TEST BANK
Introduction to Genetic Analysis
Anthony J.F. Griffiths, John Doebley, Catherine Peichel, and David A. Wassarman
12th Edition
,Table of Contents
Chapter 01 The Genetics Revolution 1
Chapter 02 Single Gene Inheritance 6
Chapter 03 Independent Assortment of Genes 26
Chapter 04 Mapping Eukaryote Chromosomes by Recombination 43
Chapter 05 Gene Interaction 63
Chapter 06 The Genetics of Bacteria and Their Viruses 84
Chapter 07 DNA Structure and Replication 104
Chapter 08 RNA Transcription Processing and Decay 117
Chapter 09 Proteins and Their Synthesis 130
Chapter 10 Gene Isolation and Manipulation 141
Chapter 11 Regulation of Gene Expression in Bacteria and Their Viruses 160
Chapter 12 Regulation of Gene Expression in Eukaryotes 175
Chapter 13 The Genetic Control of Development 184
Chapter 14 Genomes and Genomics 192
Chapter 15 DNA Damage Repair and Recombination 197
Chapter 16 The Dynamic Genome Transposable Elements 216
Chapter 17 Large Scale Chromosomal Changes 225
Chapter 18 Population Genetics 243
Chapter 19 The Inheritance of Complex Traits 258
Chapter 20 Evolution of Genes and Traits 270
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Test Bank - Introduction to Genetic Analysis, 12th Edition (Griffiths, 2021)
Chapter 01: The Genetics Revolution
1. The early 1900s was an important period for genetics due to which of the following major events?
a. the rediscovery of Gregor Mendel's scientific findings
b. Watson and Crick solving the structure of DNA
c. Walter Sutton and Theodore Boveri hypothesizing that chromosomes are the hereditary elements
d. the rediscovery of Gregor Mendel's scientific findings and Walter Sutton and Theodore Boveri
hypothesizing that chromosomes are the hereditary elements
e. All of the answer options are correct.
ANSWER: e
2. A sample of normal double-stranded DNA was found to have a guanine content of 18%. What is the expected
proportion of adenine?
a. 9%
b. 32%
c. 36%
d. 68%
e. 82%
ANSWER: b
3. In one strand of DNA, the nucleotide sequence is 5'-ATGC-3'. The complementary sequence in the other
strand must be
a. 3'-ATGC-5'.
b. 3'-TACG-5'.
c. 5'-ATCG-3'.
d. 5'-CGTA-3'.
e. 5'-TACG-3'.
ANSWER: b
4. How many different DNA molecules that are eight-nucleotide-pairs long are theoretically possible?
a. 24
b. 32
c. 64
d. 256
e. 65,536
ANSWER: e
5. Which of the following is/are TRUE about genes?
a. Genes are located on chromosomes.
b. Genes come in variants known as alleles.
c. Genes usually encode protein products.
d. All of the answer options are correct.
e. None of the answer options is correct.
ANSWER: d
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6. Wild cats (Felis silvestris) and common mice (Mus musculus) are diploid. In wild cats, 2n = 38, while in
common mice, 2n = 40. Based on this information, we can conclude that wild-cat cells have
a. less DNA than common-mouse cells.
b. smaller genomes than common-mouse cells.
c. fewer DNA molecules than common-mouse cells.
d. fewer genes than common-mouse cells.
e. fewer sets of chromosomes than common-mouse cells.
ANSWER: c
7. Which of the following is a component of DNA?
a. alanine
b. arginine
c. cysteine
d. guanine
e. tyrosine
ANSWER: d
8. Which of the following is/are TRUE of the DNA structure solved by Watson and Crick?
a. It is a double-helical structure.
b. Sugar–phosphate backbone is always toward the outside of the DNA.
c. There are two hydrogen bonds between A and T and three hydrogen bonds between C and G.
d. There are four types of nitrogenous bases.
e. All of the answer options are correct.
ANSWER: e
9. Which of the following is a CORRECT representation of the central dogma?
a. RNA → DNA → protein
b. protein → DNA → RNA
c. DNA → RNA → protein
d. DNA → protein → DNA
e. None of the answer options is correct.
ANSWER: c
10. You have come across a dog (named Cindy) that does not have a tail. Interestingly, all the puppies produced
by this dog don't have a tail. If the lack of tail is caused by a genetic mutation, where has this mutation most
likely taken place?
a. in Cindy's gametes
b. in the cells that should normally have given rise to Cindy's tail
c. in the cells that should normally have given rise to Cindy's and her puppies' tails
d. in all of Cindy's cells (including her gametes)
e. in a gamete of one of Cindy's parents
ANSWER: a
11. Which of the following features makes a species suitable as a model organism?
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Test Bank - Introduction to Genetic Analysis, 12th Edition (Griffiths, 2021)
a. small organism
b. short generation time
c. small genome
d. produce large number of offspring
e. All of the answer options are correct.
