TEST BANK for Concepts of Genetics, 12th edition.
William S. Klug, Michael R. Cummings, Charlotte A.
Spencer, Michael A. Palladino, Darrell Killian. All 26
Chapters in 341 Pages
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 1 Introduction to Genetics
1) In the 1600s, William Harvey studied reproduction and development. What is the term given
to the theory that states that an organism develops from the fertilized egg by a succession of
developmental events that lead to an adult?
A) preformation
B) sequential pattern formation
C) equational transformation
D) transduction
E) epigenesis
Answer: E
Section: 1.1
Bloom's Taxonomy: Remembering/Understanding
2) What is the term given to the theory that states that the fertilized egg 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
3) What is the term given to the theory that 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
4) What is a homunculus?
A) a large cyst or growth on a plant due to viral infection
B) a sperm or egg containing a miniature adult, perfect in size and proportion
C) the intermediate stage of the DNA after CRISPR-Cas treatment
D) when the mitochondrion grows in size before splitting into two via fission
E) during development sometimes a growing individual's cell can become mutated and one part
of the child has different characteristics than the other
Answer: B
Section: 1.1
Bloom's Taxonomy: Remembering/Understanding
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5) 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
6) 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
7) 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) monoploid; n
Answer: C
Section: 1.2
Bloom's Taxonomy: Remembering/Understanding
8) Genetics is the study of ________.
A) inheritance 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
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9) 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 of the ________.
A) Chromosome Theory of Inheritance
B) Law of Segregation
C) Law of Independent Assortment
D) First Law of Thermodynamics
E) Chromosomal Maintenance Theory
Answer: A
Section: 1.2
Bloom's Taxonomy: Remembering/Understanding
10) What is a mutation?
A) an inherited change in a DNA sequence
B) the source of all genetic variation
C) a change in DNA that leads to death
D) an inherited change in DNA sequences that is the source of all genetic variation
E) an inherited changed in DNA sequence that is always bad for an organism
Answer: D
Section: 1.2
Bloom's Taxonomy: Applying/Analyzing
11) Which of the following is TRUE about alleles?
A) An allele is a variant form of a gene.
B) Alleles come in two forms, the good form and the bad form.
C) Individuals carry both forms of each allele.
D) The phenotype of the individual will always indicate with certainty the alleles of the
individual.
E) An individual will only carry one version of an allele.
Answer: A
Section: 1.2
Bloom's Taxonomy: Applying/Analyzing
12) Until the mid-1940s, many scientists considered proteins to be the likely candidates for the
genetic material. Which of the following characteristics led scientist 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
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13) Name the individual who, while working with the garden pea in the mid-1850s,
demonstrated quantitative patterns of heredity and developed a theory involving the behavior of
hereditary factors.
A) Walter Sutton
B) Theodor Boveri
C) Barbara McClintock
D) Gregor Mendel
E) George Wallace
Answer: D
Section: 1.2
Bloom's Taxonomy: Remembering/Understanding
14) Which of the following is the subdiscipline of biology concerned with the study of heredity
and variation at the molecular, cellular, developmental, organismal, and populational levels?
A) genetics
B) cell biology
C) molecular biology
D) cytogenetics
E) biochemistry
Answer: A
Section: 1.2
Bloom's Taxonomy: Remembering/Understanding
15) 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) human beings develop freckles from being out in the sun
C) depending on the food a turtle eats, it shell may grow faster or slower
D) sometime during human's life they break a bone and it heals
E) bacteria can be effectively killed by treatment with bleach
Answer: A
Section: 1.1
Bloom's Taxonomy: Evaluating/Creating
16) What term is used to describe the fact that different genes in an organism often provide
differences in observable features?
A) phenotype
B) genotype
C) alleles
D) natural selection
E) inheritance
Answer: A
Section: 1.2
Bloom's Taxonomy: Remembering/Understanding
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17) Which of the following is an example of heredity?
A) Doberman pinschers and boxers have similar body shapes.
B) Both moths and birds have wings and can fly.
C) Dalmation dogs all have spots.
D) Flying squirrels have a different mechanism of flight than mosquitos.
E) Flies and molluscs both have eyes.
Answer: C
Section: 1.2
Bloom's Taxonomy: Evaluating/Creating
18) Which of the following is NOT an example of variation?
A) a child does not have her mother's hair color
B) cats can have long or short fur
C) giraffes have not been seen in an albino form
D) both monocotyledons and dicotyledons perform the dark reaction
E) lobsters can come in many colors including blue, read, and brown
Answer: D
Section: 1.2
Bloom's Taxonomy: Evaluating/Creating
19) What would happen if, during meiosis, the chromosome number was not halved before egg
and sperm formation?
A) nothing
B) in each successive generation, the offspring would double their chromosome number
C) n would become halved
D) each offspring would have different phenotypes than their parents
E) the spindle would be compromised
Answer: B
Section: 1.2
Bloom's Taxonomy: Applying/Analyzing
20) Alternative forms of a gene are called ________.
A) alleles
B) mutants
C) phenotypes
D) genotypes
E) meiotic products
Answer: A
Section: 1.2
Bloom's Taxonomy: Remembering/Understanding
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21) The various characteristics of organisms that result from their genetic makeup are
collectively referred to as an organism's ________.
A) genotype
B) alleles
C) phenotype
D) genome
E) proteome
Answer: C
Section: 1.2
Bloom's Taxonomy: Remembering/Understanding
22) Name the substance that serves as the hereditary material in eukaryotes and prokaryotes.
A) RNA or ribonucleic acid
B) DNA or deoxyribonucleic acid
C) protein
D) lipid
E) carbohydrate
Answer: B
Section: 1.3
Bloom's Taxonomy: Remembering/Understanding
23) 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
24) A fundamental property of DNA's nitrogenous bases that is necessary for the doublestranded nature of its structure is ________.
A) complementarity
B) anti-parallel
C) ring structure
D) sugar phosphate backbone
E) deoxyribose versus ribose
Answer: A
Section: 1.3
Bloom's Taxonomy: Applying/Analyzing
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25) Which of the following is the function of DNA?
A) DNA serves to hold the information for protein, lipid, and carbohydrate storage.
B) DNA is required when cells are using their ribosomes to translate a protein.
C) DNA is responsible for the storage and replication of genetic information.
D) DNA is involved in the expression of stored genetic information.
E) DNA is used structurally to hold the nucleus together.
Answer: C
Section: 1.3
Bloom's Taxonomy: Remembering/Understanding
26) Which of the following molecules serves the function to express the genetic material by
being translated to protein?
A) DNA
B) lipid
C) carbohydrate
D) RNA
E) cholesterol
Answer: D
Section: 1.3
Bloom's Taxonomy: Remembering/Understanding
27) Name the bases in DNA and their pairing specificities.
A) adenine:thymine, guanine:cytosine
B) adenine:uracil, guanine:cytosine
C) adenine:guanine, guanine:uracil
D) adenine:cytosine, guanine:uracil
E) adenine:guanine, thymine:cytosine
Answer: A
Section: 1.3
Bloom's Taxonomy: Remembering/Understanding
28) The ________ consists of a linear series of three adjacent nucleotides present in mRNA
molecules.
A) genetic code
B) Watson—Crick base pairing
C) chromosomal theory of inheritance
D) law of segregation
E) messenger RNA
Answer: A
Section: 1.3
Bloom's Taxonomy: Remembering/Understanding
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29) 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
30) 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
31) 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
32) Given that DNA is the genetic material in prokaryotes and eukaryotes, what other general
structures (macromolecules) and substances made by the cell are associated with the expression
of that genetic material?
A) lipids and carbohydrates
B) DNA and protein
C) RNA (messenger, ribosomal, and transfer), ribosomes, enzymes, and proteins
D) DNA and RNA
E) chromosomes
Answer: C
Section: 1.3
Bloom's Taxonomy: Evaluating/Creating
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33) Which of the following are true about codons?
A) They are complementary to RNA and specify amino acids at the ribosome.
B) They are complementary to DNA and are a two-nucleotide code for an amino acid.
C) They are complementary to DNA and specify amino acids at the ribosome.
D) They are a circular series of nucleotide triplets.
E) They are placed at random in the RNA.
Answer: C
Section: 1.3
Bloom's Taxonomy: Remembering/Understanding
34) What is another term for a biological catalyst?
