Test Bank for Concepts of Genetics 10th Edition

TEST BANK FOR CONCEPTS OF GENETICS 10TH EDITION

Essentials of Genetics, 10th Edition (Klug) Chapter 1 Introduction to Genetics 1) The CRISPR-Cas system potentially represents one of the most power techniques in genetics as a result of its role in associated with specific human disorders.

A) identifying genes B) editing genes C) producing new genes D) regulating genes E) transmitting genes Answer: B Section: Introduction Bloom's Taxonomy: Remembering/Understanding 2) In the 1600s, William Harvey studied reproduction and development. What is the term given to his theory which states that an organism develops from the fertilized egg by a succession of developmental events that lead to an adult?

A) preformation B) spontaneous generation C) cell theory D) transduction E) epigenesis Answer: E Section: 1.1 Bloom's Taxonomy: Remembering/Understanding 3) What is the term given to the theory which states that the gamete contains a complete miniature adult?

A) preformation B) transduction C) transformation D) conjugation E) cell theory Answer: A Section: 1.1 Bloom's Taxonomy: Remembering/Understanding 4) What is the term given to the theory which put forth the idea that living organisms could arise by incubating nonliving components?

A) spontaneous generation B) natural selection C) evolution D) preformation E) collective combination Answer: A Section: 1.1 Bloom's Taxonomy: Remembering/Understanding

5) Schleiden and Schwann proposed the cell theory which states that .

A) cells contain genetic information in their nucleus B) cells move from the various parts of the body to the reproductive organs to produce offspring C) cells are derived from preexisting cells D) cells propagate via asexual reproduction E) cells represent the basic units of heredity Answer: C Section: 1.1 Bloom's Taxonomy: Remembering/Understanding 6) Who, along with Alfred Wallace, formulated the theory of natural selection?

A) Gregor Mendel B) William Harvey C) Louis Pasteur D) Charles Darwin E) James Watson Answer: D Section: 1.1 Bloom's Taxonomy: Remembering/Understanding 7) Which of the following is an example of natural selection?

A) a bird's beak is able to effectively crack the seeds it encounters B) dog breeding by humans C) depending on the food a turtle eats, its shell may grow faster or slower D) sometime during human's life they break a bone E) bacteria can be effectively killed by treatment with bleach Answer: A Section: 1.1 Bloom's Taxonomy: Evaluating/Creating 8) Which of the following botanists was not involved with bringing Mendel's work to light in the 1900s?

A) Carl Correns B) Carl Linnaeus C) Hugo de Vries D) Erich Tschermak Answer: A Section: 1.1 Bloom's Taxonomy: Remembering/Understanding

9) Who was the Augustinian monk that conducted a decade of experiments on the garden pea, eventually showing that traits are passed from parents to offspring in predictable ways?

A) Francis Crick B) Alfred Wallace C) Hippocrates D) Aristotle E) Gregor Mendel Answer: E Section: 1.2 Bloom's Taxonomy: Remembering/Understanding 10) In many species, there are two representatives of each chromosome. In such species, the characteristic number of chromosomes is called the number. It is usually symbolized as .

A) haploid; n B) haploid; 2n C) diploid; 2n D) diploid; n E) polyploid; n Answer: C Section: 1.2 Bloom's Taxonomy: Remembering/Understanding 11) Genetics is defined as the branch of biology associated with A) heredity and variation B) mutation and recession C) transcription and translation D) diploid and haploid E) replication and recombination Answer: A Section: 1.2 Bloom's Taxonomy: Remembering/Understanding

.

12) Early in the twentieth century, Walter Sutton and Theodor Boveri noted that the behavior of chromosomes during meiosis is identical to the behavior of genes during gamete formation. They proposed that genes are carried on chromosomes, which led to the basis .

A) of the germ-plasm theory B) of the chromosome theory of inheritance C) of the law of independent assortment D) for the determination of DNA as genetic material E) of predicting patterns of inheritance Answer: B Section: 1.2 Bloom's Taxonomy: Remembering/Understanding

13) What is a simple definition of an allele?

Answer: An allele is an alternative form of a gene.

Section: 1.2 Bloom's Taxonomy: Remembering/Understanding 14) The observable feature of an organism is referred to as a A) genotype B) phenotype C) prototype D) karyotype E) bryophyte Answer: B Section: 1.2 Bloom's Taxonomy: Remembering/Understanding

.

15) Until the mid-1940s, many scientists considered proteins to be the likely candidates for the genetic material. Which of the following characteristics led scientists to believe DNA was NOT the genetic material?

A) DNA is more stable than protein.

B) DNA is less abundant than protein.

C) DNA has less variation than protein.

D) Protein can fold into may shapes.

E) DNA is less abundant than protein and DNA has less variation than protein.

Answer: E Section: 1.2 Bloom's Taxonomy: Applying/Analyzing 16) Which of the following is an example of heredity?

A) A man has low blood pressure due to medications.

B) Both moths and birds have wings and can fly.

C) Dalmation dogs all have spots.

D) Whales and fish both swim.

E) Flies and molluscs both have eyes.

Answer: C Section: 1.2 Bloom's Taxonomy: Evaluating/Creating 17) Which is not a component of DNA? A) mRNA B) deoxyribose sugar C) nitrogenous base D) phosphate Answer: A Section: 1.3 Bloom's Taxonomy: Remembering/Understanding

18) Distinguish the functions of DNA and RNA in a eukaryote.

Answer: DNA is responsible for the storage and replication of genetic information; RNA is involved in the expression of stored genetic information.

Section: 1.3 Bloom's Taxonomy: Applying/Analyzing 19) Genetic information contained in DNA that encodes for an amino acid is referred to as a(n) .

A) allele B) trait C) nucleotide D) codon E) genotype Answer: D Section: 1.3 Bloom's Taxonomy: Remembering/Understanding 20) Which of the following processes describes the formation of a complementary RNA molecule?

A) replication B) transcription C) translation D) mutation E) mosaicism Answer: B Section: 1.3 Bloom's Taxonomy: Remembering/Understanding 21) Reference is often made to adapter molecules when describing protein synthesis in that they allow amino acids to associate with nucleic acids. To what class of molecules does this term refer?

A) DNA B) protein C) mRNA D) amino acids E) tRNA Answer: E Section: 1.3 Bloom's Taxonomy: Remembering/Understanding

22) If a scientist changed a cell's ionic composition and complementarity between DNA strands could no longer occur, what would the scientist first detect?

A) DNA becomes single stranded B) DNA strands become shorter C) RNA would start binding to DNA D) ribosomes would move into the nucleus E) cell membranes would become less permeable Answer: A Section: 1.3 Bloom's Taxonomy: Evaluating/Creating 23) Sickle-cell anemia, which is associated with blockage of blood flow in capillaries and small blood vessels, causing severe pain and damage to the heart, brain, muscles, and kidneys is the result of .

A) a large chromosome deletion B) a small chromosome deletion C) a change in a single nucleotide D) an environmental pathogen E) a hormonal variation Answer: C Section: 1.3 Bloom's Taxonomy: Remembering/Understanding 24) Which of the following contains all the others?

A) double helix B) nucleotide C) hydrogen bond D) DNA strand E) sugar Answer: A Section: 1.3 Bloom's Taxonomy: Applying/Analyzing 25) Once a protein is made, its biochemical or structural properties play a role in producing .

A) genotype B) phenotype C) mutant D) chromosome E) DNA Answer: B Section: 1.3 Bloom's Taxonomy: Applying/Analyzing

26) A primary discovery to generate recombinant DNA molecules was the use of A) spliceosomes B) restriction enzymes C) microscopes D) bioinformatics E) x-ray diffraction Answer: B Section: 1.4 Bloom's Taxonomy: Remembering/Understanding

.

27) A is an organism produced by biotechnology that involves the transfer of hereditary traits across species.

A) transgenic organism B) mutant C) clone D) vector E) frankenfood Answer: A Section: 1.5 Bloom's Taxonomy: Remembering/Understanding 28) The study of the structure, function, and evolution of genes and genomes is referred to as .

A) genomics B) proteomics C) bioinformatics D) genetics E) cell theory Answer: A Section: 1.6 Bloom's Taxonomy: Remembering/Understanding 29) is a discipline involved in the development of both hardware and software for processing, storing, and retrieving nucleotide and protein data.

A) Bioinformatics B) Genomics C) Recombinant DNA technology D) Cloning E) Proteomics Answer: A Section: 1.6 Bloom's Taxonomy: Remembering/Understanding

30) There are various technologies to knock out genes in organisms, what is the primary rationale for performing these experiments?

Answer: To determine the function of the gene based on changes in phenotype Section: 1.6 Bloom's Taxonomy: Applying/Analyzing 31) Organisms that are well understood from a scientific standpoint and are often used in basic biological research are often called .

A) clones B) vectors C) recombinant DNA technology D) model organisms E) restriction enzymes Answer: D Section: 1.7 Bloom's Taxonomy: Remembering/Understanding 32) The most accurate depiction of the chronological order of genetic discoveries would be .

A) chromosome theory of inheritance, Mendelian genetics, recombinant DNA technology, determination of DNA structure B) determination of DNA structure, chromosome theory of inheritance, Mendelian genetics, recombinant DNA technology C) Mendelian genetics, chromosome theory of inheritance, determination of DNA structure, recombinant DNA technology D) Mendelian genetics, determination of DNA structure, chromosome theory of inheritance, recombinant DNA technology Answer: C Section: 1.8 Bloom's Taxonomy: Remembering/Understanding

Essentials of Genetics, 10th Edition (Klug) Chapter 2 Mitosis and Meiosis 1) Living organisms are categorized into two major groups based on the presence or absence of a nucleus. What group is defined by the presence of a nucleus?

A) eukaryotic organism B) virus C) eubacterium D) archaea E) bacteriophage Answer: A Section: 2.1 Bloom's Taxonomy: Remembering/Understanding 2) What is the name of the membranous structure that compartmentalizes the cytoplasm of eukaryotic organisms?

A) ribosome B) mitochondria C) cytosol D) endoplasmic reticulum E) nucleoid Answer: D Section: 2.1 Bloom's Taxonomy: Remembering/Understanding 3) Organized by the centrioles, what structures are important in the movement of chromosomes during cell division?

A) mitochondria B) chloroplasts C) cell walls D) spindle fibers E) centromeres Answer: D Section: 2.1 Bloom's Taxonomy: Remembering/Understanding 4) Which of the following is not a true statement about mitochondria?

A) the mitochondrial genome is contained in the nucleus B) the mitochondria duplicate themselves C) the mitochondria transcribe and translate their own genetic information D) mitochondria are found in plants and animals E) mitochondria are sites for cellular respiration Answer: A Section: 2.1 Bloom's Taxonomy: Applying/Analyzing

1 Copyright © 2021 Pearson Education Ltd.

5) The nucleolus organizer region (NOR) is responsible for production of what type of cell structure?

A) nucleolus B) ribosome C) chromatids D) mitochondria E) endoplasmic reticulum Answer: B Section: 2.1 Bloom's Taxonomy: Remembering/Understanding 6) Which of the following terms defines a chromosome in which the centromere is near one end, but not at the end of a chromosome?

A) acentric B) submetacentric C) metacentric D) acrocentric E) telocentric Answer: D Section: 2.2 Bloom's Taxonomy: Remembering/Understanding 7) The diploid chromosome number of an organism is usually represented as 2n. Humans have a diploid chromosome number of 46. What would be the expected haploid chromosome number in a human?

A) 92 B) 16 C) 12 D) 24 E) 23 Answer: E Section: 2.2 Bloom's Taxonomy: Applying/Analyzing 8) Which of the following is incorrect?

A) a locus is a gene site on a chromosome B) an allele is an alternate form of the same gene C) sex chromosomes are not strictly homologous D) homologous chromosomes contain identical genetic information E) a karyotype is generated from a metaphase spread Answer: D Section: 2.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

9) Which of the following is true about sex-determining chromosomes?

A) They are independent during meiosis.

B) They do not participate in meiosis.

C) They act like homologous chromosomes during meiosis so each gamete will get one sex chromosome.

D) They have the same gene configuration and same loci.

E) They are always metacentric.

Answer: C Section: 2.2 Bloom's Taxonomy: Applying/Analyzing 10) During interphase of the cell cycle, .

A) centrioles migrate B) sister chromatids move to opposite poles C) the nuclear membrane disappears D) cytokinesis occurs E) DNA content essentially doubles Answer: E Section: 2.3 Bloom's Taxonomy: Remembering/Understanding 11) What significant genetic function occurs in the S phase of the cell cycle?

A) cytokinesis B) karyokinesis C) DNA synthesis D) chromosome condensation E) centromere division Answer: C Section: 2.3 Bloom's Taxonomy: Remembering/Understanding 12) When cells withdraw from the continuous cell cycle and enter a "quiescent" phase, they are said to be in what stage?

A) G1 B) G2 C) G0 D) M E) S Answer: C Section: 2.3 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

13) A cell in the G2 phase of the cell cycle .

A) contains twice as much DNA as in the G1 phase B) contains twice as many chromosomes as in the G1 phase C) contains twice as much DNA and twice as many chromosomes as in the G1 phase D) contains the same amount of DNA as in the G1 phase E) contains the same amount of chromatids as in the G1 phase Answer: A Section: 2.3 Bloom's Taxonomy: Applying/Analyzing 14) Which of the following represents the correct sequence of events in mitosis?

A) prophase-metaphase-prometaphase-anaphase-telophase B) telophase-prophase-prometaphase-metaphase-anaphase C) prophase-prometaphase-metaphase-anaphase-telophase D) prometaphase-prophase-metaphase-anaphase-telophase E) anaphase-metaphase-prometaphase-prophase-telophase Answer: C Section: 2.3 Bloom's Taxonomy: Remembering/Understanding 15) Migration of chromosomes is made possible by the binding of the spindle to the A) kinetochore B) telomere C) centriole D) equatorial plate E) centrosome Answer: A Section: 2.3 Bloom's Taxonomy: Remembering/Understanding 16) Which protein directly holds the chromatid arms together prior to anaphase of mitosis?

A) shugoshin B) cohesin C) separase D) middle lamella E) histone Answer: B Section: 2.3 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

.

17) The event referred to as disjunction occurs during A) prophase B) prometaphase C) metaphase D) anaphase E) telophase Answer: D Section: 2.3 Bloom's Taxonomy: Remembering/Understanding

.

18) Normal diploid somatic (body) cells of the mosquito Culex pipiens contain six chromosomes.

Assuming that all nuclear DNA is restricted to chromosomes and that the amount of nuclear DNA essentially doubles during the S phase of interphase, how much nuclear DNA would be present in metaphase I of mitosis? Note: Assume that the G1 nucleus of a mosquito cell contains 3.0 × 10-12 grams of DNA. A) 3.0 × 10−12 g B) 6.0 × 10−12 g C) 1.5 × 10−12 g D) 0.75 × 10−12 g E) 12 × 10−12 g Answer: B Section: 2.3 Bloom's Taxonomy: Evaluating/Creating 19) At which stage of cell division do sister chromatids go to opposite poles?

A) mitotic anaphase and anaphase of meiosis I B) mitotic anaphase and anaphase of meiosis II C) anaphase of meiosis I only D) mitotic anaphase only E) when they don't get along Answer: B Section: 2.4 Bloom's Taxonomy: Remembering/Understanding 20) Which of the following are the areas where chromatids intertwine during meiosis?

A) synapsis B) chiasma C) tetrad D) bivalent E) nondisjunction Answer: B Section: 2.4 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

21) During meiosis, chromosome number reduction takes place in A) anaphase II B) anaphase I C) metaphase I D) prophase I E) telophase II Answer: B Section: 2.4 Bloom's Taxonomy: Remembering/Understanding 22) A bivalent at prophase I contains A) one B) two C) three D) four E) eight Answer: D Section: 2.4 Bloom's Taxonomy: Evaluating/Creating

.

chromatids.

23) In an organism with 52 chromosomes, how many tetrads would be expected to form during meiosis?

A) 13 B) 26 C) 52 D) 104 E) 208 Answer: B Section: 2.4 Bloom's Taxonomy: Evaluating/Creating 24) In an organism with 52 chromosomes, how many chromatids would be expected in each cell after the second meiotic division?

A) 52 B) 26 C) 13 D) 104 E) 208 Answer: B Section: 2.4 Bloom's Taxonomy: Evaluating/Creating

6 Copyright © 2021 Pearson Education Ltd.

25) The meiotic cell cycle involves number of cell division(s) and of DNA replication(s).

A) one; one B) one; two C) two; one D) two; two E) two; zero Answer: C Section: 2.4 Bloom's Taxonomy: Remembering/Understanding

number

26) If a typical somatic cell has 64 chromosomes, how many chromosomes are expected in each gamete of that organism?

A) 8 B) 16 C) 32 D) 64 E) 128 Answer: C Section: 2.4 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

27) The ant, Myrmecia pilosula, is found in Australia and is named bulldog because of its aggressive behavior. It is particularly interesting because it carries all its genetic information in a single pair of chromosomes. In other words, 2n = 2. (Males are haploid and have just one chromosome.) Which of the following figures would most likely represent a correct configuration of chromosomes in a metaphase I cell of a female?

A)

B)

C)

D)

E)

Answer: A Section: 2.4 Bloom's Taxonomy: Evaluating/Creating 8 Copyright © 2021 Pearson Education Ltd.

28) For the purposes of this question, assume that a G1 somatic cell nucleus in a female Myrmecia pilosula contains 2 picograms of DNA. How much DNA would be expected in a metaphase I cell of a female?

A) 16 picograms B) 32 picograms C) 8 picograms D) 4 picograms E) Not enough information is provided to answer the question.

Answer: D Section: 2.4 Bloom's Taxonomy: Evaluating/Creating 29) What is the outcome of synapsis, a significant event in meiosis?

A) side-by-side alignment of nonhomologous chromosomes B) dyad formation C) monad movement to opposite poles D) side-by-side alignment of homologous chromosomes E) chiasma segregation Answer: D Section: 2.4 Bloom's Taxonomy: Remembering/Understanding 30) Which of the following is true about the second meiotic division?

A) Sister chromatids disjoin and are pulled to opposite poles.

B) Homologous chromosomes are pulled apart resulting in dyads at each pole.

C) Nondisjunction would lead to extra bivalents forming.

D) Synapsis occurs in the second meiotic division.

E) The products are four identical gametes.

Answer: A Section: 2.4 Bloom's Taxonomy: Remembering/Understanding

9 Copyright © 2021 Pearson Education Ltd.

31) The accompanying sketch depicts a cell from an organism in which 2n = 2 and each chromosome is metacentric.

Which of the following is the correct stage for this sketch?

A) anaphase of mitosis B) anaphase of meiosis I C) anaphase of meiosis II D) telophase of mitosis E) telophase of meiosis II Answer: C Section: 2.4 Bloom's Taxonomy: Applying/Evaluating 32) The horse (Equus caballus) has 32 pairs of chromosomes, whereas the donkey (Equus asinus) has 31 pairs of chromosomes. How many chromosomes would be expected in the somatic tissue of a mule, which is a hybrid of these two animals?

A) 63 B) 64 C) 62 D) 60 E) 126 Answer: A Section: 2.4 Bloom's Taxonomy: Applying/Analyzing 33) In figure 2-10, an organism with a haploid number of 2 (diploid 4) generates 4 combinations of chromosomes at the end of meiosis. An organism with a haploid number of 4 would generate 16 chromosomal combinations, and that of 6 would generate 64 chromosomal combinations.

Based on this pattern, an organism with a haploid number of 10 will produce combinations of chromosomes at the end of meiosis.

A) 10 B) 100 C) 10,000 D) 32 E) 1024 Answer: E Section: 2.4 Bloom's Taxonomy: Evaluating/Creating 10 Copyright © 2021 Pearson Education Ltd.

34) An organism with a diploid chromosome number of 46 will produce of chromosomes at the end of meiosis.

A) 23 B) 46 C) 8388608 D) 529 E) 7.04 × 1013 Answer: C Section: 2.4 Bloom's Taxonomy: Evaluating/Creating

combinations

35) Which of the following is not a source of genetic variation in meiosis?

A) crossing over B) law of independent assortment C) the random lining up of chromosomes on the metaphase plate D) tetrad formation E) polar body formation Answer: E Section: 2.5 Bloom's Taxonomy: Remembering/Understanding 36) In a healthy male, how many sperm cells would be expected to form from (a) 400 primary spermatocytes? (b) 400 secondary spermatocytes?

A) (a) 800; (b) 800 B) (a) 1600; (b) 1600 C) (a) 1600; (b) 800 D) (a) 400; (b) 400 E) (a) 400; (b) 800 Answer: C Section: 2.5 Bloom's Taxonomy: Applying/Analyzing 37) In a healthy female, how many secondary oocytes would be expected to form from 100 primary oocytes? How many first polar bodies would be expected from 100 primary oocytes?

A) 200; 50 B) 100; 50 C) 200; 200 D) 100; 100 E) 50; 50 Answer: D Section: 2.5 Bloom's Taxonomy: Applying/Analyzing

11 Copyright © 2021 Pearson Education Ltd.

38) There is about as much nuclear DNA in a primary spermatocyte as in A) 0.5 B) 1 C) 2 D) 3 E) 4 Answer: E Section: 2.5 Bloom's Taxonomy: Applying/Analyzing

spermatids.

39) List, in order of appearance, all the cell types expected to be formed during spermatogenesis.

A) spermatogonia, primary spermatocyte, secondary spermatocyte, spermatid, spermatozoa B) primary spermatocyte, secondary spermatocyte, spermatozoa, spermatid, spermatogonia C) spermatozoa, spermatid, spermatogonia , primary spermatocyte, secondary spermatocyte D) spermatogonia, spermatozoa, spermatid, primary spermatocyte, secondary spermatocyte E) primary spermatocyte, secondary spermatocyte, spermatid, spermatozoa, spermatogonia Answer: A Section: 2.5 Bloom's Taxonomy: Remembering/Understanding 40) In plants, spores are produced by the process of A) binary fission B) mitosis C) meiosis D) oogenesis E) spormatogenesis Answer: C Section: 2.6 Bloom's Taxonomy: Remembering/Understanding 41) Mitotic chromosomes are estimated to be interphase.

A) 50 B) 100 C) 500 D) 1,000 E) 5,000 Answer: E Section: 2.7 Bloom's Taxonomy: Remembering/Understanding

.

fold more compacted than chromatin in

12 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 3 Mendelian Genetics 1) The investigative approach of genetics attributed to Gregor Mendel is A) population genetics B) cytogenetics C) transmission genetics D) molecular genetics E) biochemical genetics Answer: C Section: Introduction Bloom's Taxonomy: Remembering/Understanding

.

2) Mendel's significant contributions in genetics published in 1866 went largely unnoticed until the discovery of .

A) chromosomal patterns during meiosis B) the double helical structure of DNA C) methods to sequence DNA D) recombinant DNA technology E) confirmation with fruit fly crosses Answer: A Section: Introduction Bloom's Taxonomy: Remembering/Understanding 3) Mendel utilized the garden pea, Pisum sativum for his studies based on features that include all but .

A) easy to grow B) being able to hybridize artificially C) reaches maturity in a single season D) has visible features with a wide spectrum of intermediates E) can generate a large number of progeny Answer: D Section: 3.1 Bloom's Taxonomy: Remembering/Understanding 4) Which of the following statements was not a contributing factor in the generation of Mendel's first three postulates?

A) true-breeding plants were used for the parental generation B) all offspring of the first filial generation demonstrated the same trait C) reciprocal crosses did not change the patterns of inheritance D) the proposal that there are basic units of heredity to explain results of monohybrid crosses E) monohybrid crosses for several analyzed characteristics yielded similar patterns of inheritance Answer: B Section: 3.2 Bloom's Taxonomy: Applying/Analyzing

1 Copyright © 2021 Pearson Education Ltd.

5) The physical expression of a trait is referred to as a(n) A) allele B) genotype C) phenotype D) gene E) homozygote Answer: C Section: 3.2 Bloom's Taxonomy: Remembering/Understanding

.

6) The process that leads to the development of haploid gametes is best described as A) segregation B) independent assortment C) Mendelian inheritance D) replication E) dominance or recessiveness Answer: A Section: 3.2 Bloom's Taxonomy: Remembering/Understanding

.

7) Mendel indicated that traits were made up of unit factors. Today, we call unit factors .

A) genotypes B) phenotypes C) alleles D) characters E) genes Answer: C Section: 3.2 Bloom's Taxonomy: Remembering/Understanding 8) Assuming a typical monohybrid cross in which one allele is completely dominant to the other, what phenotypic ratio is expected if the F1s are crossed?

A) 3:1 B) 2:1 C) 3:2 D) 4:3 E) 1:1 Answer: A Section: 3.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

9) Mendel crossed two pea plants with round seeds. All seeds of the offspring were round. He then crossed a plant with round seeds to a plant with wrinkled seeds and all offspring had wrinkled seeds. Which of the following is true?

A) round is dominant B) wrinkled is dominant C) the plants he used were not true breeding D) a mutation occurred E) the trait does not breed true Answer: B Section: 3.2 Bloom's Taxonomy: Applying/Analyzing 10) A cross between two individuals with different phenotypes that resulted in approximately 50% of each type of offspring would indicate the cross was .

A) true breeding dominant to recessive B) a heterozygous dominant crossed to a heterozygous recessive C) a heterozygous dominant crossed to a homozygous recessive D) a homozygous recessive crossed to a heterozygous recessive E) true dominant to a heterozygous dominant Answer: C Section: 3.2 Bloom's Taxonomy: Applying/Analyzing 11) Mendel's Law of Segregation is supported by a A) 1:1 B) 2:1 C) 3:2 D) 3:1 E) 5:2 Answer: A Section: 3.2 Bloom's Taxonomy: Remembering/Understanding

testcross ratio.

12) If an F2 generation from a F1 self-cross always yields offspring in a 3:1 phenotypic ratio, which of the following P crosses could have occurred?

A) AA × aa B) AA × AA C) aa × aa D) Aa × AA E) aa × Aa Answer: A Section: 3.2 Bloom's Taxonomy: Evaluating/Creating

3 Copyright © 2021 Pearson Education Ltd.

13) A recessive allele in dogs causes white spots. If two solid colored dogs are mated and produce a spotted offspring, what is the percentage chance their next puppy would be solid colored?

A) 25% B) 50% C) about 66% D) 75% E) about 90% Answer: D Section: 3.2 Bloom's Taxonomy: Applying/Analyzing 14) Polydactyly is expressed when an individual has extra fingers and/or toes, and is transmitted via an autosomal dominant allele. Assume that a man with six fingers on each hand and six toes on each foot marries a woman with a normal number of digits. The couple has a son with normal hands and feet, but the couple's second child has extra digits. What is the probability that their next child will have polydactyly?

A) 1/32 B) 1/8 C) 7/16 D) 1/2 E) 3/4 Answer: D Section: 3.2 Bloom's Taxonomy: Applying/Analyzing 15) Tightly curled or wooly hair is caused by a dominant gene in humans. If a heterozygous curly-haired person marries a person with straight hair, what percentage of their offspring would be expected to have straight hair?

A) 0% B) 25% C) 50% D) 75% E) 100% Answer: C Section: 3.2 Bloom's Taxonomy: Applying/Analyzing

4 Copyright © 2021 Pearson Education Ltd.

16) A researcher crossed two plants, and informed an assistant researcher to determine the genotypes and phenotypes of the plants that were crossed by analyzing the offspring. The assistant counted 30 plants with green pods and 10 plants with yellow pods. Which of the following conclusions can accurately be made by the assistant?

A) The researcher had crossed two plants with yellow pods.

B) The researcher had crossed true-breeding plants of differing pod colors.

C) The researcher had crossed two heterozygous plants.

D) The researcher had crossed a plant with green pods to a plant with yellow pods.

E) The researcher had selfed a true-breeding plant with green pods.

Answer: C Section: 3.2 Bloom's Taxonomy: Evaluating/Creating 17) Assume that you have a garden and some pea plants have solid leaves and others have striped leaves. You conduct a series of crosses [(a) through (e)] and obtain the results given in the table. Define gene symbols and give the possible genotypes of the parents of cross e.

A) solid is dominant to striped Ss × Ss B) solid is dominant to striped SS × ss C) solid is dominant to striped Ss × ss D) striped is dominant to solid SS × ss E) striped is dominant to solid Ss × ss Answer: B Section: 3.2 Bloom's Taxonomy: Evaluating/Creating 18) Albinism, lack of pigmentation in humans, results from an autosomal recessive gene. Two parents with normal pigmentation have an albino child. What is the probability that their next two children will be albino?

A) 1/4 B) 1/3 C) 1/2 D) 3/4 E) 1/16 Answer: E Section: 3.2 Bloom's Taxonomy: Applying/Analyzing 5 Copyright © 2021 Pearson Education Ltd.

19) Albinism, lack of pigmentation in humans, results from an autosomal recessive gene. Two parents with normal pigmentation have an albino child. What is the probability that their next child will be wild type?

A) 1/2 B) 3/4 C) 1/8 D) 0 E) 3/16 Answer: B Section: 3.2 Bloom's Taxonomy: Applying/Analyzing 20) The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that their first child will have brachydactyly?

A) 1/4 B) 1/2 C) 1/8 D) 3/4 E) 2/3 Answer: B Section: 3.2 Bloom's Taxonomy: Applying/Analyzing 21) Tightly curled hair is caused by a dominant autosomal gene in humans. If a heterozygous curly-haired person marries a person with straight hair, what phenotypes (and in what proportions) are expected in the offspring?

A) 1 curly : 1 straight B) 2 curly : 1 straight C) 3 curly : 1 straight D) 1 curly : 2 straight E) 1 curly : 3 straight Answer: A Section: 3.2 Bloom's Taxonomy: Applying/Analyzing

6 Copyright © 2021 Pearson Education Ltd.

22) A certain type of congenital deafness in humans is caused by a rare autosomal dominant gene. In a mating involving a deaf man and a deaf woman, could all the children have normal hearing?

A) No, because it is dominant. Children always get the dominant alleles.

B) No, because children favor their parents.

C) Yes, assuming that the parents are heterozygotes (because the gene is rare), it is possible that all of the children could have normal hearing.

D) Yes, because traits assort independently.

E) Yes, because it must be recessive if it's rare.

Answer: C Section: 3.2 Bloom's Taxonomy: Evaluating/Creating 23) A certain type of congenital deafness in humans is caused by a rare autosomal recessive gene. In a mating involving a deaf man and a deaf woman, could some of the children have normal hearing?

A) Yes, because it's rare.

B) Yes, because traits assort independently.

C) No, since the gene in question is recessive, both of the parents are homozygous and one would not expect normal hearing in the offspring.

D) No, traits get passed down from father to child and the father is deaf.

E) No, because the parents would teach their children sign language.

Answer: C Section: 3.2 Bloom's Taxonomy: Evaluating/Creating 24) Which of the following postulates was not attainable from a monohybrid cross?

A) unit factors are in pairs B) dominance/recessiveness C) segregation D) independent assortment Answer: D Section: 3.2, 3.3 Bloom's Taxonomy: Remembering/Understanding 25) What are expected ratios for monohybrid and dihybrid testcrosses, respectively in which the parent with the dominant phenotype is heterozygous for all traits?

A) 1:1 and 9:3:3:1 B) 1:1 and 1:1:1:1 C) 3:1 and 9:3:3:1 D) 3:1 and 1:1:1:1 Answer: B Section: 3.2, 3.3 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

26) Which of the following describes the product law?

A) The probably of two linked events occurring simultaneously is equal to the probabilities of each individual event.

B) The probability of two or more independent events occurring simultaneously is equal to the product of their individual probabilities.

C) The probability of two or more independent events occurring simultaneously is equal to the sum of their individual probabilities.

D) The product of the sum of the probabilities of two simultaneous events describes the ability of those two events to occur together.

E) The probabilities of two events occurring in a specific sequence are the product of their individual probabilities squared.

Answer: B Section: 3.3 Bloom's Taxonomy: Remembering/Understanding 27) Which phenotypic ratio is likely to occur in crosses of two completely dominant, independently segregating gene pairs when both parents are fully heterozygous?

A) 2:4:6:8 B) 3:2:3:1 C) 9:3:3:1 D) 1:1:1:1 E) 3:1 Answer: C Section: 3.3 Bloom's Taxonomy: Remembering/Understanding 28) What is segregation?

A) During gamete formation, segregating pairs of unit factors assort independently of each other.

B) Fertilization is random.

C) During gamete formation, allele pairs are separated to form haploid gametes.

D) Genes lie on chromosomes.

E) Chromosomes can swap information during meiosis.

Answer: C Section: 3.2, 3.3 Bloom's Taxonomy: Remembering/Understanding 29) What is independent assortment?

A) During gamete formation, a pair of unit factors segregates randomly from another pair of unit factors.

B) Fertilization is random.

C) During gamete formation, allele pairs are separated to form haploid gametes.

D) Genes lie on chromosomes.

E) Chromosomes can swap information during meiosis.

Answer: A Section: 3.3 Bloom's Taxonomy: Remembering/Understanding 8 Copyright © 2021 Pearson Education Ltd.

30) When performing a dihybrid cross for the traits round/wrinkled and yellow/green, which of the following results best demonstrates that the unit factors from each trait was assorting independently?

A) the 9:3:3:1 ratio of the F2 generation B) finding a 3:1 ratio of round:wrinkled and a 3:1 ratio of yellow:green in the F2 generation C) finding the Punnett square and branched diagram methods yielded the same results D) finding only two visible traits in the F1 generation, but four visible traits among the F2 offspring E) finding that round is dominant to wrinkled and yellow is dominant to green Answer: B Section: 3.3 Bloom's Taxonomy: Applying/Analyzing 31) The phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body.

Assume that a cross is made between a fly the is homozygous for normal wings with a hairy body and a fly with vestigial wings that is homozygous for normal body hair. The wild-type F1 flies were crossed among each other to produce 1024 offspring. Which phenotypes would you expect among the 1024 offspring, and how many of each phenotype would you expect?

A) Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (256), vestigial (256), hairy (256), and vestigial hairy (256).

B) All wild type.

C) Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (192), vestigial (256), hairy (64), and vestigial hairy (192).

D) All vestigial hairy.

E) Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (576), vestigial (192), hairy (192), and vestigial hairy (64).

Answer: E Section: 3.3 Bloom's Taxonomy: Applying/Analyzing 32) Under what conditions does one expect a 1:1:1:1 ratio?

A) AABB × aabb B) AaBb × AaBb C) AaBb × aabb D) AAbb × aaBB E) Aabb × AABB Answer: C Section: 3.3 Bloom's Taxonomy: Remembering/Understanding

9 Copyright © 2021 Pearson Education Ltd.

33) What conditions are likely to apply if the progeny from the cross AaBb × AaBb appear in the 9:3:3:1 ratio?

A) complete dominance, independent assortment, and no gene interaction B) incomplete segregation and complete dominance C) dihybrid test cross D) gene interaction and independent assortment E) dihybrid cross with incomplete dominance Answer: A Section: 3.3 Bloom's Taxonomy: Remembering/Understanding 34) Which types of phenotypic ratios are likely to occur in a trihybrid crosses when dealing with three gene pairs with dominant and recessive alleles for which all the genotypic combinations are of equal viability if both parents are heterozygous for all three traits?

A) 27:9:9:9:3:3:3:1 B) 1:2:1, 3:1 C) 1:4:6:4:1, 1:1:1:1 D) 12:3:1, 9:7 E) 2:3, 1:2 Answer: A Section: 3.4 Bloom's Taxonomy: Remembering/Understanding 35) How many different possible gametes can be obtained from an organism with the following genotype AaBbCc? A) 3 B) 4 C) 8 D) 9 E) 16 Answer: C Section: 3.4 Bloom's Taxonomy: Remembering/Understanding 36) How many kinds of gametes will be expected from an individual with the genotype PpCcTTRr? A) 16 B) 8 C) 4 D) 2 E) 1 Answer: B Section: 3.4 Bloom's Taxonomy: Evaluating/Creating

10 Copyright © 2021 Pearson Education Ltd.

37) Using the branch diagram method, what would be the genotypic ratio of offspring from the following cross AaBBcc x AaBbCC? A) 27:9:9:9:3:3:3:1 B) 1:1:2:2:2:2:2:2:1:1 C) 1:1:4:4:2:2:1:1 D) 1:1:2:2:1:1 E) 9:3:3:1 Answer: D Section: 3.4 Bloom's Taxonomy: Applying/Analyzing 38) Using the branch diagram method, what would be the phenotypic ratio of offspring from the following cross AaBBcc × AaBbCC (assuming dominance/recessive at each locus and independent assortment)?

A) 27:9:9:9:3:3:3:1 B) 9:3:3:1 C) 1:1:2:2:1:1 D) 1:1:3:3:1:1 E) 3:1 Answer: E Section: 3.4 Bloom's Taxonomy: Applying/Analyzing 39) For the purposes of this question, assume that being Rh+ is a consequence of D and that Rh− individuals are dd. The ability to taste phenylthiocarbamide (PTC) is determined by the gene symbolized T (tt are nontasters). A female whose mother was Rh− has the MN blood group, is Rh+ and a nontaster of PTC, and is married to a man who is MM, Rh-, and a nontaster. Assume that all the loci discussed in this problem are autosomal and independently assorting. Which of the following is NOT a possible genotype of the children?

A) MMDdtt B) MMddtt C) MNDdtt D) MNddtt E) MNDDtt Answer: E Section: 3.4 Bloom's Taxonomy: Evaluating/Creating

11 Copyright © 2021 Pearson Education Ltd.

40) Two organisms, AABBCCDDEE and aabbccddee, are mated to produce an F1 that is selffertilized. If the capital letters represent dominant, independently assorting alleles. How many different genotypes will occur in the F2?

A) 35 = 243 B) 25 = 32 C) 45 = 1024 D) 43 = 64 E) 53 = 75 Answer: A Section: 3.4 Bloom's Taxonomy: Evaluating/Creating 41) Two organisms, AABBCCDDEE and aabbccddee, are mated to produce an F1 that is selffertilized. If the capital letters represent dominant, independently assorting alleles. What proportion of the F2 genotypes will be recessive for all five loci?

A) 1/1024 B) 1/256 C) 1/243 D) 1/64 E) 1/16 Answer: A Section: 3.4 Bloom's Taxonomy: Evaluating/Creating 42) Assuming independent assortment, what proportion of the offspring of the cross AaBbCcDd × AabbCCdd will have the aabbccdd genotype?

A) 0 B) 1/64 C) 1/256 D) 1/512 E) 1/1024 Answer: A Section: 3.4 Bloom's Taxonomy: Applying/Analyzing 43) How many different kinds of gametes can be produced by an individual with the genotype AABbCCddEeFf? A) 64 B) 32 C) 16 D) 8 E) 4 Answer: D Section: 3.4 Bloom's Taxonomy: Applying/Analyzing 12 Copyright © 2021 Pearson Education Ltd.

44) Assuming independent assortment and dominance/recessiveness for loci A, B and C, what proportion of offspring of the following cross between a mother of genotype AabbCc × a father of genotype AaBbCC would be phenotypically the same as the mother?

A) 27/64 B) 9/64 C) 3/8 D) 1/8 E) 1/64 Answer: C Section: 3.4 Bloom's Taxonomy: Evaluating/Creating 45) According to Mendel's model, because of the of chromosomes during meiosis, all possible combinations of gametes will be formed in equal frequency.

A) product rule B) law of segregation C) law of unit factors D) independent assortment E) chromosomal theory of inheritance Answer: D Section: 3.5 Bloom's Taxonomy: Remembering/Understanding 46) Assuming no crossing over between a gene in question and the centromere, when do alleles segregate during meiosis?

A) interphase B) prophase I C) anaphase I D) prophase II E) anaphase II Answer: C Section: 3.5 Bloom's Taxonomy: Remembering/Understanding 47) In tomato plants smooth texture (P) is dominant to peach (p), red color (R) is dominant to yellow (r), and normal leaf (B) is dominant to broad leaf (b). In the cross PpRrbb × Pprrbb what percent of the offspring will be smooth, yellow and broad?

A) 9/16 B) 3/16 C) 1/16 D) 3/8 E) 1/8 Answer: D Section: 3.4 Bloom's Taxonomy: Evaluating/Creating

13 Copyright © 2021 Pearson Education Ltd.

48) In tomato plants smooth texture (P) is dominant to peach (p), red color (R) is dominant to yellow (r), and normal leaf (B) is dominant to broad leaf (b). In the cross PpRrBb × PprrBb what proportion of the offspring will be peach, red, and normal leafed?

A) 9/64 B) 3/64 C) 3/32 D) 3/8 E) 1/16 Answer: C Section: 3.4 Bloom's Taxonomy: Evaluating/Creating 49) Which of the following groups of scientists were influential around the year 1900 in setting the stage for our present understanding of transmission genetics?

A) Beadle, Tatum, and Lederberg B) Watson, Crick, Wilkins, and Franklin C) de Vries, Correns, Tschermak, Sutton, and Boveri D) Darwin, Mendel, and Lamarck E) Hippocrates, Aristotle, and Kölreuter Answer: C Section: 3.5 Bloom's Taxonomy: Remembering/Understanding 50) The number of possible gametes, each with different chromosome compositions, is 2n, where n equals .

A) the diploid number B) the haploid number C) the number of genes D) the number of offspring E) the number of alleles Answer: B Section: 3.6 Bloom's Taxonomy: Remembering/Understanding 51) What meiotic process, relative to the number of chromosomes of a given species, accounts for a significant amount of genetic variation in gametes?

A) independent assortment of chromosomes B) trivalent formation C) bivalent formation D) pairing of homologous chromosomes E) formation of the meiotic spindle during chromosome segregation Answer: A Section: 3.6 Bloom's Taxonomy: Remembering/Understanding

14 Copyright © 2021 Pearson Education Ltd.

52) For an organism in which the diploid number of chromosomes equals 6, how many possible gamete combinations can be formed as a result of independent assortment (excluding the impact of crossing over)?

A) 3 B) 6 C) 8 D) 16 E) 64 Answer: C Section: 3.6 Bloom's Taxonomy: Applying/Analyzing 53) Albinism, lack of pigmentation in humans, results from an autosomal recessive gene (a).

Two parents with normal pigmentation have an albino child. What is the probability that their next child will be an albino girl?

A) 1/4 B) 1/3 C) 1/2 D) 3/4 E) 1/8 Answer: E Section: 3.7 Bloom's Taxonomy: Applying/Analyzing 54) Albinism, lack of pigmentation in humans, results from an autosomal recessive gene (a).

Two parents with normal pigmentation have an albino child. What is the probability that their next three children will be albino?

A) 1/8 B) 1/4 C) 3/4 D) 1/64 E) 0 Answer: D Section: 3.7 Bloom's Taxonomy: Applying/Analyzing 55) Albinism, lack of pigmentation in humans, results from an autosomal recessive gene. Two parents with normal pigmentation have an albino child. What is the probability that their next child will be a wild type girl?

A) 1/2 B) 3/4 C) 3/8 D) 0% E) 1/4 Answer: B Section: 3.7 Bloom's Taxonomy: Applying/Analyzing 15 Copyright © 2021 Pearson Education Ltd.

56) Albinism, lack of pigmentation in humans, results from an autosomal recessive gene. Two parents with normal pigmentation have an albino child. What is the probability that their next three children will be wild type?

A) 1/4 B) 3/64 C) 1/32 D) 27/64 E) 0 Answer: D Section: 3.7 Bloom's Taxonomy: Applying/Analyzing 57) The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that their first two children will have brachydactyly?

A) 1/4 B) 1/2 C) 1/8 D) 3/4 E) 2/3 Answer: A Section: 3.7 Bloom's Taxonomy: Applying/Analyzing 58) The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that their first child will be a brachydactylous girl?

A) 1/4 B) 1/2 C) 1/8 D) 3/4 E) 2/3 Answer: A Section: 3.7 Bloom's Taxonomy: Applying/Analyzing

16 Copyright © 2021 Pearson Education Ltd.

59) For which of the following questions would you use the sum law?

A) determining the chance of having a baby girl B) determining the chance of having a tall plant with purple flowers C) determining the change of rolling a 5 on a six-sided die D) determining the chance of pulling either a club or a heart from a deck of cards E) determining the chance of pulling an ace from a deck of cards Answer: D Section: 3.7 Bloom's Taxonomy: Applying/Analyzing 60) What is the probability of rolling a 2 or a 3 on a six-sided die?

A) 1/3 B) 1/6 C) 1/12 D) 1/18 E) 1/36 Answer: A Section: 3.7 Bloom's Taxonomy: Applying/Analyzing 61) The Chi-square test involves a statistical comparison between measured (observed) and predicted (expected) values. One generally determines degrees of freedom as .

A) the number of categories being compared B) one less than the number of classes being compared C) one more than the number of classes being compared D) ten minus the sum of the two categories E) the sum of the two categories Answer: B Section: 3.8 Bloom's Taxonomy: Remembering/Understanding 62) In what ways is sample size related to statistical testing?

A) By increasing sample size, one increases the reliability of the statistical test and decreases the likelihood of erroneous conclusions from chance fluctuations in the data.

B) By increasing sample size, one decreases the reliability of the statistical test and increases the likelihood of erroneous conclusions from chance fluctuations in the data.

C) There is no correlation.

D) By increasing sample size, one increases the reliability of the statistical test and increases the likelihood of erroneous conclusions from chance fluctuations in the data.

E) The larger the sample size is, the more likely you are to introduce errors into your data.

Answer: A Section: 3.8 Bloom's Taxonomy: Remembering/Understanding

17 Copyright © 2021 Pearson Education Ltd.

63) In a Chi-square analysis, what general condition causes one to reject (fail to accept) the null hypothesis?

A) when observed = expected B) when observed >> expected C) usually when the probability value is less than 0.5 D) usually when the probability value is less than 0.05 E) usually when the probability value is less than 0.005 Answer: D Section: 3.8 Bloom's Taxonomy: Remembering/Understanding 64) If one is testing a goodness of fit to a 9:3:3:1 ratio, how many degrees of freedom would be associated with the Chi-square analysis?

A) 1 B) 2 C) 3 D) 4 E) 5 Answer: C Section: 3.8 Bloom's Taxonomy: Remembering/Understanding 65) Which of the following is true?

A) Assume that a Chi-square test was conducted to test the goodness of fit to a 9:3:3:1 ratio and a Chi-square value of 10.62 was obtained. The null hypothesis should be accepted.

B) Assume that a Chi-square test was conducted to test the goodness of fit to a 3:1 ratio and that a Chi-square value of 2.62 was obtained. The null hypothesis should be accepted.

C) With the test of a 3:1 ratio, there are three degrees of freedom.

D) Assume that a Chi-square test provided a probability value of 0.02. The null hypothesis should be accepted.

E) The larger the Chi-square value, the more likely the difference of your observed results from the expected are due to chance.

Answer: B Section: 3.8 Bloom's Taxonomy: Remembering/Understanding 66) In a Chi-square test, as the value of the χ2 increases, the likelihood of rejecting the null hypothesis .

A) increases B) decreases C) stays the same D) is doubled E) increases by a factor of 10 Answer: A Section: 3.8 Bloom's Taxonomy: Applying/Analyzing 18 Copyright © 2021 Pearson Education Ltd.

67) In studies of human genetics, usually a single individual brings the condition to the attention of a scientist or physician. When pedigrees are developed to illustrate transmission of the trait, what term does one use to refer to this individual?

A) mutant B) proband C) cousin D) F1 B E) child Answer: B Section: 3.9 Bloom's Taxonomy: Remembering/Understanding 68) In a pedigree analysis a male child with a particular trait has two parents that do not exhibit the that trait. Which of the following represents the most likely scenario?

A) the trait is inherited as an autosomal dominant and the genotypes of the child and his parents can be determined B) the trait is inherited as an autosomal recessive and the genotypes of the child and his parents can be determined C) the trait is inherited as an autosomal dominant and the genotypes of the child and parents cannot be determined D) the trait is inherited as an autosomal recessive and the genotypes of the child and parents cannot be determined E) the trait is inherited as an autosomal recessive and the genotype of the child can be determined but not the genotypes of the parents Answer: B Section: 3.9 Bloom's Taxonomy: Applying/Analyzing 69) Rare autosomal dominant diseases are .

A) typically more severe in homozygous dominant individuals B) typically similar in severity in homozygous dominant and heterozygous individuals C) typically less severe in homozygous dominant individuals than heterozygous individuals D) typically more severe in homozygous recessive individuals than in homozygous dominant individuals E) typically not passed on to offspring Answer: A Section: 3.9 Bloom's Taxonomy: Applying/Analyzing

19 Copyright © 2021 Pearson Education Ltd.

70) Name the single individual whose work in the mid-1800s contributed to our understanding of the particulate nature of inheritance as well as the basic genetic transmission patterns. With what organism did this person work?

