protein H1, as histone protein H4.

D) The five types of histone proteins are small, basic proteins with a positive charge that allows them to bind to DNA. E) Within the nucleosome core particle, H1 is the most conserved and crucial histone protein.

Answer: D Section: 10.6 Skill: Applying/Analyzing LO: S10.6a 30) The flies with variegated eye color had what abnormal chromosome structure?

A) X-chromosome inversion that moves the w+ gene to a different orientation on the chromosome in some cells following X-ray exposure B) X-chromosome translocation that moves the w+ gene to a region of heterochromatin on another chromosome in some cells following X-ray exposure C) a mutation in the w+ gene following X-ray exposure that causes the gene to be overexpressed in a tissue-specific manner D) X-chromosome deletion that deleted the w+ gene in some cells following X-ray exposure E) a mutation in a somatic chromosome that causes the w+ gene to be translocated to the Xchromosomes following X-ray exposure Answer: A Section: 10.6 Skill: Remembering/Understanding LO: S10.6c 31) Which of the following did Hermann Muller determine when studying eye color variegation in Drosophila melanogaster?

A) The pattern of variegation was the same from fly to fly and that the two eyes of a single fly also had the same variegation patterns.

B) The w+ allele was expressed in the cell if the heterochromatin spread covered its new location.

C) Transcription of the w+ allele was directly influenced by the level DNA compaction.

D) The chromosome region immediately surrounding the centromere was typically a euchromatic and contains a large number of expressed genes.

E) The w+ allele had been deleted in cells that produced the white eye patches.

Answer: C Section: 10.6 Skill: Applying/Analyzing LO: S10.1c, S10.6c

10 .

10.2 Short Answer Questions 1) Karyotypes are a method for displaying chromosomes by grouping them into homologous pairs based on what two distinguishing factors?

Answer: size and banding pattern Section: 10.1 Skill: Remembering/Understanding LO: S10.1a 2) You are attempting to isolate cells for your karyotype, and you find one cell in each of the phases (prophase, metaphase, anaphase, and telophase). Which one is best suited for a karyotype?

Answer: metaphase Section: 10.1 Skill: Applying/Analyzing LO: S10.1a 3) During gametogenesis, what percentage of gametes would be aneuploid if the nondisjunction event occurs during meiosis I? Answer: 100% (4/4) Section: 10.2 Skill: Applying/Analyzing LO: S10.2a 4) During gametogenesis, what percentage of gametes would be trisomic if the nondisjunction event occurs during meiosis II? Answer: 25% (1/4) Section: 10.2 Skill: Applying/Analyzing LO: S10.2a 5) "Seedless" fruits and vegetables have how many chromosomes and what type of chromosomal distribution?

Answer: odd-numbered polyploidy Section: 10.3 Skill: Applying/Analyzing LO: S10.3a 6) Cri-du-chat syndrome is a human disorder caused by which type of chromosomal defect?

Answer: terminal deletion Section: 10.4 Skill: Remembering/Understanding LO: S10.5d

11 .

7) What structure, seen during synapsis, is indicative of a chromosome inversion?

Answer: inversion loop Section: 10.5 Skill: Remembering/Understanding LO: S10.5d 8) What is the term for the phenomenon that describes the significant reduction, or absence, of recombinant chromosomes in the gametes produced by inversion heterozygotes, when a crossover is within the inverted region?

Answer: crossover suppression Section: 10.5 Skill: Remembering/Understanding LO: S10.5a 9) How will the centromeres be distributed in the two recombinant chromosomes following a crossover in the inversion loop of a paracentric inversion heterozygote?

Answer: dicentric chromosome (with two centromeres) and an acentric fragment (no centromere) Section: 10.5 Skill: Remembering/Understanding LO: S10.5a 10) What are the two common patterns of chromosome segregation seen in the tetravalent structures found in translocation heterozygotes?

Answer: alternate and adjacent-1 Section: 10.5 Skill: Remembering/Understanding LO: S10.5b 11) In individuals heterozygous for a reciprocal balanced translocation, adjacent-2 segregation of tetravalent structures is very rare since it requires which type of centromeres to move to the same pole of the cell at anaphase I? Answer: homologous Section: 10.5 Skill: Applying/Analyzing LO: S10.5b 12) Nucleosomes are comprised of how many molecules of histones?

Answer: 8 Section: 10.6 Skill: Remembering/Understanding LO: S10.6a

12 .

13) What is the name of the structure formed when the 10 nm fibers of chromatin form a cylindrical filament of coiled nucleosomes?

Answer: solenoid structure Section: 10.6 Skill: Remembering/Understanding LO: S10.6b 14) Which proteins would you target if you wanted to disrupt the chromosome superstructure for chromatin condensation that ultimately produces the characteristic shape of the metaphase chromosome?

Answer: scaffold proteins Section: 10.6 Skill: Applying/Analyzing LO: S10.6b 15) If you want to prevent chromatin loops from being anchored to the chromosome scaffold, which regions of the DNA would you target?

Answer: MARs (matrix attachment regions) Section: 10.6 Skill: Applying/Analyzing LO: S10.6b 10.3 Fill-in-the-Blank Questions 1) The phenomenon of ________ in allopolyploids consists of more rapid growth, increased fruit and flower production, and improved disease resistance.

Answer: hybrid vigor Section: 10.3 Skill: Remembering/Understanding LO: S10.3a 2) ________ is rare and occurs most commonly when repetitive regions of homologous chromosomes misalign, resulting in partial deletions and partial duplications.

Answer: Unequal crossover Section: 10.4 Skill: Remembering/Understanding LO: S10.4a 3) ________ is used to map genes in deleted chromosome regions by a method known as deletion mapping. It is a genetic phenomenon that occurs when a normally recessive allele is "unmasked" and expressed in the phenotype because the dominant allele on the homologous chromosome has been deleted.

Answer: Pseudodominance Section: 10.5 Skill: Remembering/Understanding LO: S10.5d 13 .

4) Chromosomal translocation involves chromosome breakage and reattachment of the broken segment to a ________ chromosome.

Answer: non-homologous Section: 10.5 Skill: Remembering/Understanding LO: S10.5d 5) After DNA replication, during reassembly of nucleosomes, the ________ tetramers are bound

by H2A-H2B dimers to form complete nucleosomes.

Answer: H3-H4 Section: 10.6 Skill: Remembering/Understanding LO: S10.6a

14 .

10.4 Essay Questions 1) There is clear evidence that chromatin state is directly related to the ability of transcriptionally active proteins to initiate gene transcription. Compare and contrast euchromatin to heterochromatin in the context of transcription.

Answer: Regions of active expression are identified as euchromatin or euchromatic regions.

Most expressed genes or genes that are actively transcribed are located in euchromatic regions.

Condensation of the DNA is variable in euchromatin throughout the cell cycle.

Regions in which chromatin is tightly condensed are identified as heterochromatin or heterochromatic regions. Heterochromatic regions contain fewer expressed genes or fewer genes that are actively transcribed compared to euchromatic regions.

Section: 10.1 Skill: Applying/Analyzing LO: G7, G8, S10.1c 2) Describe two mechanisms for causing uniparental disomy, and explain how this chromosomal defect is involved in Prader-Willi syndrome and Angelman syndrome.

Answer: Uniparental disomy occurs when both copies of a homologous chromosome pair originate from a single parent. Uniparental disomy has two mechanisms of origin. First, the rare mechanism would involve nondisjunction of the same chromosome in both the sperm and egg.

Second, the more common mechanism involves nondisjunction in one parent that creates an aneuploid gamete contributing two copies of chromosome 15. The other gamete is normal and contributes a single copy of chromosome 15. Gamete union results in trisomy 15, but through the process of trisomy rescue, one of the extra copies of chromosome 15 is randomly ejected in one of the first mitotic divisions following fertilization. This results in a zygote that retains two copies of chromosome 15 from the same parent (i.e., uniparental disomy).

Both Angelman syndrome and Prader-Willi syndrome are usually the result of a partial deletion of the 15q11.12 portion of chromosome 15. Uniparental disomy produces 10-20% of cases of AS and PWS. When uniparental disomy causes AS, both copies of chromosome 15 are from the father, and there is no maternal copy of chromosome 15. In uniparental disomy cases of PWS, both copies of chromosome 15 are from the mother, and there is no copy of the paternal chromosome 15.

Section: 10.2 Skill: Evaluating/Creating LO: G2, G8

15 .

3) How is aneuploidy different from polyploidy? What are the mechanisms by which aneuploidy and polyploidy are caused, and what are the consequences of these chromosomal alterations in terms of survival and fertility of the offspring?

Answer: Aneuploidy occurs when one or more chromosomes are lost or gained relative to the normal euploid number. Chromosome nondisjunction is a principal cause of aneuploidy.

Nondisjunction can occur in somatic cells or in germ-line cells with similar results. If a single pair of homologous chromosomes fails to properly disjoin in a somatic cell during mitotic cell division, one of the resulting daughter cells carries an extra chromosome (2n + 1), and the other is missing a chromosome (2n – 1). Mitotic cells of animals that contain the wrong number of chromosomes may suffer reduced viability, although cells with abnormal numbers of chromosomes are common in cancer where other alterations of chromosome number play a central role in cancer cell survival and proliferation.

Polyploidy is the presence of three or more sets of chromosomes in the nucleus of an organism.

Polyploidy in nature can result either from the duplication of euploid chromosome sets from a single species or from the combining of chromosome sets from different species.

Polyploids whose karyotype is comprised of chromosomes derived from a single species are designated as autopolyploids (auto = "self"), and polyploids with chromosome sets from two or more species are called allopolyploids (allo = "different"). Polyploidy is common in plant species but is extremely rare in animals; it is caused by multiple fertilization, mitotic nondisjunction, or meiotic nondisjunction.

Section: 10.3 Skill: Evaluating/Creating LO: G5, G8, S10.3a

16 .

4) The figure shows the structure of a series of fruit fly deletion mutants. The gap in each line represents the segment of the chromosome that is deleted in each mutant. Stains that were heterozygous for each deletion mutation and a point mutation (A, B, C, D, E, and F) were created and analyzed for wild-type (+) or mutant (-) phenotypes. The results of the analyses are shown in the table.

Which of the following is the correct order of the six point mutations on the chromosome?

Answer: C,E,A,F,B,D Section: 10.4 Skill: Evaluating/Creating LO: G2, G3, S10.4b 5) Eukaryotic chromosomes are a combination of proteins and DNA, which allows the cell to perform which four essential functions?

Answer: 1. compacts the DNA so it can fit in the nucleus 2. stabilizes the DNA to protect it from damage 3. promotes chromosome condensation required for cell division 4. regulates DNA replication and gene transcription Section: 10.6 Skill: Remembering/Understanding LO: G2, G7

17 .

6) Histone proteins are the principal agents in chromatin packaging. The nucleosome core particle is a heterooctameric protein complex that contains two molecules each of the four histones H2A, H2B, H3, and H4. Describe the process of managing existing nucleosome core particles during replication and the process of adding these and new nucleosome core particles to DNA after the replication fork passes.

Answer: The assembly of nucleosome core particles in connection with replication is driven by the partial denaturing of old (previously used) core particles into either dimers (H2A-H2B) or tetramers (H3-H4). These old core particle components are randomly joined with other dimers and tetramers after replication to form complete nucleosome core particles. New histone proteins are also available during replication that also form dimers and tetramers. A mixture of old and newly synthesized core particles is the pool from which post-replication core particles are assembled.

Section: 10.6 Skill: Applying/Analyzing LO: G7, G8, S10.6a

18 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 11 Gene Mutation, DNA Repair, and Homologous Recombination 11.1 Multiple-Choice Questions 1) Studies of gene mutation frequencies have shown that ________.

A) mutations are rare, and genomes are generally stable B) mutation are common and adaptive C) mutation frequencies are consistent between organisms, and each region of DNA is equally susceptible to random mutations D) mutations occur only at certain nucleotides and not others E) mutations affect RNA, but do not change DNA sequence Answer: A Section: 11.1 Skill: Knowledge/Comprehension LO: S11.1a 2) Which type of mutation is possible due to the redundant nature of the genetic code?

A) silent B) missense C) nonsense D) frameshift E) splice site Answer: A Section: 11.2 Skill: Knowledge/Comprehension LO: S11.1c 3) Given the sequence of triplet codons: 5′-TAC AAA ATA CAG CGG-3′, which of these sequences represents a nonsense mutation?

A) 5′-TAG AAA ATA CAG CGG-3′ B) 5′-TAC AAA TAC AGC GGG-3′ C) 5′-TAC AAG ATA CAG CGG-3′ D) 5′-TAC AAA ATA CAC CGG-3′ E) 5′-TAC AAA ATA CAG AGG-3′ Answer: A Section: 11.2 Skill: Applying/Analyzing LO: S11.1c

1 .

4) Given the sequence of triplet codons: 5′-TAC AAA ATA CAG CGG-3′, which of these sequences represents a missense mutation?

A) 5′-TAG AAA ATA CAG CGG-3′ B) 5′-TAC AAA TAC AGC GGG-3′ C) 5′-TAC AAG ATA CAG CGG-3′ D) 5′-TAC AAA ATA CAC CGG-3′ E) 5′-TAC AAA ATA CAG AGG-3′ Answer: D Section: 11.2 Skill: Applying/Analyzing LO: S11.1c 5) Which of the following is able to cause a change in a reading frame?

A) transversion B) deletion C) missense D) transition E) transversion or missense Answer: B Section: 11.2 Skill: Remembering/Understanding LO: S11.1c 6) What type of mutation is seen here?

Wild type: 5′-TAC AAA ATA CAG CGG-3′ Mutation: 5′-TAC AAG ATA CAG CGG-3′ A) transition B) transversion C) insertion D) deletion E) nonsense Answer: A Section: 11.2 Skill: Applying/Analyzing LO: S11.1c

2 .

7) A mutant DNA polymerase that increases the frequency of strand slippage would increase the frequency of which type of mutation?

A) missense B) transposition C) splice site D) transition E) triplet-repeat expansion Answer: E Section: 11.2 Skill: Applying/Analyzing LO: S11.2a 8) Which type of mutation converts a nucleotide to an alternative structure with the same composition but a slightly different placement of rare, less stable hydrogen bonds that cause base-pair mismatch?

A) depurination B) deamination C) tautomeric shift D) transition E) transversion Answer: C Section: 11.3 Skill: Knowledge/Comprehension LO: S11.3a 9) The fluctuation test allowed Luria and Delbruck to conclude that ________.

A) mutations occur at random in response to environmental challenges B) mutations are common and thus allow natural selection C) mutations occur in the absence of environmental challenges D) chemical mutagens are present even in drinking water E) none of the above Answer: C Section: 11.1 Skill: Applying/Analyzing LO: S11.3c

3 .

10) You would like to induce a transversion mutation into a sequence of DNA. Which type of chemical mutagen would give you the best chance of inducing the correct mutation without causing transition mutations as well?

A) base analog B) deaminating agent C) oxidative agent D) alkylating agent E) intercalating agent Answer: C Section: 11.3 Skill: Applying/Analyzing LO: S11.3a 11) Transposons can integrate into the promoters of genes, what is the most likely outcome of such an event?

A) frame shift B) altered gene expression C) amino acid substitution D) point mutation E) deamination Answer: B Section: 11.7 Skill: Applying/Analyzing LO: S11.1d 12) You have conducted an Ames test on a given compound. Which of the following would be classified as a positive result on the Ames test?

A) his- strain grows on an his- plate.

B) his- strain grows on an his+ plate.

C) his+ strain grows on an his- plate.

D) his+ strain grows on an his+ plate.

E) his+ strain grows on either an his- or an his+ plate.

Answer: A Section: 11.3 Skill: Applying/Analyzing LO: S11.3c

4 .

13) A strain of E. coli is unable to use the UV repair pathway because of an apparent absence of DNA helicase activity. Which UV repair gene is likely mutated in this strain?

A) uvr-A B) uvr-B C) uvr-C D) uvr-D E) pol I Answer: D Section: 11.5 Skill: Applying/Analyzing LO: S11.4 14) What phenotype would you expect to see in a strain of E. coli with a mutation in the phr gene?

A) decrease in UV-induced mutations B) increase in UV-induced mutations C) decrease in reversion of base-pair substitution mutations D) increase in the methylation of nucleotide bases E) decrease in the methylation of nucleotide bases Answer: B Section: 11.5 Skill: Knowledge/Comprehension LO: S11.4 15) Which pathway is affected by an inherited mutation in the ATM gene?

A) nucleotide excision repair pathway B) photoreactivation C) p53 repair pathway D) reactivation repair pathway E) UV repair pathway Answer: C Section: 11.5 Skill: Applying/Analyzing LO: S11.4

5 .

16) Which of these statements best describes gene expression in a damaged cell?

A) In a damaged cell, p53 is low and BAX transcription is inactive, so Bcl-2 represses apoptosis.

B) In a damaged cell, p53 is high and BAX transcription is active, so Bcl-2 is repressed and apoptosis is induced.

C) In a damaged cell, p53 is low and BAX transcription is active, so Bcl-2 represses apoptosis.

D) In a damaged cell, p53 is low and BAX transcription is active, so Bcl-2 is repressed and apoptosis is induced.

E) In a damaged cell, p53 is high and BAX transcription is inactive, so Bcl-2 is activated and apoptosis is induced.

Answer: B Section: 11.5 Skill: Knowledge/Comprehension LO: S11.4 17) Which of the following statements is true of non-homologous end joining (NHEJ)?

A) it is a double-strand repair pathway B) it is error-prone C) it is error-free D) it utilizes the sister chromatid as a template for repair E) it is both error-prone and is a double-strand repair pathway Answer: E Section: 11.6 Skill: Knowledge/Comprehension LO: S11.5a 18) In eukaryotes, homologous recombination is initiated by ________.

A) Rad51 generating double-stranded DNA breaks B) RecA, RecB, and RecC generating single-stranded DNA breaks C) Spo11 generating double-stranded DNA breaks D) p53 generating single-stranded DNA breaks E) ATM signaling double-stranded DNA breaks Answer: C Section: 11.7 Skill: Knowledge/Comprehension LO: S11.5a 19) If one Holliday junction region is resolved by an NS cut and the other by an EW cut, the resulting chromosomes ________.

A) are nonrecombinant and carry the combinations A1 and B2 or A2 and B1 B) are nonrecombinant and carry the combinations A1 and B1 or A2 and B2 C) are A1 and A2 or B1 and B2 D) are recombinant and carry the combinations A1 and B2 or A2 and B1 E) are recombinant and carry the combinations A1 and B1 or A2 and B2 Answer: D Section: 11.7 Skill: Applying/Analyzing LO: S11.5b 6 .

20) Which enzyme is required to mobilize transposons of any type?

A) RNA helicase B) telomerase C) terminal inverted repeats D) reverse transcriptase E) transposase Answer: E Section: 11.7 Skill: Knowledge/Comprehension LO: S11.6a 11.2 Short Answer Questions 1) If you have screened 100,000 pollen grains and identified 5 mutants, what is the mutation frequency?

Answer: 5 × 10-5 Section: 11.1 Skill: Applying/Analyzing LO: S11.1f 2) You use several model organisms in your lab's research. Which of your model organisms would you expect to exhibit a higher mutation frequency: E. coli, D. melanogaster, or M. musculus?

Answer: M. musculus Section: 11.1 Skill: Applying/Analyzing LO: S11.1f 3) What term describes a gene or region of a genome where mutations occur much more often than average?

Answer: hotspots of mutation Section: 11.1 Skill: Knowledge/Comprehension LO: S11.1f 4) Localized mutations that occur at a specific location, rather than over a larger span on a gene are better known as what?

Answer: point mutations Section: 11.2 Skill: Knowledge/Comprehension LO: S11.1c 5) What are the two basic types of base-pair substitutions?

Answer: transition and transversion Section: 11.2 Skill: Knowledge/Comprehension LO: S11.1c 7 .

6) What process, which is responsible for many trinucleotide repeat disorders, alters the number of DNA repeats?

Answer: strand slippage Section: 11.3 Skill: Knowledge/Comprehension LO: S11.8a 7) What is the purpose of using liver enzyme (S9 extract) in the Ames test?

Answer: S9 extract allows you to see if a mutagenic substance can be created by digestive reactions.

Section: 11.3 Skill: Applying/Analyzing LO: S11.3c 8) What type of mutation commonly affects CpG islands, altering promoter regions of mammalian genes?

Answer: deamination Section: 11.3 Skill: Knowledge/Comprehension LO: S11.1b 9) Which type of mutagen has a structure similar to one of the DNA nucleotides and therefore can work its way into DNA, where it pairs with a nucleotide?

Answer: base analog Section: 11.4 Skill: Knowledge/Comprehension LO: S11.3a 10) What type of mutagen would you use if you wanted to study the effect of pyrimidine dimers in causing cancer?

Answer: UV radiation Section: 11.4 Skill: Applying/Analyzing LO: S11.3a 11) You have developed a new drug and have been asked to test the drug's mutagenicity before putting it on the market. Which test could you use to determine your compound's mutagenic potential?

Answer: Ames test Section: 11.4 Skill: Knowledge/Comprehension LO: S11.3c

8 .

12) Which mechanism can be used to repair UV damage in Drosophila but not in humans?

Answer: photoreactive repair Section: 11.5 Skill: Applying/Analyzing LO: S11.4 13) DNA glycosylase inhibitors are used to study which DNA repair mechanism?

Answer: nucleotide base excision repair Section: 11.5 Skill: Knowledge/Comprehension LO: S11.3a 14) How can the Ames test distinguish mutagens that cause small insertions/deletions from those that cause base substitutions?

Answer: The Ames test utilizes two different his- strains, one with a frame shift mutation whose reversion frequency is increased by a mutagen causing small insertions/deletions, and a base substitution mutant whose reversion frequency is affected by base substitution mutagens.

Section: 11.3 Skill: Applying/Analyzing LO: S11.3c 15) The alkylating agent nitrosoguanidine adds methyl groups to nucleotide bases and is repaired by which enzyme that removes the added methyl group and restores nucleotides to their normal form?

Answer: O6-methylguanine methyltransferase Section: 11.5 Skill: Knowledge/Comprehension LO: S11.3a 16) Huntington's disease is caused by what kind of mutation?

Answer: trinucleotide repeat expansion Section: 11.3 Skill: Knowledge/Comprehension LO: S11.1c 17) Which type of double-stranded break repair would you expect to be used by a cell in the G1 phase of the cell cycle?

Answer: non-homologous end joining (NHEJ) Section: 11.6 Skill: Applying/Analyzing LO: S11.4 18) What are three genes that are usually encoded by retrotransposons?

Answer: gag, pol, and env Section: 11.7 Skill: Applying/Analyzing LO: S11.6a 9 .

11.3 Fill-in-the-Blank Questions 1) A second-site mutation that compensates for the mutation in one gene by mutating a second gene and restoring the wild-type phenotype is also known as a ________ mutation.

Answer: suppressor Section: 11.2 Skill: Knowledge/Comprehension LO: S11.1f 2) ________ are structures that have the same composition and general arrangement but a slight difference in bonding and placement of a hydrogen.

Answer: Tautomers Section: 11.3 Skill: Knowledge/Comprehension LO: S11.3a 3) Agents generating mutation-inducing DNA damage are collectively referred to as ________.

Answer: mutagens.

Section: 11.4 Skill: Knowledge/Comprehension LO: S11.3a 4) The Ames test is designed to identify the rate of ________ that restore the ability of bacteria to synthesize their own histidine, thus eliminating the need for histidine supplementation of the growth medium.

Answer: reversion mutations Section: 11.4 Skill: Knowledge/Comprehension LO: S11.3c 5) DNA damage signaling processes are essential for regulating the ________ transition within the cell cycle.

Answer: G1-S Section: 11.5 Skill: Knowledge/Comprehension LO: S11.4 6) The 'activator' transposable element Ac encodes the enzyme ________ which can cause the mobilization of the 'dissociation' element Ds? Answer: transposase Section: 11.7 Skill: Knowledge/Comprehension LO: S11.8a

10 .

7) Retrotransposons are unique among transposable elements in that they require the enzyme ________ for mobilization?

Answer: reverse transcriptase Section: 11.7 Skill: Knowledge/Comprehension LO: S11.8a 11.4 Essay Questions 1) You are interested in studying the function of a gene, so you decide to create point mutations to help you identify important residues in the encoded protein.

In the following set of triplet codons, two nucleotides are underlined, which nucleotide should be mutated to achieve your goal and why?

5′-ATG CCC TGG AAC CTG-3′ Answer: the adenosine should be changed because it would allow you to change the codon to encode a different amino acid. Changing the cytosine in the third position would not allow you to change the encoded Serine to another amino acid.

Section: 11.2 Skill: Synthesis/Evaluation LO: S11.1e 2) A manufacturer of nutritional supplements proposes to sell extracts from an Antarctic algae as an ingestible remedy for sleep disorders. Your task is to determine if this extract may be mutagenic, how do you test this?

Answer: The Ames test Section: 11.3 Skill: Synthesis/Evaluation LO: S11.3c 3) Which bases would you mutate to create a nonsense transition mutation?

5′-ATG GGA TGG CAC CTG-3′ Answer: both G's of the third codon, TGG — change each G to A Section: 11.2 Skill: Synthesis/Evaluation LO: S11.1c

11 .

4) When a DNA mismatch is detected in E.coli, what mechanism allows the cell to determine which was more likely to be the original nucleotide?

Answer: The identification is accomplished by the sensitivity of mismatch repair enzymes to the methylation (the addition of CH3 groups) of specific nucleotides in the original DNA strand. In E. coli, methylation is a common feature of DNA, and prior to replication, DNA parental strands are fully methylated. However, the daughter DNA duplexes produced by replication are only hemimethylated– that is, fully methylated on just the parental strand. A period of time must pass before the newly synthesized strands are fully methylated, and it is in this window of time that the methyl-sensitive components of mismatch repair operate.

Section: 11.4 Skill: Synthesis/Evaluation LO: S11.4 5) You accidentally exposed your cell culture dish to radiation, but you are not sure of the precise wavelength of the exposure. How can you use the induced DNA damage to tell whether the radiation was UV or higher energy? What is the difference between the DNA damage caused

by UV radiation compared to higher-energy radiation? Which type of damage is likely to be the most lethal to your cells?

Answer: Any radiation energy in wavelengths shorter than 380 nm is mutagenic. Ultraviolet radiation is a commonly encountered mutagen that creates pyrimidine dimers in DNA, and these can lead to base-pair substitution mutations. If the dimer cannot be removed, it may also lead to replication stalling at the site of the lesion. Replication blockage by pyrimidine dimers may induce reinitiation of DNA synthesis at an adjacent RNA primer site. This reinitiation of replication potentially leaves gaps spanning dozens to hundreds of nucleotides in newly synthesized DNA strands, but the gaps are subsequently filled by translesion DNA synthesis, which is carried out by specialized bypass DNA polymerases that can replicate across the gaps but lack proofreading capabilities.

Higher-energy radiation can induce DNA strand breaks. DNA can be damaged in multiple ways, the most serious being the induction of DNA single-strand or double-strand breaks. These breaks potentially block DNA replication and thus pose a significant threat to the integrity and survival of affected cells. DNA damage of this type is dealt with by specialized strand-break repair mechanisms. Double-strand DNA breaks are often the most difficult to repair and thus are the most lethal to cells.

Section: 11.4 Skill: Synthesis/Evaluation LO: S11.4

12 .

6) Describe a hypothetical result of the fluctuation test that would have supported the hypothesis that mutations occur in response to environmental challenge, and explain how this result was different from the observed result described in the chapter?

Answer: If a fluctuation test were conducted and detected mutants at roughly equal frequencies from independent cultures, that would support the hypothesis that mutations occur upon environmental challenge. Instead, the test described in the chapter detected mutants at vary different frequencies from independent cultures, suggesting that the mutations occurred at random and prior to the environmental challenge.

Section: 11.1 Skill: Synthesis/Evaluation LO: S11.1b 7) Double-strand breaks cause genomic instability and are thus potentially lethal to cells. They also greatly increase the risk of cancer and the chance of chromosome structural mutations.

Compare and contrast the two mechanisms of double-strand break repair. Why would one mechanism be used over another?

Answer: Two mechanisms have evolved to carry out double-strand break repair.

Non-homologous end joining (NHEJ) is an error-prone repair process that repairs double-strand breaks occurring before DNA replication. If a double-stranded break damages a eukaryotic chromosome during G1 of the cell cycle, replication of the damaged chromosome is blocked.

Completion of NHEJ produces an intact DNA duplex and allows replication across the repaired region in the upcoming replication cycle, but the repair is often imperfect because resection removes nucleotides that cannot be replaced. While NHEJ is error prone, it prevents more extensive loss from degradation of unprotected ends. Mutations can be generated, however, when nucleotides are lost by transcribed genes.

Synthesis-dependent strand annealing (SDSA) is an error-free process that repairs double-strand breaks occurring after the completion of DNA replication. When a double-strand break (DSB) affects one sister chromatid, the other chromatid is intact. The strand invasion process displaces one strand of the duplex and creates a displacement (D) loop. DNA replication within the D loop synthesizes new DNA strands from intact template strands, and the sister chromatids are reformed by dissociation and annealing of the nascent strand to the other side of the break. By accomplishing the removal of DNA in the immediate vicinity of a double-strand break and the replacement of the excised DNA with a duplex identical to that in the sister chromatid, SDSA carries out error-free repair of double-strand breaks.

To summarize, non-homologous end joining is an error-prone system for the repair of doublestrand DNA breaks that occur before S phase. Synthesis-dependent strand annealing takes place after completion of the S phase and uses the identical sister chromatid as a source of the template strand sequence to synthesize a replacement duplex in an error-free manner.

Section: 11.6 Skill: Synthesis/Evaluation LO: S11.5a

13 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 12 Regulation of Gene Expression in Bacteria and Bacteriophages 12.1 Multiple-Choice Questions 1) Which of the following is FALSE regarding regulation of gene expression in bacteria?

A) There can be transcriptional regulation for the initiation and the amount of transcription, as well as post-transcriptional regulatory mechanisms.

B) Regulated transcription is necessary; constitutive transcription, continuously transcribing genes with no regulatory control, does not occur without mutation.

C) Negative control of transcription involves the binding of a repressor protein to a regulatory DNA sequence, with the consequence of preventing transcription.

D) Positive control of transcription involves the binding of an activator protein to a regulatory DNA sequence, with the result of initiating transcription.

E) Repressor proteins commonly contain DNA-binding and allosteric domains.

Answer: B Section: 12.1 Skill: Applying/Analyzing LO: S12.1a, S12.1b 2) In negative control, what molecule would you expect to find bound to the operator if there is no transcription?

A) activator B) repressor C) inducer D) corepressor E) RNA polymerase Answer: B Section: 12.1 Skill: Applying/Analyzing LO: S12.1a 3) You want to design a repressor protein mutant. Which protein domain is the best target for preventing binding of the corepressor?

A) DNA-binding domain B) allosteric domain C) promoter domain D) helix-turn-helix domain E) activator binding site Answer: B Section: 12.1 Skill: Applying/Analyzing LO: S12.1c

1 .

4) The presence of which combination of molecules could prevent transcription of a repressible operon?

A) activator + corepressor B) activator + repressor C) inducer + corepressor D) repressor + corepressor E) activator + inducer Answer: D Section: 12.1 Skill: Applying/Analyzing LO: S12.1b 5) In the lac operon, what acts as the inducer?

A) glucose B) permease C) allolactose D) β-galactosidase E) transacetylase Answer: C Section: 12.2 Skill: Remembering/Understanding LO: S12.2a 6) The enzyme β-galactosidase catalyzes what reaction?

A) allolactose → glucose + lactose B) lactose → glucose + fructose C) lactose → galactose + glucose D) glucose → galactose + lactose E) galactose → glucose + lactose Answer: C Section: 12.2 Skill: Remembering/Understanding LO: S12.2a 7) A bacterium is unable to transport lactose into the cell to be broken down. Which gene is likely mutated in this bacterium?

A) lacZ B) lacY C) lacI D) lacP E) lacO Answer: B Section: 12.2 Skill: Remembering/Understanding LO: S12.3b

2 .

8) Which region of the lac operon would you target if you want to disrupt the -10 and -35 consensus sequences?

A) lacZ B) lacY C) lacI D) lacP E) lacO Answer: D Section: 12.2 Skill: Applying/Analyzing LO: S12.3a 9) In the presence of glucose, where is the lac repressor bound?

A) lacZ B) lacO C) lacP D) lacI E) The lac repressor is not bound to the operon.

Answer: B Section: 12.2 Skill: Applying/Analyzing LO: S12.2b 10) Which of the following can be mutated to result in constitutive expression from the lac operon?

