, TESTBANK FOR Lewin's GENES XII Twelfth Edition Krebs
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All available appendices and Excel files (if included in the original resources) are
provided.
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,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
Chapter 1
Multiple Choice
1. The complete set of hereditary information carried by an organism is its:
A) chromosome.
B) genome.
C) double helix.
D) gene.
E) nucleotide.
Ans: B
2. Hershey and Chase tracked DNA during phage infection using a radioisotope of:
A) oxygen.
B) sulfur.
C) nitrogen.
D) phosphorus.
E) carbon.
Ans: D
3. The process of ___________ in eukaryotic cells is analogous to bacterial transformation.
A) transfection
B) transcription
C) transition
D) translation
E) translocation
Ans: A
4. In a polynucleotide, a phosphate group is linked to the ________ carbons of two pentoses.
A) 1′ and 2′
B) 1′ and 3′
C) 3′ and 5′
D) 1′ and 5′
E) 2′ and 3′
Ans: C
5. In DNA, the number of times one strand crosses over the other in space is its:
A) twisting number.
B) positive supercoiling number.
C) writhing number.
D) negative supercoiling number.
E) linking number.
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 1
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
Ans: E
6. In the B-form of DNA:
A) the two polynucleotide chains are described as parallel.
B) base pairs lie perpendicular to the sugar–phosphate backbone.
C) the width of the helix varies between A-T and G-C base pairs.
D) purine bases pair with other purines.
E) there are approximately 12 base pairs per turn of the helix.
Ans: B
7. DNA replication is semiconservative, meaning that:
A) one of the two daughter duplexes consists of both original parental strands.
B) for each daughter duplex, only one of the parental strands is used as a template for both
daughter strands.
C) each daughter duplex may differ in sequence from the parental duplex.
D) each daughter duplex consists of one parental strand and one newly synthesized daughter
strand.
E) each daughter duplex has sections of original parental duplex and sections of newly
synthesized daughter duplex.
Ans: D
8. Which of the following is a transition mutation?
A) A-T → G-C
B) A-T → C-G
C) A-T → T-A
D) G-C → C-G
E) G-C → T-A
Ans: A
9. As compared to the original mutation, a suppression mutation occurs:
A) in the homologous allele of the same gene.
B) at the same site.
C) in another nonhomologous gene.
D) in a different codon of the same gene.
E) in another site of the same codon.
Ans: C
10. A viroid is:
A) RNA encapsulated in protein.
B) double-stranded DNA encapsulated in protein.
C) a naked DNA molecule.
D) single-stranded DNA encapsulated in protein.
E) a naked RNA molecule.
Ans: E
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 2
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
11. A chromosome is best defined as:
A) the complete sequence of a gene, including sequences removed from the RNA product.
B) the functional coding sequences of a single DNA molecule.
C) the complete set of hereditary information of an organism.
D) a linear array of genes.
E) the sequence of a gene that has coding information for the polypeptide product.
Ans: D
12. Mutation is random with respect to the structure and function of a gene. As a result,
A) the effects of a particular mutation cannot be predicted.
B) mutations are likely to impair the function of a gene.
C) mutations are equally likely at every nucleotide of the gene.
D) mutations invariably destroy the function of the gene.
E) mutations are likely to improve the function of a gene.
Ans: B
13. In the complementation test, crossing two homozygous recessive individuals with similar
phenotypes to yield all wild-type offspring means that:
A) the two parents carry different mutations of the same nucleotide.
B) the two parents’ mutations are in the same gene.
C) the two parents’ mutations affect different traits.
D) the two parents carry identical mutations.
E) the two parents’ mutations are in different genes.
Ans: E
14. Mutations that have no apparent effect are called:
A) silent mutations.
B) loss-of-function mutations.
C) null mutations.
D) leaky mutations.
E) gain-of-function mutations.
Ans: A
15. Genes that are on opposite ends of a long eukaryotic chromosome:
A) never recombine.
B) show less than 50% recombination.
C) show exactly 50% recombination.
D) show between 50% and 100% recombination.
E) show 100% recombination.