A) Gregor Mendel; Pisum sativum B) George Beadle; Neurospora C) Thomas Hunt Morgan; Drosophila D) Calvin Bridges; Drosophila E) Boris Ephrussi; Ephestia Answer: A Section: 3.1 Bloom's Taxonomy: Remembering/Understanding 71) Which types of phenotypic ratios are likely to occur in crosses when dealing with a single gene pair for which all the genotypic combinations are of equal viability?

A) 9:3:3:1, 27:9:9:9:3:3:3:1 B) 1:2:1, 3:1 C) 1:4:6:4:1, 1:1:1:1 D) 12:3:1, 9:7 E) 2:3, 1:2 Answer: B Section: 3.2 Bloom's Taxonomy: Evaluating/Creating 72) Assume that a black guinea pig crossed with an albino guinea pig produced 5 black offspring. When the albino was crossed with a second black guinea pig, 4 black and 3 albino offspring were produced. What genetic explanation would apply to these data?

A) albino = recessive; black = recessive B) albino = dominant; black = incompletely dominant C) albino and black = codominant D) albino = recessive; black = dominant E) albino = dominant; black = recessive Answer: D Section: 3.2 Bloom's Taxonomy: Evaluating/Creating 73) To test Mendel's Law of Segregation, the experimenter needs A) at least 500 offspring to count B) a minimum of two contrasting forms of a gene C) a DNA sequencing apparatus D) the ability to perform a test cross E) access to several generations of data Answer: B Section: 3.2 Bloom's Taxonomy: Remembering/Understanding

20 Copyright © 2021 Pearson Education Ltd.

.

74) You have identified a plant in your garden with a new flower color. You want to determine if this phenotype is dominant or recessive. Which cross would tell you this?

A) selfing the plant B) crossing the plant to a plant of the same type of any color C) crossing the new plant to one you know has the dominant trait D) crossing the plant to one you know has the recessive trait E) sequencing the DNA for the trait Answer: D Section: 3.2 Bloom's Taxonomy: Evaluating/Creating 75) Assuming complete dominance, a phenotypic ratio of monohybrid sib or self-cross.

A) 1:1 B) 2:1 C) 3:1 D) 5:2 E) 3:2 Answer: C Section: 3.2 Bloom's Taxonomy: Remembering/Understanding

is expected from a

76) Dentinogenesis imperfecta is a rare, autosomal, dominantly inherited disease of the teeth that occurs in about one in 8000 people. The teeth are somewhat brown in color, and the crowns wear down rapidly. Assume that a male with dentinogenesis imperfecta and no family history of the disease marries a woman with normal teeth. What is the probability that their first child will have dentinogenesis imperfecta?

A) 1/8 B) 1/4 C) 1/2 D) 3/4 E) 100% Answer: C Section: 3.2 Bloom's Taxonomy: Applying/Analyzing 77) Which types of phenotypic ratios are likely to occur in crosses when dealing with two gene pairs for which all the genotypic combinations are of equal viability?

A) 9:3:3:1 B) 1:2:1, 3:1 C) 6:6:2 D) 12:2:2 E) 2:3, 1:2 Answer: A Section: 3.3 Bloom's Taxonomy: Remembering/Understanding 21 Copyright © 2021 Pearson Education Ltd.

78) Which types of phenotypic ratios are likely to occur in testcrosses when dealing with two gene pairs for which all the genotypic combinations are of equal viability?

A) 9:3:3:1, 27:9:9:9:3:3:3:1 B) 1:2:1, 3:1 C) 1:1:1:1 D) 12:3:1, 9:7 E) 2:3, 1:2 Answer: C Section: 3.3 Bloom's Taxonomy: Remembering/Understanding 79) Under what conditions does one expect a 9:3:3:1 ratio?

A) monohybrid test cross (F2) with independently assorting, completely dominant genes B) monohybrid cross (F2) with independently assorting, completely dominant genes C) dihybrid test cross (F2) with independently assorting, completely dominant genes D) dihybrid cross (F1) with independently assorting, completely dominant genes E) dihybrid cross (F2) with independently assorting, completely dominant genes Answer: E Section: 3.3 Bloom's Taxonomy: Remembering/Understanding 80) Which of the following represents the trihybrid ratio if AaBbCc is selfed?

A) 9:3:3:1 B) 27:9:9:9:3:3:3:1 C) 1:1:1:1:1:1:1:1 D) 16:9:9:6:6:3:3:1 E) 18:12:15:9:9:3:3:1 Answer: B Section: 3.4 Bloom's Taxonomy: Remembering/Understanding 81) Which of the following best describes the relationship at the molecular level between mutant alleles and phenotype?

A) A mutant allele will cause the wild-type protein not to be made.

B) A mutant allele will cause only one phenotype.

C) A mutant allele can have different effects depending on the gene product's function.

D) The wild-type allele is always dominant.

E) Mutant alleles will only cause problems when seen in the homozygous form, no matter the gene affected.

Answer: C Section: 3.10 Bloom's Taxonomy: Applying/Analyzing

22 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 4 Modification of Mendelian Ratios 1) A mutation in a gene that results in a loss of a functional product of that gene best defines what type of mutation?

A) codominance B) incomplete dominance C) gain of function D) multiple allelism E) null Answer: E Section: 4.1 Bloom's Taxonomy: Remembering/Understanding 2) Gain-of-function mutations are associated with being A) dominant B) recessive C) neutral D) wild type E) incomplete dominance Answer: A Section: 4.1 Bloom's Taxonomy: Remembering/Understanding

.

3) If the wrinkled-wing mutation in Drosophila is dominant and designated Wr, a fruit fly with the genotype Wr+ Wr+ would exhibit what phenotype?

A) wrinkled wings B) wild-type wings C) partially wrinkled wings D) no wings E) one wrinkled wing and one normal wing Answer: B Section: 4.2 Bloom's Taxonomy: Applying/Analyzing 4) With incomplete dominance, a likely ratio resulting from a monohybrid cross would be .

A) 1:2 B) 1:2:2:4 C) 1:2:1 D) 9:3:3:1 E) 3:1 Answer: C Section: 4.3 Bloom's Taxonomy: Applying/Analyzing

1 Copyright © 2021 Pearson Education Ltd.

5) Assume that a cross is made between two organisms that are both heterozygous for a gene that shows incomplete dominance. What phenotypic and genotypic ratios are expected in the offspring?

A) 1:2:1, 1:2:1 B) 3:1, 3:1 C) 3:1, 1:2:1 D) 3:1, 1:1 E) 1:1, 1:1 Answer: A Section: 4.3 Bloom's Taxonomy: Remembering/Understanding 6) Assume that a dihybrid cross for two different traits (AaBb × AaBb) is made in which the gene loci are autosomal, independently assorting, and incompletely dominant. How many different phenotypes are expected in the offspring?

A) 4 B) 5 C) 8 D) 9 E) 16 Answer: D Section: 4.3 Bloom's Taxonomy: Applying/Analyzing 7) Assume that a dihybrid cross for two different traits (AaBb × AaBb) is made in which the gene loci are autosomal, independently assorting, and both incompletely dominant. What phenotypic ratio would you expect from such a cross?

A) 9:3:3:1 B) 1:2:1 C) 1:1:2:2:1:1 D) 1:2:1:2:4:2:1:2:1 E) 1:2:1:1:2:1:1:2:1 Answer: D Section: 4.3 Bloom's Taxonomy: Evaluating/Creating 8) The MN blood group is a codominant trait as a result of which of the following?

A) presence of both M and N cell surface glycoproteins on the same blood cell B) presence of a M/N intermediate cell surface glycoprotein on blood cells C) presence of M cell surface glycoproteins on some blood cells and N glycoproteins on others D) presence of M, M/N intermediate, and N cell surface glycoproteins on blood cells E) presence of either M, M/N intermediate, or N cell surface glycoproteins on different blood cells Answer: A Section: 4.4 Bloom's Taxonomy: Remembering/Understanding 2 Copyright © 2021 Pearson Education Ltd.

9) What is the probability of having a child with the N blood type, if both parents are MN? A) 0 B) 1/4 C) 1/2 D) 3/4 E) 1.0 Answer: B Section: 4.4 Bloom's Taxonomy: Applying/Analyzing 10) Which of the following is not a characteristic of the ABO gene locus?

A) multiple alleles B) dominance/recessiveness C) codominance D) epistasis E) four possible phenotypes Answer: D Section: 4.5 Bloom's Taxonomy: Remembering/Understanding 11) An individual test his blood type by placing drops of blood in anti-A sera and anti-B sera with the results as follows:

What is the blood type of this individual?

A) A B) B C) AB D) 0 E) can't be determined Answer: B Section: 4.5 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

12) Based on the following pedigree, with blood types provided (and ignoring the Bombay phenotype), what conclusions about paternity can be drawn?

A) This cannot be the father of either child.

B) This can be the father of the daughter but not the son.

C) This can be the father of the son but not the daughter.

D) This can be the father of either child.

Answer: B Section: 4.5 Bloom's Taxonomy: Applying/Analyzing 13) In a cross in which one parent exhibits blood type O due to the Bombay phenotype, but has alleles IA and i, and the other parent is heterozygous for the Bombay phenotype but has blood type O, what would be the possibility of their first child having blood type O? A) 0 B) 1/4 C) 1/2 D) 3/4 E) 3/16 Answer: D Section: 4.5 Bloom's Taxonomy: Applying/Analyzing 14) The presence of more than two alternative forms of a given gene would be called A) multiple alleles B) alternation of generations C) codominance D) incomplete dominance E) hemizygosity Answer: A Section: 4.5 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

.

15) The following coat colors are determined by alleles at one locus in horses: palomino = golden coat with lighter mane and tail cremello = almost white chestnut = brown The following table gives ratios obtained in matings of the above varieties: Cross Parents 1 cremello × cremello 2 chestnut × chestnut 3 cremello × chestnut 4 palomino × palomino

Offspring all cremello all chestnut all palomino 1/4 = chestnut 1/2 = palomino 1/4 = cremello

Based on this data set, which of the following is the correct statement?

A) cremello is dominant to chestnut; chestnut is dominant to palomino B) palomino is dominant to chestnut; chestnut is dominant to cremello C) cremello and chestnut are codominant; palomino is recessive to both D) cremello is dominant to palomino and chestnut; palomino and chestnut are codominant E) palomino is the incomplete dominant form of chestnut and cremello alleles Answer: E Section: 4.5 Bloom's Taxonomy: Evaluating/Creating 16) Mice have a set of multiple alleles of a gene for coat color. Four of those alleles are as follows: C = full color (wild) cch = chinchilla cd = dilution c = albino Given that the gene locus is not sex-linked and that each allele is dominant to those lower in the list, what would be the phenotypic ratio of a cross between a mouse of full color (heterozygous for dilution) and a mouse with the chinchilla coat color (heterozygous for albino)?

A) 2 full color:1 chinchilla:1 dilution B) 3 full color: 1 chinchilla C) 1 full color: 1 chinchilla: 1 dilution: 1 albino D) all full color E) 2 full color: 2 chinchilla Answer: A Section: 4.5 Bloom's Taxonomy: Evaluating/Creating

5 Copyright © 2021 Pearson Education Ltd.

17) What is the ratio expected for offspring of a cross between two yellow mice?

A) 3 yellow: 1 agouti B) 2 yellow: 1 agouti C) all yellow D) 1 yellow: 1 agouti E) cannot be determined due to unknown genotypes Answer: B Section: 4.6 Bloom's Taxonomy: Remembering/Understanding 18) A couple each with blood type AB and normal pigmentation have a child with AB blood type and albinism. What is the probability that their next child will have the same phenotype as the first child?

A) 9/16 B) 3/16 C) 1/8 D) 1/4 E) 1/16 Answer: C Section: 4.7 Bloom's Taxonomy: Applying/Analyzing 19) A couple in which one parent is blood type A and the other is blood type B, and both have normal pigmentation, have a child that is blood type O and has albinism. What is the probability that their next child will have normal pigmentation and blood type A? A) 0 B) 1/16 C) 3/16 D) 1/8 E) 9/16 Answer: C Section: 4.7 Bloom's Taxonomy: Applying/Analyzing 20) Typical ratios resulting from epistatic interactions in dihybrid crosses would be A) 9:3:3:1, 1:2:1 B) 1:1:1:1, 1:4:6:4:1 C) 9:3:4, 9:7 D) 1:2:2:4:1:2:1:2:1, 1:2:1 E) 3:1, 1:1 Answer: C Section: 4.8 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

.

21) A condition in which one gene pair masks the expression of a nonallelic gene pair is called .

A) codominance B) epistasis C) dominance D) recessiveness E) additive alleles Answer: B Section: 4.8 Bloom's Taxonomy: Remembering/Understanding 22) In the mouse, gene B can produce black pigment from a colorless precursor molecule. A mouse having at least one B allele can produce black pigment, whereas the homozygous recessive mouse (bb) cannot and is albino. The agouti locus (A) can convert the black pigment to brown in the presence of at least one dominant A allele, whereas the homozygous recessive (aa) cannot convert the black pigment to brown. What color fur would a mouse have with the genotype aaBB? A) albino B) black C) brown D) patches of brown and white E) can't be determined Answer: B Section: 4.8 Bloom's Taxonomy: Applying/Analyzing 23) In the mouse, gene B can produce black pigment from a colorless precursor molecule. A mouse having at least one B allele can produce black pigment, whereas the homozygous recessive mouse (bb) cannot and is albino. The agouti locus (A) can convert the black pigment to brown in the presence of at least one dominant A allele, whereas the homozygous recessive (aa) cannot convert the black pigment to brown. What would be the probability of an albino mouse offspring if the parents were genotypes AaBb and AAbb? A) 0 B) 1/16 C) 3/16 D) 1/4 E) 1/2 Answer: E Section: 4.8 Bloom's Taxonomy: Evaluating/Creating

7 Copyright © 2021 Pearson Education Ltd.

24) Many of the color varieties of summer squash are determined by several interacting loci: AA or Aa gives white, aaBB or aaBb gives yellow, and aabb produces green. Assume that two fully heterozygous plants are crossed. Give the phenotype ratio of the offspring.

A) 12 (white):3 (yellow):1 (green) B) 12 (yellow):3 (green): 1 (white) C) 1 (green): 1 (yellow): 1 (white) D) 9 (white): 3 (yellow): 4 (green) E) 9 (white): 7 (yellow) Answer: A Section: 4.8 Bloom's Taxonomy: Evaluating/Creating 25) Comb shape in chickens represents one of the classic examples of gene interaction. Two gene pairs interact to influence the shape of the comb. The genes for rose comb (R) and pea comb (P) together produce walnut comb. The fully homozygous recessive condition (rrpp) produces the single comb. Assume that a rose-comb chicken is crossed with a walnut-comb chicken and the following offspring are produced: 17 walnut, 16 rose, 7 pea, 6 single. What are the probable genotypes of the parents?

A) RrPp x RrPp B) rrPp x RrPp C) rrpp x RrPp D) Rrpp × RrPp E) RRpp x Rrpp Answer: D Section: 4.8 Bloom's Taxonomy: Evaluating/Creating 26) Individuals from two separate true-breeding strains of white deer mice are crossed yielding all grey offspring. White is recessive to gray color based on crossing mice from each strain with a grey mouse. Which of the following would best explain this result?

A) incomplete dominance B) multiple alleles C) epistasis D) lethal alleles E) complementation Answer: E Section: 4.9 Bloom's Taxonomy: Applying/Analyzing

8 Copyright © 2021 Pearson Education Ltd.

27) The following F2 results occur from a typical dihybrid cross: purple: A_B_ 9/16 white: aaB_ 3/16 white: A_bb 3/16 white: aabb 1/1 If a double heterozygote (AaBb) is crossed with a fully recessive organism (aabb), what phenotypic ratio is expected in the offspring?

A) 3 (white): 1 (purple) B) 3 (purple): 1 (white) C) 1 (purple): 1 (white) D) 9 (purple): 7 (white) E) 9 (white): 7 (purple) Answer: A Section: 4.9 Bloom's Taxonomy: Applying/Analyzing 28) The phenomenon of a gene which has multiple phenotypic effects on an individual is referred to as .

A) continuous variation B) pleiotropy C) expressivity D) penetrance E) epistasis Answer: B Section: 4.10 Bloom's Taxonomy: Remembering/Understanding 29) With which of the following would hemizygosity most likely be associated?

A) codominance B) incomplete dominance C) autosomal dominance D) X-linked inheritance E) sex-limited inheritance Answer: D Section: 4.11 Bloom's Taxonomy: Remembering/Understanding

9 Copyright © 2021 Pearson Education Ltd.

30) Because of the mechanism of sex determination, males of many species can be neither homozygous nor heterozygous. Such males are said to be .

A) dominant B) hemizygous C) recessive D) complementary E) None of the answers listed is correct.

Answer: B Section: 4.11 Bloom's Taxonomy: Remembering/Understanding 31) The white-eye gene in Drosophila is recessive and sex-linked. Assume that a white-eyed female is mated to a wild-type male. What would be the phenotypes of the offspring?

A) 1/4 wild-type females, 1/4 white-eyed females, 1/4 wild-type males, 1/4 white-eyed males B) 1/2 wild-type females, 1/4 wild-type males, 1/4 white-eyed males C) 1/2 wild-type males, 1/4 wild-type females, 1/4 white-eyed females D) 1/2 wild-type females, 1/2 white-eyed males E) all wild type males and females Answer: D Section: 4.11 Bloom's Taxonomy: Applying/Analyzing 32) Which of the following statements would be the most likely scenario of offspring ofa woman exhibiting an X-linked recessive disorder and a man that does not exhibit the trait?

A) all offspring with have the disorder B) all offspring will not have the disorder C) all daughters will not have the disorder and all sons will have the disorder D) all daughters will not have the disorder and half the sons will have the disorder E) half the daughters and half the sons will have the disorder Answer: C Section: 4.11 Bloom's Taxonomy: Applying/Analyzing

10 Copyright © 2021 Pearson Education Ltd.

33) A cross was made between homozygous wild-type female Drosophila and yellow-bodied male Drosophila. All of the resulting offspring were phenotypically wild type. Offspring of the F2 generation had the following phenotypes: Sex male male female

Phenotype wild yellow wild

Number 96 99 197

Which is correct for the mode of inheritance of this gene?

A) autosomal dominant B) autosomal recessive C) sex-linked dominant D) sex-linked recessive E) Y-linked Answer: D Section: 4.11 Bloom's Taxonomy: Applying/Analyzing 34) A deficiency of the enzyme glucose-6-phosphate dehydrogenase (G6PD) is inherited as an X- linked recessive trait in humans. A phenotypically normal woman (whose father had G6PD) is married to a man with normal G6PD function. What fraction of their sons would be expected to have G6PD deficiency?

A) 1/2 B) 0 C) all D) 1/4 E) not enough information available to determine Answer: A Section: 4.11 Bloom's Taxonomy: Applying/Analyzing

11 Copyright © 2021 Pearson Education Ltd.

35) Below is a pedigree of a fairly common human hereditary trait in which the boxes represent males and the circles represent females. Shading symbolizes the abnormal phenotype.

Given that one gene pair is involved, what is/are the possible mode(s) of inheritance?

A) autosomal dominant B) autosomal recessive C) autosomal recessive or X-linked recessive D) X-linked recessive E) autosomal recessive, X-linked recessive, or X-linked dominant Answer: B Section: 4.11 Bloom's Taxonomy: Evaluating/Creating 36) Pattern baldness is a sex-influenced trait, with heterozygous males exhibiting the trait. What would be the probability of the sons exhibiting the trait from a woman with pattern baldness and a male without pattern baldness?

A) 100% B) 50% C) 25% D) 0% E) not enough information to determine Answer: A Section: 4.12 Bloom's Taxonomy: Applying/Analyzing

12 Copyright © 2021 Pearson Education Ltd.

37) Pattern baldness is a sex-influenced trait, with heterozygous males exhibiting the trait. What would be the probability of the daughters exhibiting the trait from a woman with pattern baldness and a male without pattern baldness?

A) 100% B) 50% C) 25% D) 0% E) not enough information to determine Answer: D Section: 4.12 Bloom's Taxonomy: Applying/Analyzing 38) A trait exhibited in one sex but not the other is referred to as A) sex-influenced B) sex-linked C) sex-limited D) genomic imprinting Answer: C Section: 4.12 Bloom's Taxonomy: Remembering/Understanding

.

39) 90% of children that inherit a mutated retinoblastoma gene develop the disease. This represents an example of .

A) penetrance B) expressivity C) pleiotropy D) a conditional mutation E) genetic imprinting Answer: A Section: 4.13 Bloom's Taxonomy: Remembering/Understanding 40) Fruit flies homozygous for the eyeless mutation demonstrate various gradations in phenotype. This represents an example of .

A) penetrance B) expressivity C) pleiotropy D) a conditional mutation E) genetic imprinting Answer: B Section: 4.13 Bloom's Taxonomy: Remembering/Understanding

13 Copyright © 2021 Pearson Education Ltd.

41) Mutations in which the level of expression is based on environmental conditions is an example of .

A) penetrance B) expressivity C) pleiotropy D) a conditional mutation E) genetic imprinting Answer: D Section: 4.13 Bloom's Taxonomy: Remembering/Understanding 42) refers to observations that a genetic disorder occurs at an earlier age in successive generations.

A) Geneticanticipation B) Genomic imprinting C) Conditional mutation D) Penetrance E) Pleiotropy Answer: A Section: 4.13 Bloom's Taxonomy: Remembering/Understanding 43) occurs when the offspring's phenotype is under the control of the mother's nuclear gene products that are present in the egg.

A) Maternal inheritance B) Maternal effect C) Maternal imprinting D) Sex-linked inheritance E) Genetic anticipation Answer: B Section: 4.14 Bloom's Taxonomy: Remembering/Understanding 44) A human disorder resulting from a mitochondrial DNA mutation such as MERFF might show various levels of expression among children of a mother with MERFF. This is most likely the result of .

A) maternal effect B) a conditional mutation C) heteroplasmy D) expressivity E) genetic anticipation Answer: C Section: 4.14 Bloom's Taxonomy: Remembering/Understanding

14 Copyright © 2021 Pearson Education Ltd.

45) The trait of medium-sized leaves in iris is determined by the genetic condition PP'. Plants with large leaves are PP, whereas plants with small leaves are P'P'. A cross is made between two plants each with medium-sized leaves. If they produce 80 seedlings, what would be the expected phenotypes, and in what numbers would they be expected? What is the term for this allelic relationship?

Answer: 20 (large leaves), 40 (medium leaves), 20 (small leaves); incomplete dominance Section: 4.4 Bloom's Taxonomy: Applying/Analyzing 46) The trait for medium-sized leaves in iris is determined by the genetic condition PP'. Plants with large leaves are PP, whereas plants with small leaves are P'P'. The trait for red flowers is controlled by the genes RR, pink by RR', and white by R'R'. A cross is made between two plants each with medium-sized leaves and pink flowers. If they produce 320 seedlings, what would be the expected phenotypes, and in what numbers would they be expected? Assume no linkage.

Answer: 20 large, red 40 medium, red 20 small, red 40 large, pink 80 medium, pink 40 small, pink 20 large, white 40 medium, white 20 small, white Section: 4.4 Bloom's Taxonomy: Applying/Analyzing

15 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 5 Sex Determination and Sex Chromosomes 1) The Protenor mode of sex determination is the .

A) scheme based on F plasmids inserted into the FMR-1 gene B) XX/XO scheme C) XO/YY scheme D) hermaphroditic scheme E) scheme based on single translocations in the X chromosome Answer: B Section: 5.1 Bloom's Taxonomy: Remembering/Understanding 2) The Lygaeus mode of sex determination is the .

A) XY/XX scheme B) XX/XO scheme C) XO/YY scheme D) hermaphroditic scheme E) scheme based on single translocations in the X chromosome Answer: A Section: 5.1 Bloom's Taxonomy: Remembering/Understanding 3) In humans, the genetic basis for determining the sex "male" is accomplished by the presence of .

A) a portion of the Y chromosome B) one X chromosome C) a balance between the number of X chromosomes and the number of haploid sets of autosomes D) high levels of estrogen E) multiple alleles scattered throughout the autosomes Answer: A Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 4) Klinefelter syndrome in humans, which leads to underdeveloped testes and sterility, is caused by which chromosomal condition?

A) 47, XXY B) 47, 21+ C) 45, X D) 47, XYY E) triploidy Answer: A Section: 5.2 Bloom's Taxonomy: Remembering/Understanding

1 Copyright © 2021 Pearson Education Ltd.

5) Which of the following elements is responsible for determining male sex in humans?

A) PARS B) Xic C) SRY D) MSY E) Xist Answer: C Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 6) Which of the following elements allows for homologous pairing in meiosis?

A) PARS B) Xic C) SRY D) MSY E) Xist Answer: A Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 7) Which of the following is not part of the Y chromosome in humans?

A) PARS B) Xic C) MSY D) Xist E) Both Xic and Xist Answer: E Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 8) The sex of birds, some insects, and other organisms is determined by a ZW chromosomal arrangement in which the males have like sex chromosomes (ZZ) and females are ZW (similar to XY in humans). Assume that a recessive lethal allele on the Z chromosome causes death of an embryo in birds. What sex ratio would result in the offspring if a cross were made between a male heterozygous for the lethal allele and a normal female?

A) 4:1 male to female B) 2:1 male to female C) 3:1 male to female D) 1:2 male to female E) 1:1 male to female Answer: B Section: 5.3 Bloom's Taxonomy: Applying/Analyzing

2 Copyright © 2021 Pearson Education Ltd.

9) If a population of humans carries an X-linked lethal recessive allele that causes spontaneous abortions in the first trimester, how would you expect this to affect the sex ratio of that population?

A) It would be 1:1 male to female ratio.

B) There would be a reduced number of females.

C) There would be a reduced number of males D) There would be more intersex individuals.

E) There will be more metafemales.

Answer: C Section: 5.3 Bloom's Taxonomy: Applying/Analyzing 10) For an individual with the XXY chromosomal composition, the expected number of Barr bodies in interphase cells is .

A) variable B) one C) two D) three E) zero Answer: B Section: 5.4 Bloom's Taxonomy: Remembering/Understanding 11) Which of the following has one Barr body?

A) Turner's B) Klinefelter's C) Normal female D) Triplo-X female E) Klinefelter's and normal female Answer: E Section: 5.4 Bloom's Taxonomy: Remembering/Understanding 12) Which of the following elements represents a large untranslated transcript involved in inactivation?

A) PARS B) Xic C) SRY D) MSY E) Xist Answer: E Section: 5.4 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

13) In Drosophila, sex is determined by a balance between the number of haploid sets of autosomes and the number of .

A) chromomeres B) centromeres C) X chromosomes D) Y chromosomes E) satellited chromosomes Answer: C Section: 5.5 Bloom's Taxonomy: Remembering/Understanding 14) Assume that you are told that a particular organism, Drosophila, has the XO chromosome complement. You are also told that the autosomal complement is a normal 2n. You know that in humans the XO complement is female determining. What would be the sex of this fly and why?

A) This fly would be male due to an X:A ratio of 1/2.

B) This fly would be female due to no Y chromosome.

C) Sex would depend on temperature dependent aromatase expression.

D) This fly would be a female due to the presence of the X. E) This fly would be considered intersex based on it's X:A ratio.

Answer: A Section: 5.5 Bloom's Taxonomy: Remembering/Understanding 15) Which gene is dependent on the X:A ratio for it's function in sex determination in Drosophila?

A) sxl B) aromatase gene C) dsx D) tra E) SRY Answer: A Section: 5.5 Bloom's Taxonomy: Remembering/Understanding 16) A fruit fly with a chromosome formulation of 2X:3A would be which of the following?

A) intersex B) metamale C) metafemale D) male E) female Answer: A Section: 5.5 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

17) A fruit fly with an X:A ratio of .33 would which sexual morphology?

A) male B) female C) intersex D) metafemale E) metamale Answer: E Section: 5.5 Bloom's Taxonomy: Remembering/Understanding 18) Which protein is responsible for converting androgens into estrogens?

A) doublesex B) sex lethal C) transformer D) testis determining factor E) aromatase Answer: E Section: 5.6 Bloom's Taxonomy: Remembering/Understanding 19) Imagine you discover an unknown species of reptile that has a male to female ratio that is very low in cool climates. As seasonal temperatures warm you notice that the male to female ratio increases. What mode of sex determination appears to be involved?

A) Lygaeus mode B) Protenor mode C) Class I temperature dependence D) Class II temperature dependence E) Class III temperature dependence Answer: C Section: 5.6 Bloom's Taxonomy: Applying/Analyzing 20) Imagine you discover an unknown species of reptile and study its sex ratio. You observe that when climate temperatures are cool the male to female ratio is very low. As the climate warms up the male to female ratio raises and then goes back down under hot temperatures. Hat mode of sex determination appears to be involved?

A) Lygaeus mode B) Protenor mode C) Class I temperature dependence D) Class II temperature dependence E) Class III temperature dependence Answer: E Section: 5.6 Bloom's Taxonomy: Evaluating/Creating

5 Copyright © 2021 Pearson Education Ltd.

21) If a reptile had a null mutation in the aromatase gene what would be the predicted outcome?

A) All females.

B) 1:1 male to female sex ratio.

C) More intersex individuals.

D) Mostly Metamales.

E) Males even in female producing temperatures.

Answer: E Section: 5.6 Bloom's Taxonomy: Evaluating/Creating 22) List three abnormalities involving numbers of X chromosomes.

Answer: Klinefelter syndrome, Turner syndrome, XXX syndrome Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 23) Individuals have been identified who have two different karyotypes, such as 45,X/46,XY, or 45,X/46,XX. Such individuals are called .

Answer: mosaics Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 24) What particular karyotype was once considered to be related to criminal predisposition?

Answer: XYY Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 25) A small part of the human Y chromosome contains the gene that is responsible for determining maleness. What is the name of this gene?

Answer: SRY (sex-determining region Y) Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 26) Under what condition might a human female have the XY sex chromosome complement?

Answer: This female would have one complete X chromosome and a Y chromosome that lacks SRY. Section: 5.2 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

27) A color-blind woman with Turner syndrome (XO) has a father who is color-blind. Given that the gene for the color-blind condition is recessive and X-linked, provide a likely explanation for the origin of the color-blind and cytogenetic conditions in the woman.

Answer: The woman inherited an Xrg chromosome from the father. Nondisjunction in the mother (either at meiosis I or II) produced an egg with no X chromosome, which, when fertilized by the Xrg-bearing sperm, produced the Turner syndrome condition.

Section: 5.2 Bloom's Taxonomy: Evaluating/Creating 28) Although triple-X human females typically have normal offspring, what kinds of gametes, with respect to the X chromosomes, would you expect from such XXX females? Draw meiotic stages that show the gametes that are expected to be produced.

Answer:

Section: 5.2 Bloom's Taxonomy: Evaluating/Creating 29) How many chromosomes do Klinefelter and Turner syndromes have, respectively?

Answer: 47, 45 Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 30) Studies done in the 1960s suggested that individuals with the XYY condition were prone to criminal behavior. What conclusions currently seem appropriate concerning this chromosomal condition?

Answer: There is a high, but not constant, correlation between the extra Y chromosome and the predisposition of males to have behavioral problems.

Section: 5.2 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

31) Describe an experiment in which transgenic mice were used to identify the male-determining region of the Y chromosome.

Answer: When DNA containing only the mouse SRY is injected into normal XX mouse eggs, most of the offspring develop into males.

Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 32) Describe three distinct genetic regions of the human Y chromosome.

Answer: PARs = pseudoautosomal regions, MSY = male specific region of the Y, SRY = sexdetermining region Section: 5.2 Bloom's Taxonomy: Remembering/Understanding 33) Glucose-6-phosphate dehydrogenase (G6PD) deficiency is inherited as an X-linked recessive gene in humans. A normal woman whose father suffered from G6PD deficiency marries a normal man. (a) What proportion of their sons is expected to be G6PD deficient? (b) If the husband were not normal but were G6PD deficient, would you change your answer in part (a)?

Answer: (a) 1/2 (b) no Section: 5.4 Bloom's Taxonomy: Applying/Analyzing 34) A color-blind, chromatin-positive male child (one Barr body) has a maternal grandfather who was color-blind. The boy's mother and father are phenotypically normal. Construct and support (using appropriately labeled diagrams) a rationale whereby the chromosomal and genetic attributes of the chromatin-positive male are fully explained.

Answer: The female (mother) must be heterozygous and undergo nondisjunction at meiosis II to produce the XrgXrgY boy.

Section: 5.4 Bloom's Taxonomy: Evaluating/Creating 35) A cross is made between a female calico cat and a male cat having the gene for black fur on his X chromosome. What fraction of the offspring would one expect to be calico?

Answer: 1/4 Section: 5.4 Bloom's Taxonomy: Applying/Analyzing

8 Copyright © 2021 Pearson Education Ltd.

36) Dosage compensation leads to a variety of interesting coat color patterns in certain mammals. For instance, a female cat that is heterozygous for two coat color alleles, say black and orange, will usually have the "calico" or mosaic phenotype. Describe the chromosomal basis for the mosaicism (calico) in the female. Explain why chromosomally normal male cats do not show the mosaic phenotype, but XXY male cats can be calico.

Answer: Because of dosage compensation, one of the X chromosomes randomly "turns off" early in development, such that one X chromosome may be active in one cell and the other X chromosome may be active in another cell. Once such a chromosome is inactivated, it remains so in daughter cells. Recessive alleles on the remaining active X chromosome are expressed because their normal allele (on the inactive X chromosome) is not capable of expression.

Because males typically have only one X chromosome, X chromosome inactivation does not occur; however, in XXY males that are heterozygous for certain coat color genes, such inactivation and mosaicism are possible.

Section: 5.4 Bloom's Taxonomy: Remembering/Understanding 37) How many Barr bodies would one expect to see in cells of Turner syndrome females and Klinefelter syndrome males?

Answer: zero and one, respectively Section: 5.4 Bloom's Taxonomy: Remembering/Understanding 38) What can cause phenotypic mosaicism for X-linked genes in female mammals?

Answer: dosage compensation involving the X chromosome Section: 5.4 Bloom's Taxonomy: Remembering/Understanding 39) Dosage compensation in mammals typically involves the random inactivation of one of the two X chromosomes relatively early in development. Such X chromosome inactivation often leads to phenotypic mosaicism. Assume that black fur in cats is due to the X-linked recessive gene b, whereas its dominant allele B produces yellow fur. A Bb heterozygote is a mosaic called "tortoise shell" or "calico." Using appropriate gene symbols, diagram a mating between a black male and a calico female. Give the phenotypes and genotypes of all the offspring.

Answer: bY × Bb = Bb (calico female), bb (black female), BY (yellow male), bY (black male) Section: 5.4 Bloom's Taxonomy: Applying/Analyzing 40) What is the composition of a Barr body?

Answer: an inactivated X chromosome that is coated with Xist transcripts Section: 5.4 Bloom's Taxonomy: Applying/Analyzing

9 Copyright © 2021 Pearson Education Ltd.

41) Assume a woman with normal vision marries a man with normal vision. If this couple has a daughter with color blindness explain cytogentically how this could occur.

Answer: The mother is heterozygous and passed on the color blindness gene to her daughter.

The daughter inactivated the normal X in the retinal tissue.

Section: 5.4 Bloom's Taxonomy: Evaluating/Creating 42) How would administration of estradiol to a developing reptile affect its sex determination?

Assume that the administration occurs at the critically relevant point during development.

Answer: Estradiol administration would circumvent the normal aromatase pathway and cause females to be produced and conflict with male production under male producing temperatures.

Section: 5.4 Bloom's Taxonomy: Evaluating/Creating 43) Give the sex-chromosome constitution (X and Y chromosomes) and possible genotypes of offspring resulting from a cross between a white-eyed female (Xw XwY) and a wild-type male (normal chromosome complement) in Drosophila melanogaster. Include all zygotic combinations whether viable or unviable.

Answer: X+XwXw = unviable (dies at third instar stage) XwXwY = white-eyed female X+Y = wild-type male YY = unviable (dies at egg stage) X+XwY = wild-type female XwYY = white-eyed male X+Xw = wild-type female XwY = white-eyed male Section: 5.5 Bloom's Taxonomy: Evaluating/Creating 44) In Drosophila, an individual female fly was observed to be of the XXY chromosome complement (normal autosomal complement) and to have white eyes as contrasted with the normal red eye color of wild type. The female's father had red eyes, and the mother had white eyes. Knowing that white eyes are X-linked and recessive, present an explanation for the genetic and chromosomal constitution of the XXY, white-eyed individual. It is important that you state in which parent and at what stage the chromosomal event occurred that caused the genetic and cytogenetic abnormality.

Answer: Nondisjunction could have occurred either at meiosis I or meiosis II in the mother, thus giving the XwXwY complement in the offspring.

Section: 5.5 Bloom's Taxonomy: Evaluating/Creating

10 Copyright © 2021 Pearson Education Ltd.

45) In Drosophila, an individual female fly was observed to be of the XXY chromosome complement (normal autosomal complement) and to have white eyes as contrasted with the normal red eye color of wild type. The female's mother and father had red eyes. The mother, however, was heterozygous for the gene for white eyes. Knowing that white eyes are X-linked and recessive, present an explanation for the genetic and chromosomal constitution of the XXY, white-eyed individual. It is important that you state in which parent and at what stage the chromosomal event occurred that caused the genetic and cytogenetic abnormality.

Answer: Nondisjunction would have occurred at meiosis II in the mother, thus giving the XwXwY complement in the offspring.

Section: 5.5 Bloom's Taxonomy: Evaluating/Creating 46) Give the sex of the following organisms assuming that the autosomes are present in the normal number.

Sex Chromosome Organism Complement Humans

Drosophila

XX XY XO XXX XXY Answer: Sex Chromosome Organism Complement Humans Drosophila XX female female XY male male XO female male XXX female female XXY male female Section: 5.5 Bloom's Taxonomy: Applying/Analyzing 47) Assuming a normal number of autosomes, what would be the sex of the following: XXY mouse, XXY Drosophila?

Answer: male and female, respectively Section: 5.5 Bloom's Taxonomy: Applying/Analyzing 48) Data produced by C. Bridges in the early part of this century indicate that sex in Drosophila is determined by .

Answer: a balance between the number of X chromosomes and the number of haploid sets of autosomes Section: 5.5 Bloom's Taxonomy: Remembering/Understanding 11 Copyright © 2021 Pearson Education Ltd.

49) In Drosophila, the sex of a fly with the karyotype XO:2A is Answer: male Section: 5.5 Bloom's Taxonomy: Remembering/Understanding

.

50) What would be the sex of fruit fly that is XX and carrying a null allele for sex lethal? Explain why.

Answer: Male because sex lethal is dosage dependent and a null allele will create a functional X:A ratio of .5 which would set the male pathway.

Section: 5.5 Bloom's Taxonomy: Applying/Analyzing

12 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 6 Chromosome Mutations: Variation in Number and Arrangement 1) The condition that exists when an organism gains or loses one or more chromosomes but not a complete haploid set is known as .

A) polyploidy B) euploidy C) aneuploidy D) triploidy E) trisomy Answer: C Section: 6.1 Bloom's Taxonomy: Remembering/Understanding 2) The condition that exists where complete haploid sets of chromosomes are present is known as .

A) polyploidy B) euploidy C) aneuploidy D) triploidy E) diploidy Answer: B Section: 6.1 Bloom's Taxonomy: Remembering/Understanding 3) Which of the following is viewed as a major cause of aneuploidy?

A) nondisjunction B) pericentric inversion C) paracentric inversion D) reciprocal tranlocations E) recombination Answer: A Section: 6.1 Bloom's Taxonomy: Remembering/Understanding 4) In meiosis a triploid would exhibit which meiotic pairing configuration?

A) trivalent B) bivalent C) tetrad D) monovalent E) triad Answer: A Section: 6.1 Bloom's Taxonomy: Evaluating/Creating

1 Copyright © 2021 Pearson Education Ltd.

5) If a species has a haploid number of 10, how many chromosomes would an individual of this species have if it were 2n + 3?

A) 10 B) 20 C) 23 D) 15 E) 25 Answer: C Section: 6.1 Bloom's Taxonomy: Applying/Analyzing 6) Trisomy 21, or Down syndrome, occurs when there is a normal diploid chromosomal complement but one (extra) chromosome 21. Although fertility is reduced in both sexes, females have higher fertility rates than males. Van Dyke et al. (1995; Down Syndrome Research and Practice 3[2]:65–69) summarize data involving children born of Down syndrome individuals.

Assume that children are born to a female with Down syndrome and a normal 46-chromosome male. What proportion of the offspring would be expected to have Down syndrome?

A) One-third of the offspring would be expected to have Down syndrome.

B) Two-thirds of the offspring would be expected to have Down syndrome.

C) All the children would be expected to have Down syndrome.

D) None of the offspring would be expected to have Down syndrome.

E) One-half of the offspring would be expected to have Down syndrome.

Answer: E Section: 6.2 Bloom's Taxonomy: Evaluating/Creating 7) If a species has a haploid number of 10, what term would we use to describe an individual with 11 chromosomes?

A) monosomic B) trisomic C) eusomic D) haploid E) triploid Answer: B Section: 6.2 Bloom's Taxonomy: Applying/Analyzing 8) Which of the following would be considered a triploid?

A) Down's syndrome B) Patau's syndrome C) Klinefelter's syndrome D) both Down's and Patau's syndrome E) none of these Answer: E Section: 6.2 Bloom's Taxonomy: Applying/Analyzing 2 Copyright © 2021 Pearson Education Ltd.

9) Which of the following is a method for harvesting fetal cells for karyotyping?

A) amniocentesis B) chorionic villus sampling C) noninvasive prenatal genetic diagnosis D) fetal ultrasound E) amniocentesis, chorionic villus sampling, and noninvasive prenatal genetic diagnosis Answer: E Section: 6.2 Bloom's Taxonomy: Remembering/Understanding 10) If a species has diploid number of 10, but gave rise to progeny with 20 chromosomes, what term would we use to describe the progeny?

A) triploid B) diploid C) haploid D) tetraploid E) aneuploid Answer: D Section: 6.3 Bloom's Taxonomy: Applying/Analyzing 11) Which of the following is technique used to create polyploidy cells?

A) heat shock B) colchicine addition C) NIPGD D) CVS E) both heat shock and colchicine addition Answer: E Section: 6.3 Bloom's Taxonomy: Remembering/Understanding 12) If a haploid cell replicates and then is treated with colchicine, what will be its ploidy (i.e. how many chromosomes will it have) after the cell cycle is complete?

A) tetraploid B) diploid C) triploid D) haploid E) aneuploid Answer: B Section: 6.3 Bloom's Taxonomy: Applying/Analyzing

3 Copyright © 2021 Pearson Education Ltd.

13) If two sperm fertilize a single ovum the resulting zygote would be considered a A) tetraploid B) diploid C) triploid D) haploid E) aneuploidy Answer: C Section: 6.3 Bloom's Taxonomy: Remembering/Understanding

.

14) A diploid gamete that is fertilized by a haploid gamete from the same species would be an example of .

A) allohaploidy B) autopolyploidy C) tetraploidy D) allopolyploidy E) amphidiploidy Answer: B Section: 6.3 Bloom's Taxonomy: Applying/Analyzing 15) What term is used when two known diploid species fertilize to create a tetraploid offspring?

A) allopolyploidy B) amphidiploidy C) autopolyploidy D) tetraploidy E) euploidy Answer: B Section: 6.3 Bloom's Taxonomy: Remembering/Understanding 16) What term most specifically describes when a polyploid organism contains an equivalent of four haploid genomes derived from separate species?

A) allotetraploidy B) aneuploidy C) autotetraploidy D) tetraploidy E) euploidy Answer: A Section: 6.3 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

17) The condition known as cri-du-chat syndrome in humans has a genetic constitution designated as .

A) 45, X B) heteroplasmy C) 46, 5pD) triploidy E) trisomy Answer: C Section: 6.5 Bloom's Taxonomy: Remembering/Understanding 18) If a Drosophila that has a white eye mutation on one homolog of chromosome 1 and a deletion in the same interval of the other homolog, what would be the phenotypic outcome and why?

A) It would have white eyes due to the deletion uncovering the recessive white mutation.

B) It would be wild type due to the heterozygous state of the deletion.

C) It would be lethal due to deletions being lethal.

D) It would have white eye due to the homozygosity of the recessive allele.

E) It would have a mosaic of wild type and white due to presence of the recessive allele and the deletion.

Answer: A Section: 6.5 Bloom's Taxonomy: Evaluating/Creating 19) Which of the following is/are not lines of evidence that support the evolution by gene duplication theory?

A) trypsin/chymotrypsin homology B) myoglobin/hemoglobin homology C) SRGAP2 genes D) CCL3L1 E) All are examples of evolution by gene duplication.

Answer: D Section: 6.6 Bloom's Taxonomy: Applying/Analyzing 20) Which of the following typically show gene redundancy?

A) rDNA B) CCL3L1 C) The trypsin gene.

D) sxl E) The white eyed mutant in Drosophila.

Answer: A Section: 6.6 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

21) What type of inversion has break points that flank the centromere?

A) paracentric B) acentric C) pericentric D) dicentric E) eucentric Answer: C Section: 6.7 Bloom's Taxonomy: Remembering/Understanding 22) What type of inversion has break points that do not proximally flank the centromere?

A) paracentric B) acentric C) pericentric D) dicentric E) eucentric Answer: A Section: 6.7 Bloom's Taxonomy: Remembering/Understanding 23) When crossover occurs within a paracentric inversion the chromatids will be A) paracentric B) acentric C) pericentric D) dicentric E) both acentric and dicentric Answer: E Section: 6.7 Bloom's Taxonomy: Remembering/Understanding

.

24) In working with Drosophila it is common to use a balancer stock to maintain mutations of interest. The balancer chromosomes are generally carrying multiple inversions on them. What effect would the inverted chromosomes have on meiosis?

A) Inversions would have no effect on meiosis.

B) Inversions would prevent synapsis.

C) Inversions would suppress recombination.

D) Balancers would help facilitate homologous pairing.

E) None of the above would be an effect of balancers.

Answer: C Section: 6.7 Bloom's Taxonomy: Evaluating/Creating

6 Copyright © 2021 Pearson Education Ltd.

25) Although the most frequent forms of Down syndrome are caused by a random error, nondisjunction of chromosome 21, Down syndrome occasionally runs in families. The cause of this form of familial Down syndrome is .

A) an inversion involving chromosome 21 B) a chromosomal aberration involving chromosome 1 C) too many X chromosomes D) a translocation between chromosome 21 and a member of the D chromosome group E) a maternal age effect Answer: D Section: 6.8 Bloom's Taxonomy: Remembering/Understanding 26) During prenatal screening a couple is told that their baby is 46XX. At birth; however, the baby is diagnosed as Down's Syndrome. Whats is possible explanation for this?

A) The baby received a 14/21, DG translocation from one of the parents.

B) The baby received a segmental deletion of chromosome 14.

C) The baby received a duplication of the p arm of chromosome 14.

D) The baby received a deletion of the DSCR1 gene.

E) The baby received a duplication of the DSCR 1 gene.

Answer: A Section: 6.8 Bloom's Taxonomy: Evaluating/Creating 27) Which of the following describes a Robertsonian translocation?

A) This is when two acrocentric chromosomes break and fuse at the centromeres.

B) This is when two chromosomes trades telomeres.

C) This is when a chromosome excises a region of DNA and re-inserts it in the opposite orientation.

D) This is when a chromosome breaks and fuses with it homolog.

E) This is when two metacentric chromosomes fuse end-to-end.

Answer: A Section: 6.8 Bloom's Taxonomy: Remembering/Understanding 28) Which chromosome pairs does familial Down's Syndrome involve?

A) 1 and 14 B) 2 and 13 C) 13 and 18 D) 21 and 13 E) 14 and 21 Answer: E Section: 6.8 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

29) Explain how an individual can have Down's Syndrome with a normal diploid set of chromosomes.

A) The individual has a deletion for all necessary genes.

B) The individual has duplication for all of the necessary genes.

C) The individual has an inversion in the DCSR 1 gene.

D) The individual has a translocation of the DSCR. E) The individual has either a translocation of the DSCR or a duplication of the necessary genes.

Answer: E Section: 6.8 Bloom's Taxonomy: Evaluating/Creating 30) If a euchromatic gene was translocated into a heterochromatic region of the new chromosome, what would be the effect on this gene activity?

A) Gene expression would be normal.

B) Gene expression would be completely turned on.

C) Gene expression would be reduced.

