TEST BANK
ESSENTIAL GENETICS: A GENOMIC PERSPECTIVE 4TH EDITION
Elizabeth W. Jones & Daniel L. Hartl
,Study Guide and Solutions Manual to accompany
s
A GENOMICS PERSPECTIVE
Fourth Edition
Daniel L. Hartl and Elizabeth W. Jones
Prepared by Elena R. Lozovsky
JONES AND BARTLETT PUBLISHERS
Sudbury, Massachusetts
BOSTON TORONTO LONDON SINGAPORE
,Contents
1 The Genetic Code of Genes and Genomes 1
2 Transmission Genetics: Heritage from Mendel 4
3 The Chromosomal Basis of Heredity......................• ............................. 9
4 Gene Linkage and Genetic Mapping.......................·.· ........................13
5 Human Chromosomes and Chromosome Behavior 18
6 DNA Structure, Replication, and Manipulation 22
7 The Genetics of Bacteria and Their Viruses 26
8 The Molecular Genetics of Gene Expression 30
9 Molecular Mechanisms of Gene Regulation 34
10 Genomics, Proteomics, and Genetic Engineering .............................. 38
11 The Genetic Control of Development 42
12 Molecular Mechanisms of Mutation and DNA Repair 45
13 Molecular Genetics of the Cell Cycle and cancer 48
14 Molecular Evolution and Population Genetics ...............•..................52
15 The Genetic Basis of Complex Inheritance 55
Appendix: Answers 58
V
,The Genetic Code of Genes
and Genomes
Key Concepts
• Inherited traits are affected by genes.
• Genes are composed of the chemical deoxyribonucleic acid (DNA).
• DNA replicates to form (usually identical) copies of itself.
• DNA contains a code specifying what types of enzymes and other proteins are made
in cells.
• DNA occasionally mutates, and the mutant forms specify altered proteins.
• A mutant enzyme is an "inborn error of metabolism" that blocks one step in a bio-
chemical pathway for the metabolism of small molecules.
• Traits are affected by environment as well as by genes.
• Organisms change genetically through generations in the process of biological
evolution.
• Because of their comm.on descent, organisms share many features of their genetics
and biochemistry.
Key Terms
1. complementary base pairing
2. antiparallel
3. central dogma
4. transcri tion
5. ribosome
6. transfer RNA (tRNA)
7. proteome
8. substrate
9. alkaptonuria
10. phenylketonuria
11. pleiotropic effect or pleiotropy
12. prokaryote
Concepts in Action
1.1. (a) false; (b) true; (c) true; (d) false.
1.2. The importance of the nucleus in inheritance was implied by its prominence in fer-
tilization. The discovery of chromosomes inside the nucleus, their behavior during
cell division, and the observation that each species has a characteristic chromosome
number made it likely that chromosomes were the carriers of the genes. Micro-
scopic studies showed that DNA and proteins are both present in chromosomes,
but whereas nearly all cells of a given species contain a constant amount of DNA,
the amount and kinds of proteins differ greatly in different cell types.
C 2006 by Jones and Bartlett Publishers
1
,1.3. It was generally believed that the genetic material must be a very complex molecule.
Proteins were chemically the most complex macromolecules known at the time.
DNA was thought to be a monotonous polymer composed of a simple repeating unit.
1.4. Because the mature T2 phage contains only DNA and protein, the labeled RNA was
left behind in material released by the burst cells.
1.5. Watson and Crick noted that the nucleotide sequence of the DNA molecule could be
replicated if each of the strands were used as a template for the formation of a new
daughter strand having a complementary sequence of bases. They also noted that
genetic information could be coded by the sequence of bases along the DNA mole-
cule, analogous to letters of the alphabet printed on a strip of paper. Finally, they
noted that changes in genetic information could result from errors in replication,
and the altered nucleotide sequence could then be perpetuated.
1.6. RNA differs from DNA in that the sugar-phosphate backbone contains ribose rather
than deoxyribose. RNA contains the base uracil (U) instead of thymine (T), and RNA
usually exists as a single strand (although any particular molecule of RNA may contain
short regions of complementary base pairs that can come together to form duplexes).
1.7. Percent C = 37.5%, so percent G = 37.5% also. Percent A+ T = 1-0.375 -0.375
=25%, but because percent A= percent T, it must be that percent A= 12.5%.
1.8. Because A# T and G i= C in this DNA, it seems likely that the DNA molecule pres-
ent in this particular virus is single-stranded.