A) lacZ and lacP B) lacZ and lacI C) lacO and lacI D) lacO and lacZ E) lacI and lacY Answer: C Section: 12.3 Skill: Applying/Analyzing LO: S12.3a 11) Which of the following mutants are noninducible?

A) lacIS B) lacI C) lacI+ D) lacOC E) lacO+ Answer: A Section: 12.3 Skill: Applying/Analyzing LO: S12.3a 3 .

12) Which part of the lac operon is cis-dominant?

A) lacA B) lacZ C) lacI D) lacY E) lacO Answer: E Section: 12.3 Skill: Applying/Analyzing LO: S12.3a 13) Which part of the lac operon produces a regulatory protein that is trans-acting?

A) lacP B) lacZ C) lacI D) lacY E) lacO Answer: C Section: 12.3 Skill: Applying/Analyzing LO: S12.3a 14) Which of these haploid strains produces β-galactosidase constitutively but does not produce permease?

A) I — P+ O+ Z+ Y+ B) I+ P+ O+ Z- YC) I- P+ O+ Z- Y+ D) I+ P+ O- Z+ Y+ E) I- P+ O+ Z+ YAnswer: E Section: 12.3 Skill: Applying/Analyzing LO: S12.3b 15) Which of these haploid strains produces permease inducibly, but does not produce βgalactosidase?

A) I — P+ O+ Z+ Y+ B) I+ P+ O+ Z — Y+ C) I — P+ O+ Z — Y — D) I+ P+ O C Z+ Y+ E) I — P+ O+ Z — Y+ Answer: B Section: 12.3 Skill: Applying/Analyzing LO: S12.3b 4 .

16) In a partial diploid of the genotype I— P+ O+ Z— Y+/ I+ P— O+ Z+ Y+, synthesis of A) β-galactosidase and permease will be inducible B) β-galactosidase will be uninducible and permease will be inducible C) β-galactosidase and permease will be constitutive D) β-galactosidase and permease will be uninducible E) β-galactosidase will be inducible and permease will be constitutive Answer: B Section: 12.3 Skill: Applying/Analyzing LO: G2, G3, S12.3c 17) Which of the following best describes expression of the lac genes in this partial diploid?

I+ P+OC Z– Y+/ I– P+ O+ Z+ Y– A) Both β-galactosidase and permease are made only when lactose is present.

B) Both β-galactosidase and permease are made both in the presence of and in the absence of lactose.

C) Both β-galactosidase and permease are not made whether lactose is present or not.

D) β-galactosidase is only made when lactose is present, but permease is made both in the presence and absence of lactose.

E) β-galactosidase is made both in the presence and absence of lactose, but permease is only made when lactose is present.

Answer: D Section: 12.3 Skill: Applying/Analyzing LO: G2, G3, S12.3c

5 .

18) The electrophoresis gel shown below is from a DNase I footprint analysis of an operon transcription control region. Separate samples of control-region DNA were exposed to DNase I and run in the following lanes: uprotected DNA is in lane 1, DNA protected by repressor is in lane 2, and DNA protected by RNA polymerase is in lane 3.

Based on the gel, what nucleotide positions are occupied by the repressor and RNA polymerase?

A) 1 through 18 by the repressor and 1 through 6 by RNA polymerase B) 19 through 29 by the repressor and 7 through 24 RNA polymerase C) 19 through 29 by the repressor and 1 through 6 by RNA polymerase D) 30 through 35 by the repressor and 7 through 24 by RNA polymerase E) 1 through 18 by the repressor and 25 through 35 by RNA polymerase Answer: B Section: 12.3 Skill: Applying/Analyzing LO: G3, G7, S12.3d

6 .

19) Northern blot analysis is performed on mRNA isolated from E. coli after growth in a glucose-only medium (lane 1) or a lactose-only medium (lane 2). Following hybridization with a probe for a portion of the lacZ sequence, which of the following northern blots (A-E) would be expected for bacteria with the genotype I+ P+ O+ Z+ Y+?

A) A B) B C) C D) D E) E Answer: A Section: 12.3 Skill: Applying/Analyzing LO: G3, G7, S12.3e

7 .

20) Northern blot analysis is performed on mRNA isolated from E. coli after growth in a glucose-only medium (lane 1) or a lactose-only medium (lane 2). Following hybridization with a probe for a portion of the lacZ sequence, which of the following northern blots (A-E) would be expected for bacteria with the genotype I+ P+ O+ Z+ Y+and a mutation that prevents binding of the CAP-cAMP complex to the CAP binding site?

A) A B) B C) C D) D E) E Answer: D Section: 12.3 Skill: Applying/Analyzing LO: G2, G3, G7, S12.3e

8 .

21) Northern blot analysis is performed on mRNA isolated from E. coli after growth in a glucose-only medium (lane 1) or a lactose-only medium (lane 2). Following hybridization with a probe for a portion of the lacZ sequence, which of the following northern blots (A-E) would be expected for bacteria with the genotype IS P+ OC Z+ Y+?

A) A B) B C) C D) D E) E Answer: C Section: 12.3 Skill: Applying/Analyzing LO: G2, G3, G7, S12.3e 22) During the attenuation of the trp operon, which stem loop leads to polycistronic mRNA synthesis during tryptophan starvation?

A) 1-3 (antitermination) stem loop B) 3-4 (termination) stem loop C) 1-2 (pause) stem loop D) 2-3 (antitermination) stem loop E) 2-4 (termination) stem loop Answer: D Section: 12.4 Skill: Remembering/Understanding LO: S12.4

9 .

23) Disrupting the formation of which stem-loop structure of trpL would decrease the efficiency of transcriptional regulation?

A) 1-3 (antitermination) stem loop B) 3-4 (termination) stem loop C) 1-2 (pause) stem loop D) 2-3 (antitermination) stem loop E) 2-4 (termination) stem loop Answer: B Section: 12.4 Skill: Remembering/Understanding LO: S12.4 24) In the absence of tryptophan, ________.

A) the inactive repressor cannot bind trpO, so operon gene transcription occurs B) the active repressor binds trpP, so operon gene transcription is repressed C) the inducer cannot bind trpO, so operon gene transcription occurs D) the active repressor cannot bind trpO, so operon gene transcription is attenuated E) the repressor binds the corepressor, and operon gene transcription occurs Answer: A Section: 12.4 Skill: Applying/Analyzing LO: S12.4 25) Which of the following statements regarding attenuation of the trp operon is NOT true?

A) It is controlled by alternative folding of the mRNA synthesized from the trpL region.

B) Translation of a trpL coding sequence into a 14-amino acid polypeptide (including the methionine of the start codon and two back-to-back tryptophan codons) is crucial to its function.

C) If the two tryptophan codons were changed to arginine codons the operon would no longer be sensitive to tryptophan levels but would instead be sensitive to arginine levels.

D) When tryptophan levels are low, the ribosome will occupy regions 1 and 2 after completing translation of the coding sequence, regions 3 and 4 will pair, and transcription will terminate.

E) When tryptophan levels are high, the formation of a 3-4 stem loop followed by the poly-U string will prevent expression of the structural genes of the operon.

Answer: D Section: 12.4 Skill: Applying/Analyzing LO: G2, S12.4

10 .

26) Which of the following genotypes and conditions will have the HIGHEST level of trp operon expression?

A) trp+ / tryptophan present B) trpR- / tryptophan present C) trpL- / tryptophan present D) trp+ / tryptophan absent E) trpP- / tryptophan absent Answer: D Section: 12.4 Skill: Applying/Analyzing LO: S12.4 27) Which of the following genotypes and conditions will have the LOWEST level of trp operon expression?

A) trp+ / tryptophan present B) trpR- / tryptophan present C) trpL- / tryptophan present D) trp+ / tryptophan absent E) trpP- / tryptophan absent Answer: E Section: 12.4 Skill: Applying/Analyzing LO: S12.4 28) Which of the following mutations of the trpL region would be least likely to affect attenuation of trp operon transcription?

A) a deletion of region 4 B) a deletion of the uracil nucleotides immediately following region 4 C) a deletion of the start (AUG) codon of the trpL polypeptide D) an insertion of two nucleotides immediately after the polypeptide stop codon E) an insertion of two nucleotides immediately following the polypeptide start codon Answer: D Section: 12.4 Skill: Evaluating/Creating LO: G2, G3, G7, S12.4

11 .

29) Which of the following is FALSE regarding how bacteria regulate the transcription of stress response genes?

A) Under extreme growth conditions, such as heat stress and starvation, bacteria can induce transcription of alternative sigma factors.

B) The rpoH gene encodes a sigma factor known as σ32 that is active at high temperatures.

C) High heat can cause a change in the chaperone proteins.

D) Transcription is activated by antitermination, which allows transcription to extend into heat shock genes.

E) Holoenzymes containing an alternative sigma subunit are able to recognize the promoters of genes not transcribed by the common bacterial RNA polymerase.

Answer: D Section: 12.5 Skill: Applying/Analyzing LO: S12.5a 30) Which of the following is FALSE regarding translational regulation in bacteria?

A) It is the most predominant mode of controlling gene expression in bacteria.

B) Translation repressor proteins regulate translation by binding mRNA near the Shine-Dalgarno sequence.

C) Antisense RNA can bind to a portion of a specific mRNA to which it is complementary.

D) The bacterial insertion sequence (IS10) uses antisense RNA to regulate translation of the mRNA that produces the enzyme transposase.

E) The binding of an mRNA by an antisense RNA can prevent ribosome attachment to the mRNA and block translation.

Answer: A Section: 12.5 Skill: Applying/Analyzing LO: S12.5a 31) Which of the following regarding riboswitches is correct?

A) They can use repressors to bind to DNA sequences called operators.

B) The binding of small regulatory molecules can generate termination stem loops that lead to increased transcription.

C) The binding of small regulatory molecules can generate Shine-Dalgarno sequester stem loops that lead to increased translation.

D) Low concentrations of a small regulatory molecule can cause cleavage of the mRNA. E) They can regulate bacterial transcription, translation, and mRNA stability.

Answer: E Section: 12.6 Skill: Remembering/Understanding LO: G2, G7

12 .

32) Which region of the λ phage genome enables the linear chromosome to circularize when it enters a host cell?

A) cohesive (cos) site B) IS10 C) early operators D) integrase (int) gene E) excisionase (xis) gene Answer: A Section: 12.7 Skill: Remembering/Understanding LO: S12.6 33) In λ phage, a mutation that inactivates which of the following genes should leave the mutated phage strain capable of establishing lysogeny?

A) cI B) cII C) cro D) int E) N Answer: C Section: 12.7 Skill: Applying/Analyzing LO: S12.6 34) In λ phage, which protein, the product of the cI gene, blocks the transcription required to initiate the lytic cycle?

A) cro B) repressor C) enhancer D) integrase E) operator Answer: B Section: 12.7 Skill: Applying/Analyzing LO: S12.6 35) What would be the phenotype of an int- λ mutant?

A) The phage would not be able to undergo the lytic cycle, but could lysogenize.

B) Both lytic and lysogenic pathways would be inactivated.

C) The phage would be able to enter the lytic cycle, but would not be able to lysogenize.

D) The phage would be able to lysogenize but would not be able to be induced to resume the lytic cycle.

E) The cell would build up high levels of λ repressor, so lysogeny would be maintained.

Answer: C Section: 12.7 Skill: Applying/Analyzing LO: S12.6 13 .

36) Which of the following would NOT occur in a phage undergoing induction?

A) UV damage to DNA can trigger the switch from lysogeny to the lytic cycle.

B) There will be activation of the protease activity of RecA. C) The transcription level of the int gene will increase, leading to production of the integrase.

D) Cleavage of the λ repressor monomers inactivates the repressor protein.

E) The transcription level of the xis gene will increase, leading to production of the enzyme excisionase.

Answer: C Section: 12.7 Skill: Applying/Analyzing LO: S12.6 12.2 Short Answer Questions 1) What are the two active sites or "domains" on a repressor protein?

Answer: DNA-binding and allosteric Section: 12.1 Skill: Remembering/Understanding 2) If a corepressor is inhibited, what effect would you expect to see on transcription in a repressible operon?

Answer: Transcription will occur.

Section: 12.1 Skill: Applying/Analyzing LO: S12.1b 3) Which proteins facilitate RNA polymerase binding at promoters to increase levels of transcription?

Answer: activator proteins Section: 12.1 Skill: Remembering/Understanding LO: S12.1a 4) Catabolite repression refers to the repression of the lac operon in the presence of which catabolite?

Answer: glucose Section: 12.2 Skill: Remembering/Understanding LO: S12.2b 5) Binding of which complex increases the ability of RNA polymerase to transcribe the lac operon?

Answer: CAP-cAMP Section: 12.2 Skill: Remembering/Understanding LO: S12.2b 14 .

6) Which enzyme is the product of the lacZ gene?

Answer: β-galactosidase Section: 12.3 Skill: Remembering/Understanding LO: S12.3a 7) When both glucose and lactose are available in the growth medium, the presence of which molecule is capable of inducing basal transcription of the lac operon?

Answer: cAMP Section: 12.2 Skill: Applying/Analyzing LO: S12.2a, S12.2b 8) In the trp operon, what gene contains the attenuator region?

Answer: trpL Section: 12.4 Skill: Knowledge/Comprehension LO: S12.4 9) Which molecule functions as the corepressor in the trp operon?

Answer: tryptophan Section: 12.4 Skill: Remembering/Understanding LO: S12.4 10) What are the names of the three possible alternative stem loops that can form in mRNA? Answer: pause, antitermination, and termination Section: 12.4 Skill: Remembering/Understanding LO: S12.4 11) Bacteria grown at 45°C initiate expression of which two types of proteins?

Answer: heat shock and chaperone Section: 12.5 Skill: Remembering/Understanding LO: S12.5a 12) Antisense control of translation involves the regulation of which enzyme produced by bacterial insertion sequence IS10?

Answer: transposase Section: 12.5 Skill: Remembering/Understanding LO: S12.5a

15 .

13) The genetic switch controlling whether a bacterium enters the lytic or lysogenic cycle relies on the binding of which two proteins?

Answer: cro and λ repressor Section: 12.7 Skill: Remembering/Understanding LO: S12.6 14) What is the process by which a bacterium switches from a lysogenic to lytic cycle?

Answer: induction Section: 12.7 Skill: Remembering/Understanding LO: S12.6 15) What protein cleaves the λ repressor monomers to inactivate the repressor protein, and is activated by DNA-damaging agents?

Answer: RecA Section: 12.7 Skill: Remembering/Understanding LO: S12.6 12.3 Fill-in-the-Blank Questions 1) Most of the regulation of gene expression in bacteria occurs at the ________ level.

Answer: transcriptional Section: 14.1 Skill: Knowledge/Comprehension 2) Changing conformation at the active site as a result of binding a substance at a different site is known as ________.

Answer: allostery Section: 12.1 Skill: Remembering/Understanding LO: S12.1a 3) Expression of a(n) ________ alters gene transcription in E. coli by activating transcription of specialized heat stress response genes.

Answer: alternative sigma factor Section: 12.5 Skill: Remembering/Understanding LO: S12.5a 4) In bacteriophage λ, if Cro succeeds in binding OR3 within PRM, the ________ cycle is established.

Answer: lytic Section: 12.7 Skill: Remembering/Understanding LO: S12.6 16 .

5) In bacteriophage λ, if λ repressor successfully binds OR1 and OR2 within PR, the ________ cycle is established.

Answer: lysogenic Section: 12.7 Skill: Remembering/Understanding LO: S12.6 12.4 Essay Questions 1) DNA-binding proteins exert control by binding DNA at specific sequences. Describe how DNA and protein interact at the molecular level. Are proteins able to identify specific DNA sequences, or do they bind to all regions with equal frequency? What motifs are commonly found in DNA-binding proteins?

Answer: DNA-protein interactions occur when the amino acid side chains of the proteins interact with the specific nucleotide bases and the sugar-phosphate backbone of DNA. The proteins make their contact with specific base pairs located in the major groove and the minor groove of the DNA helix using a unique pattern of hydrogen, nitrogen, and oxygen atoms that characterize each base pair. To achieve protein-DNA specificity in these interactions, the protein must simultaneously contact multiple nucleotides.

Protein secondary structures are a common motif in the structures of DNA-binding regulatory proteins, particularly α-helices.

Bacterial regulatory DNA sequences frequently contain inverted repeats or direct repeats, and proteins often interact with one of the inverted repeat segments. By far, the most common structural motif in these proteins in bacteria is the helix-turn-helix (HTH) motif. In the HTH motif, two α-helical regions in each of two polypeptides in a homodimer interact with inverted repeat regulatory sequences in DNA. In each of the polypeptides, one α-helical region forms the recognition helix that fits into the major groove of DNA and binds the inverted repeat sequences.

This is connected to a short string of amino acids forming the "turn," which is connected to a second α-helical region known as the stabilizing helix. The stabilizing helix lies across the major groove and contacts the sugar-phosphate backbone, ensuring a strong DNA-protein interaction and properly orienting the recognition helix to sit in the major groove.

Section: 12.1 Skill: Evaluating/Creating LO: G2, G8

17 .

2) What is the purpose of the lac operon, and under what circumstances is it activated? Describe the structure of the lac operon, including the repressor protein, structural genes, and regulatory regions of the operon. What is the role of catabolite repression in the lac operon?

Answer: The lac operon is responsible for the production of three polypeptides that permit E. coli to use the sugar lactose as a carbon source for growth and metabolic energy.

Glucose is the preferred energy source of E. coli, but lactose can serve as an alternative carbon source for bacteria. The lac operon acts as a genetic switch to begin breaking down lactose, but only if (1) lactose is present in the cell; and (2) all glucose has been depleted.

The lac operon consists of a multipart regulatory region and a structural gene region containing three genes. The regulatory region contains three protein-binding regulatory sequences: the promoter that binds RNA polymerase, the operator (lacO) that binds the lac repressor protein, and the CAP-cAMP region.

The three structural genes of the lac operon are lacZ, a gene encoding the enzyme βgalactosidase; lacY, which encodes the enzyme permease; and lacA, which encodes transacetylase. The β-galactosidase enzyme cleaves the β-galactoside linkage of lactose to yield glucose and galactose. The enzyme also converts lactose into allolactose, which acts as the inducer for the operon. The permease enzyme product of the lacY gene controls entry of lactose into the cell. Transacetylase, the product of lacA, is not essential for lactose utilization, although in bacteria it protects against potentially damaging by-products of lactose metabolism. The regulatory gene, lacI, produces the lac repressor and has its own promoter, which is not regulated and drives constitutive transcription. The lac operon promoter, lacP, contains the consensus sequence sites that are critical for RNA polymerase binding. LacO, which binds the repressor protein produced by lacI, overlaps lacP near the start of transcription.

The removal of the lac repressor by allolactose alone will not result in maximal induction of the lac operon. The binding of the CAP-cAMP complex to the CAP binding region of the promoter is also required. cAMP is not normally found in high concentrations in the presence of abundant glucose, because glucose inhibits adenylate cyclase, the enzyme that converts ATP to cAMP. Thus, the preferred catabolite, glucose, helps to prevent expression of the lac operon genes, a process called "catabolite repression." In the absence of glucose, however, cAMP levels rise, allowing this molecule to bind to the catabolite activator protein (CAP). This complex binds to the lac promoter at the CAP binding site, bending the DNA to allow increased transcription by RNA polymerase.

Section: 12.2 Skill: Evaluating/Creating LO: G8, S12.2a, S12.2b

18 .

3) In E. coli, the cya gene encodes adenylate cyclase. Will cells that are cya− be able to grow on medium that contains lactose as the sole carbon source? Why or why not?

Answer: No. Metabolically, cells that are cya− would be similar to wild-type cells that are grown in the presence of both glucose and lactose. Although negative regulation has been relieved (the small amounts of allolactose produced binds the lac repressor protein, removing it from the lac operator), there is no positive regulation by CAP-cAMP because cAMP is not synthesized. Therefore, only basal transcription occurs. This is not sufficient to drive active lactose metabolism.

Section: 12.2 Skill: Evaluating/Creating LO: G8, S12.2a, S12.2b 4) What is the role of cis-acting and trans-acting factors in transcription? Explain how these factors play a role in the lac operon, and discuss how complementation can be used to identify where the mutations have occurred (structural genes versus regulatory regions). Which type of mutations can be complemented by a second gene with normal function? Are IS and OC mutants cis- or trans-acting?

Answer: If a factor is trans-acting, genes that are not physically adjacent can be regulated together and are mediated by genes that encode regulatory proteins. The lac repressor is an example of a trans-acting element because mutations can be complemented by a second gene with normal function.

If a factor is cis-acting, the DNA sequence must be adjacent to the genes it regulates. The lac operator is an example of a cis-acting element because mutations cannot be complemented by a second normal element.

Trans-acting factors (and complementation) indicate mutations in structural genes, whereas cis-acting factors indicate regulatory sequences have been mutated.

IS is a trans-acting mutation because the mutation disrupts ligand binding of the repressor to the operator on both copies of the DNA. OC is a cis-acting mutation because only one molecule of DNA is unable to be repressed due to the mutation. The operon is constitutively active on the molecule of DNA that contains the OC allele, but the other copy of the gene is intact and functions properly when the repressor is present.

Section: 12.3 Skill: Evaluating/Creating LO: G2, S12.3a, S12.3b

19 .

5) For each of the following E. coli partial diploid strains, indicate whether expression of βgalactosidase and permease is inducible (I), constitutive (C), or noninducible (N). Assume glucose levels are very low. Which of the strains are able to use lactose as their sole carbon source?

Answer:

Only strain c is able to use lactose as its sole carbon source. Strains a and d never make permease and strain b never makes β-galactosidase.

Section: 12.3 Skill: Applying/Analyzing LO: G2, G3, S12.3c

20 .

6) Using the autoradiograph of a DNA sequencing gel below determine which of the following is the sequence of the 35-bp operon transcription control region.

Answer: The DNA sequence is determined by reading the DNA sequencing gel from the bottom to the top: 5'- ACTAGCTTTATGTCAATGTGAATGTTAGTTCTAAG -3' 3'- TGATCGAAATACAGTTACACTTACAATCAAGATTC -5' Section: 12.3 Skill: Evaluating/Creating LO: G3, G7, S12.3d

21 .

7) Northern blot analysis is performed on mRNA isolated from E. coli after growth in a glucoseonly medium (lane 1) or a lactose-only medium (lane 2). The blots are hybridized with a probe for a portion of the lacZ sequence. Indicate the expected results for the following bacterial genotypes using dark, light, or no shading to indicate full, low, or no expression, respectively. a. I+ P+ O+ Z+ Y+ b. I+ P– O+ Z+ Y+ c. I– P+ O+ Z+ Y+ d. I+ P+ OC Z+ Y+ e. I+ P+ O+ Z+ Y+ and a mutation of the CAP binding site preventing CAP-cAMP binding

Answer:

Skill: Evaluating/Creating LO: G2, G3, G7, S12.3e

22 .

8) Describe the differences between inducible and repressible operons. How does attenuation differ from inducible or repressible operons?

Would the lac operon be active if a cell has the following food sources? a. only lactose b. only glucose c. lactose and glucose Answer: Operons like lac that are involved in catabolism of alternative energy sources are typically inducible, since they are called upon only when glucose is depleted and the alternative sugar is available. The PRESENCE of the effector molecule (inducer) is needed to ACTIVATE transcription. In contrast, repressible operons are involved in anabolic pathways (pathways that synthesize compounds needed by the cell) and can be regulated by negative feedback mechanisms that operate through activity of the end product of the pathway to block operon gene transcription. For these, the PRESENCE of the effector molecule (corepressor) is needed to INACTIVATE transcription. Some repressible operons can also fine-tune the transcription to help achieve the steady-state amount of a given compound. This lets the cell make more product when needed, but stop when the cell has reached its necessary amount. So, whereas inducible and repressible are simple on-off systems, attenuation allows a wider range of transcriptional activation levels. Furthermore, the molecular mechanisms differ between inducible and repressible operons as compared to attenuation. Inducible and repressible operons are regulated through the interactions between proteins and DNA whereas attenuation involves RNA-RNA interactions in addition to the interaction between mRNA and protein (specifically RNA polymerase) a. yes b. no c. yes (basal transcription only) Section: 12.4 Skill: Evaluating/Creating LO: G8, S12.1b, S12.2a, S12.2b, S12.4 9) Suppose a λ phage infected a protease-deficient bacterial cell. Would the lytic or lysogenic life cycle be favored? Why? Answer: The lytic cycle. Protease action is responsible for degradation of the cII protein under conditions of active bacterial growth. The absence of cII prevents the formation of the cII/cIII complex and the subsequent transcription of the cI gene, which encoded the λ repressor protein.

Without the λ repressor, transcription of the lytic genes is not suppressed and the lytic cycle ensues.

Section: 12.7 Skill: Evaluating/Creating LO: G2, G8, S12.6

23 .

10) What would be the consequences to a λ phage that lysogenized in a recA− bacterial cell?

Explain why this would occur.

Answer: Induction (the termination of lysogeny and the reinitiation of the lytic cycle) would not occur, so the prophage would be unable to excise from the bacterial chromosome and re-enter the lytic cycle. Normally the RecA protein cleaves the λ repressor protein, disrupting the dimer and resulting in the loss of binding of the repressor to OR1, OR2, and OR3. Positive regulation of cI and negative regulation of cro end, resulting in a transition from lysogeny to the lytic cycle. In the absence of RecA, the λ repressor is not cleaved, positive regulation of cI and negative regulation of cro continue, so lysogeny is maintained.

Section: 12.7 Skill: Evaluating/Creating LO: S12.6

24 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 13 Regulation of Gene Expression in Eukaryotes 13.1 Multiple-Choice Questions 1) Gene regulation in eukaryotes often involves which of the following, which are not also used by prokaryotes?

A) formyl-methionine B) histone modification C) promoters D) transcription factors E) coupled transcription and translation Answer: B Section: 13.1 and 13.2 Skill: Knowledge/Comprehension LO: S13.1a 2) You have identified a mutation in a gene that seems to decrease transcription of another gene 2000 bp away from the mutation site. What regulatory sequence, which may be found within another gene, has likely been mutated in this instance?

A) core promoter B) proximal elements C) enhancer sequence D) homeodomain motif E) upstream activator sequence Answer: C Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1a 3) Which of the following is not involved in control of gene expression in eukaryotes?

A) transcriptional repression B) destruction of mRNA C) use of alternative promoters D) change to DNA sequence E) export of RNA from the nucleus Answer: D Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1a

1 .

4) Enhancers and silencers are?

A) trans-acting sequences that interact with RNA polymerase B) are proteins that control gene expression C) must be transported in to the nucleus after translation D) are cis-acting proteins E) are DNA elements where transcription factors bind Answer: E Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1b 5) Trans-acting regulators of gene expression include?

A) introns B) silencers C) histone deacetylases D) the TATA box E) promoters Answer: C Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1a 6) Which protein binds to the silencer sequence and promotes transcriptional silencing in the presence of glucose?

A) Gal2 B) Mig1 C) Gal4 D) Gal10 E) Gal80 Answer: B Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1e 7) Multicellular organisms generally utilize more complex gene regulation than unicellular organisms via mechanisms that include?

A) transcriptional repression B) polycistronic RNAs C) epigenetic modification D) polycistronic RNAs and transcriptional repression E) transcriptional repression and epigenetic modification Answer: E Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1b

2 .

8) For the following gene, MspI and HpaII restriction sites and their positions are indicated here.

Recall that HpaII is methylation sensitive, while MspI is not.

If this gene is being actively transcribed, how many fragments would you expect from this restriction digest?

A) 1 B) 3 C) 4 D) 5 E) 6 Answer: E Section: 13.2 Skill: Applying/Analyzing LO: S13.2d 9) Which of the following would you expect to show the least evolutionary conservation?

A) core promoters B) histone genes C) coding sequence D) enhancer sequence E) intergenic DNA Answer: E Section: 13.2 Skill: Applying/Analyzing LO: S13.2a 10) Chromatin from two tissues, skin and liver, was extracted and analyzed near the gene Twist.

The liver chromatin was found to be more DNAse sensitive than the chromatin from skin cells.

What might explain this difference?

A) Twist is expressed in skin and not liver.

B) Chromatin near Twist in liver cells is heterochromaitic.

C) Twist is silenced in skin and not liver.

D) Twist is not expressed in either liver or skin cells E) Twist is expressed at the same level in both liver and skin cells.

Answer: C Section: 13.2 Skill: Applying/Analyzing LO: S13.2a

3 .

11) Molecular biologists can determine experimentally whether a region of DNA contains closed chromatin or open chromatin by assessing the sensitivity of the region to ________.

A) nucleosomes B) histone deacetylase C) RNA polymerase II D) methyltransferase E) DNAse Answer: E Section: 13.2 Skill: Knowledge/Comprehension LO: S13.2b 12) Chromatin remodeling involves both sliding and relocating of the nucleosomes. Which eukaryotic chromatin remodeling complex is likely involved?

A) ISWI complex B) SWR1 complex C) SWI/SNF complex D) SHH complex E) Mig1 complex Answer: C Section: 13.2 Skill: Knowledge/Comprehension LO: S13.2d 13) If you want to affect chromatin packaging, which amino acid could you mutate to affect both histone acetylation and methylation patterns?

A) arginine B) methionine C) histidine D) lysine E) asparagine Answer: D Section: 13.2 Skill: Applying/Analyzing LO: S13.2d 14) A region of chromatin has recently become DNAse I hypersensitive. Which enzyme has been activated to cause this change in chromatin structure?

A) histone methyltransferase B) histone acetylase C) histone deacetylase D) phosphatase E) kinase Answer: B Section: 13.2 Skill: Applying/Analyzing LO: S13.2c 4 .

15) Prader-Willi syndrome is a genetic disorder involving a partial deletion of chromosome 15q on the paternal chromosome. When both copies of a gene (or chromosome) are functional but only one is expressed, this is an example of ________.

A) position effect variegation B) X-inactivation C) genomic imprinting D) histone acetylation E) chromatin modifications Answer: C Section: 13.2 Skill: Knowledge/Comprehension LO: S13.2g 16) When CpG islands are unmethylated, ________.

A) chromatin in the promoter region is open, allowing access by transcription factors and RNA polymerase B) chromatin in the promoter region is closed, preventing transcription factors and RNA polymerase from binding C) chromatin in the enhancer region is closed, so they are unable to bind regulatory proteins to initiate transcription D) DNAse hypersensitivity in that region of the chromosome is lost E) genes downstream of the CpG islands cannot be expressed, because the promoter region is blocked by histones Answer: A Section: 13.2 Skill: Applying/Analyzing LO: S13.2c 17) Typically, methylation of nucleosome N-terminal tails leads to ________.

A) removal of the protein components of the chromatin from the DNA B) relaxed packaging of the chromatin and increased transcription C) tighter packaging of the chromatin and reduced transcription D) increased amounts of euchromatin relative to heterochromatin E) activation of topoisomerase Answer: C Section: 13.2 Skill: Knowledge/Comprehension LO: S13.2d

5 .

18) Typically, acetylation of histone tails leads to ________.

A) removal of the protein components of the chromatin from the DNA B) relaxed packaging of the chromatin and increased transcription C) tighter packaging of the chromatin and reduced transcription D) increased amounts of euchromatin relative to heterochromatin E) activation of topoisomerase Answer: B Section: 13.2 Skill: Applying/Analyzing LO: S13.2d 19) dsRNA is detected and cleaved by?

A) dicer B) insulator C) miRNA D) RNA polymerase II E) RISC Answer: A Section: 13.3 Skill: Knowledge/Comprehension LO: S13.3a 20) Loss of Xist gene due to mutation leads to defects in which epigenetic process?

A) RNAi B) X-inactivation C) genomic imprinting D) position effect variegation E) methylation of CpG islands Answer: B Section: 13.2 Skill: Knowledge/Comprehension LO: S13.2f 21) During RNAi, what do miRNAs target for destruction?

A) heterochromatic regions of DNA B) CpG islands C) mRNAs D) histones E) ribosomes Answer: C Section: 13.3 Skill: Synthesis/Evaluation LO: S13.3a

6 .

22) Many types of cancer are known to overexpress the receptor protein tyrosine kinase. Which molecular technique can be used to reduce expression of tyrosine kinase in cell culture?

A) PCR B) DNAse sensitivity assay C) RNAi D) southern blotting E) western blotting Answer: C Section: 13.3 Skill: Knowledge/Comprehension LO: S13.3a 13.2 Short Answer Questions 1) Which molecules bind regulatory sequences of DNA to encourage positive regulation of transcription?

Answer: activator proteins Section: 13.1 Skill: Applying/Analyzing LO: S13.1b 2) Which protein complex directs DNA bending into loops that contact RNA polymerase and transcription factors bound at the core promoter or with protein complexes bound to proximal promoter elements?