Ans: C
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 3
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
16. Full or partial reversion of a deletion mutation may be accomplished by:
A) an insertion mutation in another gene.
B) an insertion mutation close to the original deletion.
C) a substitution mutation close to the original deletion.
D) a second deletion mutation close to the original deletion.
E) a second deletion mutation in another gene.
Ans: B
17. A closed reading frame:
A) contains frequent stop codons.
B) contains many substitution mutations.
C) is formed by the deletion of a sequence from an open reading frame.
D) begins and ends with the same sequence.
E) produces a polypeptide that serves no apparent function.
Ans: A
18. The colinearity of prokaryotic genes means that:
A) within a prokaryotic gene, both DNA strands encode the same polypeptide.
B) a prokaryotic polypeptide has exactly 3x as many amino acids as the number of nucleotides in
the gene encoding it.
C) prokaryotic proteins are identical in length to their genes.
D) prokaryotic genes contain a continuous unbroken sequence encoding a polypeptide.
E) a physical map of a gene will not match the amino acid map of its polypeptide product.
Ans: D
19. The production of RNA from a DNA template is called:
A) splicing.
B) translation.
C) RNA processing.
D) transport.
E) transcription.
Ans: E
20. A regulatory control site of a gene:
A) must be very close to the gene it affects.
B) must lie upstream of the gene it affects.
C) is cis-acting.
D) is trans-acting.
E) must be very distant from the gene it affects.
Ans: C
Short Answer
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 4
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
1. The basic building block of nucleic acids is the nucleotide. It is composed of three components.
What are they?
Ans: They are a nitrogenous base, a sugar, and a phosphate.
2. There are two types of nucleic acids in cells, DNA and RNA. List two differences between the
structure of DNA and RNA.
Ans: 1) DNA has the pyrimidine base thymine, whereas RNA contains uracil. 2) DNA has 2′-
deoxyribose, whereas RNA has ribose. The sugar in RNA has an -OH group at the 2′ position of
the pentose ring.
3. If the mole % of A residues in double-stranded DNA is 29%, what would the mole % of G
residues be?
Ans: The two strands of DNA are complementary with G residues on one strand paired with C
residues on the other strand, and with A residues paired with T residues. Therefore, if the mole %
of A is 29%, then the mole % of T is 29% and the mole % of G + C would be 42%. G must equal
C, so as a result the mole % of G and of C would be 21%.
4. The size of the smallest human chromosome is 4.7 × 107 bp. If the DNA is in the B-form helix,
what would be the length in cm of the chromosome fully extended if not packaged by proteins?
Ans: 1.6 cm. In B-form DNA, the distance between adjacent nucleotides is 3.4 Å. Thus the length
of the chromosome would be 4.7 × 107 × 3.4 Å = 16 × 107 Å. There are 108 Å per cm, so the
length would be 1.6 cm.
5. Meselson and Stahl used density labeling of DNA to show that DNA replication occurs via a
semiconservative mechanism. In their experiment, they started with an organism grown in a
heavy density label (15N). After two generations of growth in light medium (the more common 14N
isotope), if the DNA is isolated and separated by density, how many bands would be observed
and how would their density compare with the starting DNA?
Ans: In each generation, the newly synthesized DNA is composed of one parental strand and one
new daughter strand. After two generations, the DNA would yield two different density bands.
Two of the daughters would contain one original heavy parental strand and one light daughter
(hybrid density) strand, and two of the daughters would contain two light strands (light density).
6. The genetic information of organisms is always in the form of nucleic acid. Give examples of
cellular organisms or other genetic systems whose genomes are double-stranded DNA, single-
stranded DNA, double-stranded RNA, or single-stranded RNA.
Ans: Living organisms (either prokaryotic or eukaryotic cells) always have genomes that are
double-stranded DNA. There are, however, many viruses of both prokaryotes and eukaryotes
whose genomes can be single-stranded DNA, or double- or single-stranded RNA, depending on
the virus.
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 5
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
7. The central dogma of molecular biology is that genetic information flows from DNA to RNA by
transcription, and then to polypeptide by translation. Give an example of an exception to this
generally true statement.