1.9. 3'-CA-5', because the dinucleotide 3'-CA-5' pairs with 5'-GT-3, so where either
strand contains 5'-GT-3', the other contains 3'-CA-5'.
1.10. The repeating Asn results from translation in the reading frame
5'-AAUAAUAAUAAU...-3', and the repeating Ile results from translation in the
reading frame 5'-AUAAUAAUAAUA ... -3'. There is no product corresponding to
the third reading frame (5'-UAAUAAUAAUAA...-3'), because 5'-UAA-3' is a stop
codon.
1.11. Most likely the mutant protein does not fold properly and is degraded by proteases.
1.12. It is the case because each codon is exactly three nucleotides in length. In a protein-
coding region, an insertion or deletion of anything other than an exact multiple of
three nucleotides would shift the reading frame (phase) in which the mRNA is
translated. All amino acids downstream of the site of the mutation would be trans-
lated incorrectly.
1.13. 5'-TGTCGTATTTGCAAG-3
1.14. Transcription takes place from left to right, and the mRNA sequence is
5'- UGUCGUAUUUGCAAG-3'.
1.15. Cys-Arg-lle-Cys-Lys or, using the single-letter abbreviations, CRICK.
1.16. Cys-His-Ile-Cys-Lys, CHICK
1.17. The codon 5-UGG-3' codes for Trp, and in this random polymer the Trp codon is
expected with a frequency of 1/4 x 3/4 x 3/4 = 9/64. The amino acid Val could be
specified by either 5'-GUU-3' or 5'-GUG-3; the former has an expected frequency
of 3/4 x 1/4 x 1/4 = 3/64, and the latter of 3/4 x 1/4 x 3/4 = 9/64, totaling 12/64.
The amino acid Phe could be specified only by 5'-UUU-3' in this random polymer,
so Phe would have an expected frequency of 1/4 x 1/4 x 1/4 =1/64.
1.18. (a) Met-Ser-Thr-Ala-Val-Leu-Glu-Asn-Pro-Gly. (b) The mutation changes the initia-
tion codon into a noninitiation codon, so translation will not start with the first
AUG; translation will start with the next AUG farther along the mRNA or, if this is
too distant, not at all. (c) Met-Ser-Thr-Ala-Val-Leu-Glu-Asn-Pro-Gly; there is no
change, because both 5-UCC-3' and 5'-UCG-3' code for serine. (d) Met-Ser-Thr-
Ala-Ala-Leu-Glu-Asn-Pro-Gly; there is a Val->Ala amino acid replacement because
5'GUC-3' codes for Val, whereas 5'GCC-3' codes for Ala. (e) Met-Ser-Thr-Ala-Val-
Leu; translation is terminated at UAA because 5'-UAA-3' is a "stop" (termination)
code.
1.19. (a) X,Y, and z missing, W in excess; (b) Y and z missing, X in excess; (c) Z missing, Y
in excess.
€2006 by Jones and Bartlett Publishers
2 Study Guide to accompany Essential Genetics: A Genomics Perspective, Fourth Edition
, 1.20. The mfinding mthat mthe mcells mcan mgrow min mthe mpresence mof mY mimplies
mthat mstep mC mis mfunc- mtional. m The mfinding m that m the m cells m cannot
mgrow min mthe m presence m of mX mimplies mthat mstep mB m is mblocked. mThe
mresults mwith mW mimply mthat ma mdownstream mstep mis mblocked, mbut mdo
mnot mreveal mwhich m one.
Study mQuestions
1. S1. When mthe mbase mcomposition mof mdouble-stranded mDNA mfrom ma mnew
mspecies mof mbacteria mwas mdetermined, m 15% mof mthe mbases mwere mfound
mto mbe mcytosine. mWhat mis mthe mpercent- mage mof madenine min mthe mDNA
mof mthis morganism?