Answer: enhanceosome Section: 13.1 Skill: Applying/Analyzing LO: S13.1a 3) What are three proteins or groups of proteins you would expect to find bound to the core promoter region in eukaryotes?

Answer: TBP, GTFs, and Pol II Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1a 4) Enzymes that create epigenetic marks are sometimes referred to as writers, whereas enzymes that eliminate epigenetic marks are called ________.

Answer: erasers Section: 13.1 Skill: Knowledge/Comprehension LO: S13.2d

7 .

5) In the GAL gene system, which protein acts as an activator protein through its transcriptioninitiating effect?

Answer: Gal4 Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1e 6) In the GAL gene system, which protein binds to the activation domain of the activator protein, ultimately blocking transcription in the absence of galactose?

Answer: Gal80 Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1e 7) Gal4 is mutated such that it still binds to the DNA but cannot interact with Gal80. What effect would you expect to see in the absence of galactose?

Answer: transcriptional activation Section: 13.1 Skill: Applying/Analyzing LO: S13.1e 8) Gal80 is mutated such that it cannot interact with Gal3. What effect would you expect to see in the presence of galactose?

Answer: Transcription is blocked.

Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1e 9) Eukaryotes use these highly specialized enhancer elements, which regulate the transcription of multiple genes packaged in complexes of closely related genes (e.g., β-globin).

Answer: locus control regions (LCRs) Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1b 10) What are the protein-binding sequences that direct enhancers to interact with the intended promoter and that block communication between enhancers and other promoters?

Answer: insulator sequences Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1a

8 .

11) For the following gene, which mutant likely has lost its enhancer?

Mutant # Deletion Region WT None 1 1-200 2 250-400 3 500-800 4 950-1100 Answer: mutant 3 (region 500-800) Section: 13.1 Skill: Applying/Analyzing LO: S13.2a

% Transcription 100% 130% 100% 50% 0%

12) For the following gene, which type of regulatory sequence has likely been deleted in mutant 1?

Mutant # Deletion Region WT None 1 1-200 2 250-400 3 500-800 4 950-1100 Answer: silencer Section: 13.1 Skill: Applying/Analyzing LO: S13.2a

% Transcription 100% 130% 100% 50% 0%

13) For the following gene, you notice the following results.

Mutant # WT 1 2 3 4

Deletion Region None 1-200 250-400 500-800 950-1100

% Transcription (lungs) 100% 130% 100% 50% 0%

% Transcription (kidneys) 100% 130% 100% 50% 100%

What type of sequence has been mutated in mutant 4?

Answer: tissue-specific (lung) promoter Section: 13.1 Skill: Applying/Analyzing LO: S13.2a 14) What are the three mechanisms by which chromatin remodelers can move nucleosomes?

Answer: nucleosome sliding, nucleosome repositioning, and removal of nucleosomes Section: 13.2 Skill: Knowledge/Comprehension LO: S13.2c 9 .

15) A transcription factor that binds to a gene first and facilitates the binding of other transcription factors is called a ________.

Answer: pioneer factor Section: 13.1 Skill: Knowledge/Comprehension LO: S13.1c 16) RNA polymerase can bind to ________ to generate different mRNAs from the same gene.

Answer: alternative promoters Section: 13.2 Skill: Knowledge/Comprehension LO: S13.1a 17) Inherited traits that can change without affecting the sequence of DNA are examples of what?

Answer: epigenetics Section: 13.3 Skill: Knowledge/Comprehension LO: S13.2a 18) If you block histone deacetylase, what effect would you expect to see on transcriptional activity?

Answer: continued/prolonged transcription Section: 13.2 Skill: Applying/Analyzing LO: S13.2b 19) What effect does methylation of CpG islands have on human promoters?

Answer: reduced transcription Section: 13.3 Skill: Knowledge/Comprehension LO: S13.2d 20) What gene is critical for establishment of X-inactivation in mammals?

Answer: Xist Section: 13.2 Skill: Knowledge/Comprehension LO: S13.2f 21) Which protein complex binds dsRNA fragments to generate ssRNAs for RNAi? Answer: RNA-induced silencing complex (RISC) Section: 13.3 Skill: Knowledge/Comprehension LO: S13.3a

10 .

22) Which enzyme cuts dsRNA into 21-25 bp fragments during RNAi? Answer: Dicer Section: 13.3 Skill: Knowledge/Comprehension LO: S13.3a 13.3 Fill-in-the-Blank Questions 1) Viruses with dsRNA genomes evolved a long time ago and can be found in parasitic relationships with species from a variety of taxa. What mechanism do eukaryotes use to combat such viruses?

Answer: RNA-induced silencing, RNAi Section: 13.3 Skill: Knowledge/Comprehension LO: S13.3b 2) In the GAL gene system, ________ are cis-acting regulatory elements, and ________ protein is a trans-acting regulatory protein.

Answer: UAS; Gal4 Section: 13.1 Skill: Knowledge/Comprehension LO: S13.2a 3) Demethylation and acetylation lead to open chromatin structure and are associated with ________ regions of genomes.

Answer: euchromatic Section: 13.2 Skill: Knowledge/Comprehension LO: S13.2c 4) ________ in Drosophila results from the movement of the transcriptionally active w+ allele into the centromeric heterochromatin region of the fruit-fly X chromosome.

Answer: Position effect variegation (PEV) Section: 13.2 Skill: Knowledge/Comprehension LO: S13.2g 5) ________ binds to siRNAs and then either blocks translation or destroys mRNA that is complementary to the siRNA. Answer: RISC Section: 13.3 Skill: Knowledge/Comprehension LO: S13.3

11 .

13.4 Essay Questions 1) How does transcription inhibition differ between prokaryotes and eukaryotes? Describe the two mechanisms used by eukaryotes to direct the enhancers toward certain promoters and away from others.

Answer: Prokaryotes typically use a repression mechanism in which repressor proteins inhibit transcription by binding to operator sequences that overlap promoters. This blocks the binding of RNA polymerase, thus inhibiting the gene or operon.

In eukaryotes, this mechanism of transcription inhibition is not seen. Instead, eukaryotic repressors inhibit transcription through other mechanisms.

1. Eukaryotic repressors can bind to silencer sequences, cis-acting regulatory sequences that block transcription by directly preventing enhancer-mediated transcription. For example, yeast use the proteins Mig1 and Tup1 to bind to the silencer site when glucose is present, thus inhibiting transcription of the galactose utilization pathway.

2. Eukaryotic insulator sequences can block the action of the enhancer. Insulators are cis-acting sequences located between enhancers and promoters of genes that are to be insulated from the effects of the enhancer. Insulators direct enhancers to interact with the intended promoter and block communication between enhancers and other promoters. This mechanism likely involves allowing the formation of DNA loops containing enhancers and their intended promoter targets while preventing the formation of DNA loops containing an enhancer and a promoter that is not its intended target.

Section: 13.1 Skill: Synthesis/Evaluation LO: S13.1b

12 .

2) What role do chromatin remodelers play in eukaryotic gene expression?

Answer: Chromatin remodelers can enzymatically change the distribution or composition of nucleosomes. These remodeler enzymes can act by reorganizing nucleosomes to repress transcription, sliding nucleosomes along the chromosome or removing them to activate transcription, or changing histone composition to activate genes.

Section: 13.1 Skill: Synthesis/Evaluation LO: S13.2c 3) Scientists can create transgenic organisms by inserting foreign DNA randomly into the chromosomes of a host. A scientist might do this in order to express genes carried in the foreign DNA and in this case the foreign DNA is often engineered to include insulator sequences at the ends of the foreign sequence. Explain why insulators would be useful in this context.

Answer: Insulators would prevent the chromatin present in the host chromosome from influencing the expression of the DNA that the scientist inserts into the host chromosome. For instance, insulators could prevent the influence of local enhancers/silencers on the foreign DNA, which would affect expression of the transgenic DNA. Section: 13.2 Skill: Synthesis/Evaluation LO: S13.1a 4) The enhancers that are present near a gene are present in every cell in an organism, yet these enhancers can have tissue-specific effects on gene expression. Explain how enhancers could have tissue-specific effects on gene expression.

Answer: Enhancers can have tissue-specific effects when the transcription factors that they bind are either expressed or activated in a tissue-specific pattern.

Section: 13.2 Skill: Synthesis/Evaluation LO: S13.1b 5) Histone deacetylase (HDAC) inhibitors are commonly used as mood stabilizers or in the treatment of neurodegenerative diseases. What effect would a HDAC inhibitor have on the target cells?

Answer: A histone deacetylase removes acetyl groups from the histone tails. This process results in tighter chromosome compaction and thus less transcription of the genes in that region of the chromosome. HDAC inhibitors block this enzyme, resulting in hyperacetylation of histones and an increase in gene expression in the target cells, which can increase transcription (and translation) of gene products that may be decreased in the diseased tissues.

Section: 13.2 Skill: Synthesis/Evaluation LO: S13.2d

13 .

6) Explain how genomic imprinting can effect gene expression in mammalian embryos?

Answer: Imprinting, the placement of epigenetic marks on chromatin of paternal or maternal alleles can influence the relative expression of those alleles. For instance, higher expression of the maternal allele of a gene relative to the paternal allele may occur if a paternal allele carries more silencing marks (e.g. methylation) than the maternal allele.

Section: 13.2 Skill: Synthesis/Evaluation LO: S13.3 7) The gene encoding coat color is X-linked, and male cats have a single copy of the X chromosome. They can therefore express either the O or o allele, resulting in orange or black coat, respectively. Female cats, on the other hand, have two copies of the X chromosome. Thus, they can be heterozygous and express either orange or black in a pattern called tortoiseshell.

Using this information, list the possible genotypes for these individuals: • orange male • orange female • black male • black female • tortoiseshell Answer: O-, OO, o-, oo, Oo (females only) Section: 13.2 Skill: Applying/Analyzing LO: S13.2f

14 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 14 Forward Genetics and Recombinant DNA Technology 14.1 Multiple-Choice Questions 1) Before Muller's discovery that radiation induces mutation, scientists had to work on spontaneous mutants that were found solely by phenotype differences in natural populations.

Which of the features of Drosophila made it a fortuitous choice for Morgan and his colleagues?

A) having a long life cycle B) especially high rate of mutation C) large number of visible phenotypes D) well-known biochemical pathways E) both sexual and asexual reproduction Answer: C Section: 14.1 Skill: Synthesis/Evaluation LO: S14.1a 2) Genes for regulatory elements, such as creb, were found to be important in animal learning.

Which of the following would increase the probability that such an element is studied throughout the animal kingdom?

A) its unique to Drosophila B) the finding that creb is highly conserved C) the large number of copies of the creb gene within a genome D) the interaction of creb with cAMP E) the finding that creb activates odor perception Answer: B Section: 14.1 Skill: Applying/Analyzing LO: S14.3a 3) A researcher wants to mutagenize an organism. She is not in need of a large number of mutants but is more concerned with being able to find and then to amplify the mutated sequence.

Which of the following would therefore be more useful?

A) ionizing radiation B) UV radiation C) transposon insertion D) a chemical mutagen E) search for spontaneous mutant Answer: C Section: 14.1 Skill: Applying/Analyzing LO: S14.1a

1 .

4) In order to identify genes that function together in genetic pathways researchers can perform modifier screens, looking for second site enhancers or suppressors of a mutant. Which of the following is the most reasonable strategy for carrying out a suppressor screen in particular?

A) analyze the DNA sequence near the mutation of interest, looking for sequences that may act as silencers B) isolate mutants with phenotypes similar to the original mutant phenotype and then cross the mutants to identify the affected gene(s) C) identify mutations that, when crossed with the original mutant, result in synthetic lethality D) screen for enhancer-trap strains that show gene expression differences in the original mutant.

E) mutagenize the original mutant and look for new mutations that reduce the severity of the original mutant phenotype Answer: E Section: 14.1 Skill: Applying/Analyzing LO: S14.1a 5) In Drosophila, mutation screening usually involves use of a balancer chromosome that includes three elements: a set of overlapping inversions, an easily recognized dominant mutation, and a recessive lethal mutation that prevents balancer homozygotes from surviving. Which one, or combination, of these elements is necessary and sufficient to suppress crossovers?

A) inversions B) dominants C) lethal recessive D) inversions plus dominant E) lethal plus inversions Answer: A Section: 14.1 Skill: Applying/Analyzing LO: S14.1b 6) A mutagenesis screen provides 24 alleles shown by complementation analysis to be in the same gene, which of the following is true?

A) These mutants can be used to identify the function of this gene.

B) The gene being studied must be present in more than one copy per haploid genome.

C) The gene, in whole or in part, must be involved in transformation.

D) The gene must be highly conserved in evolution.

E) The gene must be involved in regulating a signal pathway.

Answer: A Section: 14.1 Skill: Applying/Analyzing LO: S14.2a

2 .

7) EMS (ethylmethane sulfonate) is mutagenic because ________.

A) it breaks phosphodiester bonds B) it adds alkyl groups to bases C) it forms cross-links between DNA strands D) it makes DNA more susceptible to radiation E) it causes chromosomal deletions Answer: B Section: 14.1 Skill: Knowledge/Comprehension LO: S14.1a 8) Besides the obvious fact that chemical mutagens are dangerous to handle, another major disadvantage to their use is that ________.

A) most of these mutations are immediately lethal B) most chemically induced changes are detectable only by genetic mapping and sequencing C) chemical mutagens activate transposons which cause widespread genetic damage D) they often result in only dramatic alterations of gene function E) they alter all A-T and C-G base pairing Answer: B Section: 14.1 Skill: Applying/Analyzing LO: S14.3d 9) The reasons for using transposon-based mutagenesis include all of the following EXCEPT ________.

A) likely to cause only minimal effects on gene function B) antibiotic selection can be employed during mutagenesis.

C) mutations are often 'tagged' by the nearby transposon D) dramatic mutations which often cause null alleles E) transposons are found in nearly all organisms Answer: A Section: 14.1 Skill: Synthesis/Evaluation LO: S14.3d

3 .

10) A researcher is interested in testing the role for a highly-conserved amino acid in the function of a mouse protein. She has been unable to locate any pre-existing variants in this codon among a large number of available mouse strains. What might be her approach to generating such mutants?

A) Use RNAi to knockdown the expression of the target gene.

B) Use transposons to generate insertion alleles in the gene.

C) Measure the expression of this gene in mice as they develop.

D) Use CRISPR-cas to cause DNA cleavage at the codon encoding this amino acid.

E) Use RNAi to interfere with translation of the gene.

Answer: D Section: 14.2 Skill: Applying/Analyzing LO: S14.3c 11) In a disease associated with a nucleotide repeat, anticipation is often noticed in subsequent generations, usually associated with a more severe phenotype. Which of the following is a more likely cause of anticipation?

A) Alleles with a large number of repeats are unstable and change size from generation to generation.

B) Children of affected parents receive more attention from physicians than other children.

C) Family members are exposed to the same mutagens.

D) Highly repeated sequences are expressed more often.

E) The sequences have greater stability in later generations.

Answer: A Section: 14.2 Skill: Synthesis/Evaluation LO: S14.1c 12) If you wish to use RNAi to reduce the expression of a gene in an organism. What would you inject into the organism?

A) Dicer with double-stranded (ds) DNA B) ssRNA complementary to cDNA C) ssRNA complementary to introns of the target gene D) dsRNA homologous to the gene E) dsRNA including transposase Answer: D Section: 14.3 Skill: Applying/Analyzing LO: S14.3a

4 .

13) To introduce dsRNA into C. elegans, it is enough just to dip the worms into media that includes E. coli containing the dsRNA because ________.

A) E. coli infect C. elegans and then the RNA is released B) E. coli DNA integrates into the genome of C. elegans C) C. elegans feed on the transgenic bacteria D) C. elegans emit toxins that paralyze the bacteria E) E. coli emit chemicals that cause pores to open in the nematode body wall Answer: C Section: 14.3 Skill: Knowledge/Comprehension LO: S14.3a 14) Which of the following tools/techniques would allow a researcher to measure the expression of a gene of interest in live organisms?

A) a lacZ reporter fused to the gene promoter B) RNAi targeting the promoter of the gene of interest C) a transgene that fuses GFP to the gene of interest D) measuring the frequency of mutations by DNA sequencing E) homologous recombination to replace the promoter of the gene of interest with a bacterial promoter Answer: C Section: 14.3 Skill: Knowledge/Comprehension LO: S14.4b 15) In reverse genetics, what is the correct order in which the experimenter proceeds?

A) silencing the genes in question using RNAi, followed by mapping to locate the gene B) screening individuals by PCR/DNA sequencing to associate altered genes with the trait of interest C) selection of mutants with a phenotype of interest and then mapping to locate the effected gene(s) D) identification of a gene of interest followed by screening for mutations in that gene and studies of the phenotypes caused those mutations.

E) random bombardment of the DNA with a known mutagen, followed by observation of offspring for newly acquired traits Answer: D Section: 14.3 Skill: Applying/Analyzing LO: S14.3a

5 .

16) If a gene in Drosophila is expressed only in imaginal disc (A) when it begins to differentiate, it is referred to as a homeotic gene. Recessive mutations in such genes often result in loss of function so that the structure does not develop. If a transgenic organism is made in which the gene is expressed in a different imaginal, disc (B), what is the predicted result?

A) loss of function of disc B's primary gene B) gain of function of several inappropriate genes in disc A C) gain of function in disc B of some of the structures usually seen in disc A D) deregulation of gene expression throughout disc A E) deregulation of gene expression throughout disc B Answer: C Section: 14.4 Skill: Applying/Analyzing LO: S14.1c 17) A gene, VTE4, from the sunflower of known sequence and well-understood transcriptional regulation, is available from another lab. The product of this gene enables sunflowers to to produce a significantly increased amount of vitamin E from a substrate produced in all plants.

The plant Brassica napus is harvested for canola oil. To produce canola oil with increased vitamin E, how would you proceed?

A) Select Brassica plants with the highest titer of the vitamin, isolate the genes responsible, and cause these genes to duplicate.

B) Use a yeast system to engineer yeasts that produce the vitamin, and get manufacturers to add this to the canola oil during bottling.

C) Introduce the gene and necessary cis elements into Brassica, along with a reporter gene, then measure the vitamin E produced.

D) Clone VTE4 in A. tumefaciens and use these bacteria to produce transgenic Brassica.

E) Introduce VTE4 into tomato plants or another food that humans eat more than Brassica.

Answer: C Section: 14.4 Skill: Applying/Analyzing LO: S14.4a

6 .

14.2 Short Answer Questions 1) Identifying an autosomal recessive mutation in a mutagenic screen in Drosophila requires identification of a mutant in the F3 generation. If testing for a sex-linked recessive lethal mutation (e.g., cn l + using a balancer chromosome such as cnCyO), in which generation can lines with mutations be identified?

Answer: F2 Section: 14.1 Skill: Applying/Analyzing LO: S14.1a 2) Which portion of a gene might you target with the CRISPR-cas system to interfere with splicing?

Answer: intron/exon junction or splice site Section: 14.1 Skill: Knowledge/Comprehension LO: S14.3c 3) What kind of analysis is needed to determine whether two mutations are in the same gene or in different genes?

Answer: complementation Section: 14.1 Skill: Knowledge/Comprehension LO: S14.1c 4) In a given situation, reduced or missing function in gene A results in a viable but noticeable phenotype. The same is true for gene B. However, when both gene A and gene B products are reduced or missing in the same organism, the result is lethality. What is this called?

Answer: synthetic lethality or synthetic enhancement Section: 14.1 Skill: Knowledge/Comprehension LO: S14.1c 5) Suppose a scientist works with a mouse strain B75 and isolates an interesting recessive mutant that they call glyph. The researcher wants to identify the gene affected in glyph by complementation cloning. Which strain would the researcher use as the source of DNA for transformation?

Answer: the wild type strain B75 Section: 14.2 Skill: Applying/Analyzing LO: S14.2a

7 .

6) In positional cloning, the researcher begins with a phenotype. To move toward identification of its DNA sequence, she must begin with what step?

Answer: genetic mapping Section: 14.2 Skill: Knowledge/Comprehension LO: S14.2a 7) DNA markers that are found to segregate with a mutation of interest can be used to determine the mutation's distance from the markers by looking at what?

Answer: recombination frequency Section: 14.2 Skill: Knowledge/Comprehension LO: S14.2a 8) Which technique can be used to replace an endogenous gene with one constructed in vitro?

Answer: homologous recombination Section: 14.2 Skill: Knowledge/Comprehension LO: S14.4b 9) Libraries of insertion strains can be screened, primarily by which common technique, to identify mutations in a gene of interest?

Answer: polymerase chain reaction or PCR Section: 14.2 Skill: Knowledge/Comprehension LO: S14.3a 10) In an organism, X, it is found to be impossible to generate loss-of-function mutants for most genes. There is an alternative method of introducing random mutations that can later be screened for the gene(s) in question. This can be accomplished by using insertions due to what genetic entities?

Answer: transposons or T-DNA Section: 14.3 Skill: Applying/Analyzing LO: S14.3d 11) What method allows researchers to reduce the expression of a target gene, even in wildtype organisms that have not previously been genetically modified?

Answer: RNA interference or RNAi Section: 14.3 Skill: Knowledge/Comprehension LO: S14.3a

8 .

12) In nature, RNAi protects cells against infection by what kind of pathogen?

Answer: dsRNA viruses or double-stranded RNA viruses Section: 14.3 Skill: Knowledge/Comprehension LO: S14.3b 13) Reporter genes such as GFP can be integrated into a native gene to study the expression pattern of a target gene. If a researcher wishes to make a fusion of the GFP protein to the protein expressed by the target gene, where must the GFP gene integrate?

Answer: into the coding region or exon of the target gene Section: 14.3 Skill: Applying/Analyzing LO: S14.4b 14) The GFP gene is useful as a reporter gene because the cells and tissues with it emit fluorescence when treated with what, which functions as a substrate?

Answer: UV, purple, or blue light Section: 14.4 Skill: Knowledge/Comprehension LO: S14.4a 15) If an experimenter wants to use the GFP method but needs to detect the presence or absence of several proteins at the same time, he can take advantage of mutational variants of GFP that emit what?

Answer: other colors Section: 14.4 Skill: Applying/Analyzing LO: S14.4b 14.3 Fill-in-the-Blank Questions 1) A genetic strategy that starts with knowing the DNA sequence and works toward identifying an associated phenotype is known as ________.

Answer: reverse genetics Section: 14.1 Skill: Knowledge/Comprehension LO: S14.3a 2) Chemical mutagens are preferred to ________ because they have a broader mutation spectrum, including generating alleles that affect only a single codon.

Answer: transposons or radiation Section: 14.1 Skill: Knowledge/Comprehension LO: S14.3d

9 .

3) A balancer chromosome includes a recessive lethal allele, a marker dominant allele, and ________.

Answer: inversion(s) Section: 14.1 Skill: Knowledge/Comprehension LO: S14.1a 4) A single nucleotide in a genome can be targeted for mutagenesis using the ________ technique, adapted from a bacterial defense system.

Answer: CRISPR-cas Section: 14.3 Skill: Knowledge/Comprehension LO: S14.3b 14.4 Essay Questions 1) You have performed a mutagenesis screening in a population of organisms. Describe how you would identify mutations that are dominant. Describe how you would identify mutations that are recessive.

Answer: To identify dominant mutations, breed the mutant individual to a wild-type individual.

Dominant mutations will be visible in the F1 generation. These will be rare, since they represent gain-of-function mutations. Although more common, recessive mutations cannot be identified until the F3 generation. To identify recessive mutations, mate-mutagenized individuals to wildtype individuals — the F1 will have a wild-type phenotype. Back-cross individual F1 progeny to a wild-type parent to produce separate F2 families; half the individuals will carry newly induced mutations. Interbreed the F2 family siblings to produce individuals that are homozygous for each mutation.

Section: 14.1 Skill: Applying/Analyzing LO: S14.1c 2) Why are conditional alleles useful in studying genes with functions in essential cell processes?

Answer: By definition, essential functions are those required for an organism's survival.

Mutations that affect these functions are thus often inviable and can be difficult to isolate, study or maintain. Conditional mutations such as temperature-sensitive mutations allow researchers to isolate mutants based on their relevant phenotype (e.g. failure to complete cell division) at the restrictive condition, and then maintain those mutants at the permissive condition. Another particular advantage of conditional alleles is that they allow researchers to investigate multiple possible roles for genes in a sequential process, such as cell division, by transfer between permissive and restrictive conditions over the course of the relevant biological process.

Section: 14.1 Skill: Synthesis/Evaluation LO: S14.1b

10 .

3) Why is only a fraction of an organism's genes represented in any cDNA library?

Answer: cDNA libraries are comprised of complementary DNAs that are reverse transcribed from existing mRNA transcripts. Thus, only genes that are being actively transcribed in a given organism/tissue/organ/developmental stage will be represented in the cDNA library.

Section: 14.2 Skill: Synthesis/Evaluation LO: S14.3a 4) You have performed an analysis of a gene promoter using a genetic reporter and a series of deletions. From this analysis, you identify a small deletion upstream of the start codon that causes increased expression of the reporter gene, compared to the expression of the reporter when it is fused to the wild type promoter. How do you explain the increased expression of the reporter in the mutant?

Answer: The small deletion eliminates a silencer, a place where a repressor usually binds in the wild type promoter. There is more reporter expression in the mutant because the repressor no longer interacts with the mutant promoter sequence.

Section: 14.4 Skill: Synthesis/Evaluation LO: S14.4b 5) A friend of yours claims to have identified a gene that causes a human disease! Your friend says that they have sequenced the DNA of someone with the disorder and compared that to a healthy individual. They found a nucleotide difference in a gene called Hsr12 that encodes a protein that functions cell-to-cell communication. What is wrong with your friends claim?

Answer: There are likely to be many thousands of nucleotide differences between the sequences from one diseased individual to one healthy individual, so concluding that the difference in the Hsr12 gene is what causes the disease is not a reasonable conclusion.

Section: 14.3 Skill: Synthesis/Evaluation LO: S14.1b

11 .

6) In TILLING, a population of inbred organisms is mutagenized to saturation. DNA from mutagenized lines is isolated and amplified by PCR. PCR products are denatured and allowed to reanneal and then treated with an endonuclease. Why is the denaturation/re-annealing step important for the detection of mutant alleles?

Answer: Wild-type individuals will be homozygous and will produce a single PCR product.

When wild-type PCR products reanneal after denaturation, all single strands have a perfect complement so the reannealed duplexes will contain no mismatches and will not be susceptible to nicking by the endonuclease. By contrast, mutants can be homozygous or heterozygous, so there will potentially be two different template DNAs in the mutant, which would result in two different PCR products. When the PCR products from the mutants are mixed and reannealed with those of the wild type, they can form heteroduplexes (where one strand is wild type and the other mutant). Heteroduplex strands will contain mismatches, which will be susceptible to nicking by the endonuclease and the products of the endonuclease are detectable upon gel electrophoresis.

Section: 14.2 Skill: Applying/Analyzing LO: S14.3a 7) Transgenic fusions between a gene and a reporter are used to determine whether a gene is expressed. Which of these components allows the researcher to work with living specimens, and why?

Answer: The reporter genes are extremely helpful for tracking gene expression, but it is important that the reporter product does not harm the organism. For instance, the presence of the lacZ protein is often used to screen cells in culture through blue-white screening, but requires the substrate X-gal to produce the blue product. Injecting X-gal into a living organism may have toxic effects and is, therefore, not commonly used when researchers want to examine living organisms. Alternatively, reporter genes such as green fluorescent protein (GFP) are not toxic and require no substrate. They have been used as markers in the cloning of transgenic pigs and have had no toxic effects. The other concern is the effect of the reporter gene fusion on wild-type function of the protein. Because protein conformation and proper folding of the protein is so important to the function of the protein, fusing another gene product to the protein can have unintended consequences for protein function. It may be necessary to test N-terminal versus Cterminal fusions, as well as using protein reporters that are different sizes. Additionally, some organisms and microenvironments may not be conducive to certain reporter genes. Fluorescent protein reporters such as GFP are not.

Section: 14.4 Skill: Applying/Analyzing LO: S14.4b

12 .

8) Design an experiment that uses enhancer trapping to drive the expression of a gene of interest.

Answer: Enhancer trap constructs include a reporter gene that is fused to a weak promoter that, by itself, is insufficient to drive expression of the reporter gene. The construct also requires a mobile element, typically a P element, which allows this basal transcription unit to be inserted into the genome. When the reporter gene is integrated near an enhancer region, the enhancer is "trapped" and the reporter gene is expressed, which allows you to identify spatial regulation by enhancers.

In E. coli, a common enhancer trap is P[lacZ], while yeast often uses a P[GAL4] enhancer trap.

Enhancer trapping also uses a visible marker to help with easy identification of new insertions.

For example, the white gene, which is responsible for wild-type red eye color in Drosophila, is disrupted by an insertion event, creating a white-eyed phenotype. Drosophila also has the benefit of having a large number of readily available fly stocks containing GAL4 insertions, which act as the "drivers." Using one of these stocks, plus a second fly stock containing a UAS DNA sequence followed by your gene of interest, you can simply cross these two lines of flies. This eliminates the need to directly clone transgenic flies with the enhancer linked to your gene and saves a lot of time and resources, making enhancer trapping a relatively easy process.

Section: 14.4 Skill: Applying/Analyzing LO: S14.4a

13 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 15 Recombinant DNA Technology and Its Applications 15.1 Multiple-Choice Questions 1) Half of the palindromic restriction site that the endonuclease PstI recognizes, binds, and cleaves is 5'-CTG________-3'. What is the complete sequence?

A) 5'-CTGCTG-3' B) 5'-CTGGTC-3' C) 5'-CTGCAG-3' D) 5'-CTGGAG-3' E) 5'-CTGGAC-3' Answer: C Section: 15.1 Skill: Applying/Analyzing LO: S15.1a 2) The human genome is 3 × 109 bp in length. Calculate how many fragments would result from the complete digestion of the human genome with EcoRI (5'-GꜜAATTC-3').

A) 9.72 × 105 B) 4.58 × 104 C) 4.32 × 105 D) 1.17 × 107 E) 7.32 × 105 Answer: E Section: 15.1 Skill: Applying/Analyzing LO: G4, S15.1a

1 .

3) What size fragments you would expect from an EcoRI/ XhoI double digest of the plasmid below?

A) 1.25 kb, 2 kb, 2.5 kb, 3 kb B) 1.25 kb, 2.5 kb, 3 kb, 3.25 kb C) 2.5 kb, 3 kb, 3.25 kb D) 2 kb, 3.75 kb, 4.25 kb E) 2 kb, 8 kb Answer: D Section: 15.1 Skill: Applying/Analyzing LO: G3, G4, S15.1a 4) Assume that a plasmid is 4000 base pairs in length and has restriction sites at the following locations: 600, 1000, 2200, and 3800. Give the expected sizes of the restriction fragments following complete digestion.

A) 200 bp, 600 bp, 1200 bp, 1600 bp B) 200 bp, 800 bp, 1200 bp, 1800 bp C) 200 bp, 400 bp, 600 bp, 1200 bp, 1800 bp D) 400 bp, 800 bp, 1200 bp, 1600 bp E) 400 bp, 800 bp, 1000 bp, 1800 bp Answer: D Section: 15.1 Skill: Applying/Analyzing LO: G3, G4, S15.1a

2 .

5) You have a linear DNA fragment of 1600 bp. To characterize it, you decide to make a restriction map. Cutting with one enzyme, you show by electrophoresis that the fragments produced are 400 and 1200 bp. Cutting the same DNA with another enzyme you recover 700 and 900 bp fragments. After treatment with both enzymes, fragments were 300, 400, and 900 bp.

Which of the following order of fragments, if any, best represents the restriction map?

A) 900/ 400/ 300 B) 400/ 900/ 300 C) 300/ 400/ 900 D) 400/ 300/ 900 E) It is not possible to determine the order.

Answer: D Section: 15.1 Skill: Applying/Analyzing LO: G3, G4, S15.1b 6) In producing a recombinant plasmid to be used to clone a given donor insert, it is possible to cut both the donor and plasmid with the same restriction enzyme, resulting in complementary sticky ends. Assuming plenty of plasmid DNA is available, why is further selection necessary before the introduction of the plasmid into a cellular system?

A) Contamination will have introduced other donor inserts.

B) Some donor inserts will be single stranded and deteriorate.

C) Some donor inserts will be sensitive to particular antibiotics.

D) Some donor pieces will remain uninserted.

E) Some donor strands will be inserted with an incorrect orientation.

Answer: E Section: 15.1 Skill: Applying/Analyzing LO: S15.1c, S15.1d 7) In selecting recombinant bacteria, cells are chosen that are resistant to a specific antibiotic.

How are the bacteria made resistant?