D) Gene expression would slightly increase.

E) Gene expression would either be normal or slightly increased.

Answer: C Section: 6.8 Bloom's Taxonomy: Evaluating/Creating 31) A genomic condition that may be responsible for some forms of fragile-X syndrome, as well as Huntington disease, involves .

A) plasmids inserted into the FMR-1 gene B) various lengths of trinucleotide repeats C) multiple breakpoints fairly evenly dispersed along the X chromosome D) multiple inversions in the X chromosome E) single translocations in the X chromosome Answer: B Section: 6.9 Bloom's Taxonomy: Remembering/Understanding 32) Recently, a gene located on chromosome 3 in humans, FHIT, has been shown to be associated with .

A) several types of cancer B) Huntington disease C) "mad-cow" disease D) Klinefelter syndrome E) XYY/XY mosaicism Answer: A Section: 6.9 Bloom's Taxonomy: Remembering/Understanding

8 Copyright © 2021 Pearson Education Ltd.

33) Fragile-X Syndrome is caused by which of the following genes?

A) DSCR1 B) FMR1 C) EGFR D) CCL3L1 E) SRGAP2 Answer: B Section: 6.9 Bloom's Taxonomy: Remembering/Understanding 34) A male patient, that desperately wants to have a son, with a family history of Fragile-X comes in to have his CGG copy number tested and finds that it is 102. He knows that when his mother was tested that hers was 51. What genetic phenomenon is at work here and is it possible for him to pass Fragile-X to his son?

A) This is an example of interference and yes he can pass it on to his son.

B) This is an example of anticipation and yes he can pass it on to his son.

C) This is an example of anticipation and no he cannot pass it on to his son.

D) This is an example of interference and no he cannot pass it on to his son.

E) This is an example of haploinsufficiency and no he cannot pass it on to his son.

Answer: C Section: 6.9 Bloom's Taxonomy: Applying/Analyzing 35) What explanation is generally given for lethality of monosomic individuals?

Answer: Monosomy may unmask recessive lethals that are tolerated in heterozygotes carrying the wild-type allele.

Section: 6.2 Bloom's Taxonomy: Remembering/Understanding 36) Describe the maternal age effect associated with Down syndrome.

Answer: For unknown reasons, the nondisjunctional event that produces Down syndrome occurs more frequently during oogenesis in women older than age 35.

Section: 6.2 Bloom's Taxonomy: Applying/Analyzing 37) In what way might gene duplication play a role in evolution?

Answer: In 1970, Ohno proposed that gene duplication provides a way in which new genes arise. By duplicating a gene, the duplicated copy or the original gene is able to mutate without necessarily having an adverse influence on the phenotype.

Section: 6.6 Bloom's Taxonomy: Remembering/Understanding 38) Name two methods used in genetic prenatal diagnostic testing in humans.

Answer: amniocentesis and chorionic villus sampling (CVS) Section: 6.2 Bloom's Taxonomy: Remembering/Understanding 9 Copyright © 2021 Pearson Education Ltd.

39) Trisomics are observed in humans; with the exception of XO females, monosomics are not.

Why? Answer: Monosomics are inviable. Such haploinsufficiency combines the loss of multiple genes.

Section: 6.2 Bloom's Taxonomy: Remembering/Understanding 40) Name the polyploid condition that is formed from the addition of one or more extra sets of chromosomes identical to the normal haploid complement of the same species.

Answer: autopolyploidy Section: 6.3 Bloom's Taxonomy: Remembering/Understanding 41) Given that a human normally contains 46 chromosomes, give the chromosome number for each of the following conditions: Turner syndrome (female, no Barr bodies) Klinefelter syndrome (male, one Barr body) triploid Down syndrome (trisomic) trisomy 13 Answer: Turner syndrome (female, no Barr bodies) 45 Klinefelter syndrome (male, one Barr body) 47 Triploid 69 Down syndrome (trisomic) 47 Trisomy 13 47 Section: 6.2 Bloom's Taxonomy: Applying/Analyzing 42) Colchicine is an alkaloid derived from plants. What is its effect on chromosome behavior?

Answer: By interfering with spindle formation, replicated chromosomes fail to migrate to the poles at anaphase; thus, sister chromatids end up in the same nucleus.

Section: 6.3 Bloom's Taxonomy: Remembering/Understanding 43) Provide an example of gene redundancy that occurs in both eukaryotes and prokaryotes.

Answer: rDNA Section: 6.6 Bloom's Taxonomy: Remembering/Understanding 44) Describe Bar mutations in Drosophila melanogaster.

Answer: Bar mutations are duplications in portions of the X chromosome.

Section: 6.6 Bloom's Taxonomy: Remembering/Understanding

10 Copyright © 2021 Pearson Education Ltd.

45) Clearly illustrate the pairing configuration of an inversion (paracentric) heterokaryotype.

Answer: The pairing of homologous chromosomes of an inversion heterokaryotype is typically one of an "outside" loop filled by an "inside loop."

Section: 6.7 Bloom's Taxonomy: Evaluating/Creating 46) What is meant by the terms acentric and dicentric?

Answer: A chromosome without a centromere is acentric; a chromosome with two centromeres is dicentric.

Section: 6.7 Bloom's Taxonomy: Remembering/Understanding 47) Under what circumstance can an individual with Down syndrome have 46 chromosomes?

Answer: if he or she carries a D/G translocation, 14/21, for example Section: 6.8 Bloom's Taxonomy: Remembering/Understanding 48) Fragile-X syndrome (or Martin-Bell syndrome) is the most common form of inherited mental retardation in humans. Is it more common in males or females? What is FMR1?

Answer: males (1/4000 compared to 1/8000 in females); FMR1 is one of a growing number of genes in which a sequence of three nucleotides is repeated many times, expanding the size of the gene.

Section: 6.9 Bloom's Taxonomy: Remembering/Understanding

11 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 7 Linkage and Chromosome Mapping in Eukaryotes 1) In the early 1900s, two scientists noted there were many more genes than chromosome pairs, thus setting the stage for the suggestion that some gene loci might be linked during meiotic processes. Who were these two scientists?

A) Sturtevant and Morgan B) Watson and Crick C) Sutton and Boveri D) Lederberg and Tatum E) Klug and Cummings Answer: C Section: Introduction Bloom's Taxonomy: Remembering/Understanding 2) When two genes fail to assort independently, the term normally applied is A) discontinuous inheritance B) Mendelian inheritance C) linkage D) tetrad analysis E) dominance and/or recessiveness Answer: C Section: 7.1 Bloom's Taxonomy: Remembering/Understanding

.

3) Assume that a cross is made between AaBb and aabb plants and that the offspring fall into approximately equal numbers of the following groups: AaBb, Aabb, aaBb, aabb. These results are consistent with .

A) independent assortment B) alternation of generations C) complete linkage D) incomplete dominance E) hemizygosity Answer: A Section: 7.1 Bloom's Taxonomy: Remembering/Understanding 4) Assume that a cross is made between AaBb and aabb plants and that all the offspring are either AaBb or aabb. These results are consistent with .

A) complete linkage B) alternation of generations C) codominance D) incomplete dominance E) hemizygosity Answer: A Section: 7.1 Bloom's Taxonomy: Remembering/Understanding 1 Copyright © 2021 Pearson Education Ltd.

5) Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are most consistent with .

A) sex-linked inheritance with 30 percent crossing over B) linkage with 50 percent crossing over C) linkage with approximately 33.3 map units between the two gene loci D) independent assortment E) 100 percent recombination Answer: C Section: 7.2 Bloom's Taxonomy: Applying/Analyzing 6) Which scientist proposed that physical exchange due to crossing over between chromosomes leads to genetic recombination?

A) Walter Sutton B) Thomas Morgan C) F.A. Janssens D) James Watson E) James German Answer: B Section: 7.2 Bloom's Taxonomy: Remembering/Understanding 7) The cross GE/ge × ge/ge produces the following progeny: GE/ge 404, ge/ge 396, gE/ge 97, Ge/ge 103. From these data, one can conclude that the recombinant progeny are .

A) gE/ge and Ge/ge B) GE/ge and Ge/ge C) ge/ge. gE/ge, and Ge/ge D) GE/ge and ge/ge Answer: A Section: 7.2 Bloom's Taxonomy: Applying/Analyzing

2 Copyright © 2021 Pearson Education Ltd.

8) Phenotypically wild-type F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw) bristles, produced the following offspring when crossed to homozygous lightstraw males: Phenotype Number light-straw 22 wild-type 18 light 990 straw 970 Total: 2000 Compute the map distance between the light and straw loci.

A) 0.02 map units B) 2.0 map units C) 20 map units D) 0.98 map units E) 98 map units Answer: B Section: 7.2 Bloom's Taxonomy: Applying/Analyzing 9) Assume that two genes are 80 map units apart on chromosome II of Drosophila and that a cross is made between a doubly heterozygous female and a homozygous recessive male. What percent recombination would be expected in the offspring of this type of cross?

A) 20 percent B) 40 percent C) 50 percent D) 80 percent E) 0 percent Answer: C Section: 7.2 Bloom's Taxonomy: Applying/Analyzing 10) Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with which arrangement of alleles for the heterozygous parent plants?

A) AB/ab B) Ab/aB C) ab/ab D) can't be determined Answer: A Section: 7.2 Bloom's Taxonomy: Applying/Analyzing

3 Copyright © 2021 Pearson Education Ltd.

11) Assume that the genes for the recessive traits of tan body and bare wings are 15 map units apart on chromosome II in Drosophila. Assume also that a tan-bodied, bare-winged female was mated to a wild-type male and that the resulting F1 phenotypically wild-type females were then mated to tan-bodied, bare-winged males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected?

A) wild type = 75; tan-bare = 75; tan = 425; bare = 425 B) wild type = 425; tan-bare = 425; tan = 75; bare = 75 C) wild type = 350; tan-bare = 350; tan = 150; bare = 150 D) wild type = 150; tan-bare = 150; tan = 350; bare = 350 E) wild type = 500; tan-bare = 200; tan = 200; bare = 100 Answer: B Section: 7.2, 7.3 Bloom's Taxonomy: Evaluating/Creating 12) The genes for the recessive traits of mahogany eyes and ebony body are approximately 25 map units apart on chromosome III in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and that the resulting F1 phenotypically wild-type females were then mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected?

A) mahogany = 375; ebony = 375; wild-type = 125; mahogany-ebony = 125 B) wild-type = 375; mahogany-ebony = 375; mahogany = 125; ebony = 125 C) wild-type = 250; mahogany-ebony = 250; mahogany = 250; ebony = 250 D) mahogany = 400; ebony = 400; wild-type = 100; mahogany-ebony = 100 E) wild-type = 400; mahogany-ebony = 400; mahogany = 100; ebony = 100 Answer: A Section: 7.2, 7.3 Bloom's Taxonomy: Evaluating/Creating 13) In a three-point mapping experiment, how many different genotypic classes are expected?

How many phenotypic classes?

A) 16; 4 B) 16; 8 C) 8; 8 D) 8; 4 E) 3; 3 Answer: C Section: 7.3 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

14) Assume that there are 12 map units between two loci in the mouse and that you are able to microscopically observe meiotic chromosomes in this organism. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between the two loci?

A) 6 B) 12 C) 24 D) 48 E) 98 Answer: D Section: 7.2 Bloom's Taxonomy: Applying/Analyzing 15) The phenomenon in which one crossover decreases the likelihood of crossovers in nearby regions is called .

A) chiasma B) negative interference C) reciprocal genetic exchange D) positive interference E) mitotic recombination Answer: D Section: 7.4 Bloom's Taxonomy: Remembering/Understanding 16) Genes A and B are 10 map units apart and genes B and C are 20 map units apart, with gene B located between genes A and C. What would be the expected amount of double crossovers for a trihybrid test cross?

A) 0.2% B) 2.0% C) 15% D) 20% E) 30% Answer: B Section: 7.4 Bloom's Taxonomy: Remembering/Understanding 17) Methods for determining the linkage group and genetic map in humans involve A) DNA markers B) twin spots and tetrad analysis C) tetrad analysis and bromodeoxyuridine D) gamete genotyping E) dihybrid test crosses Answer: A Section: 7.5 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

.

18) What is the unit of distance between genes based on a physical map?

A) map units B) centiMorgans C) lod scores D) base pairs E) inches Answer: D Section: 7.5 Bloom's Taxonomy: Remembering/Understanding 19) Which of the following statements is not associated with sister chromatid exchanges?

A) provide new allelic combinations B) occur during mitosis C) are observable as patch-like patterns after two rounds of replication in the presence of BrdU D) are found elevated in the human disorder Bloom syndrome E) increase with exposure to mutagenic agents Answer: A Section: 7.6 Bloom's Taxonomy: Remembering/Understanding 20) Which of the following data were generated by Creighton and McClintock that supported crossing over in homologous chromosomes?

A) BrdU staining of replicating chromsomes B) electron microscopy images C) observable exchange of cytological features D) mobile DNA elements E) increase of recombinants by exposure to mutagens Answer: C Section: 7.6 Bloom's Taxonomy: Remembering/Understanding 21) Under what circumstance might two loci be on the same chromosome but behave as if independently assorting in crosses?

Answer: If the genes are far apart, they may show independent assortment.

Section: 7.1 Bloom's Taxonomy: Remembering/Understanding 22) What is a linkage group?

Answer: A set of genes that show linkage to one another and therefore are on the same chromosome Section: 7.1 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

23) In theory, how many linkage groups are in an organism?

Answer: The number of linkage groups corresponds to the haploid number of chromosomes in an organism.

Section: 7.1 Bloom's Taxonomy: Remembering/Understanding 24) At what stage of the meiotic cell cycle and during what chromosomal configuration does crossing over occur?

Answer: at the four-strand stage of meiosis, after synapsis of homologous chromosomes, and before the end of prophase I Section: 7.2 Bloom's Taxonomy: Remembering/Understanding 25) A dihybrid test cross is necessary to assess linkage of two genes on the same chromosome.

When analyzing the offspring, what do the different phenotypes represent?

Answer: proportion of gametes from the dihybrid parent.

Section: 7.2 Bloom's Taxonomy: Applying/Analyzing 26) What is the relationship between the degree of crossing over and the distance between two genes?

Answer: It is direct: as the distance increases, the frequency of recombination increases.

Section: 7.2 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

27) Assume that investigators crossed a strain of flies carrying the dominant eye mutation Lobe on the second chromosome with a strain homozygous for the second chromosome recessive mutations smooth abdomen and straw body. The F1 Lobe females were then backcrossed with homozygous smooth abdomen, straw-body males, and the following phenotypes were observed: smooth abdomen, straw body Lobe smooth abdomen, Lobe straw body smooth abdomen Lobe, straw body

820 780 42 58 148 152

(a) Give the arrangement of alleles of the F1 Lobe females (b) Which gene is in the middle? (c) Determine the distances in map units for these three loci. (d) What is the coefficient of coincidence and interference values? (e) Is there positive, negative, total or no interference?

Answer: (a) + + L/ sa sb + (b) Lobe is in the middle. (c) smooth abdomen---5---Lobe-----------15 ------------- straw body (d) CC = 0; I = 1.0 (e) total interference (explains the observance of only six instead of the eight possible phenotypes in the offspring) Section: 7.3, 7.4 Bloom's Taxonomy: Evaluating/Creating

8 Copyright © 2021 Pearson Education Ltd.

28) In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eyes) and hairless (no thoracic bristles). Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and the following progeny (1000 total) were observed: PhenotypesNumber Observed spineless 321 wild-type 38 claret, spineless 130 claret 18 claret, hairless 309 hairless, claret, spineless 32 hairless 140 hairless, spineless 12 (a) With respect to the three genes mentioned in the problem, what are the genotypes of the homozygous parents used in making the phenotypically wild-type F1 heterozygote? (b) Which gene is in the middle? (c) What are the map distances for the three genes? A correct formula with the values "plugged in" for each distance will be sufficient. (d) What is the coefficient of coincidence? A correct formula with the values "plugged in" will be sufficient. (e) What is the value for interference? Is there positive, negative, total, or no interference?

Answer: (a) cl h +/cl h + and + + sp/+ + sp (b) ) hairless (c) cl 30 h 10 sp (d) 0.03/0.03 = 1 (e) I = 0; no interference Section: 7.3, 7.4 Bloom's Taxonomy: Evaluating/Creating 29) Provide a brief definition for positive interference.

Answer: A crossover in one region decreases the likelihood of crossovers in nearby regions.

Section: 7.4 Bloom's Taxonomy: Remembering/Understanding 30) Describe a convenient method for determining gene order from three-point cross results.

Answer: Compare the double-crossover class with the parental class and ask which gene has switched places. The gene that switched places is in the middle.

Section: 7.3 Bloom's Taxonomy: Applying/Analyzing 31) If interference is complete, what is the frequency of double crossovers?

Answer: zero Section: 7.4 Bloom's Taxonomy: Applying/Analyzing 9 Copyright © 2021 Pearson Education Ltd.

32) Three loci, mitochondrial malate dehydrogenase with alleles MDHa and MDHb, glucouronidase with alleles GUS1 and GUS 2 and a histone gene with alleles H+ and H- are located on chromosome 7 in humans. Assume that the MDH locus is at position 35, GUS at position 45, and H at position 75. A female whose mother was homozygous for MDHa, GUS2, and H+ and whose father was homozygous for MDHb, GUS1, and H- produces a sample of 1000 egg cells. Give the genotypes and expected numbers of the various types of cells she would produce. Assume no chromosomal interference.

Answer: MDHa GUS2 H+ = 315 MDHa GUS2 H= 135 MDHb GUS1 H= 315 MDHb GUS1 H+ = 135 MDHa GUS1 H= 35 MDHa GUS1 H+ = 15 MDHb GUS2 H+ = 35 MDHb GUS2 H= 15 Section: 7.3, 7.4 Bloom's Taxonomy: Evaluating/Creating 33) What are two commonly used DNA landmarks for mapping human genes?

Answer: microsatellites and restriction fragment length polymorphisms (RFLPs) Section: 7.5 Bloom's Taxonomy: Remembering/Understanding 34) What advantage does BrdU (bromodeoxyuridine) have in the study of chromosome structure and recombination?

Answer: Chromatids stained with BrdU in both DNA strands are distinguishable from those with BrdU in only one strand of the double helix.

Section: 7.6 Bloom's Taxonomy: Remembering/Understanding

10 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 8 Genetic Analysis and Mapping in Bacteria and Bacteriophages 1) Which of the following is a true statement?

A) a bacterial colony found on minimal media consists of various auxotroph and prototroph cells B) a bacterial colony found on minimal media consists of various prototroph cells C) a bacterial colony found on minimal media consists of cells derived from a single prototroph cell D) a bacterial colony found on minimal media consists of cells derived from a single auxotroph cell E) a bacterial colony found on minimal media consists of a single prototroph bacterial cell Answer: C Section: 8.1 Bloom's Taxonomy: Remembering/Understanding 2) A stock of a bacterial colony in liquid media was diluted by a factor of 105, and 500 μl of this dilution was spread on a Petrie dish of solidified media, 40 colonies were observed. What was the count of bacteria of the initial stock?

A) 20 × 105 per mL B) 40 × 105 per mL C) 80 × 105 per mL D) 40 × 106 per mL E) 40 × 10-5 per mL Answer: C Section: 8.1 Bloom's Taxonomy: Applying/Analyzing 3) What occurs when a F+ cell is mated to a F- cell?

A) F- cells become F+ cells with rare transfer of chromosomal genes B) F- cells remain F- cells with rare transfer of chromosomal genes C) F- cells become Hfr cells D) F- cells become F+ cells with a high rate of chromosomal gene transfer E) The F+ cells become F- cells, whereas the F- cells become F+ cells Answer: A Section: 8.2 Bloom's Taxonomy: Remembering/Understanding

1 Copyright © 2021 Pearson Education Ltd.

4) An Hfr strain (met+,his+, arg+, strs) is mated with an F- strain (met-, his-, arg-, strr), grown in complete media, then plated on minimal media supplemented with streptomycin, histidine, and arginine. The few colonies that grow on the plate indicate that .

A) the wild-type methionine gene has been transferred from Hfr to FB) the wild-type histidine and arginine genes have been transferred from Hfr to FC) the wild-type histidine and arginine genes and streptomycin resistance gene have been transferred from Hfr to FD) all the Hfr genes have been transferred to the F- strain E) no mating has occurred, the colonies are just Hfr strains Answer: A Section: 8.2 Bloom's Taxonomy: Applying/Analyzing 5) A mating of an Hfr strain of E. coli with an F+ strain will yield A) all F+ B) Hfr and transformed F+ C) all Hfr D) Hfr and F' E) nothing, the mating cannot occur Answer: E Section: 8.2 Bloom's Taxonomy: Remembering/Understanding 6) A mating between an Hfr strain of E. coli with an F- strain will yield A) high rate of recombination and rare transfer of the F factor B) high rate of recombination and frequent transfer of the F factor C) low rate of recombination and frequent transfer of the F factor D) frequent transfer of the F factor and no transfer of chromosomal genes E) no exchange of genetic information Answer: A Section: 8.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

.

.

7) From one F+ strain the following three Hfr strains were derived, each shown with the first three markers transferred in an Hfr X F- cross.

Hfr1...DAG-> Hfr2...GBE-> Hfr3...ECD-> The order of genes on the bacterial chromosome circle must be ends to represent circularity) A) ADCEBGA B) ABCDGEA C) ACDGEBA D) AEGBCDA E) AGBDECA Answer: A Section: 8.2 Bloom's Taxonomy: Applying/Analyzing

. (A is shown at both

8) What is a form of recombination in bacteria that involves the F plasmid?

A) transduction B) transformation C) conjugation D) vertical transfer E) meiosis Answer: C Section: 8.2 Bloom's Taxonomy: Remembering/Understanding 9) An F' cell is generated by .

A) partial transfer of the F factor B) imprecise excision of the F factor from the bacterial chromosome C) an Hfr x F- mating D) conjugation between two F+ cells E) interrupted mating Answer: B Section: 8.2 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

10) A partially diploid bacterial cell is referred to as a(n) A) hemizygote B) heterozygote C) merozygote D) hemiploid E) auxotroph Answer: C Section: 8.2 Bloom's Taxonomy: Remembering/Understanding

.

11) F'-lac+ is "mated" with an F- strain with a mutated lac gene. The F- strain will become .

A) F+ B) F'-lac+/lacC) F+ lacD) Hfr lac+ E) remain F-, lacAnswer: B Section: 8.2 Bloom's Taxonomy: Remembering/Understanding 12) Name the general category into which double-stranded circular extrachromosomal DNA elements such as F factors, ColE1, and R would fall.

A) capsid B) r-determinant C) plaque D) partial diploid E) plasmid Answer: E Section: 8.3 Bloom's Taxonomy: Remembering/Understanding 13) Which of the following is not a feature of a Col plasmid?

A) toxic to bacterial cells that don't harbor the same plasmid B) typically present in 10-20 copies in a cell C) is autonomously replicating D) is not usually transmissible to other cells E) confers antibiotic resistance to recipient cells Answer: E Section: 8.3 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

14) R plasmids often contain .

A) genes for antibody production B) genes for antibiotic resistance C) genes for protection to colicins D) genes associated with the fertility factor E) foreign genes Answer: B Section: 8.3 Bloom's Taxonomy: Remembering/Understanding 15) What defines a competent bacterial cell?

A) the ability to transfer DNA to recipient cells B) capability of uptake of foreign DNA from their environment C) capability to grow in minimal media D) immunity to phage infection E) ability to outcompete non-competent cells for nutrients Answer: B Section: 8.4 Bloom's Taxonomy: Remembering/Understanding 16) Two labs perform cotransformation experiments for genes A and B. Lab 1 finds these genes cotransformed into 8% of recipient cells; whereas lab 2 observes 3% cotransformation of these genes. Which lab shows genes A and B to be more closely linked?

A) lab 1 B) lab 2 Answer: A Section: 8.4 Bloom's Taxonomy: Applying/Analyzing

5 Copyright © 2021 Pearson Education Ltd.

17) In a transformation experiment involving a wild type bacterial strain with a recipient strain with mutations in genes a, b, c, and d, pairs of genes were analyzed for cotransformation with the following results: Gene Pair a+ b+ a+ c+ a+ d+ b+ c+ b+ d+ c+ d+

Cotransformation no yes no no yes yes

What is the linear order of these genes relative to each other?

A) a-b-c-d B) d-a-c-b C) a-d-b-c D) a-c-d-b E) c-a-d-b Answer: D Section: 8.4 Bloom's Taxonomy: Evaluating/Creating 18) Bacteriophages engage in two interactive cycles with bacteria. What are these cycles?

A) lytic and lysogenic B) lytic and prototrophic C) auxotrophic and prototrophic D) heteroduplex and homoduplex E) negative and positive Answer: A Section: 8.5 Bloom's Taxonomy: Remembering/Understanding 19) A bacteriophage that is capable of entering either a lytic or lysogenic cycle is called a(n) .

A) temperate bacteriophage B) virulent bacteriophage C) plasmid D) episome E) plaque-forming unit Answer: A Section: 8.5 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

20) Viral DNA that has integrated into a bacterial chromosome is referred to as a A) prophage B) prephage C) plaque D) temperate phage E) virulent phage Answer: A Section: 8.5 Bloom's Taxonomy: Remembering/Understanding

.

21) The clearing made by bacteriophages in a "lawn" of bacteria on an agar plate is called a .

A) phage zone B) lysogenic zone C) prophage D) plaque E) host range Answer: D Section: 8.5 Bloom's Taxonomy: Remembering/Understanding 22) Viral-mediated transfer of bacterial DNA is referred to as A) conjugation B) transformation C) transduction D) lysogeny E) endosymbiosis Answer: C Section: 8.6 Bloom's Taxonomy: Remembering/Understanding

.

23) How does an auxotroph differ from a prototroph?

Answer: Auxotrophs have lost, through mutation, the ability to grow on minimal medium.

Prototrophs are capable of growth on minimal medium.

Section: 8.1 Bloom's Taxonomy: Remembering/Understanding 24) What are prototrophs?

Answer: Prototrophs are bacteria that can grow on minimal medium and are assumed to be wild type.

Section: 8.1 Bloom's Taxonomy: Remembering/Understanding 25) In general, what two methods are used to grow bacteria in the laboratory?

Answer: liquid and semisolid (agar) media Section: 8.1 Bloom's Taxonomy: Remembering/Understanding 7 Copyright © 2021 Pearson Education Ltd.

26) Explain the composition and use of minimal medium in the study of bacterial genetics.

Answer: Minimal medium consists of an organic carbon source such as glucose or lactose and a variety of inorganic ions: Na+, K+, Mg++, Ca++, and NH4+. It is useful in isolating bacterial strains (auxotrophs) that are incapable of synthesizing more complex nutritional requirements.

Section: 8.1 Bloom's Taxonomy: Remembering/Understanding 27) Contrast horizontal and vertical gene transfer.

Answer: Vertical transfer is the transmission of genetic information from one generation to the next and therefore within the same species. Horizontal transfer is the process of genetic exchange between unrelated cells, including that of one species to another.

Section: 8.2 Bloom's Taxonomy: Remembering/Understanding 28) In the experiment by Lederberg and Tatum, why was it necessary to first combine the two auxotrophic strains in complete media?

Answer: To be certain that both these strains are capable of growth and cell division.

Section: 8.2 Bloom's Taxonomy: Applying/Analyzing 29) In what way was the interrupted mating technique used to generate a genetic map in E. coli based on time?

Answer: The interrupted mating technique showed that genes were passed in a linear fashion from the Hfr bacterial strain to an F- strain. By interrupting the mating tube, it could be determined when (in minutes) genes were transferred.

Section: 8.2 Bloom's Taxonomy: Remembering/Understanding 30) Jacob, Wollman, and others developed a linkage map of E. coli that is based on time. What form of recombination is involved in generating a linkage map based on time?

Answer: Conjugation; an Hfr bacterium conjugating with an F- strain allows the generation of a map that is dependent on the passage of the donor chromosome across a conjugation tube.

Section: 8.2 Bloom's Taxonomy: Applying/Analyzing 31) Using interrupted mating experiments with three different Hfr strains, the following results were obtained (times of transfer for various genes): Hfr1 x F- : X (15 minutes); Y (25 minutes); Z (32 minutes) Hfr2 x F-: Q (8 minutes); H (16 minutes): Z (19 minutes); Y (26 minutes) Hfr3 x F-: Y (7 minutes); X (17 minutes); Q (22 minutes); H (30 minutes) Starting and ending with gene X, write out the order and distances between the genes.

Answer: X — 10 — Y - 7- Z -3 — H — 8 — Q — 5 — X Section: 8.2 Bloom's Taxonomy: Evaluating/Creating 8 Copyright © 2021 Pearson Education Ltd.

32) What is the role of the F factor in bacterial recombination?

Answer: The F factor contains the genes such as for sex pili and the production of a conjugation tube that confer the capacity to donate genes to a recipient cell.

Section: 8.2 Bloom's Taxonomy: Remembering/Understanding 33) Describe how different strains of E. coli can reveal different linkage arrangements of genes in Hfr crosses.

Answer: F factors can integrate into one of several possible sites and in different orientations leading to the potential of different initiation points (genes) and transfer in different (e.g. clockwise vs counter-clockwise) directions.

Section: 8.2 Bloom's Taxonomy: Applying/Analyzing 34) What is meant by the term cotransformation?

Answer: Cotransformation occurs when several linked genes are transformed simultaneously.

Section: 8.4 Bloom's Taxonomy: Remembering/Understanding 35) Explain what is meant by the term heteroduplex in the context of bacterial transformation.

Answer: If transforming DNA is derived from a genetically distinct bacterium, incoming DNA may be different from the host DNA. During initial stages of integration into the bacterial chromosome, the recombinant region contains one strand of DNA that may have a different base sequence than the other strand. Because these strands are not genetically identical (complementary), this double-stranded region is called a heteroduplex.

Section: 8.4 Bloom's Taxonomy: Applying/Analyzing 36) In a transformation experiment involving a donor (a+b+) and a recipient strain of genotype (a-b-) the following results were obtained: Transformants (%) a+ b- (2%) a- b+ (5%) a+ b+ (0.1%) Do genes a and b appear to be linked? Why or why not?

Answer: Genes a and b appear to be linked since they are cotransformed more frequently than expected if independent transformation. Locus a is transformed at a frequency of 0.02 and locus b at a frequency of 0.05, therefore by chance these loci will be cotransformed 0.05 × 0.02 or 0.0001 or 0.01% which is less than the observed value of 0.1% Section: 8.4 Bloom's Taxonomy: Evaluating/Creating

9 Copyright © 2021 Pearson Education Ltd.

37) Assume that one counted 67 plaques on a bacterial plate where 0.1 ml of a 10-5 dilution of phage was added to bacterial culture. What is the initial concentration of the undiluted phage?

Answer: 67 × 105 × 10 = 6.7 × 107 pfu/ml (pfu = plaque-forming units) Section: 8.5 Bloom's Taxonomy: Applying/Analyzing 38) What is a bacteriophage?

Answer: A bacteriophage is a virus that has a bacterium as its host.

Section: 8.5 Bloom's Taxonomy: Remembering/Understanding 39) Lysogeny is an important phenomenon in bacteria and phages. Briefly describe lysogeny.

Answer: Lysogeny is the process in which a temperate bacteriophage infects a bacterial cell and subsequently integrates its chromosome into the bacterial chromosome.

Section: 8.5 Bloom's Taxonomy: Remembering/Understanding 40) Compare and contrast bacteriophage lysis and lysogeny.

Answer: Both lysis and lysogeny involve bacterial/phage interactions related to the production of phage progeny. Lysis occurs when progeny phage burst from the bacterial cell; lysogeny involves the incorporation of the phage chromosome into the bacterial chromosome.

Section: 8.5 Bloom's Taxonomy: Remembering/Understanding 41) A donor strain with genes a+ b+ c+ is infected with a bacteriophage, which transfers genes to a recipient cell that is a-b-c-. Upon selecting for recipient cells that contain c+, cotransduction is analyzed for a+ and for b+. The lab finds 10% of the c+ recipient cells also contain a+; whereas 6% of the c+ recipient cells also contain b+. Which gene (a or b) is more closely linked to c and why?

Answer: a is more closely linked to c, which would explain the greater frequency of a and c cotransduction relative to b and c cotransduction Section: 8.6 Bloom's Taxonomy: Applying/Analyzing 42) In the Lederberg-Zinder experiment using a U-tube that prevents cell-cell contact, they identified genetic exchange. How did they rule out the possibility of this occurring via transformation?

Answer: The investigators incorporated DNase into the media which would enzymatically degrade free floating (unprotected) DNA in the media Section: 8.6 Bloom's Taxonomy: Applying/Analyzing

10 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Ediiton (Klug) Chapter 9 DNA structure and Analysis 1) In the 1860s, a Swiss chemist, Friedrich Miescher, isolated an acidic substance from cell nuclei. What was the name of this substance?

A) cistron B) nuclein C) hyaluronic acid D) chromatin E) acetic acid Answer: B Section: 9.2 Bloom's Taxonomy: Remembering/Understanding 2) In the experiment by Avery et al. (1944), smooth bacterial cells were identified when live rough cells were mixed with an extract of heat-killed smooth cells treated with .

A) DNase B) either RNase or protease C) either DNase, RNase, or protease D) chloramphenicol Answer: B Section: 9.3 Bloom's Taxonomy: Applying/Analyzing 3) In 1927 Frederick Griffith combined an extract from heat-killed smooth bacterial cells with live rough cells and injected them into mice. The findings and conclusions of the experiment were .

A) mice survived as a result of immunity from rough cells B) mice died as a result of a transforming agent picked up by the rough cells from the smooth cell extract C) mice survived as a result of a transforming agent picked up by the rough cells from the smooth cell extract D) mice died as a result of incorporation of DNA into the mouse genome from heat-killed smooth cells E) mice died due to random mutations in the rough cells to become smooth cells Answer: B Section: 9.3 Bloom's Taxonomy: Remembering/Understanding

1 Copyright © 2021 Pearson Education Ltd.

4) The classic Hershey and Chase (1952) experiment that offered evidence in support of DNA being the genetic material in bacteriophages made use of which of the following labeled component(s)?

A) phosphorus and sulfur B) nitrogen and oxygen C) tritium D) hydrogen E) carbon and nitrogen Answer: A Section: 9.3 Bloom's Taxonomy: Remembering/Understanding 5) What was the Avery et al. (1944) contribution to the understanding of molecular biology?

A) demonstration that dead bacterial cells can come to life B) demonstration that DNA is the transforming principle from dead bacteria that can convert the phenotype of live bacteria C) demonstration that RNA is the intermediate in the Central Dogma of Molecular Biology D) demonstration that chromosomes contain genetic material E) demonstration that viruses can transform bacterial cells Answer: B Section: 9.3 Bloom's Taxonomy: Remembering/Understanding 6) In the classic experiment conducted by Hershey and Chase, why was the pellet radioactive in the centrifuge tube that contained bacteria with viruses?

A) The bacteria were in the pellet, and they had incorporated radioactive proteins into their cell membranes.

B) The radioactive viruses (coats plus DNA) were in the pellet.

C) The bacteria were in the pellet, many of which contained the radioactive viral DNA. D) The radioactive protein coats of the viruses were in the pellet.

E) The radioactive viruses were in the pellet, and the bacteria were in the supernatant.

Answer: C Section: 9.3 Bloom's Taxonomy: Applying/Analyzing 7) Which of the following is not a true statement regarding retroviruses?

A) RNA serves as genetic material.

B) RNA is converted to DNA by DNA polymerase.

C) The human immunodeficiency virus (HIV) is an example of a retrovirus.

D) DNA generated from retroviruses can potentially integrate into the host's genome.

Answer: B Section: 9.5 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

8) Reverse transcriptase is an enzyme found in association with retroviral activity. It has the property of .

A) synthesis of DNA from an RNA template B) synthesis of RNA from a DNA template C) requiring no template D) translation E) synthesis of miRNA to cDNA Answer: A Section: 9.5 Bloom's Taxonomy: Remembering/Understanding 9) Given the structure below, please answer the following questions. Is the accompanying figure DNA or RNA? Is the arrow closest to the 5' or 3' end?

Spleen diesterase is an enzyme that breaks the covalent bond that connects the phosphate to the 5' carbon. Assume that the dinucleotide is digested with spleen diesterase. To which base and to which carbon on the sugar is the phosphate now attached, A or T?

A) DNA; 3' end; T, 5' B) RNA; 3' end; A, 3' C) DNA; 5' end; A, 5' D) DNA; 3' end; A, 3' E) RNA; 3' end; T, 3' Answer: D Section: 9.6 Bloom's Taxonomy: Evaluating/Creating 10) The basic structure of a nucleotide includes A) amino acids B) phosphorus and zinc C) base, sugar, and phosphate D) mRNA, rRNA, and tRNA E) phosphorus and sulfur Answer: C Section: 9.6 Bloom's Taxonomy: Remembering/Understanding

.

3 Copyright © 2021 Pearson Education Ltd.

11) Regarding the structure of DNA, the covalently arranged combination of a deoxyribose and a nitrogenous base would be called a(n) .

A) nucleotide B) ribonucleotide C) monophospate nucleoside D) oligonucleotide E) nucleoside Answer: E Section: 9.6 Bloom's Taxonomy: Remembering/Understanding 12) Which is not a major feature of the Watson and Crick DNA model?

A) a complete turn is 20 angstroms B) chains run antiparallel C) contains alternating major and minor grooves D) strands are paired by hydrogen bonds between complementary bases E) coiled strands form a right-handed double helix Answer: A Section: 9.6 Bloom's Taxonomy: Remembering/Understanding 13) Considering the structure of double-stranded DNA, which kind(s) of bonds hold one complementary strand to the other?

A) ionic B) covalent C) van der Waals D) hydrogen E) hydrophobic and hydrophilic Answer: D Section: 9.6 Bloom's Taxonomy: Remembering/Understanding 14) Which type of bond links nucleotides in a single strand of DNA? A) hydrogen B) phosphodiester C) hydrophobic and hydrophilic D) electrostatic Answer: B Section: 9.6 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

15) If 15 percent of the nitrogenous bases in a sample of DNA from a particular organism is thymine, what percentage should be cytosine?

A) 15 percent B) 30 percent C) 35 percent D) 40 percent E) 70 percent Answer: C Section: 9.6 Bloom's Taxonomy: Applying/Analyzing 16) In an analysis of the nucleotide composition of double-stranded DNA to see which bases are equivalent in concentration, which of the following would be true?

A) A = C B) A = G and C = T C) A + C = G + T D) A + T = G + C E) A = G and C = T and A + C = G + T are both true.

Answer: C Section: 9.6 Bloom's Taxonomy: Applying/Analyzing 17) Which of the following clusters of terms accurately describes DNA as it is generally viewed to exist in prokaryotes and eukaryotes?

A) double-stranded, parallel, (A + T)/(C + G) = variable, (A + G)/(C + T) = 1.0 B) double-stranded, antiparallel, (A + T)/(C + G) = variable, (A + G)/(C+ T) = 1.0 C) single-stranded, antiparallel, (A + T)/(C + G) = 1.0, (A + G)/(C + T) = 1.0 D) double-stranded, parallel, (A + T)/(C + G) = 1.0, (A + G)/(C + T) = 1.0 E) double-stranded, antiparallel, (A + T)/(C + G) = variable, (A + G)/(C + T) = variable Answer: B Section: 9.6 Bloom's Taxonomy: Applying/Analyzing 18) Which of the following would represent a complete turn of a DNA double helix?

A) 5' end to the 3' end B) major groove to minor groove C) minor groove to minor groove D) one nucleotide to the next nucleotide Answer: C Section: 9.6 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

19) If the GC content of a DNA molecule is 60 percent, what is the molar percentage of adenine?

A) 40% B) 30% C) 20% D) 10% E) can't be determined Answer: C Section: 9.6 Bloom's Taxonomy: Applying/Analyzing 20) Assume that the molar percentage of thymine in a double-stranded DNA is 20. What is the molar percentage of cytosine?

A) 20% B) 30% C) 40% D) 80% E) can't be determined Answer: B Section: 9.6 Bloom's Taxonomy: Applying/Analyzing 21) Assume that A + T/G + C equals 0.5 in one strand of DNA. What is the ratio of these bases in the complementary strand?

A) 0.25 B) 0.50 C) 1.0 D) 2.0 E) can't be determined Answer: B Section: 9.6 Bloom's Taxonomy: Evaluating/Creating 22) If A + G/T + C equals 0.5 in one strand, what is the ratio of these bases in the complementary strand?

A) 0.25 B) 0.50 C) 1.0 D) 2.0 E) can't be determined Answer: D Section: 9.6 Bloom's Taxonomy: Evaluating/Creating

6 Copyright © 2021 Pearson Education Ltd.

23) During the polymerization of nucleic acids, covalent bonds are formed between neighboring nucleotides. Which carbons are involved in such bonds?

A) C-2' and C-3' B) C-1' and C-2' C) C-3' and C-5' D) C-3' and C-3' E) C-1' and C-3' Answer: C Section: 9.6 Bloom's Taxonomy: Remembering/Understanding 24) Purines in RNA include .

A) adenine and guanine B) adenine and uracil C) adenine, guanine, and uracil D) guanine and uracil E) cytosine and uracil Answer: A Section: 9.6 Bloom's Taxonomy: Remembering/Understanding 25) Is the accompanying figure DNA or RNA? Is the circle closer to the 5' or 3' end?

A) RNA; 3' B) RNA; 5' C) DNA; 3' D) DNA; 5' Answer: A Section: 9.6 Bloom's Taxonomy: Applying/Analyzing

7 Copyright © 2021 Pearson Education Ltd.

26) Is the accompanying figure DNA or RNA? Is the circle closer to the 5' or 3' end?

A) RNA; 3' B) RNA; 5' C) DNA; 3' D) DNA; 5' Answer: D Section: 9.6 Bloom's Taxonomy: Applying/Analyzing 27) Suppose that the dinucleotide in the accompanying figure were cleaved with the enzyme spleen diesterase, which breaks the covalent bond connecting the phosphate to C-5'. After such cleavage, to which nucleoside is the phosphate now attached?

A) deoxythymidine B) deoxyadenosine C) neither D) both Answer: B Section: 9.6 Bloom's Taxonomy: Applying/Analyzing

8 Copyright © 2021 Pearson Education Ltd.

28) Which of the following was not required for Watson and Crick to determine the doublehelical structure of DNA? A) X-ray diffraction analysis B) Chargaff's base composition studies C) nucleic acid chemistry D) visualization on agarose gels E) hydrophobic nature of nitrogenous bases Answer: D Section: 9.6 Bloom's Taxonomy: Remembering/Understanding 29) Adenosine is an example of a ; whereas adenine is an example of a A) nucleoside; nucleotide B) nucleoside; nitrogenous base C) nitrogenous base; nucleoside D) sugar; nitrogenous base E) DNA molecule; RNA molecule Answer: B Section: 9.6 Bloom's Taxonomy: Remembering/Understanding 30) The most abundant type of RNA in a cell is A) rRNA B) mRNA C) tRNA D) snRNA E) miRNA Answer: A Section: 9.8 Bloom's Taxonomy: Remembering/Understanding

.

.

31) Which of the following segments of double-stranded DNA requires the highest temperature to separate the strands?

A) 5'-AAAATTTT-3' 3'-TTTTAAAA-5' B) 5'-CGAATAGC-3' 3'-GCTTATCG-5' C) 5'-ATGCATGC-3' 3'-TACGTACG-5' D) 5'-CGATTAGC-3' 3'-GCTAATCG-5' E) 5'-GGCAGCTG-3' 3'-CCGTCGAC-5' Answer: E Section: 9.9 Bloom's Taxonomy: Applying/Analyzing 9 Copyright © 2021 Pearson Education Ltd.

32) Gel electrophoresis provides a tool to separate fragments of DNA by size. Which of the following is a true statement?

A) DNA moves toward the negative electrode; larger fragments run faster B) DNA moves toward the negative electrode; smaller fragments run faster C) DNA moves toward the positive electrode; larger fragments run faster D) DNA moves toward the positive electrode; smaller fragments run faster Answer: D Section: 9.9 Bloom's Taxonomy: Remembering/Understanding 33) Describe four major properties that genetic material must exhibit.

Answer: replication = duplication of genetic material; expression = production of a phenotype; storage = stable maintenance and passage of information; variation = capable of alteration Section: 9.1 Bloom's Taxonomy: Remembering/Understanding 34) In a eukaryotic cell, if DNA is stored in the nucleus and protein synthesis occurs in the cytoplasm, then how does the information in DNA become functional (i.e. processed into protein)?

Answer: A messenger RNA molecule is transcribed from the DNA and carries the information to the ribosome where it is translated into protein Section: 9.1 Bloom's Taxonomy: Remembering/Understanding 35) Provide reasons as to why prior to the 1940s protein was considered genetic material.

Answer: Proteins are highly diverse and abundant in cells, whereas DNA was postulated to be invariable based on the tetranucleotide hypothesis.

Section: 9.2 Bloom's Taxonomy: Remembering/Understanding 36) Experiments conducted in the 1920s by Frederick Griffith involving the bacterium Diplococcus pneumoniae demonstrated that a substance from one bacterial strain could genetically transform other bacterial strains. What was the name of the substance capable of such transformation, and who finally determined its identity?

Answer: deoxyribonucleic acid (DNA); Avery et al. (1944) Section: 9.3 Bloom's Taxonomy: Remembering/Understanding 37) Briefly define transformation and describe the relationship between the phenomenon of transformation and the discovery that DNA is the genetic material in bacteria.

Answer: Transformation is the process whereby one organism is genetically altered by exposure to DNA from another organism. Because DNA can carry heritable "traits" from one organism to another, it must be the genetic material.

Section: 9.3 Bloom's Taxonomy: Remembering/Understanding

10 Copyright © 2021 Pearson Education Ltd.

38) The strongest direct evidence that DNA is the genetic material comes from which contemporary methodology? Provide an example.

Answer: Recombinant DNA technology. The recombinant DNA introduced into bacteria is replicated and passed down to daughter cells.

Section: 9.4 Bloom's Taxonomy: Remembering/Understanding 39) Gametes of various eukaryotic organisms were shown to have half the amount of DNA than somatic cells. Explain the relationship of this find with Mendel's Law of Segregation and the Chromosomal Theory of Inheritance.

Answer: Mendel's Law of Segregation indicates that unit factors are in pairs with each randomly segregated into gametes, therefore half the number of unit factors corresponds with half the amount of DNA. The Chromosomal Theory of Inheritance demonstrates that members of homologous pairs are randomly segregated into gametes each having a haploid set of chromosomes, which would be half the number of somatic cells.

Section: 9.4 Bloom's Taxonomy: Evaluating/Creating 40) At what approximate wavelengths do DNA, RNA, and proteins maximally absorb light?

Answer: 260 nm, 260 nm, and 280 nm, respectively Section: 9.4 Bloom's Taxonomy: Remembering/Understanding 41) List two major differences between RNA and DNA at the level of the nucleotide.

Answer: ribose in RNA, deoxyribose in DNA; uracil in RNA replaces thymine in DNA Section: 9.6 Bloom's Taxonomy: Remembering/Understanding 42) The base content of a sample of nucleic acid is as follows: A = 31%, G = 31%, T = 19%, C = 19%. What conclusion can be drawn from this information?

Answer: The nucleic acid is single-stranded DNA. Section: 9.6 Bloom's Taxonomy: Applying/Analyzing 43) What is the difference between a polynucleotide and an oligonucleotide?

Answer: Polynucleotides are polymers longer than 20 nucleotides; oligonucleotides are shorter than polynucleotides.

Section: 9.6 Bloom's Taxonomy: Remembering/Understanding 44) What is meant by the term antiparallel?

Answer: The sugar-phosphate backbones of the two chains in a double-stranded nucleic acid are connected in opposite directions in terms of 5'-3' orientation.

Section: 9.6 Bloom's Taxonomy: Remembering/Understanding

11 Copyright © 2021 Pearson Education Ltd.

45) List three forms of DNA. Answer: A-DNA, B-DNA, and Z-DNA Section: 9.7 Bloom's Taxonomy: Remembering/Understanding 46) Provide an overview of the structure of Z-DNA. Answer: left-handed helix with two antiparallel complementary strands, 1.8 nm in diameter, 12 bases per turn, zigzag configuration, and shallow major groove Section: 9.7 Bloom's Taxonomy: Remembering/Understanding 47) What is fluorescent in situ hybridization (FISH) and what information is gained?

Answer: hybridization of a fluorescently-labeled segment of DNA to a fixed metaphase spread to potentially determine the chromosomal location(s) housing specific genetic information Section: 9.9 Bloom's Taxonomy: Remembering/Understanding

12 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 10 DNA Replication 1) Which of the following terms accurately describes the replication of DNA in vivo?