Ans: Retroviruses have genomes that are single-stranded RNA, and as part of their replication
cycle they convert the RNA into first a single-stranded DNA (using an enzyme called reverse
transcriptase) and then a double-stranded DNA, in a process called reverse transcription. The
double-stranded DNA copy of its genome integrates into the DNA of the host cell it infects and
functions as any other gene, directing the synthesis of both RNA and polypeptide. In this
example, genetic information is flowing from RNA to DNA.
8. Is the enzyme reverse transcriptase an example of an RNA polymerase or a DNA polymerase?
Ans: Reverse transcriptases are RNA-dependent DNA polymerases that use deoxynucleotides as
substrates.
9. In the process of DNA replication, the two parental strands separate and serve as templates for
the new daughter strands. During this process, are covalent bonds being broken? Explain.
Ans: No. The separation of the two parental strands involves breakage of the hydrogen (H) bonds
(not covalent bonds) between the complementary base pairs. The covalent phosphodiester bonds
between adjacent nucleotides on each strand are not broken.
10. The separation of the double helical DNA in cells is mediated by enzymes called helicases. In
the laboratory, however, what method is commonly used to denature (separate) the individual
strands of a double helical DNA?
Ans: DNA may be denatured by heating the DNA above its Tm, usually by boiling.
11. A scientist at the San Diego Zoo is sent DNA samples to analyze. The samples were
collected from two different bacterial species from different parts of the world. She determines
that the G-C content of one of them is 32% and the G-C content of the other is 62%. The labels
have fallen off the tubes, however, so she does not know which sample is which. One of the
organisms lived deep in the ocean at temperatures around 10ºC and the other was isolated from
the Mojave desert, where temperatures can reach 45ºC. Which sample is likely to be which?
Ans: The bacteria from the Mojave desert is likely to have a G-C content of 62%, whereas the
ocean bacteria is likely to have a G-C content of 32%. The stability of the double-stranded DNA is
greater with higher G-C content because G-C base pairs are more stable than A-T base pairs.
Organisms that thrive at elevated temperatures have genomes with a higher fraction of G-C base
pairs to stabilize the double helical structure of their DNA.
12. Once a mutation has occurred in a particular gene, several types of events can occur to
reverse the effects of the original mutation. Give three of these types.
Ans: 1) There may be a true reversion whereby the original base change is reversed, or the
insertion of genetic material, such as a transposon, is deleted. 2) There may be a second site
reversion whereby a second mutation in the same gene changes the amino acid sequence at a
distinct site from the first mutation and allows the polypeptide to regain function. 3) There may be
a suppressor mutation in another gene, altering a different polypeptide. This change bypasses
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 6
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
the need for the original polypeptide or enables it to interact with the mutant form of the original
polypeptide.
13. Why is the modified base 5-methylcytosine a hotspot for mutations?
Ans: The oxidative deamination of 5-methylcytosine converts this base to thymine (not uracil, as
for ordinary cytosine). The DNA repair enzyme uracil-DNA-glycosidase, which recognizes
deoxyuracil as abnormal and removes it, cannot distinguish the incorrect thymine from other
correct thymines, and therefore the new T-A pair is not repaired.
14. One of the most common causes of mutations in DNA results from the oxidative deamination
of cytosine. Why does this lead to a mutation?
Ans: Due to oxidation, the amino group of cytosine is converted to a keto group, changing the
cytosine to uracil. During replication, the new uracil now pairs with adenine instead of thymine,
resulting in a C-G pair being replaced by a T-A pair when the A now pairs with T in the next
replication cycle.
15. Not all infectious agents that cause disease contain nucleic acid. What is the infectious agent
that causes scrapie, a disease of sheep and goats?
Ans: It is a prion—specifically, an abnormally folded form of a protein called PrP. This protein can
exist in cells in various conformations, and those that are resistant to degradation by proteases
not only cause disease, but “convert” normally folded PrP proteins to the abnormal conformation.
16. What is the definition of an allele?
Ans: An allele is an alternative sequence of DNA for a gene at a specific locus.