A) protein
B) enzyme
C) codon
D) ribosome
E) lipid
Answer: B
Section: 1.3
Bloom's Taxonomy: Remembering/Understanding
35) A protein's shape and chemical behavior are determined by ________.
A) its linear sequence of amino acids
B) the type of cell in which it resides
C) the environment of an organism
D) the cholesterol makeup of the lipid membrane
E) the cell's age
Answer: A
Section: 1.3
Bloom's Taxonomy: Applying/Analyzing
36) 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
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37) When mutation alters a gene, it may modify or even eliminate the encoded protein's usual
________ and cause an altered ________.
A) function; phenotype
B) function; genotype
C) structure; genotype
D) cell type; genotype
E) ribosome; phenotype
Answer: A
Section: 1.3
Bloom's Taxonomy: Applying/Analyzing
38) Recombinant DNA technology is dependent on a particular class of enzymes, known as
________ that cuts DNA at specific nucleotide sequences.
A) restriction enzymes
B) clones
C) recombinant DNA technology
D) genomes
E) vectors
Answer: A
Section: 1.4
Bloom's Taxonomy: Remembering/Understanding
39) What represents an organism's genome?
A) all the RNA in a cell
B) an organism's genome can be defined as the complete haploid nuclear DNA content of an
organism.
C) all the protein in a cell
D) the nuclear and mitochondrial DNAs
E) a catalog of mutations in a cell
Answer: B
Section: 1.4
Bloom's Taxonomy: Remembering/Understanding
40) 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
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41) What term is applied to a variety of projects whereby genome sequences are deposited in
databases for research purposes?
A) proteomics
B) bioinformatics
C) genetics
D) cloning
E) genomics
Answer: E
Section: 1.6
Bloom's Taxonomy: Remembering/Understanding
42) 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
43) ________ 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
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Concepts of Genetics, 12e, Global 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) bacterium
D) prokaryotic organism
E) mitochondrial organism
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) You have identified a mutant in human cells that when shifted to 37°C, the microfilaments
depolymerize (fall apart). Which of the following would be true about this mutant at 37°C?
A) The mitochondria would no longer work.
B) The sister chromatids would no longer be attached to each other.
C) The cells would no longer be able to produce ATP.
D) The cells would change shape.
E) The endoplasmic reticulum could still import polypeptides but could no longer synthesize
lipids.
Answer: D
Section: 2.1
Bloom's Taxonomy: Evaluating/Creating
4) Name two cellular organelles, each containing genetic material, which are involved in either
photosynthesis or respiration.
A) rough and smooth endoplasmic reticula
B) chloroplast and endoplasmic reticulum
C) peroxisomes and mitochondria
D) lysosome and chloroplast
E) chloroplasts and mitochondria
Answer: E
Section: 2.1
Bloom's Taxonomy: Remembering/Understanding
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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) 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
7) Which chromosome has a telomere but the p arm is much shorter than the q arm?
A) metacentric
B) submetacentric
C) acrocentric
D) telocentric
E) sex chromosome
Answer: C
Section: 2.2
Bloom's Taxonomy: Remembering/Understanding
8) 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
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9) 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
10) During interphase of the cell cycle, ________.
A) DNA recombines
B) sister chromatids move to opposite poles
C) the nuclear membrane disappears
D) RNA replicates
E) DNA content essentially doubles
Answer: E
Section: 2.3
Bloom's Taxonomy: Remembering/Understanding
11) The house fly, Musca domestica, has a haploid chromosome number of 6. How many
chromatids should be present in a diploid, somatic, metaphase cell?
A) 3
B) 6
C) 12
D) 18
E) 24
Answer: E
Section: 2.3
Bloom's Taxonomy: Applying/Analyzing
12) How many haploid sets of chromosomes are present in a diploid individual cell with a
chromosome number of 32?
A) 2
B) 1
C) 8
D) 16
E) 32
Answer: A
Section: 2.3
Bloom's Taxonomy: Applying/Analyzing
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13) How many haploid sets of chromosomes are present in an individual cell that is pentaploid
(5n)?
A) 2
B) 3
C) 4
D) 5
E) It is impossible to tell with the information given.
Answer: D
Section: 2.3
Bloom's Taxonomy: Applying/Analyzing
14) You may have heard through various media of an animal alleged to be the hybrid of a rabbit
and a cat. Given that the cat (Felis domesticus) has a diploid chromosome number of 38 and a
rabbit (Oryctolagus cuniculus) has a diploid chromosome number of 44, what would be the
expected chromosome number in the somatic tissues of this alleged hybrid?
A) 38
B) 41
C) 44
D) 82
E) 40
Answer: B
Section: 2.3
Bloom's Taxonomy: Evaluating/Creating
15) Which of the follow could occur if a cell cycle checkpoint was missed?
A) An unreplicated chromosome could be put through mitosis.
B) DNA would mutate during G2.
C) The cell cycle would be arrested until the error could be corrected.
D) The spindle apparatus would not form.
E) Cohesin could not function correctly.
Answer: A
Section: 2.3
Bloom's Taxonomy: Evaluating/Creating
16) In which stage of the cell cycle is G0 located?
A) G1
B) S
C) G2
D) M
E) anaphase
Answer: A
Section: 2.3
Bloom's Taxonomy: Remembering/Understanding
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17) 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
18) A typical G1 nucleus is 2n and contains 2C (two complements) of DNA. Which of the
following is true?
A) A prophase cell is 4n and contains 4C of DNA.
B) A cell in prophase is 2n and contains 2n of DNA.
C) A cell in prophase is 2n and contains 4C of DNA.
D) A cell in metaphase is 2n and contains 2C of DNA.
E) A cell in G2 is 4n and contains 2C of DNA.
Answer: C
Section: 2.3
Bloom's Taxonomy: Applying/Analyzing
19) Which part of interphase does DNA duplication take place?
A) G1
B) G2
C) S
D) G0
E) M
Answer: C
Section: 2.3
Bloom's Taxonomy: Remembering/Understanding
20) The centromere of a chromosome separates during ________.
A) interphase
B) prometaphase
C) prophase
D) anaphase
E) telophase
Answer: D
Section: 2.3
Bloom's Taxonomy: Remembering/Understanding
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21) 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
22) 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: Applying/Analyzing
23) In an organism with 60 chromosomes, how many bivalents would be expected to form
during meiosis?
A) 15
B) 30
C) 60
D) 120
E) 240
Answer: B
Section: 2.4
Bloom's Taxonomy: Evaluating/Creating
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24) 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
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25) A G1 somatic cell nucleus in a female diploid Myrmecia pilosula (bulldog ant) 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: Applying/Analyzing
26) Myrmecia pilosula (the bulldog ant) actually consists of several virtually identical, closely
related species, with females having chromosome numbers of 18, 20, 32, 48, 60, 62, and 64.
Assume one crossed a female of species (A) with 32 chromosomes and a male of species (B)
with 9 chromosomes (males are haploid, and each gamete contains the n complement). How
many chromosomes would one expect in the body (somatic) cells of the female offspring?
A) 4.5
B) 9
C) 25
D) 32
E) 41
Answer: C
Section: 2.4
Bloom's Taxonomy: Evaluating/Creating
27) 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
28) Which of the following is true about the second meiotic division?
A) Sister chromatids are pulling apart.
B) Homologous chromosomes are pulling apart.
C) Nondisjunction would lead to extra bivalents forming.
D) Synapsis occurring in the second meiotic division.
E) The products are four identical gametes.
Answer: A
Section: 2.4
Bloom's Taxonomy: Remembering/Understanding
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29) Which if 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: 2.4
Bloom's Taxonomy: Remembering/Understanding
30) 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
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31) Given that each G1 nucleus from this organism contains 16 picograms of DNA, how many
picograms of chromosomal DNA would you expect in the cell shown below?
A) 2
B) 4
C) 8
D) 16
E) 32
Answer: D
Section: 2.4
Bloom's Taxonomy: Evaluating/Creating
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) 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
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34) After meiosis II, ________ would be formed.
A) dyads
B) tetrads
C) monads
D) synapsis
E) chiasma
Answer: C
Section: 2.4
Bloom's Taxonomy: Applying/Evaluating
35) Which of the following would occur if there was no chiasma formation in prophase I?
A) In a heterozygote, there would only be a 2:2 formation after meiosis II, never a 1:1:1:1.
B) All gametes would have the same genotype.
C) Mosaic chromosomes would form.
D) All gametes would have the same phenotype.