ANSWER: e
12. Using molecular techniques, researchers have knocked out both copies of gene G in a series of genetically
identical mouse embryos. These mice develop normally, except for their forelimbs, which are missing several
small bones. What can be concluded from the results of this experiment?
a. Gene G encodes a protein that is a crucial component of the forelimbs' small bones in mice.
b. Gene G encodes a protein that is normally only present in the forelimb cells of developing mice.
c. Gene G is necessary for proper development of the forelimbs' small bones in mice.
d. Gene G is normally only present in the forelimb cells of developing mice.
e. Gene G is normally only transcribed in the forelimb cells of developing mice.
ANSWER: c
13. Who originated the one-gene–one-enzyme hypothesis?
a. Tatum and Beadle
b. Gregor Mendel
c. Watson and Crick
d. Franklin and Wilkins
e. Hershey and Chase
ANSWER: a
14. What are alleles?
a. gene variants
b. enzymes
c. regulatory elements
d. de novo mutations
e. quantitative trait loci
ANSWER: a
15. Which enzyme cuts DNA at a specific location?
a. polymerase
b. ligase
c. nuclease
d. allele
e. ribosome
ANSWER: c
16. Which type of mutation is a unique DNA variant that exists in a child but in neither of its parents?
a. point mutation
b. de novo mutation
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Test Bank - Introduction to Genetic Analysis, 12th Edition (Griffiths, 2021)
c. quantitative trait locus
d. single nucleotide polymorphism
e. dominant allele
ANSWER: b
17. Which enzyme is responsible for DNA replication?
a. polymerase
b. ligase
c. nuclease
d. allele
e. ribosome
ANSWER: a
18. Which scientists offered the first compelling experimental evidence that genes are made of
deoxyribonucleic acid (DNA)?
a. Oswald Avery, Colin MacLeod, and Maclyn McCarty
b. John Gurdon and Shinya Yamanaka
c. François Jacob and Jacques Monod
d. James Watson and Francis Crick
e. Barbara McClintock and Erwin Chargoff
ANSWER: a
19. The Central Dogma describes
a. the hypothesis of how DNA is packaged into small molecules.
b. the process by which RNA is processed within a cell.
c. the flow of genetic information within cells from DNA to RNA to protein.
d. how model organisms are used in experiments.
e. the method of gene transfer between organisms.
ANSWER: c
20. The process of inserting foreign DNA molecules into the genomes of a recipient organism is called
a. replication.
b. transformation.
c. transcription.
d. translation.
e. ligation.
ANSWER: b
21. Adenine and thymine are held together by two hydrogen bonds, while guanine and cytosine are held
together by three hydrogen bonds. If you were to slowly heat a piece of DNA rich in GC base pairs—in order to
denature it—would you expect the melting temperature to be higher or lower than a piece of DNA rich in AT
base pairs?
ANSWER: The melting temperature would be higher for DNA rich in GC, owing to the three hydrogen bonds
that must be broken in order for it to denature.
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Test Bank - Introduction to Genetic Analysis, 12th Edition (Griffiths, 2021)
22. Arabidopsis thaliana is a diploid plant model organism with 2n = 10.
a) How many copies of each gene does each Arabidopsis thaliana cell have?
b) How many sets of chromosomes does the nucleus of an Arabidopsis thaliana leaf cell contain?
c) How many pairs of homologous chromosomes does the nucleus of an Arabidopsis thaliana leaf cell contain?
ANSWER: a) two b) two c) five
23. Explain what it means to say that the genetic code is redundant. How does this redundancy help protect
against mutations?
ANSWER: The genetic code is redundant because some of the amino acids are encoded by more than one triplet
(codon). This protects against the effects of mutation since a change in the nucleotide base may not
cause a different amino acid to be inserted.
24. Mutations are often viewed as negative events, and they are nearly always bad for an organism.
Paradoxically, without mutations there would be no evolution, and so they are essential. Explain how this is so.
ANSWER: Variation is introduced. So even though mutations are often viewed as negative events, all variation
that we see around us originally came from mutations.
25. Describe the purpose and function of DNA polymerase, nuclease, and ligase.
ANSWER: DNA polymerase, nuclease, and ligase are tools for characterizing and manipulating DNA, RNA,
and proteins. They each have a different function. DNA polymerase copies DNA, nucleases cut
DNA molecules in specific locations or degrade an entire DNA molecule into single nucleotides,
and ligases join two DNA molecules.
26. Why does the age of the father matter, while that of the mother seems to have no effect on the frequency of
new point mutations?
ANSWER: Eggs are made prior to a woman's birth, while sperm production occurs throughout a man's life.
From the point of conception until formation of egg cells, there are about 23 rounds of cell division
and DNA replication. Because egg formation occurs prior to birth, as a woman ages there is no
chance for additional point mutations. By comparison, the cell divisions that produce sperm
continue throughout a man's life and with each cell division there is greater risk of introducing new
point mutations.