A) They are pre-selected for the experiment on this basis.

B) The antibiotic resistance gene is encoded in the vector.

C) The antibiotic resistance gene is encoded on the donor insert.

D) Resistance is activated by the recombination event.

E) Resistance is activated when the cells are provided with the antibiotic Answer: B Section: 15.1 Skill: Applying/Analyzing LO: S15.1c

3 .

8) If bacteria transformed with a recombinant pUC18 plasmid produce white colonies when grown on plates containing ampicillin and Xgal, which of the following is least likely?

A) They produce β-lactamase that provides resistance to ampicillin.

B) They produce functional β-galactosidase that cleaves X-gal.

C) They carry a vector with the bla gene used as the selectable marker.

D) They carry a vector that contains a DNA fragment inserted into the multiple cloning site.

E) They carry a vector with a lacZ gene that has been disrupted and rendered nonfunctional.

Answer: B Section: 15.1 Skill: Applying/Analyzing LO: G2, G7, G9, S15.1c, S15.1d 9) Which of the following lists the steps for cDNA library preparation in the correct order?

I. Partially degrade the mRNA with RNase H II. Isolate mRNA from a specific cell/ tissue with an oligo-dT primer that hybridizes to poly-A tail III. Synthesize a strand of cDNA using DNA polymerase with mRNA fragments as primers IV. Ligate cDNAs into a cloning vector V. Create a single-stranded DNA molecule complementary to the mRNA using reverse transcriptase A) I, II, III, IV, V B) II, V, I, III, IV C) II, III, V, I, IV D) V, III, I, IV, II E) II, V, IV, III, I Answer: B Section: 15.1 Skill: Applying/Analyzing LO: G7, G9, S15.1e 10) Which of the following is true of the sequences found in DNA libraries from human tissues?

A) Those in a muscle genomic library should not be found within a brain genomic library.

B) Those in a brain cDNA library should not be found within a muscle genomic library.

C) Those in a muscle cDNA library should all be present in a brain genomic library.

D) Those in a brain cDNA library should all be present a muscle cDNA library.

E) Those in a muscle cDNA library should all be present in a brain cDNA library.

Answer: C Section: 15.1 Skill: Applying/Analyzing LO: G2, G7, G9, S15.1e

4 .

11) Efficient production of eukaryotic proteins in E. coli may require all of the following except?

A) an E. coli expression vector with a promoter sequence that binds RNA polymerase B) an E. coli expression vector with a Shine-Dalgarno sequence for efficient translation C) altered codon usage within the heterologous sequences to approximate the codon bias in E. coli D) the use of cDNAs, that are free of introns, as eukaryotic transgenes E) a negative selectable marker to select against nonhomologous recombination events Answer: E Section: 15.1 Skill: Applying/Analyzing LO: G2, G7, S15.1e 12) Which of the following best lists the correct order of the steps taken to produce human insulin in E.coli in the 1970s?

I. Each insulin chain was cloned into an expression vector as a continuation of the lacZ reading frame, resulting in a fusion protein for each chain.

II. Disulfide bonds were induced to form between the cysteine residues of the purified insulin A and B chains.

III. Both human insulin genes were reverse translated to create synthetic genes.

IV. EcoRI and BamHI restriction sites were added to the ends of the DNA to facilitate directional cloning of each into a separate plasmid vector.

A) I, II, III, IV B) III, IV, I, II C) IV, III, I, II D) III, I, II, IV E) I, III, IV, II Answer: B Section: 15.2 Skill: Applying/Analyzing LO: G5, G7, G9, S15.2a, S15.2d 13) Which of the following is a possible advantage for the environment of having farmers use plants that genetically produce Bt toxin?

A) Plants with genes for the toxin are eaten less often by insects, and therefore less insecticide is needed.

B) Plants that produce Bt toxin must be grown from new seeds each season, and therefore fewer of them are planted.

C) Bt toxin is advantageous to many species of invertebrates that grow in the same fields.

D) Harvesting plants with these genes is less disturbing to the soil layers where they are planted.

E) Plants with these genes must be sprayed with the organism, but need much less than other plants.

Answer: A Section: 15.2 Skill: Applying/Analyzing LO: G5, G7, S15.2a 5 .

14) In using Agrobacterium tumefaciens to transfer genes into plants, what is transferred from bacterium to plant?

A) Ti plasmid B) T-DNA C) opines D) only the recombining sequence E) the entire bacterium Answer: B Section: 15.2 Skill: Remembering/Understanding LO: G5, S15.2c 15) Which of the following is FALSE regarding P element-mediated transformation?

A) The gene of interest is typically inserted in the vector plasmid between the P element inverted repeat end sequences.

B) The selectable marker is typically inserted in the vector plasmid between the P element inverted repeat end sequences.

C) A second plasmid can be used to supply the transposase activity in trans.

D) The transposase activity inserts P elements into the genome at specific positions in the genome.

E) P elements transpose only in the germ-line cells of Drosophila.

Answer: D Section: 15.2 Skill: Applying/Analyzing LO: G7, G9, S15.2e 16) Suppose you have engineered a plasmid vector in order to produce a functional animal protein in E. coli. But the product you collect, while it has the correct amino acid sequence, is in fact nonfunctional. What might you try that would have a greater chance of success?

A) a eukaryotic expression vector that has sequences for regulating transcription B) using a host other than E. coli that does not have the same codon bias C) a eukaryotic expression vector that includes sequences for controlling posttranslational effects D) treatment of the posttranslation proteins to add signal sequences E) using yeast artificial chromosomes that can hold longer sequences Answer: D Section: 15.2 Skill: Applying/Analyzing LO: G7, G9, S15.2a

6 .

17) Why might the expression of a transgene be especially abnormal in vertebrates with larger average size of genes and amount of heterochromatin in their genomes?

A) position effect B) homologous recombination C) vector insufficiency D) mitotic recombination E) unequal crossing over Answer: A Section: 15.2 Skill: Applying/Analyzing LO: G7, S15.2e 18) Which of the following is FALSE regarding the formation of transgenic animals?

A) DNA is usually inserted into eggs or embryos since totipotency is not characteristic of most animal cells.

B) Homologous recombination occurs much more frequency than illegitimate recombination.

C) If some cells of the embryo do not receive the introduced DNA, the result is an organism with cells of different genotypes; the organism is known as a chimera.

D) For a transgenic animal line to be produced from a chimera, the transgene must be carried by germ line or germ cells.

E) Either integration of multiple copies of the transgene or position effect can lead to variability in the expression of the transgene.

Answer: B Section: 15.2 Skill: Applying/Analyzing LO: G2, G7, S15.2e 19) Which of the following is true regarding the Cre-lox recombination system?

A) Cre recombinase initiates conjugative transfer of the T-DNA in Agrobacterium tumefaciens.

B) It can be used to remove selectable markers in transgenic organisms.

C) Cre recombinase can recognize the P element inverted repeat end sequences.

D) It is limited to manipulating DNA sequences in vitro.

E) This site-specific recombination system is derived from Drosophila.

Answer: B Section: 15.2 Skill: Applying/Analyzing LO: S15.2c

7 .

20) Which lists the potential steps for genetic therapy for mice with sickle cell disease (βS/βS) in the correct order?

I. Differentiation of iPS cells into hematopoietic progenitor (HP) cells by infecting with retrovirus expressing HoxB4.

II. Reprogramming tail fibroblasts into induced pluripotent stem (iPS) cells by infecting them with viral vectors containing transcription factors.

III. Infection with a viral vector expressing Cre recombinase to remove transcription factor(s) from from iPS cells.

IV. Transplantation of corrected (βA/βA) hematopoietic cells back into irradiated mice.

V. Homologous recombination with βS and wild-type βA allele via CRISPR-Cas9 genome editing.

A) I, II, III, V, IV B) II, III, V, I, IV C) V, III, IV, I, II D) V, III, I, IV, II E) II, III, IV, V, I Answer: B Section: 15.3 Skill: Applying/Analyzing LO: G2, G5, G7, G9, S15.3 21) Which of the following statements is FALSE regarding gene therapy?

A) Both somatic and germinal gene therapy have been successful in animal systems; but for ethical reasons, only somatic gene therapy has been attempted in humans.

B) The integration of a transgene into the genome of the target cell by a viral vector is not permanent, and therefore it will require repeated treatments.

C) The insertion of a vector may cause a detrimental mutation.

D) Approaches with iPS cells avoid problems of immune system incompatibility, but are limited to diseases, like blood disorders, in which cells can be isolated, corrected, and reintroduced.

E) iPS cells that continue to express the Yamanaka factors are predisposed to become cancerous.

Answer: B Section: 15.3 Skill: Applying/Analyzing LO: G2, G5, G7, G9, S15.3 22) Which of the following is FALSE regarding grape cultivation?

A) phenotypes are usually maintained through self-fertilization or crossing with another cultivar B) clonal propagation helps to maintain highly heterozygous cultivars C) the vines are usually chimeric D) the shoots are often genetically identical and chosen on the basis of their fruit phenotype E) the roots are often genetically identical and chosen for being well adapted to soil conditions Answer: A Section: 15.4 Skill: Applying/Analyzing LO: G5,G7, S15.4 8 .

23) Cloning animals by nuclear implantation involves all of the following except?

A) removal of somatic cells and extraction of a diploid donor nucleus from the animal to be cloned B) removal of an egg cell, extraction of the nucleus, and injection of the diploid donor nucleus C) fertilization of the cell egg with a sperm cell followed by implantation in a surrogate mother D) induction of cell division and implantation of blastocyst in the surrogate mother E) a low frequency of success likely due to variations in the completeness of epigenetic programming of the somatic cell nucleus Answer: C Section: 15.4 Skill: Applying/Analyzing LO: G7, S15.4 24) Snuppy, a male Afghan hound, was created by transferring the nucleus from an ear cell of the male Afghan hound, Tai, into an enucleated egg from a donor. The embryo was transplanted into a surrogate mother to develop to term. Which of the following animals would be a genetic match for Snuppy if nuclear DNA were tested?

A) Tai B) the egg donor C) the surrogate D) both Tai and the egg donor E) both Tai and the surrogate Answer: A Section: 15.4 Skill: Applying/Analyzing LO: S15.4 25) Snuppy, a male Afghan hound, was created by transferring the nucleus from an ear cell of the male Afghan hound, Tai, into an enucleated egg from a donor. The embryo was transplanted into a surrogate mother to develop to term. Which of the following animals would be a genetic match for Snuppy if mitochondrial DNA were tested?

A) Tai B) the egg donor C) the surrogate D) both Tai and the egg donor E) both Tai and the surrogate Answer: B Section: 15.4 Skill: Applying/Analyzing LO: G7, S15.4

9 .

15.2 Short Answer Questions 1) What are the short single-stranded overhangs created by cleavage of DNA by specific restriction enzymes?

Answer: sticky ends Section: 15.1 Skill: Remembering/Understanding LO: S15.1a 2) The ends left by some restriction enzymes that do not have 5' or 3' overhangs are called what?

Answer: blunt ends Section: 15.1 Skill: Remembering/Understanding LO: S15.1a 3) In the pUC plasmids, a polylinker region that includes multiple restriction sites is embedded in the lacZ component. If a sequence of interest is inserted into this region, what is the effect on lacZ? Answer: inactivation Section: 15.1 Skill: Applying/Analyzing LO: S15.1d 4) If bacteria containing non-recombinant pUC18 are plated on X-gal medium, would colonies appear blue or white colonies or both?

Answer: blue Section: 15.1 Skill: Applying/Analyzing LO: S15.1c 5) In creating a recombinant DNA molecule, what are two methods that prevent re-ligation of the cloning vector that contains no insert DNA? Answer: 1) removal of the 5' phosphate 2) directional cloning Section: 15.1 Skill: Applying/Analyzing LO: S15.1c 6) Which type of DNA library (genomic or cDNA) would include introns and promoters?

Answer: genomic Section: 15.1 Skill: Remembering/Understanding LO: S15.1e

10 .

7) What technique of genomic library construction allows the genome to be cut by a given restriction enzyme at some but not all recognition sequences?

Answer: partial digestion or using less of the enzyme Section: 15.1 Skill: Remembering/Understanding LO: S15.1a 8) In a cDNA library, not all expressed genes will be represented. On what does this depend?

Answer: on the tissue the mRNA comes from or on the source of mRNA Section: 15.1 Skill: Remembering/Understanding LO: S15.1e 9) In producing an E. coli plasmid to be an expression vector, an experimenter includes genes for selectable markers and a promoter for the expression of the introduced genes. She uses a cDNA sequence to ensure the insert contains no introns. The gene inserted is transcribed, but not translated. What sequence did she omit from the plasmid?

Answer: Shine-Dalgarno (sequence) Section: 15.2 Skill: Applying/Analyzing LO: G2, S15.2a, S15.2e 10) If a gene is replaced with a heterologous sequence such that the gene is then nonfunctional, what is the result usually called?

Answer: a knockout Section: 15.2 Skill: Remembering/Understanding LO: S15.2c 11) Agrobacterium tumefaciens is often used in the production of transgenic plants because of its Ti plasmid. This plasmid is usually "disarmed" by deleting its tumor genes along with the genes for producing what products that would have been advantageous to the bacterium?

Answer: opines Section: 15.2 Skill: Remembering/Understanding LO: S15.2e 12) By what method is T-DNA integrated into the plant genome?

Answer: non-homologous recombination Section: 15.2 Skill: Remembering/Understanding LO: S15.2c

11 .

13) To ensure that a human gene is translated properly in a bacterial cell, what should you clone instead of the genomic DNA fragment that contains the gene?

Answer: cDNA Section: 15.2 Skill: Remembering/Understanding LO: S15.2a 14) In the case of Jesse Gelsinger, a young man who died during gene therapy, the therapy involved delivery via a viral vector. Was this germ-line or somatic therapy?

Answer: somatic Section: 15.3 Skill: Applying/Analyzing LO: S15.3 15) Genetic therapy for mice with Duchenne muscular dystrophy demonstrated that which method could be used to delete a mutant exon of the dystrophin gene in muscle cells?

Answer: CRISPR-Cas9 genome-editing Section: 15.3 Skill: Remembering/Understanding LO: S15.3 15.3 Fill-in-the-Blank Questions 1) A hypothetical restriction sequence is 5′-TCTAGA-3′. The palindromic sequence is 3′________-5′.

Answer: AGATCT (must be in this orientation) Section: 15.1 Skill: Applying/Analyzing LO: S15.1a 2) The binary approach uses both a transformation vector that contains the T-region flanked by border sequences and a disarmed ________ with the genes required for virulence and conjugative transfer.

Answer: T plasmid Section: 15.2 Skill: Remembering/Understanding LO: S15.2c 3) If recombination results in random integration of a DNA sequence so that it produces a nonhomologous region, the recombination is ________.

Answer: illegitimate Section: 15.2 Skill: Remembering/Understanding LO: S15.2c

12 .

4) Suppose a gene from a cow is introduced into a frog. This results in a ________ transgene.

Answer: heterologous Section: 15.2 Skill: Remembering/Understanding LO: S15.2e 5) In species X, the UUU codon for phenylalanine is used significantly more frequently than other phenylalanine codons, and knowing this allows the scientist to manipulate translation efficiency. This unequal frequency is known as ________.

Answer: codon bias Section: 15.2 Skill: Remembering/Understanding LO: S15.2a

13 .

15.4 Essay Questions 1) You have isolated a plasmid from E. coli. You get the following results, using three restriction enzymes, 1 XhoI, 2 BamHI, and 3 EcoRI to perform six different digestions: single digests using each enzyme alone and double digests using each combination of two enzymes. enzyme XhoI BamHI EcoRI XhoI + BamHI XhoI + EcoRI

# fragments 1 2 2 3 3

BamHI + EcoRI 4

sizes (in kb) 10 3.75, 6.25 4.25, 5.75 1.25, 3.75, 5 2, 3.75, 4.25 1.25, 2.5, 3, 3.25

Draw a restriction map of the plasmid.

Answer:

XhoI must cut the 10 kb plasmid only once, while BamHI and EcoRI each cut the plasmid at two sites. The XhoI site must cut the 6.25 kb BamHI fragment into 1.25 and 5 kb fragments, and not cut within the 3.75 kb fragment. The XhoI site must also cut the 5.75 kb EcoRI fragment into 2 and 3.75 kb fragments, and not cut within the 4.25 kb fragment. Finally, the digest with BamHI and EcoRI indicates that EcoRI must cut the 6.25 kb BamHI fragment into 3 and 3.25 kb fragments, as well as the 3.75 kb BamHI fragment into 1.25 and 2.5 kb.

Section: 15.1 Skill: Evaluating/Creating LO: G2, G3, G4, S15.1b

14 .

2) Your student is trying to create a "Gene X" knockout mouse by introducing a recombinant construct into embryonic stem (ES) cells, selecting transformants, and introducing the transformed cells into mouse blastocysts. He designed his construct as follows: NeoR----region of homology to Gene X---HSV tk--- region of homology to Gene X where NeoR is the neomycin-resistance gene and HSV tk is the Herpes Simplex virus thymidine kinase gene. He transforms the ES cells and exposes them to G418 and ganciclovir. Explain to your student he failed to recover transformed ES cells.

Answer: If his construct incorporated into the mouse DNA by illegitimate recombination, both the neomycin resistance gene and the thymidine kinase gene would be incorporated into the genome. These cells would be resistant to neomycin, but sensitive to ganciclovir, and would die.

If the construct incorporated into the mouse DNA by homologous recombination, only the gene flanked by the regions homologous to Gene X would be incorporated. In your student's construct, this gene was the tk gene. The neomycin resistance gene, outside the regions of homology, would not be incorporated. These cells would be sensitive to both G418 and ganciclovir, and would die.

Section: 15.2 Skill: Evaluating/Creating LO: G2, G7, G8, G9, S15.2e 3) Discuss the ethical and societal reasons for and against germ-line gene therapy.

Answer: Answers may vary, but it is important to distinguish between somatic and germ-line gene therapy. Somatic gene therapy will treat the disorder in individuals receiving therapy, but they still have the potential to pass the same mutation to their progeny. Germ-line therapy corrects the mutation in all cells, essentially changing not only the individuals receiving therapy but all their subsequent generations. There may be differing opinions about the ethics of the two types of therapy, since somatic gene therapy can be performed on individuals old enough to consent to treatment, whereas germ-line gene therapy can affect all potential offspring before fertilization has even occurred.

Section: 15.3 Skill: Evaluating/Creating LO: G2, G5, G8, S15.3

15 .

4) In the 1990s, gene therapy was seen as a probable cure for many genetic diseases. List three or four reasons that it has so far been of limited success.

Answer: Gene therapy is technically challenging, particularly in the mechanisms of gene delivery and activation in the correct target tissue. The gene must be targeted to the correct tissue to ensure proper wild-type expression. If the gene is targeted to the wrong tissue, it would be inefficient, but could also result in severe health problems by introducing a gene into a tissue which normally does not express that gene product. Similarly, if a gene is intended to be targeted to one particular tissue and is, instead, introduced into the germ line, it may be passed on to offspring with unintended consequences. Instead of expressing the gene only in the targeted tissue, it would now affect all cells in the patient's offspring, which may prove to be fatal or have severe side effects.

Within the individual, gene therapy may inadvertently result in the activation or deactivation of other genes based on where the DNA incorporates into the genome. The viral vector itself can also elicit a severe immune reaction in some individuals, which was the case in early gene therapy trials. In addition, if a construct has been incorporated, it is difficult to control levels of gene expression and the amount of expression may be higher or lower than anticipated.

Section: 15.3 Skill: Evaluating/Creating LO: G2, G5, G8, S15.3 5) What characteristics of plants allow them to be cloned relatively easily, whereas cloning in animals is rare and difficult to achieve?

Answer: There are several reasons why cloning is easier in plants than in animals. First, plants can be easily propagated through vegetative reproduction. In addition, plant cells are often totipotent, which makes the cloning process much easier because there is no need to "reprogram" the nucleus. Animal cloning can be accomplished through parthenogenesis in rare instances, but is primarily accomplished through reproductive cloning and somatic cell nuclear transfer. This process requires the transfer of a nucleus from a donor cell to a recipient egg, and is often complicated by the incomplete reprogramming of the genome because donor cells are no longer totipotent as in plant cells. Cell differentiation in animals therefore limits the success rate of somatic cell nuclear transfer and the formation of viable organisms from such techniques.

Section: 15.4 Skill: Evaluating/Creating LO: G2, G7, G8, S15.4

16 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 16 Genomics: Genetics from a Whole-Genome Perspective 16.1 Multiple-Choice Questions 1) Which type of sequencing involves first creating a physical map?

A) whole-genome shotgun sequencing B) clone-by-clone sequencing C) paired-end sequencing D) metagenomic sequencing E) annotation sequencing Answer: B Section: 16.1 Skill: Knowledge/Comprehension LO: S16.1a 2) Previously unidentifiable microorganisms have now been identified through ________.

A) proteomics B) transcriptomics C) metagenomics D) evolutionary genomics E) reverse transcription Answer: C Section: 16.1 Skill: Knowledge/Comprehension LO: S16.1c 3) Shotgun sequencing requires all of the following except?

A) a computer B) a genetic map C) a DNA sequencing technology D) software capable of comparing DNA sequence E) many overlapping DNA sequences Answer: B Section: 16.1 Skill: Knowledge/Comprehension LO: S16.1a

1 .

4) Groups of genes that are evolutionarily related are called ________.

A) gene families B) gene duplications C) SNPs D) gene domains E) metagenomes Answer: A Section: 16.2 Skill: Knowledge/Comprehension LO: S16.3a 5) It has been predicted that genes may be involved in smoking behavior. Previously, large twin and family studies have provided evidence of this and an estimate of heritability (genetic contribution) of 46-59%, depending on sex. A large group of researchers from several institutions composed a pathway- or-systems-based series of studies including SNP analysis, cellular location and protein interaction, replication with other samples, statistical analysis, and bioinformatics. What is a likely result?

A) The study is unlikely to produce meaningful results because it is too diffuse in its methods.

B) The samples used also correlate with behaviors like depression, often associated with smoking, and they may be finding depression-associated genes.

C) Two specific genes, one for a transporter protein and one for an opioid receptor, are clearly shown to be causative in all members of each sample.

D) A group of genes that tend to code for proteins with similar biological function have been identified to contribute significantly.

E) Genome-wide association studies are of no practical use in studying human behaviors.

Answer: D Section: 16.2 Skill: Synthesis/Evaluation LO: S16.4a 6) Which of the following would likely exclude a sequence from being considered as an active gene?

A) a promoter with signals identical to the promoter of another gene B) exons in different reading frames from one another C) an exon followed by 3′ processing signals D) only one long exon E) introns missing the GT-AG splice junctions compared to other genes in this family Answer: E Section: 16.2 Skill: Applying/Analyzing LO: S16.2a

2 .

7) Interspecific genome comparisons can help to identify all but of the following except?

A) biological fathers in cases of paternity B) the farms where produce at the supermarket comes from C) differences in nucleotide sequence between identical twins D) variation in nucleotides across a species breeding range E) conserved nucleotides within a genus Answer: E Section: 16.3 Skill: Synthesis/Evaluation LO: S16.3d 8) Celiac disease (CD) is a common, usually inherited human disorder of intestinal inflammation that is triggered by eating gluten, a major protein in wheat and other cereal grains. It is the most frequent food intolerance. One of the three genes involved encodes SH2B3, a protein involved in innate, nonspecific immune response to bacterial pathogens. Celiac sufferers frequently have a particular allele of this gene, an allele so common in the human population that which of the following might be hypothesized?

A) The gene must be common to all primate genomes.

B) The gene must be common in most mammalian genomes.

C) The allele must have been positively selected for during a period of especially frequent bacterial infections.

D) The non-celiac allele must have been selected against in previous human populations.

E) The gene must code for a T-cell receptor for a common bacterial component.

Answer: C Section: 16.3 Skill: Synthesis/Evaluation LO: S16.3b 9) Although whole-genome duplications that result in polyploidy are fairly common in plant species, which of the following are common in other eukaryotes?

A) duplication of whole chromosome arms B) segmental duplications of small gene regions C) deletion of an entire haploid set of chromosomes D) whole chromosome loss E) miscellaneous aneuploidies in living members of a species Answer: B Section: 16.3 Skill: Knowledge/Comprehension LO: S16.3c

3 .

10) To study the maximum amount of genome diversity in humans, which of the following provides the most information?

A) karyotyping B) SNP variation C) DNA fingerprinting D) sequencing of one or more specific genes E) ortholog comparisons Answer: B Section: 16.1 Skill: Knowledge/Comprehension LO: S16.2b 11) Subtelomeres are blocks of homologous DNA sequences close to the ends of chromosomes.

When sequences of chromosomes 4q and 10q subtelomeres were analyzed from a world-wide population, only seventeen 4q haplotypes and and eight 10q haplotypes were found among African, European, and Asian groups, making these the least diverse subtelomeres shared by these ethnic groups. Which conclusion do these data support?

A) finding that few or no interchromosomal exchanges can occur among these chromosomes B) a conclusion that all haplotypes were present before the human population migrated from Africa C) reporting that haplotype polymorphism shows no response to natural selection D) humans have made multiple migrations out of Africa E) a conclusion that assigning chromosomal origin is only possible in a population with the same ethnic origin Answer: B Section: 16.3 Skill: Applying/Analyzing LO: S16.3a 12) Recombination between two Alu elements can result in deletions of chromosome segments.

In a recent study, researchers compared Alu recombination-mediated deletions of humans and chimpanzees. They identified 492 human-specific deletions of this type. Why is this finding important?

A) It shows that humans are not as genetically close to chimpanzees as researchers previously thought.

B) It shows how much the chimpanzee genome has changed since the divergence of these two lineages.

C) It shows that the time of divergence of the two lineages is much older than was previously thought.

D) It shows that more deletions have occurred in humans than in chimps.

E) It demonstrates the importance of Alu recombination in shaping genome divergence.

Answer: E Section: 16.3 Skill: Applying/Analyzing LO: S16.3a

4 .

13) When authors discuss gene annotation, to what do they refer?

A) describing protein expression patters B) gene expression measurements for all coding genes in a genome C) using DNA sequencing to detect genes of unknown function D) comparison of conserved sequences E) use of computers to guess at gene location and function based on comparison to known genes Answer: E Section: 16.4 Skill: Knowledge/Comprehension LO: S16.2b 14) Which of the following describes genome tiling arrays?

A) use of cDNA arrays B) use of RNA-based arrays C) use of one or more arrays that together contain all sequences of a genome D) aligning chromosomes based the sequence of short pieces of DNA E) use of high-throughput DNA sequencing technology Answer: C Section: 16.4 Skill: Knowledge/Comprehension LO: S16.4a 15) What is the purpose of the two-hybrid system?

A) to study the nature of two hybrid organisms at a time B) to study the interaction between an enhancer and a promoter C) to explore a cDNA library to find interacting DNA sequences D) to discover how and whether two proteins interact E) to locate the chromosomal position of hybridized genes Answer: D Section: 16.4 Skill: Knowledge/Comprehension LO: S16.4c 16) Which of the following defines the essential gene set of an organism such as a species of yeast?

A) genes that, when mutated, reduce the growth of a heterozygous organism B) genes whose deletion alleles are lethal C) the fewest genes that an organism needs to divide D) genes whose mutations result in haploinsufficiency E) genes that, when mutated, are conditionally lethal in special environments Answer: B Section: 16.4 Skill: Applying/Analyzing LO: S16.4b

5 .

17) When Warnecke and colleagues were looking for genes that are involved in the digestion of lignocellulose by termites, they needed genes ________.

A) from the sequences of the termite genome B) of the termites' mitochondria C) for the bacterial enzymes found in the lignocellulose-digesting part of the gut D) expressed when a termite is eating E) genes found in cells of the termites mouth Answer: C Section: 16.4 Skill: Applying/Analyzing LO: S16.4b 18) Which is more likely to be the case for gene A found at the hub of a genetic network, compared to gene B, in the same network but not in a hub?

A) gene A is highly expressed B) gene B encodes a protein that interacts with many different proteins C) gene A encodes a transcription factor that controls the expression of one gene D) gene A is a nuclear gene, while gene B is mitochondrial E) gene A controls the expression of many genes Answer: E Section: 16.4 Skill: Applying/Analyzing LO: S16.4a 19) Transcomplementation is a name sometimes given for a test to see whether a gene from one organism can "rescue" a loss-of-function mutant of a different organism with a homologous gene. To which of the following processes might this term apply?

A) replacement of a nonfunctional globin gene with a globin pseudogene B) substitution of a drought-resistant gene from potato plants with a drought-resistant gene from a cactus species C) introduction of a human WEE1 gene into a yeast culture with wee1- mutant yeast D) replacement of a cdc25- with a cdc13+ gene to rescue a cell cycle dysfunction E) use of Drosophila heat shock gene promoters to regulate heat shock genes in the model fish, fugu Answer: C Section: 16.4 Skill: Applying/Analyzing LO: S16.4b

6 .

20) A researcher measures the quantity of cDNA that corresponds to the exons of a particular gene in mice using RNA-seq. They find that RNA for this gene isolated from the brain contains 4 exons, while RNA from the liver heart and skin contains 5 exons. What phenomenon likely explains this result?

A) cells of the brain splice this mRNA differently than other cells in the mouse B) the mRNA in the brain is a pseudogene C) the brain cell chromosome lacks the exon that is present in the other cells of the mouse D) the skin, liver and heart cells differentiate before the brain and so have different exons E) there is a brain-specific allele compared to the other cells in the mouse Answer: A Section: 16.4 Skill: Applying/Analyzing LO: S16.4a 21) A researcher is annotating a newly-sequenced chromosome that they constructed from contigs sequenced from a sample of Antarctic ice. The cell/organism that the chromosome came from is not known, however it yielded a single circular chromosome. You are asked to run a BLAST search to help this researcher identify the most related organism in GenBank. What would be the best strategy?

A) Compare the most unusual portion of this newly-discovered chromosome to al known animal genomes.

B) Perform many careful one-by-one comparisons between the newly-discovered DNA sequence and species selected from GenBank, one at a time.

C) Use several ~1000bp portions of the DNA sequence to search for the closest matches from among the entire available database.

D) Look for a nearly identical match to a very small part (20-40 bp) of the chromosome from all known ice bacteria.

E) Look for an exact match to the whole chromosome.

Answer: C Section: 16.4 Skill: Synthesis/Evaluation LO: S16.2b

7 .

16.2 Short Answer Questions 1) ________ is a strategy that compiles genome sequence from many overlapping DNA sequences each corresponding to a random location in the target genome.

Answer: Shotgun sequencing Section: 16.1 Skill: Applying/Analyzing LO: S16.1a 2) A researcher has produced groups of contigs that are linked via paired-end sequences, although these include some sequence gaps. What are these groups called?

Answer: scaffolds Section: 16.1 Skill: Knowledge/Comprehension LO: S16.1a 3) In 2000, when then-President Clinton along with Francis Collins of the Human Genome Project and Craig Venter of Celera announced the completion of a "draft" of the human genome, the event did not, in fact, represent true completion because most of what types of sequences were not included?

Answer: repetitive Section: 16.1 Skill: Knowledge/Comprehension LO: S16.1b 4) Often, in the absence of experimental data, computerized algorithms are used to predict gene structures from large sequences. What is this approach known as?

Answer: bioinformatics Section: 16.2 Skill: Knowledge/Comprehension LO: S16.2b 5) Human genes often occupy a larger part of the chromosome compared to the homologous gene in yeast, yet the number of amino acids in more homologous human and yeast proteins are not very different. What structure explains this difference?

Answer: introns Section: 16.2 Skill: Applying/Analyzing LO: S16.2a 6) In humans, a large number of genes associated with immune function share sequence similarity and are evolutionarily related. Together these genes are referred to as what kind of group?

Answer: a gene family Section: 16.2 Skill: Applying/Analyzing LO: S16.2a 8 .

7) In a simple bioinformatics exercise, students are asked to locate the beginning of an ORF among the six possible reading frames for a hypothetical DNA molecule. The first step in their process should involve looking for what sequence?

Answer: ATG Section: 16.2 Skill: Applying/Analyzing LO: S16.2a 8) A gene product might be initially hypothesized from comparative genomics but should be confirmed how?

Answer: by experimenting or experimentally Section: 16.2 Skill: Knowledge/Comprehension LO: S16.2b 9) In general, which part of a coding gene shows the most evolutionary conservation?

Answer: exons or codons or coding sequence Section: 16.2 Skill: Knowledge/Comprehension LO: S16.2a 10) Species whose ancestors might once have been free living may now be present as parasites.

Would their genomes now tend to be larger, smaller, or approximately the same in size and complexity?