A) conservative B) dispersive C) semiconservative D) nonlinear E) continuous Answer: C Section: 10.1 Bloom's Taxonomy: Remembering/Understanding 2) Which terms accurately reflect the nature of replication of the chromosome in E. coli?

A) bidirectional and fixed point of initiation B) unidirectional and reciprocal C) unidirectional and fixed point of initiation D) multirepliconic and telomeric E) bidirectional and multirepliconic Answer: A Section: 10.1 Bloom's Taxonomy: Applying/Analyzing 3) Assume that a culture of E. coli was grown for approximately 50 generations in 15N (provided in the medium in the ammonium ion), which is a heavy isotope of nitrogen (14N). You extract the DNA from the culture, and it has a density of 1.723 gm/cm3 (water = 1.00 gm/cm3).

From the literature, you determine that DNA containing only the common form of nitrogen, 14N, has a density of 1.700 gm/cm3. Bacteria from the 15N culture were washed in buffer and transferred to 14N medium for one generation immediately after which the DNA was extracted and its density determined. What would be the expected approximate density of the extracted DNA? A) 1.712 gm/cm3 B) 1.723 gm/cm3 C) 1.700 gm/cm3 D) 1.700 and 1.712 gm/cm3 E) 1.723 and 1.700 gm/cm3 Answer: A Section: 10.1 Bloom's Taxonomy: Evaluating/Creating

1 Copyright © 2021 Pearson Education Ltd.

4) Assume that a culture of E. coli was grown for approximately 50 generations in 15N (provided in the medium in the ammonium ion), which is a heavy isotope of nitrogen (14N). You extract the DNA from the culture, and it has a density of 1.723 gm/cm3 (water = 1.00 gm/cm3).

From the literature, you determine that DNA containing only the common form of nitrogen, 14N, has a density of 1.700 gm/cm3. Bacteria from the 15N culture were washed in buffer and transferred to 14N medium for one generation immediately after which the DNA was extracted and its density determined. After you heat the extracted DNA until it completely denatures (95°C for 15 minutes), what would you expect the density of the DNA in the denatured extract to be?

For the purposes of this question, assume that DNA has the same density regardless of whether it is single- or double-stranded.

A) 1.712 gm/cm3 B) 1.723 gm/cm3 C) 1.700 gm/cm3 D) 1.700 and 1.712 gm/cm3 E) 1.723 and 1.700 gm/cm3 Answer: E Section: 10.1 Bloom's Taxonomy: Evaluating/Creating 5) Assume that a culture of E. coli was grown for approximately 50 generations in 15N (provided in the medium in the ammonium ion), which is a heavy isotope of nitrogen (14N). You extract the DNA from the culture, and it has a density of 1.723 gm/cm3 (water = 1.00 gm/cm3).

From the literature, you determine that DNA containing only the common form of nitrogen, 14N, has a density of 1.700 gm/cm3. Bacteria from the 15N culture were washed in buffer and transferred to 14N medium for one generation immediately after which the DNA was extracted and its density determined. Assuming that the molar percentage of adenine in the extracted DNA was 20 percent, what would be the expected molar percentages of the other nitrogenous bases in this DNA? A) Thymine = 30%, Guanine = 30%, Cytosine = 20% B) Thymine = 20%, Guanine = 20%, Cytosine = 20% C) Thymine = 20%, Guanine = 30%, Cytosine = 30% D) Thymine = 20%, Guanine = 20%, Cytosine = 30% E) Thymine = 20%, Guanine = 30%, Cytosine = 20% Answer: C Section: 10.1 Bloom's Taxonomy: Evaluating/Creating

2 Copyright © 2021 Pearson Education Ltd.

6) Which of the following researchers provided evidence for semiconservative replication in prokaryotes?

A) Meselson and Stahl B) Kornberg C) DeLucia and Cairns D) Okazaki E) Taylor, Woods and Hughes Answer: A Section: 10.1 Bloom's Taxonomy: Remembering/Understanding 7) Which cluster of terms accurately reflects the nature of DNA replication in prokaryotes?

A) fixed point of initiation, bidirectional, conservative B) fixed point of initiation, unidirectional, conservative C) random point of initiation, bidirectional, semiconservative D) fixed point of initiation, bidirectional, semiconservative E) random point of initiation, unidirectional, semiconservative Answer: D Section: 10.1 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

8) Given the diagram below, assume that a G1 chromosome (left) underwent one round of replication in 3H-thymidine and the metaphase chromosome (right) had both chromatids labeled.

Which of the following replicative models (conservative, dispersive, semiconservative) could be eliminated by this observation?

A) conservative B) semiconservative C) dispersive D) continuous E) discontinuous Answer: A Section: 10.1 Bloom's Taxonomy: Evaluating/Creating

4 Copyright © 2021 Pearson Education Ltd.

9) Which of the following is believed to be responsible for removing the primer and filling in the gaps left behind during replication?

A) DNA polymerase I B) DNA polymerase II C) DNA polymerase III D) DNA polymerase IV E) DNA polymerase V Answer: A Section: 10.2 Bloom's Taxonomy: Remembering/Understanding 10) DNA polymerase III adds nucleotides .

A) to the 3' end of the primer B) to the 5' end of the primer C) in the place of the primer after it is removed D) to both ends of the primer E) to internal sites in the DNA template Answer: A Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 11) DNA polymerase I is thought to add nucleotides A) to the 5' end of the primer B) to the 3' end of the primer C) in the place of the primer after it is removed D) on single-stranded templates without need for an primer E) in a 5' to 5' direction Answer: C Section: 10.3 Bloom's Taxonomy: Remembering/Understanding

.

12) The discontinuous aspect of replication of DNA in vivo is caused by A) polymerase slippage B) trinucleotide repeats C) the 5' to 3' polarity restriction.

D) topoisomerases cutting the DNA in a random fashion E) sister-chromatid exchanges Answer: C Section: 10.3 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

.

13) Which of the following relaxes supercoiling by single or double strand cuts in the DNA ahead of the replication fork?

A) DNA gyrase B) DNA helicase C) single stranded binding protein D) primase E) polymerase Answer: A Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 14) Which of the following is a short stretch of DNA on the lagging strand?

A) Okazaki fragment B) primer C) discontinuous strand D) continuous strand E) oriC Answer: A Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 15) Which of the following is the side of the replication fork where synthesis is discontinuous?

A) lagging strand B) leading strand C) origin of replication D) template strand Answer: A Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 16) Which of the following indicates that from the point of initiation, replication occurs in both directions?

A) bidirectional B) unidirectional C) polydirectional D) discontinuous E) dispersive Answer: A Section: 10.3 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

17) Which term refers to the complex of proteins that is involved in the replication of DNA? A) replicon B) spliceosome C) transcriptome D) polymerase holoenzyme E) Okazaki complex Answer: D Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 18) Which protein is responsible for the initial step in unwinding the DNA helix during replication of the bacterial chromosome?

A) Dna A B) helicase C) topoisomerase D) telomerase E) gyrase Answer: A Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 19) If we created a null mutation that prevented polymerases from being loaded onto the replicating strand, in which structure would we most likely find the mutation?

A) 4 B) 3 C) 7 D) 6 E) 1 Answer: A Section: 10.4 Bloom's Taxonomy: Evaluating/Creating 7 Copyright © 2021 Pearson Education Ltd.

20) The structure labeled 6 would represent which of the following?

A) DNA polymerase I B) DNA polymerase II C) DNA polymerase III D) DNA polymerase IV E) DNA polymerase V Answer: C Section: 10.4 Bloom's Taxonomy: Applying/Analyzing 21) Which of the following genes encodes DNA helicase in e. coli?

A) polA B) dnaE C) dnaG D) rep E) rpoB Answer: D Section: 10.5 Bloom's Taxonomy: Remembering/Understanding 22) Replication origins in yeast are called .

A) oriC B) autonomous replicating sequences C) nuclear organizing regions D) chromatin associated factors E) none of these are names for yeast origins Answer: B Section: 10.6 Bloom's Taxonomy: Remembering/Understanding

8 Copyright © 2021 Pearson Education Ltd.

23) Which eukaryotic enzyme is responsible for replication of the ends of chromosomes?

A) pol α B) helicase C) topoisomerase D) telomerase E) shelterin Answer: D Section: 10.7 Bloom's Taxonomy: Remembering/Understanding 24) Structures located at the ends of eukaryotic chromosomes are called A) centromeres B) telomerases C) recessive inversions D) telomeres E) permissive mutations Answer: D Section: 10.7 Bloom's Taxonomy: Remembering/Understanding

.

25) Which of the following can be found in prokaryotes?

A) TERC B) TERT C) telomerase D) shelterin E) none of these can be found in prokaryotes Answer: E Section: 10.7 Bloom's Taxonomy: Applying/Analyzing 26) Assume that you are microscopically examining mitotic metaphase cells of an organism with a 2n chromosome number of 4 (one pair metacentric and one pair telocentric). Assume also that the cell passed through one S phase labeling (innermost phosphate of dTTP radioactive) just prior to the period of observation. Explain the autoradiographic pattern you would expect to see.

Answer: We would see two metacentric labeled chromosomes and two telocentric labeled chromosomes. Each unlabeled old strand would be complexed with and labeled new strand which would give us two labeled sister chromatids joined at the centromere to form our fours chromosomes, two metacentric and two telocentric each with labeled sister chromatids.

Section: 10.1 Bloom's Taxonomy: Evaluating/Creating 27) What three possible models were suggested to originally describe the nature of DNA replication?

Answer: conservative, semiconservative, dispersive Section: 10.1 Bloom's Taxonomy: Remembering/Understanding 9 Copyright © 2021 Pearson Education Ltd.

28) The Meselson and Stahl experiment provided conclusive evidence for the semiconservative replication of DNA in E. coli. What pattern of bands would occur in a CsCl gradient for conservative replication?

Answer: After one generation in the 14N, there would be two bands, one heavy and one light (no intermediate). After the second generation in the 14N, there would also be two bands, one heavy and one light (no intermediate).

Section: 10.1 Bloom's Taxonomy: Remembering/Understanding 29) Meselson and Stahl determined that DNA replication in E. coli is semiconservative. What additive did they initially supply to the medium in order to distinguish "new" from "old" DNA? Answer: 15N Section: 10.1 Bloom's Taxonomy: Remembering/Understanding 30) Briefly describe what is meant by the term autoradiography and identify a classic experiment that used autoradiography to determine the replicative nature of DNA in eukaryotes.

Answer: Autoradiography is a technique that allows an isotope to be detected within a cell; the Taylor, Woods, and Hughes (1957) experiment used 3H-thymidine.

Section: 10.1 Bloom's Taxonomy: Remembering/Understanding 31) What primary ingredients, coupled with DNA polymerase I, are needed for the in vitro synthesis of DNA? Answer: dNTP, DNA template, primer DNA or RNA, Mg++ (appropriate buffering, temperature, and salt concentrations might be considered "secondary" ingredients).

Section: 10.2 Bloom's Taxonomy: Applying/Analyzing 32) DNA replication in vivo requires a primer with a free 3' end. What molecular species provides this 3' end, and how is it provided?

Answer: The free 3' end is provided by an RNA primer; it is provided by the enzymatic activity of RNA primase.

Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 33) List four enzymes known to be involved in the replication of DNA in bacteria.

Answer: Appropriate answers would include any four of the following: DNA polymerase I, DNA polymerase III, ligase, RNA primase, helicase, gyrase Section: 10.3 Bloom's Taxonomy: Remembering/Understanding

10 Copyright © 2021 Pearson Education Ltd.

34) Which structural circumstance in DNA sets up the requirement for the semidiscontinuous nature of its replication?

Answer: 5' > 3' polarity restrictions of DNA synthesis and the antiparallel orientation of the DNA strands in DNA Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 35) As unwinding of the helix occurs during DNA replication, tension is created ahead of the replication fork. Describe the nature of this tension and state the manner in which it is resolved.

Answer: supercoiling; DNA gyrase Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 36) Given that the origin of replication is fixed in E. coli, what signals the location of the origin?

Answer: a region called oriC, which consists of about 250 base pairs characterized by repeating sequences of 9 and 13 bases (9mers and 13mers) Section: 10.3 Bloom's Taxonomy: Remembering/Understanding 37) If you had a deletion of the structure labeled 5, in the figure above, that led to a loss of function, what would you anticipate would be the effect on replication?

Answer: This is the sliding clamp which improves processivity and keeps the polymerase from dissociating. Loss of this would mean slower replication speeds and premature dissociation.

Section: 10.4 Bloom's Taxonomy: Evaluating/Creating

11 Copyright © 2021 Pearson Education Ltd.

38) What is the protein labeled 3 and what is its function? Why is this function necessary?

Answer: This is DNA gyrase which relieves supercoiling. It is needed because as DNA unwinds it creates coiling strain in the rest of the DNA molecule which needs to be resolved.

Section: 10.4 Bloom's Taxonomy: Applying/Analyzing 39) If you had a deletion of the structure labeled 1, in the figure above, that led to a loss of function, what would you anticipate would be the effect on replication?

Answer: This is the single-stranded binding protein which keeps unwound strands apart and keeps them from annealing long enough so that replication can occur. A loss of this function would mean that unwound DNA would re-anneal quickly and block access to polymerase and the clamp loader.

Section: 10.4 Bloom's Taxonomy: Evaluating/Creating

12 Copyright © 2021 Pearson Education Ltd.

40) Explain why structure 7 is needed in replication.

Answer: This is the RNA primer. To replicate polymerase needs a 3'OH to add the next nucleotide to by dehydration synthesis. Template alone has no free 3'OH to add to so an RNA primer is needed to provide that free OH. Section: 10.4 Bloom's Taxonomy: Evaluating/Creating 41) Compare the rate of DNA replication in prokaryotes and eukaryotes.

Answer: Eukaryotic DNA polymerases synthesize DNA at a rate 25 times slower (about 2000 nucleotides per minute) than do prokaryotes.

Section: 10.6 Bloom's Taxonomy: Remembering/Understanding 42) Given that the nature of DNA replication in eukaryotes is not as well understood as in prokaryotes, (a) present a description of DNA (chromosome) replication as presently viewed in eukaryotes and (b) state the differences known to exist between prokaryotic and eukaryotic DNA replication.

Answer: (a) Eukaryotic DNA is replicated in a manner very similar to that in E. coli: bidirectional, continuous on one strand and discontinuous on the other, and similar requirements for synthesis (four deoxyribonucleoside triphosphates, divalent cation, template, and primer). (b) Okazaki fragments are about one-tenth the size of those in bacteria. Different portions of the chromosome (euchromatin, heterochromatin) replicate at different times. There are multiple replication origins in eukaryotic chromosomes.

Section: 10.6 Bloom's Taxonomy: Applying/Analyzing 43) What is the name of the replication unit in prokaryotes, and how does it differ in eukaryotes?

Answer: replicon; one replicon in prokaryotes, multiple replicons in eukaryotes Section: 10.6 Bloom's Taxonomy: Applying/Analyzing

13 Copyright © 2021 Pearson Education Ltd.

44) Describe the DNA base sequence arrangement at the end of the Tetrahymena chromosome and the resolution of DNA replication at the end of a linear DNA strand.

Answer: Telomeres terminate in a 5'-TTGGGG-3' sequence, and telomerase is capable of adding repeats to the ends, thus allowing the completion of replication without leaving a gap and shortening the chromosome following each replication.

Section: 10.7 Bloom's Taxonomy: Applying/Analyzing 45) Explain how prokaryotes deal with the end-replication problem.

Answer: Prokaryotes don't have ends or telomeres. They have circular chromosomes and therefore can replicate all the way through their chromosomes without encountering an end.

Telomeres are a consequence of linear chromosomes.

Section: 10.7 Bloom's Taxonomy: Applying/Analyzing

14 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 11 Chromosome Structure and DNA Sequence Organization 1) Viral chromosomes exist in a variety of conformations and can be made up of A) protein- or lipid-coding sequences B) DNA only C) DNA or RNA D) RNA only E) DNA, RNA, or protein Answer: C Section: 11.1 Bloom's Taxonomy: Remembering/Understanding 2) In E. coli, the genetic material is composed of A) circular, double-stranded DNA B) linear, double-stranded DNA C) RNA and protein D) circular, double-stranded RNA E) polypeptide chains Answer: A Section: 11.1 Bloom's Taxonomy: Remembering/Understanding

.

.

3) Eukaryotic chromosomes contain two general domains that relate to the degree of condensation. These two regions are .

A) called heterochromatin and euchromatin B) uniform in the genetic information they contain C) separated by large stretches of repetitive DNA D) each void of typical protein-coding sequences of DNA E) void of introns Answer: A Section: 11.4 Bloom's Taxonomy: Remembering/Understanding 4) Chromatin of eukaryotes is organized into repeating interactions with protein octamers called nucleosomes. Nucleosomes are composed of which class of molecules?

A) histones B) glycoproteins C) lipids D) H1 histones E) nonhistone chromosomal proteins Answer: A Section: 11.4 Bloom's Taxonomy: Remembering/Understanding

1 Copyright © 2021 Pearson Education Ltd.

5) Some organisms contain much larger amounts of DNA than are apparently "needed." This is more typical in .

A) viruses than in bacteria B) RNA viruses than in DNA viruses C) eukaryotes than in prokaryotes D) haploids rather than diploids E) prokaryotes than in eukaryotes Answer: C Section: 11.5 Bloom's Taxonomy: Remembering/Understanding 6) In human chromosomes, satellite DNA sequences of about 170 base pairs in length are present in tandem arrays of up to 1 million base pairs. Found mainly in centromere regions, these DNA sequences are called .

A) telomeres B) primers C) alphoid sequences D) euchromatic regions E) telomere-associated sequences Answer: C Section: 11.5 Bloom's Taxonomy: Remembering/Understanding 7) Which of the following middle repetitive sequences are not categorized as tandem repeats?

A) STRs B) rRNA genes C) Microsatellites D) SINEs E) VNTRs Answer: D Section: 11.5 Bloom's Taxonomy: Remembering/Understanding 8) A particular variant of the lambda bacteriophage has a DNA double-stranded genome of 51,365 base pairs. Estimate the length of the nucleic acid?

Answer: One base pair is 0.34 nm; therefore, 51,365 bp × 0.34 nm/bp = 17,464 nm, or 17.46 μm.

Section: 11.4 Bloom's Taxonomy: Evaluating/Creating

2 Copyright © 2021 Pearson Education Ltd.

9) Bacterial chromosomes generally consist of which of the following characteristics?

A) single-stranded, linear DNA B) single-stranded, circular RNA C) double-stranded, linear DNA D) double-stranded, circular RNA E) double-stranded, circular DNA Answer: E Section: 11.1 Bloom's Taxonomy: Remembering/Understanding 10) Estimate the number of base pairs in the Escherichia coli genome.

Answer: The Escherichia coli genome measures approximately 1200 μm in length. One base pair is 0.34 nm; therefore, 1200000 nm / 0.34 nm/bp = approximately 3.5 x 106 bp.

Section: 11.1, 11.4 Bloom's Taxonomy: Evaluating/Creating 11) The appearance of puffs within polytene chromosomes is thought to be a visible manifestation of .

A) DNA replication B) transcription of RNA C) translation of proteins D) posttranscriptional modifications E) posttranslational modifications Answer: B Section: 11.3 Bloom's Taxonomy: Remembering/Understanding 12) Although mutations have been observed in many different genes, they have not been isolated in histones. Why does this seem reasonable? If one wanted to produce antibodies to histones, would it be an easy task? Explain your answer.

Answer: Histones represent one of the most conserved molecules in nature because they are involved in a fundamental and important function relating to chromosome structure. Mutations are probably lethal. Because all antibody-producing organisms have essentially the same histones, it would be difficult to find an organism that produces histone antibodies, for to do so would be self-destructive.

Section: 11.4 Bloom's Taxonomy: Evaluating/Creating 13) What is unusual about the amino acid composition of histones? How is the function of histones related to the amino acid composition? Of which histones are nucleosomes composed?

Answer: Histones contain large amounts of positively charged amino acids such as lysine and arginine. Thus, they can bind electrostatically to the negatively charged phosphate groups of nucleotides. Nucleosomes are composed of all histones except H1.

Section: 11.4 Bloom's Taxonomy: Evaluating/Creating

3 Copyright © 2021 Pearson Education Ltd.

14) Which if the following histones is not part of the octamer found within a nucleosome?

A) H1 B) H2A C) H2B D) H3 E) H4 Answer: A Section: 11.4 Bloom's Taxonomy: Remembering/Understanding 15) Compare and contrast the chromosome structure of viruses, bacteria, and eukaryotes.

Answer: The amount of DNA per structure (virus particle, bacterium, cell) increases as one goes from viruses to eukaryotic cells. Viral chromosomes may be composed of single-stranded or double-stranded RNA or DNA, whereas bacterial and eukaryotic DNA is double-stranded.

Bacterial DNA is considered to be a covalently closed circle; the "global" structure of eukaryotic chromosomes is uncertain. Although some proteins are associated with viral and bacterial DNA, the regularly spaced histones of eukaryotic chromosomes are unique.

Section: 11.4 Bloom's Taxonomy: Applying/Analyzing 16) Describe how histone acetylation affects gene expression.

Answer: One of the most well-studied histone modifications involves acetylation by the action of the enzyme histone acetyltransferase (HAT). The addition of an acetyl group to the positively charged amino group present on the side chain of the amino acid lysine effectively changes the net charge of the protein by neutralizing the positive charge. It appears that high levels of acetylation open up, or remodel, the chromatin fiber, an effect that increases in regions of active genes and decreases in inactive regions (Chapter 11, Page 211).

Section: 11.4 Bloom's Taxonomy: Applying/Analyzing 17) Which of the following characteristics do bacterial chromosomes share with eukaryotic chromosomes? Both are composed of .

A) single-stranded DNA B) single-stranded RNA C) double-stranded DNA D) linear RNA E) circular DNA Answer: C Section: 11.4 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

18) Explain what are histones, how are they arranged in nucleosomes, and why nucleosomes are often repositioned.

Answer: Histones include five main classes of relatively small basic proteins containing relatively large amounts of lysine and arginine. Nucleosomes are made of two each of four types of histones. The chromatin fiber, when complexed with histones and folded into various levels of compaction, makes the DNA inaccessible to interaction with important nonhistone proteins. chromatin must be induced to change its structure, a process called chromatin remodeling. In the case of replication and gene expression, chromatin must relax its compact structure but be able to reverse the process during periods of inactivity (Chapter 11, Page 210).

Section: 11.4 Bloom's Taxonomy: Applying/Analyzing 19) Which of the following statements regarding mitochondria and chloroplasts is false?

A) They both typically contain circular, double-stranded DNA. B) They are inherited through maternal cytoplasm in most organisms.

C) Mitochondria typically contain many more introns than chloroplasts.

D) They contain genetic information remarkably similar to bacteria and viruses.

E) The size if cpDNA is typically much larger than mtDNA. Answer: C Section: 11.2 Bloom's Taxonomy: Remembering/Understanding 20) With which of the following structures does histone H1 directly associate?

A) nucleosome core particle B) H2A/H2B tetramer C) the histone octamer D) linker DNA E) H3/H4 tetramer Answer: D Section: 11.4 Bloom's Taxonomy: Remembering/Understanding 21) Explain how histone modifications can affect gene expression.

Answer: Gene expression can be affected by acetylation, methylation, and phosphorylation of histones. High levels of acetylation open up, or remodel, the chromatin fiber, an effect that increases in regions of active genes and decreases in inactive regions. Methyl groups may be added to both arginine and lysine residues of histones, and these changes can either increase or decrease transcription depending on which amino acids are methylated. Phosphate groups can be added to the hydroxyl groups of the amino acids serine and histidine, introducing a negative charge on the protein. (Chapter 11, Page 211).

Section: 11.4 Bloom's Taxonomy: Evaluating/Creating

5 Copyright © 2021 Pearson Education Ltd.

22) DNA in bacterial chromosomes is found to be associated with which of the following proteins?

A) H1 B) H2A C) HU D) H3 E) H4 Answer: C Section: 11.1 Bloom's Taxonomy: Remembering/Understanding 23) If the linker DNA between nucleosomes is 103 base pairs in length, how many H4 proteins are expected in a stretch of DNA 30,000 base pairs long?

Answer: The nucleosome core particle ( 147 bp) combined with a linker region (103 bp) would total 250 bp. The total DNA length (30,000 bp) divided by 250 bp equals 120. Each of the 120 regions would contain one nucleosome with two H4 proteins each for a total of 240 H4 proteins.

Section: 11.4 Bloom's Taxonomy: Applying/Analyzing 24) Describe the role of chemical modification in the generation of CpG islands. Predict where CpG islands are likely to be found within the genome.

Answer: Not to be confused with histone methylation, the nitrogenous base cytosine within the DNA itself can also be methylated, forming 5-methyl cytosine. Cytosine methylation is usually negatively correlated with gene activity and occurs most often when the nucleotide cytidylic acid is next to the nucleotide guanylic acid, forming what is called a CpG island (Chapter 11, Page 211). Since CpG islands play a role in regulating expression, they are likely to be located in relative proximity to genes and regulatory regions such as promoters.

Section: 11.4 Bloom's Taxonomy: Applying/Analyzing 25) Which of the following repetitive sequences was the original basis for the forensics technique referred to as DNA fingerprinting?

A) Alu sequences B) rRNA genes C) LINEs D) SINEs E) VNTRs Answer: E Section: 11.5 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

26) What percent of the human genome is composed of highly repetitive DNA? A) 1% B) 5% C) 10% D) 13% E) 21% Answer: B Section: 11.5 Bloom's Taxonomy: Remembering/Understanding 27) Which of the following statements regarding VNTRs is false?

A) VNTRs may be 15 to 100 base pairs long.

B) VNTRs may be found within and between genes.

C) VNTRs are referred to as retrotransposons.

D) VNTRs may be repeated to give regions 1000 to 20,000 bp in length.

E) VNTRs are dispersed throughout the genome.

Answer: C Section: 11.5 Bloom's Taxonomy: Remembering/Understanding 28) Which of the following characteristics is not associated with pseudogenes?

A) They are duplicated copies of genes.

B) They have undergone significant mutational alterations.

C) They share homology with parent genes.

D) They are usually activated to produce proteins.

E) They often contain insertions and deletions.

Answer: D Section: 11.6 Bloom's Taxonomy: Remembering/Understanding 29) Which of the following statements regarding the E. coli chromosome is true? The E. coli chromosome is composed of .

A) circular, double-stranded DNA B) circular, double-stranded RNA C) linear, double-stranded DNA D) circular, single-stranded DNA E) linear, single-stranded RNA Answer: A Section: 11.1 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

30) Which of the following characteristics is not associated with histones?

A) Histones octamers are found within nucleosomes.

B) Histones are negatively charged.

C) There are five types of histones.

D) Histones play a role in DNA folding.

E) Histones can be chemically modified.

Answer: B Section: 11.4 Bloom's Taxonomy: Remembering/Understanding 31) In situ hybridization is a molecular technique that may be utilized to A) determine rates of gene transcription B) identify proteins C) locate of nucleotide sequences on chromosomes D) identify DNA/protein interactions E) sequence DNA Answer: C Section: 11.5 Bloom's Taxonomy: Remembering/Understanding 32) Which of the following statements regarding telomeres is false?

A) They are composed of heterochromatin.

B) They stain more lightly during interphase.

C) They replicate late during S phase.

D) They are not related to the structural integrity of chromosomes.

E) They contain regions of DNA that are genetically inert.

Answer: D Section: 11.4, 11.5 Bloom's Taxonomy: Remembering/Understanding 33) Which of the following characteristics is true of heterochromatin?

A) include regions of DNA that remain largely uncoiled B) found in genetically active regions of the chromosome C) stain more deeply during interphase D) replicates early during S phase E) not located within centromeres or telomeres Answer: C Section: 11.4 Bloom's Taxonomy: Remembering/Understanding

8 Copyright © 2021 Pearson Education Ltd.

.

34) What percentage of the human genome is estimated to consist of protein-coding genes?

A) 2% B) 5% C) 10% D) 13% E) 21% Answer: A Section: 11.6 Bloom's Taxonomy: Remembering/Understanding

9 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 12 The Genetic Code and Transcription 1) Which of the following is not a characteristic of the genetic code?

A) The genetic code is universal.

B) The genetic code is overlapping.

C) The genetic code is degenerate.

D) The genetic code is commaless.

E) The genetic code is unambiguous.

Answer: B Section: 12.1 Bloom's Taxonomy: Remembering/Understanding 2) The genetic code is said to be triplet, meaning that there .

A) are three amino acids per base in mRNA B) are three bases in mRNA that code for an amino acid C) may be three ways in which an amino acid may terminate a chain D) are three "nonsense" triplets E) none of the answers listed is correct Answer: B Section: 12.2 Bloom's Taxonomy: Remembering/Understanding 3) A class of mutations that results in multiple contiguous (side-by-side) amino acid changes in proteins is probably caused by which one of the following types of mutations?

A) frameshift B) transversion C) transition D) base analog E) recombinant Answer: A Section: 12.2 Bloom's Taxonomy: Applying/Analyzing 4) When scientists were attempting to determine the structure of the genetic code, Crick and coworkers found that when three base additions or three base deletions occurred in a single gene, the wild-type phenotype was sometimes restored. These data supported the hypothesis that .

A) there are three amino acids per base B) the code contains internal punctuation C) AUG is the initiating triplet D) the code is overlapping E) the code is triplet Answer: E Section: 12.2 Bloom's Taxonomy: Remembering/Understanding 1 Copyright © 2021 Pearson Education Ltd.

5) How many five letter codons could be made from a four nucleotide code?

A) 16 B) 64 C) 256 D) 54 E) 1024 Answer: E Section: 12.2 Bloom's Taxonomy: Evaluating/Creating 6) Significant in the deciphering of the genetic code was the discovery of the enzyme polynucleotide phosphorylase. What is this enzyme used for?

A) manufacture of synthetic RNA for cell-free systems B) ribosomal translocation C) peptide bond formation D) production of ribosomal proteins E) degradation of RNA Answer: A Section: 12.3 Bloom's Taxonomy: Remembering/Understanding 7) In 1964, Nirenberg and Leder used the triplet binding assay to determine specific codon assignments. A complex of which of the following components was trapped in the nitrocellulose filter?

A) ribosomes and DNA B) free tRNAs C) charged tRNA, RNA triplet, and ribosome D) uncharged tRNAs and ribosomes E) sense and antisense strands of DNA Answer: C Section: 12.3 Bloom's Taxonomy: Remembering/Understanding 8) What is the name given to the three bases in a messenger RNA that bind to the anticodon of tRNA to specify an amino acid placement in a protein?

A) protein B) anti-anticodon C) cistron D) rho E) codon Answer: E Section: 12.3 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

9) In a mixed heteropolymer experiment A and C are added in a ratio of 1A:5C. What would be the frequency of getting an ACA triplet?

A) 2.3% B) 6.9% C) 11.6% D) 57.9% E) 34.8% Answer: A Section: 12.3 Bloom's Taxonomy: Evaluating/Creating 10) What would be the probability of occurrence of the triplet GCG in a heteropolymer experiment with a ratio of 2G:1C? A) 14.8% B) 67.2% C) 50.1% D) 25.6% E) 10.5% Answer: A Section: 12.3 Bloom's Taxonomy: Evaluating/Creating 11) Which of the following researchers deciphered the first specific coding sequences?

A) Watson and Crick B) Hershey and Chase C) Nirenberg and Matthaei D) Khorana E) Weiss Answer: C Section: 12.3 Bloom's Taxonomy: Remembering/Understanding 12) What is the initiator triplet in both prokaryotes and eukaryotes? Which amino acid does this triplet recruit?

A) UAA; no amino acid called in B) UAA or UGA; arginine C) AUG; arginine D) AUG; methionine E) UAA; methionine Answer: D Section: 12.4 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

13) When examining the genetic code, it is apparent that .

A) there can be more than one amino acid for a particular codon B) AUG is a terminating codon C) there can be more than one codon for a particular amino acid D) the code is ambiguous in that the same codon can code for two or more amino acids E) there are 44 stop codons because there are only 20 amino acids Answer: C Section: 12.4 Bloom's Taxonomy: Remembering/Understanding 14) Which of the following is the stop codon?

A) AUG B) UAG C) GUU D) AUC E) AUA Answer: B Section: 12.4 Bloom's Taxonomy: Remembering/Understanding 15) Which of the following explains the pattern of degeneracy at the third position of the codon?

A) The Wobble hypothesis.

B) The redundancy of the code.

C) The universal nature of the code.

D) RNA editing.

E) RNA splicing.

Answer: A Section: 12.4 Bloom's Taxonomy: Remembering/Understanding 16) The genetic code is fairly consistent among all organisms. The term often used to describe such consistency in the code is .

A) universal B) exceptional C) trans-specific D) overlapping E) none of the answers listed is correct Answer: A Section: 12.6 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

17) In paramecium the codon UAG encodes glutamine. What does this encode in humans?

A) glutamine B) glycine C) glutamate D) start E) stop Answer: E Section: 12.6 Bloom's Taxonomy: Remembering/Understanding 18) The relationship between a gene and a messenger RNA is that A) genes are made from mRNAs B) mRNAs are made from genes C) mRNAs make proteins, which then code for genes D) genes are structurally identical to mRNAs E) mRNA is directly responsible for making Okazaki fragments Answer: B Section: 12.8 Bloom's Taxonomy: Remembering/Understanding

.

19) When considering the initiation of transcription, one often finds consensus sequences located in the region of the DNA where RNA polymerase(s) binds. Which of the following is a common consensus sequence?

A) TATAAT B) GGTTC C) TTTTAAAA D) any trinucleotide repeat E) satellite DNAs Answer: A Section: 12.9, 12.10 Bloom's Taxonomy: Remembering/Understanding 20) Which of the following subunits confers promoter specificity in prokaryotes?

A) α B) β C) ω D) β' E) σ Answer: E Section: 12.9 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

21) Which of the following includes the three posttranscriptional modifications often seen in the maturation of mRNA in eukaryotes?

A) 5'-capping, 3'-poly(A) tail addition, splicing B) 3'-capping, 5'-poly(A) tail addition, splicing C) removal of exons, insertion of introns, capping D) 5'-poly(A) tail addition, insertion of introns, capping E) heteroduplex formation, base modification, capping Answer: A Section: 12.10, 12.11 Bloom's Taxonomy: Remembering/Understanding 22) In which cellular organelle do the three posttranscriptional modifications often seen in the maturation of mRNA in eukaryotes occur?

A) nucleus B) cytoplasm C) mitochondrion D) lysosome E) Golgi Answer: A Section: 12.10 Bloom's Taxonomy: Applying/Analyzing 23) Which of the following has the capacity to inhibit transcription?

A) silencers B) enhancer mutations C) chromatin structure D) methylated cytosines E) all of these can inhibit transcription Answer: E Section: 12.10 Bloom's Taxonomy: Evaluating/Creating 24) Which of the following protects the 3' end of the mRNA in eukaryotes from nuclease degradation?

A) 7-methylguanosine B) N-formylmethionine C) poly-A polymerase D) poly-A binding protein E) 3' branch site Answer: D Section: 12.10 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

25) Imagine you wanted to sequence the introns of a gene that you were interested in. Which of the following molecules would you need to isolate?

A) tRNA B) mRNA C) mtDNA D) snRNA E) genomic DNA Answer: E Section: 12.11 Bloom's Taxonomy: Applying/Analyzing 26) It has been recently determined that the gene for Duchenne muscular dystrophy (DMD) is more than 2000 kb (kilobases) in length; however, the mRNA produced by this gene is only about 14 kb long. What is a likely cause of this discrepancy?

A) The exons have been spliced out during mRNA processing.

B) The DNA represents a double-stranded structure, whereas the RNA is single-stranded.

C) More amino acids are coded for by the DNA than by the mRNA. D) The introns have been spliced out during mRNA processing.

E) When the mRNA is produced, it is highly folded and therefore less long.

Answer: D Section: 12.11 Bloom's Taxonomy: Applying/Analyzing 27) Introns are known to contain termination codons (UAA, UGA, or UAG), yet these codons do not interrupt the coding of a particular protein. Why? A) UAA, UGA, and UAG are initiator codons, not termination codons.

B) Exons are spliced out of mRNA before translation.

C) These triplets cause frameshift mutations, but not termination.

D) More than one termination codon is needed to stop translation.

E) Introns are removed from mRNA before translation.

Answer: E Section: 12.11 Bloom's Taxonomy: Applying/Analyzing 28) An intron is a section of .

A) protein that is clipped out posttranslationally B) RNA that is removed during RNA processing C) DNA that is removed during DNA processing D) transfer RNA that binds to the anticodon E) carbohydrate that serves as a signal for RNA transport Answer: B Section: 12.11 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

29) If one compares the base sequences of related genes from different species, one is likely to find that corresponding are usually conserved, but the sequences of are much less well conserved.

A) exons; introns B) introns; exons C) introns; chaperons D) chaperons; exons E) introns; proteins Answer: A Section: 12.11 Bloom's Taxonomy: Evaluating/Creating 30) What was the name of the molecule that Jacob and Monod (1961) postulated carried genetic information from DNA to the site of protein manufacture?

Answer: messenger RNA (mRNA) Section: 12.2 Bloom's Taxonomy: Remembering/Understanding 31) From the late 1950s to the mid-1960s, numerous experiments using in vitro cell-free systems provided information on the nature of the genetic code. Briefly outline significant experiments in the determination of the genetic code.

Answer: Use of polynucleotide phosphorylase for the random assembly of nucleotides provided for the assembly of RNA homopolymers and random heteropolymers, which when placed in the cell-free protein-synthesizing system provided products (polypeptide chains) for analysis. The triplet binding assay along with the use of repeating copolymers was used to verify information provided earlier and to establish the ordered codon assignments.

Section: 12.3 Bloom's Taxonomy: Evaluating/Creating 32) Suppose that in the use of polynucleotide phosphorylase, nucleotides A and C are added in a ratio of 1A:5C. What is the probability that an AAA sequence will occur?

Answer: 1/6 × 1/6 × 1/6 Section: 12.3 Bloom's Taxonomy: Applying/Analyzing 33) What two experimental procedures allowed deciphering of the ordered triplet assignments of the genetic code?

Answer: the triplet binding assay and the use of repeating RNA polymers with known sequence Section: 12.3 Bloom's Taxonomy: Remembering/Understanding 34) A base at the first position of an anticodon on the tRNA would pair with a base at the position of the mRNA. Answer: third Section: 12.4 Bloom's Taxonomy: Applying/Analyzing 8 Copyright © 2021 Pearson Education Ltd.

35) The relationship between codon and anticodon can be characterized as involving .

Answer: hydrogen bonds between complementary bases (usually) in typical antiparallel fashion Section: 12.4 Bloom's Taxonomy: Applying/Analyzing 36) Describe the function of N-formylmethionine in prokaryotes.

Answer: N-formylmethionine is a modified amino acid, which serves as the starting amino acid in protein synthesis.

Section: 12.4 Bloom's Taxonomy: Remembering/Understanding 37) What is meant by punctuation in terms of the genetic code?

Answer: Certain triplets (AUG) commonly signal the starting point for protein synthesis; other triplets (UAA, UGA, UAG) typically signal stop.

Section: 12.4 Bloom's Taxonomy: Remembering/Understanding 38) Describe an observation that supports the view that the genetic code is ordered.

Answer: Certain amino acids may be grouped according to the middle base; for example, U or C in the second position often specifies hydrophobic amino acids. Also, codons with the same two starting letters frequently encode the same amino acid.

Section: 12.4 Bloom's Taxonomy: Remembering/Understanding 39) List four base triplets that are clearly responsible for punctuation (initiation, termination).

Answer: AUG (rarely GUG) for initiation; UAA, UGA, and UAG for termination Section: 12.4 Bloom's Taxonomy: Remembering/Understanding 40) Referring to the genetic code, what is meant by "wobble"?

Answer: relaxed pairing specificities in the third-base position of a codon Section: 12.4 Bloom's Taxonomy: Remembering/Understanding 41) The finding that virtually all organisms use the same genetic code provides the basis for declaring that the code is universal. Name at least two exceptions to such universality.

Answer: mitochondrial DNA, Mycoplasma capricolum, some protozoans Section: 12.6 Bloom's Taxonomy: Remembering/Understanding 42) In the context of molecular genetics, how does one reconcile the terms overlapping genes and nonoverlapping code?

Answer: The genetic code contains codons that are nonoverlapping; however, overlapping genes are observed in some viruses in which, due to differential use of AUG to initiate translation, the same mRNA can yield different protein products.

Section: 12.7 Bloom's Taxonomy: Remembering/Understanding 9 Copyright © 2021 Pearson Education Ltd.

43) What is polycistronic mRNA? Answer: Polycistronic mRNA is seen primarily among prokaryotes in which one mRNA carries coding information and internal punctuation for the translation of more than one protein.

Section: 12.9 Bloom's Taxonomy: Remembering/Understanding 44) Describe how the sigma subunit (factor) of E. coli RNA polymerase participates in transcription.

Answer: The sigma subunit may give specificity to the RNA polymerase and play a regulatory function. It may be involved in the recognition of initiation sites or promoters.

Section: 12.9 Bloom's Taxonomy: Remembering/Understanding 45) Imagine that you are studying a gene in C. elegans and find that a single mRNA that you have isolated generates more than one protein. Explain what mechanism(s) could be causing this.

Answer: This could be an example of a polycistronic mRNA since C. elegans have some of this in their genome or it could be an example of different isoforms produced by alternative splicing.

Section: 12.9 Bloom's Taxonomy: Evaluating/Creating 46) Regarding the efficient initiation of transcription by RNA polymerase II, which specific "upstream" signals appear to be involved?

Answer: TATA and CAAT base sequences and enhancers Section: 12.10 Bloom's Taxonomy: Applying/Analyzing 47) Describe a difference between the RNA polymerases of eukaryotes and prokaryotes.

Answer: In eukaryotes, three polymerases (I, II, III) have been identified; only one has been described in prokaryotes.

Section: 12.10 Bloom's Taxonomy: Applying/Analyzing 48) In eukaryotes, which three factors appear to encourage the specific association of RNA polymerase(s) to a specific region of DNA? Answer: promoters, enhancers, and transcription factors Section: 12.10 Bloom's Taxonomy: Remembering/Understanding 49) What would be the expected outcome if an organism had a mutation in TFIID? Answer: TFIID binds to the TATA box to recruit the polymerase to the promoter. TFIID deficiency would cause a failure of the pre-initiation complex formation.

Section: 12.10 Bloom's Taxonomy: Evaluating/Creating

10 Copyright © 2021 Pearson Education Ltd.

50) Originally, some of the theories of molecular evolution favor DNA as the first molecule to emerge since DNA is required to make proteins. Others theorized that proteins were the first to emerge since proteins are required to make DNA and possess catalytic activity that DNA lacks.

How did the discovery of ribozymes add to this discussion?

Answer: Ribozymes are catalytic RNA. This allows catalysis to occur in the absence of protein which provides a third possibility where RNA was the first molecule to emerge. This led to the "RNA World" idea.

Section: 12.11 Bloom's Taxonomy: Evaluating/Creating 51) Imagine that you discovered a mutation in a gene where the 5' donor site of the second intron was mutated from a GU to a GG. What would you anticipate would be the effect of this mutation?

Answer: The U1snRNP would not recognize the 5' site and would either result in the inclusion of the second intron in the mature transcript or there would be use of an alternative or cryptic 5' site upstream or downstream leading to potential deletions of coding sequence.

Section: 12.11 Bloom's Taxonomy: Evaluating/Creating

11 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 13 Translation and Proteins 1) A protein is 300 amino acids long. Which of the following could be the number of nucleotides in the section of DNA that codes for this protein? (Remember: DNA is double-stranded.)

A) 3 B) 100 C) 300 D) 500 E) 1800 Answer: E Section: 13.1 Bloom's Taxonomy: Applying/Analyzing 2) A short segment of an mRNA molecule is shown below. The polypeptide it codes for is also shown: 5'-AUGGUGCUGAAG : methionine-valine-leucine-lysine Assume that a mutation in the DNA occurs so that the fourth base (counting from the 5' end) of the messenger RNA now reads A rather than G. What sequence of amino acids will the mRNA now code for? (You do not need a copy of the genetic code to answer the question.)

A) methionine-valine-leucine-lysine B) methionine-lysine-leucine-lysine C) methionine-leucine-leucine-lysine D) methionine-valine-methionine-lysine E) methionine-methionine-leucine-lysine Answer: E Section: 13.1 Bloom's Taxonomy: Applying/Analyzing 3) Which of the following are among the major components of prokaryotic ribosomes?

A) 12S rRNA, 5.8S rRNA, and proteins B) 16S rRNA, 5.8S rRNA, and 28S rRNA C) 16S rRNA, 5S rRNA, and 23S rRNA D) lipids and carbohydrates E) 18S rRNA, 5.8S rRNA, and proteins Answer: C Section: 13.1 Bloom's Taxonomy: Remembering/Understanding

1 Copyright © 2021 Pearson Education Ltd.

4) What is the function of aminoacyl tRNA synthetase?

A) It binds to the mRNA cap during initiation.

B) It adds an anticodon to tRNA. C) It folds tRNA into a clover leaf structure.

D) It binds an amino acid to its tRNA. E) It guides tRNA into the A site of a ribosome.

Answer: D Section: 13.1 Bloom's Taxonomy: Remembering/Understanding 5) Which of the following is not a part of the tRNA structure?

A) variable loop B) acceptor stem C) D stem D) TψC stem E) Kozak stem Answer: E Section: 13.1 Bloom's Taxonomy: Remembering/Understanding 6) Which one of the following is a sequence in bacteria that binds to the 16S rRNA to initiate translation?

A) Shine-Delgarno sequence B) Kozak sequence C) initiation codon D) initiation factor binding site E) TATA bix Answer: A Section: 13.2 Bloom's Taxonomy: Remembering/Understanding 7) Which of the following initiation factors acts as a GTPase?

A) IF1 B) IF2 C) IF3 D) EF-Tu E) Both IF2 and EF-Tu Answer: E Section: 13.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

8) In bacteria the catalytic activity of peptide transfer is a function of A) 16S rRNA B) 50S subunit C) snRNA D) 23S rRNA E) 40S subunit Answer: D Section: 13.2 Bloom's Taxonomy: Remembering/Understanding

.

9) In e. coli which one of the following binds GTP and brings aminoacyl tRNA to the A site of a ribosome?

A) EF-Tu B) EF-Ts C) RF-1 D) IP-2 E) EF-G Answer: A Section: 13.2 Bloom's Taxonomy: Remembering/Understanding 10) During translation, which triplets signal chain termination?

A) UAA B) AUG C) AUA D) UGA E) UAA and UGA Answer: E Section: 13.2 Bloom's Taxonomy: Remembering/Understanding 11) It is now known that the 23S rRNA performs the function that was originally attributed to the enzyme peptidyl transferase. What is that function?

A) base additions during mRNA synthesis B) peptide bond formation during protein synthesis C) elongation factors binding to the large ribosomal subunit D) discontinuous strand replication E) 5' capping of mRNA Answer: B Section: 13.2 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

12) Which one of the following is found in eukaryotic mRNAs upstream of the AUG and increases initiation efficiency?

A) Shine-Delgarno sequence B) Kozak sequence C) initiation sequence D) poly A tail E) TATA box Answer: B Section: 13.4 Bloom's Taxonomy: Remembering/Understanding 13) What part of the mRNA is involved in initiation of translation in eukaryotes?

A) 3' poly A tail B) m7cap C) introns D) both 3' poly A tail and m7cap E) Shine-Delgarno sequence Answer: D Section: 13.4 Bloom's Taxonomy: Remembering/Understanding 14) Early in the 1900s, Sir Archibald Garrod studied a number of metabolic defects in humans.

One particular disorder involved the inability to metabolize homogentisic acid. What is the name of this disorder?

A) phenylketonuria B) alkaptonuria C) yyorsinemia D) albinism E) homogentinuria Answer: B Section: 13.5 Bloom's Taxonomy: Remembering/Understanding 15) The one-gene:one-enzyme hypothesis emerged from work on which two organisms?

A) E. coli and yeast B) Drosophila and humans C) Neurospora and Drosophila D) E. coli and humans E) All of the answers listed are correct Answer: C Section: 13.6 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

16) By their experimentation using the Neurospora fungus, Beadle and Tatum were able to propose the far-reaching hypothesis that .