17. If a gene is duplicated in the genome so that the two copies reside at different chromosomal
locations, are these genes alleles?
Ans: No. Alleles refer to alternative forms of a gene at a specific locus in the genome, so that
copies even of nearly identical sequences that reside at different loci are not alleles of one
another (instead they are called paralogs).
18. Why is the original genetic concept of one gene-one enzyme not entirely true?
Ans: Although it is true that genes encode mRNAs, which in turn encode polypeptides, some
enzymes are composed of more than one polypeptide chain and are therefore encoded by more
than one gene. Secondly, some genes can produce alternative mRNA species (a process termed
alternative splicing) and therefore can generate multiple (related) polypeptides.
19. How is genetic complementation used to determine whether two different mutations occur in
the same gene or in different genes?
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, Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
Ans: A genetic cross is performed with individuals that each are homozygous (both copies of the
gene have the same mutation) for the two different mutations. If the mutations occur in the same
gene (but possibly at different sites within the gene), all of the progeny will exhibit a mutant
phenotype, because they have no wild-type copy of the gene. If the mutations occur in different
genes, then the progeny will inherit one wild-type allele from each parent and will be
phenotypically wild-type.
20. There are many types of changes in DNA that lead to mutations. Give a molecular
explanation for the following types of mutations:
A. a null mutation
B. a loss-of-function mutation
C. a gain-of-function mutation
Ans: A. A null mutation is caused either by a complete or partial deletion of a gene or a change
that renders the polypeptide product completely nonfunctional, such as a premature stop codon
that truncates the polypeptide. B. A loss-of-function mutation is a base change in DNA that alters
an amino acid such that the polypeptide has less activity than the wild-type polypeptide. DNA
changes that affect regulatory regions that decrease the level of the transcript, and therefore the
protein, can also be loss-of-function. C. Gain-of-function mutations are DNA changes resulting in
either an altered amino acid sequence that results in the polypeptide having a novel activity, or
changes in regulatory regions that increase the level of the polypeptide or cause it to be
expressed in tissues or times in development that normally lack the polypeptide. Some
polypeptides can be produced at higher or lower levels than wild-type without causing a mutant
phenotype.
21. Not every difference in DNA sequence results in a mutant allele. Sometimes genes don’t have
a single “wild-type” allele, such as the gene specifying the blood group antigens A and B. Given
the basis for the three types of blood groups (O, A, and B), explain why people with blood type O
are considered universal donors and people with the blood type AB are universal recipients of
blood transfusions.
Ans: The allele for O lacks a functional galactosyltransferase, and adds no sugar to the O antigen
on blood cells. The allele for A produces one that adds an N-acetylgalactose sugar. The allele for
B produces one that adds a galactose sugar. An individual’s immune system attacks blood cells
that have different sugars than those present on that person’s own blood cells. People with O
blood have no sugars and therefore are accepted by O, A, and B recipients, whereas people with
AB blood produce both sugars themselves and therefore will accept blood that has either sugar or
both sugars.
22. The genotypes of two parents are both AB/ab. One of their children, however, has the
genotype AB/Ab. Given that A is normally linked to the B allele, explain how this child inherited a
chromosome with A linked to the b allele.
Ans: At meiosis in either the father or the mother, there was a crossing over event that
exchanged the DNA between the A and B loci, giving rise to chromosomes of the Ab and aB
genotypes. One of these, Ab, was combined by fertilization with an AB chromosome from the
other parent.
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 8
Important Notes
The file includes the complete test bank, organized chapter by chapter.
A sample of selected pages has been provided for preview.
All available appendices and Excel files (if included in the original resources) are
provided.
Quizzes, Midterm and final exams are included (if available in the original resources).
We continuously update our files to ensure you receive the latest and most accurate
editions.
New editions are added regularly – stay connected for updates!
⚠️Note on Answer Keys: If the answer key is not included within the chapter
questions, you will find the complete answers and solutions at the end of each
chapter.
✅ Why Buy From Us?
📚 Complete & organized chapter-by-chapter – no missing content, no guessing.