A) 15% 15/ m C
B) 25% /51G7 m b
@5% 35/A
D) 45% 3571 m m7% >
1.S2. m - m A mduplex mDNA mmolecule mcontains ma mrandom msequence mof mthe mfour
mnucleotides min mequal mproportions. mWhat mis mthe maverage mspacing
mbetween mconsecutive moccurrences mof mthe msequence m5'-GGGG-3'2
4
� / m )' � m P""-"\\i-t--j mof m e,ACM m vu.<cu..olid..L m oc_cu.n-itj
c) m2'
D) m4
1. S3. The msequence mof mone mstrand mof mDNA mis m5'-CCATATGC-3'. mThe
msequence mof mthe mcom- mplementary mstrand mwould mbe mwhat? 3'-
GGTA1CG~
A) 5'-CCATATGC-3'
B) 5'-GGTATACG-3' S'- mGCtAGG3'
@s-6CA1ATGG-3
_DJ m 5'-AAGCGCTA-3'
E) 5'-GCAUAUGC-3' m m N
1.S4. An mRNA mmolecule mfolds mback mupon mitself mto mform ma m"hairpin"
mstructure mheld mtogether mby m a m region m of m base mpairing. m One m segment
mof mthe m molecule m in m the m paired m region mhas mthe mbase msequence m5'-
UCAAUGC-3'. mWhat mis mthe mbase msequence mwith mwhich mthis mseg-
ment mis mpaired? -uuACG-5'
@s'-GCAUUGA-3'Su..•-
. ,•••
Cua 1-
m
A ,,. m .,._
m0
E)-CCUt9
C) 5'-TCAATGC-3°
D) 5'-GCATTGA-3'
What mcodon mwould mpair mwith mthe manticodon mof mtRNA" m5'-UAG-32 msequences mcan mencode m
1.55.
A) 5-CAU-' \&&&I mamino macid msequence m
mvO mPro-Met-Arg?
B) 5'-GAU-3' 3'-AUC-S' ol
C) 5'-UAG-3' l 1 (
P!'-ATG-3'
(E-CUA-3'
1.56. Traits m are m affected m by m environment m as m well m as m by m 0&RA'_.
1.$7. An menzyme mcalled mreverse m transcriptase m can mproduce m a m complementary
mDNA m strand mfrom man N template.
1. S8. If mDNA mfrom mBaker's myeast mhas ma mguanine mcontent mof m25%, mthen mwhat
mis mthe mcontent mof mthe madenine? m 25 7
If ma mparticular mpiece mof mRNA mhas ma muracil mcontent mof m25%, mthen
mwhat mis mits mguanine mcontent? Sh AU N[4 le
1.S10. @N AN\k,vs mHow mmany mdifferent mDNA
46
ESSENTIAL GENETICS: A GENOMIC PERSPECTIVE 4TH EDITION
Elizabeth W. Jones & Daniel L. Hartl
,Study Guide and Solutions Manual to accompany
s
A GENOMICS PERSPECTIVE
Fourth Edition
Daniel L. Hartl and Elizabeth W. Jones
Prepared by Elena R. Lozovsky
JONES AND BARTLETT PUBLISHERS
Sudbury, Massachusetts
BOSTON TORONTO LONDON SINGAPORE
,Contents
1 The Genetic Code of Genes and Genomes 1
2 Transmission Genetics: Heritage from Mendel 4
3 The Chromosomal Basis of Heredity......................• ............................. 9
4 Gene Linkage and Genetic Mapping.......................·.· ........................13
5 Human Chromosomes and Chromosome Behavior 18
6 DNA Structure, Replication, and Manipulation 22
7 The Genetics of Bacteria and Their Viruses 26
8 The Molecular Genetics of Gene Expression 30
9 Molecular Mechanisms of Gene Regulation 34
10 Genomics, Proteomics, and Genetic Engineering .............................. 38
11 The Genetic Control of Development 42
12 Molecular Mechanisms of Mutation and DNA Repair 45
13 Molecular Genetics of the Cell Cycle and cancer 48
14 Molecular Evolution and Population Genetics ...............•..................52
15 The Genetic Basis of Complex Inheritance 55
Appendix: Answers 58
V
,The Genetic Code of Genes
and Genomes
Key Concepts
• Inherited traits are affected by genes.
• Genes are composed of the chemical deoxyribonucleic acid (DNA).
• DNA replicates to form (usually identical) copies of itself.
• DNA contains a code specifying what types of enzymes and other proteins are made
in cells.
• DNA occasionally mutates, and the mutant forms specify altered proteins.
• A mutant enzyme is an "inborn error of metabolism" that blocks one step in a bio-
chemical pathway for the metabolism of small molecules.
• Traits are affected by environment as well as by genes.
• Organisms change genetically through generations in the process of biological
evolution.
• Because of their comm.on descent, organisms share many features of their genetics
and biochemistry.