A) prototrophs will grow only if provided with nutritional supplements B) several different enzymes may be involved in the same step in a biochemical pathway C) the role of a specific gene is to produce a specific enzyme D) genetic recombination occurred in Neurospora E) more than one codon can specify a given amino acid Answer: C Section: 13.6 Bloom's Taxonomy: Remembering/Understanding 17) The β chain of adult hemoglobin is composed of 146 amino acids of a known sequence. In comparing the normal β chain with the β chain in sickle-cell hemoglobin, what alteration is one likely to find?

A) valine instead of glutamic acid in the sixth position B) glutamic acid replacing valine in the first position C) extensive amino acid substitutions D) trinucleotide repeats E) frameshift substitutions Answer: A Section: 13.7 Bloom's Taxonomy: Remembering/Understanding 18) Studies of Neurospora led to the statement, whereas studies of human hemoglobin led to the statement.

A) one-gene:one enzyme;one-gene polypeptide B) one-gene:one-polypeptide;one-gene:one enzyme C) one-gene:one protein; one-gene:one enzyme D) one-gene:one enzyme; one-gene:one protein E) one-gene:one polycistron; one-gene: onepolypeptide Answer: A Section: 13.7 Bloom's Taxonomy: Remembering/Understanding 19) A procedure that is often used to separate molecules by using their molecular charges is called .

A) heteroduplex analysis B) southern blot C) affinity chromatography D) electrophoresis E) density sedimentation Answer: D Section: 13.7 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

20) Electrophoretic separation of HbA from HbS is based on a difference in their A) number of base pairs B) size of proteins C) pH D) amino acid sequence E) charges Answer: E Section: 13.7 Bloom's Taxonomy: Remembering/Understanding 21) What is the expected proportion of offspring with sickle-cell anemia if both partners are heterozygous for the sickle-cell anemia gene?

A) 25% B) 50% C) 100% D) 75% E) 10% Answer: A Section: 13.7 Bloom's Taxonomy: Applying/Analyzing 22) Which of the following researchers established the α helix model of protein secondary structure?

A) Pauling, Branson, Corey B) Beadle and Tatum C) Watson and Crick D) Garrod and Bateson E) Ramakrishnan and Noller Answer: A Section: 13.8 Bloom's Taxonomy: Remembering/Understanding 23) Assuming that an amino acid sequence is 250 amino acids long, how many different molecules, each with a unique sequence, could be formed?

A) 20250 B) 225 C) 2025 D) 2250 E) 2502 Answer: A Section: 13.8 Bloom's Taxonomy: Applying/Analyzing

6 Copyright © 2021 Pearson Education Ltd.

.

24) The primary structure of a protein is determined by .

A) the sequence of amino acids B) hydrogen bonds formed between the components of the peptide linkage C) a series of helical domains D) pleated sheets E) covalent bonds formed between fibroin residues Answer: A Section: 13.8 Bloom's Taxonomy: Remembering/Understanding 25) The secondary structure of a protein includes A) gamma and delta B) alpha and gamma C) α-helix and β-pleated sheet D) hydrophobic clusters E) disulfide bridges Answer: C Section: 13.8 Bloom's Taxonomy: Remembering/Understanding

.

26) Side groups of amino acids are typically classified under which of the following?

A) polar, nonpolar B) linear, circular C) alpha, omega D) long, short E) primary, secondary Answer: A Section: 13.8 Bloom's Taxonomy: Remembering/Understanding 27) Which of the following mediates the protein folding process by excluding the formations of alternative incorrect folding patterns?

A) ubiquitins B) chaperones C) proteosome D) initiation factors E) prions Answer: B Section: 13.8 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

28) Which of the following is not a disorder caused by the aggregation of misfolded protein?

A) Huntinton's Disease B) Alzheimer's Disease C) Parkinson's Disease D) sickle-cell anemia E) alkaptonuria Answer: E Section: 13.8 Bloom's Taxonomy: Applying/Analyzing 29) Which of the the following functions as a "tag" that marks misfolded proteins to be moved to the proteosome?

A) ubiquitins B) chaperones C) prions D) initiation factors E) release factors Answer: A Section: 13.8 Bloom's Taxonomy: Remembering/Understanding 30) Three major types of RNAs are mRNA, rRNA, and tRNA. For each of the conditions below, predict the consequences in terms of the population of proteins being synthesized in a particular cell. What qualitative and quantitative changes, if any, are expected in the individual protein involved (if one is involved) and in the population of proteins in this cell? (a) a frameshift mutation in the heterozygous state (b) a deletion of a significant proportion(but not all) of ribosomal genes (rDNA) Answer: (a) Population of proteins: Half of the protein products of that gene will be defective, and the other half will be normal. Individual protein: The protein should show multiple amino acid substitutions "downstream" from the point of the mutation. If a nonsense triplet is introduced, the protein will be shortened in the substituted region. (b) Population of proteins: There would be an overall reduction in protein synthesis. Individual proteins: All of the proteins would be made in their normal form but at reduced levels.

Section: 13.2 Bloom's Taxonomy: Evaluating/Creating 31) What are polyribosomes?

Answer: clusters of ribosomes held together by an mRNA Section: 13.2 Bloom's Taxonomy: Remembering/Understanding

8 Copyright © 2021 Pearson Education Ltd.

32) The following table presents the effect of different media on the growth response of tryptophan mutations in Salmonella typhimurium (+ = growth, - = no growth).

Strain trp-8 trp-2 trp-3 trp-1

No Supplement -

Medium Supplemented with IGP AA IN TRY + + + + + - + + - + + - +

(a) Construct the biochemical pathway for the compounds IGP, AA, IN, and TRY based on these data. (b) Place strains of bacteria (mutations) in the appropriate steps in the pathway.

Answer: (a) → + → >AA → + → >IGP → + → >IN → + → Try (b) → + → >AA → + → >IGP → + → >IN → + → Try ↑ ↑ ↑ ↑ 8 2 3 1 Section: 13.6 Bloom's Taxonomy: Evaluating/Creating 33) Describe the conceptual basis for constructing biochemical pathways using nutrient supplement experiments with Neurospora.

Answer: The substance that when added to minimal medium "cures" the largest number of strains must be toward the end of the pathway. A supplement that fails to "cure" many strains must be early in the pathway.

Section: 13.6 Bloom's Taxonomy: Evaluating/Creating 34) Below is a set of experimental results relating the growth (+) of Neurospora on several media (MM = minimal medium). Based on the information provided, present the biochemical pathway and the locations of the metabolic blocks.

Strain Medium MM MM+A s111 + + t60 + + + s211 + Answer: →+→A→+→B

MM+B

s111 s211 Section: 13.6 Bloom's Taxonomy: Evaluating/Creating

9 Copyright © 2021 Pearson Education Ltd.

35) The problem below relates to the synthesis of several intermediates in the citric acid cycle, which is essential in the production of ATP through aerobic respiration. A set of experimental results relating the growth (+) of Neurospora on several media is given in the table. Based on the information provided, present the biochemical pathway for the substances oxaloacetate, fumarate, malate, and succinate and the locations of the metabolic blocks produced by the various strains.

Strain

Medium MM MM MM + oxaloacetate

MM l62 l36 l41

-

+ + +

+

+ +

MM + fumarate

+ malate

+ succinate

-

MM = minimal medium Answer: succinate → + → fumarate→ + → malate → + → oxaloacetate ↑ ↑ ↑ 141 162 136 Section: 13.6 Bloom's Taxonomy: Evaluating/Creating 36) Much has been learned about the relationship between genes and gene products through the use of the mold Neurospora. What specific attributes make Neurospora a good organism for such studies?

Answer: knowledge of its biochemistry, its haploid ascospores, relative ease of isolating nutritional mutations Section: 13.6 Bloom's Taxonomy: Remembering/Understanding 37) Nutritional mutants in Neurospora can be "cured" by treating the medium with substances in the defective metabolic pathway. What determines whether the mutant strain (auxotroph) is "cured" by a particular substance?

Answer: The substance needs to be added after the metabolic block in the biochemical pathway.

Section: 13.6 Bloom's Taxonomy: Applying/Analyzing 38) Imagine you performed an irradiation experiment on Neurospora conidia and find that normal growth is only restored when tyrosine is added to minimal media and not when leucine, alanine or phenylalanine are added. Explain why tyrosine restores growth and the others do not.

Answer: Tyrosine addition rescues the tyrosine deficiency and restores growth. In the others the added amino acids were deficient and tyrosine remained deficient; therefore there was no growth.

Section: 13.6 Bloom's Taxonomy: Applying/Analyzing 10 Copyright © 2021 Pearson Education Ltd.

39) Imagine you irradiate some Neurospora conidia and analyze them for nutritional deficiencies. You find that normal growth is only restored when adding either tyrosine to minimal media or phenylalanine to minimal media. Explain what these results tell you about the biochemical pathway.

Answer: Phenylalanine→Tyrosine. If tyrosine deficiency produces no growth then adding tyrosine will restore growth. If phenylalanine is a precursor to make tyrosine then the absence of phenylalanine would also produce no growth. Adding phenylalanine would restore the initial substrate and allow the enzyme (if functional) to make tyrosine thereby restoring growth.

Section: 13.6 Bloom's Taxonomy: Evaluating/Creating 40) Describe the basic structure of normal adult hemoglobin and the abnormality observed in sickle-cell hemoglobin.

Answer: The predominant form of adult hemoglobin is composed of two α and two β chains. In sickle-cell hemoglobin, the sixth amino acid in the β chain is valine instead of glutamic acid.

Section: 13.7 Bloom's Taxonomy: Remembering/Understanding 41) In what ways do the amino acid side chains interact to influence protein function?

Answer: Higher-level folding of proteins is dependent on a variety of interactions (ionic, covalent, hydrogen, hydrophobic, hydrophilic, etc.), which determine the functional threedimensional structure of proteins.

Section: 13.8 Bloom's Taxonomy: Applying/Analyzing 42) Below are several phenomena relating to protein structure. Clearly describe each phenomenon, the conditions under which each occurs, and the probable influence each has on protein structure. (a) hydrophobic interactions (b) hydrogen bonds (c) disulfide bridges Answer: (a) Hydrophobic interactions: These are nonpolar side chains of amino acids that tend to associate to form hydrophobic clusters usually away from the protein surface. (b) Hydrogen bonds: Such bonds may occur between the components of the peptide bond, the side chains, or a combination of the two. They are responsible for helical and pleated sheet structures of proteins. (c) Disulfide bridges: Such bonds are formed between two cysteine side chains and, because of their covalent nature, represent relatively strong attractive forces between different (sometimes distant) regions of proteins.

Section: 13.8 Bloom's Taxonomy: Evaluating/Creating

11 Copyright © 2021 Pearson Education Ltd.

43) Regarding the protein structure, how are β-pleated sheets arranged and stabilized?

Answer: Several chains run in parallel or antiparallel fashion stabilized by hydrogen bonds formed between components of the peptide linkage.

Section: 13.8 Bloom's Taxonomy: Remembering/Understanding 44) Considering the types of side chains on amino acids and their relationship to protein structure, where are the amino acids with hydrophobic side chains most likely to be located?

Answer: away from the water environment and in the interior portion of the molecule Section: 13.8 Bloom's Taxonomy: Evaluating/Creating 45) Knowing that the base sequence of any given messenger RNA is responsible for precisely ordering the amino acids in a respective protein, present two mechanisms by which intrinsic properties of mRNA may regulate the "net output" of a given gene.

Answer: (a) Final output of a given gene may be influenced by the stability of an mRNA, and stability of an mRNA is determined in part by its base content and sequence. (b) Differential splicing of mRNA (actually mRNA precursors) can influence how much of a given product will be made from a gene.

Section: 13.9 Bloom's Taxonomy: Evaluating/Creating

12 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 14 Gene Mutation, DNA Repair, and Transposition 1) Nutritional mutations can be described as .

A) those mutations that do not allow a bacterium or fungus to synthesize an essential amino acid B) those mutations that change the composition of the medium C) those mutations belonging to the group called prototrophs D) those mutations caused by site-specific mutagenesis E) all strains that are not auxotrophic Answer: A Section: 14.1 Bloom's Taxonomy: Remembering/Understanding 2) A class of mutations that results in multiple contiguous amino acid changes in proteins is likely to be which of the following?

A) base analog B) transversion C) transition D) frameshift E) recombinant Answer: D Section: 14.1 Bloom's Taxonomy: Remembering/Understanding 3) Two formal terms used to describe categories of mutational nucleotide substitutions in DNA are .

A) base analogs and frameshift B) error prone and spontaneous C) transversions and transitions D) euchromatic and heterochromatic E) sense and antisense Answer: C Section: 14.1 Bloom's Taxonomy: Remembering/Understanding 4) Phenylalanine hydroxylase is an enzyme that catalyzes the reaction whereby the amino acid phenylalanine is converted to the amino acid tyrosine. A mutation in this enzyme would fall under which of the following classifications?

A) morpholgical mutation B) nutritional or biochemical mutation C) behavioral mutation D) regulatory mutation E) lethal mutation Answer: B Section: 14.1 Bloom's Taxonomy: Applying/Analyzing 1 Copyright © 2021 Pearson Education Ltd.

5) Which of the following is a type of mutation where one wild-type allele in a heterozygote is not able to produce the wild-type phenotype?

A) haploinsufficiency B) loss-of-function C) suppressor mutation D) null mutation E) transition Answer: A Section: 14.1 Bloom's Taxonomy: Remembering/Understanding 6) Which of the following is a type of mutation that relieves the effects of a previous mutation?

A) haploinsufficiency B) loss-of-function C) suppressor mutation D) null mutation E) transition Answer: C Section: 14.1 Bloom's Taxonomy: Remembering/Understanding 7) A mutation that relieves the effects of a previous mutation is called a(n) A) spontaneous mutation B) induced mutation C) dominant negative mutation D) loss of function mutation E) suppressor mutation Answer: E Section: 14.2 Bloom's Taxonomy: Remembering/Understanding 8) Mutations that arise in nature, from no particular artificial agent, are called A) oblique mutations B) induced mutations C) spontaneous mutations D) chromosomal aberrations E) cosmic mutations Answer: C Section: 14.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

.

.

9) Conditional mutations are more likely to result from a mutation caused by which of the following alterations to the coding region of a gene?

A) four bases added within a short region of a gene B) base addition C) X rays D) deletion E) tautomeric shift Answer: E Section: 14.3 Bloom's Taxonomy: Applying/Analyzing 10) Apurinic sites (AP sites) involve a spontaneous loss of a DNA molecule.

A) purine B) pyrimidine C) nucleotide D) cytosine E) thymine Answer: A Section: 14.3 Bloom's Taxonomy: Remembering/Understanding

in an intact double-helix

11) Recent discoveries on causes of fragile-X syndrome, and Huntington Disease indicate which type of genetic alteration?

A) tautomeric shifts B) thymine dimers C) depurination D) deamination E) changes in trinucleotide repeats Answer: E Section: 14.3 Bloom's Taxonomy: Remembering/Understanding 12) The process of error correction of mismatched bases carried out by DNA polymerases is called .

A) postreplication repair B) base excision repair C) nucleotide excision repair D) proofreading E) mismatch repair Answer: D Section: 14.3 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

13) Which of the following refers to the loss of guanine in an intact double-helical DNA molecule?

A) tautomer B) deamination C) depurination D) replication slippage E) oxidative damage Answer: C Section: 14.3 Bloom's Taxonomy: Remembering/Understanding 14) When an amino group in cytosine or adenine is converted to a keto group it is called .

A) a tautomer B) deamination C) depurination D) replication slippage E) oxidative damage Answer: B Section: 14.3 Bloom's Taxonomy: Remembering/Understanding 15) The mutagen that is an analog of thymine and which anomalously pairs with guanine is/are .

A) 5-bromouracil B) ultraviolet light C) ethylmethane sulfonate D) free radicals E) acetaldehyde Answer: A Section: 14.4 Bloom's Taxonomy: Applying/Analyzing 16) The mutagen that cause thymine dimers is/are A) 5-bromouracil B) ultraviolet light C) ethylmethane sulfonate D) free radicals E) acetaldehyde Answer: B Section: 14.4 Bloom's Taxonomy: Applying/Analyzing

.

4 Copyright © 2021 Pearson Education Ltd.

17) The mutagen that donates a CH2CH3 to nucleotides is/are A) 5-bromouracil B) ultraviolet light C) ethylmethane sulfonate D) free radicals E) acetaldehyde Answer: C Section: 14.4 Bloom's Taxonomy: Applying/Analyzing 18) The mutagen that is a strong oxidizer is/are A) 5-bromouracil B) ultraviolet light C) ethylmethane sulfonate D) free radicals E) acetaldehyde Answer: D Section: 14.4 Bloom's Taxonomy: Applying/Analyzing

.

.

19) Which of the following is a base analog?

A) acridine orange B) ethylmethane sulfonate C) ultraviolet light D) 5-bromouracil E) hydroxyurea Answer: D Section: 14.4 Bloom's Taxonomy: Remembering/Understanding 20) Considering the electromagnetic spectrum, identify likely mutagens from the following.

A) radio waves B) microwaves C) gamma rays D) X-rays E) Xrays and gamma rays Answer: E Section: 14.4 Bloom's Taxonomy: Applying/Analyzing

5 Copyright © 2021 Pearson Education Ltd.

21) What is the common influence of ultraviolet light on DNA? A) tautomeric shifting B) adduct formation C) depurination D) deamination E) generation of pyrimidine dimers Answer: E Section: 14.4 Bloom's Taxonomy: Remembering/Understanding 22) When X rays penetrate cells, electrons are ejected from atoms of molecules. Stable molecules can be transformed into which two types of hazardous materials?

A) free radicals and reactive ions B) heterocylic amines and acetaldehyde C) 6-ethylguanine and 5-bromouracil D) bromodeoxyuridine and ehtylmethane sulfonate E) thymine dimers and superosxides Answer: A Section: 14.4 Bloom's Taxonomy: Remembering/Understanding 23) Assuming one mutational event in a gene, on average, which of the following mutagens or mutational conditions would be expected to cause the most damage to a protein synthesized by such a mutagenized gene?

A) 5-bromouracil B) ethylmethane sulfonate C) 2-amino purine (base analog) D) frameshift E) heterocyclic amines Answer: D Section: 14.4 Bloom's Taxonomy: Applying/Analyzing 24) Which of the following would be considered an alkylating agent?

A) ethane B) heterocyclic amines C) acetaldehyde D) free radicals E) ethylmethane sulfonate Answer: E Section: 14.4 Bloom's Taxonomy: Evaluating/Creating

6 Copyright © 2021 Pearson Education Ltd.

25) Ethidium bromide is a chemical commonly used to visualize DNA in research due to its ability to fluoresce under UV lighting and to wedge between DNA base pairs. It however is very mutagenic. What class of mutagen would Ethidium bromide fall under?

A) base analog B) adduct-forming agent C) intercalating agent D) ionizing agent E) UV radiation Answer: C Section: 14.4 Bloom's Taxonomy: Applying/Analyzing 26) Ultraviolet light causes pyrimidine dimers to form in DNA. Some individuals are genetically incapable of repairing some dimers at "normal" rates. Such individuals are likely to suffer from .

A) xeroderma pigmentosum B) SCID C) phenylketonuria D) muscular dystrophy E) Huntington Disease Answer: A Section: 14.6 Bloom's Taxonomy: Remembering/Understanding 27) Recombinational repair is activated when damaged DNA has escaped repair and the distortion disrupts the process of replication. Which gene is responsible for recombinational repair?

A) uvrA B) uvrB C) uvrC D) recA E) mutH Answer: D Section: 14.6 Bloom's Taxonomy: Remembering/Understanding 28) Which human condition is caused by unrepaired UV-induced lesions?

A) xeroderma pigmentosum B) fragile-x syndrome C) Huntington's D) Marfan syndrome E) cystic fibrosis Answer: A Section: 14.6 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

29) Imagine that using Drosophila you recover two mutant alleles from a mutation screen where you induced transitions using EMS. What would you initially do to characterize the mutations?

A) Do a complementation test to determine if they are polygenic.

B) Map the alleles to nearby transposable elements.

C) Find the transition by sequencing the gene.

D) Cross the mutants to wild-type flies.

E) Cytoligcially study the chromosomes.

Answer: A Section: 14.6 Bloom's Taxonomy: Evaluating/Creating 30) Which of the following is a gene that encodes a protein that cuts out lesions in nucleotide excision repair?

A) uvr B) mutH C) recA D) mutS E) c-myc Answer: A Section: 14.6 Bloom's Taxonomy: Remembering/Understanding 31) Which of the following repair pathways focusses on double-strand break repair?

A) base excision repair B) nucleotide excision repair C) nonhomologous end joining D) post replication repair E) photoreactivation Answer: C Section: 14.6 Bloom's Taxonomy: Remembering/Understanding 32) Which of the following repair pathways uses an enzyme that absorbs blue light to cleave thymine dimers?

A) base excision repair B) nucleotide excision repair C) nonhomologous end joining D) post replication repair E) photoreactivation Answer: E Section: 14.6 Bloom's Taxonomy: Remembering/Understanding

8 Copyright © 2021 Pearson Education Ltd.

33) Transposons, or jumping genes, are DNA elements that move within the genome. In which organisms are transposons found?

A) bacteria B) eukaryotes C) mammals D) ancient bacteria E) all organisms Answer: E Section: 14.8 Bloom's Taxonomy: Remembering/Understanding 34) All DNA transposons contain two features that are essential for their movement. What are these two elements?

A) transposase and inverted terminal repeats B) integrase and pseudogenes C) integrase and oncogenes D) proto-oncogenes and oncogenes E) transposase and oncogenes Answer: A Section: 14.8 Bloom's Taxonomy: Remembering/Understanding 35) Barbara McClintock discovered mobile elements in corn by analyzing the genetic behavior of two elements, Ds and Ac. The interplay between these two elements has become one of the most interesting stories of discovery in the field of genetics. How do Ds and Ac interact?

A) Ds causes a deletion next to the insertion site of Ac. B) Ac causes a deletion next to the insertion site of Ds. C) Ds moves only if Ac is present in the genome; Ac is capable of autonomous movement.

D) The movement of Ac is dependent on two forms of Ds. E) Both elements can move only within chromosome 9.

Answer: C Section: 14.8 Bloom's Taxonomy: Applying/Analyzing 36) Which of the following makes staggered cuts in chromosomal DNA into which transposable elements can insert?

A) ligase B) AP endonuclease C) gyrase D) reverse transcriptase E) transposase Answer: E Section: 14.8 Bloom's Taxonomy: Remembering/Understanding

9 Copyright © 2021 Pearson Education Ltd.

37) What are LINES? A) long interspersed elements B) short interspersed elements C) retrotransposon D) copia-like elements E) LTR elements Answer: A Section: 14.8 Bloom's Taxonomy: Remembering/Understanding 38) How are retrotransposon different from other transposons?

A) They use an RNA intermediate.

B) They use integrase.

C) They use reverse transcriptase.

D) They use a "cut and paste" mechanism.

E) All of these are differences.

Answer: E Section: 14.8 Bloom's Taxonomy: Evaluating/Creating 39) Which of the following is not a potential use of transposable elements in research?

A) They could be used to deliver modified sequence to target site.

B) They could be used to disrupt other genes and create mutants.

C) They could be used to create double strand breaks that lead to translocations.

D) They can be used as markers to help map nearby genes.

E) They can be used to study dosage effects of gene expression.

Answer: C Section: 14.8 Bloom's Taxonomy: Evaluating/Creating 40) List five general categories of mutation.

Answer: induced, spontaneous, morphological, nutritional/biochemical, behavioral, regulatory, lethal, conditional Section: 14.1 Bloom's Taxonomy: Remembering/Understanding 41) Under which condition(s) might one have an amino acid substitution in a protein that does not result in an altered phenotype?

Answer: The possibility of a change in protein function, therefore phenotype, depends on the location and chemical properties of the involved amino acid(s). For example a silent mutation would produce no resultant effect and theoretically a nonpolar to nonpolar amino substitution would be less impactful than a nonpolar to polar.

Section: 14.1 Bloom's Taxonomy: Remembering/Understanding

10 Copyright © 2021 Pearson Education Ltd.

42) Three major types of RNAs are mRNA, rRNA, and tRNA. For each of the conditions below, predict the consequences in terms of the population of proteins being synthesized in a particular cell. What qualitative and quantitative changes, if any, are expected in the individual protein involved (if one is involved) and in the population of proteins produced in that cell? (a) A frameshift mutation in mRNA. The condition is heterozygous in the involved cell. (b) A deletion (homozygous) that removes approximately half of each type of rRNA genes.

Answer: (a) Population of proteins: Half of the protein products of that gene will be defective, and the other half will be normal. Individual protein: The protein should show multiple amino acid substitutions "downstream" from the point of the mutation. If a nonsense triplet is introduced, the protein would be shortened in the substituted region. (b) Population of proteins: There would be an overall reduction in protein synthesis. Individual protein: All the proteins would be made in their normal form, but at reduced levels.

Section: 14.1 Bloom's Taxonomy: Evaluating/Creating 43) Assume that a new mutation occurs in the germ line of an individual. What finding would suggest that the new mutation is dominant rather than recessive?

Answer: If dominant and if passed to the next generation, it would be expressed. New recessive mutations are not normally expressed in the next generation unless, through a combination with a like mutation from the other parent, they are homozygous.

Section: 14.1 Bloom's Taxonomy: Applying/Analyzing 44) One type of mutation involves the replacement of a purine with a purine, whereas another causes the replacement of a pyrimidine with a purine or the reverse. What general terms are associated with these two mutational phenomena?

Answer: transition and transversion, respectively Section: 14.1 Bloom's Taxonomy: Remembering/Understanding 45) In general, mutation rates in humans are .

Answer: determined by defining the likelihood that a gene will undergo a mutation in a single generation or in forming a single gamete Section: 14.2 Bloom's Taxonomy: Remembering/Understanding 46) DNA may be damaged from the by-products of normal cellular aerobic respiration. Name three of these electrophilic oxidants that are generally classified as reactive oxidants.

Answer: superoxides (O2-), hydroxyl radicals (∙OH), and hydrogen peroxide (H2O2) Section: 14.3 Bloom's Taxonomy: Remembering/Understanding

11 Copyright © 2021 Pearson Education Ltd.

47) Three human disorders—fragile-X syndrome and Huntington disease—are conceptually linked by a common mode of molecular aberration. Describe the phenomena that link these disorders.

Answer: Both are caused by disparate genes, but each gene was found to contain repeats of a unique trinucleotide sequence. In addition, the number of repeats may increase in each subsequent generation (genetic anticipation) due to replication slippage.

Section: 14.3 Bloom's Taxonomy: Evaluating/Creating 48) Imagine that you want to generate a point mutation for the study of meiosis in Drosophila.

What would you use to induce your mutations?

Answer: We could use an alkylating agent like EMS which would induce transitions by making 6-ehtylguanine. Alternatively 5-bromouracil could be used to induce transitions.

Section: 14.4 Bloom's Taxonomy: Evaluating/Creating 49) Explain how carrying a mutation in the BRCA1 gene can cause a familial predisposition for breast cancer.

Answer: BRCA1 has been found to be involved in nonhomologous end joining as part of the recognition, digest and ligation activity. Absence of BRCA1 in one allele makes it more likely acquire cancer since only one more mutation needs to occur to create homozygosity.

Section: 14.6 Bloom's Taxonomy: Evaluating/Creating 50) In organisms that lack photolyase explain the mechanics by which they can repair DNA in a region where there is a pyrimidine dimer.

Answer: They could us postreplication repair where polymerase skips over the lesion leaving the affected region with a gap. Then the undamaged strand can donate the correct sequence to the damaged strand and then have its own consequent gap filled in.

Section: 14.6 Bloom's Taxonomy: Evaluating/Creating 51) What is meant by the term photoreactivation repair?

Answer: Photoreactivation repair, discovered in 1949, is a process described in E. coli in which UV-induced DNA damage can be partially reversed if cells are briefly exposed to light in the blue range of the visible spectrum.

Section: 14.6 Bloom's Taxonomy: Remembering/Understanding 52) What is the Ames test, and how does it work?

Answer: Four tester strains of Salmonella typhimurium are used to test for sensitivity and specificity of mutagenesis.

Section: 14.7 Bloom's Taxonomy: Remembering/Understanding

12 Copyright © 2021 Pearson Education Ltd.

53) Imagine that an Ames test was performed on a new red dye to determine if it will be safe for consumers. For this his- mutants are grown in growth media and the disk is soaked in the red dye. The results show that the reversion rate is not significantly above the spontaneous rate.

Would you conclude that this dye is safe? Explain why or why not.

Answer: No because we don't know if the liver metabolizes this dye into an unsafe mutagenic metabolite. We would need to incubate in liver enzymes to determine safety accurately.

Section: 14.7 Bloom's Taxonomy: Evaluating/Creating

13 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 15 Regulation of Gene Expression in Bacteria 1) Which term would be applied to a regulatory condition that occurs when protein is associated with a particular section of DNA and greatly reduces transcription?

A) negative control B) positive control C) induction D) reduction E) compression Answer: A Section: 15.1 Bloom's Taxonomy: Remembering/Understanding 2) Which term refers to a contiguous genetic complex that is under coordinated control?

A) lysogen B) prototroph C) operon D) allosteric E) attenuation Answer: C Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 3) Which term most appropriately refers to a regulatory protein in prokaryotes?

A) translation B) RNA processing C) trans-acting factor D) gyrase action E) helicase activation Answer: C Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 4) In the lac operon, the product of structural gene lacZ is capable of A) nonautonomous replication B) forming lactose from two glucose molecules C) replacing hexokinase in the early steps of glycolysis D) splitting the glycosidic bond of lactose E) forming ATP from pyruvate Answer: D Section: 15.2 Bloom's Taxonomy: Remembering/Understanding

1 Copyright © 2021 Pearson Education Ltd.

.

5) Which of the following is the gene that codes for transacetylase in the lac operon?

A) LacI B) LacZ C) LacY D) LacA E) LacOc Answer: D Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 6) Which of the following elements of the lac operon encodes a protein that facilitates the entry of lactose into the bacterial cell?

A) LacI B) LacZ C) LacY D) LacA E) LacOc Answer: C Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 7) Which of the following regions of the lac operon represents a mutated form of the element?

A) LacI B) LacZ C) LacY D) LacA E) LacOc Answer: E Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 8) Which of the following is considered a gratuitous inducer of the lac operon?

A) glucose B) lactose C) isoporpylthiogalactoside D) galactose E) glucose and galactose Answer: C Section: 15.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

9) Which of the following is a product of the protein encoded by lacZ? A) glucose B) lactose C) isoporpylthiogalactoside D) galactose E) glucose and galactose Answer: E Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 10) If lactose is accumulating outside of E. coli and is unable to enter, then which structural gene of the lac operon is most likely mutated?

A) lacZ B) lacY C) lacA D) lacX E) lacI Answer: B Section: 15.2 Bloom's Taxonomy: Evaluating/Creating 11) lacI codes for a(n) , and when lacI is mutated, the lac operon is A) repressor; constitutively expressed B) repressor; never expressed C) activator; constitutively expressed D) activator; never expressed E) structural gene; never expressed Answer: A Section: 15.2 Bloom's Taxonomy: Evaluating/Creating

.

12) Some lac operon mutations allow for beta galactosidase to be expressed constitutively even in the absence of lactose. Which of the following lac genotypes would allow for this constitutive expression?

A) I+ P+ O+ Z+ Y+ A+ B) I+ P- O+ Z+ Y+ A+ C) I+ P+ O+ Z- Y+ A+ D) I+ P+ Oc Z+ Y+ A+ E) Is P+ O+ Z+ Y+ A+ Answer: D Section: 15.2 Bloom's Taxonomy: Evaluating/Creating

3 Copyright © 2021 Pearson Education Ltd.

13) A constitutive mutation in the lac operon may be of several types. Name two types of constitutive mutations.

A) LacI- and LacOc B) LacZ- and LacIC) LacOc and LacZD) LacY- and LacAE) LacY- and LacIAnswer: A Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 14) Mutations in the lacI and lacO genes of the lac regulatory system in E. coli often lead to full transcription of the three structural genes, even when no lactose is available to the organism.

Such mutations would be called .

A) constitutive B) activated C) repressive D) induced E) attentuated Answer: A Section: 15.2 Bloom's Taxonomy: Applying/Analyzing 15) A molecule that can change shape, and therefore function, in its interactions with other molecules is called .

A) allosteric B) an inducer C) a repressor D) an enhancer E) an attenuator Answer: A Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 16) Which of the following is involved in catabolite repression of the lac operon?

A) glucose B) lactose C) isoporpylthiogalactoside D) galactose E) glucose and galactose Answer: A Section: 15.3 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

17) Which of the following is not necessary for CAP associated transcription?

A) ATP B) adenyl cyclase C) catabolite activating protein D) glucose E) LacI Answer: E Section: 15.3 Bloom's Taxonomy: Applying/Analyzing 18) The enzyme responsible for making cAMP is A) adenyl cyclase B) tryptophan synthetase C) ATP synthetase D) cAMP syntetase E) transacetylase Answer: A Section: 15.3 Bloom's Taxonomy: Remembering/Understanding

.

19) Which of the following terms best characterizes catabolite repression associated with the lac operon in E. coli?

A) inducible system B) repressible system C) negative control D) positive control E) constitutive Answer: D Section: 15.3 Bloom's Taxonomy: Remembering/Understanding 20) Which of the following is not a structural gene of the trp operon?

A) TrpD B) TrpC C) TrpB D) TrpE E) TrpR Answer: E Section: 15.4 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

21) Regarding regulation of the trp operon, what might one appropriately call the amino acid tryptophan?

A) corepressor B) coactivator C) inducer D) an enhancer E) an attenuator Answer: A Section: 15.4 Bloom's Taxonomy: Remembering/Understanding 22) Which of the following elements of the trp operon is involved in a process that "weakens or impairs" expression of the operon?

A) attenuator B) promoter C) leader sequence D) repressor E) operator Answer: A Section: 15.5 Bloom's Taxonomy: Remembering/Understanding 23) When referring to attenuation in regulation of the trp operon, it would be safe to say that when there are high levels of tryptophan available to the organism, .

A) the trp operon is being transcribed at relatively high levels B) translational termination is likely C) transcriptional termination is likely D) tryptophan is inactivating the repressor protein E) ribosomes are stalling during translation of the attenuator region Answer: C Section: 15.5 Bloom's Taxonomy: Remembering/Understanding 24) Which of the following is the ligand binding site of a riboswitch?

A) aptamer B) expression platform C) terminator hairpin D) antiterminator hairpin E) ribosome-binding site Answer: A Section: 15.5 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

25) Which of the following is masked by the binding of sRNA to mRNA's?

A) aptamer B) expression platform C) terminator hairpin D) antiterminator hairpin E) RBS Answer: E Section: 15.5 Bloom's Taxonomy: Remembering/Understanding 26) Formation of the terminator structure in riboswitches is accomplished by the A) aptamer B) expression platform C) terminator hairpin D) antiterminator hairpin E) ribosome-binding site Answer: B Section: 15.5 Bloom's Taxonomy: Remembering/Understanding 27) In the presence of excess tryptophan which of the following is formed?

A) aptamer B) expression platform C) terminator hairpin D) antiterminator hairpin E) ribsosome-binding site Answer: C Section: 15.5 Bloom's Taxonomy: Remembering/Understanding 28) When tryptophan is scarce in the cell what mRNA secondary structure forms?

A) aptamer B) expression platform C) terminator hairpin D) antiterminator hairpin E) ribosome-binding site Answer: D Section: 15.5 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

.

29) Which one of these genes is a negative regulator of gene expression in e. coli, and responds to low iron levels?

A) DsrA B) RyhB C) TrpD D) TrpE E) TrpR Answer: B Section: 15.5 Bloom's Taxonomy: Remembering/Understanding 30) Usually, bacteria only make tryptophan when tryptophan is absent or available in low concentration. However, a particular bacterial mutation makes tryptophan all the time whether or not tryptophan is present. What could explain this phenotype?

A) the terminator hairpin is unable to form B) the antiterminator hairpin is unable to form C) trpE is mutated D) trpD is mutated E) trpA is mutated Answer: A Section: 15.5 Bloom's Taxonomy: Evaluating/Creating 31) A mutation in the bacterial gene Cas-1 would result in what phenotype?

A) It would cause an inability to transcribe crRNA. B) It would cause an inability to acquire new spacers.

C) It would cause an inability to make Cas-9.

D) It would cause constitutive nuclease activity.

E) It would cause a deficiency in target interference.

Answer: B Section: 15.6 Bloom's Taxonomy: Applying/Analyzing 32) How does CRISPR behave as a form of bacterial adaptive immunity?

A) Spacer sequences provide a molecular memory of past viral attacks.

B) CRISPR loci encodes antibody proteins.

C) CRISPR nucleases trigger apoptosis.

D) CRISPR repeats provide molecular memory of infection.

E) Cas genes encode complementary DNA to phage genomes.

Answer: A Section: 15.6 Bloom's Taxonomy: Applying/Analyzing

8 Copyright © 2021 Pearson Education Ltd.

33) If you were studying the evolution of bacteria and specifically the history of viral infections of bacteria, what type of RNA would help you see that history?

A) miRNA B) siRNA C) sRNA D) crRNA E) rRNA Answer: D Section: 15.6 Bloom's Taxonomy: Applying/Analyzing 34) Compare and contrast positive and negative control of gene expression in bacteria.

Answer: Both forms of control result from an interaction of a molecule (usually considered to be a protein) with the genetic material (either RNA or DNA). Positive control results when the interaction stimulates transcription, whereas negative control occurs when the interaction inhibits transcription.

Section: 15.1 Bloom's Taxonomy: Applying/Analyzing 35) Certain mutations in the regulator gene of the lac system in E. coli result in maximal synthesis of the Lac proteins (β-galactosidase, etc.) even in the absence of the inducer (lactose).

Provide an explanation for this observation.

Answer: There has been a mutation in the gene that produces the repressor, or the operator is mutated so that it will not interact with the repressor.

Section: 15.2 Bloom's Taxonomy: Evaluating/Creating 36) Present a detailed description of the actions of the regulatory proteins in inducible and repressible enzyme systems.

Answer: Inducible system: The repressor is normally active, but the inducer inactivates the repressor. Repressible system: The repressor is inactive but is activated by the corepressor.

Active repressors turn off transcription.

Section: 15.1 Bloom's Taxonomy: Remembering/Understanding

9 Copyright © 2021 Pearson Education Ltd.

37) The table below lists several genotypes associated with the lac operon in E. coli. For each, indicate with a "+" or a "—" whether functional β-galactosidase would be expected to be produced at induced levels.

Genotype (a) I + O+ Z+/F' I - O+ Z+ (b) I - Oc Z+/F' I - O+ Z(c) I s Oc Z+/F' I + O+ Z+ (d) I - O+ Z+/F' I - O+ Z+

β-Galactosidase Production No Lactose With Lactose

I+ = wild-type repressor I= mutant repressor (unable to bind to the operator) Is = mutant repressor (insensitive to lactose) O+ = wild-type operator Oc = constitutive operator (insensitive to repressor) Answer: (a) - + (b) + + (c) + + (d) + + Section: 15.2 Bloom's Taxonomy: Evaluating/Creating

10 Copyright © 2021 Pearson Education Ltd.

38) The table below lists several genotypes associated with the lac operon in E. coli. For each, indicate with a "+" or a "—" whether β-galactosidase would be expected to be produced at induced levels. β-Galactosidase Production Genotype No Lactose With Lactose (a) I + O+ Z+/F' I - O+ Z+ (b) I - Oc Z -/F' I - Oc Z (c) I - Oc Z+/F' I - O+ Z+ (d) I s Oc Z -/F' I s O+ Z+ I+ = wild-type repressor I= mutant repressor (unable to bind to the operator) Is = mutant repressor (insensitive to lactose) O+ = wild-type operator Oc = constitutive operator (insensitive to repressor) Answer: (a) - + (b) - (c) + + (d) - Section: 15.2 Bloom's Taxonomy: Evaluating/Creating 39) Describe what is meant by a gratuitous inducer. Give an example.

Answer: A gratuitous inducer is a chemical analog of a natural inducer. It serves as an inducer but is not a substrate for the reactions related to the natural inducer. Isopropylthiogalactoside (IPTG) is a gratuitous inducer of the lactose operon.

Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 40) What is the function of the lacY gene in the lac operon?

Answer: The lacY gene codes for permease, a membrane-bound protein that transports lactose into the bacterial cell.

Section: 15.2 Bloom's Taxonomy: Remembering/Understanding 41) Explain why lacOc mutations are cis-acting, whereas lacI mutations can be trans-acting.

Answer: The operator region does not produce a diffusible product, whereas the lacI gene does.

Section: 15.2 Bloom's Taxonomy: Applying/Analyzing

11 Copyright © 2021 Pearson Education Ltd.

42) What experimental results would indicate that the mutation lacIs is dominant to lacI+?

Answer: In lacIs / lacI+ partial diploids, the lac operon is in a repressed state in the presence of lactose.

Section: 15.2 Bloom's Taxonomy: Applying/Analyzing 43) Describe the positive control exerted by the catabolite activating protein (CAP). Include a description of catabolite repression.

Answer: Regarding regulation of the lac operon, in the absence of glucose, CAP (dependent on cAMP and adenyl cyclase) binds to the CAP site and facilitates transcription (positive control).

Transcription of the operon is inhibited in the presence of glucose (catabolite repression).

Section: 15.3 Bloom's Taxonomy: Applying/Analyzing 44) The catabolite repression system in E. coli essentially represses the lac operon when glucose is present. What evolutionary advantage would favor evolution of such a system?

Answer: Glucose can enter glycolysis "as is," whereas lactose must first be split into glucose and galactose. To do so, the energy-requiring synthesis of β-galactosidase is required. It is more energy efficient to burn glucose than lactose.

Section: 15.3 Bloom's Taxonomy: Evaluating/Creating 45) Monod discovered that if tryptophan is present in relatively high quantities in the growth medium, the enzymes necessary for its synthesis are repressed. How does this occur?

Answer: Tryptophan acts as a corepressor to activate the tryptophan repressor protein; it also is involved in an attenuation system, which causes termination of transcription.

Section: 15.4, 15.5 Bloom's Taxonomy: Remembering/Understanding 46) Regarding the trp operon, trpR- maps to a considerable distance from the structural genes.

The mutation either inhibits the interaction with tryptophan or inhibits repressor formation entirely. In the presence of tryptophan in the medium, would you expect the trp operon with the trpR- mutation to be transcriptionally active? Explain.

Answer: With either of the two scenarios mentioned in the problem, the absence of repressor function in a repressible system means that there would be no repression of the operon. The operon would be transcriptionally active.

Section: 15.4 Bloom's Taxonomy: Applying/Analyzing 47) Within the control region of the trp operon is a section of DNA that is sensitive to levels of tryptophan in the system. What is the name of this region?

Answer: leader or attenuator region Section: 15.5 Bloom's Taxonomy: Remembering/Understanding

12 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 16 Regulation of Gene Expression in Eukaryotes 1) At what level is genetic regulation considered most likely in prokaryotes?

A) transcriptional B) capping C) polyadenylation of the 3' end of the mRNAs D) intron processing E) exon processing Answer: A Section: 16.1 Bloom's Taxonomy: Applying/Analyzing 2) What type of genetic regulation seems to be the most similar between prokaryotes and eukaryotes?

A) transcriptional regulation B) RNA splicing regulation C) intron/exon shuffling D) 5'-capping regulation E) poly-A tail addition Answer: A Section: 16.1 Bloom's Taxonomy: Applying/Analyzing 3) DNA methylation may be a significant mode of genetic regulation in eukaryotes. Methylation refers to .

A) altering RNA polymerase activity by methylation B) changes in DNA-DNA hydrogen binding C) altering translational activity, especially of highly methylated tRNAs D) alteration of DNA polymerase activity by addition of methyl groups to glycine residues E) addition of methyl groups to the cytosine of CG doublets Answer: E Section: 16.2 Bloom's Taxonomy: Remembering/Understanding 4) Gene regulation in eukaryotes is more complicated than bacterial gene regulation. Which of the following describes a gene regulation event that occurs in eukaryotes but not in bacteria?

A) In eukaryotes, transcription and translation occur in the same cellular compartment.

B) In eukaryotes, mRNA does not need to be modified.

C) In eukaryotes, histones must be added or removed to regulate gene expression.

D) In eukaryotes, mRNA degrades quickly compared to bacterial mRNA that is more stable.

E) In eukaryotes, proteins are post-translationally modified, whereas bacterial proteins are never post-translationally modified.

Answer: C Section: 16.2 Bloom's Taxonomy: Evaluating/Creating 1 Copyright © 2021 Pearson Education Ltd.

5) Which of the following is true about histone acetyltransferase (HAT)?

A) HATs will remove acetyl groups from histone proteins.

B) HATs will add acetyl groups to DNA. C) HATs will make the chromatin more compacted and less available to transcription regulatory proteins.

D) HATs are usually linked to increasing gene expression.

E) HATs have an overall effect of increasing the positive charges on DNA. Answer: D Section: 16.2 Bloom's Taxonomy: Applying/Analyzing 6) Affinity of histones for the negative charges on the backbone phosphates of DNA is reduced by .

A) histone acetyltransferase B) histone deacetylase C) chromatin remodeling complexes D) DNA methylation E) acetyl kinases Answer: A Section: 16.2 Bloom's Taxonomy: Remembering/Understanding 7) Which of the following clusters of terms applies when addressing enhancers as elements associated with eukaryotic genetic regulation?

A) cis-acting, variable orientation, variable position B) trans-acting, fixed position, fixed orientation C) cis-acting, fixed position, fixed orientation D) cis-acting, variable position, fixed orientation E) trans- and cis-acting, variable position Answer: A Section: 16.3 Bloom's Taxonomy: Remembering/Understanding 8) Two modular elements that appear as consensus sequences upstream from RNA polymerase II transcription start sites are .

A) microsatellites and transposons B) rDNA and nucleolar organizers C) TATA and CAAT D) TTAA and CCTT E) enhancers and telomeres Answer: C Section: 16.3 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

9) Which of the following would be an example of a cis acting eukaryotic gene regulatory element?

A) RNA polymerase B) metallothionein C) general transcription factors D) TATA binding protein E) enhancer Answer: E Section: 16.3 Bloom's Taxonomy: Remembering/Understanding 10) A regulatory sequence of DNA that is 10,000 base pairs away from the gene it regulates is mutated. The result is that the gene being regulated is now expressed at a higher rate compared to when this regulatory sequence was not mutated. What would this sequence of DNA best be called?

A) insulator B) activator protein C) enhancer D) silencer E) zinc finger motif Answer: D Section: 16.3 Bloom's Taxonomy: Evaluating/Creating 11) A mutation in which promoter region would result in a reduction in transcription level?

A) TATA B) GC C) CAAT D) silencer E) TATA, GC and CAAT Answer: E Section: 16.3 Bloom's Taxonomy: Remembering/Understanding 12) A mutation in which promoter region would result in an increase in transcription level?

A) TATA B) GC C) CAAT D) silencer E) TATA, GC and CAAT Answer: D Section: 16.3 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

13) What do we call an element that directs initiation from a number of weak transcription start sites?

A) focused core promoter B) dispersed core promoter C) basal element D) GC box E) CAAT box Answer: B Section: 16.3 Bloom's Taxonomy: Remembering/Understanding 14) Which term describes an element that directs initiation of transcription at a single specific start site?

A) focused core promoter B) dispersed core promoter C) basal element D) GC box E) CAAT box Answer: A Section: 16.3 Bloom's Taxonomy: Remembering/Understanding 15) Which is not considered to be an enhancer of transcription in the metallothionein 2A gene?

A) MRE B) ARE C) GRE D) Pz120 E) BLE Answer: D Section: 16.4 Bloom's Taxonomy: Applying/Analyzing 16) Which is not part of the metallothionein 2A gene promoter region?

A) MRE B) GC C) ARE D) ERE E) GRE Answer: D Section: 16.4 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

17) In the metallothionein 2A gene AP2 binds to which promoter element?

A) MRE B) ARE C) GRE D) GC E) BLE Answer: B Section: 16.4 Bloom's Taxonomy: Remembering/Understanding 18) In the metallothionein 2A gene TFIID binds to which promoter element?

A) MRE B) GRE C) Inr D) TATA E) BLE Answer: D Section: 16.4 Bloom's Taxonomy: Remembering/Understanding 19) In the metallothionein 2A gene SP1 binds to which promoter element?

A) MRE B) ARE C) GRE D) GC E) BLE Answer: D Section: 16.4 Bloom's Taxonomy: Remembering/Understanding 20) In the metallothionein 2A gene BLE binds to which transcription factor?