⚡ Instant digital delivery – get your file the moment you pay, no waiting.
📅 Always up to date – we track new editions so you always get the latest version.
💬 Friendly support – real humans ready to help, anytime you need us.
🔒 Safe & secure – thousands of satisfied students trust us every semester.
🛡️Our Guarantees
💰 Money-Back Guarantee: Not satisfied? We offer a full refund – no questions asked.
🔄 Wrong File? No Problem: Contact us and we will replace it immediately with the
correct version, free of charge.
⏰ 24/7 Support: We are always here – reach out anytime and expect a fast response.
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
Chapter 1
Multiple Choice
1. The complete set of hereditary information carried by an organism is its:
A) chromosome.
B) genome.
C) double helix.
D) gene.
E) nucleotide.
Ans: B
2. Hershey and Chase tracked DNA during phage infection using a radioisotope of:
A) oxygen.
B) sulfur.
C) nitrogen.
D) phosphorus.
E) carbon.
Ans: D
3. The process of ___________ in eukaryotic cells is analogous to bacterial transformation.
A) transfection
B) transcription
C) transition
D) translation
E) translocation
Ans: A
4. In a polynucleotide, a phosphate group is linked to the ________ carbons of two pentoses.
A) 1′ and 2′
B) 1′ and 3′
C) 3′ and 5′
D) 1′ and 5′
E) 2′ and 3′
Ans: C
5. In DNA, the number of times one strand crosses over the other in space is its:
A) twisting number.
B) positive supercoiling number.
C) writhing number.
D) negative supercoiling number.
E) linking number.
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 1
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
Ans: E
6. In the B-form of DNA:
A) the two polynucleotide chains are described as parallel.
B) base pairs lie perpendicular to the sugar–phosphate backbone.
C) the width of the helix varies between A-T and G-C base pairs.
D) purine bases pair with other purines.
E) there are approximately 12 base pairs per turn of the helix.
Ans: B
7. DNA replication is semiconservative, meaning that:
A) one of the two daughter duplexes consists of both original parental strands.
B) for each daughter duplex, only one of the parental strands is used as a template for both
daughter strands.
C) each daughter duplex may differ in sequence from the parental duplex.
D) each daughter duplex consists of one parental strand and one newly synthesized daughter
strand.
E) each daughter duplex has sections of original parental duplex and sections of newly
synthesized daughter duplex.
Ans: D
8. Which of the following is a transition mutation?
A) A-T → G-C
B) A-T → C-G
C) A-T → T-A
D) G-C → C-G
E) G-C → T-A
Ans: A
9. As compared to the original mutation, a suppression mutation occurs:
A) in the homologous allele of the same gene.
B) at the same site.
C) in another nonhomologous gene.
D) in a different codon of the same gene.
E) in another site of the same codon.
Ans: C
10. A viroid is:
A) RNA encapsulated in protein.
B) double-stranded DNA encapsulated in protein.
C) a naked DNA molecule.
D) single-stranded DNA encapsulated in protein.
E) a naked RNA molecule.
Ans: E
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 2
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
11. A chromosome is best defined as:
A) the complete sequence of a gene, including sequences removed from the RNA product.
B) the functional coding sequences of a single DNA molecule.
C) the complete set of hereditary information of an organism.
D) a linear array of genes.
E) the sequence of a gene that has coding information for the polypeptide product.
Ans: D
12. Mutation is random with respect to the structure and function of a gene. As a result,
A) the effects of a particular mutation cannot be predicted.
B) mutations are likely to impair the function of a gene.
C) mutations are equally likely at every nucleotide of the gene.
D) mutations invariably destroy the function of the gene.
E) mutations are likely to improve the function of a gene.
Ans: B
13. In the complementation test, crossing two homozygous recessive individuals with similar
phenotypes to yield all wild-type offspring means that:
A) the two parents carry different mutations of the same nucleotide.
B) the two parents’ mutations are in the same gene.
C) the two parents’ mutations affect different traits.
D) the two parents carry identical mutations.
E) the two parents’ mutations are in different genes.
Ans: E
14. Mutations that have no apparent effect are called:
A) silent mutations.