Answer: smaller Section: 16.1 Skill: Knowledge/Comprehension LO: S16.1b 11) The BLAST program from NIH enables a researcher to look for homologous genes. It does so in general by searching for identity in what?

Answer: nucleotide sequence Section: 16.3 Skill: Knowledge/Comprehension LO: S16.2b 12) Interspecific (between species) genome studies show us conserved sequences, but intraspecific (within species) comparisons identify what?

Answer: (sequence) differences or polymorphisms Section: 16.3 Skill: Knowledge/Comprehension LO: S16.3a

9 .

13) Researchers look for segments of chromosomes, larger than genes, that are similar in arrangement between distant species. Such segments may reveal relatedness and chromosome rearrangements that have happened in the evolutionary past. What are such segments identified by?

Answer: synteny or homology Section: 16.3 Skill: Applying/Analyzing LO: S16.3a 14) An inherited human disorder, familial dysautonomia, results from a nucleotide mutation in the gene IKAP that is expressed in the nervous system. The decreased IKAP protein leads to abnormal development, and the resulting disease is usually fatal by age 30. The nucleotide change alters splicing. If this change affects only the nervous system and not the immune system, in which the gene is also expressed, what feature must be found in this gene?

Answer: alternative splicing Section: 16.3 Skill: Applying/Analyzing LO: S16.2a 15) A gene in a sheep-like species has been found to differ from that in related species by the inclusion of a variable number of tandem repeats of a short sequence. These repeats or duplications arise from what process?

Answer: unequal crossing over or strand slippage Section: 16.3 Skill: Applying/Analyzing LO: S16.1b 16) Mammalian genomes all have a number of genes for globin polypeptides. Several of these genes are functional, but several are never expressed. What are the nonfunctional ones known as?

Answer: pseudogenes Section: 16.3 Skill: Knowledge/Comprehension LO: S16.3b 17) Conserved noncoding sequences (CNGs) are usually sequences involved in what kinds of functions?

Answer: regulatory or regulation Section: 16.3 Skill: Knowledge/Comprehension LO: S16.3b

10 .

18) ________ can lead to the evolution of a gene with new functions while maintaining the ancestral function.

Answer: gene duplication Section: 16.3 Skill: Knowledge/Comprehension LO: S16.3c 16.3 Fill-in-the-Blank Questions 1) The set of transcripts present in a cell or organism is called the ________.

Answer: transcriptome Section: 16.4 Skill: Knowledge/Comprehension LO: S16.4a 2) Genome-wide expression patterns can be studied using ________.

Answer: microarrays or RNA sequencing Section: 16.4 Skill: Knowledge/Comprehension LO: S16.4a 3) If an entire genome is fragmented and then a large number of pieces are sequenced, the approach is known as the ________ approach.

Answer: shotgun Section: 16.1 Skill: Knowledge/Comprehension LO: S16.1a 4) Conservation of DNA sequence within an intron across wide taxa suggests that this sequence is important for ________.

Answer: Gene expression or gene regulation Section: 16.3 Skill: Knowledge/Comprehension LO: S16.3a 5) A genome sequence is considered to be complete when all its ________ sequences are included, as well as reasonable amounts of repetitive sequences.

Answer: euchromatic Section: 16.1 Skill: Knowledge/Comprehension LO: S16.2a

11 .

6) Sequencing of DNA from a community of organisms rather than from a single individual is known as ________.

Answer: metagenomics Section: 16.1 Skill: Knowledge/Comprehension LO: S16.1c 7) The process of finding the location of genes and functional (e.g., regulatory) sequences and their functions within a genome sequence is called ________.

Answer: annotation Section: 16.2 Skill: Knowledge/Comprehension LO: S16.2b 8) The use of computational approaches to understand DNA sequence information is called ________.

Answer: bioinformatics Section: 16.2 Skill: Knowledge/Comprehension LO: S16.2b 9) The movement of genes from one species into the genome of another is referred to as ________.

Answer: lateral gene transfer Section: 16.3 Skill: Knowledge/Comprehension LO: S16.3b 10) Finding only the coding sequence of an ancestral gene, no introns or promoter, in a new location in a genome suggests what mechanism of gene birth gave rise to this sequence?

Answer: reverse transcription Section: 16.3 Skill: Knowledge/Comprehension LO: S16.3b

12 .

16.4 Essay Questions 1) Compare and contrast the two approaches to genome sequencing (whole-genome shotgun sequencing and clone-by-clone sequencing).

Answer: Both approaches seek to completely cover an organism's entire genome and both require piecing together overlapping sequences.

WGS overlaps sequences after random fragmentation and sequencing of those fragments.

CBC use sequences from already overlapping chromosomal fragments.

Section: 16.1 Skill: Applying/Analyzing LO: S16.1a 2) Explain the role that gene duplication may play in genome evolution.

Answer: Gene duplication results in the generation of redundant gene function, which can lead to a variety of later events. For instance, duplicated genes can retain their original function, however now with more gene product available to the cell. Duplicated genes can also evolve new functions, sometimes with the original function maintained by one of the duplicated genes.

One duplicate can lose its function, becoming a pseudogene. Gene duplication can increase the size of gene families.

Section: 16.3 Skill: Synthesis/Evaluation LO: S16.3b 3) What evidence has been found from genomics that support the conclusion that whole genome duplication has happened in the past?

Answer: Genomics has identified large sections of chromosomes that have homologous sections on other chromosomes. These segments are larger than a few genes and can be the size of whole chromosomes, but they can also show rearrangements in their orientation and gene order. There is also a phylogenetic signature of such duplications, where more closely related species share more similarity in the arrangement and number of chromosome segments, meaning that this phenomenon is a product of evolution.

Section: 16.3 Skill: Synthesis/Evaluation LO: S16.3c 4) Provide a reasonable explanation for the discovery of a gene in the human genome that has no homolog in chimpanzees, but does have homologs with E. coli and other bacteria.

Answer: This gene was likely transferred into the human genome from bacterial via lateral gene transfer, in which, by some mechanism, E. coli DNA was introduced to human cells and integrated into the nuclear DNA. Possible routes for E. coli to human cells include via transduction with a human virus that integrated bacterial DNA. Section: 16.1 Skill: Synthesis/Evaluation LO: S16.3b

13 .

5) How and why is the two-hybrid system used to find where two proteins interact?

Answer: Two-hybrid screening is used to detect protein-protein or protein-DNA interactions by testing for physical binding. You will look for the activation of a reporter gene when a transcription factor binds to an upstream activating sequence. In a two-hybrid screen, the transcription factor is broken into two fragments: the binding domain (BD), which binds to the UAS, and the activation domain (AD), which activates transcription of the reporter gene. For instance, the yeast two-hybrid screen uses a protein that is translationally fused to a Gal4 BD and another protein to a Gal4 AD. If the proteins interact, the two Gal4 elements are brought together and transcription occurs.

Section: 16.3 Skill: Synthesis/Evaluation LO: S16.4c 6) Under what circumstances is a metagenomic analysis preferable to analysis of a gene or genes from a single species?

Answer: Metagenomic analyses are preferred when you are interested in studying the diversity in a given environment, and they enable researchers to look for sequences found in organisms that make up a small part of a biotic community. Analysis of a gene from a single species is helpful when you wish to study the role of that specific gene, but does not provide insight into genetic diversity or homologous genes in other species.

Section: 16.3 Skill: Synthesis/Evaluation LO: S16.1c

14 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 17 Cytoplasmic Inheritance and the Evolution of Organelle Genomes 17.1 Multiple-Choice Questions 1) Sigma virus infection in Drosophila melanogaster causes flies to become paralyzed and then die in high concentrations of CO2. The virus is passed on extrachromosomally. Two isogenic strains of Drosophila cultures are maintained: R (resistant to CO2 or wild type) and S (susceptible to CO2, infected with sigma). After crosses between male R and female S flies, subsets of offspring of each sex are tested for CO2 sensitivity. Which of the following results do you expect?

A) All the offspring are sensitive.

B) All the offspring are female.

C) All the offspring are male.

D) All the offspring are sensitive, but only males pass it on.

E) Only male offspring backcrossed to the female parent are sensitive.

Answer: A Section: 17.1 Skill: Applying/Analyzing LO: S17.2b 2) A cell or organism in which all copies of the cytoplasmic organelle gene are the same is said to be ________.

A) homozygous B) heteroplasmic C) homoplasmic D) variegated E) uniparental Answer: C Section: 17.1 Skill: Knowledge/Comprehension LO: S17.1a 3) Random segregation of organelles during cell division in a multicellular species is likely to result in an increased rate of which of the following?

A) incomplete penetrance B) dominant loss-of-function mutations C) dominant gain-of-function mutations D) chromosomal duplication mutations E) chromosomal deletion mutations Answer: A Section: 17.2 Skill: Synthesis/Evaluation LO: S17.2c

1 .

4) The human egg has about 2000 mitochondrial genomes, but somatic cells have a range of hundreds to thousands. Which of the following is most likely to account for the difference?

A) Heteroplastic cells can give rise to homoplastic cells.

B) Large mitochondria in oocytes can divide into many smaller mitochondria, and smaller mitochondria can replicate their DNA. C) Somatic cells replicate their nuclear DNA at a faster rate than the oocyte.

D) Mutant mitochondria replicate at different frequencies than do wild type.

E) Haploid organelle genomes can replicate to form diploid or polyploidy genomes.

Answer: B Section: 17.2 Skill: Applying/Analyzing LO: S17.1b 5) In certain Latin American countries during periods of political volatility, children were removed from their families and their parents were executed (then referred to as being "disappeared"). The methods with the best chance of reuniting those children with members of their extended family have used mtDNA matching. Why and how?

A) Remains of the children's mothers can be used to match their mtDNA with that of their children.

B) Children's mitochondria will have genomes that are 1/4 identical to each of their grandparents.

C) Children's mitochondria will have genomes that are 1/2 identical to their maternal grandparents.

D) Children's mitochondria will have genomes that are identical to those of their maternal grandmothers.

E) Children's mtDNA will be identical to those of their first cousins.

Answer: D Section: 17.2 Skill: Applying/Analyzing LO: S17.5b 6) The noncoding regions of mammalian mitochondrial genomes have little to no selective pressure. Which of the following are consequences of this condition?

A) a rate of mtDNA repair is much lower than that of nuclear sequences B) a high rate of loss of mitochondrial genomes relative to nuclear sequences C) a high rate of natural selection for genomes identical to those of the mother D) noncoding mitochondrial DNA increases in nucleotide diversity more quickly than coding DNA E) a rate of mutation that can increase with successive generations in a maternal line Answer: D Section: 17.2 Skill: Applying/Analyzing LO: S17.5b

2 .

7) In Leber's disease (LHON), degeneration of the optic nerve results in blindness due to a defect in NADH dehydrogenase affecting electron transport. If this is due to a mutation in a mitochondrial gene, which combination of the following effects would you expect to see when doing a pedigree analysis of a family with LHON? i. variable expressivity ii. incomplete penetrance iii. progressive blindness over time iv. maternal inheritance v. diminished muscle mass A) i and ii only B) ii and iii only C) iii and iv only D) i, ii, iii, and v only E) i, ii, iii, and iv only Answer: E Section: 17.2 Skill: Applying/Analyzing LO: S17.3 8) In 1718, the tsar of Russia, his wife the tsarina, and their five children were presumably assassinated and buried in an unmarked grave. Later, several women came forward and claimed that they were in fact Anastasia, one of the daughters whose bones had not been found. More recently, forensic scientists have extracted DNA, including mtDNA, from the skeletal remains.

Which of the following samples of mtDNA had to be analyzed to discern which, if any, of the claimants was Anastasia?

A) the tsar, tsarina, and the other assassinated children B) each of the claimants or their living children C) the tsarina and the claimants D) the tsar and tsarina only E) the tsar and tsarina and living relatives of each of them Answer: C Section: 17.2 Skill: Applying/Analyzing LO: S17.2b

3 .

9) In a cross between a petite yeast mutant and a wild-type strain, results showed half petite and half wild-type progeny. Which of these statements must be true of this petite mutation?

A) It is suppressive.

B) It is a nuclear or segregational mutation.

C) It is a neutral mitochondrial mutation.

D) It replicates faster than wild type.

E) It has small deletions of mitochondrial DNA. Answer: B Section: 17.2 Skill: Applying/Analyzing LO: S17.2a 10) Chlamydomonas, a unicellular green algae, is sensitive to an antibiotic, erythromycin. If mtmating-type alga that is sensitive is mated with an mt+ cell that is resistant, and if the mt+ strain donates chloroplasts, what results do you expect and why?

A) All progeny are sensitive because sensitivity is the dominant trait.

B) Approximately half the progeny are sensitive because this is the recessive trait.

C) All the progeny are sensitive because this is the recessive trait.

D) All progeny are resistant because cpDNA is responsible for the resistant phenotype.

E) Approximately half the progeny are sensitive because cpDNA is responsible.

Answer: D Section: 17.2 Skill: Applying/Analyzing LO: S17.2a 11) Myoclonal epilepsy and ragged red fiber disease (MERRF) is a human condition named for the ragged red fibers of skeletal muscle cells and myoclonic epilepsy in affected individuals.

People with this disorder have a mutation in a mitochondrial gene for a tRNA, specifically that for lysine. Affected individuals are heteroplastic. Why? A) Interference with translation is not fully rescued by importing lysine from the cytoplasm.

B) Homoplastic organelles would be inviable due to failures in translation.

C) Lysine is required in translation of every one of the mitochondrial genes.

D) Mutations of this gene affect individuals before birth.

E) Mutations of this gene are dominant.

Answer: B Section: 17.3 Skill: Applying/Analyzing LO: S17.1a

4 .

12) All of the following are evidence in support of the endosymbiotic origin of mitochondria, except?

A) Mitochondria maintain their own DNA. B) Yeast cells can live without mitochondria.

C) Mitochondiral genetic codes (codons) are related to each other across phyla, and are distinct from nuclear genetic codes.

D) Transcription in mitochondria is performed by an enzyme similar to bacterial RNA polymerase.

E) Mitochondria have a double membrane.

Answer: B Section: 17.5 Skill: Applying/Analyzing LO: S17.5a 13) Two databases, OMIM and Mitomap, are especially useful in exploring human mitochondrial mutations. If you are interested in learning whether aminoglycoside-induced deafness is inherited maternally, and what gene(s) is/are involved, which of the following would give you the most information?

A) the Mitomap position map with gene names B) the sequence with specifically altered amino acid coding C) the OMIM description of symptoms and diagnosis D) the OMIM description of the TRMU gene at 22q13 E) the Mitomap loci of genes associated with deafness Answer: E Section: 17.3 Skill: Applying/Analyzing LO: S17.2a 14) There is a hypothetical gene in mice that produces a substance that induces twitchiness in hind leg muscles. A female mouse of a true-breeding twitchy strain is mated with a male of a true-breeding non-twitchy strain. All progeny are twitchy. Twitchiness may simply be a dominant trait caused by a nuclear allele, or it could be due to mtDNA. Which of these procedures would you use to provide evidence for your hypothesis that twitchiness is due to mtDNA? A) Sequence the mtDNA of the twitchy mother.

B) Breed twitchy male progeny with normal females to show no paternal inheritance.

C) Breed twitchy female with twitchy male F1 mice to verify there is maternal inheritance.

D) Backcross twitchy female offspring with normal fathers.

E) Sequence the mtDNA of the twitchy male offspring.

Answer: B Section: 17.3 Skill: Synthesis/Evaluation LO: S17.2b

5 .

15) The accompanying hypothetical pedigree is for a family where some members have symptoms of Leigh syndrome, which is caused by an mtDNA mutation that reduces ATPase activity.

Which of the following in the youngest generation might transmit the condition in the future?

A) any of the women B) any of those affected C) any of the members of this generation D) only those with an affected mother E) any of the affected women Answer: E Section: 17.3 Skill: Applying/Analyzing LO: S17.2b 16) The human genome contains DNA that is similar to mtDNA. Some of this DNA shows significant levels of nucleotide divergence from human mtDNA, while other nuclear DNA shows very high levels of nucleotide identity with mtDNA. Which of the following hypotheses best explains these data?

A) mtDNA and nuclear DNA share many sequences by chance.

B) Nuclear DNA is older than mtDNA. C) Nuclear DNA has a low level of mutation compared to the mtDNA. D) mtDNA is derived from the nuclear DNA with higher levels of homology.

E) mtDNA has integrated into nuclear chromosomes several times over a long span of time.

Answer: E Section: 17.5 Skill: Synthesis/Evaluation LO: S17.5b

6 .

17) Organisms have been created by nuclear transfer cloning, in which a diploid nucleus from a somatic cell of an adult animal is injected into an enucleated egg cell to create an embryo.

Among the many species thus cloned, there has been a prevalence of mitochondrial defects.

Which of the following is the most likely explanation?

A) The transferred nucleus may destroy the egg mitochondria.

B) The egg cell may be heteroplastic before injection.

C) The somatic cell's mitochondria may have been transferred as well.

D) The transferred nucleus may have mutations that are incompatible with the egg cell's mitochondrial function.

E) The egg nucleus may have had one or more mutations that alter mitochondrial function.

Answer: D Section: 17.3 Skill: Synthesis/Evaluation LO: S17.2b 18) Giardia is eukaryotic yet is long thought to lack a mitochondira. Recently, researchers have found that each Giardia cell contains double membrane-bound organelles similar to a mitochondria called mitosomes that lack mtDNA. What is the best explanation for mitosomes?

A) Mitosomes are the ancestral form of the mitochondria.

B) Mitochondria in Giardia evolved into mitosomes and have lost or transferred their DNA to the Giardia nuclear genome.

C) Mitosomes evolved from another Giardia organelle that lacked DNA. D) Mitosomes are more advantageous to eukaryotes than mitochondria.

E) Giardia are asexual and thus do not require mitochondria with its own DNA. Answer: B Section: 17.5 Skill: Synthesis/Evaluation LO: S17.5a 19) A consequence of organelle heredity is that sequence changes in mtDNA vary more than nuclear DNA in response to reduction in population size. Which of the following does this imply about human evolutionary history?

A) Current human populations must have a much higher mutation rate than ancient populations.

B) Mitochondrial variants in people descending from originally small populations show enough diversity to be used to construct an evolutionary tree.

C) The original human population must have grown to be quite large before its members began migrating to other lands and other continents.

D) Nuclear gene sequencing must be used to resolve the question of human ancestral origins.

E) A single human population, living about 200,000 years ago, could not have given rise to all current human populations.

Answer: B Section: 17.5 Skill: Synthesis/Evaluation LO: S17.5b

7 .

17.2 Short Answer Questions 1) Among the discoveries that Correns and Bauer (1908) made in plants was that, sometimes, reciprocal crosses resulted in what pattern of inheritance?

Answer: maternal inheritance Section: 17.1 Skill: Knowledge/Concepts LO: S17.2a 2) When a cell divides, chloroplasts are randomly distributed. What is this called?

Answer: replicative segregation Section: 17.1 Skill: Knowledge/Concepts LO: S17.1a 3) Variation in numbers of mutated chloroplast genomes can give rise to green, white, or variegated leaves. What color leaves do you expect in offspring of pollen from a white-leafed plant and ovules from a green-leafed plant?

Answer: green Section: 17.1 Skill: Applying/Analyzing LO: S17.1a 4) What is the name of the area of the organelle in which organelle DNA is packaged?

Answer: nucleoid Section: 17.1 Skill: Knowledge/Comprehension LO: S17.4 5) Which factor in organelle replication immediately precedes "organelle division"?

Answer: nucleoid division Section: 17.1 Skill: Knowledge/Comprehension LO: S17.4 6) In mammals, the number of mitochondria per cell is highly variable. Which of the following would have the largest number: skin cells, red blood cells, or skeletal muscle cells?

Answer: skeletal muscle (cells) Section: 17.1 Skill: Applying/Analyzing LO: S17.1b

8 .

7) Chloroplast mRNAs are not 5'-capped or polyadenylated, making them more like RNAs of the domain ________.

Answer: bacteria Section: 17.2 Skill: Knowledge/Concepts LO: S17.4 8) Is the rate of mutation in mitochondrial DNA of mammals greater or less than that of the species' nuclear DNA? Answer: greater than Section: 17.2 Skill: Knowledge/Concepts LO: S17.5b 9) Allan Wilson showed that the mtDNA of Africans is more diverse than that of other human populations. Does this finding provide evidence for the multiregional (MRE) model or the recent African origin (RAO) model?

Answer: RAO Section: 17.2 Skill: Knowledge/Concepts LO: S17.5b 10) Three of the four modes of inheritance of organelle genes are • maternal inheritance as in mammals, • paternal inheritance as in gymnosperms, and • selective degradation of one source during mating.

What is the fourth?

Answer: biparental inheritance Section: 17.2 Skill: Knowledge/Concepts LO: S17.2a 11) Mating in Chlamydomonas occurs between haploid mt+ and mt- individuals. The mtchloroplast genome is selectively degraded 95% of the time. In one such mating, the mt+ genome is strR and mt- is strS. Will offspring live or die if they are exposed to streptomycin?

Answer: live Section: 17.2 Skill: Applying/Analyzing LO: S17.2b 12) Mitochondrial genome transcription, translation, and replication occur where?

Answer: in the (mitochondrial) matrix Section: 17.3 Skill: Knowledge/Concepts LO: S17.4 9 .

13) In one or two words, give two characteristics of mitochondrial genomes.

Answer: Any two of the following: Circular or linear no histones/chromatin anchored (to the mitochondrial membrane) variable in size Section: 17.3 Skill: Knowledge/Concepts LO: S17.4 14) What name is given to the alteration of an RNA sequence after transcription?

Answer: RNA editing Section: 17.4 Skill: Knowledge/Concepts LO: S17.3 15) The protein-coding genes of chloroplasts are involved in what part of photosynthesis?

Answer: oxidative phosphorylation Section: 17.4 Skill: Knowledge/Concepts LO: S17.4 16) Do the protein-coding parts of organelle genes more closely resemble those of Bacteria (Eubacteria) or Archaea?

Answer: bacteria Section: 17.5 Skill: Knowledge/Concepts LO: S17.5a 17) What are the closest "relatives" of chloroplasts still found as free-living organisms today?

Answer: cyanobacteria Section: 17.5 Skill: Knowledge/Concepts LO: S17.5a 18) Sequencing of genomes of eukaryotes, both mitochondrial and nuclear, have revealed evidence of sequence transfer from one to the other. Are such transfers ancient, recent, or both?

Answer: both Section: 17.5 Skill: Knowledge/Concepts LO: S17.5a

10 .

19) If a chloroplast coding sequence has been transferred to the nuclear genome, but the protein product is needed in the chloroplast, what must be added to the original polypeptide sequence?

Answer: signal sequence Section: 17.5 Skill: Knowledge/Concepts LO: S17.4 20) DNA transfer has occurred many times between organelle and nuclear genomes. Two pairs of nuclear and mitochondrial DNA sequences are analyzed. Pair A has more similarities than pair B. Which of these is the most ancient transfer?

Answer: Pair B Section: 17.5 Skill: Applying/Analyzing LO: S17.5a 17.3 Fill-in-the-Blank Questions 1) Any form of inheritance that does not follow Mendelian patterns and that involves most of the cytoplasm being contributed to the embryo by one of the parents is called ________.

Answer: extrachromosomal (inheritance) or uniparental inheritance Section: 17.1 Skill: Knowledge/Comprehension LO: S17.1a 2) The organelles involved in organelle heredity are ________ and ________.

Answer: chloroplast; mitochondria (the reverse order is also correct) Section: 17.1 Skill: Knowledge/Comprehension LO: S17.2a 3) If a cell includes a mixture of variable numbers of wild-type and mutant organelles, this condition is known as ________.

Answer: heteroplasmy Section: 17.1 Skill: Knowledge/Comprehension LO: S17.1a 4) In the 1950s, mitochondrial genomes were discovered by differential staining for ________.

Answer: DNA Section: 17.1 Skill: Knowledge/Comprehension LO: S17.1a

11 .

5) Mitochondrial mutations in mammals are passed to the offspring by the ________ parent.

Answer: female or maternal Section: 17.2 Skill: Knowledge/Comprehension LO: S17.2b 6) Transcriptional regulation of mitochondrial genes closely resembles regulation of ________ operons.

Answer: bacterial Section: 17.3 Skill: Knowledge/Comprehension LO: S17.4 7) The plastids can differentiate into ________ that carry out photosynthesis.

Answer: chloroplasts Section: 17.4 Skill: Knowledge/Comprehension LO: S17.4 8) Chloroplast RNA undergoes ________, which is the alteration of the RNA sequence after transcription.

Answer: RNA editing Section: 17.4 Skill: Knowledge/Comprehension LO: S17.3

12 .

17.4 Essay Questions 1) If the mother and grandmother of a given family both exhibit phenotypic effects of a mitochondrial disorder, how is it possible that the mother's three children each have different phenotypes from one another and different degrees of severity?

Answer: The mother's and children's cells are heteroplasmic. Phenotypes are directly dependent on the number or percentage of mutated mitochondria per cell. Thus, cells that inherit more of the mitochondria with the mutation will show a more severe phenotype than those with enough normal mitochondria to maintain the wild-type function of the gene. If, during embryonic development, the cells that gave rise to a particular organ or tissue are derived from a cell with more of the mutant mitochondria, that individual will inherit the phenotype associated with that particular tissue defect. Similarly, if the cell that gave rise to a particular tissue or organ had mostly normal mitochondria, that individual may be phenotypically normal with regard to that tissue or organ.

Section: 17.2 Skill: Synthesis/Evaluation LO: S17.2b 2) What makes it possible for human mitochondria, which have only 22 sequences that code for tRNA, to translate mitochondrial polypeptides that require 32 tRNAs? Answer: Thanks to the third-base wobble, many of the tRNAs can recognize codons using the first two bases of the codon. In addition, many nuclear genes can be imported into mitochondria and the exchange of genetic information between the nucleus and mitochondria is required for proper mitochondrial function. Therefore, some of mitochondrial mRNAs are translated using nuclear-encoded tRNAs. Section: 17.3 Skill: Synthesis/Evaluation LO: S17.3 3) Contrast the mechanism that gives rise to tortoise-shell cats (a mosaic coat color) with the mechanism that gives rise to variegated plants.?

Answer: Tortoise shell cats and variegated plants are both genetic mosaics, however they are caused by two different mechanisms. Tortoise shell cats are female cats that are heterozygous for X-linked alleles that influence coat color. Early in development, one of each pair of X chromosomes forms a Barr body and expression of genes from that chromosome is diminished, leaving the coat color determined by the allele on the other X-chromosome. Patches of epidermis in these cats arise as the progeny of cells with alternative X chromosomes inactivated.

Variegated plants are caused by heteroplasmic cells, which have plastids green plastics and nongreen plastids. Segregation of these two populations of plasmids during somatic cell division leads to patches of plant tissue that are either green or white, depending on which type of plastid becomes homoplasmic.

Section: 17.3 Skill: Applying/Analyzing LO: S17.2a

13 .

4) How can you establish that, in a newly described species of plant, a particular phenotype is due to a chloroplast rather than a nuclear gene?

Answer: Mutations in chloroplast genes violate Mendel's Law of Segregation in that they show uniparental inheritance patterns. In addition, the irregular segregation of chloroplasts during cell division can result in variable segregation patterns, producing a variegated phenotype due to unequal chloroplast distribution in leaf cells.

Section: 17.4 Skill: Applying/Analyzing LO: S17.2a 5) In a plant, analysis of the Rubisco protein complex reveals that, in some complexes, rbcL is deficient while rbcS is normal. In other Rubisco complexes, both rbcL and rbcS are normal.

What does this indicate about the two genes encoding these subunits?

Answer: rbcS is encoded by a nuclear gene while rbcL is encoded by a chloroplast gene. The plant is heteroplasmic in that in some chloroplasts, there is a mutation in the gene encoding the rbcL and in other chloroplasts, the rbcL gene is normal.

Section: 17.4 Skill: Synthesis/Evaluation LO: S17.5a 6) Explain the distinction between primary and secondary endosymbiosis.

Answer: Primary endosymbiosis is when descendants of a free-living species without endosymbionts evolves into the endosymbiont of another species. Secondary endosymbiosis is where a free-living species with its own endosymbiont(s) is enveloped by a second host species.

After a period of gene transfer, the endosymbionts from the once free-living species become the endosymbionts of the second host, resulting in the transfer of endosymbiont genomes (mitochondria or chloroplasts) across phylogenetically-distinct host taxa.

Section: 17.5 Skill: Synthesis/Evaluation LO: S17.5a 7) What is the argument that, although human mitochondria all descend from a single ancestral population, this may not be true for nuclear genomes?

Answer: Human mitochondria are inherited maternally while nuclear DNA is inherited biparentally, leading to sexual exchange of nuclear chromosomes and recombination. It is essentially impossible for mitochondrial DNA to undergo recombination due to sexual exchange and is thus more likely to become fixed (monomorphic) in a population compared to the nuclear chromosomes which can more readily undergo admixture wherever distinct genotypes interbreed.

Section: 17.5 Skill: Synthesis/Evaluation LO: S17.5b

14 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 18 Developmental Genetics 18.1 Multiple-Choice Questions 1) There is much controversy in the news—both scientific and popular—about the use of embryonic stem cells for research. These cells are grown as cell lines in culture, created by taking the inner cell mass from 5-day-old embryos. What is the center of the controversy?

A) the fact that the embryos are mostly those unused by in vitro clinics B) the knowledge that these cell lines have reverted spinal cord injuries in mice C) the unproven usefulness of stem cells to treat human disorders D) the realization that these cells are pluripotent rather than totipotent E) the difference of opinion about whether these embryos are persons Answer: E Section: 18.1 Skill: Synthesis/Evaluation LO: S18.1 2) Which of the following is not a mechanism of differentiation?

A) induction B) inhibition C) apoptosis D) nuclear migration E) morphogen secretion Answer: C Section: 18.1 Skill: Knowledge/Comprehension LO: S18.2a 3) What is the name of the Drosophila multinucleated cell formed early in development?

A) syncytium B) blastocoel C) meristem D) homeobox E) blastoderm Answer: A Section: 18.1 Skill: Knowledge/Comprehension LO: S18.2b

1 .

4) In Drosophila, a developmental mutation called ssa (spineless aristapedia) results in flies with mini leg parts on the antennae. What kind of gene does this mutation affect?

A) a gap gene B) a pair-rule gene C) a homeotic gene D) a segment polarity gene E) a maternal effect gene Answer: C Section: 18.2 Skill: Applying/Analyzing LO: S18.2b 5) A pair-rule gene in Drosophila called runt controls nervous system formation. In the mouse, runt controls blood cell formation and genital structures. In humans, the analog of runt, when mutated, causes children to lack collar bones, and the opening at the top of the skull fails to close. What kind of experiment could tell you whether these genes are conserved in function?

A) sequencing of the runt genes from each of the three species B) an in vivo experiment that replaces the coding sequence of Drosophila runt gene with the coding sequence from the homologous mouse runt gene to see if the mouse gene allows the nervous system to develop.

C) an in vitro experiment that identifies the DNA binding sites of the Drosophila runt gene.

D) sequencing the proteins encoded for by each of the species' runt genes E) use of immunological methods to identify the target cells of the species' runt transcription factors Answer: B Section: 18.2 Skill: Synthesis/Evaluation LO: S18.2a 6) An Antennapedia (Antp) mutant in Drosophila has legs in place of antennae. This is caused by ________.

A) transcription of the Antp gene in the head region instead of in the thorax B) the product of a head segment gene being over-expressed in the head C) a chromosomal error that shifts this Antp gene to the anterior end of the complex D) the interaction of an inhibitor gene of the bithorax complex with the Antp mutation E) loss of function of the Antp gene Answer: A Section: 18.2 Skill: Applying/Analyzing LO: S18.2c

2 .

7) Human Hox genes are in four clusters: HOXA, HOXB, HOXC, and HOXD, for a total of 39 genes. Besides conserving the clustering of genes, evolution has also conserved ________.

A) their chromosomal locations B) the order of the genes within the clusters C) their expression along the dorsal-ventral axis D) the number of HOX genes E) their bilateral symmetry Answer: B Section: 18.2 Skill: Applying/Analyzing LO: S18.2b 8) Segmentation genes in Drosophila act in which of the following orders?

A) segmental, helical, spherical B) pair-rule, transdeterminal, gap genes C) gap, pair-rule, segment polarity D) pair-rule, segment polarity, gap genes E) pair-rule, segment polarity, Hox genes Answer: C Section: 18.2 Skill: Applying/Analyzing LO: S18.2b 9) As a Drosophila larva molts and becomes a pupa, most of the larval tissues die and adult structures are formed from cells in clusters known as imaginal discs that have remained unchanged throughout the larval stages. Therefore, the disc cells are ________.