E) In a heterozygote, there would only be a 1:1:1:1 formation after meiosis II, never a 2:2.
Answer: A
Section: 2.4
Bloom's Taxonomy: Evaluating/Creating
36) Which term describes meiosis I?
A) middling
B) confrontational
C) multiplicative
D) reducational
E) equinational
Answer: D
Section: 2.4
Bloom's Taxonomy: Remembering/Understanding
37) Which if the following is true?
A) A chromosome may contain one or two chromatids in different phases of the mitotic or
meiotic cell cycle.
B) A chromosome always contains the same number of chromatids, regardless of phase of the
mitotic or meiotic cell cycle.
C) Cells are considered to be 2n after meiosis I.
D) Cells are 4n after S phase.
E) Sister chromatids in mitosis are not identical.
Answer: B
Section: 2.4
Bloom's Taxonomy: Remembering/Understanding
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38) If a typical G1 nucleus contains 2C (two complements) of DNA, a gamete that is haploid (n)
contains ________ of DNA.
A) 1C
B) 0.5C
C) 2C
D) 3C
E) 4C
Answer: A
Section: 2.4
Bloom's Taxonomy: Applying/Analyzing
39) 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
40) A bivalent at prophase I contains ________ chromatids.
A) one
B) two
C) three
D) four
E) eight
Answer: D
Section: 2.4
Bloom's Taxonomy: Evaluating/Creating
41) The meiotic cell cycle involves ________ number of cell division(s) and ________ number
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
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42) 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
43) An organism with a diploid chromosome number of 46 will produce ________ combinations
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
44) The stage at which "sister chromatids go to opposite poles" immediately follows which of
the stages listed below?
A) mitotic metaphase
B) metaphase of meiosis I
C) metaphase of meiosis II
D) A and B
E) A and C
Answer: E
Section: 2.4
Bloom's Taxonomy: Remembering/Understanding
45) Drosophila melanogaster, the fruit fly, has a 2n chromosome number of 8. Assuming that a
somatic G2 nucleus from one of the individuals in this scenario contains about 8.0 picograms of
DNA, how much nuclear DNA would you expect in a fly egg?
A) 8 pg
B) 4 pg
C) 2 pg
D) 1 pg
E) 16 pg
Answer: B
Section: 2.4
Bloom's Taxonomy: Evaluating/Creating
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46) 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; 300
D) 100; 100
E) 50; 50
Answer: D
Section: 2.5
Bloom's Taxonomy: Evaluating/Creating
47) In a healthy male, how many sperm cells would be expected to be formed 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) 100; (b) 800
Answer: C
Section: 2.5
Bloom's Taxonomy: Evaluating/Creating
48) There is about as much nuclear DNA in a primary spermatocyte as in ________ spermatids.
A) 0.5
B) 1
C) 2
D) 3
E) 4
Answer: E
Section: 2.5
Bloom's Taxonomy: Applying/Analyzing
49) 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: 2.4
Bloom's Taxonomy: Remembering/Understanding
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50) List, in order of appearance, all the cell types expected to be formed during oogenesis.
A) primary oocyte, secondary oocyte and first polar body, oogonium, second polar body and
ootid
B) oogonium, primary oocyte, secondary oocyte and first polar body, ootid and second polar
body
C) primary oocyte, secondary oocyte and first polar body, ootid, second polar body, oogonium
D) primary oocyte, secondary oocyte and first polar body, second polar body, ootid, oogonium
E) oogonium, primary oocyte, second polar body and ootid, secondary oocyte and first polar
body
Answer: B
Section: 2.5
Bloom's Taxonomy: Remembering/Understanding
51) In plants, which stage is haploid?
A) gametophyte
B) sporophyte
C) polar body
D) spermatozoa
E) germ cell
Answer: A
Section: 2.6
Bloom's Taxonomy: Remembering/Understanding
52) Which of the following is diploid?
A) egg
B) sperm
C) zygote
D) megaspore
E) gametophyte
Answer: C
Section: 2.6
Bloom's Taxonomy: Remembering/Understanding
53) Electron microscopy of metaphase chromosomes demonstrated various degrees of coiling.
What was the name of the model that depicted this process?
A) folded-fiber
B) double-stranded
C) chromatid folding
D) packing
E) condensation
Answer: A
Section: 2.7
Bloom's Taxonomy: Remembering/Understanding
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54) During the transition from interphase to metaphase chromosome, the DNA undergoes how
much compaction?
A) 2 fold
B) 10 fold
C) 50 fold
D) 500 fold
E) 5000 fold
Answer: E
Section: 2.7
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 3 Mendelian Genetics
1) 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
2) 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
3) Polydactyly is expressed when an individual has extra fingers and/or toes. 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. Having extra digits is caused by a dominant allele. 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
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4) Tightly curled or wooly hair is caused by a dominant gene in humans. If a heterozygous curlyhaired person marries a person with straight hair, what percentage of their offspring would be
expected to have straight hair?
A) 25% curly
B) 50% straight
C) 75% curly
D) 100% straight
E) It is impossible to predict the outcome.
Answer: B
Section: 3.2
Bloom's Taxonomy: Applying/Analyzing
5) 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
6) 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
7) 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
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8) The process that leads to development haploid gamete 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
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 ________ testcross ratio.
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
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12) 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
13) If an F2 generation from a 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
14) Assuming complete dominance, a phenotypic ratio of ________ is expected from a
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
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15) 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: C
Section: 3.2
Bloom's Taxonomy: Evaluating/Creating
16) 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 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
17) 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
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18) 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
19) Assuming a typical monohybrid cross in which one allele is completely dominant to the
other, what 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
20) 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: E
Section: 3.2
Bloom's Taxonomy: Applying/Analyzing
21) 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
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22) 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
23) 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
24) 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
25) What are two typical testcross ratios?
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
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26) 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
27) 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
28) 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
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29) 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 with normal wings and a hairy body and a fly with
vestigial wings and 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
30) 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: Evaluating/Creating
31) 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: 3.3
Bloom's Taxonomy: Remembering/Understanding
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32) What is independent assortment?
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: A
Section: 3.3
Bloom's Taxonomy: Remembering/Understanding
33) 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
34) 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
35) 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
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36) Which types of phenotypic ratios are likely to occur in crosses when dealing with three gene
pairs for which all the genotypic combinations are of equal viability?
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: 3.3
Bloom's Taxonomy: Remembering/Understanding
37) According to Mendel's model, because of the ________ of chromosomes during meiosis, all
possible combinations of gametes will be formed in equal frequency.
A) the product rule
B) law of segregation
C) law of unit factors
D) independent assortment
E) chromosomal theory of inheritance
Answer: D
Section: 3.3
Bloom's Taxonomy: Remembering/Understanding
38) 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
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
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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) 243
B) 32
C) 1024
D) 64
E) 25
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: C
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
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44) 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
45) Assuming no crossing over between the 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
46) According to Charles Darwin and Alfred Wallace, variations followed a more continuous
pattern. According to Mendel's model, variation due to dominance-recessive relationships
followed a more ________ form.
A) abstract
B) smooth
C) discontinuous
D) normalized
E) transferred
Answer: C
Section: 3.5
Bloom's Taxonomy: Remembering/Understanding
47) 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
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48) 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
49) 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
50) 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
51) 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
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52) Dentinogenesis imperfecta is a rare, autosomal, dominantly inherited disease of the teeth that
occurs in about one in 8000 people (Witkop 1957). 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 two children will have dentinogenesis imperfecta?
A) 1/8
B) 1/4
C) 1/2
D) 3/4
E) 100%
Answer: B
Section: 3.7
Bloom's Taxonomy: Applying/Analyzing
53) Dentinogenesis imperfecta is a rare, autosomal, dominantly inherited disease of the teeth that
occurs in about one in 8000 people (Witkop 1957). 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 be a girl with dentinogenesis imperfecta?
A) 1/8
B) 1/4
C) 1/2
D) 3/4
E) 100%
Answer: B
Section: 3.7
Bloom's Taxonomy: Applying/Analyzing
54) 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: C
Section: 3.7
Bloom's Taxonomy: Applying/Analyzing
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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
three children will be wild type?
A) 1/4
B) 3/8
C) 1/32
D) 3/64
E) 0%
Answer: D
Section: 3.7
Bloom's Taxonomy: Applying/Analyzing
56) 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
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 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
58) 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 ten-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
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59) 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
60) 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
61) 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
62) 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
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63) 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 large the Chi-square value, the more likely your results are real.