B) loss-of-function mutations.
C) null mutations.
D) leaky mutations.
E) gain-of-function mutations.
Ans: A
15. Genes that are on opposite ends of a long eukaryotic chromosome:
A) never recombine.
B) show less than 50% recombination.
C) show exactly 50% recombination.
D) show between 50% and 100% recombination.
E) show 100% recombination.
Ans: C
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 3
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
16. Full or partial reversion of a deletion mutation may be accomplished by:
A) an insertion mutation in another gene.
B) an insertion mutation close to the original deletion.
C) a substitution mutation close to the original deletion.
D) a second deletion mutation close to the original deletion.
E) a second deletion mutation in another gene.
Ans: B
17. A closed reading frame:
A) contains frequent stop codons.
B) contains many substitution mutations.
C) is formed by the deletion of a sequence from an open reading frame.
D) begins and ends with the same sequence.
E) produces a polypeptide that serves no apparent function.
Ans: A
18. The colinearity of prokaryotic genes means that:
A) within a prokaryotic gene, both DNA strands encode the same polypeptide.
B) a prokaryotic polypeptide has exactly 3x as many amino acids as the number of nucleotides in
the gene encoding it.
C) prokaryotic proteins are identical in length to their genes.
D) prokaryotic genes contain a continuous unbroken sequence encoding a polypeptide.
E) a physical map of a gene will not match the amino acid map of its polypeptide product.
Ans: D
19. The production of RNA from a DNA template is called:
A) splicing.
B) translation.
C) RNA processing.
D) transport.
E) transcription.
Ans: E
20. A regulatory control site of a gene:
A) must be very close to the gene it affects.
B) must lie upstream of the gene it affects.
C) is cis-acting.
D) is trans-acting.
E) must be very distant from the gene it affects.
Ans: C
Short Answer
Copyright © 2018 by Jones & Bartlett Learning, LLC, an Ascend Learning Company 4
,Lewin’s GENES XII
Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
1. The basic building block of nucleic acids is the nucleotide. It is composed of three components.
What are they?
Ans: They are a nitrogenous base, a sugar, and a phosphate.
2. There are two types of nucleic acids in cells, DNA and RNA. List two differences between the
structure of DNA and RNA.
Ans: 1) DNA has the pyrimidine base thymine, whereas RNA contains uracil. 2) DNA has 2′-
deoxyribose, whereas RNA has ribose. The sugar in RNA has an -OH group at the 2′ position of
the pentose ring.
3. If the mole % of A residues in double-stranded DNA is 29%, what would the mole % of G
residues be?
Ans: The two strands of DNA are complementary with G residues on one strand paired with C
residues on the other strand, and with A residues paired with T residues. Therefore, if the mole %
of A is 29%, then the mole % of T is 29% and the mole % of G + C would be 42%. G must equal
C, so as a result the mole % of G and of C would be 21%.
4. The size of the smallest human chromosome is 4.7 × 107 bp. If the DNA is in the B-form helix,
what would be the length in cm of the chromosome fully extended if not packaged by proteins?
Ans: 1.6 cm. In B-form DNA, the distance between adjacent nucleotides is 3.4 Å. Thus the length
of the chromosome would be 4.7 × 107 × 3.4 Å = 16 × 107 Å. There are 108 Å per cm, so the
length would be 1.6 cm.
5. Meselson and Stahl used density labeling of DNA to show that DNA replication occurs via a
semiconservative mechanism. In their experiment, they started with an organism grown in a
heavy density label (15N). After two generations of growth in light medium (the more common 14N
isotope), if the DNA is isolated and separated by density, how many bands would be observed
and how would their density compare with the starting DNA?
Ans: In each generation, the newly synthesized DNA is composed of one parental strand and one
new daughter strand. After two generations, the DNA would yield two different density bands.
Two of the daughters would contain one original heavy parental strand and one light daughter
(hybrid density) strand, and two of the daughters would contain two light strands (light density).