Key Terms
1. complementary base pairing
2. antiparallel
3. central dogma
4. transcri tion
5. ribosome
6. transfer RNA (tRNA)
7. proteome
8. substrate
9. alkaptonuria
10. phenylketonuria
11. pleiotropic effect or pleiotropy
12. prokaryote
Concepts in Action
1.1. (a) false; (b) true; (c) true; (d) false.
1.2. The importance of the nucleus in inheritance was implied by its prominence in fer-
tilization. The discovery of chromosomes inside the nucleus, their behavior during
cell division, and the observation that each species has a characteristic chromosome
number made it likely that chromosomes were the carriers of the genes. Micro-
scopic studies showed that DNA and proteins are both present in chromosomes,
but whereas nearly all cells of a given species contain a constant amount of DNA,
the amount and kinds of proteins differ greatly in different cell types.
C 2006 by Jones and Bartlett Publishers
1
,1.3. It was generally believed that the genetic material must be a very complex molecule.
Proteins were chemically the most complex macromolecules known at the time.
DNA was thought to be a monotonous polymer composed of a simple repeating unit.
1.4. Because the mature T2 phage contains only DNA and protein, the labeled RNA was
left behind in material released by the burst cells.
1.5. Watson and Crick noted that the nucleotide sequence of the DNA molecule could be
replicated if each of the strands were used as a template for the formation of a new
daughter strand having a complementary sequence of bases. They also noted that
genetic information could be coded by the sequence of bases along the DNA mole-
cule, analogous to letters of the alphabet printed on a strip of paper. Finally, they
noted that changes in genetic information could result from errors in replication,
and the altered nucleotide sequence could then be perpetuated.
1.6. RNA differs from DNA in that the sugar-phosphate backbone contains ribose rather
than deoxyribose. RNA contains the base uracil (U) instead of thymine (T), and RNA
usually exists as a single strand (although any particular molecule of RNA may contain
short regions of complementary base pairs that can come together to form duplexes).
1.7. Percent C = 37.5%, so percent G = 37.5% also. Percent A+ T = 1-0.375 -0.375
=25%, but because percent A= percent T, it must be that percent A= 12.5%.
1.8. Because A# T and G i= C in this DNA, it seems likely that the DNA molecule pres-
ent in this particular virus is single-stranded.
1.9. 3'-CA-5', because the dinucleotide 3'-CA-5' pairs with 5'-GT-3, so where either
strand contains 5'-GT-3', the other contains 3'-CA-5'.
1.10. The repeating Asn results from translation in the reading frame
5'-AAUAAUAAUAAU...-3', and the repeating Ile results from translation in the
reading frame 5'-AUAAUAAUAAUA ... -3'. There is no product corresponding to
the third reading frame (5'-UAAUAAUAAUAA...-3'), because 5'-UAA-3' is a stop
codon.
1.11. Most likely the mutant protein does not fold properly and is degraded by proteases.
1.12. It is the case because each codon is exactly three nucleotides in length. In a protein-
coding region, an insertion or deletion of anything other than an exact multiple of
three nucleotides would shift the reading frame (phase) in which the mRNA is
translated. All amino acids downstream of the site of the mutation would be trans-
lated incorrectly.
1.13. 5'-TGTCGTATTTGCAAG-3
1.14. Transcription takes place from left to right, and the mRNA sequence is
5'- UGUCGUAUUUGCAAG-3'.
1.15. Cys-Arg-lle-Cys-Lys or, using the single-letter abbreviations, CRICK.
1.16. Cys-His-Ile-Cys-Lys, CHICK
1.17. The codon 5-UGG-3' codes for Trp, and in this random polymer the Trp codon is
expected with a frequency of 1/4 x 3/4 x 3/4 = 9/64. The amino acid Val could be
specified by either 5'-GUU-3' or 5'-GUG-3; the former has an expected frequency
of 3/4 x 1/4 x 1/4 = 3/64, and the latter of 3/4 x 1/4 x 3/4 = 9/64, totaling 12/64.
The amino acid Phe could be specified only by 5'-UUU-3' in this random polymer,
so Phe would have an expected frequency of 1/4 x 1/4 x 1/4 =1/64.