Answer: B
Section: 3.8
Bloom's Taxonomy: Remembering/Understanding
64) 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
65) 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
66) 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
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 4 Extensions of Mendelian Genetics
1) A mutation in a gene often results in a reduction of the product of that gene. The term for this
type of mutation is ________.
A) codominance
B) incomplete dominance
C) gain of function
D) multiple allelism
E) loss of function or null (in the case of complete loss)
Answer: E
Section: 4.1
Bloom's Taxonomy: Remembering/Understanding
2) Which mutations are generally dominant since one copy in a diploid organism is sufficient to
alter the normal phenotype?
A) loss of function
B) gain of function
C) null
D) neutral
E) conditional
Answer: B
Section: 4.1
Bloom's Taxonomy: Applying/Analyzing
3) Assume that a mutation occurs in the gene responsible for the production of hexosaminidase
A, such that only about 50% of the enzyme activity is found in the heterozygote compared with a
homozygous normal individual. If heterozygotes are phenotypically normal, we would say that
the mutant allele is ________ to its normal allele.
A) recessive
B) dominant
C) gain of function
D) epistatic
E) auxotrophic
Answer: A
Section: 4.1
Bloom's Taxonomy: Remembering/Understanding
4) What is the most prevalent form of an allele called?
A) normal
B) dominant
C) recessive
D) wild type
E) neutral
Answer: D
Section: 4.1
Bloom's Taxonomy: Remembering/Understanding
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5) Which of the following is true?
A) A neutral allele leads to a change in function in a protein that neither harms nor helps an
organism.
B) A neutral mutation leads to a phenotype that can help or hurt the organism depending on the
environment.
C) A neutral mutation leads to no discernable phenotype.
D) Each biochemical pathway is governed by one gene.
E) A mutation in the DNA can be observed at the macroscopic level even if there's no change in
phenotype.
Answer: C
Section: 4.1
Bloom's Taxonomy: Remembering/Understanding
6) Typically, when one wishes to represent a gene, the symbol used is ________.
A) in italics
B) in all lowercase letters
C) in all upper case letters
D) in bold print
E) underlined
Answer: A
Section: 4.2
Bloom's Taxonomy: Remembering/Understanding
7) With incomplete dominance, a likely ratio resulting from a monohybrid cross would be
________.
A) 3:3
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: Remembering/Understanding
8) Assume that a dihybrid cross (AaBb × AaBb) is made in which the gene loci are autosomal,
independently assorting, and incompletely dominant. What phenotypic ratio would you expect
from such a cross?
A) 1:2:2:1:4:2:1:2:1
B) 2:1:1:2:2:1:4:2:2
C) 1:2:1:2:2:2:1:4:1
D) 2:2:2:2:2:2:2:2
E) 1:2:1:2:4:2:1:2:1
Answer: E
Section: 4.3
Bloom's Taxonomy: Evaluating/Creating
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9) 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) phenotypic 3:1; genotypic 1:2:1
B) phenotypic 1:2:1; genotypic 1:3
C) phenotypic 1:2:1; genotypic 1:2:1
D) phenotypic 1:1:1:1; genotypic 1:2:1
E) phenotypic 1:2:1; genotypic 1:1:1:1
Answer: C
Section: 4.3
Bloom's Taxonomy: Applying/Analyzing
10) Assume that a dihybrid cross is made in which the genes' loci are autosomal, independently
assorting, and incompletely dominant. How many different phenotypes are expected in the
offspring?
A) 2
B) 4
C) 6
D) 9
E) 12
Answer: D
Section: 4.3
Bloom's Taxonomy: Applying/Analyzing
11) A ________ ratio indicates incomplete dominance.
A) 1:1:1:1
B) 1:2:1
C) 9:7
D) 3:1
E) 2:2:2
Answer: B
Section: 4.3
Bloom's Taxonomy: Remembering/Understanding
12) 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. What is the term for this allelic relationship?
A) incomplete dominance
B) codominance
C) multiple alleles
D) epistasis
E) positional effect
Answer: A
Section: 4.3
Bloom's Taxonomy: Remembering/Understanding
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13) 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?
A) 10 (large leaves), 60 (medium leaves), 10 (small leaves)
B) 20 (large leaves), 40 (medium leaves), 20 (small leaves)
C) 20 (large leaves), 20 (medium leaves), 40 (small leaves)
D) 40 (large leaves), 40 (medium leaves), 0 (small leaves)
E) 40 (large leaves), 0 (medium leaves), 40 (small leaves)
Answer: B
Section: 4.3
Bloom's Taxonomy: Applying/Analyzing
14) 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.
A) 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
B) 40 large, red; 20 medium, red; 40 small, red; 20 large, pink; 80 medium, pink; 20 small, pink;
40 large, white; 20 medium, white; 40 small, white
C) 35 large, red; 35 medium, red; 35 small, red; 35 large, pink; 35 medium, pink; 35 small, pink;
35 large, white; 35 medium, white; 35 small, white
D) 80 large, red; 40 medium, red; 20 small, red; 40 large, pink; 20 medium, pink; 40 small, pink;
20 large, white; 40 medium, white; 20 small, white
E) 20 large, red; 40 medium, red; 20 small, red; 80 large, pink; 40 medium, pink; 40 small, pink;
20 large, white; 40 medium, white; 20 small, white
Answer: A
Section: 4.3
Bloom's Taxonomy: Evaluating/Creating
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15) The following coat colors are known to be 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
1
2
3
4
Parents
Offspring
cremello x cremello
chestnut x chestnut
cremello x chestnut
palomino x palomino
all cremello
all chestnut
all palomino
1/4 = chestnut
1/2 = palomino
1/4 cremello
Assign gene symbols for the genetic control of coat color on the basis of these data.
A) C1C1 = chestnut, C2C2 = cremello, C1C2 = palomino
B) C1C1 = palomino, C2C2 = chestnut, C1C2 = cremello
C) C1C1 = chestnut, C2C2 = palomino, C1C2 = cremello
D) C1C1 = cremello, C2C2 = chestnut, C1C2 = palomino
E) C1C1 = palomino, C2C2 = cremello, C1C2 = chestnut
Answer: D
Section: 4.3
Bloom's Taxonomy: Remembering/Understanding
16) Which of the following statements is true?
A) With both incomplete dominance and codominance, one expects heterozygous and
homozygous classes to be phenotypically identical.
B) An individual won't show incomplete dominance if one allele is a loss of function allele.
C) The threshold effect is only seen in incomplete dominance, not in codominance.
D) Even in a seemingly clear-cut example of complete dominance, at the protein or enzyme
level, one can detect about half the activity or gene product.
E) In the threshold effect, one would expect the gain of function mutant to be recessive.
Answer: D
Section: 4.3, 4.4
Bloom's Taxonomy: Applying/Analyzing
17) Which of the following is an example of codominance?
A) flowers of a certain plant can come in purple, blue, or yellow
B) a scorpion's venom is more potent when it's younger
C) in roan cattle, you can see a mix or both red and white furs
D) there are many possible mutations that lead to white eyes in flies
E) a certain plant can have jagged, rough, or smooth edges on its leaves
Answer: C
Section: 4.4
Bloom's Taxonomy: Remembering/Understanding
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18) What is the blood type of individuals who cannot add the terminal sugar to the H substance?
A) O
B) B
C) A
D) AB
E) Bombay phenotype
Answer: A
Section: 4.5
Bloom's Taxonomy: Remembering/Understanding
19) A situation in which there are 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
20) In mice, there is a set of multiple alleles of a gene for coat color. Four of those alleles are as
follows:
C = full color (wild type)
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, give the phenotypic ratios expected from the following cross.
wild (heterozygous for dilution) × chinchilla (heterozygous for albino)
A) 2 full color : 1 chinchilla : 1 dilution
B) 1 chinchilla : 2 dilution : 1 albino
C) 1 full color : 1 chinchilla : 1 dilution : 1 albino
D) 2 full color : 2 chinchilla : 1 dilution
E) 1 full color : 1 chinchilla : 2 dilution : 1 albino
Answer: A
Section: 4.5
Bloom's Taxonomy: Applying/Analyzing
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21) In mice, there is a set of multiple alleles of a gene for coat color. Four of those alleles are as
follows:
C = full color (wild type)
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, give the phenotypic ratios expected from the following cross.
chinchilla (heterozygous for albino) × albino
A) 1 dilution : 1 chinchilla : 1 albino
B) 1 full color : 1 chinchilla : 1 albino
C) 2 chinchilla : 1 albino
D) 1 chinchilla : 2 albino
E) 1 chinchilla :1 albino
Answer: E
Section: 4.5
Bloom's Taxonomy: Applying/Analyzing
22) In a mating between individuals with the genotypes IAi × ii, what percentage of the offspring
are expected to have the O blood type?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: C
Section: 4.5
Bloom's Taxonomy: Applying/Analyzing
23) In a mating between individuals with the genotypes IAIB × ii, what percentage of the
offspring are expected to have the O blood type?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: A
Section: 4.5
Bloom's Taxonomy: Applying/Analyzing
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24) With ________, there will be more than two genetic alternatives for a given locus.