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Test Bank - Introduction to Genetic Analysis, 12th Edition (Griffiths, 2021)
Chapter 02: Single Gene Inheritance
1. If a plant of genotype A/a is selfed, and numerous offspring are scored, what proportion of the progeny is
expected to have homozygous genotypes?
a. 0
b. 25%
c. 50%
d. 75%
e. 100%
ANSWER: c
2. What is the maximum number of heterozygous genotypes that could be produced by monohybrid self?
a. 1
b. 2
c. 3
d. 4
e. 6
ANSWER: a
3. A plant is heterozygous at three loci. How many different gamete genotypes can it theoretically produce with
respect to these three loci?
a. 2
b. 3
c. 4
d. 8
e. 16
ANSWER: d
4. In mountain rabbits, the EL-1 gene is located on chromosome 3. Four alleles of this gene have been identified
in the population. With respect to EL-1, what is the maximum number of genotypes in the progeny of a
SINGLE CROSS between two mountain rabbits?
a. 1
b. 2
c. 3
d. 4
e. 6
ANSWER: d
5. A wild-type strain of haploid yeast is crossed to a mutant strain with phenotype d. What phenotypic ratios
will be observed in the progeny?
a. All wild type
b. 75% wild type and 25% mutant (d)
c. 50% wild type and 50% mutant (d)
d. 25% wild type and 75% mutant (d)
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e. All mutant (d)
ANSWER: c
6. Mice (Mus musculus) have 40 chromosomes per diploid cell (2n = 40). How many double- stranded DNA
molecules and how many chromosomes are there in a mouse cell that is in the G2 stage of the cell cycle?
a. 40 DNA molecules and 20 chromosomes
b. 40 DNA molecules and 40 chromosomes
c. 40 DNA molecules and 80 chromosomes
d. 80 DNA molecules and 40 chromosomes
e. 80 DNA molecules and 80 chromosomes
ANSWER: d
7. A mutation occurs in a germ cell of a pure-breeding, wild-type male mouse prior to DNA replication. The
mutation is not corrected, and the cell undergoes DNA replication and a normal meiosis producing four
gametes. How many of these gametes will carry the mutation?
a. 1
b. 2
c. 3
d. 4
e. It is impossible to predict.
ANSWER: b
8. What is the mechanism that ensures Mendel's first law of segregation?
a. formation of chiasmata
b. formation of the kinetochore
c. pairing of homologous chromosomes
d. segregation of homologous chromosomes during meiosis I
e. segregation of sister chromatids during meiosis II
ANSWER: d
9. A laboratory mouse homozygous for an RFLP marker is mated to a wild mouse that is heterozygous for that
marker. One of the heterozygous individuals resulting from this cross is mated back to the wild parent. What
proportion of the offspring will have the same RFLP pattern as the original laboratory mouse?
a. None of the offspring
b. 1/4
c. 1/2
d. 3/4
e. All of the offspring
ANSWER: c
10. The diagram below shows a part of the biochemical pathway responsible for fruit color in peppers
(Caspicum annuum). Enzyme 1 is responsible for catalyzing the reaction that turns the colorless precursor into
yellow pigment, whereas Enzyme 2 catalyzes the step that turns the yellow pigment into red pigment. A breeder
crosses a pure-breeding plant that makes yellow peppers to a pure-breeding plant that makes red peppers. What
proportion of the offspring will make red peppers?
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Test Bank - Introduction to Genetic Analysis, 12th Edition (Griffiths, 2021)
a. All of the offspring
b. 3/4
c. 1/2
d. 1/4
e. None of the offspring
ANSWER: a
11. The wild-type eye color in the fruit fly Drosophila melanogaster is dark red, as a result of a mixture of
bright red and brown pigments. Enzyme A is encoded by the a gene and is required to synthesize the bright red
pigment. A lack of red pigment results in a somewhat brown eye color. You cross two fruit flies who are
heterozygous for a recessive mutation that completely inactivates the a gene. What proportion of their offspring
will have a recessive eye color phenotype?
a. All of the offspring
b. 3/4
c. 1/2
d. 1/4
e. None of the offspring
ANSWER: d
12. In pet rabbits, brown coat color is recessive to black coat color. A black female rabbit gives birth to four
black-coated and three brown-coated baby rabbits. What can be deduced about the genotype of the baby rabbits'
father?
a. He could be heterozygous black/brown or homozygous brown.
b. He could be heterozygous black/brown or homozygous black.
c. He must be heterozygous black/brown.
d. He must be homozygous black.
e. He must be homozygous brown.
ANSWER: a
13. "Dumpy" is a commonly used mutant phenotype in the nematode worm C. elegans. Two dumpy individuals
are crossed to each other, and this cross produces 210 dumpy and 68 wild-type individuals. If one of the dumpy
individuals used in this cross was mated with a wild type, what dumpy:wild-type ratio would we observe in the
offspring?
a. 0:1
b. 1:0
c. 1:1
d. 1:3
e. 3:1
ANSWER: c
14. A female rabbit of phenotype c′ is crossed to a male rabbit with cch. The F1 is comprised of five rabbits with
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