A) TFIID B) SP1 C) AP1 D) AP4 E) Both AP1 and AP4 Answer: E Section: 16.4 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

21) The pre-initiation complex (PIC) contains several proteins. What would the direct consequence be if PIC failed to form?

A) Replication would not be initiated.

B) Translation would not be initiated.

C) Transcription would not be initiated.

D) mRNA splicing would not be initiated.

E) Protein would not fold properly.

Answer: C Section: 16.5 Bloom's Taxonomy: Applying/Analyzing 22) TATA binding protein is a component of which transcription factor?

A) TFIID B) TFIIB C) TFIIF D) TFIIH E) TFIIE Answer: A Section: 16.5 Bloom's Taxonomy: Remembering/Understanding 23) Which of these transcription factors binds to BRE's?

A) TFIID B) TFIIB C) TFIIF D) TFIIH E) TFIIE Answer: B Section: 16.5 Bloom's Taxonomy: Remembering/Understanding 24) Imagine you recover a mutation that contains a transversion in the T of the GT consensus sequence in 5' splice site of exon 2 of your gene of interest. Upon sequencing you also notice that there is a cryptic or alternative GT splice site 5bp upstream. What would you predict would be the outcome of the mutation on the transcript that is produced?

A) Use of the alternative splice site would create a 5 bp deletion of exon 2.

B) Mutation in intron 2 would cause exon 2 to be skipped.

C) Use of the alternative splice site would cause exon 3 to be deleted.

D) With a mutated splice site intron 2 would be included.

E) Mutation in intron 2 would cause exon 3 to be skipped.

Answer: A Section: 16.6 Bloom's Taxonomy: Evaluating/Creating

6 Copyright © 2021 Pearson Education Ltd.

25) Spliceforms with different 3' ends are produced by A) alternative 3' splice sites B) mutually exclusive exons C) alternative polyadenylation D) alternative promoters E) intron retention Answer: C Section: 16.6 Bloom's Taxonomy: Remembering/Understanding

.

26) In regards to mRNA stability, what is the effect of decapping enzymes?

A) The m7G cap of mRNAs is stabilized allowing mRNAs to be protected at the 5' end.

B) The m7G cap of mRNAs is removed causing mRNAs to be degraded from the 5' end.

C) The polyadenylation of mRNAs is removed and mRNAs are degraded from the 3' end.

D) The polyadenylation of mRNAs is stabilized allowing mRNAs to be protected at the 3' end.

E) The splicing of mRNAs is blocked, which destabilizes mRNAs. Answer: B Section: 16.7 Bloom's Taxonomy: Remembering/Understanding 27) A researcher is studying plant development. She isolates a mutant and discovers that the piece of DNA that is mutated does not code for a protein. However, this piece of DNA is complimentary to a gene known to function in early embryonic plant development. The mutant she identified most likely functions as a(n) .

A) ribosomal RNA (rRNA) B) transfer RNA (tRNA) C) intron D) general transcription factor known to regulate many different genes E) microRNA (miRNA) Answer: E Section: 16.8 Bloom's Taxonomy: Evaluating/Creating 28) If the protein Dicer is not functioning, then which kind of posttranscriptional gene regulation is most likely affected?

A) alternative splicing B) alternative adenylation C) only microRNA (miRNA), but not small interfering RNA (siRNA) D) small interfering RNA (siRNA) E) microRNA (miRNA) and small interfering RNA (siRNA) Answer: E Section: 16.8 Bloom's Taxonomy: Evaluating/Creating

7 Copyright © 2021 Pearson Education Ltd.

29) microRNA (miRNA) and small interfering RNA (siRNA) use a similar process and proteins to regulate gene expression. Which of the following is more commonly associated with the miRNA pathway?

A) The protein Drosha is found in the miRNA pathway but not the siRNA pathway.

B) The protein Dicer is found in the miRNA pathway but not the siRNA pathway.

C) The protein complex RISC is found in the miRNA pathway but not the siRNA pathway.

D) The protein AGO is found in the miRNA pathway but not the siRNA pathway.

E) In animals, the miRNA pathway usually leads to mRNA degradation, and in the siRNA pathway, the mRNA is not degraded.

Answer: A Section: 16.8 Bloom's Taxonomy: Evaluating/Creating 30) Imagine you discover an mRNA that is unable to properly localize due to a deletion of part of the transcript. What region of the mRNA would likely find the deletion?

A) 3' UTR B) One of the exons C) 5' splice site D) 5' UTR E) Branch site one of the introns Answer: A Section: 16.9 Bloom's Taxonomy: Evaluating/Creating 31) A mutation results in the failure of fibroblasts to move to the site of infection. It is discovered that this mutation causes a particular protein to be absent, which leads to premature actin mRNA translation. Which protein is most likely absent due to the mutation?

A) Src B) ZBP1 (zip code-binding protein) C) CPEB D) general transcription factor E) Either Src or ZBP1 Answer: E Section: 16.9 Bloom's Taxonomy: Evaluating/Creating 32) are multi-subunit structures that possess protease activity.

A) Ribosomes B) Mitochondria C) Golgi apparatus D) Proteasomes E) Processing bodies Answer: D Section: 16.10 Bloom's Taxonomy: Remembering/Understanding

8 Copyright © 2021 Pearson Education Ltd.

33) Which protein modification is most closely linked to protein degradation?

A) phosphorylation B) ubiquitination C) acetylation D) methylation E) glycosylation Answer: B Section: 16.10 Bloom's Taxonomy: Remembering/Understanding 34) Imagine you induce a null mutation in a ubiquitin ligase gene. What would you predict would be the phenotypic outcome of this mutation?

A) Increased protein stability.

B) Decreased protein stability.

C) Increase in misfolded proteins.

D) Aggregation of misfolded proteins.

E) No effect due to multiple copies of the gene.

Answer: E Section: 16.10 Bloom's Taxonomy: Evaluating/Creating 35) Name at least three different levels of regulation in eukaryotes.

Answer: pretranscriptional, transcriptional, processing, transport, translational, posttranslational Section: 16.1 Bloom's Taxonomy: Applying/Analyzing 36) Approximately 5 percent of the cytosine residues are methylated in the genome of any given eukaryote. In what way is DNA methylation related to genetic regulation?

Answer: There is an inverse relationship between the degree of methylation of eukaryotic DNA and the degree of gene expression.

Section: 16.2 Bloom's Taxonomy: Remembering/Understanding 37) In what way is chromosomal organization within the nucleus of an interphase cell thought to be related to gene activity?

Answer: Transcriptionally active genes are located at the edges of chromosome territories next to the channels of the interchromatin compartments. This organization brings actively transcribed genes into closer association with each other and with the transcriptional machinery, thereby facilitating their coordinated expression. Transcripts produced at the edge of chromosome territories move into the adjacent interchromatin compartments, which house RNA processing machinery and are contiguous with nuclear pores.

Section: 16.2 Bloom's Taxonomy: Applying/Analyzing

9 Copyright © 2021 Pearson Education Ltd.

38) Describe how nucleosomes may influence gene transcription.

Answer: The binding of transcription factors requires accessing nucleosomal DNA, and such factors may displace nucleosomes. The removal of a nucleosome from a stretch of DNA exposes it to binding by transcription factors and RNA polymerase, allowing transcription to proceed.

Section: 16.2 Bloom's Taxonomy: Applying/Analyzing 39) Describe three characteristics of enhancers.

Answer: position need not be fixed, orientation may be inverted without significant effect, not gene specific Section: 16.3 Bloom's Taxonomy: Remembering/Understanding 40) Describe the function and general nature of promoters in eukaryotes.

Answer: Promoter regions are necessary for the initiation of transcription. Promoters that interact with RNA polymerase II are usually located within 100 bp upstream of a gene and usually contain a TATA box and a CAAT box.

Section: 16.3 Bloom's Taxonomy: Remembering/Understanding 41) What is the location of an enhancer in relation to the gene it affects?

Answer: The position of an enhancer need not be fixed; it can be upstream, downstream, or within the gene it regulates.

Section: 16.3 Bloom's Taxonomy: Remembering/Understanding 42) Name two consensus sequences or modular DNA sequences that may be found upstream from coding regions of some eukaryotic genes.

Answer: TATA and CAAT sequences (boxes) Section: 16.3 Bloom's Taxonomy: Remembering/Understanding 43) Describe the general structure of a transcription factor.

Answer: Transcription factors are modular proteins with at least two functional domains: one binds to DNA in promoters and another initiates transcription.

Section: 16.5 Bloom's Taxonomy: Remembering/Understanding 44) In what ways are eukaryotic activators thought to function?

Answer: By binding the DNA, they appear to influence the structural organization of chromatin and possibly the looping out of DNA that separates the enhancer from the transcription initiation complex (transcription factors, promoter, RNA polymerase, etc.).

Section: 16.5 Bloom's Taxonomy: Remembering/Understanding

10 Copyright © 2021 Pearson Education Ltd.

45) What type of genetic control, positive or negative, typically results when transcription factors interact with DNA? Answer: positive control Section: 16.5 Bloom's Taxonomy: Applying/Analyzing 46) What is alternative splicing?

Answer: Alternative splicing is a process that allows, through differential splicing of premRNA, a variety of different forms of mRNA. This process allows a number of different proteins to be formed from a single gene.

Section: 16.6 Bloom's Taxonomy: Remembering/Understanding 47) Alternative RNA splicing is a method that apparently evolved for the production of many different polypeptides from the same pre-mRNA. Provide an example of alternative splicing.

Answer: Various splicing schemes occur in the CT/CGRP gene in different tissues. In thyroid tissue, only the first four exons remain, whereas in the brain, exons 5 and 6 are included, but not exon 4.

Section: 16.6 Bloom's Taxonomy: Remembering/Understanding 48) What is the function of RNA interference?

Answer: RNA interference appears to be a sequence-specific posttranscriptional regulatory mechanism that employs short RNA sequences.

Section: 16.8 Bloom's Taxonomy: Applying/Analyzing

11 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 17 Recombinant DNA Technology 1) Words such as did, mom, and pop have something in common with the fundamental tool of recombinant DNA technology. In the context of recombinant DNA technology, which term would be used to describe such words?

A) lysogenic B) prototrophic C) palindromic D) conjugation E) insertion Answer: C Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 2) Restriction endonucleases are especially useful if they generate "sticky" ends. What makes an end sticky?

A) single-stranded complementary tails B) blunt ends C) poly-A sequences D) 5' cap E) 3' cap Answer: A Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 3) Which of the following statements include two useful characteristics of cloning vectors?

A) replication within host cells and antibiotic resistance gene(s) B) virulence and lysogenicity C) ability to integrate into the host chromosome and then causing a lytic cycle D) nonautonomous replication and transposition E) reverse transcriptase and ligase activities Answer: A Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 4) Commercially prepared cloning vectors such as pUC18 are designed to contain several useful features. An example of one of these features is .

A) high GC content B) consensus sequence C) multiple cloning sites D) β-galactosidase E) complementation Answer: C Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 1 Copyright © 2021 Pearson Education Ltd.

5) Expression vectors differ from cloning vectors by .

A) a reduced number of cloning sites B) the ability to be introduced into host cells C) the presence of a selectable marker gene D) the presence of necessary sequences to initiate transcription and translation E) the presence of sequences necessary to initiate replication Answer: D Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 6) Assume that a circular plasmid is 3200 base pairs in length and has restriction sites at the following locations: 400, 700, 1400, 2600. Give the expected sizes of the restriction fragments following complete digestion.

A) 400, 800, 1000 (2 of these) B) 400, 1200, 1600 C) 300, 700, 2200 D) 700, 400, 1400, 2600 E) 300, 700, 1000, 1200 Answer: E Section: 17.1 Bloom's Taxonomy: Applying/Analyzing 7) Place the following processes of most cloning experiments in the order in which they would most likely occur.

I. transforming bacteria II. plating bacteria on selective medium III. cutting DNA with restriction endonucleases IV. ligating DNA fragments A) I, II, III, IV B) III, IV, I, II C) IV, III, II, I D) III, I, II, IV E) II, I, III, IV Answer: B Section: 17.1 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

8) One of the primary reasons for generating a large number of clones in a eukaryotic genomic library is that .

A) each clone replicates in coordination with the host chromosome B) lysogenic phages continue to integrate their DNA into the host chromosome, thus reducing the number of desired recombinant clones C) each vector can take up only a relatively small fraction of the eukaryotic DNA D) each ligation product is sequence specific E) the host range of the vector is limited Answer: C Section: 17.2 Bloom's Taxonomy: Remembering/Understanding 9) Which of the following includes an appropriate use of an RNA or DNA probe?

A) locate a specific gene or nucleotide sequence in a DNA library B) direct restriction enzymes to their cut site C) insert nucleotides into specific areas of the genome or library D) induce point mutations in the target DNA Answer: A Section: 17.2 Bloom's Taxonomy: Remembering/Understanding 10) What is a cDNA molecule?

A) a DNA copy of an RNA molecule produced by reverse transcription B) copies of introns with exons excluded C) a molecule rich in restriction cut sites D) DNA sequences that are highly conserved among organisms E) a product of polymerase chain reaction Answer: A Section: 17.2 Bloom's Taxonomy: Remembering/Understanding 11) Which of the following is an accurate comparison of a cDNA library and a genomic library?

A) a cDNA library is used to study enome structure and a genomic library is used to study expression patterns B) a cDNA library is used to study regulatory regions and a genomic library is used to study coding regions C) a cDNA library includes both coding and noncoding DNA and a genomic library includes only genes that have been expressed D) a cDNA library includes only genes that have been expressed and a genomic library includes both coding and noncoding DNA Answer: D Section: 17.2 Bloom's Taxonomy: Applying/Analyzing

3 Copyright © 2021 Pearson Education Ltd.

12) What is the specific application of reverse transcriptase in the preparation of cDNA? A) ligate RNA into cloning vectors B) ligate DNA into cloning vectors C) make DNA copies of mRNA present in the cell D) recognize and cut specific E) generate DNA probes Answer: C Section: 17.2 Bloom's Taxonomy: Remembering/Understanding 13) The invention of yeast artificial chromosomes (YACs) allowed for A) cloning of small DNA inserts in high copy number B) ligate DNA into cloning vectors C) the study of telomeric regions of chromosomes D) the cloning of large DNA inserts E) the expression human genes in yeast Answer: D Section: 17.1 Bloom's Taxonomy: Remembering/Understanding

.

14) How can cDNA libraries be used to gain insight into the genetic basis of the diseased state of a cell?

A) detect mutations in regulatory regions of the genome B) finding genes that have been deleted or duplicated in the genome C) determine which genes are differentially expressed in diseased and normal cells D) determining the location of a gene of interest in the genome Answer: C Section: 17.2 Bloom's Taxonomy: Applying/Analyzing 15) In the context of molecular genetics, reverse transcription PCR (RT-PCR) refers to .

A) assembling a DNA sequence from an mRNA template B) assembling an RNA sequence from a DNA template C) translating in the 3' to 5' direction D) transcribing first, then translating E) making an amino acid sequence from a DNA sequence Answer: A Section: 17.3 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

16) The PCR (polymerase chain reaction) protocol that is currently used in laboratories was facilitated by the discovery of a bacterium called Thermus aquaticus in a hot spring inside Yellowstone National Park, in Wyoming. This organism contains a heat-stable form of DNA polymerase known as Taq polymerase, which continues to function even after it has been heated to 95°C. Why would such a heat-stable polymerase be beneficial in PCR? A) Each cycle includes a "hot" saturation phase (95°C), which allows the primers to anneal to the target DNA. B) Each cycle includes a "hot" denaturation phase (95°C), which serves to sterilize the culture.

C) Each cycle includes a "hot" denaturation phase (95°C), which activates the Taq polymerase.

D) Each cycle includes a "hot" denaturation phase (95°C), which separates the hydrogen bonds that hold the strands of the template DNA together.

E) More than one of the above are correct.

Answer: D Section: 17.3 Bloom's Taxonomy: Remembering/Understanding 17) In what way is the method of PCR similar to that of cloning using vectors?

A) easily introduced into host cells B) generates large copy numbers of a specific segment of DNA C) requires the use of restriction enzymes D) primers take advantage of palindromic sequences Answer: B Section: 17.3 Bloom's Taxonomy: Applying/Analyzing 18) How does one target specific DNA sequences of the template to be amplified by polymerase chain reaction?

A) by designing probes complementary to regions on both sides of the specific target B) by designing primers complementary to regions on both sides of the specific target C) cutting areas on both sides of the target sequence with the same restriction enzyme D) including ddNTPs in the PCR reaction Answer: B Section: 17.3 Bloom's Taxonomy: Applying/Analyzing 19) Nucleic acid blotting is widely used in recombinant DNA technology. In a Southern blot, one generally .

A) hybridizes filter-bound DNA with a DNA probe B) hybridizes filter-bound RNA with a DNA probe C) examines amino acid substitutions with radioactive probes D) cleaves RNA with restriction endonucleases E) ligates DNA with DNA ligase Answer: A Section: 17.4 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

20) A Northern blot involves .

A) hybridizing filter-bound DNA with a DNA probe B) hybridizing filter-bound RNA with a DNA probe C) examines amino acid substitutions with radioactive probes D) cleaves RNA with restriction endonucleases E) ligates DNA with DNA ligase Answer: B Section: 17.4 Bloom's Taxonomy: Remembering/Understanding 21) Fluorescence in situ hybridization (FISH) would most likely be used to study A) tissue specific expression patterns B) the phenotypic effects of a non-functional mutant gene C) DNA sequence variation among individuals D) gene specific mutation rates E) nonhomologous end-joining repair Answer: A Section: 17.4 Bloom's Taxonomy: Applying/Analyzing

.

22) In which of the following biochemical reactions is it common to use ddNTPs (dideoxyribonucleoside triphosphates)?

A) immunoprecipitation B) Sanger sequencing C) restriction digestion D) transfection E) polymerase chain reaction (PCR) Answer: B Section: 17.5 Bloom's Taxonomy: Remembering/Understanding 23) A ddNTP, used often in DNA sequencing, lacks a(n) A) OH; 3' B) methyl; 3' C) carboxyl; 3' D) phosphate; 5' E) none of the listed answers is correct Answer: A Section: 17.5 Bloom's Taxonomy: Applying/Analyzing

6 Copyright © 2021 Pearson Education Ltd.

at the

carbon.

24) What is the function of a ddNTP in DNA sequencing?

A) terminate polynucleotied elongation by polymerase B) produce fluorescent signal as it is incorporated into the polynucleotide C) facilitate reverse transcription D) enhance the efficiency of modified polymerases used in sequencing reactions E) provide targets for sequencing probes Answer: A Section: 17.5 Bloom's Taxonomy: Remembering/Understanding 25) Which of the following is an advantage of next generation sequencing (NGS) over first generation Sanger sequencing?

A) increase the per read accuracy of data B) production of large amounts of data at a cheaper cost C) elimination of reactions involving polymerase D) the ability to sequence RNA directly without producing cDNA first E) the ability to produce much longer sequencing read data Answer: B Section: 17.5 Bloom's Taxonomy: Remembering/Understanding 26) Third generation sequencing methods are unique from other sequencing method in that .

A) produce sequencing data with highest levels of accuracy yet B) they sequence single molecules of single stranded DNA C) elimination of reactions involving polymerase D) elimination of the use of fluorescently tagged nucleotides Answer: B Section: 17.5 Bloom's Taxonomy: Remembering/Understanding 27) A conditional knockout organism is useful tool to study .

A) upregulation of gene expression under different environmental conditions B) downregulation of gene expression under different environmental conditions C) gene specific loss-of-function mutations in selected tissues D) gene and tissue specific mutation rates E) epistatic interactions Answer: C Section: 17.6 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

28) The production of both knockout or transgenic organisms often involves A) preventing homologous recombination B) inserting expression vectors into the host cell C) delivering a vector to embryonic stem cells D) delivering a vector to adult cells in the specific tissue of interest E) bombarding the organism with chemical mutagens Answer: C Section: 17.6 Bloom's Taxonomy: Remembering/Understanding

.

29) What is the function of the CRISPR-Cas system in nature?

A) a key component of the mutation repair pathway B) an enzyme complex involved in homologous recombination C) part of the bacterial immune response against viral infection D) a molecule involved in regulation of expression E) bacterial sex Answer: C Section: 17.7 Bloom's Taxonomy: Remembering/Understanding 30) Which of the following is needed in order to direct CRISPR-Cas9 to cut the specific area of the genome you are trying to manipulate?

A) a specific palindromic sequence in the that area of the genome B) an sgRNA molecule designed to be complementary to the targeted area of the genome C) a PAM motif specific to the targeted area of the genome D) multiple restriction sites in the targeted area of the genome Answer: B Section: 17.7 Bloom's Taxonomy: Applying/Analyzing 31) What is recombinant DNA technology? What are the safety issues related to recombinant DNA technology?

Answer: Recombinant DNA technology refers to the creation of new combinations of DNA molecules that are not normally found in nature. Safety issues generally center on the creation and release (accidental or intentional) of genetically engineered organisms that are a threat to human health or to animals and plants in the environment. Many organisms that are "genetically engineered" carry genes for antibiotic resistance.

Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 32) In the context of recombinant DNA technology, what is meant by the term vector and what are examples of commonly use vectors?

Answer: A vector is a vehicle to carry recombinant DNA molecules into the host cells where independent replication can occur. Most common vectors are plasmids, bacteriophages, and cosmids.

Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 8 Copyright © 2021 Pearson Education Ltd.

33) Molecular biologists rely on many, often sophisticated, techniques to pursue their discipline.

Ultracentrifugation, electron microscopy, X-ray diffraction, electrophoresis, and computer interfacing are fundamental tools. Model organisms provide the raw materials for study. List three organisms (or organismic groups) often used by recombinant DNA technologists and describe a major advantage of each group.

Answer: Bacteriophage: useful vector, relatively simple genome, short generation time.

Bacteria: relatively simple, short generation time, simple growth requirements, well understood genetics, transformable. Viruses: capable of carrying recombinant DNA and infecting eukaryotic cells. Yeast: relatively simple for a eukaryote, short generation time, simple growth requirements, transformable.

Section: 17.1 Bloom's Taxonomy: Applying/Analyzing 34) Assume that a given plasmid vector to be used in a cloning experiment contains 4000 base pairs of DNA. Assume also that the restriction endonuclease Cuj cuts this plasmid at the following sites (starting from an arbitrary zero point): 1000, 1500, and 3000. Given complete digestion of the plasmid with the endonuclease so that only linear fragments are produced, which of the following sizes of DNA are not expected to be produced?

Answer: 500 bp, 1500 bp, and 2000 bp Section: 17.1 Bloom's Taxonomy: Applying/Analyzing 35) Some restriction enzymes cleave DNA in such a manner as to produce blunt ends. Ligation of blunt end fragments is most often enhanced by the use of the enzyme terminal deoxynucleotidyl transferase. Speculate on the function of deoxynucleotidyl transferase in terms of using blunt end fragments in cloning.

Answer: Terminal deoxynucleotidyl transferase extends single-stranded ends by the addition of nucleotide tails. If complementary tails are added, the fragments can hybridize and the recombinant molecules can be ligated.

Section: 17.1 Bloom's Taxonomy: Applying/Analyzing 36) Over the years, sophisticated plasmid vectors have been developed for use in recombinant DNA technology. List at least two features that have been introduced in particularly useful vectors.

Answer: small size to allow large inserts, high copy number, large numbers of unique restriction sites (multiple cloning sites), variety of selection schemes (pigmented colonies, antibiotic resistance) Section: 17.1 Bloom's Taxonomy: Remembering/Understanding

9 Copyright © 2021 Pearson Education Ltd.

37) List, in order, the steps usually followed in producing recombinant DNA molecules in a plasmid vector.

Answer: isolation of DNA (foreign and plasmid), digestion of both foreign and plasmid DNAs with the same restriction endonuclease, ligation of fragments, transformation of host cells Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 38) What might be a reasonable function of restriction endonucleases in a bacterium, distinct from their use by molecular biologists?

Answer: Isolated from bacteria, restriction endonucleases restrict or prevent viral infection by degrading the invading nucleic acid of the virus.

Section: 17.1 Bloom's Taxonomy: Evaluating/Creating 39) Name at least two typical characteristics of a DNA cloning plasmid.

Answer: small size, high copy number, multiple cloning site in a selectable marker Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 40) Which term refers to the process in which DNA can be introduced into host bacterial cells?

Answer: transformation (or transfection in some cases) Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 41) Of what advantage is it to have a multiple cloning site (multiple unique restriction sites) embedded in the lacZ component of a plasmid?

Answer: An insert of DNA in the multiple cloning site inactivates the lacZ component and allows identification of recombinant plasmids under proper genetic and environmental conditions. In addition, it provides the researcher with multiple options when searching for restriction sites in the target gene or adding adapters with restriction sites.

Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 42) Assume that a researcher conducted a cloning experiment using a typical plasmid, transformed an appropriate host bacterial strain, and plated the bacteria on an appropriate X-gal medium. Blue and white colonies appeared. Which of the two types of colonies, blue or white, would more likely contain the recombinant plasmid? Why? Answer: the white colonies because of insertional activation of the lacZ component Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 43) When propagating a clone in the lambda phage, would you have more immediate success if the phage entered the lysogenic cycle or the lytic cycle?

Answer: lytic cycle Section: 17.1 Bloom's Taxonomy: Evaluating/Creating 10 Copyright © 2021 Pearson Education Ltd.

44) If a researcher wishes to clone a gene using typical restriction endonucleases, how does the restriction endonuclease recognize genes in the genome?

Answer: Restriction endonucleases do not recognize functional regions in the genome (genes).

They can recognize only relatively short DNA sequences specific to each type of endonuclease, which have no relationship to functionality.

Section: 17.1 Bloom's Taxonomy: Remembering/Understanding 45) Under ideal conditions, how many copies of all the sequences of the host genome should be represented in a genomic library?

Answer: At least one should be represented. Typically, library construction includes a severalfold greater number of clones than necessary for one representative of each fragment in order to increase the likelihood of cloning difficult fragments and stochastic loss.

Section: 17.2 Bloom's Taxonomy: Applying/Analyzing 46) Compare information contained in a genomic library to that in cDNA library.

Answer: genomic libraries used to produce at least one copy overlapping segments over the entire genome of the organism (coding and noncoding). cDNA libraries produce copies of the regions of the genome that are actively being transcribed or expressed when the library is made.

To this end, cDNA libraries provide information on coding portion of the genome.

Section: 17.2 Bloom's Taxonomy: Applying/Analyzing 47) In the polymerase chain reaction, what is the purpose of the initial high temperature? What is the purpose of cooling in the second step?

Answer: to denature the target (template) DNA; to anneal the primer to the target.

Section: 17.3 Bloom's Taxonomy: Remembering/Understanding 48) Nucleic acid blotting is commonly used in molecular biology. Two types, Southern blots and Northern blots, involve gel electrophoresis and a filter, which holds the nucleic acid. Briefly describe the procedure of "blotting" in this context and differentiate between Southern and Northern blots.

Answer: In a Southern blot, the DNA to be "probed" is cut with a restriction enzyme(s); then the fragments are separated by gel electrophoresis. Alkali treatment of the gel denatures the DNA, which is then "blotted" onto the filter (nylon or nitrocellulose). A labeled probe (RNA or DNA) is then hybridized to complementary fragments on the filter. In a Northern blot, RNA is separated in the gel and "probed" with the labeled DNA. Section: 17.4 Bloom's Taxonomy: Remembering/Understanding

11 Copyright © 2021 Pearson Education Ltd.

49) Assume that you have cut λ DNA with the restriction enzyme HindIII. You separate the fragments on an agarose gel and stain the DNA with ethidium bromide. You notice that the intensity of the stain is less in the bands that have migrated closer to the "+" pole. Give an explanation for this finding.

Answer: Because the smaller fragments migrate toward the "+" pole, away from the origin, they bind less stain than the larger fragments near the origin.

Section: 17.4 Bloom's Taxonomy: Applying/Analyzing 50) Discuss differences between first and second generation sequencing technology with respect to data production.

Answer: First generation or Sanger sequencing produces sequence data at a relatively slower rate than next generation sequencing (NGS). Sanger sequencing produces sequence read length typically under 1000bp with up to 100 sequences produced simultaneously. NGS on the other hand produces tens of thousands of sequences simultaneously producing much larger amounts of data at a reduced cost.

Section: 17.5 Bloom's Taxonomy: Evaluating/Creating 51) Compare the traditional methods of creating knock-out organism to that of transgenic (knock-in) organism.

Answer: In creating a KO organism a targeting vector is used to insert a non-functional for of the gene of interest at a specific locus. The nonfunctional gene is created by inserting a selectable marker to disrupt the reading frame of the gene. Embryonic stem cells or fertilized eggs are transformed with the targeting vector and recombinant cells are selected. ES cells are microinjected into blastocysts. Successfully transformed organisms are used to create heterozygotes which are then mated to produce KO mutants. Transgenic organisms are produced in the same basic manner except a targeting vector is not necessary as the location of the gene insertion does not matter and a promoter must be included with the inserted gene to allow induction of expression.

Section: 17.6 Bloom's Taxonomy: Evaluating/Creating 52) Describe the relationship between homology directed repair (HDR) and precise genome editing using CRISPR-Cas9.

Answer: In CRISPR-Cas9 a double stranded break is induced in a targeted area of the genome that is complementary to an sgRNA molecule designed by the scientist. This double stranded break activates the cells homology directed repair (HDR) machinery. The HDR molecules are manipulated into using a researcher designed donor template as the template for repair. This donor template includes the desired edit which is incorporated into the gene or genome.

Section: 17.7 Bloom's Taxonomy: Evaluating/Creating

12 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 18 Genomics, Bioinformatics, and Proteomics 1) A newly sequenced gene is cloned and expressed in a bacterial cell. The bioinformatics databases suggest that this gene produces an enzyme that catalyzes the conversion of carbon dioxide to bicarbonate. After weeks of experiments, a scientist determined the enzyme has a completely different function. This is an example of incorrect .

A) sequencing B) annotation C) BLAST parameters D) enzyme experiments E) qPCR Answer: B Section: 18.2 Bloom's Taxonomy: Applying/Analyzing 2) Which E-value represents the highest probability of a BLAST result not being chance?

A) 1 B) 0.1 C) 0.01 D) 1 × 10-3 E) 1 × 10-4 Answer: E Section: 18.2 Bloom's Taxonomy: Evaluating/Creating 3) Predict the number of potential reading frames in a double-stranded DNA within a protein coding gene.

A) 1 B) 2 C) 4 D) 5 E) 6 Answer: E Section: 18.2 Bloom's Taxonomy: Evaluating/Creating

1 Copyright © 2021 Pearson Education Ltd.

4) A new gene is discovered during the sequencing of a bacterial genome. After a BLAST analysis, the new gene aligns with a known sequence from the same bacterium. The sequence identity is 95%. Which of the following statements best describes the relationship of the newly discovered gene and the known gene?

A) It is an ortholog to the known gene and is a result of a recent gene duplication.

B) It is a paralog to the known gene and is a result of a recent gene duplication.

C) It is an ortholog to the known gene and is a result of an older gene duplication.

D) It is a paralog to the known gene and is a result of an older gene duplication.

Answer: B Section: 18.2 Bloom's Taxonomy: Evaluating/Creating 5) The Human Genome Project, which got under way in 1990, was an international effort to .

A) determine the base sequence of the human genome and to identify all the genes within it B) collect samples of cells from all parts of the world in order to preserve human genetic diversity C) collect plant seeds in order to reduce the impact of human activity on plant extinction D) clone deleterious genes from humans and study their mode of action E) clone beneficial genes from humans for eventual use in gene therapy Answer: A Section: 18.3 Bloom's Taxonomy: Remembering/Understanding 6) The sequencing of the human genome led to the realization that chromosome 19 contains many genes while chromosomes 13 and Y contain relatively few. What does this finding imply about the density of genes in a genome?

A) Genes are uniformly distributed throughout the genome.

B) Genes are uniformly distributed on chromosomes but not through the entire genome.

C) Genes are not uniformly distributed and appear in clusters separated by gene deserts.

D) Genes have no organization and randomly appear throughout the genome and chromosomes.

Answer: C Section: 18.3 Bloom's Taxonomy: Evaluating/Creating 7) Under the umbrella of the Human Genome Project, a program was set up to involve scientists, health professionals, policy makers, and the public to consider the types of issues arising from the project and to ensure that personal genetic information would be safe guarded. What is it called?

A) ELSI (Ethical, Legal, and Social Implications) B) HGSI (Human Genome Sequencing Initiative) C) WGS (Whole Genome Systems) D) HGPC (Human Genome Project Consortium) Answer: A Section: 18.3 Bloom's Taxonomy: Remembering/Understanding 2 Copyright © 2021 Pearson Education Ltd.

8) What is one cause for the 21,000 protein encoding sequences in the human genome producing between 200,000 and 1 million different proteins?

A) restriction enzymes B) CRISPR-Cas systems C) alternative splicing D) CNVs E) SNPs Answer: C Section: 18.3 Bloom's Taxonomy: Remembering/Understanding 9) Numerous scientists around the world have proposed to sequence 10,000 vertebrate genomes in five years. What is the name of this plan?

A) Genome 10K B) Bigger Than Life Plan C) 10K or Bust D) Vertebrate Beginnings E) Vertebrate Enlightenment Answer: A Section: 18.4 Bloom's Taxonomy: Remembering/Understanding 10) Which of the following sequences is translated into the amino acid sequence of a protein in eukaryotes?

A) promoter B) enhancer C) silencer D) exon E) TATA box Answer: D Section: 18.2 Bloom's Taxonomy: Remembering/Understanding 11) Splice sites between exons and introns can be predicted since most introns begin with and end with .

A) UA; GC B) CA; AT C) AG; CT D) AT; CA E) CT; AG Answer: E Section: 18.2 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

12) How many genes did the J. Craig Venter Institute determine were necessary for a minimal bacterial genome?

A) 149 B) 256 C) 473 D) 525 Answer: C Section: 18.9 Bloom's Taxonomy: Remembering/Understanding 13) Prokaryotic gene regulation utilizes which of the following sequences?

A) introns B) promoters C) exons D) polyadenylation sequences Answer: B Section: 18.2 Bloom's Taxonomy: Remembering/Understanding 14) The typical gene density in euchromatic regions of the Drosophila genome is estimated to be about 1 gene per .

A) 116.8 Mb B) 20.7 Mb C) 70,000 bp D) 8730 bp Answer: D Section: 18.9 Bloom's Taxonomy: Remembering/Understanding 15) What overlapping fragments are utilized to align individual sequences of a continuous chromosome during shotgun sequencing?

A) transcripts B) telomeres C) contigs D) exons E) pseudogenes Answer: C Section: 18.1 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

16) Which of the following is not an emerging "-omics" field?

A) pharmacogenomics B) metagenomics C) glycomics D) chromosomics E) transcriptomics Answer: D Section: 18.4 Bloom's Taxonomy: Remembering/Understanding 17) Personal genome sequencing assists with the recognition and determination of what?

A) inheritable diseases and their alleles B) silent mutations in the genome C) somatic genome mosaicism D) locations for introns Answer: C Section: 18.4 Bloom's Taxonomy: Remembering/Understanding 18) Somatic genome mosaicism is best described as .

A) differences in genome sequence between cells of an organism due to errors in sequencing B) differences in genome sequence between cells of an organism due to errors incurred during division C) differences in gene location in the genome D) differences in proteins expressed between cells of an organism Answer: B Section: 18.4 Bloom's Taxonomy: Remembering/Understanding 19) Describe the difference between whole-genome sequencing and whole-exome sequencing.

Answer: Whole-genome sequencing sequences the entire genome of an organism. Whole-exome sequencing generates deeper reads of only the exons of an organism's genome.

Section: 18.1, 18.4 Bloom's Taxonomy: Applying/Analyzing 20) The project seeking to determine transcriptional start sites, promoters, enhancers, and other functional elements of the genome is called .

A) PANTHER B) GenBank C) NCBI D) NIH E) ENCODE Answer: E Section: 18.4 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

21) Nutrigenomics is the study of .

A) the vitamins and minerals in the genome B) the relationship between diet and the genome C) the effect of genomics on dietary desires D) the effect of nutrition on genome sequences Answer: B Section: 18.4 Bloom's Taxonomy: Remembering/Understanding 22) Several years ago, scientists compared the genome of a woolly mammoth to that of an African elephant and found them to be 98.5% identical. This is an example of what type of "omics"?

A) stone-age genomics B) space-age genomics C) proteomics D) exomics E) extinctomics Answer: A Section: 18.4 Bloom's Taxonomy: Remembering/Understanding 23) One major difference between prokaryotic and eukaryotic genes is that eukaryotic genes can contain internal sequences, called , which get removed in the mature message.

A) introns B) exons C) promoters D) enhancers E) silencers Answer: A Section: 18.3 Bloom's Taxonomy: Remembering/Understanding 24) The dog (Canis familiaris) genome has recently been sequenced. About how many of the dog's genes are shared with humans?

A) <25% B) ~50% C) ~75% D) >95% Answer: C Section: 18.5 Bloom's Taxonomy: Applying/Analyzing

6 Copyright © 2021 Pearson Education Ltd.

25) Aligning genome sequences from different organisms to study gene and genome evolution is referred to as .

A) comparative genomics B) metagenomics C) evogenomics D) stone-age genomics Answer: A Section: 18.5 Bloom's Taxonomy: Remembering/Understanding 26) The advent of genomic sequencing and annotations using bioinformatics has provided data about prokaryotic genomes. Which of the following claims could be refuted using this data?

A) Bacterial genomes are always smaller than eukaryotic genomes.

B) All bacterial genomes are circular.

C) Prokaryotes have a single chromosome.

D) Prokaryote genomes contain introns.

Answer: D Section: 18.2 Bloom's Taxonomy: Applying/Analyzing 27) Which of the following is a major contributor to variation in gene density in eukaryotes?

A) repetitive sequences B) splice sites C) restriction sites D) heterochromatin Answer: A Section: 18.3 Bloom's Taxonomy: Evaluating/Creating 28) Describe how comparative genomics is aiding in the discovery and treatment of human disease.

Answer: Comparative genomics is revealing orthologous genes in lower eukaryotes that provide opportunity to study disease states without the threat to human life.

Section: 18.5 Bloom's Taxonomy: Applying/Analyzing 29) Which of the following organisms shares the highest percentage of genes with humans?

A) yeast B) fruit fly C) pig D) mouse Answer: C Section: 18.5 Bloom's Taxonomy: Applying/Analyzing

7 Copyright © 2021 Pearson Education Ltd.

30) Genomes from several primates have been sequenced and annotated. These primates range from chimpanzee to Rhesus monkey to Neanderthal and modern humans. What is a potential advantage of having so many primate sequences available to researchers?

A) Phenotypes can be determined from the genomes.

B) Comparative genomics allows for determining evolutionary relationships.

C) The primate sequences can be compared to learn about social behavior.

D) The primate sequences can be compared to learn about dietary trends.

Answer: B Section: 18.5 Bloom's Taxonomy: Applying/Analyzing 31) With the rise of drug-resistant bacteria in hospitals there has been an effort to develop new antibiotics as well as novel methods to combat resistant infections. The sequencing of entire communities of microbes including environmental samples has led to the discovery of millions of uncharacterized bacteria that could provide new drugs. This sequencing method is called .

A) whole-community sequencing B) biome genomics C) environmental genomics D) shotgun genomics Answer: C Section: 18.6 Bloom's Taxonomy: Remembering/Understanding 32) A microbiome is essential for human infant development. Where does an infant obtain the origins of its microbiome?

A) the mother B) the father C) the hospital D) The infant has a microbiome at conception.

Answer: A Section: 18.6 Bloom's Taxonomy: Remembering/Understanding 33) The data from metagenomic experiments is commonly displayed in a Venn diagram where there is overlap between datasets that suggest correlation with disease states. If a patient exhibits genomic sequences from multiple overlapping regions of the diagram what happens to their predisposition to the disease?

A) It goes up.

B) It goes down.

C) It stays the same.

D) Genetics do not play a role in disease predisposition.

Answer: A Section: 18.6 Bloom's Taxonomy: Applying/Analyzing

8 Copyright © 2021 Pearson Education Ltd.

34) Transcriptomics is .

A) the quantification of gene expression B) measuring how much DNA is in the genome C) determining epigenetic markers D) determining phenotype Answer: A Section: 18.7 Bloom's Taxonomy: Remembering/Understanding 35) Transcriptomics requires the isolation of RNA to generate sequence data. If a scientist wants to use next-generation sequencing to perform their experiment which of the following is necessary?

A) reverse transcription B) protein quantification C) heterochromatin D) euchromatin Answer: A Section: 18.7 Bloom's Taxonomy: Evaluating/Creating 36) DNA microarray analysis is excellent for .

A) studying one gene's expression very closely B) studying all of a sample's expressed genes simultaneously C) studying all of a sample's genes simultaneously D) studying all of a sample's unexpressed genes simultaneously Answer: B Section: 18.7 Bloom's Taxonomy: Remembering/Understanding 37) A drawback to using microarrays is .

A) commercial microarrays have high variability in quality B) researcher-made arrays have high variability C) the spots are very tiny D) the arrays do not cover enough of the genome Answer: B Section: 18.7 Bloom's Taxonomy: Remembering/Understanding 38) Using RNA-seq generates data in situ. What data is not provided by RNA-seq?

A) gene expression levels B) sequence of the RNA C) location of transcripts in cells and tissues D) amount of protein produced by RNA Answer: D Section: 18.7 Bloom's Taxonomy: Applying/Analyzing 9 Copyright © 2021 Pearson Education Ltd.

39) Proteomics is the "-omics" study of proteins, investigating the complete set of proteins in a cell. What data does proteomics provide?

A) amount of protein in a cell B) posttranslational modification to proteins C) the similarity between genes coding for a protein D) the gene sequence for the protein Answer: B Section: 18.8 Bloom's Taxonomy: Remembering/Understanding 40) A common first separation technique in 2D gels is isoelectric focusing. What property is isoelectric focusing taking advantage of?

A) size B) charge C) amino acid content D) the pH at which the overall charge is zero Answer: D Section: 18.8 Bloom's Taxonomy: Evaluating/Creating 41) Proteins separated on 2D gels provide information on their pH and size but not their identity.

MALDI-mass spectrometry provides this missing information by which of the following?

A) generating sequence data for the protein B) generating a m/z ratio "fingerprint" for the protein C) generating a sequence "fingerprint" for the protein D) generating a gene "fingerprint" for the protein Answer: B Section: 18.8 Bloom's Taxonomy: Remembering/Understanding 42) What is the purpose of the Human Genome Project-Write (HGP-Write)?

A) recoding a genome to create cells that are resistant to cancer B) to generate an organism that can degrade pollutants C) recoding a genome to create cells that are resistant to infection by viruses D) to generate an organism that can produce pharmaceuticals Answer: C Section: 18.9 Bloom's Taxonomy: Remembering/Understanding 43) A proteome is defined as .

A) a catalog of known structures for all proteins expressed in a cell B) the complete set of proteins encoded by a given genome C) the complete set of all posttranslational modifications occurring in a cell D) a catalog of what proteins are being expressed from a particular genome Answer: B Section: 18.8 Bloom's Taxonomy: Remembering/Understanding 10 Copyright © 2021 Pearson Education Ltd.

44) What does MALDI stand for?

A) Matrix-Assisted Laser Desorption Ionization B) Maser-Assisted Laser Desorption Ionization C) Matrix-Assisted Ligand Desorption Ionization D) Maser-Assisted Ligand Desorption Ionization Answer: A Section: 18.8 Bloom's Taxonomy: Remembering/Understanding 45) Propose a reason why the human proteome is larger than the human genome.

Answer: Many human genes produce more than one protein through alternative splicing, thus enabling human cells to produce a much larger number of proteins (perhaps as many as 200,000) from only _20,000 genes (Chapter 18, Page 354).

Section: 18.3 Bloom's Taxonomy: Evaluating/Creating

11 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 19 The Genetics of Cancer 1) Driver mutations provide a growth advantage to a tumor cell. Which type of mutation is known to accumulate in cancer cells but has no direct contribution to the cancer phenotype?

A) alteration mutations B) passenger mutations C) carrier mutations D) indirect mutations E) insignificant mutations Answer: B Section: 19.1 Bloom's Taxonomy: Remembering/Understanding 2) A derangement in which of the following general mechanisms appears to be fundamental in the formation of cancer cells?

A) DNA repair B) inversion C) methylation D) protein phosphorylation E) differentiation Answer: A Section: 19.2 Bloom's Taxonomy: Remembering/Understanding 3) What protein combines with cyclins to exert local control of the cell cycle?

A) cyclin-dependent kinase B) phosphatase C) ATPase D) integrase E) hexokinase Answer: A Section: 19.3 Bloom's Taxonomy: Remembering/Understanding 4) Which of the following proteins is a transcription factor that can induce cell death (apoptosis) within damaged cells?

A) CDK B) ABL C) cyclin D) p53 E) phosphokinase Answer: D Section: 19.4 Bloom's Taxonomy: Remembering/Understanding

1 Copyright © 2021 Pearson Education Ltd.

5) Mutations in the ras gene family induce normally quiescent cells to proceed into the replication cycle. This converts the ras gene from a/an gene to a/an gene.

A) proto-oncogene; oncogene B) oncogene; proto-oncogene C) mutant; oncogene D) tumor suppressor; proto-oncogene E) pseudooncogene; proto-oncogene Answer: A Section: 19.4 Bloom's Taxonomy: Applying/Analyzing 6) Mutant versions of genes involved in promoting the development of cancer are known as .

A) tumor suppressors B) proto-oncogenes C) oncogenes D) malignant genes E) attenuators Answer: C Section: 19.4 Bloom's Taxonomy: Remembering/Understanding 7) In sporadic cases of retinoblastoma, only two gene mutations are thought to be necessary in the same cell for a tumor to develop. This happens when an organism inherits a mutant allele and a wild-type allele from the parents. When the wild-type allele is mutated, resulting in tumorigenesis, this process is called, .

A) loss of heterozygosity B) recombination C) attenuation D) gain of heterozygosity E) metastasizing Answer: A Section: 19.6 Bloom's Taxonomy: Applying/Analyzing 8) What is the name of the protein that appears to regulate the entry of cells into an S phase? This protein is also known as the "guardian of the genome."

A) p34 B) p102 C) cyclin D) p53 E) phosphokinase Answer: D Section: 19.4 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

9) Provide a definition of cancer at the genetic level.

Answer: Cancer is a genetic disorder that can result from mutation of a given gene or genes, which may produce a defective gene product or a change in the timing or amount of gene expression. Such mutations alter cell-cycle control. Some cancers show familial distributions.

Section: 19.1 Bloom's Taxonomy: Remembering/Understanding 10) Which of the following terms refers to the spreading of cancerous cells to colonize other parts of the body?

A) proliferation B) apoptosis C) metastasis D) clonal expansion Answer: C Section: 19.1 Bloom's Taxonomy: Remembering/Understanding 11) Chronic myelogenous leukemia (CML) appears to be associated with a chromosomal rearrangement that involves the ABL gene whose product is involved in which of the following mechanisms?

A) methylation B) phosphorylation C) acetylation D) hydroxylation Answer: B Section: 19.2 Bloom's Taxonomy: Remembering/Understanding 12) Signal transduction is best described as .

A) communication between cells through physical interaction B) transmission of external signals to the nucleus of a cell C) transmission of signals from cancer cells to normal cells D) communication between nuclei of two organisms Answer: B Section: 19.3 Bloom's Taxonomy: Remembering/Understanding 13) Why do cancer researchers study molecular events associated with mitosis?

Answer: While mitosis is a basic process related to genetic and general biological studies, it is also a significant event in the cell cycle. The cell cycle is regulated by a variety of gene products, which, when altered by mutation, may lead to cancer.

Section: 19.3 Bloom's Taxonomy: Evaluating/Creating

3 Copyright © 2021 Pearson Education Ltd.

14) What is the significance of CDK/cyclin interactions with respect to cancer cell proliferation?

Answer: Many CDK/cyclin interactions serve as checkpoints for progression through the cell cycle. In cancerous cells mutations in genes encoding either CDK or its partner cyclin could deregulate key checkpoints resulting in cancer cell proliferation.

Section: 19.3 Bloom's Taxonomy: Evaluating/Creating 15) A loss-of-function mutation in caspase genes could inhibit the apoptotic response in cells.

Develop an argument for why this type of mutation could be considered a passenger mutation rather than a driver mutation.

Answer: The loss of function of the proapoptotic genes are a passenger mutation as they do not contribute to tumorigenesis but confer a selective advantage by eliminating the apoptotic defense mechanism.

Section: 19.1, 19.3 Bloom's Taxonomy: Evaluating/Creating 16) Describe why cancer cells are more susceptible to chemotherapies and radiation therapies than normal cells.