A) codominance
B) gain of function mutant
C) incomplete dominance
D) epistasis
E) multiple alleles
Answer: E
Section: 4.5
Bloom's Taxonomy: Remembering/Understanding
25) The ABO blood group locus in humans provides an example of ________.
A) multiple alleles
B) epistasis
C) conditional alleles
D) auxotrophy
E) incomplete dominance
Answer: A
Section: 4.5
Bloom's Taxonomy: Remembering/Understanding
26) What would be a typical phenotypic monohybrid ratio in which a lethal allele is involved?
A) 2:1
B) 9:3:3:1
C) 6:2:3:1
D) 13:5
E) 3:1
Answer: A
Section: 4.6
Bloom's Taxonomy: Remembering/Understanding
27) Achondroplasia is a form of human dwarfism. If two individuals with achondroplasia have
children with the following ratio, 2 dwarf to 1 wild type, what is the means of inheritance of this
phenotype?
A) simple dominance
B) simple recessive
C) dominant lethal
D) recessive lethal
E) epistasis
Answer: C
Section: 4.6
Bloom's Taxonomy: Evaluating/Creating
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28) An allele of RNA polymerase in Drosophila demonstrates a recessive lethal inheritance
pattern. Which of the following is true?
A) Two heterozygotes will have offspring with a 2:1 pattern.
B) Two homozygous recessive parents will have 100% homozygous offspring.
C) Homozygous recessive offspring will never be seen.
D) The threshold effect is not seen in recessive lethality.
E) Homozygous lethal genes will always demonstrate a gain of function pattern.
Answer: C
Section: 4.6
Bloom's Taxonomy: Evaluating/Creating
29) A snapdragon plant has red, pink, or white flowers, inherited in an incomplete dominance
pattern. It can come in tall or dwarf (homozygous recessive) varieties. What is the ratio expected
if you cross a pink tall (heterozygous) plant with a pink short plant?
A) 1 red tall : 1 red short : 2 pink tall : 2 pink short : 1 white tall : 1 white short
B) 1 red tall : 1 red short : 1 pink tall : 1 pink short : 1 white tall : 1 white short
C) 3 red tall : 3 red short : 1 white tall : 1 white short
D) 2 red tall : 2 red short : 1 pink tall : 1 pink short : 1 white tall : 1 white short
E) 2 red tall : 1 red short : 1 pink tall : 1 pink short : 2 white tall : 1 white short
Answer: A
Section: 4.7
Bloom's Taxonomy: Evaluating/Creating
30) Rabbits can come in an allelic series where brown > himilayan > chinchilla > albino, where
> means dominant over. They also can come in long (recessive) and short fur. You wish to
obtain rabbit babies that are albino with long fur. Which of the following crosses would you do?
A) albino with long fur × homozygous himilayan with short fur (heterozygote)
B) chinchilla heterozygous with albino with short fur (heterozygous) × brown heterozygous with
albino with long fur
C) albino with long fur × albino with short fur (homozygous)
D) albino with long fur × brown chinchilla heterozygote with long fur
E) homozygous brown with long fur × albino with short fur (heterozygote)
Answer: B
Section: 4.7
Bloom's Taxonomy: Evaluating/Creating
31) What term is used to express the idea that several genes exert influence over the same
characteristic?
A) coextension
B) co-adhesion
C) terminal interaction
D) gene interaction
E) transformation
Answer: D
Section: 4.8
Bloom's Taxonomy: Remembering/Understanding
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32) 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
33) 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
34) A mutant gene that produces brown eyes (bw) is located on chromosome #2 of Drosophila
melanogaster, whereas a mutant gene producing bright red eyes, scarlet (st), is located on
chromosome #3. Phenotypically, wild-type flies (with dull red eyes), whose mothers had brown
eyes and whose fathers had scarlet eyes, were mated. The 800 offspring possessed the following
phenotypes: wild type (dull red), white, scarlet (bright red), and brown. Most of the 800
offspring had wild-type eyes, whereas those with white eyes were the least frequent.
From the information presented above, how many white-eyed flies would you expect in the F2
generation?
A) 450
B) 150
C) 200
D) 350
E) 50
Answer: E
Section: 4.8
Bloom's Taxonomy: Evaluating/Creating
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35) In the mouse, gene A allows pigmentation to be deposited in the individual coat hairs; its
allele a prevents such deposition of pigment, resulting in an albino. Gene B gives agouti (wildtype fur); its allele b gives black fur. What would the expected ratio of the progeny be in a cross
of a doubly heterozygous agouti mouse mated with a doubly homozygous recessive white
mouse?
A) 1 (agouti) :1 (black) :1 (albino)
B) 1 (agouti) :1 (black) : 2 (albino)
C) 9 (agouti) :3 (black) : 4 (albino)
D) 3 (agouti) :4 (black) : 9 (albino)
E) 9 (black) : 27(albino)
Answer: B
Section: 4.8
Bloom's Taxonomy: Applying/Analyzing
36) Which of the following statements is false?
A) Epistasis refers to a gene (or genes) of one pair masking the expression of a gene (or genes) at
a different locus.
B) Dominance refers to the form of expression of a gene in relation to its allele (or alleles).
C) When an allele is dominant, the heterozygous combination is the same phenotypically as one
of the homozygotes.
D) Epistasis is a nonallelic interaction; dominance is an allelic interaction.
E) Epistasis would not be expected to occur in a biochemical pathway.
Answer: E
Section: 4.8
Bloom's Taxonomy: Applying/Analyzing
37) 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) 12 purple : 4 white
B) 9 purple : 7 white
C) 1 purple : 3 white
D) 3 purple : 1 white
E) 1 purple : 1 white
Answer: C
Section: 4.8
Bloom's Taxonomy: Applying/Analyzing
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38) 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 phenotypes (with frequencies) of the offspring.
A) 3 white : 9 yellow : 4 green
B) 4 white :9 yellow : 3 green
C) 12 white : 1 yellow : 3 green
D) 12 white : 3 yellow : 1 green
E) 9 white : 3 yellow : 4 green
Answer: D
Section: 4.8
Bloom's Taxonomy: Applying/Analyzing
39) Which of the following ratio is not likely to occur in crosses (F2) when one is dealing with
two interacting, epistatic gene pairs?
A) 9:7
B) 3:1
C) 9:3:4
D) 12:3:1
E) 15:1
Answer: B
Section: 4.8
Bloom's Taxonomy: Remembering/Understanding
40) Assume that a dihybrid F2 ratio, resulting from epistasis, was 9:3:4. If a double heterozygote
was crossed with the fully recessive type, what phenotypic ratio is expected among the
offspring?
A) 1:1:2
B) 9:3:4
C) 12: 3: 1
D) 1:1:1
E) 1:2:1
Answer: A
Section: 4.8
Bloom's Taxonomy: Evaluating/Creating
41) Assume that a dihybrid F2 ratio, resulting from epistasis, was 15:1. If a double heterozygote
was crossed with the fully recessive type, what phenotypic ratio is expected among the
offspring?
A) 1:2
B) 1:3
C) 5:7
D) 3:1
E) 1:1
Answer: D
Section: 4.8
Bloom's Taxonomy: Evaluating/Creating
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42) 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, and 6 single. What
are the probable genotypes of the parents?
A) RRPP × Rrpp
B) RrPp × RRPP
C) Rrpp × RrPp
D) rrpp × RrPp
E) rrPP × RRpp
Answer: C
Section: 4.8
Bloom's Taxonomy: Evaluating/Creating
43) 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. Crosses among
heterozygotes give a 12:3:1 ratio. What type of gene interaction would account for these results?