6. The genetic information of organisms is always in the form of nucleic acid. Give examples of
cellular organisms or other genetic systems whose genomes are double-stranded DNA, single-
stranded DNA, double-stranded RNA, or single-stranded RNA.
Ans: Living organisms (either prokaryotic or eukaryotic cells) always have genomes that are
double-stranded DNA. There are, however, many viruses of both prokaryotes and eukaryotes
whose genomes can be single-stranded DNA, or double- or single-stranded RNA, depending on
the virus.
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Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
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7. The central dogma of molecular biology is that genetic information flows from DNA to RNA by
transcription, and then to polypeptide by translation. Give an example of an exception to this
generally true statement.
Ans: Retroviruses have genomes that are single-stranded RNA, and as part of their replication
cycle they convert the RNA into first a single-stranded DNA (using an enzyme called reverse
transcriptase) and then a double-stranded DNA, in a process called reverse transcription. The
double-stranded DNA copy of its genome integrates into the DNA of the host cell it infects and
functions as any other gene, directing the synthesis of both RNA and polypeptide. In this
example, genetic information is flowing from RNA to DNA.
8. Is the enzyme reverse transcriptase an example of an RNA polymerase or a DNA polymerase?
Ans: Reverse transcriptases are RNA-dependent DNA polymerases that use deoxynucleotides as
substrates.
9. In the process of DNA replication, the two parental strands separate and serve as templates for
the new daughter strands. During this process, are covalent bonds being broken? Explain.
Ans: No. The separation of the two parental strands involves breakage of the hydrogen (H) bonds
(not covalent bonds) between the complementary base pairs. The covalent phosphodiester bonds
between adjacent nucleotides on each strand are not broken.
10. The separation of the double helical DNA in cells is mediated by enzymes called helicases. In
the laboratory, however, what method is commonly used to denature (separate) the individual
strands of a double helical DNA?
Ans: DNA may be denatured by heating the DNA above its Tm, usually by boiling.
11. A scientist at the San Diego Zoo is sent DNA samples to analyze. The samples were
collected from two different bacterial species from different parts of the world. She determines
that the G-C content of one of them is 32% and the G-C content of the other is 62%. The labels
have fallen off the tubes, however, so she does not know which sample is which. One of the
organisms lived deep in the ocean at temperatures around 10ºC and the other was isolated from
the Mojave desert, where temperatures can reach 45ºC. Which sample is likely to be which?
Ans: The bacteria from the Mojave desert is likely to have a G-C content of 62%, whereas the
ocean bacteria is likely to have a G-C content of 32%. The stability of the double-stranded DNA is
greater with higher G-C content because G-C base pairs are more stable than A-T base pairs.
Organisms that thrive at elevated temperatures have genomes with a higher fraction of G-C base
pairs to stabilize the double helical structure of their DNA.
12. Once a mutation has occurred in a particular gene, several types of events can occur to
reverse the effects of the original mutation. Give three of these types.
Ans: 1) There may be a true reversion whereby the original base change is reversed, or the
insertion of genetic material, such as a transposon, is deleted. 2) There may be a second site
reversion whereby a second mutation in the same gene changes the amino acid sequence at a
distinct site from the first mutation and allows the polypeptide to regain function. 3) There may be
a suppressor mutation in another gene, altering a different polypeptide. This change bypasses
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Jocelyn E. Krebs, Elliott S. Goldstein, Stephen T. Kilpatrick
Test Bank
the need for the original polypeptide or enables it to interact with the mutant form of the original
polypeptide.
13. Why is the modified base 5-methylcytosine a hotspot for mutations?
Ans: The oxidative deamination of 5-methylcytosine converts this base to thymine (not uracil, as
for ordinary cytosine). The DNA repair enzyme uracil-DNA-glycosidase, which recognizes
deoxyuracil as abnormal and removes it, cannot distinguish the incorrect thymine from other
correct thymines, and therefore the new T-A pair is not repaired.
14. One of the most common causes of mutations in DNA results from the oxidative deamination
of cytosine. Why does this lead to a mutation?