Answer: Cancer cells suffer from mutations that affect DNA repair enzymes and are constantly proliferating resulting in a much higher mutation accumulation rate that could lead to apoptosis or necrosis. Normal cells are nonproliferating and have functional DNA repair mechanisms that prevent apoptosis.

Section: 19.3 Bloom's Taxonomy: Applying/Analyzing 17) Which of the following factors is typically monitored during the M checkpoint?

A) cell size B) the presence of DNA damage C) the completion of DNA replication D) attachment of spindle fibers to kinetochores Answer: D Section: 19.3 Bloom's Taxonomy: Remembering/Understanding 18) Which of the following genes is an example of a tumor-suppressor?

A) c-myc B) c-kit C) BRCA1 D) RAR Answer: C Section: 19.4 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

19) Which molecule pair combines to directly regulate progression through the cell cycle?

A) p21 and MDM2 B) CDK and cyclin C) E-cadherin and TIMP D) pRB and GDP Answer: B Section: 19.3 Bloom's Taxonomy: Remembering/Understanding 20) Describe the general relationship that may exist between mutations and cancer.

Answer: Control of the cell cycle is dependent on a variety of gene-produced proteins such as kinases and cyclins as well as other related factors. Mutations in genes that encode these proteins may disrupt normal cell-cycle control. The G1 checkpoint is altered in some forms of cancer, and mutant cyclins have been shown to be related to a gene product that is overexpressed in some forms of leukemia. Nonmutant genes often suppress the formation of cancer by exerting control over the cell cycle. When mutated, such control may be lost.

Section: 19.4 Bloom's Taxonomy: Applying/Analyzing 21) What translocation is present in white blood cells from patients with chronic myelogenous leukemia?

A) t(5;17) B) t(2;8) C) t(9;22) D) t(1;11) Answer: C Section: 19.2 Bloom's Taxonomy: Remembering/Understanding 22) Which gene is correctly paired with the normal function of its protein product?

A) c-myc; DNA repair B) RB1; binds E2F C) TP53; tyrosine kinase D) c-kit; transcription factor Answer: B Section: 19.4 Bloom's Taxonomy: Remembering/Understanding 23) Which statement about the ras gene family and its products is false?

A) Ras genes are tumor-suppressors.

B) Ras genes encode signal transduction molecules.

C) Ras genes are mutated in more than 30% of human tumors.

D) Ras proteins are active when bound to GTP. Answer: A Section: 19.4 Bloom's Taxonomy: Remembering/Understanding 5 Copyright © 2021 Pearson Education Ltd.

24) Which statement best describes an example of a mutation in a tumor-suppressor gene?

A) overexpression of a signal transduction molecule that promotes cell growth B) a mutation that places a receptor tyrosine kinase in a continuous "on" state C) a mutation that causes cyclins to be expressed at all times D) a mutation that makes a cell unable to undergo programmed cell death Answer: D Section: 19.4 Bloom's Taxonomy: Applying/Analyzing 25) What type of gene encodes products that normally promote cellular division?

A) proto-oncogene B) post-oncogene C) oncogene D) cancer gene Answer: A Section: 19.4 Bloom's Taxonomy: Remembering/Understanding 26) Which category of genetic changes does not lead to the formation of oncogenes?

A) point mutations B) translocations C) copy number changes D) correct DNA repair Answer: D Section: 19.4 Bloom's Taxonomy: Remembering/Understanding 27) The following events occur when p53 regulates the cell cycle. Place these events in the correct sequence.

1) p53 is activated by posttranslational modifications.

2) The cell cycle is arrested, and apoptosis is initiated.

3) p53 regulates the transcription of multiple target genes.

4) DNA damage causes MDM2 to dissociate from p53.

A) 4, 1, 3, 2 B) 4, 3, 1, 2 C) 3, 4, 2, 1 D) 2, 1, 4, 3 Answer: A Section: 19.4 Bloom's Taxonomy: Evaluating/Creating

6 Copyright © 2021 Pearson Education Ltd.

28) p53 is best described as .

A) an oncogene B) a proto-oncogene C) a tumor-suppressor gene D) a signal transduction molecule Answer: C Section: 19.4 Bloom's Taxonomy: Remembering/Understanding 29) Mutation in which gene is the first step in the development of colorectal cancer?

A) APC B) KRAS C) PI3K D) TGFAnswer: A Section: 19.1 Bloom's Taxonomy: Remembering/Understanding 30) Individuals with xeroderma pigmentosum have an extreme sensitivity to A) human-made carcinogens B) nuclear radiation C) UV light D) tobacco smoke Answer: C Section: 19.2 Bloom's Taxonomy: Remembering/Understanding

.

31) Which of the following genes is most likely to become an oncogene when mutated?

A) RB1 B) TP53 C) BRCA2 D) RAR Answer: D Section: 19.4 Bloom's Taxonomy: Remembering/Understanding 32) Which mechanism is least likely to inactivate a tumor-suppressor gene?

A) chemical modification B) mutation C) gene duplication D) loss of heterozygosity Answer: C Section: 19.4 Bloom's Taxonomy: Applying/Analyzing

7 Copyright © 2021 Pearson Education Ltd.

33) In what way can loss of heterozygosity lead to cancer?

Answer: Loss of heterozygosity occurs when a cancer-producing gene that exists in the heterozygous state becomes exposed through deletion, mutation, recombination, or chromosomal aberration. Such a gene now is either homozygous or hemizygous.

Section: 19.6 Bloom's Taxonomy: Applying/Analyzing 34) Explain the genetic difference between familial retinoblastoma and sporadic retinoblastoma.

Answer: Those with the familial form start out being heterozygous for the retinoblastoma gene, whereas those with the sporadic form start out being homozygous wild-type.

Section: 19.6 Bloom's Taxonomy: Evaluating/Creating 35) Which human virus is not a significant contributor to virus-induced cancers?

A) Epstein-Barr B) HIV-1 C) hepatitis C D) influenza A Answer: D Section: 19.7 Bloom's Taxonomy: Remembering/Understanding 36) Many viruses are known to insert or integrate their own genomes into host genomes, either disrupting or upregulating host genes in the process. In what way might viral gene integration contribute to cancer formation?

Answer: Viral infection could alter the expression of a proto-oncogene, generating an oncogene.

A virus may also introduce an oncogene during an infection. In either case, the cell cycle can be upregulated by a virus.

Section: 19.7 Bloom's Taxonomy: Applying/Analyzing 37) All of these are known to cause cancer EXCEPT A) radiation B) chronic infections C) some viruses D) DNA repair Answer: D Section: 19.7 Bloom's Taxonomy: Remembering/Understanding

.

38) Provide a simple definition of a carcinogen.

Answer: A cancer-causing agent Section: 19.1 Bloom's Taxonomy: Remembering/Understanding

8 Copyright © 2021 Pearson Education Ltd.

39) As more is learned about cancer, it has become clear that cancer, with few exceptions, .

A) has no genetic basis B) is 100% dependent on inherited genetics C) is only dependent on the inherited genetics D) is a result of genetics and environmental factors Answer: D Section: 19.1 Bloom's Taxonomy: Remembering/Understanding 40) Many of the known cancers are a result of A) the genetic stability of the human genome B) the genetic instability of the human genome C) the non-carcinogenic nature of our environment D) exercising regularly Answer: B Section: 19.2 Bloom's Taxonomy: Remembering/Understanding

.

41) Any agent that causes damage to DNA is a potential A) carcinogen B) oncogene C) pollutant D) proto-oncogene Answer: A Section: 19.1 Bloom's Taxonomy: Remembering/Understanding 42) All of the following are checkpoints in the cell cycle EXCEPT A) G1/S B) G2/M C) M D) M/G1 Answer: D Section: 19.3 Bloom's Taxonomy: Remembering/Understanding 43) Which mechanism is directly dependent upon caspases?

A) apoptosis B) DNA replication C) G1/S checkpoint D) chromosome condensation Answer: A Section: 19.3 Bloom's Taxonomy: Remembering/Understanding

9 Copyright © 2021 Pearson Education Ltd.

.

.

44) A tumor-suppressor gene normally functions to A) suppress cell division B) induce cell division C) repair cancer cells D) create mutations forming cancer cells Answer: A Section: 19.4 Bloom's Taxonomy: Remembering/Understanding

.

45) Human-made, industrial chemicals are thought to contribute to approximately percent of all cancers.

A) 1 B) 5 C) 10 D) 25 Answer: C Section: 19.7 Bloom's Taxonomy: Remembering/Understanding 46) When considered as a root cause for cancer, which of the following is linked to at least 17 types of human cancer?

A) eating a low-fat diet B) smoking C) eating a high-fat diet D) not exercising regularly Answer: B Section: 19.7 Bloom's Taxonomy: Remembering/Understanding 47) Examined cancer cells from the lungs of smokers demonstrate which of the following characteristics?

A) normal mutation rates compared to non-smokers B) change in DNA methylation patterns C) mutation rates that are similar to those in a smoker's heart D) the occurrence of tumor-suppressor mutations is less in the lung than the throat Answer: B Section: 19.7 Bloom's Taxonomy: Remembering/Understanding

10 Copyright © 2021 Pearson Education Ltd.

48) After entering the cardiovascular system or the lymphatic system, what percentage of circulating cancer cells survive to establish metastatic tumors?

A) 0.01% B) 0.1% C) 1.0% D) 10% Answer: A Section: 19.6 Bloom's Taxonomy: Remembering/Understanding 49) Propose a reason why most inherited cancers are due to mutations in genes that control DNA repair rather than oncogenes.

Answer: Loss of cell cycle control during developmental stages would be more likely to produce a nonviable embryo.

Section: 19.2 Bloom's Taxonomy: Evaluating/Creating

11 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 20 Quantitative Genetics and Multifactorial Traits 1) Environmental factors typically influence inheritance of A) multiple alleles B) codominance C) trihybrid crosses D) polygenic traits E) dominantly inherited traits Answer: D Section: 20.1 Bloom's Taxonomy: Remembering/Understanding

.

2) Quantitative inheritance involves the interaction of a number of gene loci. The pattern of genetic transmission typical of quantitative inheritance is .

A) discontinuous distributions such as 3:1 B) typical of Mendelian inheritance C) continuous variation of phenotypic expression D) a 9:3:3:1 ratio E) usually a pattern that clearly reflects dominance and recessiveness Answer: C Section: 20.1 Bloom's Taxonomy: Remembering/Understanding 3) Assume that a cross is made between tall and dwarf tobacco plants. The F1 generation showed intermediate height, whereas the F2 generation showed a distribution of height ranging from tall to dwarf, like the original parents, and many heights between the extremes. These data are consistent with which one of the following modes of inheritance?

A) multiple-factor inheritance B) alternation of generations C) codominance D) incomplete dominance E) hemizygosity Answer: A Section: 20.1 Bloom's Taxonomy: Applying/Analyzing 4) Bell-shaped distributions produced by plotting results of F2 and F3 crosses are typical of which type of inheritance?

A) multiple-factor inheritance B) alternation of generations C) codominance D) incomplete dominance E) hemizygosity Answer: A Section: 20.1 Bloom's Taxonomy: Remembering/Understanding 1 Copyright © 2021 Pearson Education Ltd.

5) Characteristics exhibited by continuously varying traits include A) sex-linked traits only B) autosomal traits only C) quantitative traits D) a 9:3:3:1 ratio E) 3:1 and 1:1 ratios Answer: C Section: 20.1 Bloom's Taxonomy: Applying/Analyzing

.

6) In an experiment, you cross a line homozygous for additive alleles with a line homozygous for nonadditve alleles. The F1s are crossed to produce the F2 generation in which there are seven phenotypic categories. How many gene pairs are likely involved in this trait?

A) 15 B) 8 C) 7 D) 3 E) 2 Answer: D Section: 20.1 Bloom's Taxonomy: Applying/Analyzing 7) The 9:3:3:1 ratio is typical of a dihybrid cross in which complete dominance and independent assortment occur. What is the dihybrid ratio with independent assortment of polygenes with additive and nonadditive alleles?

A) 9:3:3:1 B) 1:4:6:4:1 C) 1:1:1:1 D) 1:2:1 E) 1:2:4:2:1 Answer: B Section: 20.1 Bloom's Taxonomy: Applying/Analyzing 8) How many gene pairs are involved in generating a typical 1:8:28:56:70:56:28:8:1 ratio in the F2 generation of an experimental cross?

A) 2 B) 3 C) 4 D) 5 E) 6 Answer: C Section: 20.1 Bloom's Taxonomy: Applying/Analyzing

2 Copyright © 2021 Pearson Education Ltd.

9) In the early part of the twentieth century, Nilsson-Ehle and others described experiments showing that multiple loci may be involved in the inheritance of certain traits. Such patterns are often called .

A) polygenic B) epigenetic C) epistatic D) nonallelic E) multiallelic Answer: A Section: 20.1 Bloom's Taxonomy: Remembering/Understanding 10) Offspring produced by an experimental cross have the following heights (cm): 10, 10, 12, 12, 14, 14, 16, 16, 18, 18. Calculate the mean.

A) 10 B) 12 C) 14 D) 16 E) 18 Answer: C Section: 20.2 Bloom's Taxonomy: Evaluating/Creating 11) Offspring produced by an experimental cross have the following heights (cm): 10, 10, 12, 12, 14, 14, 16, 16, 18, 18. Calculate the variance.

A) 11.4 B) 10 C) 8.9 D) 12 E) 18 Answer: C Section: 20.2 Bloom's Taxonomy: Applying/Analyzing 12) What statistic is used for describing sample variability by averaging the squared distance of all measurements from the mean?

A) standard deviation B) standard error of the mean C) covariance D) variance E) correlation coefficient Answer: D Section: 20.2 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

13) To estimate how much the means of a variety of like samples drawn from the same population might vary, which statistic is often used?

A) standard deviation B) standard error of the mean C) covariance D) variance E) correlation coefficient Answer: B Section: 20.2 Bloom's Taxonomy: Remembering/Understanding 14) You are studying the phenotypic traits of bone density and muscle mass in cattle and find that the two traits have a correlation coefficient of 0.97. What is the interpretation of this statistic?

A) cattle with higher bone density tend to have larger muscle mass B) cattle with higher bone density tend to have lower muscle mass C) there is no relationship between bone density and muscle mass D) there is a cause and effect relationship between the two traits E) the two traits are controlled by the same genes Answer: A Section: 20.2 Bloom's Taxonomy: Applying/Analyzing 15) You are studying the phenotypic traits of bone density and muscle mass in cattle and find that the two traits have a correlation coefficient of -1.0. What is the interpretation of this statistic?

A) cattle with higher bone density tend to have larger muscle mass B) cattle with higher bone density tend to have lower muscle mass C) there is no relationship between bone density and muscle mass D) there is a cause and effect relationship between the two traits E) the two traits are controlled by the same genes Answer: B Section: 20.2 Bloom's Taxonomy: Applying/Analyzing 16) You are studying the phenotypic traits of bone density and muscle mass in cattle and find that the two traits have a correlation coefficient of 0. What is the interpretation of this statistic?

A) cattle with higher bone density tend to have larger muscle mass B) cattle with higher bone density tend to have lower muscle mass C) there is no relationship between bone density and muscle mass D) there is a cause and effect relationship between the two traits E) the two traits are controlled by the same genes Answer: C Section: 20.2 Bloom's Taxonomy: Applying/Analyzing

4 Copyright © 2021 Pearson Education Ltd.

17) The term heritability is used to describe .

A) the extent to which an individual's phenotype is due to their genotype B) how much of a trait is due to genetics C) how related any two individuals are in a population D) the proportion of phenotypic variation in a population that is due to genetic factors E) the proportion of alleles that are shared between siblings Answer: D Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 18) Broad-sense heritability (H2) represents .

A) an estimate of the contribution of additive genetic variance to phenotypic variance B) an estimate of the contribution of dominance genetic variance to phenotypic variance C) an estimate of the contribution of interactive genetic variance to phenotypic variance D) the contribution of genotype by environment interaction variance to the total phenotypic variance E) the contribution of total genotypic variance to the total phenotypic variance Answer: E Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 19) Narrow-sense heritability (h2) represents .

A) an estimate of the contribution of additive genetic variance to phenotypic variance B) an estimate of the contribution of dominance genetic variance to phenotypic variance C) an estimate of the contribution of interactive genetic variance to phenotypic variance D) the contribution of genotype by environment interaction variance to the total phenotypic variance E) the contribution of total genotypic variance to the total phenotypic variance Answer: A Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 20) You would like to breed chickens to increase the number of eggs produced. To assess the effectiveness of a breeding program would have on this phenotype, it would be most useful to know .

A) the broad sense heritability of egg number B) the narrow sense heritability of egg number C) the QTL for egg number D) the contribution of dominance genetic variance to the total phenotypic variance for egg number E) the contribution of total genotypic variance to the total phenotypic variance for egg number Answer: B Section: 20.3 Bloom's Taxonomy: Evaluating/Creating

5 Copyright © 2021 Pearson Education Ltd.

21) You are attempting to increase the number of flowers per plant through artificial selection in a controlled greenhouse environment. Your population of plants has a mean of 7 flowers per plant. You select a number of individuals to breed, these individuals have a mean of 15 flowers per plant. The offspring produced have a mean of 9 flowers per plant. What is the realized heritability of flower number?

A) 1 B) .75 C) .5 D) .25 E) 0 Answer: D Section: 20.3 Bloom's Taxonomy: Evaluating/Creating 22) What is the relationship between the narrow-sense heritability (h2) index and the impact of selection?

A) an h2 approaching 1 indicates selection would have a significant impact in altering the phenotypic variation of the population B) an h2 approaching 1 indicates selection would have a little impact in altering the phenotypic variation of the population C) an h2 approaching 0 indicates selection would have a significant impact in altering the phenotypic variation of the population D) h2 estimates are a strong indicator of the influence of environmental factors on phenotype not the impacts of selection Answer: A Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 23) What is the name of the process of selecting a specific group of organisms from an initially heterogeneous population for future breeding purposes?

A) natural selection B) sexual selection C) response to selection D) realized heritability E) artificial selection Answer: E Section: 20.3 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

24) Traits such as height, general body structure, and skin color that show population frequencies distributed in a bell-shaped curve are most likely caused by genes that behave in a(n) manner.

A) codominant B) epistatic C) additive D) dominant Answer: C Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 25) Genotypic variance in quantitative traits is summarized in the equation VG = VA + VD + VI. What component of genotypic variance is represented by VA? A) variance due to interaction of additive alleles B) variance due to nonadditvie phenotypic expression of heterozygotes C) variance due to epistatic interactions among loci D) variance due to interaction between genotype and environment E) variance due to non-genetic factors Answer: A Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 26) Genotypic variance in quantitative traits is summarized in the equation VG = VA + VD + VI. What component of genotypic variance is represented by VD? A) variance due to interaction of additive alleles B) variance due to nonadditvie phenotypic expression of heterozygotes C) variance due to epistatic interactions among loci D) variance due to interaction between genotype and environment E) variance due to non-genetic factors Answer: B Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 27) Genotypic variance in quantitative traits is summarized in the equation VG = VA + VD + VI. What component of genotypic variance is represented by VI? A) variance due to interaction of additive alleles B) variance due to nonadditvie phenotypic expression of heterozygotes C) variance due to epistatic interactions among loci D) variance due to interaction between genotype and environment E) variance due to non-genetic factors Answer: C Section: 20.3 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

28) If for a given trait both members of a twin pair express the trait, they are A) discordant B) concordant C) acordant D) monozygotic E) dizygotic Answer: B Section: 20.4 Bloom's Taxonomy: Remembering/Understanding

.

29) Which of the following is not included in the general conclusions of large-scale twin studies?

A) the interactions between genotype and environment are often trait specific B) a significant component of trait variation in a population has a genetic basis C) most complex traits are primarily controlled by one gene of large effect and secondarily by many genes of small effect D) a large percentage of traits follow a simple additive model Answer: C Section: 20.4 Bloom's Taxonomy: Remembering/Understanding 30) What is a quantitative trait locus (QTL)?

A) a region of a chromosome that accounts for all of the additive genetic variation contributing to a quantitative trait B) a region of a chromosome that contains one or more genes contributing to a quantitative trait C) a specific gene known to control the expression of a phenotype D) a region of a chromosome highly influenced by environmental (nongenetic) factors E) a set of genes that show a strong response to selection Answer: B Section: 20.5 Bloom's Taxonomy: Remembering/Understanding 31) What is an expression quantitative trait locus (eQTL)?

A) a region of a chromosome that accounts for all of the additive genetic variation contributing to a quantitative trait B) a region of a chromosome that contains one or more genes contributing to a quantitative trait C) a region of a chromosome that controls the expression of genes involved in quantitative traits D) a region of a chromosome highly influenced by environmental (nongenetic) factors E) sets of genes that show correlated expression of phenotypic variation Answer: C Section: 20.5 Bloom's Taxonomy: Remembering/Understanding

8 Copyright © 2021 Pearson Education Ltd.

32) Continuous inheritance is often related to the term quantitative inheritance. Why? Answer: In continuous inheritance, each involved locus has a quantitative input on the production of a single characteristic of the phenotype. In addition, although it may not always be the case, each gene product would be considered to be qualitatively similar.

Section: 20.1 Bloom's Taxonomy: Remembering/Understanding 33) Some traits are polygenic but do not show typical continuous distributions. For example, type II diabetes is determined by a number of genes, but one is either diagnosed as diabetic or not diabetic based on blood sugar levels. How might one explain such a phenomenon?

Answer: In some cases, a physiological threshold may be crossed such that in the presence of a given proportion of related genes expression occurs. Often, the environment plays a role in expression and therefore, along with genetic makeup, diagnosis of the condition.

Section: 20.1 Bloom's Taxonomy: Evaluating/Creating 34) Provide a brief description of discontinuous inheritance and continuous inheritance. How are the two related? How are they different?

Answer: At the transmission level, one sees "stepwise" distributions in discontinuous inheritance, but "smoother" or more bell-shaped distributions in continuous inheritance, as shown in the following figures:

Both patterns are formed from normal Mendelian principles of segregation and independent assortment. The differences are in the manner in which the gene products interact; continuous inheritance involves additive effects.

Section: 20.1 Bloom's Taxonomy: Remembering/Understanding

9 Copyright © 2021 Pearson Education Ltd.

35) If the proportion of F2 individuals resembling either of the two most extreme parental phenotypes can be determined, which formula can be applied to determine the number of gene pairs involved?

Answer: 1/4n (where n is the number of gene pairs) = ratio of F2 individuals expressing either extreme phenotype Section: 20.1 Bloom's Taxonomy: Remembering/Understanding 36) Which formula can be used to determine the number of categories (phenotypes) possible in the F2 results of a polygenic system?

Answer: 2n + 1 (where n is the number of gene pairs) = number of phenotypic categories Section: 20.1 Bloom's Taxonomy: Remembering/Understanding 37) Assume that in the F2 of a series of crosses, 1/64 of the offspring resemble one of the parents (P). How many gene pairs are involved in producing these results?

Answer: three Section: 20.1 Bloom's Taxonomy: Applying/Analyzing 38) Assume that four polygenic gene pairs are involved in determining phenotypes of F2. How many phenotypic classes are expected?

Answer: nine Section: 20.1 Bloom's Taxonomy: Applying/Analyzing 39) What is the term given to a random subset of individuals who are selected for measurement in a particular study?

Answer: sample Section: 20.2 Bloom's Taxonomy: Remembering/Understanding 40) Provide a formal equation for h2 (narrow-sense heritability).

Answer: h2 = VA/VP Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 41) Inbred strains with individuals of a relatively homogeneous or constant genetic background are often used to determine the influence of genes or the environment on phenotypic variation.

Variation observed between different inbred strains reared in a constant or homogeneous environment would likely be caused by genetic factors. What would be the source of variation observed among members of the same inbred strain reared under varying environmental conditions?

Answer: nongenetic factors generally categorized as "environmental" Section: 20.3 Bloom's Taxonomy: Applying/Analyzing 10 Copyright © 2021 Pearson Education Ltd.

42) What is the formal expression used to examine the relative importance of genetic versus environmental factors?

Answer: heritability index (H2) = VG/VP Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 43) Name the three components of phenotypic variance.

Answer: environmental variance (VE), genetic variance (VG), and variance resulting from the interaction between genetics and environment (VG × E) Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 44) Interpret the meaning of an H2 (broad-sense heritability) value that approaches 1.0.

Answer: Almost all the phenotypic variation is determined by heredity.

Section: 20.3 Bloom's Taxonomy: Applying/Analyzing 45) Interpret the meaning of an H2 (broad-sense heritability) value that approaches 0.0.

Answer: Almost all the phenotypic variation is determined by the environment.

Section: 20.3 Bloom's Taxonomy: Applying/Analyzing 46) Provide a brief definition of the terms additive variance, dominance variance, and interactive variance.

Answer: Additive variance results from the average effect of additive genes. Dominance variance accounts for variation when phenotypic expression in heterozygotes is not precisely intermediate between the two homozygotes. Interactive variance occurs when two or more loci behave epistatically.

Section: 20.3 Bloom's Taxonomy: Remembering/Understanding 47) What is meant by the term heritability? Describe the components of heritability and provide a brief explanation of each. Of what interest is heritability to animal and plant breeders?

Answer: Heritability refers to the degree to which observed phenotypic variation for a given trait is inherited within a population in a certain environment. Components include environmental variance, genetic variance, and interaction between the two. From information on heritability, breeders can determine the degree of improvement to expect from selective breeding.

Section: 20.3 Bloom's Taxonomy: Applying/Analyzing

11 Copyright © 2021 Pearson Education Ltd.

48) As used in twin studies in mammals, distinguish between the terms concordant and discordant.

Answer: Concordant refers to a given trait expressed similarly between twins; discordant refers to a case in which one member of the twin pair expresses the trait but the other does not.

Section: 20.4 Bloom's Taxonomy: Remembering/Understanding 49) Describe the value of using twins in the study of questions relating to the relative impact of heredity versus environment.

Answer: Monozygotic twins are derived from the splitting of a single fertilized egg and are therefore of identical genetic makeup. When such twins are raised in the same versus different settings, an estimate of relative hereditary and environmental influences can often be made.

Section: 20.4 Bloom's Taxonomy: Applying/Analyzing 50) How would the use of a large series of monozygotic and dizygotic twins enhance studies on the genetic basis of human behavior?

Answer: Monozygotic twins are genetically identical, and when reared under the same versus different environments, one can estimate the degree to which variation in behavior is determined by heredity. Dizygotic twins are genetically different but by having the same intrauterine and developing environment (if reared in the same household), one can again estimate the influence of heredity on behavioral traits.

Section: 20.4 Bloom's Taxonomy: Applying/Analyzing 51) Typically, one thinks of identical (MZ) twins being genetically identical. However, genomic differences do exist between MZ twins by epigenetic processes. What are epigenetic processes?

Answer: Epigenetic processes involve the chemical modification of DNA and associated histones.

Section: 20.4 Bloom's Taxonomy: Applying/Analyzing 52) In the analysis of quantitative traits, positions on chromosomes called quantitative trait loci (QTLs) are often discussed. In the same context, restriction fragment length polymorphisms (RFLPs) are also discussed. What is the relationship between QTLs and RFLPs? Answer: In many organisms, traditional genetic markers are not available for the mapping of regions of chromosomes containing genes responsible for determining quantitative traits (QTLs).

DNA polymorphisms generate molecular markers (RFLPs), which can serve as reference points in mapping QTLs. Section: 20.5 Bloom's Taxonomy: Evaluating/Creating

12 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Chapter 21 Population and Evolutionary Genetics 1) is a group of individuals belonging to the same species that live in a defined geographic area and actually or potentially interbreed.

A) Population B) Cohort C) Hybrid D) Pool E) Pod Answer: A Section: 21.1 Bloom's Taxonomy: Remembering/Understanding 2) Which term is given to the total genetic information carried by all members of a population?

A) gene pool B) genome C) chromosome complement D) breeding unit E) race Answer: A Section: 21.1 Bloom's Taxonomy: Remembering/Understanding 3) A number of mechanisms operate to maintain genetic diversity in a population. Why is such diversity favored?

A) Homozygosity is an evolutionary advantage.

B) Diversity leads to inbreeding advantages.

C) Genetic diversity may better adapt a population to inevitable changes in the environment.

D) Greater genetic diversity increases the chances of haploidy.

E) Genetic diversity helps populations avoid diploidy.

Answer: C Section: 21.1 Bloom's Taxonomy: Remembering/Understanding 4) The diversity of alleles and most loci reflect .

A) the strength of natural selection B) population size and the accumulation of neutral mutations C) levels of inbreeding in the population D) the strength of sexual selection in the population E) the number of pre-reproductive age individuals in the population Answer: B Section: 21.1 Bloom's Taxonomy: Remembering/Understanding

1 Copyright © 2021 Pearson Education Ltd.

5) A neutral mutation is one that .

A) produces an allele that is functionally equivalent to the original allele B) is heavily favored by selection C) is quickly eliminated by selection D) is found only in coding regions of the genome E) is found only in regulatory regions of the genome Answer: A Section: 21.1 Bloom's Taxonomy: Remembering/Understanding 6) In a population of 100 individuals, 49 percent are of the NN blood type. What percentage is expected to be MN assuming Hardy-Weinberg equilibrium conditions?

A) 9 percent B) 21 percent C) 42 percent D) 51 percent E) There is insufficient information to answer this question.

Answer: C Section: 21.2 Bloom's Taxonomy: Applying/Analyzing 7) Albinism is an autosomal recessive trait in humans. Assume that there are 100 albinos (aa) in a population of 1 million. How many individuals would be expected to be homozygous normal (AA) under equilibrium conditions?

A) 100 B) 10,000 C) 19,800 D) 980,100 E) 999,900 Answer: D Section: 21.2 Bloom's Taxonomy: Applying/Analyzing 8) In the Hardy-Weinberg equation the term (2pq) represents .

A) the observed number of heterozygotes an equilibrium population with allele frequencies p and q B) the expected number of heterozygotes an equilibrium population with allele frequencies p and q C) the observed number of homozygotes an equilibrium population with allele frequencies p and q D) the expected number of homozygotes in an equilibrium population with allele frequencies p and q Answer: B Section: 21.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

9) Assume that a trait is caused by the homozygous state of a gene that is recessive and autosomal. Nine percent of the individuals in a given population express the phenotype caused by this gene. What percentage of the individuals would be heterozygous for the gene? Assume that the population is in Hardy-Weinberg equilibrium.

A) 49% B) 21% C) 42% D) 25% E) 50% Answer: C Section: 21.2 Bloom's Taxonomy: Applying/Analyzing 10) Which of the following is not included in the list of assumptions that pertain to a population in a Hardy-Weinberg equilibrium?

A) large population size and no genetic drift B) random mating C) some genotypes may have a selective advantage D) no migration E) no new alleles are generated by mutation Answer: C Section: 21.2 Bloom's Taxonomy: Remembering/Understanding 11) Assume that in a Hardy-Weinberg population, 9 percent of the individuals are of the homozygous recessive phenotype. What percentage are homozygous dominant?

A) 91% B) 25% C) 42% D) 30% E) 49% Answer: E Section: 21.2 Bloom's Taxonomy: Applying/Analyzing 12) In a population that meets the Hardy-Weinberg equilibrium assumptions, 81 percent of the individuals are homozygous for a recessive allele. What percentage of the individuals would be expected to be heterozygous for this locus in the next generation?

A) 9% B) 18% C) 1% D) 81% E) 50% Answer: B Section: 21.2 Bloom's Taxonomy: Applying/Analyzing 3 Copyright © 2021 Pearson Education Ltd.

13) In a population of cattle, the following color distribution was noted: 36 percent red (RR), 48 percent roan (Rr), and 16 percent white (rr). Which of the following statements is most accurate concerning this population?

A) the percentage of red cattle will likely increase in the next generation B) the population is in Hardy-Weinberg equilibrium C) the population has a high coefficient of inbreeding (F) D) the frequency of the r allele is .16 E) the population is not in Hardy-Weinberg equilibrium Answer: B Section: 21.2 Bloom's Taxonomy: Applying/Analyzing 14) In a population of 10,000 individuals, in which 3600 are MM, 1600 are NN, and 4800 are MN, what are the frequencies of the M alleles and the N alleles?

A) M = 0.36; N = 0.16 B) M = 0.36; N = 0.64 C) M = 0.64; N = 0.36 D) M = 0.6; N = 0.4 E) M = 0.4; N = 0.6 Answer: D Section: 21.2 Bloom's Taxonomy: Applying/Analyzing 15) In the Hardy-Weinberg equation the term p2 represents: A) The observed number of heterozygotes an equilibrium population with allele frequencies p and q.

B) The expected number of heterozygotes an equilibrium population with allele frequencies p and q.

C) The observed number of one category of homozygotes an equilibrium population with allele frequencies p and q.

D) The expected number of one category of homozygotes in an equilibrium population with allele frequencies p and q.

Answer: D Section: 21.2 Bloom's Taxonomy: Remembering/Understanding 16) Given that p = 0.3 for a population in Hardy-Weinberg equilibrium, what would be the expected frequency of heterozygotes for the involved allelic pair?

A) .42 B) .09 C) .49 D) .60 E) .50 Answer: A Section: 21.2 Bloom's Taxonomy: Applying/Analyzing 4 Copyright © 2021 Pearson Education Ltd.

17) What is meant by the term equilibrium in the context of population genetics?

A) a condition in which the genotype and allele frequencies change drastically from generation to generation B) the situation where alleles migrate into and out of a population at equal frequencies C) the situation where rates of inbreeding and outbreeding are equal D) a condition in which the genotype and allele frequencies remains constant from generation to generation E) the situation where rates of positive and negative assortative mating are equal Answer: D Section: 21.2, 21.7 Bloom's Taxonomy: Remembering/Understanding 18) A certain form of albinism in humans is recessive and autosomal. Assume that 1 percent of the individuals in a given population are albino. Assuming that the population is in HardyWeinberg equilibrium, what is the likelihood that a person sampled from that population a carrier for albinism?

A) .01 B) .18 C) .81 D) .10 E) .90 Answer: B Section: 21.2, 21.3 Bloom's Taxonomy: Applying/Analyzing 19) Which method is often used to analyze proteins and nucleic acids by physical separation when estimating genetic variation in populations?

A) electrophoresis B) centrifugation C) absorption spectrophotometry D) fluorometry E) in situ hybridization Answer: A Section: 21.3 Bloom's Taxonomy: Remembering/Understanding 20) What is the interpretation of the equation p + q + r = 1.0?

A) only 3 alleles can exist in any population B) the sum of the relevant individual alleles in a population is equal to 100 percent of those alleles C) the probability of sampling all three alleles from a population is 1 D) there are high levels of inbreeding in the population E) mutation rates are high in the population Answer: B Section: 21.3 Bloom's Taxonomy: Remembering/Understanding 5 Copyright © 2021 Pearson Education Ltd.

21) When all genotypes within a population do not have equal rates of survival, then allele frequencies may change from one generation to the next by a process called .

A) positive assortative mating B) negative assortative mating C) genetic drift D) natural selection E) Hardy-Weinberg equilibrium Answer: D Section: 21.4 Bloom's Taxonomy: Remembering/Understanding 22) Which of the following is not a factor that contribute to the phenomenon of natural selection?

A) genetic variation among members of an interbreeding population B) overpopulation C) competition D) differential reproductive success based on heritable phenotype E) random fluctuations in allele frequency due to small population size Answer: E Section: 21.4 Bloom's Taxonomy: Remembering/Understanding 23) In a particular environments, a heritable trait and at one end of the phenotypic spectrum has a higher probability to survive and reproduce than other phenotypes. This is a case of .

A) directional selection B) stabilizing selection C) disruptive selection D) relative selection E) artificial selection Answer: A Section: 21.4 Bloom's Taxonomy: Remembering/Understanding 24) In a particular environments, a heritable trait which is intermediate in the phenotypic spectrum has a higher probability to survive and reproduce than other phenotypes. This is a case of .

A) directional selection B) stabilizing selection C) disruptive selection D) relative selection E) artificial selection Answer: B Section: 21.4 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

25) In a particular environments, a heritable trait which is intermediate in the phenotypic spectrum has a lower probability to survive and reproduce relative to phenotypes at either extreme. This is a case of .

A) directional selection B) stabilizing selection C) disruptive selection D) relative selection E) artificial selection Answer: C Section: 21.4 Bloom's Taxonomy: Remembering/Understanding 26) Under normal conditions, a population is in Hardy-Weinberg equilibrium at 'gene A' and the frequency of alleles A = 0.5. Under drought conditions, the aa genotype is lethal (waa = 0). What will be the frequency of the a allele if the population is exposed to drought conditions for 4 generations?

A) .50 B) .33 C) .25 D) .17 E) .14 Answer: D Section: 21.4 Bloom's Taxonomy: Applying/Analyzing 27) Which term is given to the measure of an individual's genetic contribution to the next generation?

A) directional selection B) genetic bottleneck C) fitness D) relative selection E) reproductive effort Answer: C Section: 21.4 Bloom's Taxonomy: Remembering/Understanding 28) What is the original source of genetic variation in a population?

A) crossing over B) mutation C) independent assortment D) random mating E) natural selection Answer: B Section: 21.5 Bloom's Taxonomy: Remembering/Understanding

7 Copyright © 2021 Pearson Education Ltd.

29) Suppose that a given gene undergoes a mutation to a dominant allele such that 2 out of 100,000 offspring exhibit the new mutant phenotype. Assuming that these offspring are heterozygous, what is the mutation rate for the gene?

A) 1/50,000 B) 1/100,000 C) 1/25,000 D) 1/1,000,000 E) 1/10,000 Answer: B Section: 21.5 Bloom's Taxonomy: Applying/Analyzing 30) In the absence of selection, migration and gene flow between two populations will result in .

A) a genetic bottleneck B) inbreeding C) similar allele frequencies in the two populations D) a reduction in the level of genetic variation E) assortative mating Answer: C Section: 21.6 Bloom's Taxonomy: Remembering/Understanding 31) In small isolated populations, allele frequencies can fluctuate considerably. The term that applies to this circumstance is .

A) genetic isolation B) allelic separation C) natural selection D) stabilizing selection E) genetic drift Answer: E Section: 21.7 Bloom's Taxonomy: Remembering/Understanding 32) Small population size, founder effect, and genetic bottlenecks can all give rise to A) directional selection B) stabilizing selection C) disruptive selection D) genetic drift E) assortative mating Answer: D Section: 21.7 Bloom's Taxonomy: Remembering/Understanding

8 Copyright © 2021 Pearson Education Ltd.

.

33) Negative assortative mating would result in .

A) more homozygous individuals than would be expected under Hardy-Weinberg equilibrium B) more heterozygous individuals than would be expected under Hardy-Weinberg equilibrium C) high levels of inbreeding D) genetic drift E) disruptive selection Answer: B Section: 21.7 Bloom's Taxonomy: Remembering/Understanding 34) A population has a coefficient of inbreeding of 1. What is the interpretation of this value?

A) the population is in Hardy-Weinberg equilibrium B) genetic variation in the population is high C) random mating in the population is common D) there is a higher number of heterozygotes in the population E) all of the individuals in the population are descended from a single ancestor Answer: E Section: 21.8 Bloom's Taxonomy: Evaluating/Creating 35) In zoo animals, inbreeding often occurs because of a lack of a sufficient pool of breeding individuals. Under such conditions, which is often exhibited among inbred organisms?

A) strong directional selection B) higher frequency of heterozygous genotypes than expected by chance C) disruptive selection D) higher frequency of aberrant recessive phenotypes E) assortative mating Answer: D Section: 21.8 Bloom's Taxonomy: Remembering/Understanding 36) Inbreeding by itself changes .

A) allele frequencies, but not genotype frequencies B) genotype frequencies, but not allele frequencies C) both allele and genotype frequencies D) neither allele or genotype frequencies Answer: B Section: 21.8 Bloom's Taxonomy: Evaluating/Creating

9 Copyright © 2021 Pearson Education Ltd.

37) Which general term is used to group various biological and behavioral properties of organisms that act to prevent or reduce interbreeding?

A) phyletic evolution B) allopatric speciation C) reproductive isolating mechanisms D) inbreeding E) genetic divergence Answer: C Section: 21.9 Bloom's Taxonomy: Remembering/Understanding 38) The process of speciation in sexually reproducing organisms always involves A) geographic separation B) prezygotic isolating mechanisms C) postzygotic isolating mechanisms D) reproductive isolation E) phenotypic differentiation Answer: D Section: 21.9 Bloom's Taxonomy: Remembering/Understanding 39) Which of the following is considered a prezygotic isolating mechanism?

A) weakness of offspring produced by parents originating from different populations B) incompatible gametes C) developmental hybrid sterility D) segregational hybrid sterility E) phenotypic differentiation Answer: B Section: 21.9 Bloom's Taxonomy: Remembering/Understanding 40) Which of the following is considered a postzygotic isolating mechanism?

A) populations inhabiting different habitats B) incompatible mating behaviors between C) sterility of interpopulation hybrids D) incompatible gametes E) phenotypic differentiation Answer: C Section: 21.9 Bloom's Taxonomy: Remembering/Understanding

10 Copyright © 2021 Pearson Education Ltd.

.

41) The application of molecular clocks to the study of speciation suggests A) all speciation events require millions of years of isolation B) some speciation events can occur rapidly C) all areas of the genome follow the same molecular clock D) speciation events require periods of geographic isolation E) mutation rates are constant Answer: B Section: 21.9, 21.10 Bloom's Taxonomy: Evaluating/Creating

.

42) Which of the following is true with respect to phylogenetic trees?

A) each node represents a common ancestor of two lineages B) DNA sequences are the only data that can be used to construct a phylogenetic tree C) the mechanism driving speciation events is demonstrated D) all areas of the genome are equally usefully for producing phylogenetic trees Answer: A Section: 21.10 Bloom's Taxonomy: Remembering/Understanding 43) Define the term speciation.

Answer: the process of splitting a genetically homogeneous population into two or more populations that undergo genetic differentiation and eventual reproductive isolation Section: 21.9 Bloom's Taxonomy: Remembering/Understanding 44) Which single event is probably common to all occurrences of speciation?

Answer: The generation of reproductive isolating mechanisms–geographic, behavioral, physiological, and mechanical–provides common factors in the speciation process.

Section: 21.9 Bloom's Taxonomy: Remembering/Understanding 45) Present a rationale for using DNA sequence polymorphisms as an index of genetic diversity.

Is genetic diversity directly proportional to evolutionary (phylogenetic) diversity?

Answer: It is natural to expect that the forces that bring about speciation are dependent upon reproductive isolation. As groups of organisms become reproductively isolated (whether through development of geographic or genetic barriers), it is natural to expect that through the accumulation of mutations, DNA sequences will diverge. It is also natural to expect that some DNA sequences will diverge more rapidly than others and that some differences in DNA sequences may be more evolutionarily important than others. Therefore, whereas there may be a general relationship between genetic diversity and evolutionary (phylogenetic) diversity, it is likely that there are exceptions.

Section: 21.10 Bloom's Taxonomy: Evaluating/Creating

11 Copyright © 2021 Pearson Education Ltd.

46) Why might mitochondrial DNA be used in construction of phylogenetic trees rather than nuclear DNA? Answer: Mitochondrial DNA evolves relatively quickly, so that if a "fast clock" is needed to estimate relatively short evolutionary time periods, mitochondrial DNA can be especially useful.

Also, it is not subject to recombination, so the separate mitochondrial genetic lineages remain distinct with little mixing.

Section: 21.10 Bloom's Taxonomy: Evaluating/Creating 47) What are phylogenetic trees?

Answer: diagrammatic representations of evolutionary relationships among organisms Section: 21.10 Bloom's Taxonomy: Remembering/Understanding 48) Would one expect a linear relationship between DNA sequence divergence and phylogenetic distance? Why or why not?

Answer: Not necessarily, because some sections of DNA evolve more slowly if they code for products that serve important cellular functions, whereas other sections without such constraints evolve more quickly.

Section: 21.10 Bloom's Taxonomy: Evaluating/Creating 49) In an evolutionary sense, what is meant by the term molecular clock?

Answer: Molecular clocks are amino acid or nucleic acid sequences in which evolutionary changes accumulate over time. Ideally, such changes accumulate at a constant rate, so that the degree of differentiation is correlated to the length of time of divergence.

Section: 21.9, 21.10 Bloom's Taxonomy: Remembering/Understanding

12 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Special Topics 1 Epigenetics 1) is the study of phenomena and mechanisms that cause chromosome-associated heritable changes to gene expression that are not dependent on changes in DNA sequence.

A) Population genetics B) Transmission genetics C) Epigenetics D) Mutagenesis E) Preformation Answer: C Section: ST 1.1 Bloom's Taxonomy: Remembering/Understanding 2) Which nitrogenous bases are most likely to be methylated?

A) cytosines adjacent to guanines B) cytosines adjacent to other cytosines C) guanines adjacent to cytosines D) guanines adjacent to other guanines E) cytosines base paired to guanines Answer: A Section: ST 1.1 Bloom's Taxonomy: Remembering/Understanding 3) In terms of epigenetics, which statement best describes the promoter for a gene that is not expressed in lung tissue but is expressed in heart tissue?

A) In lung cells, the guanines in the promoter would be hypomethylated, but would be hypermethylated in heart cells.

B) In lung cells, the guanines in the promoter would be hypermethylated, and would be hypomethylated in heart cells.

C) In lung cells, the cytosines of the promoter would be hypomethylated, and would be hypermethylated in heart cells.

D) In lung cells, the cytosines of the promoter would be hypermethylated, and would be hypomethylated in heart cells.

E) The promoters should appear the same in both types of cells, but the mRNA produced would be hypermethylated more in lung cells compared to heart cells.

Answer: D Section: ST 1.1 Bloom's Taxonomy: Applying/Analyzing

1 Copyright © 2021 Pearson Education Ltd.

4) Which carbon atom is methylated on cytosine?

A) 2' B) 4' C) 5' D) 6' Answer: C Section: ST 1.1 Bloom's Taxonomy: Remembering/Understanding 5) The sum of histone modifications and their interactions is called the A) genetic code B) histone code C) protein code D) expression code E) transcription code Answer: B Section: ST 1.1 Bloom's Taxonomy: Remembering/Understanding

.

6) Which of the following genes is known to cause instability in repetitive microsatellite sequences when hypermethylated?

A) BRCA1 B) CDKN2A C) MLH1 D) IGF2 E) H19 Answer: C Section: ST 1.3 Bloom's Taxonomy: Remembering/Understanding 7) Which statement about X inactivation is false?

A) X inactivation was the first example of epigenetic allele-specific regulation to be identified.

B) All embryonic cells randomly inactivate paternal X chromosomes during early development.

C) Xist is a sense lncRNA expressed on the inactivated X chromosome.

D) Tsix is an antisense lncRNA expressed on the active X chromosome.

E) Xist and Tsix are transcripts of the same gene that are transcribed in opposite directions.

Answer: B Section: ST 1.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

8) A particular small noncoding RNA that is linked to spermatogenesis is most likely classified as what kind of noncoding RNA? A) microRNAs B) short interfering RNAs C) antisense lncRNAs D) piwi-interacting RNAs E) bidirectional lncRNAs Answer: D Section: ST 1.1 Bloom's Taxonomy: Applying/Analyzing 9) Which of the following mechanisms is an example of genomic imprinting in humans?

A) The maternal and paternal alleles of a gene pair are both expressed.

B) One of the two X chromosomes in females is randomly expressed and the other is repressed.

C) A random pattern of autosomal allele inactivation is observed.

D) Human males have only one Y chromosome and one X chromosome.

E) In some allele pairs, only the paternal allele is expressed, and in others, only the maternal allele is expressed.

Answer: E Section: ST 1.2 Bloom's Taxonomy: Remembering/Understanding 10) Which of the following genotypes would most likely lead to a low birth weight?

A) Igf2+/Igf2+ where neither maternal nor paternal allele is mutated.

B) Igf2+/Igf2- where the mutated Igf2- allele is maternal.

C) Igf2+/Igf2- where the mutated Igf2- allele is paternal.

D) Igf2+/Igf2- where the mutated Igf2- allele is either maternal or paternal.

E) Igf2+/Igf2- but only in female progeny as the Igf2 gene is on the X chromosome.

Answer: C Section: ST 1.2 Bloom's Taxonomy: Applying/Analyzing 11) What DNA regions regulate the expression of a cluster of genes through methylation?