A) epistasis
B) conditional lethal
C) multiple alleles
D) complementation
E) incomplete dominance
Answer: A
Section: 4.8
Bloom's Taxonomy: Remembering/Understanding
44) A particular cross gives a modified dihybrid ratio of 9:7. What phenotypic ratio would you
expect in a testcross of the fully heterozygous F1 crossed with the fully recessive type?
A) 3:1
B) 2:3
C) 2:1
D) 1:3
E) 1:1
Answer: D
Section: 4.8
Bloom's Taxonomy: Evaluating/Creating
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45) When two squash plants with disc-shaped fruit are crossed, depending on the genotype of the
parents, disc, sphere, and long fruits may be seen in the offspring. This is an example of
________.
A) epistasis
B) complementation
C) heterogenous trait
D) two gene interaction with novel phenotypes
E) dominant lethal
Answer: D
Section: 4.8
Bloom's Taxonomy: Remembering/Understanding
46) What is a heterogenous trait?
A) a trait in which the phenotype differs based on the environment
B) a trait that differs, but not based on environmental conditions
C) a trait in which different sexes have different manifestations of severity
D) a trait in which offspring show a more severe phenotype depending on which parent donates
the allele
E) a trait in which many genes can potentially be mutated to lead to the same phenotype
Answer: E
Section: 4.8
Bloom's Taxonomy: Remembering/Understanding
47) Alleles that are masked by an epistatic locus are said to be ________ to the genes at that
locus.
A) hypostatic
B) auxotrophic
C) conditional
D) complementary
E) functional
Answer: A
Section: 4.8
Bloom's Taxonomy: Remembering/Understanding
48) Multiple mutations that are found to be present in a single gene are said to belong to the
same ________ group.
A) phenotypic
B) allelic
C) transfer
D) complementation
E) expression
Answer: D
Section: 4.9
Bloom's Taxonomy: Remembering/Understanding
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49) Which of the following crosses would indicate that the mutants were in the same
complementation group?
A) pink eye fly × white eye fly -> all white eyes
B) pink eye fly × pink eye fly -> all white eye offspring
C) red eye fly × red eye fly -> 3 : 1 ratio
D) pink eye fly × pink eye fly -> purple eye flies
E) pink eye fly × pink eye fly -> pink eye fly
Answer: E
Section: 4.9
Bloom's Taxonomy: Applying/Analyzing
50) What is the term for a gene mutation that leads to a myriad of disparate effects?
A) complementation
B) lethal allele
C) pleiotropy
D) multiple allele
E) expressivity
Answer: C
Section: 4.10
Bloom's Taxonomy: Remembering/Understanding
51) With which of the following would hemizygosity most likely be associated?
A) codominance
B) incomplete dominance
C) trihybrid crosses
D) X-linked inheritance
E) sex-limited inheritance
Answer: D
Section: 4.11
Bloom's Taxonomy: Remembering/Understanding
52) 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 are correct.
Answer: B
Section: 4.11
Bloom's Taxonomy: Remembering/Understanding
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53) The white-eyed 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) females wild type, males white eyed
B) all wild type
C) females half wild type, half white eyed; males white eyed
D) females white eyed; males wild type
E) all white eyed
Answer: A
Section: 4.11
Bloom's Taxonomy: Applying/Analyzing
54) Two forms of hemophilia are determined by genes on the X chromosome in humans.
Assume that a phenotypically normal woman whose father had hemophilia is married to a
normal man. What is the probability that their first son will have hemophilia?
A) zero
B) 1/3
C) 1/4
D) 1/2
E) 100%
Answer: D
Section: 4.11
Bloom's Taxonomy: Applying/Analyzing
55) Two forms of hemophilia are determined by genes on the X chromosome in humans.
Assume that a phenotypically normal woman whose father had hemophilia is married to a
normal man. What is the probability that their first daughter will have hemophilia?
A) zero
B) 1/3
C) 1/4
D) 1/2
E) 100%
Answer: A
Section: 4.11
Bloom's Taxonomy: Applying/Analyzing
56) The deficiency of the enzyme glucose-6-phosphate dehydrogenase (G6PD) is inherited as a
recessive gene on the X chromosome in humans. A phenotypically normal woman (whose father
had G6PD) is married to a normal man. What fraction of their sons would be expected to have
G6PD?
A) zero
B) 1/3
C) 1/4
D) 1/2
E) 100%
Answer: D
Section: 4.11
Bloom's Taxonomy: Applying/Analyzing
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57) 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 Number
wild
96
yellow
99
wild
197
Based on this information, how is the yellow allele inherited?
A) dominant, autosomal
B) recessive, autosomal
C) dominant, X linked
D) recessive, X linked
E) sex-influenced inheritance
Answer: D
Section: 4.11
Bloom's Taxonomy: Applying/Analyzing
58) If an X-linked disorder is lethal to the affected individual prior to the age at which one
reaches reproductive maturation, the lethality will be expressed only in males. Why is this so?
A) The only sources of the lethal allele in the population are heterozygous females who are
"carriers" and do not express the disorder.
B) Because it's sex-limited inheritance, females do not demonstrate the phenotype.
C) As it's a lethal allele, it will kill the homozygous recessive first. These are the males.
D) At the cellular level, the protein will work less well in the male's cells.
E) This is not true. We will see the lethality in females as well.
Answer: A
Section: 4.11
Bloom's Taxonomy: Evaluating/Creating
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59) The accompanying figure 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 the inheritance pattern?
A) dominant, autosomal
B) recessive, autosomal
C) dominant, X linked
D) recessive, X linked
E) sex-influenced inheritance
Answer: B
Section: 4.11
Bloom's Taxonomy: Applying/Analyzing
60) The genes for zeste eyes and forked bristles are located on the X chromosome in Drosophila
melanogaster. Both genes are recessive. A cross is made between a zeste-eyed female and a
forked-bristled male. If 200 offspring from this cross were obtained, what is the expected
offspring's phenotype and sex?
A) 100 wild female; 100 zeste male
B) 100 forked female; 100 zeste male
C) 100 zeste female; 100 zeste forked male
D) all wild type
E) 50 wild female, 50 forked female; 50 zeste forked male, 50 zeste male
Answer: A
Section: 4.11
Bloom's Taxonomy: Applying/Analyzing
61) Which of the following is true about X-linkage?
A) One result of X-linkage is a crisscross pattern of inheritance in which sons express recessive
genes of their fathers and daughters express recessive genes of their mothers.
B) One result of X-linkage is a crisscross pattern of inheritance in which daughters express
recessive genes of their fathers and sons express recessive genes of their mothers.
C) Males are more liable to express recessive alleles as they are hemizygous.
D) Females will only express dominant alleles.
E) X linkage demonstrates that only the wild-type allele is the dominant allele.
Answer: C
Section: 4.11
Bloom's Taxonomy: Applying/Analyzing
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62) What is a significant difference between X-linked and sex-influenced inheritances?
A) X-linked inheritance affects females and sex-influenced inheritance affects males.
B) There are no differences. They are the same.
C) In X-linked inheritance, the gene in question is on the X chromosome; in sex-influenced
inheritance, the gene is autosomal.
D) Sex-influenced inheritance requires epistasis. X-linked inheritance does not.
E) X-limited inheritance is only recessive. Sex-influenced inheritance can be recessive or
dominant.
Answer: C
Section: 4.11, 4.12
Bloom's Taxonomy: Remembering/Understanding
63) Name three modes of inheritance that are influenced by the sex of individuals.
A) X-linked, sex-influenced, sex-limited
B) sex-influenced, epistasis, sex-limited
C) conditional alleles, sex-limited, X-linked
D) epistasis, conditional alleles, expressivity
E) penetrance, X-limited inheritance, X-linked inheritance
Answer: A
Section: 4.11, 4.12
Bloom's Taxonomy: Remembering/Understanding
64) Which of the following is an example of sex-limited inheritance?
A) human baldness
B) cat fur length
C) beard formation in humans
D) hemophilia
E) hair color in humans
Answer: C
Section: 4.12
Bloom's Taxonomy: Remembering/Understanding
65) A ________ is one whose expression is influenced by some environmental condition.
A) X-linked allele
B) conditional mutant
C) positional effect
D) multiple allele
E) expressive allele
Answer: B
Section: 4.13
Bloom's Taxonomy: Remembering/Understanding
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66) Which of the following statements is true?
A) Penetrance specifically refers to the expression of lethal genes in heterozygotes.