1.18. (a) Met-Ser-Thr-Ala-Val-Leu-Glu-Asn-Pro-Gly. (b) The mutation changes the initia-
tion codon into a noninitiation codon, so translation will not start with the first
AUG; translation will start with the next AUG farther along the mRNA or, if this is
too distant, not at all. (c) Met-Ser-Thr-Ala-Val-Leu-Glu-Asn-Pro-Gly; there is no
change, because both 5-UCC-3' and 5'-UCG-3' code for serine. (d) Met-Ser-Thr-
Ala-Ala-Leu-Glu-Asn-Pro-Gly; there is a Val->Ala amino acid replacement because
5'GUC-3' codes for Val, whereas 5'GCC-3' codes for Ala. (e) Met-Ser-Thr-Ala-Val-
Leu; translation is terminated at UAA because 5'-UAA-3' is a "stop" (termination)
code.
1.19. (a) X,Y, and z missing, W in excess; (b) Y and z missing, X in excess; (c) Z missing, Y
in excess.
€2006 by Jones and Bartlett Publishers
2 Study Guide to accompany Essential Genetics: A Genomics Perspective, Fourth Edition
, 1.20. The mfinding mthat mthe mcells mcan mgrow min mthe mpresence mof mY mimplies
mthat mstep mC mis mfunc- mtional. m The mfinding m that m the m cells m cannot
mgrow min mthe m presence m of mX mimplies mthat mstep mB m is mblocked. mThe
mresults mwith mW mimply mthat ma mdownstream mstep mis mblocked, mbut mdo
mnot mreveal mwhich m one.
Study mQuestions
1. S1. When mthe mbase mcomposition mof mdouble-stranded mDNA mfrom ma mnew
mspecies mof mbacteria mwas mdetermined, m 15% mof mthe mbases mwere mfound
mto mbe mcytosine. mWhat mis mthe mpercent- mage mof madenine min mthe mDNA
mof mthis morganism?
A) 15% 15/ m C
B) 25% /51G7 m b
@5% 35/A
D) 45% 3571 m m7% >
1.S2. m - m A mduplex mDNA mmolecule mcontains ma mrandom msequence mof mthe mfour
mnucleotides min mequal mproportions. mWhat mis mthe maverage mspacing
mbetween mconsecutive moccurrences mof mthe msequence m5'-GGGG-3'2
4
� / m )' � m P""-"\\i-t--j mof m e,ACM m vu.<cu..olid..L m oc_cu.n-itj
c) m2'
D) m4
1. S3. The msequence mof mone mstrand mof mDNA mis m5'-CCATATGC-3'. mThe
msequence mof mthe mcom- mplementary mstrand mwould mbe mwhat? 3'-
GGTA1CG~
A) 5'-CCATATGC-3'
B) 5'-GGTATACG-3' S'- mGCtAGG3'
@s-6CA1ATGG-3
_DJ m 5'-AAGCGCTA-3'
E) 5'-GCAUAUGC-3' m m N
1.S4. An mRNA mmolecule mfolds mback mupon mitself mto mform ma m"hairpin"
mstructure mheld mtogether mby m a m region m of m base mpairing. m One m segment
mof mthe m molecule m in m the m paired m region mhas mthe mbase msequence m5'-
UCAAUGC-3'. mWhat mis mthe mbase msequence mwith mwhich mthis mseg-
ment mis mpaired? -uuACG-5'
@s'-GCAUUGA-3'Su..•-
. ,•••
Cua 1-
m
A ,,. m .,._
m0
E)-CCUt9
C) 5'-TCAATGC-3°
D) 5'-GCATTGA-3'
What mcodon mwould mpair mwith mthe manticodon mof mtRNA" m5'-UAG-32 msequences mcan mencode m
1.55.
A) 5-CAU-' \&&&I mamino macid msequence m
mvO mPro-Met-Arg?
B) 5'-GAU-3' 3'-AUC-S' ol
C) 5'-UAG-3' l 1 (
P!'-ATG-3'
(E-CUA-3'
1.56. Traits m are m affected m by m environment m as m well m as m by m 0&RA'_.
1.$7. An menzyme mcalled mreverse m transcriptase m can mproduce m a m complementary
mDNA m strand mfrom man N template.
1. S8. If mDNA mfrom mBaker's myeast mhas ma mguanine mcontent mof m25%, mthen mwhat
mis mthe mcontent mof mthe madenine? m 25 7
If ma mparticular mpiece mof mRNA mhas ma muracil mcontent mof m25%, mthen
mwhat mis mits mguanine mcontent? Sh AU N[4 le
1.S10. @N AN\k,vs mHow mmany mdifferent mDNA
46