A) maternal B) zygotic C) embryonal D) differentiated E) transcriptionally activated Answer: C Section: 18.2 Skill: Applying/Analyzing LO: S18.1

3 .

10) In an organism such as a sea urchin, administration of actinomycin D prevents all mRNA synthesis. However, early development up to gastrula formation can take place in the presence of actinomycin D but gastrulation cannot. This implies that ________.

A) development beyond gastrulation requires actinomycin D B) actinomycin D is part of the normal environment of a sea urchin and serves as a developmental signal molecule C) mRNA synthesis is not required after gastrulation D) maternal effect genes regulate development until gastrulation E) immature sea urchin embryos produce actinomycin D to regulate further development Answer: D Section: 18.2 Skill: Applying/Analyzing LO: S18.2c 11) Many mutations in the bithorax gene are known, such as postbithorax, contrabithorax and bithoriaxoid. Each mutation has a different effect on development of the reduced wing-like T3 appendage, the haltere. None of these mutations occurs in the coding sequence of bithorax, instead, where are these mutations and how do they effect development?

A) These mutations alter the slice sites for bithorax and affect the splicing of the bithorax mRNA. Their different effects on development are cause by affects on alternative splicing.

B) Each mutant alters a different codon within the bitorax gene, which affects the DNA-binding behavior of the bithorax protein.

C) These mutations each effect silent sites and effect the stability of the bithorax mRNA. D) Bithorax is inhibited by zygotic genes from the mother and these mutations affect maternal genes.

E) These mutations each effect a different enhancer(s) near the bithorax gene and the resulting misexpression of bithorax in each mutant causes patterning defects in the haltere.

Answer: E Section: 18.2 Skill: Applying/Analyzing LO: S18.2c 12) Polycomb group (PcG) gene mutants cause ectopic expression of Antennapedia (Antp) and other homeotic genes. The PcG protein encoded by the PgG gene has a domain that binds to methylated histone H3. This finding supports which of the following hypotheses?

A) that PcG binds to Antp genes and serves as a transcription activator B) that PcG mutants lack Antp mRNA compared to the wildtype C) that PcG repress transcription by interacting with chromatin D) that PcG activates Antp genes in tissues E) that PcG is allosteric activator of RNA polymerase II Answer: C Section: 18.2 Skill: Synthesis/Evaluation LO: S18.2b

4 .

13) Mutations of regulatory elements of developmental genes may have either dominant or recessive effects. If a mutation deletes a silencer upstream of a HOX gene, which do you expect?

A) a recessive effect in which cis HOX genes are up-regulated B) a recessive effect in which cis HOX genes are down-regulated C) a dominant effect which down-regulates HOX gene activity D) a dominant effect in which cis HOX gene activity is up-regulated E) a dominant trans effect on both alleles of the HOX gene Answer: D Section: 18.2 Skill: Synthesis/Evaluation LO: S18.3b 14) The abdominal segments of Drosophila have no appendages. Loss of function of bithorax genes results in appendages forming in these abdominal segments. Therefore, ________.

A) bithorax mutations are necessarily recessive B) bithorax genes normally activate appendage-forming genes C) ultrabithorax usually acts to oppose bithorax genes D) bithorax normally activates a gene that negatively regulates appendages in the abdomen E) expression of bithorax wild-type genes suppresses all appendage formation Answer: D Section: 18.2 Skill: Synthesis/Evaluation LO: S18.3b 15) Since mice embryos with loss-of-function Hox mutations result in homeotic alterations, then introduction of such a mutation might cause ________.

A) ribs on normally rib-less vertebrae B) pointed rather than rounded ears C) a change in fur coloration along the dorsal midline D) behavioral alterations causing limb shaking E) paralysis of hindlimbs but not forelimbs Answer: A Section: 18.2 Skill: Applying/Analyzing LO: S18.3b

5 .

16) Drosophila mutations such as bcd, if homozygous in the mother, produce offspring with lethal defects, no matter what the genotype of the offspring. For some other genes, the mutant phenotype can be "rescued" by introducing a wild-type allele into the embryo. What is the reason for these different outcomes?

A) Non-rescuable mutations are of paternal origin.

B) Rescuable mutations act as dominants.

C) bcd is a maternal effect mutation.

D) Rescue requires recombination between the maternal and introduced alleles.

E) bcd is a Hox gene.

Answer: C Section: 18.2 Skill: Applying/Analyzing LO: S18.3a 17) Aniridia is a human condition in which the eye has no iris. The protein encoded by the gene responsible for aniridia is very nearly identical to the protein product of the fly Eyeless protein.

What experiment could provide evidence that the two genes are functionally equivalent?

A) Replace the fly eyeless gene with the coding sequence of the mouse aniridia gene. Look for rescue compared to an eyeless gene mutant.

B) Introduce the aniridia mutation into Drosophila embryos to look for iris formation.

C) Sequence the Eyeless and aniridia DNA sequences and regulators.

D) Use the Eyeless mRNA as a probe in other invertebrate and non-mammalian species.

E) Introduce the mouse aniridia wild-type sequence into the fly egg to see whether mouse eyes develop.

Answer: A Section: 18.2 Skill: Applying/Analyzing LO: S18.3a 18) In early vertebrate embryonic development, a number of pathways, such as Wnt, Hedgehog, and Notch, regulate differentiation and organ formation. They operate by ________.

A) repressing translation of developmental RNAs B) transcription activation due to interaction among transcription factors C) cell signaling mediated by receptor binding D) establishing anterior-posterior and dorsal-ventral symmetry E) regulating the expression of MADS box genes Answer: C Section: 18.3 Skill: Applying/Analyzing LO: S18.2b

6 .

19) Use of double mutants has been invaluable in studying developmental pathways. In C. elegans, sex is determined by the ratio of sets of autosomes to X chromosomes. A diploid male has one X, and diploid hermaphrodites have two X's. The loss-of-function gene her-1 produces XO worms that are hermaphrodites, but XX worms are unaffected. Mutation of a second gene, tra, causes the XX to develop as male. Given this information, what would be the phenotype of a tra, her-1 double mutant?

A) a male nematode B) a female nematode C) a hermaphrodite D) an intersex variant E) lethality Answer: A Section: 18.3 Skill: Applying/Analyzing LO: S18.3a 20) Suppose that a homeotic gene from Drosophila is introduced into an Arabidopsis embryo that has a pre-existing mutation in one of its homeotic flower genes. What effect would you expect and why?

A) rescue (or negating) of the phenotype of the mutant because the fly Hox gene supplies the missing protein B) correction of the sequence in the mutant by recombination with the Drosophila gene sequence C) expression of the Drosophila homeotic gene correcting the mRNA of the mutant plant D) a mixture of roughly equal numbers of Drosophila and plant gene product, producing an intermediate phenotype E) no effect, because Hox genes are different from MADS box genes Answer: E Section: 18.4 Skill: Applying/Analyzing LO: S18.5 21) A homeotic mutant flower has an arrangement (from outside in) of carpels, stamens, stamens, carpels instead of the usual sepal, petals, stamens, carpels. In general, Class A genes specify sepals, Class A and B genes specify petals, Class B and C genes specify stamens, and Class C specifies carpels. This mutant must therefore have a mutation in ________.

A) a Class A gene B) a Class B gene C) a Class C gene D) both A and B E) both B and C Answer: A Section: 18.4 Skill: Applying/Analyzing LO: S18.5

7 .

22) Holoprosencephaly is a highly variable condition in mammals, including humans. Its effects range from an absence of brain areas, mid-face disorders, and cyclopia to slight dental abnormalities. One form, HPE3, is caused by null mutations in Shh. Familial cases are inherited as a dominant trait. Which of the following do you expect?

A) HPE3 is actually a multigene complex.

B) Each type of malformation is due to a different HPE3 allele.

C) HPE3 expression is modified by genetic background.

D) HPE3 individuals also share a common environmental exposure.

E) All mutations in Shh that result in HPE3 are due to frameshifts.

Answer: C Section: 18.5 Skill: Applying/Analyzing LO: S18.3b 23) Flower homoeotic genes encode: ________.

A) MADS box genes B) homeobox genes C) pair-rule genes D) housekeeping genes E) morphogen genes Answer: A Section: 18.5 Skill: Knowledge/Comprehension LO: S18.5 18.2 Short Answer Questions 1) An embryonic stem cell has the potential to differentiate into any cell or tissue type. Is it differentiated, pluripotent, or totipotent?

Answer: totipotent Section: 18.1 Skill: Knowledge/Comprehension LO: S18.1 2) What term describes the process by which a cell acquires a certain fate from its neighboring cell?

Answer: induction Section: 18.1 Skill: Knowledge/Comprehension LO: S18.2b 3) What substance, in different concentrations, directs developmental fates?

Answer: morphogen Section: 18.1 Skill: Knowledge/Comprehension LO: S18.1 8 .

4) Cells called ________ influence development by controlling other cells in their local environment?

Answer: organizers or organizer cells Section: 18.2 Skill: Knowledge/Comprehension LO: S18.1 5) Cell autonomous genes affect only those cells in which they are expressed. Which group of early developmental genes are not cell autonomous?

Answer: segment polarity (genes) Section: 18.2 Skill: Knowledge/Comprehension LO: S18.2b 6) Homeotic genes in virtually all animals are said to show colinearity between the spatial arrangement of the tissues they act on as well as what other feature?

Answer: gene location Section: 18.2 Skill: Knowledge/Comprehension LO: S18.2a 7) Null mutations in ________ genes would be expected to result in the failure to establish the anterior-posterior poles of an embryo Answer: coordinate Section: 18.2 Skill: Knowledge/Comprehension LO: S18.2c 8) If a cell from the blastoderm of the Drosophila embryo is committed to form the sixth abdominal segment of the adult and this cell is transplanted into the thorax-committed region of another embryo, will it develop properties of a thoracic or an abdominal segment?

Answer: abdominal Section: 18.2 Skill: Applying/Analyzing LO: S18.2c 9) Certain mutations in non-Hox genes actually produce homeotic mutants due to a failure to modulate chromatin acetylation appropriately. This influence on cellular "memory" is called what?

Answer: epigenetic Section: 18.2 Skill: Knowledge/Comprehension LO: S18.2b

9 .

10) A highly conserved protein domain of 60 amino acids, the homeodomain, is found in a large number of animals. This domain the proteins that have them to bind to what?

Answer: regulatory DNA (sequences) Section: 18.2 Skill: Knowledge/Comprehension LO: S18.2b 11) Hunchback protein occurs along a gradient, where the anterior embryo has high protein levels and the posterior has low levels, yet the hunchback mRNA is found at the same concentration throughout the embryo. These observations suggest that hunchback must be regulated post-________ Answer: transcriptionally Section: 18.2 Skill: Knowledge/Comprehension LO: S18.2a 12) Which group of genes in Drosophila embryos must be mutated if the result is elimination of a significantly sized, contiguous region of segmentation?

Answer: gap genes Section: 18.2 Skill: Applying/Analyzing LO: S18.2b 13) The Antennapedia complex and the bithorax complex of genes all encode transcription factors that have a 180-bp DNA-binding domain. What is this domain called?

Answer: homeobox Section: 18.2 Skill: Knowledge/Comprehension LO: S18.2b 14) You are interested in the homeobox genes that may regulate the development of the common clover plant Trifolium subterraneum. To identify clones of the T. subterraneum genome that contain these homeotic genes, you would probe with a rice gene that contains what domain?

Answer: MADS box Section: 18.2 Skill: Knowledge/Comprehension LO: S18.5 15) In the development of C. elegans, an embryo of 1090 cells then loses 131 cells due to what process?

Answer: apoptosis Section: 18.3 Skill: Knowledge/Comprehension LO: S18.1

10 .

16) Cell signaling during much of C. elegans development occurs along concentration gradients.

What kind of signaling does this represent?

Answer: inductive or induction or morphogen Section: 18.3 Skill: Knowledge/Comprehension LO: S18.2a 17) During the evolution of development in animals, it is found that many regulatory genes have been co-opted for the development of different structures, such as limbs or digits. This is an example of what current area of research?

Answer: evo-devo Section: 18.4 Skill: Knowledge/Comprehension LO: S18.4 18) In Arabidopsis thaliana, a model organism in plant genetics, there are three classes of genes that control the development of floral structures: A, B, and C. Mutations of these genes cause organs to form in inappropriate places. What category of genes are they?

Answer: homeotic Section: 18.4 Skill: Knowledge/Comprehension LO: S18.5 19) In Drosophila and other animals, Hox genes encode transcription factors that control development; but in Arabidopsis, a different family of transcription factors is produced with a different set of conserved amino acids. These are encoded by what group of genes?

Answer: MADS (box genes) Section: 18.4 Skill: Knowledge/Comprehension LO: S18.5 20) Which structure in plants acts most like embryonic stem cells in animals?

Answer: meristem Section: 18.5 Skill: Knowledge/Comprehension LO: S18.1 21) Certain species of lilies, if eaten by pregnant ewes, can cause the ewes to produce lambs with severe birth defects because the cyclopamine that the plant produces can block embryonic response to Shh. This type of birth defect is a phenocopy of mutations in which genes?

Answer: Shh Section: 18.5 Skill: Applying/Analyzing LO: S18.2d

11 .

22) Mouse eye genes can be inserted into Drosophila and can regulate the flies' eye formation.

Differences in the size of horses are controlled by a gene also found to influence size variation in zebrafish. What do these examples illustrate?

Answer: conserved genes or conservation or conserved function Section: 18.5 Skill: Applying/Analyzing LO: S18.4 18.3 Fill-in-the-Blank Questions 1) A mutation in a fly causes an antenna to develop where a leg should be. This is an example of a ________ mutation.

Answer: homeotic Section: 18.1 Skill: Applying/Analyzing LO: S18.2a 2) Molecules whose concentration and position in an embryo determine some aspect of development are called ________.

Answer: morphogens Section: 18.1 Skill: Knowledge/Comprehension LO: S18.2b 3) Genes that influence development in a manner that is independent of their paternal genotype are said to have ________ effect.

Answer: maternal Section: 18.2 Skill: Knowledge/Comprehension LO: S18.2a 4) If a cell is no longer able to differentiate into any type of tissue, it has become ________.

Answer: differentiated Section: 18.1 Skill: Knowledge/Comprehension LO: S18.2a 5) Most of the genes controlling development encode either signal molecules or ________.

Answer: transcription factors Section: 18.1 Skill: Knowledge/Comprehension LO: S18.2b

12 .

6) Coordinate genes in an organism such as Drosophila determine the ________ of the early embryo.

Answer: polarity Section: 18.2 Skill: Knowledge/Comprehension LO: S18.2a 7) In limb-positioning in tetrapods, the ________ acts as an organizer to promote digit formation at the distal ends of limb buds.

Answer: ZPA (zone of polarizing activity) Section: 18.3 Skill: Knowledge/Comprehension LO: S18.2b 8) Epistasis analysis in single and double mutants, such as in the genes specifying genital development in C. elegans has helped to identify genes that function in developmental ________.

Answer: pathways Section: 18.3 Skill: Knowledge/Comprehension LO: S18.2a

13 .

18.4 Essay Questions 1) Use a description of C. elegans development to distinguish clearly between differentiation and determination.

Answer: The developmental potential of a cell describes the range of different cell types it has the potential to become. As development progresses, the potential of individual cells decreases until, ultimately, their genetic fate is determined. Thus, determination implies a stable change at the level of gene expression that means a cell is committed to a certain developmental fate.

Differentiation is the process that follows determination in which the cell develops its cellspecific expression profile. Differentiation results in distinct cell types with clearly identifiable functions, such as a muscle or nerve cell.

The C. elegans example from the text involved vulva formation. The vulva forms during the last larval stage from six precursor cells, called vulval precursor cells (VPCs). Three of these larval cells give rise to structures of the vulva itself, the 1° cell and two 2° cells of the vulva. The other three differentiate into hypodermis (3° cells). Initially, each of the six VPCs has the potential to differentiate along any of the pathways–1°, 2°, or 3°. If the anchor cell is destroyed, no vulva will form because all six VPCs differentiate with a 3° fate and become hypodermis. Thus, these six cells are determinate, while the anchor cell must be present to induce VPCs to differentiate with 1° or 2° fates and thus form the vulva. Because all the VPCs are capable of the same kinds of differentiation, they are said to form an equivalence group.

Section: 18.1 Skill: Synthesis/Evaluation LO: S18.1 2) Describe the genetic evidence that demonstrates that a developmental gene has maternal effect.

Answer: Maternal effect genes are those whose effect on development is determined by the maternal genotype and not by the paternal genotype. For instance, a recessive maternal effect gene would cause the recessive phenotype in any cross involving a homozygous recessive mother and a father of any genotype. The children of mothers that are either heterozygous or homozygous for the dominant allele will all show the dominant phenotype.

Section: 18.2 Skill: Synthesis/Evaluation LO: S18.2a

14 .

3) Describe aspects of plant and animal development that are shared and then give specific evidence that plants and animals independently evolved mechanisms of multicellular development?

Answer: Plants and animals both control cell differentiation to generate their multicellular body plans. Both plants and animals make use of several transcriptional and signaling cascades to specify polarity and cell specification (differentiation), however in plants, cell differentiation is rarely terminal as opposed to animals. The independent evolution of plant and animal development is easily demonstrated by describing MADS and HOX gene families that control development in plants and animals respectively. MADS and HOX gene families both encode proteins with DNA-binding domains that act as homeotic transcription factors. MADS and HOX genes are not homologous but their gene functions are.

Section: 18.2 and 18.4 Skill: Synthesis/Evaluation LO: S18.5 4) How and when does the Antennapedia complex change antennae into legs?

Answer: The Antennapedia gene complex is normally expressed in parasegments 4 and 5, where it is necessary for thoracic structures that produce legs. If it is expressed inappropriately in head segments (usually in parasegment 3), it promotes development of legs instead of antennae by regulating the transcriptional cascade that leads to leg development.

Section: 18.2 Skill: Synthesis/Evaluation LO: S18.2b 5) How can transplantation and ablation be used to find out if a cell in the source of a developmental signal?

Answer: Transplantation of that cell to a new location in an undifferentiated embryo. If the transplanted cells are able to influence the differentiation of nearby cells, then the cell is generating a local signal that acts by either induction or inhibition. Similarly if ablation of the cell in question in its native environment alters the development of nearby cells, then the cell was involved in generating a developmental signal.

Section: 18.3 Skill: Synthesis/Evaluation LO: S18.2a

15 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 19 Genetic Analysis of Quantitative Traits 19.1 Multiple-Choice Questions 1) Which of the following is FALSE regarding human multifactorial traits?

A) They are influenced by genetic and environmental factors.

B) Their phenotype can be maximized by environmental factors to reach a genetic potential.

C) They are influenced by multiple genes.

D) They usually exhibit discontinuous variation.

E) They are usually quantitative rather than qualitative.

Answer: D Section: 19.1 Skill: Remembering/Understanding LO: G5, S19.1a 2) It is clearly established that obesity has greatly increased in children over the last several decades in the United States. While several genes have been identified that increase the potential for obesity, environmental influences are also involved. Which of the following is most likely?

A) There is little or no gene-environment interaction.

B) Phenotypic variation is discontinuous.

C) There is a small overlap of phenotypic ranges of different genotypes.

D) There is a wide range of phenotypes with each genotype.

E) There is a demonstrable threshold of environmental effects.

Answer: D Section: 19.1 Skill: Remembering/Understanding LO: G5, S19.1a 3) How many distinct phenotypic categories should be observed for a polygenic trait in humans that results from segregation of additive alleles for 10 genes?

A) 21 B) 20 C) 10 D) 5 E) 4 Answer: A Section: 19.1 Skill: Applying/Analyzing LO: G4, S19.1d

1 .

4) A pure-breeding tree that produces fruit with 12 seeds, the smallest number of seeds possible, is mated with a pure-breeding tree that produces fruit with 36 seeds, the largest number of seeds possible. This polygenic trait is controlled by two alleles, which both contribute to the phenotype, present at multiple loci. When two F1 individuals are mated, 7 phenotypes result.

Which of the following statements is FALSE? A) There are 3 genes that control this polygenic trait.

B) The F1 offspring all have 24 seeds.

C) One of the additive alleles contributes 2 seeds to the seed content of the fruit.

D) One of the additive alleles contributes 6 seeds to the seed content of the fruit.

E) One of the F2 phenotypic classes has 18 seeds.

Answer: E Section: 19.1 Skill: Evaluating/Creating LO: G2, G3, G4, S19.1b 5) A hypothetical condition in some domesticated fowl is identified phenotypically by patchy loss of feathers on the animals' backs. Although several factors, genetic and environmental, are found to be involved, only about 3% of any population with any of these factors are found with feather loss. Which of the following is a likely cause?

A) continuous variation B) threshold of liability C) low mutation frequency D) high number of alleles in one of the genes E) high number of environmental effects Answer: B Section: 19.1 Skill: Applying/Analyzing LO: S19.1a 6) A scientist is studying insect wing length. An individual from a pure-breeding parental line (P1), for which the VP is 40 mm2, is mated with an individual from a pure-breeding parental line (P2), for which the VP is 80 mm2. This results in F1 progeny that have a VP of 54 mm2. A cross between two F1 progeny results in offspring with a VP of 92 mm. What is the genotypic variance?

A) 14 mm B) 32 mm C) 34 mm D) 38 mm E) 58 mm Answer: C Section: 19.2 Skill: Applying/Analyzing LO: G3, G4, S19.2b

2 .

7) Which of these statements about heritability is FALSE? A) Narrow sense heritability measures the contribution of additive genotypic variance to phenotypic variance.

B) Twin studies can offer insight into broad sense heritability of human traits.

C) High narrow sense heritability values are correlated with a greater degree of response to artificial selection.

D) Heritability measures the extent to which environmental variation contributes to the total phenotypic variation.

E) Broad sense heritability measures the ratio of genetic variance to phenotypic variance.

Answer: D Section: 19.3 Skill: Applying/Analyzing LO: G2, S19.2b, S19.3a 8) Which of the following formulas can be used to calculate broad sense heritability (H2)?

A) S(h2) B) VE / VP C) R/S D) (VA + VD + VI)/ VP E) VA / VP Answer: D Section: 19.3 Skill: Remembering/Understanding LO: S19.2b, S19.3a 9) The brain mass in a population of mice has a phenotypic variance of 0.56 g2 and an environmental variance of 0.14 g2. What is the broad sense heritability (H2)?

A) 0.25 B) 0.42 C) 0.75 D) 1.54 E) The broad sense heritability cannot be calculated with the data provided Answer: C Section: 19.3 Skill: Remembering/Understanding LO: G3, G4, S19.3a

3 .

10) A sexually reproducing species of algae has been engineered to produce biofuels with varying yields. A colony with the highest yield is stimulated to reproduce with a colony with the lowest yield and the resulting F1 progeny are allowed to grow. Yields among the F1 generation have a variance of 7.8 g2. The F1 are allowed to mate and the resulting progeny are sampled and found to have a variance of 9.1 g2. Assuming that the F1 progeny are genetically uniform, determine the broad sense heritability (H2) and interpret if genetic heritability is low or high.

A) H2 = 0.14, genetic variability is low B) H2 = 0.17, genetic variability is low C) H2 = 0.86, genetic variability is low D) H2 = 0.14, genetic variability is high E) H2 = 0.86, genetic variability is high Answer: E Section: 19.3 Skill: Applying/Analyzing LO: G3, G4, G9, S19.2b, S19.3a 11) As broad sense heritability (H2) approaches 0.0, which of the following must be true?

A) Almost all the variation in phenotype is VG. B) Almost all the variation in phenotype is VE. C) Identical (MZ) twins are virtually all discordant.

D) Fraternal (DZ) twins are virtually all concordant.

E) Narrow sense heritability (h2) approaches 1.

Answer: B Section: 19.3 Skill: Applying/Analyzing LO: S19.3a, S19.3d 12) For a specific trait, monozygotic twins have a phenotypic variance of 15% and dizygotic twins have a phenotypic variance of 40%. Which of the following can be concluded for this trait based on this data from a twin study?

A) VG for monozygotic twins is 15% B) VE for monozygotic twins is 30% C) VG for dizygotic twins is 50% D) VG for dizygotic twins is 25% E) VE for dizygotic twins is 25% Answer: C Section: 19.3 Skill: Applying/Analyzing LO: G3, G4, G5, S19.2b

4 .

13) For the following traits, the concordances of MZ and DZ twins are given as percentages.

Which trait is unlikely to have a strong genetic influence?

Traits % concordance between MZ % concordance between DZ A) 95 65 B) 54 36 C) 59 62 D) 90 10 E) 37 6 Answer: C Section: 19.3 Skill: Applying/Analyzing LO: G2, G3, S19.3d 14) An animal breeder asks your advice about which, of a number of traits, could most effectively be selected for in his herd. A trait with which of the following would you most correctly suggest?

A) a high VE B) a high VA relative to VP C) a high VP relative to VG D) a low h2 value E) a low response to selection (R) Answer: B Section: 19.3 Skill: Applying/Analyzing LO: G3, G5, S19.3a 15) A cultivator is breeding a crop for increased fruit production. The mean fruit weight in the population in 500g and the plants selected for breeding have an average fruit weight of 450g.

The response to selection (R) is -25g. What is the narrow sense heritability (h2)?

A) h2 = 0.50 B) h2 = 0.05 C) h2 = 0.20 D) h2 = 0.06 E) h2 = 2.00 Answer: A Section: 19.3 Skill: Applying/Analyzing LO: G3, G4, S19.3a

5 .

16) A scientist is researching oil content of seeds from a new plant species named R. tificial. The average oil content from plants in her lab population is 52%, and she has determined the narrow sense heritability (h2) is 0.60. If she randomly mates individuals with 70%, 71%, 72%, 73%, and 74% seed oil content what is the expected response to selection (R)?

A) 6% B) 12% C) 24% D) 31.2% E) 72% Answer: B Section: 19.3 Skill: Applying/Analyzing LO: G3, G4, S19.3b 17) A scientist is researching oil content of seeds from a new plant species named R. tificial. She has determined the narrow sense heritability (h2) is 0.80. She mates plants with an average seed oil content of 32%. If the response to selection (R) was 16%, what is the estimated average oil content of seeds in the entire lab population?

A) 12% B) 20% C) 32% D) 44% E) 50% Answer: A Section: 19.3 Skill: Applying/Analyzing LO: G2, G3, G4, S19.3b 18) A group is experimenting with selection in an insect species due to exposure to a newly developed insecticide, cpd AW143. The group has found that after 10 generations of exposure, the population is now bimodal in its response to the drug. Close to 30% of the population are completely normal while 70% live only about 1/6 of the normal lifetime. This is an example of ________.

A) threshold effect B) stabilizing selection C) additive genes D) directional selection E) disruptive selection Answer: E Section: 19.3 Skill: Applying/Analyzing LO: G3, S19.3c

6 .

19) A group is experimenting with selection in a mouse model for a disease that is lethal in midlife. After 30 generations of exposure, the population's average life span is near that of a wildtype mouse population. What type of selection did the model mouse population undergo?

A) threshold effect B) stabilizing selection C) additive genes D) directional selection E) disruptive selection Answer: D Section: 19.3 Skill: Applying/Analyzing LO: G3, S19.3c 20) QTL mapping resembles linkage analysis primarily because ________.

A) it uses statistics to analyze its data B) it uses the location of a known DNA marker C) it uses the three-point mapping method D) the distances are reported in cM E) the results are quantifiable Answer: B Section: 19.4 Skill: Remembering/Understanding LO: G7, S19.4a 21) In which order should the following steps be done to identify a quantitative trait locus (QTL) gene that contributes to the phenotypic variation in a trait?

I. DNA and protein sequence analyses are used to determine if there is a correlation with phenotypic variation II. Each of the offspring is individually backcrossed with the same parent, usually the one expressing the phenotype more strongly III. Lod scores are calculated for each genetic marker to find QTLs IV. The genotype of the progeny at genetic markers located every 5-10 cM is determined for each introgression line V. Two individuals which exhibit the extremes of a specific phenotype are crossed VI. DNA sequence variants among introgression lines are analyzed to find candidate genes A) V, II, IV, III, VI, I B) I, V, III, II, IV, VI C) V, IV, II, III, I, VI D) I, II, IV, III, VI, V E) V, IV, III, II, VI, I Answer: A Section: 19.4 Skill: Applying/Analyzing LO: G7, G9, S19.4a 7 .

22) Mean platelet volume (MPV) in humans increases with heart attacks and strokes and is used to predict recurrence as well as mortality. MPV has high heritability. A recent GWA study found MPV to be associated with SNPs on chromosomes 12, 3, and 17, respectively. These three QTLs accounted for 4-5% of MPV variance. Which of the following is a reasonable explanation?

A) Other loci may also account for significant variance of MPV. B) These SNPs must be translocated or transposed alleles.

C) Monitoring of these SNPs should identify those at highest risk.

D) The three SNPs, when sequenced, should show high conservation.

E) The three SNPs will be found in genes responsible for stroke.

Answer: A Section: 19.4 Skill: Applying/Analyzing LO: G3, G5, S19.4a 23) Although no specific gene alteration has yet been demonstrated as a cause for schizophrenia (SZ), it is known to have a strong genetic component. A recent study has found a strong association between SZ and a locus at chromosome 10q22-23 in Ashkenazi Jewish and Taiwanese Han populations. Which of the following is probable?

A) Other populations will have the same association with 10q22-23.

B) The neuregulin-3 gene at this locus is mutated in all or most SZ cases.

C) SZ exhibits significant genetic heterogeneity in different populations.

D) The region on 10q must include several genes associated with SZ. E) People with SZ being characterized differently in various medical communities results in this association between the Jewish and Han populations.

Answer: C Section: 19.4 Skill: Evaluating/Creating LO: G5, S19.4b 24) Parkinson disease (PD) is a neurodegenerative disease that has been associated with a significant number of different genes, therefore showing it to be genetically heterogenic. One recent study in 278 families found evidence of two such possible loci. Evidence for linkage to chromosome 18q11 was presented by Gao et al. in 2009. Which of the following would be considered reputable evidence for this?

A) lack of quantitative difference for the occurrence of both loci in individuals B) use of persons from multiple ethnic and geographic backgrounds in the study C) finding the locus to be in Hardy-Weinberg equilibrium in the population D) finding the lod score for association between the chromosomal location and PD to be 4.1 E) analysis of all other possible SNPs with no such association to chromosome 18 Answer: D Section: 19.4 Skill: Applying/Analyzing LO: G1, G5, G7, G9, S19.4a

8 .

25) Once a chromosomal position for a putative gene associated with Parkinson disease is verified, which of the following is a next step?

A) developing a FISH screening test to identify pre-Parkinson patients B) determining whether this locus is mutated in all Parkinson patients C) using a genome-wide linkage SNP panel to identify more minor genes D) ascertaining all living and at-risk individuals who are family members of the affected test subjects E) fine mapping of the linkage region to identify particular variants for influencing risk Answer: E Section: 19.4 Skill: Evaluating/Creating LO: G1, G5, G7, G9, S19.4a 26) Which of the following is true?

A) GWAS can scan the entire genome for QTLs by statistically testing for marker variants.

B) QTL mapping approaches can analyze organisms in random mating populations.

C) GWAS requires controlled crosses and the formation of introgression lines.

D) QTL mapping can directly identify the gene responsible for the associated phenotypic variation.

E) GWAS is able to determine a single gene that plays the sole role in the development of Crohn's disease.

Answer: A Section: 19.4 Skill: Evaluating/Creating LO: G1, G5, G9, S19.4b 19.2 Short Answer Questions 1) While most additive polygenic traits show continuous distribution, some can be divided into two or more categories based on their contribution to genetic liability. Such traits are said to have what separation between affected and unaffected members of a population?

Answer: threshold Section: 19.1 Skill: Remembering/Understanding LO: S19.1a 2) A strain of a cereal grain can be either dark tan, medium-dark tan, medium tan, light tan, or cream colored. When a dark tan and a cream plant are crossed, all F1 are medium tan. The F2 are distributed in a ratio of 1:4:6:4:1 from darkest to lightest. How many genes are involved in this coloration?

Answer: 2 Section: 19.1 Skill: Applying/Analyzing LO: S19.1b

9 .

3) A strain of a cereal grain can be either dark tan, medium-dark tan, medium tan, light tan, or cream colored. When plants pure-breeding for dark tan grains and cream grains are crossed, all F1 have medium tan grains. If the F1 are crossed, what proportion of the F2 will be light tan?

Answer: 4/16 Section: 19.1 Skill: Applying/Analyzing LO: G2, G3, G4, S19.1b 4) Of mean, median, and modal, what value is the most common value in a given population?

Answer: modal Section: 19.2 Skill: Remembering/Understanding LO: G3, S19.2a 5) What two measures describe the distribution of a trait in a sample?