B) Expressivity is the term used to describe the balanced genetic output from a hemizygous
condition.
C) Genomic imprinting occurs when one allele converts another.
D) Genomic anticipation refers to observations that a genetic disorder occurs at an earlier age in
successive generations, whereas genetic imprinting occurs when gene expression varies
depending on parental origin.
E) The term expressivity defines the percentage of individuals who show at least some degree of
expression of a mutant genotype.
Answer: D
Section: 4.13
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global 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) 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: 5.1
Bloom's Taxonomy: Evaluating/Creating
4) In humans, the male is the ________ sex.
A) heterogametic
B) heteromorphic
C) homogametic
D) homomorphic
E) monogametic
Answer: A
Section: 5.1
Bloom's Taxonomy: Remembering/Understanding
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5) 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
D) sets of autosomes
E) high levels of estrogen
F) multiple alleles scattered throughout the autosomes
Answer: A
Section: 5.2
Bloom's Taxonomy: Remembering/Understanding
6) Jacobs syndrome in humans, which is manifested by a higher than average stature and
potential behavioral problems, 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
7) Which of the following human genetic conditions is missing a chromosome?
A) Klinefelter syndrome
B) Jacob syndrome
C) Turner syndrome
D) XXXX syndrome
E) Down's syndrome
Answer: C
Section: 5.2
Bloom's Taxonomy: Remembering/Understanding
8) Individuals have been identified who have two different karyotypes, such as 45, X/46, XY or
45, X/46, XX. Such individuals are called ________.
A) heterogametes
B) heteromorphic
C) homogametic
D) trisomic
E) mosaics
Answer: E
Section: 5.2
Bloom's Taxonomy: Remembering/Understanding
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9) Which of the following karyotypes would lead to male characteristics in humans?
A) XO
B) XXY
C) XYY
D) XO and XYY
E) XXY and XYY
Answer: E
Section: 5.2
Bloom's Taxonomy: Remembering/Understanding
10) A small part of the human Y chromosome contains the gene that is responsible for
determining maleness. What is the name of this gene?
A) medulla
B) SRY
C) Xist
D) Dax
E) NRY
Answer: B
Section: 5.2
Bloom's Taxonomy: Remembering/Understanding
11) Under what condition might a human female have the XY sex chromosome complement?
A) She has an XX chromosome and a Y chromosome.
B) This is not possible.
C) The X chromosome has a mutation such that it is able to overpower the Y chromosome.
D) This female would have one complete X chromosome and a Y chromosome that lacks SRY.
E) The Y chromosome has an active Xist gene on it.
Answer: D
Section: 5.2
Bloom's Taxonomy: Applying/Analyzing
12) Klinefelter and Turner syndromes have how many chromosomes, respectively?
A) 46, 45
B) 47, 45
C) 45, 47
D) 46, 46
E) 47, 46
Answer: B
Section: 5.2
Bloom's Taxonomy: Remembering/Understanding
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13) Which region(s) of the X chromosome recombine during meiosis?
A) the pseudoautosomal regions
B) the heterochromain
C) the centromere
D) the male-specific region of the Y
E) SRY
Answer: A
Section: 5.3
Bloom's Taxonomy: Remembering/Understanding
14) What does the nonrecombining region of the Y contain?
A) the genes that have homologues on the X chromosome
B) SRY
C) the centromere
D) A, B, and C
E) B and C
Answer: E
Section: 5.3
Bloom's Taxonomy: Remembering/Understanding
15) 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.
A) nondisjunction in the father at meiosis I
B) nondisjunction in the father in meiosis I or II
C) nondisjunction in the other in meiosis I or II
D) nondisjunction in the mother in meiosis I and in the father in meiosis II
E) no abnormalities had to occur to produce this daughter
Answer: C
Section: 5.2
Bloom's Taxonomy: Evaluating/Creating
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16) 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? Which of the
following is an example of metaphase of meiosis I?
A)
B)
C)
D)
E)
Answer: A
Section: 5.2
Bloom's Taxonomy: Applying/Analyzing
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17) A recessive gene for red-green color blindness is located on the X chromosome in humans.
Assume that a woman with normal vision (her father is color blind) marries a color-blind male.
What is the likelihood that this couple's first son will be color blind?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: C
Section: 5.2
Bloom's Taxonomy: Evaluating/Creating
18) One form of hemophilia is caused by a sex-linked recessive gene. Assume that a man with
hemophilia marries a phenotypically normal woman whose father had hemophilia. What is the
probability that they will have a daughter with hemophilia? (Note: In this problem, you must
include the probability of having a daughter in your computation of the final probability.)
A) 1/16
B) 1/8
C) 1/4
D) 1/2
E) 3/4
Answer: C
Section: 5.2
Bloom's Taxonomy: Evaluating/Creating
19) One form of hemophilia is caused by a sex-linked recessive gene. Assume that a man with
hemophilia marries a phenotypically normal woman whose father had hemophilia. What is the
probability that their first son will have hemophilia?
A) 1/16
B) 1/8
C) 1/4
D) 1/2
E) 3/4
Answer: D
Section: 5.2
Bloom's Taxonomy: Evaluating/Creating
20) Assume that a man who carries an X-linked gene has children. Assuming normal meiosis and
random combination of gametes, the man would pass this gene to ________.
A) half of his daughters
B) all of his daughters
C) all of his sons
D) half of his sons
E) all of his children
Answer: B
Section: 5.2
Bloom's Taxonomy: Applying/Analyzing
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21) Glucose-6-phosphate dehydrogenase (G6PD) deficiency is inherited as an X-linked recessive
gene in humans. A woman whose father suffered from G6PD marries a normal man. What
percentage of their sons is expected to be G6PD?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: C
Section: 5.2
Bloom's Taxonomy: Applying/Analyzing
22) The accompanying figure is a pedigree of a fairly common human hereditary trait; the boxes
represent males and the circles represent females. Filled in symbols indicate the abnormal
phenotype. Given that one gene pair is involved, what is the inheritance pattern of the trait?
A) autosomal dominant
B) autosomal recessive
C) X-linked dominant
D) X-linked recessive
E) epistasis
Answer: B
Section: 5.2
Bloom's Taxonomy: Applying/Analyzing
23) Which of the following statements below is false?
A) Normally in humans, all the sons of a male showing an X-linked phenotype will inherit the
trait.
B) Normally in humans, all the sons of a female homozygous for an X-linked recessive gene will
inherit that trait.
C) Normally in humans, a male that is carrying an X-linked dominant trait will pass it to all his
daughters.
D) At meiosis I, the X and Y chromosomes line up as if they were homologs.
E) Normally in humans, females are carriers of X-linked recessive traits if they are heterozygous.
Answer: A
Section: 5.2
Bloom's Taxonomy: Applying/Analyzing
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24) What is the ratio of male to female conceptions in a population called?
A) secondary sex ratio
B) primary sex ratio
C) conception birth order number
D) anomalous gender reorder calculation
E) gender order
Answer: B
Section: 5.3
Bloom's Taxonomy: Applying/Analyzing
25) 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: Applying/Analyzing
26) 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. What needed to occur to
have this genetic outcome?
A) The mother must be heterozygous and underwent nondisjunction at meiosis I.
B) The mother is homozygous recessive and underwent nondisjunction in meiosis I or II.
C) The mother must be heterozygous and underwent nondisjunction at meiosis II.
D) Nondisjunction in meiosis I of the father occurred.
E) The chromosomal doubling effect of meiosis II occurred in the father.
Answer: C
Section: 5.4
Bloom's Taxonomy: Evaluating/Creating
27) 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?
A) 1/2
B) 1/4
C) none
D) 3/4
E) 2/3
Answer: B
Section: 5.4
Bloom's Taxonomy: Applying/Analyzing
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28) What region is the major control center of X inactivation in mammals?
A) SRY
B) NRY
C) Xic
D) Barr Body
E) MSY
Answer: C
Section: 5.4
Bloom's Taxonomy: Remembering/Understanding
29) How many Barr bodies would one expect to see in cells of Turner syndrome females and
Klinefelter syndrome males?
A) one and zero
B) one and one
C) zero and two
D) zero and zero
E) zero and one
Answer: E
Section: 5.4
Bloom's Taxonomy: Applying/Analyzing
30) X-linked genes in female mammals often demonstrate ________.