Ans: Due to oxidation, the amino group of cytosine is converted to a keto group, changing the
cytosine to uracil. During replication, the new uracil now pairs with adenine instead of thymine,
resulting in a C-G pair being replaced by a T-A pair when the A now pairs with T in the next
replication cycle.
15. Not all infectious agents that cause disease contain nucleic acid. What is the infectious agent
that causes scrapie, a disease of sheep and goats?
Ans: It is a prion—specifically, an abnormally folded form of a protein called PrP. This protein can
exist in cells in various conformations, and those that are resistant to degradation by proteases
not only cause disease, but “convert” normally folded PrP proteins to the abnormal conformation.
16. What is the definition of an allele?
Ans: An allele is an alternative sequence of DNA for a gene at a specific locus.
17. If a gene is duplicated in the genome so that the two copies reside at different chromosomal
locations, are these genes alleles?
Ans: No. Alleles refer to alternative forms of a gene at a specific locus in the genome, so that
copies even of nearly identical sequences that reside at different loci are not alleles of one
another (instead they are called paralogs).
18. Why is the original genetic concept of one gene-one enzyme not entirely true?
Ans: Although it is true that genes encode mRNAs, which in turn encode polypeptides, some
enzymes are composed of more than one polypeptide chain and are therefore encoded by more
than one gene. Secondly, some genes can produce alternative mRNA species (a process termed
alternative splicing) and therefore can generate multiple (related) polypeptides.
19. How is genetic complementation used to determine whether two different mutations occur in
the same gene or in different genes?
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Test Bank
Ans: A genetic cross is performed with individuals that each are homozygous (both copies of the
gene have the same mutation) for the two different mutations. If the mutations occur in the same
gene (but possibly at different sites within the gene), all of the progeny will exhibit a mutant
phenotype, because they have no wild-type copy of the gene. If the mutations occur in different
genes, then the progeny will inherit one wild-type allele from each parent and will be
phenotypically wild-type.
20. There are many types of changes in DNA that lead to mutations. Give a molecular
explanation for the following types of mutations:
A. a null mutation
B. a loss-of-function mutation
C. a gain-of-function mutation
Ans: A. A null mutation is caused either by a complete or partial deletion of a gene or a change
that renders the polypeptide product completely nonfunctional, such as a premature stop codon
that truncates the polypeptide. B. A loss-of-function mutation is a base change in DNA that alters
an amino acid such that the polypeptide has less activity than the wild-type polypeptide. DNA
changes that affect regulatory regions that decrease the level of the transcript, and therefore the
protein, can also be loss-of-function. C. Gain-of-function mutations are DNA changes resulting in
either an altered amino acid sequence that results in the polypeptide having a novel activity, or
changes in regulatory regions that increase the level of the polypeptide or cause it to be
expressed in tissues or times in development that normally lack the polypeptide. Some
polypeptides can be produced at higher or lower levels than wild-type without causing a mutant
phenotype.
21. Not every difference in DNA sequence results in a mutant allele. Sometimes genes don’t have
a single “wild-type” allele, such as the gene specifying the blood group antigens A and B. Given
the basis for the three types of blood groups (O, A, and B), explain why people with blood type O
are considered universal donors and people with the blood type AB are universal recipients of
blood transfusions.
Ans: The allele for O lacks a functional galactosyltransferase, and adds no sugar to the O antigen
on blood cells. The allele for A produces one that adds an N-acetylgalactose sugar. The allele for
B produces one that adds a galactose sugar. An individual’s immune system attacks blood cells
that have different sugars than those present on that person’s own blood cells. People with O
blood have no sugars and therefore are accepted by O, A, and B recipients, whereas people with
AB blood produce both sugars themselves and therefore will accept blood that has either sugar or
both sugars.
22. The genotypes of two parents are both AB/ab. One of their children, however, has the
genotype AB/Ab. Given that A is normally linked to the B allele, explain how this child inherited a
chromosome with A linked to the b allele.
Ans: At meiosis in either the father or the mother, there was a crossing over event that
exchanged the DNA between the A and B loci, giving rise to chromosomes of the Ab and aB
genotypes. One of these, Ab, was combined by fertilization with an AB chromosome from the
other parent.
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