Answer: variance and standard deviation Section: 19.2 Skill: Remembering/Understanding LO: G3, S19.2a 6) What is the mean, variance (s2), and standard deviation (s) of the following sample: 1.00, 2.00, 3.00, 5.00, 7.00, 9.00, 10.00, and 11.00.

Answer: mean = 6.00, variance = 14.57, standard deviation = 3.82 Section: 19.2 Skill: Remembering/Understanding LO: G4, S19.2a 7) Genetic variance can be partitioned into which three types of variance?

Answer: additive variance (VA), dominance variance (VD), and interactive variance (VI) Section: 19.2 Skill: Remembering/Understanding LO: S19.2b 8) Calculate the narrow sense heritability (h2) for the following data on a trait in turkeys: VA = 11.0 VE = 94.4 VP = 114.2 VD = 13.8 Answer: 0.096 Section: 19.3 Skill: Applying/Analyzing LO: G3, G4, S19.3a

10 .

9) Which component of variance in a trait such as stamina in horses is the best indicator of whether the trait will respond to selection?

Answer: VA Section: 19.3 Skill: Applying/Analyzing LO: S19.3b 10) Which will exhibit the most shared maternal effects (uterine environment, etc.), identical or fraternal twins?

Answer: identical Section: 19.3 Skill: Remembering/Understanding LO: S19.3d 11) If H2 for club foot is estimated as 80%, would you expect a small difference in concordances between MZ and DZ twins, such as 30 versus 27, or a large difference, such as 30 versus 2?

Answer: large difference or 30 versus 2 Section: 19.3 Skill: Applying/Analyzing LO: G3, S19.3d 12) QTLs are most often mapped relative to DNA markers by using what scores as measurement?

Answer: lod (scores) Section: 19.4 Skill: Remembering/Understanding LO: S19.4a 13) If the odds ratio for a particular QTL/DNA marker pair (e.g., Q10M10) is 2.2, this indicates that the organism is how many times more likely to get Q10M10 gametes?

Answer: 2.2 Section: 19.4 Skill: Applying/Analyzing LO: G3, S19.4a 14) What is the statistically acceptable lod score that provides evidence of genetic linkage?

Answer: +3 or 3 Section: 19.4 Skill: Remembering/Understanding LO: G3, S19.4a 15) After locating a genome region likely to include a QTL, what must be found in this region?

Answer: candidate gene Section: 19.4 Skill: Remembering/Understanding LO: S19.4a 11 .

19.3 Fill-in-the-Blank Questions 1) If a trait results from genetic variation as well as environmental factors, it is known as a ________ trait.

Answer: multifactorial Section: 19.1 Skill: Remembering/Understanding LO: S19.1a 2) In some traits, different phenotypes result from genes whose alleles each contribute a specific increment to the whole. The phenotypes then have a ________ phenotypic distribution.

Answer: continuous Section: 19.1 Skill: Remembering/Understanding LO: S19.1a 3) The part of the genetic variance that is attributed to epistatic interactions is called the ________ variance.

Answer: interactive Section: 19.2 Skill: Remembering/Understanding LO: S19.2b 4) After successful QTL mapping, ________ lines, which result from the repeatedly backcrossing to form highly inbred lines, can help identify QTL genes.

Answer: introgression Section: 19.4 Skill: Remembering/Understanding LO: S19.4a 5) A GWA study relies on the correlation between a QTL and a ________.

Answer: SNP or DNA marker Section: 19.4 Skill: Remembering/Understanding LO: S19.4a

12 .

19.4 Essay Questions 1) Describe how quantitative traits can be explained in Mendelian terms.

Answer: The multiple-gene hypothesis is the proposal that alleles at each of the contributing genes obeyed the principles of segregation and independent assortment and had an additive effect in the production of phenotypic variation. Often, these effects can be attributed to having multiple genes (polygenic inheritance), or some traits can have additive effects such that several mutations can have differing effects than single mutations.

Section: 19.1 Skill: Evaluating/Creating LO: S19.1b 2) In a newly discovered diploid animal species, tail length is determined solely by 2 additive gene loci that can have the allele s, which contributes 4mm, and the allele l, which contributes 15mm. Two lines with the genotypes AsAsBsBs and AlAlBsBs are mated to create the F1 generation. a. What is the predicted genotype and phenotype of the F1 progeny? b. If two individuals from the F1 generation are crossed, each phenotypic category will occur in what proportion?

Answer: a. The genotype of all F1 individuals will be AsAlBsBs and the phenotype will be 3(4mm) + 1(15mm) = 27mm. b. Since an F1 individual's B locus cannot contribute an l allele, the B locus can be disregarded in these calculations. An F1 individual can contribute 0 or 1 l alleles from the A locus, so there are 3 possible F2 phenotypic categories containing 0, 1, or 2 l alleles (or 4, 3, or 2 s alleles, respectively). In F1 individuals, there is a ½ chance of transmitting zero or one l alleles from the A locus. In F2 individuals, the probability of obtaining zero l alleles from both parents is (1/2)(1/2) = 1/4. The probability of obtaining one l allele from either parent is (1/2)(1/2)+(1/2)(1/2) = 2/4. The probability of obtaining two l alleles from both parents is (1/2)(1/2) = 1/4.

Section: 19.1 Skill: Evaluating/Creating LO: G3, G4, S19.1b, S19.1d

13 .

3) In a newly discovered diploid animal species, tail length is determined solely by 3 additive gene loci that can have the allele s, which contributes 4mm, and the allele l, which contributes 15mm. Two lines with the genotypes AsAsBsBsClCl and AlAlBsBsCsCs are mated to create the F1 generation. a. What is the predicted genotype and phenotype of the F1 progeny? b. If two individuals from the F1 generation are crossed, each phenotypic category will occur in what proportion?

Answer: a. The genotype of all F1 individuals will be AsAlBsBsCsCl and the phenotype will be 4(4mm) + 2(15mm) = 46mm. b. The F1 individuals' B loci will contribute two s alleles to the F2 progeny. Therefore, the proportions will only depend on the 4 alleles at the A and C loci. An F1 individual can contribute 0, 1, or 2 l alleles from the A and/or C loci, so there are 5 possible F2 phenotypic categories containing 0, 1, 2, 3, or 4 l alleles (or 4, 3, 2, 1, or 0 s alleles, respectively). Using binomial probability, the probability of each of the phenotypic categories is: [4!/(0!4!)](1/2)0(1/2)4 = 1/16 for zero l alleles (four s alleles + two s alleles from B loci) [4!/(1!3!)](1/2)1(1/2)3 = 4/16 for one l alleles (three s alleles + two s alleles from B loci) [4!/(2!2!)](1/2)2(1/2)2 = 6/16 for two l alleles (two s alleles + two s alleles from B loci) [4!/(3!1!)](1/2)3(1/2)1 = 4/16 for three l alleles (one s alleles + two s alleles from B loci) [4!/(4!0!)](1/2)4(1/2)0 = 1/16 for four l alleles (zero s alleles + two s alleles from B loci) Section: 19.1 Skill: Evaluating/Creating LO: G3, G4, S19.1b, S19.1d 4) Distinguish between broad sense heritability and narrow sense heritability, and explain how either or both of these measures can be used to inform those using artificial selection.

Answer: Narrow sense heritability refers to the proportion of total phenotypic variation that can be attributed to the additive effects of genes, whereas broad sense heritability is the proportion of total phenotypic variation caused by all genetic effects. However, narrow sense heritability is the only component of variation that natural selection can act on and, thus, provides more accurate selection information.

Section: 19.3 Skill: Evaluating/Creating LO: S19.3a 5) Given a list of concordances (as percentages) for monozygotic and dizygotic twins for a selection of traits, describe how you would evaluate the relative input of environmental versus genetic factors for each trait.

Answer: If the differences in concordances are high between MZ and DZ twins, this would suggest a higher genetic component because MZ twins share the genome. On the other hand, if there are no differences in concordances between MZ and DZ twins, this could indicate that environmental factors (either in utero or after birth) may contribute to these particular traits.

Section: 19.3 Skill: Evaluating/Creating LO: S19.3d 14 .

6) Many human behavioral traits have both genetic and environmental components. Bipolar disorder has long been known to be familial, and more recently a number of genes and/or SNPs have been found to be involved. What are three or more reasons that this research has been hampered?

Answer: Answers will vary, but several examples are included here. Variations in the diagnosis criteria for bipolar disorder may make differing degrees of the disorder difficult to distinguish. In addition, there may be certain variants, or genes with major versus minor input, so the combination of mutations in an individual may affect the severity of their phenotype. This also may affect the age of onset variability, which can make diagnosis difficult and could be affected by environmental factors such as hormones, stress, etc. There may also be variable response to medications, which could be a function of the particular mutation or the other genes that the individual possesses that may allow a better response to a particular treatment.

Section: 19.4 Skill: Evaluating/Creating LO: S19.4a 7) Recent work in Drosophila demonstrated that the genes known as the chromatin remodeling complex co-localize at loci for hsp genes that respond to heat shock. Outline a procedure you could use to substantiate this finding.

Answer: A genome-wide association study (GWAS) can be used to tie the presence of a sequence variant of a DNA marker to a QTL influencing a specific phenotype. Thus, the relationship between an inherited genetic marker variant and the phenotype is by "association," which means organisms that carry a particular variant are more likely to have a certain phenotype than are organisms that carry a different variant. The assessment of association is quantitative; that is, it expresses the percentage of organisms with a genetic marker that also display a certain phenotype versus the percentage that have the phenotype but not the genetic marker. The advantage of GWAS over other QTL mapping approaches is that GWAS can scan the entire genome for QTLs by statistically testing for markers. Thus, GWAS can be used to correlate known genes or SNPs with hsp loci.

Section: 19.4 Skill: Evaluating/Creating LO: G9, S19.4a, S19.4b

15 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Chapter 20 Population Genetics and Evolution at the Population, Species, and Molecular Levels 20.1 Multiple-Choice Questions 1) You want to calculate the frequency of the allele causing sickle cell disease in your patients, and you know the genotypes of every individual. Assuming the population is not in HardyWeinberg equilibrium due to population substructure, which method can be used to calculate allele frequency?

A) allele-counting method B) genotype proportion method C) square root method D) binomial expansion E) chi-square test Answer: B Section: 20.1 Skill: Applying/Analyzing LO: S20.1a 2) In the same population of patients you examined for sickle cell allele frequencies, you now want to test whether the disease symptoms are related to blood type. Since blood type is a codominant trait and the alleles are easily identifiable by the phenotype (blood type) of the individuals, which method is most helpful for identifying blood type allele frequency?

A) allele-counting method B) genotype proportion method C) square root method D) binomial expansion E) chi-square test Answer: A Section: 20.1 Skill: Applying/Analyzing LO: S20.1a 3) Phenylketonuria (PKU) is extremely common in Ireland, affecting approximately 1 in 4500 live births. Assuming that the population is in Hardy-Weinberg equilibrium, what is the frequency of PKU carriers?

A) 0.022% B) 98.5% C) 2.9% D) 1.49% E) 2.2% Answer: C Section: 20.1 Skill: Applying/Analyzing LO: G4, G5, S20.1c 1 .

4) The incidence of phenylketonuria (PKU), an autosomal recessive genetic disorder, is approximately 1 in 4,500 in Ireland. Assuming Hardy-Weinberg equilibrium, what is the probability that a phenotypically normal Irish female, with no family history, and a phenotypically normal Irish male, with an affected sister, will have a child with PKU? A) 0.028% B) 0.49% C) 0.55% D) 1.96% E) 2.2% Answer: B Section: 20.1 Skill: Evaluating/Creating LO: G2, G4, G5, G7, S20.1c 5) The incidence of Tay Sachs, an autosomal recessive genetic disorder, is approximately 1 in 3,500 in a certain population of Ashkenazi Jews. Assuming Hardy-Weinberg equilibrium, what is the frequency of carriers for the Tay Sachs allele in this population?

A) 1.5% B) 1.7% C) 2.9% D) 3.3% E) 98.5% Answer: D Section: 20.1 Skill: Applying/Analyzing LO: G4, G5, S20.1c 6) The incidence of Tay-Sachs, an autosomal recessive genetic disorder, is approximately 1 in 3,500 in a certain population of Ashkenazi Jews. Assuming that this population is in HardyWeinberg equilibrium, what is the probability that a phenotypically normal female, with no family history, and a phenotypically normal male, who had an affected sister, will have a child with Tay-Sachs?

A) 0.028% B) 0.49% C) 0.55% D) 1.96% E) 2.2% Answer: C Section: 20.1 Skill: Evaluating/Creating LO: G2, G4, G5, G7, S20.1c

2 .

7) In a group of 500 people, the frequency of genotype NN is 40%. Assuming both autosomal inheritance and that the population is in Hardy-Weinberg equilibrium, how many individuals would you expect to have the MN genotype?

A) 68 B) 232 C) 300 D) 200 E) 258 Answer: B Section: 20.1 Skill: Remembering/Understanding LO: G4, S20.1c 8) During a University of Nevada, Reno blood donation campaign, the frequencies of alleles determining the ABO blood groups were 0.70 for i, 0.2 for IA and 0.1 for IB. Assuming random mating, what is the expected frequency of blood type B? A) 0.04 B) 0.15 C) 0.18 D) 0.32 E) 0.49 Answer: B Section: 20.1 Skill: Applying/Analyzing LO: G2, G4, G5, G7, S20.1c 9) You are studying ABO blood groups, and know that 6.25% of the population has genotype IAIA and 42.25% of the population has Type O blood. What is the expected frequency of blood type B? A) 1% B) 6.5% C) 13% D) 14% E) 51.5% Answer: D Section: 20.1 Skill: Applying/Analyzing LO: G2, G4, G5, G7, S20.1c

3 .

10) If an X-linked recessive trait has a frequency of 5% in males, what percentage of females are carriers? (Assume that the population is in Hardy-Weinberg equilibrium.)

A) 9.5% B) 5% C) 95% D) 4.2% E) 10% Answer: A Section: 20.1 Skill: Applying/Analyzing LO: G4, G5, G7, S20.1c 11) Jeffrey Pines from Mt. Rose were genotyped at a resistance gene for a type of fungi that affects pines. Genotypes of 1000 of these Jeffrey Pines were: 300 R1R1, 500 R1R2, and 200 R2R2. Use a chi-square (χ2) test to determine if this population is in Hardy-Weinberg equilibrium? The critical chi-square value at P=.05 for 1 degree of freedom is 3.84.

A) 0.10; fail to reject B) 0.10; reject C) 10; fail to reject D) 10; reject E) 37.5; reject Answer: A Section: 20.1 Skill: Evaluating/Creating LO: G4, G5, G7, S20.1b, S20.1d 12) Given the following genetic profile, what is the mean fitness of the population?

Genotype Frequency Number Relative fitness (w)

A1A1 0.55 2750 1.0

A1A2 0.20 1000 0.75

A2A2 0.25 1250 0.50

A) 1 B) 0.75 C) 0.825 D) 0.3025 E) 0.55 Answer: C Section: 20.2 Skill: Applying/Analyzing LO: G2, G4, S20.2b

4 .

13) In an insect species, survival in agricultural fields regularly sprayed with insecticide is determined by the genotype for a detoxification enzyme encoded by a gene with two alleles, A and D. Which of the following statements is consistent with the data?

Genotype AA AD DD

Relative Fitness 1.00 0.70 0.05

A) This will likely lead to the frequency of the A allele being eliminated (f=0.0) and the D allele being fixed (f=1.0).

B) Eventually, natural selection should lead to a balanced polymorphism.

C) The D allele frequency will change faster as it gets less frequent.

D) Both the A and D alleles should reach stable equilibrium frequencies that are maintained in a steady state.

E) Directional selection will increase the frequency of the A allele at a pace determined by the intensity of natural selection.

Answer: E Section: 20.2 Skill: Applying/Analyzing LO: G2, G4, S20.2b, S20.2c 14) Mexican hairless dogs are heterozygous (Hh), carrying a single copy of the lethal allele. In a given population, the Hh genotypic frequency is 60% and the frequency of each homozygote is 20%. Assuming w = 1 for the Hh and hh dogs, what is the estimated genotypic frequency of the lethal HH genotype after reproduction?

A) 20% B) 14% C) 10% D) 39% E) 46% Answer: B Section: 20.2 Skill: Applying/Analyzing LO: G4, G5, G7, S20.2b

5 .

15) Data for an autosomal recessive trait shows that heterozygous individuals have the highest relative fitness. Those who are unaffected and homozygous for the dominant A allele have a relative fitness of 84% compared to heterozygous individuals, but only about 30% of those who are affected survived to reproduce. What is the estimated equilibrium frequency for allele A in this population?

A) 0.160 B) 0.186 C) 0.357 D) 0.700 E) 0.814 Answer: B Section: 20.2 Skill: Applying/Analyzing LO: G4, G5, S20.2b 16) In a theoretical population where f (A1) = 0.95 and f (A2) = 0.05, µ = 1 × 10-6, and v = 1 × 10-7, what is Δq?

A) 5.00 × 10-9 B) 9.45 × 10-7 C) 9.50 × 10-7 D) 9.55 × 10-7 E) 5.26 × 10-3 Answer: B Section: 20.3 Skill: Applying/Analyzing LO: G4, S20.3 17) Researchers collect data on an island population of frogs and find a mutation rate of the recessive allele, d, is 1 × 10-6. The relative fitness of the individuals homozygous from this allele is 0.7. What is the allele frequency of this recessive allele at equilibrium?

A) 7.0 × 10-7 B) 1.8 × 10-3 C) 1.2 × 10-3 D) 3.0 × 10-7 E) 1.4 × 10-6 Answer: B Section: 20.3 Skill: Remembering/Understanding LO: G4, S20.3

6 .

18) Which aspect of population genetics can lead to changes in allele frequencies in a new mixed population following migration?

A) natural selection B) mutation C) gene flow D) founder effect E) bottleneck effect Answer: C Section: 20.4 Skill: Remembering/Understanding LO: S20.4 19) A researcher is studying a newly discovered gene that causes increased body weight in domesticated chickens. In a mainland population, the frequency of the A1 allele is 0.2, for this gene with two alleles. If 100 of these mainland chicken are transported on a ship to an isolated island with a population of 200 A1A1 chickens, 400 A1A2 chickens, and 400 A2A2 chickens, what would the frequency of the A1 allele in the admixed population?

A) 0.745 B) 0.618 C) 0.382 D) 0.764 E) 0.236 Answer: C Section: 20.4 Skill: Applying/Analyzing LO: G2, G4, S20.4 20) All Old Order Amish families from Lancaster County with Ellis-van Creveld syndrome can trace their genealogies to Mr. and Mrs. Samuel King, who immigrated to Lancaster County in 1744. This is an example of ________.

A) founder effect B) mutation C) gene flow D) Hardy-Weinberg equilibrium E) bottleneck effect Answer: A Section: 20.5 Skill: Remembering/Understanding LO: S20.5b

7 .

21) Both the founder effect and the bottleneck effect are mechanisms that produce large allele frequency sampling errors in a small population due to ________.

A) random mutation B) gene flow C) genetic drift D) lethal alleles E) random mating Answer: C Section: 20.5 Skill: Remembering/Understanding LO: S20.5a, S20.5b, S20.5c 22) Cheetahs exhibit significant inbreeding and loss of heterozygosity due to overhunting and loss of habitat experienced at some point in their evolutionary history. Which genetic mechanism occurs when a large population is substantially reduced at random, acting independently of natural selection?

A) founder effect B) mutation C) gene flow D) Hardy-Weinberg equilibrium E) bottleneck effect Answer: E Section: 20.5 Skill: Remembering/Understanding LO: S20.5c 23) The coefficient of inbreeding (F) can be used to estimate the proportion of loci that are ________.

A) fixed in a population B) homozygous identical by descent C) heterozygous by descent D) lethal alleles E) recessive Answer: B Section: 20.6 Skill: Remembering/Understanding LO: S20.6a

8 .

24) How would you calculate the coefficient of inbreeding for individual IV-1 in the pedigree below?

A) F= 2(1/2)2 B) F= 4(1/2)2 C) F= 2(1/2)3 D) F= 4(1/2)3 E) F= 2(1/2)4 Answer: E Section: 20.6 Skill: Applying/Analyzing LO: G4, S20.6a

9 .

25) Given the following pedigree, what is the coefficient of inbreeding for individual G?

Use Key Analytic Tool pedigree in Chapter 20, page 402 from SGSM A) 1/4 B) 1/8 C) 1/16 D) 1/32 E) 1/64 Answer: C Section: 20.6 Skill: Applying/Analyzing LO: G3, G4, S20.6a 26) The process by which an original species is transformed into a new species over an extended period of time that spans many generations is known as ________.

A) anagenesis B) cladogenesis C) morphogenesis D) neogenesis E) founder effect Answer: A Section: 20.7 Skill: Remembering/Understanding LO: S20.7a

10 .

27) Physical separation of a segment of a large population by a physical barrier that prevents gene flow can lead to ________.

A) sympatric speciation B) allopatric speciation C) hybrid sterility D) temporal isolation E) gene flow Answer: B Section: 20.7 Skill: Remembering/Understanding LO: S20.7a 28) Which of the following is an example of a postzygotic mechanism for reproductive isolation?

A) behavioral isolation B) gametic isolation C) habitat separation D) mechanical isolation E) hybrid sterility Answer: E Section: 20.7 Skill: Remembering/Understanding LO: S20.7a 29) When populations share a single habitat but are isolated by genetic or postzygotic mechanisms that prevent gene flow, what process can cause populations to diverge?

A) random mating B) allopatric speciation C) sympatric speciation D) random mutation E) gene duplication Answer: C Section: 20.7 Skill: Remembering/Understanding LO: S20.7a

11 .

20.2 Short Answer Questions 1) What type of statistical analysis is used to compare observed and expected results in order to evaluate the validity of an estimate based on the Hardy-Weinberg equilibrium?

Answer: chi-square test Section: 20.1 Skill: Remembering/Understanding LO: S20.1d 2) When looking at the MN blood group with two alleles, how many degrees of freedom will be used to compare observed versus expected results?

Answer: 1 Section: 20.1 Skill: Remembering/Understanding LO: S20.1b 3) If the frequency of PKU, an autosomal recessive disease, is 1% in males, what is the frequency of females with PKU? (You may assume that the population is in Hardy-Weinberg equilibrium.)

Answer: 1% Section: 20.1 Skill: Applying/Analyzing LO: S20.1c 4) If the relative fitness value is w = 0.6, what is the selection coefficient used to identify the differences between the fitnesses of organisms with different traits?

Answer: 0.4 |1-w|; i.e., the absolute value of the difference between w and the population mean fitness (0.6) Section: 20.2 Skill: Remembering/Understanding LO: S20.2a 5) What is the ultimate source of genetic variation in populations?

Answer: mutation Section: 20.3 Skill: Remembering/Understanding LO: S20.3 6) What is the consequence of natural selection favoring the heterozygous genotype (as exhibited by the sickle cell allele in malaria-prone regions), in which alleles reach stable equilibrium frequencies that are maintained by the ongoing action of selection against the homozygous genotypes?

Answer: balanced polymorphism Section: 20.2 Skill: Remembering/Understanding LO: S20.2c 12 .

7) When examining the effects of forward and reverse mutation rates, what equation is used to calculate the change in the frequency of A2, or Δq?

Answer: Δq = µp — vq Section: 20.3 Skill: Remembering/Understanding LO: S20.3 8) Which mechanism maintains the biological ability of populations to interbreed and can thus prevent evolutionary divergence of populations?

Answer: gene flow Section: 20.4 Skill: Remembering/Understanding LO: S20.4 9) A difference in allelic frequencies that is produced when a new population is established by a few individuals is known as what type of genetic drift?

Answer: founder effect Section: 20.5 Skill: Remembering/Understanding LO: S20.5b 10) Of the five aspects of Hardy-Weinberg equilibrium, which one directly affects phenotypic frequencies rather than genotypic frequencies?

Answer: nonrandom mating Section: 20.6 Skill: Remembering/Understanding LO: S20.6a 11) What is a common type of nonrandom mating?

Answer: inbreeding Section: 20.6 Skill: Remembering/Understanding LO: S20.6a 12) Inbreeding depression, the reduction in fitness of inbred organisms, often results from the reduced level of what (within the gene pool)?

Answer: genetic heterozygosity Section: 20.6 Skill: Remembering/Understanding LO: S20.6a

13 .

13) What type of geographic separation resulted in diversification of an ancestral species of ground squirrel into two species, Ammospermophilus leucurus, found on the north rim of the Grand Canyon, and members of the Ammospermophilus harrisii on the south rim of the canyon?

Answer: allopatric speciation Section: 20.7 Skill: Remembering/Understanding LO: S20.7a 14) Using the allele frequencies in the table below, what is the probability that a random person would be heterozygous 12/14 for VAW?

Answer: 2*0.015*0.131 = 0.00393 Section: 20.8 Skill: Applying/Analyzing LO: G4, G5, G7

14 .

15) Using the allele frequencies in the table below, what is the probability that a random person is heterozygous 12/14 for VAW and homozygous 18/18 for FAG? (Assume that the two loci are unlinked.)

Answer: 2*0.015*0.131*0.015^2 = 8.84 × 10-7 Section: 20.8 Skill: Applying/Analyzing LO: G4, G5, G7 20.3 Fill-in-the-Blank Questions 1) ________ is an evolutionary mechanism that favors the reproductive success of certain members of a population over others as a result of differences in anatomical, physiological, or behavioral traits they possess.

Answer: Natural selection Section: 20.2 Skill: Remembering/Understanding LO: S20.2c 2) In the pattern of natural selection called ________ the favored phenotype has a homozygous genotype, while the consequence of natural selection favoring the heterozygote is a balanced polymorphism.

Answer: directional natural selection or directional selection Section: 20.2 Skill: Remembering/Understanding LO: S20.2c

15 .

3) The coefficient of inbreeding quantifies the probability that two alleles in a homozygous individual are ________, having descended from the same copy of the allele carried by a common ancestor of the inbred individual.

Answer: identical by descent (IBD) Section: 20.6 Skill: Remembering/Understanding LO: S20.6a 4) The pattern of species evolution known as ________ is one of branching in which an ancestral species gives rise to two or more new species.

Answer: cladogenesis Section: 20.7 Skill: Remembering/Understanding LO: S20.7a 5) Burkitt's lymphoma is a cancer of the lymphatic system that is more common in individuals who are homozygous βAβA that succumb more easily to malaria. βSβS individuals suffer from anemia. βAβS have the highest relative fitness. Resistance to malaria and Burkitt's lymphoma are both an example of ________ advantage.

Answer: heterozygous Section: 20.8 Skill: Applying/Analyzing LO: S22.8

16 .

20.4 Essay Questions 1) In the general population, 1 in 300 individuals is a carrier for Tay-Sachs disease, while 1 in 30 individuals of Ashkenazi Jew descent are carriers. Tay-Sachs also affects 1 in 30 individuals with French-Canadian ancestry, although two completely unique mutations are responsible for the Ashkenazi and French-Canadian mutations. What are the chances of two individuals who are carriers having a child with Tay-Sachs if a. both individuals are not of Ashkenazi or French-Canadian descent? b. one individual is of Ashkenazi descent and the other is from the general population? c. both individuals are of Ashkenazi descent?

Answer: For each question, you will calculate [Chance of mom being a carrier] × [Chance of dad being a carrier] × [Chance of having an affected child if both parents are carriers] a. (1/300) × (1/300) × 1/4 = 2.77 × 10-6 b. (1/30) × (1/300) × 1/4 = 2.77 × 10-5 c. (1/30) × (1/30) × 1/4 = 2.77 × 10-4 Section: 20.1 Skill: Evaluating/Creating LO: G3, G4, G5, G7, S20.1c 2) In an island population, the frequencies of the three MN blood types are given here: M: 550 MN: 725 N: 225 a. Calculate the allele frequency of M and N. b. Assuming random mating, what are the expected frequencies for each blood type? c. Use chi-square analysis to test your hypothesis that mating is random for this population.

Answer: a. p = f (M) = [2(550) + 725]/3000 = 0.608 q = f (N) = [2(225) + 725]/3000 = 0.392 b. f (M) = (0.608)2(1500) = 555 f (MN) = (2)(0.608)(0.392)(1500) = 715 f (N) = (0.392)2(1500) = 230 c. χ2 = [(550 – 555)2/555] + [(725 –715)2/715] + [(225 – 230) 2/230] = 0.2936 Section: 20.1 Skill: Evaluating/Creating LO: G4, S20.1a, S20.1c, S20.1d

17 .

3) What are the three key elements for determining F, a measure that a particular allele is IBD, using pedigree analysis?

Answer: The three key elements to determining F from pedigrees are (1) the number of alleles of a gene carried by common ancestors, (2) the number of transmission events required to produce a genotype that is homozygous IBD, and (3) the probability of transmission for each event linking the allele in a common ancestor to the inbred individual. First-cousin mating is a form of inbreeding that is relatively common in many human societies and is common in mammals in general.

Section: 20.6 Skill: Evaluating/Creating LO: G2, G8, S20.6a 4) Explain why rare recessive alleles are more likely to produce a recessive phenotype as a result of inbreeding.

Answer: Negative genetic outcomes in the form of infants with recessive conditions due to inbreeding occur when a recessive allele is rare in a population. In such cases, there is a meaningful increase in the likelihood that a consanguineous mating will produce a child with a recessive phenotype. When the recessive allele frequency is q = 0.01, for example, the chance of producing a recessive homozygote from a first-cousin mating is only a few times more likely than the chance of producing a recessive homozygote by random mating.

Section: 20.6 Skill: Evaluating/Creating LO: G2, G8, S20.6a 5) Draw a pedigree for an inbreeding coefficient of F= 2(1/2)4.

Answer: Should include a closed loop consisting of two generations and only one common ancestor

Section: 20.6 Skill: Evaluating/Creating LO: G2, G4, S20.6b

18 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Application A Human Hereditary Disease and Genetic Counseling Application A.1 Multiple-Choice Questions 1) A genetic counselor ________.

A) performs genetic testing and reports their findings to a medical doctor B) chooses the best course of action for patients based on genetic testing C) helps patients choose whether or not they will undergo genetic testing, explains the results of genetic tests, and helps patients make their own medical decisions D) designs genetic tests that can be used in population-level testing programs E) performs gene therapy Answer: C Section: A.1 Skill: Knowledge/Comprehension LO: SA.3b 2) Pedigrees are useful for ________.

A) determining the paternity of a child B) discovering patterns of inheritance that may suggest familial disease C) doctors so that they can treat all affected members of a family D) parents who want to determine how many kids to have E) for genetics classes but they have little applicability to modern medical care Answer: B Section: A.1 Skill: Applying/Analyzing LO: SA.2a 3) A patient with monosomy is most easily diagnosed using what technique?

A) PCR B) a karyotype C) a pedigree D) the expression of recessive traits E) questioning the parents of that patient Answer: B Section: A.1 Skill: Synthesis/Evaluation LO: SA.2b

1 .

4) Genetic testing includes all of the following except ________.

A) prenatal testing B) presymptomatic testing C) newborn testing D) carrier testing E) all of these are forms of genetic testing Answer: E Section: A.1 Skill: Applying/Analyzing LO: SA.2b 5) In genetics, Baysian analysis uses ________.

A) court testimony to determine the chance of disease in an individual B) electrophoresis to determine the genotype of an individual C) DNA sequencing to determine the genotype of an individual D) numerical information to predict the chance of disease E) known or inferred genotype information, along with information from related offspring to make predictions about the probability of a certain genotype in an individual Answer: E Section: A.2 Skill: Applying/Analyzing LO: SA.5a 6) The chance of each of two independent events happening are ________ to determine the probability that both events would happen together.

A) added together B) divided by two C) multiplied by two D) multiplied by each E) squared Answer: D Section: A.2 Skill: Knowledge/Comprehension LO: SA.5c 7) A testing outcome in which a disease is detected, but the sample/patient in reality does not have the disease is called a ________.

A) false negative B) true positive C) false positive D) a standard error E) a true negative Answer: C Section: A.2 Skill: Knowledge/Comprehension LO: SA.2b 2 .

8) A maternal serum screen can be performed to detect ________which provide information about the genotype of a fetus.

A) RNA fragments B) maternal DNA C) precancerous cells D) blood cells E) circulating compounds Answer: E Section: A.2 Skill: Applying/Analyzing LO: SA.2b 9) Consider a particular hypothetical case in which the Baysian probability that an unaffected parent is heterozygous for a particular recessive allele is 20% and the other parent has the recessive condition; what is the chance that their first child would have the condition?