A) promoters B) insulators C) epimutations D) imprinting control regions E) enhancers Answer: D Section: ST 1.2 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

12) How would you interpret the histone code, H2K4me3? This code represents a A) trimethylated lysine at position 2 from the N-terminus of histone H4 B) dimethylated lysine at position 4 from the N-terminus of histone H3 C) trimethylated lysine at position 4 from the N-terminus of histone H2 D) dimethylated lysine at position 3 from the N-terminus of histone H3 E) trimethylated lysine at position 3 from the N-terminus of histone H4 Answer: C Section: ST 1.1 Bloom's Taxonomy: Applying/Analyzing

.

13) Which of the following statements best characterizes mice pups that experienced low levels of maternal nurturing?

A) As adults they have high levels of glucocorticoid receptor expression.

B) As adults they have increased ability to adapt to stress.

C) As adults they have increased promoter methylation.

D) As adults they are more likely to nurture their offspring.

E) As adults their brains are heavily mutated.

Answer: C Section: ST 1.4 Bloom's Taxonomy: Remembering/Understanding 14) Consider the genes IGF2 and H19 and the imprinting control region (ICR) between them. If the paternal ICR region was mutated in such a way that it could not be methylated but the maternal ICR was not mutated, how would that effect growth? In your explanation describe the expression of IGF2 and H19 from both the maternal and paternal alleles.

Answer: There would be reduced growth. The maternal copy of ICR is usually not methylated and so would remain unmethylated allowing for the expression of the maternal H19. Expressed H19 produces a noncoding RNA that suppresses IGF2. Typically, the paternal copy of ICR is hypermethylated so that H19 is not expressed and thus IGF2 can be expressed. In this mutant, the ICR is unable to be methylated and thus H19 will be expressed and then the resulting noncoding RNA will repress the IGF2. This would result in the repression of both maternal and paternal copies of IGF2 and will result in slower growth.

Section: ST 1.2 Bloom's Taxonomy: Applying/Analyzing 15) Explain the roles of Xist and Tsix during X inactivation in females.

Answer: Both Xist and Tsix are long noncoding RNAs (lncRNA). Xist is expressed on the inactivated X-chromosomes. The Xist lncRNA will repress the genes on the inactivated Xchromosomes. Tsix is expressed on the active X chromosomes. When Tsix is expressed, its lncRNA will repress Xist expression, thus preventing the active X chromosome from being silenced.

Section: ST 1.2 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

16) Explain why artificial reproductive technologies (ART) such as in vitro fertilization are associated with an increased risk of epimutations.

Answer: There is an increase in imprinting errors, and maternal specific methylation is reduced when using ART. ART is conducted when epigenetic reprogramming is occurring and may interfere with that reprogramming. Some examples are an increase in Beckwith–Wiedemann syndrome, Angelman syndrome, and Prader-Willi syndrome.

Section: ST 1.2 Bloom's Taxonomy: Remembering/Understanding 17) Explain why selected regions of the genome are hypermethylated in cancer cells even though genomes of cancer cells are typically characterized by global hypomethylation.

Answer: Hypermethylation of promoters that regulate tumor-suppressor genes can lead to reduced expression of proteins involved in DNA repair. Loss of DNA repair mechanisms can lead to loss of cell cycle control and cancer. Hypermethylation of these genes is known to reduce their expression.

Section: ST 1.3 Bloom's Taxonomy: Remembering/Understanding 18) Based on rat studies that show how maternal nurturing helps offspring respond better to stress, propose a model to help explain why human children who are abused have a higher suicide rate.

Answer: From rat studies, it could be proposed that human children that are abused will have decreased histone acetylation and increased DNA methylation associated with the genes for glucocorticoid receptors. These receptors in the hypothalamic region of the brain are important in mediating stress.

Section: ST 1.4 Bloom's Taxonomy: Evaluating/Creating 19) What are the two main goals of the NIH Roadmap Epigenomics Project?

Answer: The goals are to (1) provide a set of at least 1000 reference epigenomes in a range of cell types from healthy and diseased individuals and (2) delineate the epigenetic differences in healthy and diseased states.

Section: ST 1.5 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Special Topics 2 Genetic Testing 1) Which of the following tests is not commonly performed on samples obtained by amniocentesis?

A) karyotyping B) sex determination C) sequencing of cfDNA D) biochemical analysis E) chromosome analysis Answer: C Section: ST 2.2 Bloom's Taxonomy: Remembering/Understanding 2) Explain what a haplotype is and the advantage of haplotype analysis on cfDNA. Answer: Haplotypes are contiguous segments of DNA that do not undergo recombination during gamete formation and are therefore inherited together. Haplotype analysis of cfDNA can distinguish which cfDNA fragments are maternal and which fragments are from the fetus.

Section: ST 2.2 Bloom's Taxonomy: Applying/Analyzing 3) RFLPs can be used to identify point mutations through which of the following mechanisms?

A) The mutation can occur in a restriction site, causing a change in the cutting pattern for restriction enzymes.

B) The mutation can be silent in the coding for protein resulting in a different amino acid being inserted.

C) The mutation can occur outside of a restriction site causing a change in the cutting pattern for restriction enzymes.

D) The mutation can occur causing a shortening of the mRNA that is then compared to wild type.

Answer: A Section: ST 2.3 Bloom's Taxonomy: Evaluating/Creating 4) Whole exome sequencing (WES) is a powerful alternative to WGS and expected to replace some applications involving microarrays. Predict a major disadvantage of WES. Answer: WES analyzes only the protein coding genes within a genome. Any mutations in regulatory regions that contribute to disease would not be identified using this method.

Section: ST 2.4, ST 2.5 Bloom's Taxonomy: Evaluating/Creating

1 Copyright © 2021 Pearson Education Ltd.

5) A DNA microarray (also called a DNA chip) can be used to A) generate mutations in a gene of interest B) detect mutations that may indicate a risk of disease C) detect RFLPs D) determine protein translation levels Answer: B Section: ST 2.4 Bloom's Taxonomy: Remembering/Understanding

.

6) Probes that bind under very stringent hybridization conditions resulting in the differential binding over one nucleotide are called .

A) generation-specific probes B) short, variable repeats C) VNTRs D) microsatellites E) allele-specific oligonucleotides (ASOs) Answer: E Section: ST 2.3 Bloom's Taxonomy: Remembering/Understanding 7) Using single-cell RNA sequencing a clinician can now determine instead of two.

A) heterozygosity B) the presence of genes in the genome and their relative expression levels C) gene expression and protein levels associated with those genes D) the ratio of transcripts to proteins Answer: B Section: ST 2.5 Bloom's Taxonomy: Remembering/Understanding

in one experiment

8) Personal genomics has led to the discovery of many new genes associated with disease states.

For example, GWAS for schizophrenia have revealed genetic loci contributing to disease risk.

A) 1 B) 5 C) 63 D) >100 Answer: D Section: ST 2.6 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

9) Genome-wide association studies are identifying many genes that may affect disease risk.

Describe how GWAS is used to determine if a mutation or gene may be a risk factor for disease.

Answer: The genomes of several hundred to several thousand diseased individuals are subject to individual genome sequencing as well as several thousand of non-diseased individuals. The genomes are then compared to determine if there are patterns or associations between certain genes and the disease state. This does not mean a carrier of the correlated gene will develop the disease, just that there is a risk factor.

Section: ST 2.6 Bloom's Taxonomy: Evaluating/Creating 10) Assume that results are obtained from a two-channel microarray in which cDNA samples from normal cells (fluorescently labeled green) and cancer cells (fluorescently labeled red) are competing for binding to the same probe sets. What would dots with intermediate colors represent?

A) Genes that are expressed only in the normal cells.

B) Genes that are expressed only in the cancer cells.

C) Genes that are expressed in both normal cells and cancer cells.

D) Genes that are not expressed in both normal cells and cancer cells.

Answer: C Section: ST 2.4 Bloom's Taxonomy: Applying/Analyzing 11) Describe how microarrays have led to improved diagnosis of two types of diffuse large Bcell lymphomas.

Answer: Diffuse large B-cell lymphoma patients predominantly fell into two groups. One responded well to chemotherapy while the second group responded poorly. Previously there was" no way to know which group the patient fell into before treatment. Gene-expression microarrays demonstrated that there are different gene-expression patterns between the two groups and can be used to spare patients whose cancer would not respond well to chemotherapy and the horrendous side effects of the treatment.

Section: ST 2.4 Bloom's Taxonomy: Evaluating/Creating 12) In the past decades, direct-to-consumer (DTC) genetic tests have become widespread. Which U. S. federal agency is responsible for the future regulation of these tests?

A) EPA B) FDA C) DEA D) FBI Answer: B Section: ST 2.7 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

13) In 2010, the American Civil Liberties Union filed suit against Myriad Genetics. The court ruled in favor of the ACLU. What precedent did this court case set with regard to biotechnologies?

A) confirmed that human genes can be patented B) confirmed human genes cannot be patented C) confirmed that synthetic genes cannot be patented D) confirmed that any gene-related biotechnologies cannot be patented Answer: B Section: ST 2.7 Bloom's Taxonomy: Remembering/Understanding 14) There is a fear that preconception genetic testing may lead to "designer babies." This practice is feared because it is dangerously similar to which of the following historically unethical practices?

A) eugenics B) gene driving C) gene editing D) inbreeding Answer: A Section: ST 2.7 Bloom's Taxonomy: Remembering/Understanding 15) With the multitudes of genetic tests and bioinformatic databases available to clinicians and researchers, diagnosis via genetic information is becoming more common. Which of the following is a major concern with regard to genetic testing?

A) There are genetic tests for diseases that currently have no treatment.

B) A negative result in a genetic test means a person will never get the disease.

C) Genetic discrimination is easy to define and prevent.

D) Parents and clinicians will use prenatal genome sequencing in a responsible manner.

Answer: A Section: ST 2.7 Bloom's Taxonomy: Remembering/Understanding 16) Which of the following factors does not raise a significant ethical concern, during a period of medical genetics where most WGS studies of individuals happen in a largely unregulated environment?

A) variability of handling protocols B) variability in data collection methods C) variability in the genomes sequenced D) confidentiality of genetic information Answer: C Section: ST 2.7 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

17) The use of ASOs in the process of in vitro fertilization to determine the genetic risk factors present is ideal for .

A) next-generation sequencing B) preimplantation genetic diagnosis C) RFLP analysis D) genome scanning Answer: B Section: ST 2.3 Bloom's Taxonomy: Remembering/Understanding 18) WGS, NGS, and TGS are sequencing methods being used to help public health officials to accomplish which of the following tasks?

A) pathogen identification B) cancer prevention C) tracking impact of pathogens on host genomes D) isolating cancers after they have spread Answer: A Section: ST 2.5 Bloom's Taxonomy: Remembering/Understanding 19) Explain the significance of the threshold p value indicated on a Manhattan plot.

Answer: Manhattan plots show the results of a genetic association test along the y-axis. This is often represented by the negative log of p values for loci determined to be significantly associated with a diseased condition. A marker sequence with a significance value that exceeds the threshold p value is likely a disease-related sequence (ST 2.6, Page 462).

Section: ST 2.6 Bloom's Taxonomy: Applying/Analyzing 20) Current DTC genetic tests provide all of the following EXCEPT .

A) information regarding risk factors for disease states B) allele-specific information with regard to an individual's genome C) a reliable substitute for a trained healthcare professional D) an overwhelming amount of information regarding an individual's genetic risk factors Answer: C Section: ST 2.7 Bloom's Taxonomy: Applying/Analyzing

5 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Special Topics 3 Gene Therapy 1) Gene therapy strives to .

A) mitigate the symptoms of a disease state B) cure the disease state C) provide new gene sequences to make people stronger D) cut out defective genes Answer: B Section: ST 3.1 Bloom's Taxonomy: Remembering/Understanding 2) What is required in order for Cas9 to cut a target sequence?

A) a protospacer adjacent motif (PAM) B) a zinc-finger nuclease (ZFN) C) a chimeric antigen receptor (CAR) D) an RNA-induced silencing complex (RISC) Answer: A Section: ST 3.6 Bloom's Taxonomy: Remembering/Understanding 3) Minimizing the delivery of therapeutic genes to only the intended tissues is a major challenge when .

A) treating with ex vivo gene therapy B) treating with in vivo gene therapy C) culturing cells from a patient D) isolating tissue samples Answer: B Section: ST 3.2 Bloom's Taxonomy: Applying/Analyzing 4) A patient experiencing a disease state due to a gene deletion was treated with adenovirus gene therapy. The gene therapy was successful and the gene deletion was corrected. However, the patient began to develop cancer not long after the gene therapy. Develop a theory as to why the cancer presented so rapidly after gene therapy.

Answer: The cancer is most likely a direct result of the gene therapy. The use of adenovirus as a gene therapy vector results in the therapeutic gene inserting at random into the host cell genome.

In this case, the therapeutic gene most likely inserted into the host genome in such a way as to be tumorigenic.

Section: ST 3.2 Bloom's Taxonomy: Evaluating/Creating

1 Copyright © 2021 Pearson Education Ltd.

5) Adeno-associated viral vectors are preferred over lentiviral vectors because A) the inserted DNA cannot be reproduced when cells divide B) the inserted DNA forms episomes, which prevent insertion mutations C) the lentivirus cannot infect nondividing cells D) they do not have to be engineered by taking out their replicative genes Answer: B Section: ST 3.2 Bloom's Taxonomy: Applying/Analyzing

.

6) Viral vectors are not the only method being studied for gene therapy purposes. Which of the following nonviral delivery methods is being explored for gene therapy?

A) gene pills B) DNA covered with protein C) DNA bound to the surface of liposomes D) electrically stimulating cells to take up DNA Answer: A Section: ST 3.2 Bloom's Taxonomy: Remembering/Understanding 7) Hematopoietic stem cells (HSCs) are particularly useful in gene therapy because A) they are short lived in vitro B) they differentiate into all tissue types C) they are useable across many patients without rejection fears D) they are easy to obtain from patients Answer: D Section: ST 3.2 Bloom's Taxonomy: Remembering/Understanding

.

8) Treatment of ADA-SCID is often considered the first success for gene therapy. Ashanti DeSilva is a young girl that suffered from an autosomal form of SCID and was treated using retroviral gene therapy. Explain why the treatment is not an undisputed success.

Answer: While Ashanti's immune system was bolstered by the gene therapy, she also received regular injections of adenosine deaminase enzyme. This made it unclear whether the adenosine deaminase found in her system was due to the active gene delivered by the gene therapy or the injections of purified adenosine deaminase.

Section: ST 3.3 Bloom's Taxonomy: Evaluating/Creating 9) What is the function of the RISC complex in RNA interference? The RISC complex .

A) transports double-stranded RNA molecules across the lipid bilayer B) cleaves double-stranded RNAs into 21- to 25-nt-long pieces C) shuttles the siRNAs to the target mRNA D) inhibits transcription of the target gene Answer: C Section: ST 3.6 Bloom's Taxonomy: Remembering/Understanding 2 Copyright © 2021 Pearson Education Ltd.

10) Gene therapy to treat Leber congenital amaurosis has recently shown some promise in revitalizing the gene therapy field. This is due in large part to injecting adeno-associated virus loaded with RPE65 just under the retina. What does the product of gene RPE65 metabolize?

A) rod cell waste B) cone cell waste C) retinol D) retinal Answer: C Section: ST 3.5 Bloom's Taxonomy: Remembering/Understanding 11) Explain how RNA-based therapeutics could be used to inhibit cancer cell growth.

Answer: Antisense RNAs or vectors encoding miRNAs could be used to interfere with the translation of mRNAs expressed from oncogenes within cancer cells, inhibiting their growth.

Section: ST 3.6 Bloom's Taxonomy: Evaluating/Creating 12) ZFNs and TALENs have high specificity for their target cleavage site. How is this specificity achieved?

Answer: Both ZFNs and TALENs require dimerization to work. This means that each monomer must recognize and bind its target sequence and be close enough for dimerization to occur. ZFNs have a spacing of 5-7 nucleotides, whereas TALENs have a spacing of about 13 base pairs.

Section: ST 3.6 Bloom's Taxonomy: Applying/Analyzing 13) Explain how treating Timothy Brown's acute myeloid leukemia with stem cells from an HIV-resistant donor cured him of HIV. Answer: Because mutant CCR5 T cells in the stem cell transplant to treat Timothy Brown's leukemia, the HIV was not capable of spreading to the new T cells. This resulted in the decline of HIV-positive T cells and the eventual curing of his or her HIV. Section: ST 3.6 Bloom's Taxonomy: Evaluating/Creating 14) What disease did MIT researchers cure using the CRISPR-Cas system to edit genes in vivo?

A) Type 1 tyrosinemia B) HIV C) RDEB D) -thalassemia Answer: A Section: ST 3.6 Bloom's Taxonomy: Remembering/Understanding

3 Copyright © 2021 Pearson Education Ltd.

15) CTL019 was the first gene edited CAR-T cell therapy approved by the FDA. Initial treatments of patients with CTL019 demonstrated over remission rate after the first treatment. Exemplifying the power of gene editing in treating acute lymphoblastic leukemia.

A) 99% B) 95% C) 90% D) 80% Answer: D Section: ST 3.6 Bloom's Taxonomy: Remembering/Understanding 16) The gene PD-1 plays an important role in immune response to tumor cells. How would using gene editing to knock out this gene help T cells recognize and kill cancer cells?

A) PD-1 will not be expressed on the T-cell surface. This circumvents one line of the cancer cells defense.

B) PD-1 will be expressed in the cytosol of the T cell. This protects PD-1 from cancer cells.

C) PD-1 will not be expressed on the cancer cell surface. This prevents T cell recognition.

D) PD-1 will be expressed in the cytosol of the cancer cell. This protects PD-1 from the T cell.

Answer: A Section: ST 3.6 Bloom's Taxonomy: Remembering/Understanding 17) Which cell type was used in the first human gene therapy trial to treat ADA-SCID? A) red blood cells B) HSCs C) T cells D) fibroblasts Answer: C Section: ST 3.3 Bloom's Taxonomy: Remembering/Understanding 18) In 2015, a group of Chinese researchers used CRISPR-Cas gene editing to edit the HBB gene in 86 human embryos. What were the conclusions form this study?

A) CRISPR-Cas is safe for use in human patients.

B) CRISPR-Cas was not ready for use in human patients.

C) 71 of the 86 embryos contained the desired gene edit.

D) CRISPR-Cas gene editing had no negative effects on the surviving embryos.

Answer: B Section: ST 3.6 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

19) One gene therapy that does not rely on gene editing is RNA interference. This therapy uses to interfere with the translation of mRNAs. A) tRNA B) miRNA C) mRNA D) mtRNA Answer: B Section: ST 3.6 Bloom's Taxonomy: Remembering/Understanding 20) Which of the following gene therapies is currently approved in the United States?

A) enhancement gene therapy B) germ-line gene therapy C) somatic gene therapy D) stem cell gene therapy Answer: C Section: ST 3.7 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Special Topics 4 Advances in Neurogenetics: The Study of Huntington Disease 1) Neurogenetics is best described as .

A) the study of the protein basis for neurodegenerative disease B) the study of the genetic basis for neurodegenerative disease C) the study of genetics that makes human brains unique D) the study of proteins associated with the mitotic division of neurons Answer: B Section: ST 4.1 Bloom's Taxonomy: Remembering/Understanding 2) Huntington disease (HD) is the model neurodegenerative disease because A) it is not 100% penetrant B) only a few analytical approaches have been used to study it C) it is monogenic D) it does not have a defined set of symptoms and disease progression Answer: C Section: ST 4.1 Bloom's Taxonomy: Remembering/Understanding

.

3) HD was the first example of complete dominance in human inheritance. Describe the impact this has on inheriting the disease state.

Answer: As HD exhibits complete dominance in the inheritance pattern, it does not matter if the individual is heterozygous or homozygous for the mutant allele. The disease will develop and progress regardless of the number of mutant alleles present.

Section: ST 4.1 Bloom's Taxonomy: Applying/Analyzing 4) Huntington's observations of the disease that later took his name were published prior to the rediscovery of Gregor Mendel's work on inheritance. Support the argument that Huntington was describing a phenotype with complete dominance.

Answer: Huntington's work described a situation where one parent does not carry the mutant allele and the other could be heterozygous; thus implying that the heterozygous parent passes the mutant allele to the offspring, which in turn suffer from the disease with only one mutant allele in their genotype. The presentation of the disease with only one mutant allele in the genotype is complete dominance.

Section: ST 4.1 Bloom's Taxonomy: Evaluating/Creating

1 Copyright © 2021 Pearson Education Ltd.

5) Early mapping of the HD gene was done using A) RFLP mapping B) SNP mapping C) WGS D) northern blotting Answer: A Section: ST 4.1 Bloom's Taxonomy: Remembering/Understanding

.

6) James Gusella used an RFLP marker on to determine if there was linkage of disease phenotype to the new G8 marker.

A) the same population of HD patients as his first experiment B) a large Venezuelan population that has HD C) a small Venezuelan population that has HD D) a specific subset of an American HD family Answer: B Section: ST 4.1 Bloom's Taxonomy: Remembering/Understanding 7) Gusella demonstrated four haplotypes for HD based on the restriction enzyme HindIII. Which haplotype demonstrated linkage to the disease phenotype?

A) A B) B C) C D) D Answer: C Section: ST 4.1 Bloom's Taxonomy: Remembering/Understanding 8) The success of Gusella, Wexler, and their colleagues in mapping both the G8 marker and HD gene to chromosome 4 launched a new branch of genetics called .

A) neurogenetics B) reverse genetics C) personalized medicine D) genomics Answer: B Section: ST 4.1 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

9) Explain how researchers finally brought the search for the HD gene to a close as reported by the Huntington's Disease Collaborative Research Group.

Answer: Previous efforts had narrowed the location of the Huntington gene to the end of Chromosome 4. One of the genes expressed in this region, called IT15, encoded a previously unknown protein that contained a CAG repeat sequence. Populations without HD had many alleles of this gene with CAG repeats varying from 11 to 34 copies. Individuals with HD had alleles containing 42–66 repeats of the CAG trinucleotide. HD was known to be a trinucleotide repeat disease and the research group proposed HTT as the gene responsible for HD and named its gene product huntingtin.

Section: ST 4.1, ST 4.2 Bloom's Taxonomy: Evaluating/Creating 10) Since the discovery of the HTT gene, genetic testing for the number of CAG repeats has been developed. However, there are serious ethical questions raised by the prospect of prenatal testing as well as the testing of minors. Develop an argument for why testing minors for HD raises some ethical issues.

Answer: Minors should not be tested for HD because there is currently no cure for the disease.

This will have a direct impact on their quality of life as they will have to live with the knowledge that they will develop HD. Section: ST 4.1 Bloom's Taxonomy: Applying/Analyzing 11) HD is known as a polyQ disease. How does "polyQ" relate to HD? A) There are many Q nucleotides added to the HTT gene in the disease state.

B) There are many Q amino acids added to the huntingtin protein in the disease state.

C) There are many CAG repeats, which code for glutamic acid residues in the disease state.

D) There are many trinucleotide repeats, which are called "Qs" in the disease state.

Answer: B Section: ST 4.2 Bloom's Taxonomy: Remembering/Understanding 12) Explain why the mHTT gene is likely to have wide-ranging effects in the disease state?

Answer: The mHTT causes a wide range of effects because of the ubiquitous nature of huntingtin. Huntingtin is pivotal in protein-protein interactions with more than 180 different proteins and is found in almost all cellular compartments from nucleus to mitochondria and cytosol. When mHTT begins to aggregate, it results in the disruption of many cellular processes giving rise to the multitude of disease phenotypes.

Section: ST 4.3 Bloom's Taxonomy: Applying/Analyzing

3 Copyright © 2021 Pearson Education Ltd.

13) The favored model organism for studying the development of HD is A) C. elegans B) D. melanogaster C) Human D) Mouse Answer: D Section: ST 4.4 Bloom's Taxonomy: Remembering/Understanding

.

14) An important study of HD conducted by A. Yamamoto demonstrated that turning off the HTT gene early in the progression of the disease results in the aggregates being degraded and prevention of the disease. Explain how Yamamoto designed his experiment.

Answer: Yamamoto constructed a transgenic mouse with the mHTT gene behind an inducible promoter that responded to doxycycline. When the mice were fed water containing doxycycline, the mHTT gene was turned on and disease progression commenced. When doxycycline was removed from the water, the mHTT gene was turned off and disease progression halted.

Yamamoto monitored the aggregates of mHTT and determined that if the gene expression was arrested early enough, the disease state could be reversed by normal cellular functions.

Section: ST 4.4 Bloom's Taxonomy: Evaluating/Creating 15) Gene silencing has been effective in mouse models of HD that are heterozygous for the mHTT gene. The results are promising as .

A) the study demonstrated the ZFN only targeted the mHTT allele and had no effect on the expression of other genes containing CAG repeats.

B) both mHTT and HTT were down regulated C) the injection into both sides of the mouse brain resulted in reduced mHTT D) the injection into one side of the mouse brain affected both sides of the brain Answer: A Section: ST 4.5 Bloom's Taxonomy: Remembering/Understanding 16) Researcher Xiao-Jiang Li has used to edit exon 1 of the human mutant HTT allele in mice. Three weeks after injection, the production of mutant human HTT protein was suppressed and the mice demonstrated a reduction in protein aggregates.

A) CRISPR-Cas B) ZFNs C) TALENs D) lentiviral transfection Answer: A Section: ST 4.5 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

17) What is the probability that an affected father, who had an unaffected mother, will pass the mutant HD allele to his offspring?

A) 0% B) 25% C) 50% D) 100% Answer: C Section: ST 4.1 Bloom's Taxonomy: Applying/Analyzing 18) How was synaptic transmission affected in transgenic Drosophila strains carrying a mutant human HD allele?

A) Synaptic vesicles carrying neurotransmitter were smaller in transgenic flies.

B) Transgenic flies did not produce neurotransmitters.

C) Rates of synaptic transmission increased in transgenic flies.

D) Neurons in transgenic mice began releasing different neurotransmitters.

Answer: A Section: ST 4.3 Bloom's Taxonomy: Remembering/Understanding 19) Which of the following statements regarding the effects of mHTT on mitochondrial function is false?

A) mHTT increases the rate of ATP synthesis.

B) mHTT binds to the outer mitochondrial membrane.

C) mHTT leads to increased levels of reactive oxygen species.

D) mHTT disrupts the electron transport chain.

Answer: A Section: ST 4.3 Bloom's Taxonomy: Remembering/Understanding 20) In normal cells, misfolded proteins typically get A) sent to the mitochondria for proper folding B) tagged by ubiquitin for degradation in the proteasome C) localized to a lysosome by a chaperone protein D) released from the cell by exocytosis Answer: B Section: ST 4.3 Bloom's Taxonomy: Remembering/Understanding

.

5 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Special Topics 5 DNA Forensics 1) Which of the following is not possible using DNA profiling?

A) identifying victims of mass disasters B) establishing probability of paternity C) connecting an individual to a biological sample D) identification of plant or animal materials E) confirming guilt of an individual Answer: E Section: Introduction Bloom's Taxonomy: Remembering/Understanding 2) Alleles of both VNTRs and STRs are identified based on A) the different phenotypes they express B) differing numbers of repeated nucleotide patterns C) the different amino acid sequences they code for D) single nucleotide differences Answer: B Section: ST 5.1 Bloom's Taxonomy: Remembering/Understanding 3) STRs are useful for DNA profiling because they .

A) are strongly influenced by selection B) code for proteins C) are highly variable in human populations D) typically have two alleles per locus in human populations E) are easily connected to human phenotype Answer: C Section: ST 5.1 Bloom's Taxonomy: Remembering/Understanding 4) Why is PCR a useful tool in DNA forensics?

A) it allows very small amounts of DNA to be useful as evidence B) it eliminates the need to know anything about the genome C) it targets protein coding regions of the genome D) it eliminates the possibility of sample contamination Answer: A Section: ST 5.1 Bloom's Taxonomy: Applying/Analyzing

1 Copyright © 2021 Pearson Education Ltd.

.

5) Why is electrophoresis an appropriate tool to determine STR genotypes?

A) no PCR is necessary B) it allows visualization of the nucleotide sequence or the allele C) mixed evidence samples can be separated into individual genotypes D) STR alleles are based on DNA fragment size E) it facilitates restriction enzyme reactions Answer: D Section: ST 5.1 Bloom's Taxonomy: Remembering/Understanding 6) In an electropherogram showing the results of an STR profile, two peaks indicate A) the individual is homozygous at the STR locus B) the individual is heterozygous at the STR locus C) the sample includes DNA from more than one individual D) the STR is not useful for identifying this individual E) two alleles exist in the population Answer: B Section: ST 5.1 Bloom's Taxonomy: Remembering/Understanding 7) Which of the following is a limitation of Y-chromosome STR profiling?

A) there are less than 10 STRs on the human Y chromosome B) they are only found in males C) they cannot be heterozygous D) individuals from the same paternal lineage will have the same profile E) they are similar to STR loci on the X chromosome Answer: D Section: ST 5.1 Bloom's Taxonomy: Remembering/Understanding 8) Which of the following is a limitation of mitochondrial DNA profiling?

A) there are multiple copies of the mitochondrial genome per cell B) mitochondrial DNA can be amplified from degraded DNA C) low variability in the population D) individuals from the same maternal lineage will have the same profile E) both sexes possess mitochondrial DNA Answer: D Section: ST 5.1 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

.

9) What is DNA Phenotyping?

A) a SNP profile B) a genotypic profile based on combined mitochondrial and Y-chromosome STR genotypes C) a controversial method using DNA sequence information to predict identifying phenotypes D) a controversial method using known phenotypes to predict a suspects DNA profile E) the correct method by which DNA evidence should be stored for future use Answer: C Section: ST 5.1 Bloom's Taxonomy: Remembering/Understanding 10) Currently the most commonly used tool for DNA profiling is A) autosomal STRs B) mitochondrial DNA C) Y-chromosome STRs D) single nucleotide polymorphisms (SNPs) E) DNA phenotyping Answer: A Section: ST 5.1 Bloom's Taxonomy: Remembering/Understanding

.

11) Why is it necessary to use a large number of SNP loci to create a useful DNA profile?

A) SNPs are typically found clustered together in specific regions of the genome B) SNPs are randomly distributed throughout the genome C) SNPs have a low number of alleles per locus in human populations D) most SNPs are not in Hardy-Weinberg equilibrium E) SNPs are more variable in males than in females Answer: C Section: ST 5.1 Bloom's Taxonomy: Evaluating/Creating 12) Which of the following statements is most true with respect to DNA profiling?

A) it is not necessary to know anything about the allele frequencies in the populations of the loci used B) DNA profiles can exclude an individual with 100% confidence, but identification is based on probability C) DNA profiles can identify an individual with 100% confidence, but exclusion is based on probability D) loci used in DNA profiling are most useful if their frequencies are E) SNPs are more variable in males than in females Answer: B Section: ST 5.2 Bloom's Taxonomy: Evaluating/Creating

3 Copyright © 2021 Pearson Education Ltd.

13) Random match probability or profile probability in DNA profiling assumes A) allele and genotype frequencies in the populations are influenced by selection B) allele and genotype frequencies in the populations are influenced by genetic drift C) estimates of allele frequencies for the population are unknown D) loci used in DNA profiling are most useful if their frequencies are E) genotype frequencies are in Hardy-Weingberg proportions Answer: E Section: ST 5.2 Bloom's Taxonomy: Remembering/Understanding

.

14) Which of the following would lower the estimate of the uniqueness of a DNA profile?

A) originating from large population of random mating individuals B) creating the profile using a large number of loci C) using loci with a large number of alleles in the population D) the number of related individuals E) using neutral loci Answer: D Section: ST 5.2 Bloom's Taxonomy: Applying/Analyzing 15) DNA profiles included in the CODIS database include those that are from A) anyone who has ever been arrested B) school teachers and others working with children C) law enforcement officers D) convicted offenders and crime scene evidence E) anyone born after 1986 Answer: D Section: ST 5.2 Bloom's Taxonomy: Remembering/Understanding

.

16) How many STR loci do law enforcement agencies in North America use to create DNA profiles used in legal proceedings?

A) 1 B) 5 C) 10 D) 15 E) 20 Answer: E Section: ST 5.1 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

17) You have been genotype for locus D5S818 used in the CODIS STR database and are heterozygous with alleles 12,13. These alleles have a frequency in the population of 0.3841 and 0.1407 respectively. The probability that another person shares your genotype at this locus is approximately .

A) 1 in 10.8 B) 1 in 18.5 C) 1 in 9.25 D) 1 in 30.5 E) 1 in 3.5 Answer: C Section: ST 5.2 Bloom's Taxonomy: Applying/Analyzing 18) You have been genotyped for five loci used in the CODIS STR database. The genotypes and allele frequencies from the population database are in the following table: Locus D5S818 TPOX D8S1179 CSF1PO D19S433

Genotype 12 13 8 8 12 14 10 10 11 15

Allele Frequency 0.3841 0.1407 0.5348 0.1854 0.1656 0.2169 0.3394 0.048

What is the probability a person chosen at random from the population will have the same multilocus genotype?

A) 1 in 1,000,000 B) 1 in 2,749,391 C) 1 in 39,865 D) 1 in 343,673 E) 1 in 3,189,240 Answer: D Section: ST 5.2 Bloom's Taxonomy: Applying/Analyzing

5 Copyright © 2021 Pearson Education Ltd.

19) The Lukis Anderson case highlights the issue that .

A) two individuals can have the same DNA profile even when using 20 STR loci B) DNA can be transferred from one location to another C) the ethics of using familial DNA D) bias introduced by maintaining DNA profile databases E) frequency of contamination in crime labs Answer: B Section: ST 5.3 Bloom's Taxonomy: Remembering/Understanding 20) What is familial DNA testing?

A) using mitochondrial DNA profiles to limit the suspect pool to those of a specific maternal lineage B) using Y-chromosome DNA profiles to limit the suspect pool to those of a specific paternal lineage C) using partial matches to an individual's DNA profile in a database to justify investigation of relatives of that individual D) a special case of identical twin DNA profiling Answer: C Section: ST 5.3 Bloom's Taxonomy: Remembering/Understanding

6 Copyright © 2021 Pearson Education Ltd.

Essentials of Genetics, 10th Edition (Klug) Special Topics 6 Genetically Modified Foods 1) Describe the difference between genetically modified (GM) food and foods that are selectively bred over centuries.

Answer: GM foods are modified using biotechnology as opposed to non-GM foods, which are grown and mated selectively for many generations to promote desired phenotypes.

Section: ST 6.1 Bloom's Taxonomy: Applying/Analyzing 2) When GMOs are generated through the transfer of genetic material between unrelated species, the GMO is said to be .

A) cisgenic B) transgenic C) isogenic D) exogenic Answer: B Section: ST 6.1 Bloom's Taxonomy: Remembering/Understanding 3) Describe how genetically modifying plants could provide farmers with a boost in productivity of their crops.

Answer: Modifying the genome of crops planted by the farmer to have favorable traits such as improved yield, improved drought tolerance, and herbicide resistance would increase the farmers' productivity.

Section: ST 6.1 Bloom's Taxonomy: Evaluating/Creating 4) What is a major concern surrounding GMOs that critics cite as an argument against GMOs? They could escape containment and .

A) outcompete non-GMOs resulting in extinction of the GMO B) breed with non-GMOs, conferring undesirable phenotypes C) decrease genetic diversity D) produce even more useful genetic variants Answer: B Section: ST 6.3 Bloom's Taxonomy: Remembering/Understanding 5) Bt crops are used mainly because the Cry proteins, which confer the insecticidal properties, are shown to .

A) pollute the ground water B) break down slowly in sunlight C) have few negative effects on humans D) only target a few pest species Answer: C Section: ST 6.1 Bloom's Taxonomy: Remembering/Understanding 1 Copyright © 2021 Pearson Education Ltd.

6) Genetic modification of rice to generate beta-carotene in the endosperm has great potential to alleviate vitamin A deficiency in third-world countries. However, there is still much controversy surrounding its growth. Provide supporting arguments for the planting and consumption of Golden Rice 2.

Answer: The GM food crop Golden Rice 2 could alleviate the vitamin A deficiencies found in over 190 million children, 19 million pregnant women, and between 250,000 and 500,000 children. The Golden Rice 2 seed will be made available at the same price as non-GM seed and farmers will be allowed to keep and replant seed from their own crops.

Section: ST 6.1 Bloom's Taxonomy: Evaluating/Creating 7) Current methods used in the generation of GM plants are still in development. The chance of a productive GM plant being generated by biolistic methods is dependent on .

A) in vitro cells taking up DNA-coated gold beads B) being infected by a particular bacterium C) integration of the Ti plasmid D) random mutations making a proper promoter Answer: A Section: ST 6.2 Bloom's Taxonomy: Remembering/Understanding 8) Selection of a successfully transformed plant cell requires several steps including growth in a selective media, checking for the gene of interest, and checking for phenotype. Propose a method of detecting a transfected gene's presence in a developing organism.

Answer: PCR primers that specifically target the gene of interest can be designed and the genomic DNA isolated from the GM plant used as the template for PCR. If the gene is present in the genome, a PCR product will be formed.

Section: ST 6.2 Bloom's Taxonomy: Evaluating/Creating 9) The use of a positive selection marker in the development of Golden Rice 2 allowed the transgenic plants to grow on , which is usually unusable by plants.

A) mannose 6-phosphate B) fructose 6-phosphate C) glucose 6-phosphate D) xylose 6-phosphate Answer: A Section: ST 6.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

10) Growing GM crops containing a bacterial gene that confers tolerance to glyphosate is an example of .

A) negative marker B) positive marker C) insecticide resistance D) agrobacterium resistance Answer: B Section: ST 6.1 Bloom's Taxonomy: Remembering/Understanding 11) Which of the following statements regarding GM plants is false?

A) GM plants can be developed to alleviate vitamin deficiencies in consumers.

B) Genes can be incorporated to make GM plants resistant to insect predation.

C) Genetically modified crop varieties are accepted by all countries.

D) Genes conferring herbicide resistance are most common among GM crops.

Answer: C Section: ST 6.1 Bloom's Taxonomy: Remembering/Understanding 12) Describe how Ti plasmids are used by scientists to develop GM crops.

Answer: Agrobacterium tumefaciens is a soil microbe that can infect plant cells. They contain Ti plasmids that integrate a segment of their DNA, known as transfer DNA (T-DNA), into random locations within the plant genome. To use the Ti plasmid as a transformation vector, scientists remove the T-DNA segment and replace it with cloned DNA of the genes to be introduced into plant cells. The newly introduced gene is cloned next to an appropriate promoter sequence that will direct transcription in the required plant tissue.

Section: ST 6.2 Bloom's Taxonomy: Evaluating/Creating 13) Why is it that gene-edited organisms do not have the same regulatory oversight as other GM foods?

A) Gene-edited organisms are safer.

B) Gene-edited organisms contain no transgene material.

C) Transgenic organisms are more complicated and require more testing.

D) Transgenic organisms are simpler than gene-edited organisms.

Answer: B Section: ST 6.2 Bloom's Taxonomy: Applying/Analyzing

3 Copyright © 2021 Pearson Education Ltd.

14) GM foods are controversial for many reasons. Which of the following does not represent a controversy between advocates and critics of GM foods?

A) GM foods have an unknown impact on consumers.

B) GM foods have an unknown impact on the environment.

C) GM foods increase yields on farm land.

D) GM foods cause disease in consumers.

Answer: C Section: ST 6.1, ST 6.3 Bloom's Taxonomy: Remembering/Understanding 15) Why should GM foods be considered on a case-by-case basis instead of making general statements about all GM foods?

Answer: Each GM food is developed for a specific purpose. Each one contains different genes from different organisms and it is possible that those genes result in different phenotypes in varying host organisms. Therefore, one cannot make blanket statements about GMOs when each is a unique creation.

Section: ST 6.3 Bloom's Taxonomy: Applying/Analyzing 16) GM foods are considered safe by advocates who cite .

A) 20 years of consumption with no reliable reports of adverse effects B) the lack of reliable data due to relaxed governmental oversight C) robust toxicity testing that is done in humans shows no adverse effects D) the GM food is exactly the same as the non-GM food Answer: A Section: ST 6.3 Bloom's Taxonomy: Remembering/Understanding 17) Critics of GM foods often cite the as reason to be skeptical of the safety of GM foods.

A) vast number of studies conducted on the consumption of GM foods B) fact that most GM foods are not directly consumed by humans C) the extreme governmental oversight of GM food production D) the in depth testing of DNA that makes it through the food refinement process Answer: B Section: ST 6.3 Bloom's Taxonomy: Remembering/Understanding 18) Scientist have introduced a DNA sequence into chickens that encodes a hairpin RNA that inhibits the .

A) spreading avian influenza B) reproduction of chickens C) chickens from laying less than three eggs D) chickens from contracting avian influenza Answer: A Section: ST 6.4 Bloom's Taxonomy: Remembering/Understanding 4 Copyright © 2021 Pearson Education Ltd.

19) Why is the cassava plant difficult to engineer without the help of modern biotechnology?

A) The plant is long lived making selective breeding ineffective.

B) The plant does not undergo recombination events, which limits genetic drift.

C) The plant has a very low mutation rate.

D) The plant is integral to African agriculture and is already optimized.

Answer: A Section: ST 6.4 Bloom's Taxonomy: Remembering/Understanding 20) GM plants that are in various developmental pipelines include those that are A) tolerant of increased salinity B) tolerant of heavy metals C) tolerant of radiation exposure D) incapable of recombination Answer: A Section: ST 6.4 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.

.

Essentials of Genetics, 10th Edition (Klug) Special Topics 7 Genomics and Precision Medicine 1) Precision medicine is .

A) an approach that applies precise treatment to a large population B) an approach that uses individual and molecular information to direct treatment C) an approach that uses "-omics" data to determine treatment for large populations D) an approach that maximizes the cost/benefit to a sizable population of patients Answer: B Section: ST 7.1 Bloom's Taxonomy: Remembering/Understanding 2) Which individual would be most efficient at metabolizing acetaminophen? An individual that is .

A) homozygous for the wild-type CYP2D6 gene B) heterozygous for the VKORC1 gene C) carrying a duplicated wild-type CYP2C19 gene D) homozygous for the SLCO1B1 gene Answer: A Section: ST 7.1 Bloom's Taxonomy: Applying/Analyzing 3) Mutations in the proteins that are involved in drug metabolism are of immense importance in personalized medicine. Explain how being aware of mutations in the gene CYP2D6 can affect a physician's treatment plan for an individual.

Answer: CYP2D6 is involved in the metabolism of a significant number of current pharmaceuticals. Mutations in CYP2D6 can result in altered metabolism rates for the drugs. A physician that is aware of the mutations and their affect on drug metabolism will be able to adjust the dosage so as to avoid over/under dosing of the patient.

Section: ST 7.1 Bloom's Taxonomy: Evaluating/Creating 4) The drug Herceptin inhibits .

A) enzyme oxidation reactions in breast tissue B) cholesterol transporters in the liver C) excessive signaling through tyrosine receptor kinases D) cell growth by blocking G protein-coupled receptors Answer: C Section: ST 7.1 Bloom's Taxonomy: Applying/Analyzing

1 Copyright © 2021 Pearson Education Ltd.

5) Targeted drugs are therapeutics designed to .

A) treat a large class of diseases B) treat a specific disease based on the efficacy of the drug in a large population C) treat a specific disease based on the efficacy of the drug in a small population D) treat only one person Answer: C Section: ST 7.1 Bloom's Taxonomy: Remembering/Understanding 6) One method for determining if a targeted drug will be effective is to perform a FISH assay on a biopsy of the diseased tissue. How does FISH demonstrate potential efficacy of the drug?

Answer: A FISH assay determines the copy number of the gene for which the fluorescent probe was designed to target. If the target gene is present in a higher than normal copy number, the drug target is likely to be expressed at high levels as well. This suggests that the drug target is present and will respond to the targeted therapeutic.

Section: ST 7.1 Bloom's Taxonomy: Applying/Analyzing 7) Immunotherapies seek to take advantage of A) a patient's own cancer cells B) a patient's own immune system C) a patient's own genome D) a patient's own SNPs Answer: B Section: ST 7.2 Bloom's Taxonomy: Remembering/Understanding

in treating cancers.

8) A patient suffering from an unknown cancer is being treated with standard chemotherapy.

Describe how personalized medicine could potentially help treat this individual.

Answer: By sequencing the entire genome of both their normal and cancer cells, mutations and expression profiles can be determined for the cancer cells. These mutations and expression profiles can then be compared to databases such as that developed in the Cancer Genome Atlas Project. The patient could be matched to a targeted drug based on their tumor's genomic profile.

Section: ST 7.2 Bloom's Taxonomy: Evaluating/Creating 9) Neoantigens are .

A) mutations that result in cancer B) nonself antigens contained within the mutated proteins expressed in cancer cells C) mutated proteins expressed on the outer membrane of cancer cells D) silent mutations occurring within cancer cells Answer: B Section: ST 7.2 Bloom's Taxonomy: Remembering/Understanding

2 Copyright © 2021 Pearson Education Ltd.

10) Cancer cells have developed a mechanism to repress T-cell response to neoantigens. Present a method that would allow restoration of the T-cell response when the response is halted at a checkpoint.

Answer: To restore T-cell function when the response is halted at a checkpoint, one would design a checkpoint inhibitor. This will prevent the cancer cell from suppressing the T-cell response at the checkpoint.

Section: ST 7.2 Bloom's Taxonomy: Applying/Analyzing 11) Adoptive cell transfer (ACT) therapies involve removing tumor-infiltrating lymphocytes from the patient's tumor and for reactivity, culturing , and reintroducing them back into the patient.

A) screening; in vivo B) screening; in vitro C) mutating; in vivo D) mutating; in vitro Answer: B Section: ST 7.2 Bloom's Taxonomy: Remembering/Understanding 12) One method to target a patient's nonresponsive immune system to a cancer cell is through the use of .

A) CARs B) WGS C) SNPs D) VNTRs Answer: A Section: ST 7.2 Bloom's Taxonomy: Remembering/Understanding 13) Chimeric antigen receptors are engineered surface receptors on autologous T cells. Describe how CARs provide a method of targeting cancer cells.

Answer: CARs allow the engineered targeting of cancer cells by recombinantly expressing genetically engineered cell surface receptors that are designed to activate T cells upon binding neoantigens on the cancer cell surface.

Section: ST 7.2 Bloom's Taxonomy: Evaluating/Creating 14) Which of the following side effect is not associated with current CAR T-cell therapies?

A) inflammatory response B) neurotoxicity C) eventual tumor resistance D) headaches Answer: D Section: ST 7.2 Bloom's Taxonomy: Remembering/Understanding 3 Copyright © 2021 Pearson Education Ltd.

15) The age of precision medicine comes with many legal and ethical issues. Evaluate some of the current technical hurdles to overcome before precision medicine becomes a standard part of medical care.

Answer: One technical hurdle is the logistics of storing and readily accessing the medical records and genomic information of patients. Very few hospitals and practitioners currently have the information technology to handle the data. Second is the legal and ethical issues that surround patient confidentiality and the sharing of records to improve the accuracy and usefulness of precision medicine.

Section: ST 7.4 Bloom's Taxonomy: Evaluating/Creating 16) In the future, medical schools will have to adapt their education to include better equip physicians to interpret precision medicine data.

A) bioinformatics B) anatomy C) pharmacology D) biochemistry Answer: A Section: ST 7.4 Bloom's Taxonomy: Remembering/Understanding

to

17) Describe the structural differences between endogenous T-cell receptors and CAR T-cell receptors.

Answer: Endogenous T-cell receptors are very complex and contain multiple transmembrane domains and signaling domains. They also have constant regions as well as the variable regions that bind antigens. CAR T-cell receptors are much simpler; comprised of a signaling domain, a single transmembrane domain, and two variable regions connected with a linker sequence.

Section: ST 7.2 Bloom's Taxonomy: Applying/Analyzing 18) Which of the following strategies are not utilized by tumors to avoid immune system activity?

A) use of checkpoint molecules B) the presence T-regs C) abnormal expression of MHC molecules D) T-cell destruction through endocytosis Answer: D Section: ST 7.2 Bloom's Taxonomy: Remembering/Understanding

4 Copyright © 2021 Pearson Education Ltd.

19) A disease that leads to the loss of thymus function is most likely to inhibit maturation of which cell type?

A) B cells B) macrophages C) T cells D) dendritic cells Answer: C Section: ST 7.2 Bloom's Taxonomy: Applying/Analyzing 20) Immune system cells that are responsible for recognizing neoantigens and targeting tumors are called .

A) macrophages B) tumor-infiltrating lymphocytes C) regulatory T cells D) monocytes Answer: B Section: ST 7.2 Bloom's Taxonomy: Remembering/Understanding

5 Copyright © 2021 Pearson Education Ltd.