A) 10% B) 25% C) 100% D) 0% E) 20% Answer: A Section: A.2 Skill: Applying/Analyzing LO: SA.5c 10) A pregnant woman with recessive X-linked hemophilia and whose partner is a normal (non hemophiliac) man wants to know the chances that she will have an affected child. From an ultrasound, she knows that the child is a boy. What are the chances that her son will have hemophilia?

A) the chance that their son will be affected is 0% B) the chance that their son will be affected is 50% C) the chance that their son will be affected is 25% D) the chance that their son will be affected is 75% E) the chance that their son will be affected is 100% Answer: E Section: A.2 Skill: Synthesis/Evaluation LO: SA.5a

3 .

11) Two unaffected parents each have a parent with the same recessive condition and they want to know the chances that they will have a child who has the condition. What do you tell them?

A) the chance that their child will be affected is 0% B) the chance that their child will be affected is 50% C) the chance that their child will be affected is 25% D) the chance that their child will be affected is 75% E) the chance that their child will be affected is 100% Answer: C Section: A.2 Skill: Applying/Analyzing LO: SA.5b 12) The ethical issues guiding genetic counseling include all except ________.

A) respect for individual autonomy B) justice — the equal treatment of all individuals C) the likelihood of medical benefit D) the benefits to society of having an individual tested E) preventing any coercion regarding patient decisions Answer: D Section: A.2 Skill: Applying/Analyzing LO: SA.3a 13) ________ allows doctors to intervene in cases of rare disease to mitigate disease, often by simply prescribing dietary supplementation or restricting certain substances from the diet of affected individuals A) Direct-to-consumer genetic testing B) Carrier testing C) Population testing D) Newborn testing E) Paternity testing Answer: A Section: A.2 Skill: Applying/Analyzing LO: SA.2b 14) The only form of genetic testing that is mandated in the United States is ________.

A) prenatal testing B) carrier testing C) population testing D) newborn testing E) paternity testing Answer: D Section: A.2 Skill: Applying/Analyzing LO: SA.2b 4 .

15) A couple comes to you for advice about genetic testing for a rare early-onset dominant condition. The paternal grandfather had the condition, but no one else in the family has had it including the male partner in this couple. Do you recommend testing? Why? A) yes, the male partner is definitely a carrier B) no, the male partner is definitely a carrier and so the outcome is certain C) no, both parents would have to have the disease allele to have an affected child D) no, we can deduce that neither partner has a disease allele E) yes, testing is expensive, but worth the price, even when unnecessary Answer: D Section: A.2 Skill: Applying/Analyzing LO: SA.3a 16) A couple has just been informed that one of them has Huntington's disease, and that the other spouse does not have a disease allele. With this information, what is the chance that at least one of their two young children has a disease allele?

A) the chance that at least one of the unaffected children has a disease allele is 75% B) the chance that at least one of the unaffected children has a disease allele is 0% C) the chance that at least one of the unaffected children has a disease allele is 25% D) the chance that at least one of the unaffected children has a disease allele is 50% E) the chance that at least one of the unaffected children has a disease allele is 100% Answer: A Section: A.2 Skill: Applying/Analyzing LO: SA.5b 17) A family with only one affected parent and a child with a dominant disease would like to know what the probability is that one of their unaffected children might be a carrier. What do you tell them?

A) the chance that an unaffected child is a carrier is 0% B) the chance that an unaffected child is a carrier is less than 50% C) the chance that an unaffected child is a carrier is 2/3 D) the chance that an unaffected child is a carrier is 75% E) the chance that an unaffected child is a carrier is 50% Answer: A Section: A.2 Skill: Applying/Analyzing LO: SA.5b

5 .

18) A family with two unaffected parents and a child with a recessive disease would like to know what the probability is that one of their unaffected children might be a carrier. What do you tell them?

A) the chance that an unaffected child is a carrier is 0% B) the chance that an unaffected child is a carrier is less than 50% C) the chance that an unaffected child is a carrier is 2/3 D) the chance that an unaffected child is a carrier is 75% E) the chance that an unaffected child is a carrier is 50% Answer: C Section: A.2 Skill: Synthesis/Evaluation LO: SA.5b 19) A couple comes to you for genetic counseling and the male partner has a rare mitochondrial disease. What do you tell them about the chance that their first child would have the disease?

A) the first child will only have the disease is they are male B) the first child will have no chance of inheriting the disease C) the first child will only have the disease is they are female D) the first child can be screened using in vitro fertilization, so no testing is necessary E) the child will have the disease because mitochondria are inherited paternally Answer: B Section: A.2 Skill: Synthesis/Evaluation LO: SA.4b 20) A boy with recessive X-linked hemophilia is born from an unaffected mother and a father with hemophilia. From which parent was the disease allele inherited?

A) the mother B) the father C) both parents, the trait is recessive D) any of these answers is equally likely E) the father or the mother with an equal chance of either parent Answer: A Section: A.2 Skill: Synthesis/Evaluation LO: SA.5a

6 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Application B Human Genetic Screening Application B.1 Multiple-Choice Questions 1) Testing to determine if someone is heterozygous for a recessive allele is called ________.

A) carrier testing B) newborn genetic testing C) presymptomatic testing D) karyotyping E) direct-to-consumer testing Answer: A Section: B.1 Skill: Knowledge/Comprehension LO: SB.4b 2) If both parents are carriers of a recessive disease allele, what is the chance that one of their unaffected children is also a carrier?

A) the chance that an unaffected child is a carrier is 0% B) the chance that an unaffected child is a carrier is less than 50% C) the chance that an unaffected child is a carrier is 2/3 D) the chance that an unaffected child is a carrier is 75% E) the chance that an unaffected child is a carrier is 50% Answer: C Section: B.1 Skill: Applying/Analyzing LO: SB.4b 3) Two parents with a dominant trait have a child who shows the normal phenotype. What is the chance that that child is a carrier of the trait?

A) the chance that an unaffected child is a carrier is 0% B) the chance that an unaffected child is a carrier is less than 50% C) the chance that an unaffected child is a carrier is 2/3 D) the chance that an unaffected child is a carrier is 75% E) the chance that an unaffected child is a carrier is 50% Answer: A Section: B.1 Skill: Synthesis/Evaluation LO: SB.4b

1 .

4) The male partner has a rare autosomal-recessive trait and the female partner does not, their first child however has the trait. What do you tell them about the chance that their next child will have the trait?

A) the chance is 0% B) the chance is 50% C) the chance is 25% D) the chance is 75% E) the chance is 2/3 Answer: B Section: B.1 Skill: Synthesis/Evaluation LO: SB.4b 5) The child of a normal parent and a parent who develops Huntington disease has a ________ chance having a disease allele.

A) 25% B) 0% C) 100% D) 75% E) 50% Answer: E Section: B.1 Skill: Applying/Analyzing LO: SB.1a 6) Phenylketonuria (PKU) is detected in newborns when ________ are measured in blood samples.

A) single nucleotide polymorphisms (SNPs) B) chromosome abnormalities C) metabolites D) triplet repeat expansions E) maternal cells Answer: C Section: B.2 Skill: Knowledge/Comprehension LO: SB.4a

2 .

7) The recommended uniform screening panel (RUSP) includes tests for conditions that have all of the following characteristics except ________.

A) the condition can be treated by medical intervention B) treating the condition is not possible C) the condition must be reliably detected by the test D) the condition can be dominant E) the test can be performed with a blood sample Answer: B Section: B.2 Skill: Synthesis/Evaluation LO: SB.3b 8) Fetal cell sorting can be performed to collect ________ from ________which provide information about the genotype of a fetus.

A) fetal cells; maternal blood B) metabolic compounds; maternal blood C) maternal cells; amniotic fluid D) metabolic compounds; amniotic fluid E) fetal cells; a heel stick Answer: A Section: B.2 Skill: Applying/Analyzing LO: SB.6a 9) Unlike many other genetic disorders, Huntington's disease is caused by a full penetrance disease allele. Full penetrance means what?

A) The longer the allele (more triplet repeats), the more likely the disease is to occur in an individual.

B) There are other genes that are also required for an individual to get Huntington's disease.

C) There are environmental factors beyond age that are also required for an individual to get Huntington's disease.

D) Individuals with a disease allele are almost certain to get Huntington's disease.

E) Heterozygotes are just as likely to get Huntington's disease as homozygotes.

Answer: D Section: B.2 Skill: Applying/Analyzing LO: SB.1a

3 .

10) Polymerase chain reaction (PCR) is used to detect which type of genetic variation?

A) aneuploidy B) translocations C) monosomy D) triplet repeat expansions E) inversions Answer: D Section: B.2 Skill: Synthesis/Evaluation LO: SB.1b 11) Karyotyping is used to detect which type of genetic variation?

A) single nucleotide polymorphisms (SNPs) B) translocations C) metabolic conditions D) triplet repeat expansions E) SNPs and translocations Answer: B Section: B.2 Skill: Applying/Analyzing LO: SB.4a 12) The only form of genetic testing that is mandated in the United States is ________.

A) prenatal testing B) carrier testing C) community-based testing D) newborn testing E) paternity testing Answer: D Section: B.2 Skill: Applying/Analyzing LO: SB.3a 13) Genetic screening to assess disease risk among individuals in a subpopulation is called ________.

A) presymptomatic testing B) karyotyping C) in vitro fertilization D) prenatal screening E) community-based screening Answer: E Section: B.3 Skill: Applying/Analyzing LO: SB.7a

4 .

14) Genetic testing includes all of the following except ________.

A) prenatal testing B) presymptomatic testing C) newborn testing D) carrier testing E) all of these are forms of genetic testing Answer: E Section: B.3 Skill: Applying/Analyzing LO: SB.8a 15) Pharmacogenetic screening aims to reveal what kind of trait?

A) susceptibility to infection B) sensitivity to drugs or adverse effects of drugs C) disorders that cause imbalance of metabolic compound in the blood D) susceptibility to disease E) susceptibility to autoimmune disease Answer: B Section: B.3 Skill: Applying/Analyzing LO: SB.6b 16) Non-invasive prenatal genetic screening capable of analyzing the nucleotide sequence of DNA includes which technique(s)?

A) amniocentesis B) chorionic villus sampling C) fetal cell sorting D) karyotying E) preimplantation screening Answer: C Section: B.4 Skill: Applying/Analyzing LO: SB.6a 17) Inspecting a fetus by ultrasound can reveal which of the following?

A) morphological abnormalities such as cleft lip.

B) translocations C) Phenylketonuria (PKU) D) cystic fibrosis E) karyotype Answer: A Section: B.4 Skill: Applying/Analyzing LO: SB.6b

5 .

18) Genetic association between a genetic variant and a trait means ________.

A) that the associated genetic variant causes the trait/condition B) that the trait/condition is caused by a single allele C) that individuals with an associated allele are certain to have the trait D) that the genetic variant is not relevant to the trait E) that individuals with an associated allele are more likely than those without the allele to have the trait Answer: E Section: B.5 Skill: Applying/Analyzing LO: SB.4b 19) Alzheimer disease is associated with variation in the gene APOE. Thus, individuals with APOE mutations are ________.

A) more likely to respond to drugs that treat Alzheimer disease than individuals without APOE mutations B) less likely to have parents who have/had Alzheimer disease C) less likely to have children who are susceptible to Alzheimer disease D) certain to have Alzheimer disease E) more likely to get Alzheimer disease than individuals without APOE mutations Answer: E Section: B.5 Skill: Applying/Analyzing LO: SB.4b 20) Direct-to-consumer testing offers clients the chance to ________.

A) assess their risk of several genetic conditions B) have their results evaluated by a medical professional C) determine their chances of having a miscarriage D) have genetic testing performed on people without their consent E) none of these options is available by direct-to-consumer testing Answer: A Section: B.5 Skill: Knowledge/Comprehension LO: SB.8a

6 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Application C The Genetics of Cancer Application C.1 Multiple-Choice Questions 1) Most cancers are caused by ________ mutations whereas a minority of cancers is caused by ________ mutations.

A) germline; somatic B) inherited; somatic C) germline; inherited D) somatic; germline E) somatic; random Answer: D Section: C.1 Skill: Knowledge/Comprehension LO: SC.2b 2) Age is the greatest of all risk factors for cancer because ________.

A) older people have weaker immune systems B) germline mutations accumulate with age C) cell division is rare in older people D) tumor suppressors do not work as well in older people E) somatic mutations accumulate with age Answer: E Section: C.1 Skill: Applying/Analyzing LO: SC.1a 3) A proto-oncogene has what characteristics?

A) a gene that promotes cell division and when functioning normally, leads to cancer progression B) a gene that inhibits apoptosis and when mutated prevents tumor metastasis C) a gene that regulates the cell cycle in response to hormones D) a gene that inhibits cell division and when functioning normally, leads to cancer progression E) a gene that promotes cell division and, when mutated, leads to cancer progression Answer: E Section: C.2 Skill: Synthesis/Evaluation LO: SC.3a

1 .

4) Cells whose normal function involves preventing cell division are generally characterized as ________.

A) tumor suppressor genes B) oncogenes C) transcription factors D) pseudogenes E) apoptotic genes Answer: A Section: C.2 Skill: Applying/Analyzing LO: SC.3a 5) The modern view of cancer describes the cells of a malignant tumor as ________.

A) clonal, each being genetically identical to the other cells in the tumor B) genetically identical to the surrounding somatic cells from the same patient C) a complex mixture of cells, some malignant and some not D) abnormal cells confined to a tissue within the patient E) normal cells which have been converted to cancer cells by hormones Answer: C Section: C.2 Skill: Applying/Analyzing LO: SC.2a 6) Benign cell masses are known as ________.

A) metastatic cells B) dysplasia C) malignant tumors D) neoplasia E) apoptosis Answer: B Section: C.2 Skill: Knowledge/Comprehension LO: SC.2b 7) All of the following are hallmarks of cancer except ________.

A) sustained cell production B) evasion of growth suppression C) metastasis D) apoptosis E) angiogenesis Answer: D Section: C.2 Skill: Knowledge/Comprehension LO: SC.4a

2 .

8) Cancer cells can divide indefinitely, which is also known as ________.

A) cellular differentiation B) resistance to cell death C) mitotic arrest D) checkpoint function E) cellular immortality Answer: E Section: C.2 Skill: Applying/Analyzing LO: SC.2a 9) Inflammation can aide tumor growth by promoting which of the following?

A) angiogenesis B) drug resistance C) mutagenesis D) genome stability E) altered metabolism Answer: A Section: C.2 Skill: Applying/Analyzing LO: SC.4a 10) Cancer susceptibility can run in families due to ________.

A) population migration B) somatic mutation C) germline mutation D) inbreeding E) sporadic causes Answer: C Section: C.3 Skill: Synthesis/Evaluation LO: SC.1b 11) Chronic myelogenous leukemia results from a translocation which allows cells to express ________.

A) a missing chromosome arm B) too much protein from a normal gene C) a recessive allele D) a fusion protein E) a loss-of-function allele Answer: D Section: C.3 Skill: Applying/Analyzing LO: SC.5a

3 .

12) Burkitt's lymphoma results from a/an ________, which causes the misexpression of the protein cMyc. A) chromosomal translocation B) heat shock C) chemotherapy drug D) point mutation E) deletion Answer: A Section: C.3 Skill: Applying/Analyzing LO: SC.2a 13) Sporadic retinoblastoma is often confined to one eye, not both eyes, because homozygous RB1 mutations ________.

A) are rare B) affect only one eye C) require one other mutation, a 'second-hit' to cause cancer D) are embryonic-lethal E) prevent cell division Answer: A Section: C.3 Skill: Applying/Analyzing LO: SC.5b 14) The most common cause of cancer formation is ________.

A) mutation in a single gene B) inheriting a tumor C) germline mutations D) recessive alleles E) accumulation of mutations in a large number of genes Answer: E Section: C.3 Skill: Applying/Analyzing LO: SC.5e 15) Familial cancer often includes the inheritance of ________.

A) developmental abnormalities B) predisposition to cancer C) somatic mutations D) homozygous mutations E) missing chromosomes Answer: B Section: C.3 Skill: Applying/Analyzing LO: SC.3b

4 .

16) Common mutational targets affecting cancer susceptibility include all of the following except ________.

A) the cell cycle B) checkpoint genes C) DNA repair genes such as BRCA1 D) mitochondrial function E) growth-limiting genes such as APC Answer: D Section: C.3 Skill: Applying/Analyzing LO: SC.5e 17) Cancer therapy includes which of the following?

A) drug treatments targeting dividing cells B) drugs targeting specific enzymes/proteins C) modified immune cells D) radiation targeting cancerous cells E) All of these are possible treatment strategies.

Answer: E Section: C.4 Skill: Synthesis/Evaluation LO: SC.6a 18) Mutation in the DNA repair checkpoint gene TP53 underlie Li-Fraumeni syndrome which can be diagnosed based on ________.

A) inherited breast cancer B) melanoma C) lack of mutations over many generations D) recurrence of cancer in the same organ in multiple family members E) many different types of cancer in multiple members of the same family Answer: E Section: C.3 Skill: Applying/Analyzing LO: SC.5c 19) The 'two-hit' hypothesis suggests that ________.

A) once one allele of a gene is mutated, the other allele must be mutated in order to result in cancer B) mutations in one gene promote cancer and mutations to other genes prevent cancer C) mutations in two genes in the same cell is usually required for cancer progression D) two mutations in the same gene are required for that gene to become oncogenic E) both eyes must contain mutations in the RB1 gene in order to develop retinoblastoma Answer: C Section: C.3 Skill: Synthesis/Evaluation LO: SC.5b 5 .

20) Which of the following genes is responsible for pausing the cell cycle long enough for radiation-induced DNA damage to be repaired?

A) PD-L1 B) TP53 C) KRAS D) Both PD-L1 and KRAS E) Both PD-L1 and TP53 Answer: B Section: C.4 Skill: Synthesis/Evaluation LO: SC.3b

6 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Application D Human Evolutionary Genetics Application D.1 Multiple-Choice Questions 1) On what continent did hominids evolve?

A) Europe B) Asia C) North America D) South America E) Africa Answer: E Section: D.1 Skill: Knowledge/Comprehension LO: SD.2b 2) On which continent would you expect to find the most human genetic diversity and why?

A) Australia, because it has had a relatively small population size for a relatively short amount of time.

B) Africa, because it has the oldest human populations, containing alleles that originated long ago and have evolved since.

C) Australia, because it has the oldest human populations.

D) South America because it was one of the last continents to be inhabited by humans.

E) Europe, because it had waves of human migration from other parts of the world over the course of many thousands of years.

Answer: B Section: D.1 Skill: Knowledge/Comprehension LO: SD.5a 3) In which regions of the human genome would you expect to find the most genetic diversity and why?

A) In non-coding regions, because there is less selective pressure to maintain non-coding nucleotide sequence.

B) In non-coding regions, because there is selective pressure to alter non-coding sequence.

C) In coding regions, because there is selective pressure to alter amino acid sequence in proteins.

D) In coding regions, because there is selective pressure to preserve amino acid sequence in proteins.

E) There is no meaningful difference between coding and non-coding regions in terms of genetic diversity.

Answer: A Section: D.2 Skill: Knowledge/Comprehension LO: SD.2b

1 .

4) Mitochondrial DNA is useful to studies of human history because ________.

A) it is inherited maternally and its sequence changes very little over time B) it is inherited maternally and its sequence is typically more diverse than nuclear DNA C) it is inherited paternally D) it can be used to identify the mother of any human E) it is inherited by both parents, allowing researchers to trace DNA from maternal and paternal origins Answer: B Section: D.2 Skill: Knowledge/Comprehension LO: SD.2a 5) Phylogenetics, the study of evolutionary relationships between organisms predicts that closely related species will share which of the following when compared to species that diverged longer ago?

A) more traits B) more DNA sequence C) fewer CNVs D) all of these statements are assumptions made by phylogenetics E) none of these statements are assumptions made by phylogenetics Answer: D Section: D.2 Skill: Applying/Analyzing LO: SD.2a 6) All of the following are expected to increase genetic diversity in a population except?

A) immigration B) random mating C) small population size D) low selective pressure E) large population size Answer: C Section: D.2 Skill: Knowledge/Comprehension LO: SD.1 7) An allele that is shared by both chimpanzees and humans is said to be ________.

A) a human-specific gene B) ancestral C) conserved D) derived E) both conserved and ancestral Answer: E Section: D.3 Skill: Knowledge/Comprehension LO: SD.3 2 .

8) The discovery of non-coding sequences that are conserved in a wide range of related species suggests that these sequences are important for what?

A) migration B) gene regulation C) cell division D) the differences between those related species E) cancer Answer: B Section: D.3 Skill: Applying/Analyzing LO: SD.4b 9) The presence of similar DNA sequences in sharks, whales and humans, but not in chimpanzees suggests?

A) the DNA was gained in elephants, mice and humans, but not chimpanzees ________.

B) the DNA was lost in chimpanzees C) the DNA codes for a protein D) that humans are derived from chimpanzees E) none of these hypotheses is supported by this observation Answer: B Section: D.4 Skill: Applying/Analyzing LO: SD.3 10) In which of the following continents would you expect the human population to have the fewest alleles derived from Denisovans?

A) African B) Europe C) Asia D) South America E) North America Answer: A Section: D.4 Skill: Applying/Analyzing LO: SD.4a

3 .

11) Portions of the nuclear DNA of many modern humans have similarity to either Denisovan or Neanderthal nuclear DNA, however human mitochondrial DNA (mtDNA) is more similar to Denisovan mtDNA than to Neanderthal mtDNA. Why might this be the case?

A) Modern humans are derived from matings between Neanderthal females and human males B) Modern humans are derived from matings between Denisovan females and human males C) Modern humans are derived from matings between Denisovan males and human females D) Modern humans are derived from matings between Denisovan females and Neanderthal males E) Modern humans are the common ancestors of both Denisovans and Neanderthals Answer: B Section: D.4 Skill: Applying/Analyzing LO: SD.4a 12) Human-specific traits are found by comparing humans to what?

A) whales B) plants C) other humans D) apes E) Denisovans Answer: D Section: D.4 Skill: Applying/Analyzing LO: SD.3 13) The loss of genetic diversity at linked neutral loci near a selected site is evidence of what?

A) migration B) a haplotype C) inbreeding D) hitchhiking E) genetic drift Answer: D Section: D.4 Skill: Knowledge/Comprehension LO: SD.6a 14) Mutations affecting ________ can alter the expression of a gene without affecting its coding sequence.

A) non-coding sequence B) coding sequence C) exons D) amino acid sequence E) mutations in any of these regions could affect a gene's coding sequence Answer: A Section: D.4 Skill: Applying/Analyzing LO: SD.6b 4 .

15) Selection for an allele is expected to have what effect on neutral alleles located on the same chromosome as the selected allele?

A) the neutral sites will decrease in frequency in the population B) the neutral sites will not change in frequency in the population, because they are neutral C) the neutral sites will increase in frequency in the population D) the neutral alleles will become selectively advantageous E) there will not be an effect of the selected allele on the neutral allele because alleles of different gene assort independently Answer: C Section: D.4 Skill: Knowledge/Comprehension LO: SD.6a 16) The selection for different alleles affecting the expression of the LCT gene in milk-drinking human populations is an example of?

A) genetic drift B) hitchhiking C) common ancestry D) a selective sweep E) convergent evolution Answer: E Section: D.6 Skill: Applying/Analyzing LO: SD.6b 17) Selection for two different alleles of the hypoxia inducible factor pathway gene (EPAS1), one in Tibetan and the other in Ethiopian populations suggests that these two population share what?

A) recent common ancestry B) no common ancestry C) selective pressure for high gene expression D) This result suggests that these two populations have nothing in common.

E) selective pressure related to altitude Answer: E Section: D.6 Skill: Synthesis/Evaluation LO: SD.5a

5 .

18) Darker skin pigmentation in the ancestors to all humans may have been an adaptation to increase UV-resistance that followed the loss of fir in early hominids. The subsequent loss of pigmentation in at least two human populations may have conferred what advantage in those populations, both of which inhabited areas with less sun exposure?

A) less pigmentation increased vitamin D synthesis B) less pigmentation increased melanin synthesis C) less pigmentation increased energy needs D) less pigmentation occurred not by selection, but by genetic drift E) less pigmentation decreased vitamin D synthesis Answer: A Section: D.6 Skill: Applying/Analyzing LO: SD.5a 19) Domestication had what genetic effects on Maize compared to Teosinte?

A) domestication increased the diversity of Maize B) domestication decreased the diversity of Maize C) domestication had no genetic effect on Maize D) domestication increased the number of genes in Maize E) domestication decreased the number of genes in Maize Answer: B Section: D.7 Skill: Synthesis/Evaluation LO: SD.7b 20) The ability to travel rapidly over the Earth is expected to have what effect on human populations?

A) the differentiation between populations with decrease B) the differentiation between populations with increase C) the differentiation between populations with remain steady D) human populations will diverge from each other E) none of these is expected to result from rapid global travel Answer: A Section: D.8 Skill: Applying/Analyzing LO: SD.1

6 .

Genetic Analysis: An Integrated Approach, 3e (Sanders) Application E Forensic Genetics Application E.1 Multiple-Choice Questions 1) Genetic markers used for forensic analysis are usually ________.

A) dominant, so the presence of one of the two alleles is obvious B) codominant, where each allele can be detected separately C) codominant, where only the dominant allele can be detected D) recessive, where an individual is positive if they are homozygous E) codominant, where only the recessive allele can be detected Answer: B Section: E.1 Skill: Knowledge/Comprehension LO: SE.1c 2) CODIS markers are chosen that do not occur in coding regions of genes so that they do not disrupt what?

A) promoters B) intergenic regions C) introns D) gene expression E) codons Answer: E Section: E.1 Skill: Knowledge/Comprehension LO: SE.1c 3) CODIS markers occur at no greater than 20% for any one allele so each marker has at least 5 alleles in a population, these features help ensure high levels of what?

A) heritability B) inbreeding C) heterozygosity D) homozygosity E) similarity between individuals Answer: C Section: E.1 Skill: Knowledge/Comprehension LO: SE.1c

1 .

4) CODIS markers are analyzed ________.

A) by agarose gel electrophoresis B) only by the Federal Bureau Investigation (FBI) in certified crime labs C) by DNA sequencing D) using low-resolution electrophoresis E) by capillary electrophoresis Answer: E Section: E.1 Skill: Knowledge/Comprehension LO: SE.1b 5) Applying the exclusion principle to forensic genetics allows investigators to ________.

A) include all suspects with at least one allele that matches a crime scene B) include all suspects until guilt is established C) exclude any suspects with at least one allele that matches a crime scene D) exclude any suspects whose DNA does not contain an allele that was detected at the crime scene E) exclude any samples not found at the crime scene Answer: D Section: E.1 Skill: Applying/Analyzing LO: SE.4a 6) If the frequencies of two alleles in a population are 0.1 and 0.2, what is the chance that an individual randomly chosen from this population is heterozygous for these two alleles?

A) 0.1 * 0.2 = 2% B) 2 * 0.1 * 0.2 = 4% C) 0.1 * 0.1 = 1% D) 0.1 + 0.2 = 30% E) 2 * 0.1 + 0.2 = 60% Answer: B Section: E.1 Skill: Knowledge/Comprehension LO: SE.4a 7) If the frequencies of two alleles in a population are 0.1 and 0.2, what is the chance that an individual in this population is homozygous for the allele with the frequency of 0.1?

A) 0.1 / 0.1 = 1 B) 0.1 + 0.1 = 20% C) 0.1 * 0.1 = 1% D) 0.1 + 0.2 = 30% E) 2 * 0.1 * 0.1 = 2% Answer: C Section: E.1 Skill: Applying/Analyzing LO: SE.4a 2 .

8) For paternity testing to work, which of the following must be true?

A) A child must inherit half of their alleles from their father, the other half come from the mother.

B) A child can inherit all of their alleles from their father.

C) The true mother must be known.

D) The father must be homozygous for any allele that is tested.

E) The father and the mother cannot be identical twins.

Answer: A Section: E.1 Skill: Applying/Analyzing LO: SE.5d 9) Applying the exclusion principle to paternity testing allows investigators to ________.

A) exclude any father with at least one allele that does not match the child's non-maternal allele B) include all father with at least one allele that matches the child's non-maternal allele C) exclude all father with at least one allele that matches the child's non-maternal allele D) include any father with at least one allele that matches both of the child's alleles E) include any father with at least one gene that matches both of the child's alleles Answer: A Section: E.1 Skill: Applying/Analyzing LO: SE.5b 10) Which of the following could not be used for paternity testing?

A) DNA from hair B) DNA from the Y-chromosome C) CODIS markers D) mitochondrial DNA E) The child's DNA Answer: D Section: E.1 Skill: Applying/Analyzing LO: SE.6a 11) To help the Abuelas de Plaza de Mayo (Grandmothers of May Square) Mary-Claire King analyzed ________ to identify illegally adopted children?

A) DNA from hair B) DNA from the Y-chromosome C) CODIS markers D) mitochondrial DNA E) paternity testing Answer: D Section: E.1 Skill: Applying/Analyzing LO: SE.6a

3 .

12) First-degree relatives share 50% of their alleles, second-degree relatives share 25% of their alleles, and third-degree relatives share 12.5% of their alleles. Which of the following are third degree relatives?

A) the aunts or uncles of an individual B) parents and their children C) siblings D) grandparents of an individual E) none of these relationships is third-degree Answer: A Section: E.2 Skill: Applying/Analyzing LO: SE.7b 13) Genealogical analysis considers all of the following except?

A) the extent of shared SNPs between individuals B) place of birth C) the genotype of the tested individual D) date of birth E) the frequency of alleles in reference populations Answer: D Section: E.2 Skill: Knowledge/Comprehension LO: SE.7b 14) What is the statistical principle underlying genetic health risk assessment is called?

A) causality B) the placebo effect C) case/control D) association E) determinism Answer: D Section: E.2 Skill: Applying/Analyzing LO: SE.7e

4 .

15) The Paternity Index (PI) is calculated by dividing ________ by ________.

A) the probability that the nonmaternal allele comes from the putative father; the probability that the allele comes from another male in the population B) the probability putative father has the child's allele; the probability that the allele comes from another male in the population C) the probability that the nonmaternal allele comes from a random male; the probability that the allele comes from another male in the father's population D) the probability that the childs allele comes from the mother; the probability that the childs allele comes from the father E) the probability that the putative father is the real father; the probability that putative father is not the real father Answer: A Section: E.2 Skill: Knowledge/Comprehension LO: SE.5b 16) A putative father in paternity case is heterozygous for alleles 20 and 23 of the D21S11 gene.

The frequency of allele 20 is 0.1 and the frequency of allele 23 is 0.2. A child is born with a nonmaternal 23 allele at its D21S11 gene. What is the Paternity Index (PI) for this putative father in this case?

A) 1 / 0.2 = 5 B) 0.2 * 0.1 = 0.02 C) 0.2 / 0.2 = 1 D) 0.5 / 0.2 = 2.5 E) 0.2 / 0.1 = 2 Answer: D Section: E.2 Skill: Applying/Analyzing LO: SE.5b 17) A putative father in paternity case is homozygous for allele 23 of the D21S11 gene. The frequency of allele 23 is 0.1. A child is born with a nonmaternal 23 allele at its D21S11 gene.

What is the Paternity Index (PI) for this putative father in this case?

A) 1 / 0.1 = 10 B) 0.1 * 0.1 = 0.01 C) 0.1 / 0.1 = 1 D) 0.5 / 0.1 = 5 E) 1 * 0.1 = 0.1 Answer: A Section: E.2 Skill: Applying/Analyzing LO: SE.5b

5 .

18) Calculate the Combined Paternity Index (CPI) for a father with each of the following PIs: FGA10 = 1.25; D3S135816 = 3.0; TPOX33 = 4.4 A) 1.25 + 3.0 + 4.4 = 8.65 B) 1.25 + 3.0 + 4.4/100 = 0.0865 C) 1.25 * 3.0 * 4.4 = 16.5 D) (1) 1.25 + (0.5) 3.0 + (1) 4.4 = 7.15 E) 1.25 / 3.0 / 4.4 = 0.095 (rounded) Answer: A Section: E.2 Skill: Applying/Analyzing LO: SE.7b 19) Siblings are expected to share ________% of their alleles with their nieces or nephews.

A) 50% B) 5% C) 25% D) 0% E) 12.5% Answer: C Section: E.2 Skill: Synthesis/Evaluation LO: SE.7b 20) Genetic health risk assessment data can be used to report all of the following except?

A) the presence of a disease-associated allele in an individual B) an individuals exposure to tobacco C) a parents risk of passing on a disease-associated allele to their children D) the contribution of the tested alleles to an individuals risk of developing a particular disorder E) the frequency of the disease causing allele in the population Answer: B Section: E.2 Skill: Applying/Analyzing LO: SE.7d

6 .