A) incomplete dominance
B) phenotypic mosaicism
C) underrepresentation in the genome
D) heterochromatin formation
E) nonrecombination during meiosis
Answer: B
Section: 5.4
Bloom's Taxonomy: Remembering/Understanding
31) 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." A mating between a black male and a calico female occurred. Give
the phenotypes of the offspring.
A) all black regardless of sex
B) all calico regardless of sex
C) calico females and yellow males
D) calico females, yellow females, black males, and yellow males
E) calico females, black females, yellow makes, and black males
Answer: E
Section: 5.4
Bloom's Taxonomy: Evaluating/Creating
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32) Where is the general location of a Barr body?
A) in the nucleolus
B) attached to the nuclear envelope
C) attached to the other X chromosome
D) adjacent to the Y chromosome
E) stuck in a nuclear pore
Answer: B
Section: 5.4
Bloom's Taxonomy: Remembering/Understanding
33) Which of the following statements is false?
A) An individual with Klinefelter syndrome generally has one Barr body.
B) An individual with Turner Syndrome has no Barr bodies.
C) A typical XX human female has one Barr body.
D) Dosage compensation is accomplished in humans by inactivation of the Y chromosome.
E) In a cell with X chromosomes lacking the XIC, there is still X inactivation.
Answer: D
Section: 5.4
Bloom's Taxonomy: Applying/Analyzing
34) Which of the following is not a tenant of the Lyon hypothesis?
A) X inactivation occurs early in development.
B) X inactivation is passed down during mitosis.
C) X inactivation occurs on the maternal X chromosome.
D) X inactivation of descendant cells is the same as the progenitor cell.
E) initial X inactivation is random.
Answer: C
Section: 5.4
Bloom's Taxonomy: Remembering/Understanding
35) In Drosophila, sex is determined by a balance between the number of haploid sets of
autosomes and the number of ________.
A) telomeres
B) centromeres
C) X chromosomes
D) Y chromosomes
E) nucleolar organizers
Answer: C
Section: 5.5
Bloom's Taxonomy: Remembering/Understanding
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36) 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. Would you be correct in assuming that the
Drosophila sex for XO is also female? Choose the answer that includes the correct explanation.
A) Yes, because sex determination in humans and insects is essentially the same.
B) No, sex determination in Drosophila is dependent on the presence or absence of the Y
chromosome.
C) No, the chromosomal basis for sex determination in Drosophila is based on the balance
between the number of X chromosomes and haploid sets of autosomes.
D) Yes, the presence of an X chromosome determines a female in both organisms.
E) No, it takes two X chromosomes to produce a female in humans and a Y chromosome to
produce a male in Drosophila.
Answer: C
Section: 5.5
Bloom's Taxonomy: Evaluating/Creating
37) What are the sex-chromosome constitution (X and Y chromosomes) of the inviable offspring
resulting from a cross between a white-eyed female (Xw XwY) and a wild-type male (normal
chromosome complement) in Drosophila melanogaster?
A) X+XwXw and YY
B) XwXwY and YY
C) X+Y
D) YY and XwYY
E) X+XwY and X+XwXw
Answer: A
Section: 5.5
Bloom's Taxonomy: Evaluating/Creating
38) 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.
A) nondisjunction in meiosis II of father
B) nondisjunction in meiosis I of mother
C) nondisjunction in either meiosis I or meiosis II of the mother
D) nondisjunction in meiosis II of mother
E) nondisjunction in meiosis I of mother and meiosis II of father
Answer: C
Section: 5.5
Bloom's Taxonomy: Evaluating/Creating
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39) 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.
A) nondisjunction in meiosis II of the mother
B) nondisjunction in meiosis I of the father
C) nondisjunction in meiosis I or II in father
D) nondisjunction in meiosis I of both the mother and father
E) nondisjunction in meiosis I of the mother
Answer: A
Section: 5.5
Bloom's Taxonomy: Evaluating/Creating
40) Which of the following statements is true regarding Drosophila?
A) The Y chromosome determines male flies.
B) The Y chromosome plays no role in fly sex determination or fly reproduction.
C) The ratio of X to Y chromosomes determines fly sex.
D) The Y chromosome plays no role on fly sex determination but does contain genes for fly
reproduction.
E) The ratio of autosomes to fly Y and X chromosomes determines fly sex.
Answer: D
Section: 5.5
Bloom's Taxonomy: Remembering/Understanding
41) Give the sex of the following Drosophila assuming that the autosomes are present in the
normal number. XO; XY; XXY. Answers are in order (respectively).
A) male; male; male
B) female; male; male
C) male; male; female
D) female; male; female
E) male; female; female
Answer: C
Section: 5.5
Bloom's Taxonomy: Applying/Analyzing
42) Give the sex of the following people assuming that the autosomes are present in the normal
number. XO; XY; XXY. Answers are in order (respectively).
A) male; male; male
B) female; male; male
C) male; male; female
D) female; male; female
E) male; female; female
Answer: D
Section: 5.5
Bloom's Taxonomy: Applying/Analyzing
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43) Assuming a normal number of autosomes, what would be the sex of the following: XXY
mouse, XXY Drosophila?
A) male and female, respectively
B) female and female, respectively
C) male and male, respectively
D) female and male, respectively
E) both neuter
Answer: A
Section: 5.5
Bloom's Taxonomy: Applying/Analyzing
44) Data produced by C. Bridges in the early part of this century indicate that sex in Drosophila
is determined by ________.
A) the number of X chromosomes
B) a balance between the number of X chromosomes and the number of haploid sets of
autosomes
C) the number of Y chromosomes
D) the SRY gene
E) the ratio of X to Y chromosomes
Answer: B
Section: 5.5
Bloom's Taxonomy: Remembering/Understanding
45) In Drosophila, the sex of a fly with the karyotype ________ is male.
A) XXX:2A
B) XYY:4A
C) XX:2A
D) XXY:2A
E) XO:2A
Answer: E
Section: 5.5
Bloom's Taxonomy: Applying/Analyzing
46) In Drosophila, the female is the ________ sex.
A) heterogametic
B) homogametic
C) mosaic
D) isogametic
E) genic balance
Answer: B
Section: 5.5
Bloom's Taxonomy: Remembering/Understanding
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47) In ________, sex is determined by the ratio of the number of X chromosomes to the number
of haploid sets of autosomes.
A) Drosophila
B) C. elegans
C) humans
D) reptiles
E) Drosophila and C. elegans
Answer: E
Section: 5.5
Bloom's Taxonomy: Remembering/Understanding
48) Genotypic sex determination does not in ________.
A) flies
B) round worms
C) mammals
D) birds
E) reptiles
Answer: E
Section: 5.6
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 6 Chromosomal 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) Having a complete set or sets of chromosomes is called ________.
A) euploid
B) monoploid
C) ploidy
D) diploid
E) aneuploidy
Answer: A
Section: 6.1
Bloom's Taxonomy: Remembering/Understanding
3) ________ is viewed as a major cause of aneuploidy.
A) Colchicine treatment
B) Segmental deletions
C) Heat treatment
D) Nondisjunction
E) X-ray mutations
Answer: D
Section: 6.1
Bloom's Taxonomy: Remembering/Understanding
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4) 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: Applying/Analyzing
5) Trisomy 21, or Down syndrome, occurs when there is a normal diploid chromosomal
complement but one (extra) chromosome 21. While there is reduced fertility in both sexes,
females have higher fertility 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.
Given the fact that conceptuses with 48 chromosomes (four #21 chromosomes) are not likely to
survive early development, what percentage of surviving offspring would be expected to have
Down syndrome if both parents have Down syndrome?
A) One-third of the surviving offspring would be expected to have Down syndrome.
B) All the children would be expected to have Down syndrome.
C) None of the surviving offspring would be expected to have Down syndrome.
D) Two-thirds of the surviving offspring would be expected to have Down syndrome.
E) One-half of the surviving offspring would be expected to have Down syndrome.
Answer: D
Section: 6.2
Bloom's Taxonomy: Applying/Analyzing
6) What explanation is generally given for lethality of monosomic individuals?
A) Monosomy may unmask recessive lethals that are tolerated in heterozygotes carrying the
wild-type allele.
B) The gametes of monosomic individuals cannot undergo meiosis, and this is lethal.
C) Cells count the number of chromosomes they have and will undergo apoptosis when the
chromosome number is incorrect.
D) Monosomic chromosomes cannot undergo mitosis correctly.
E) The loss of a single chromosome is not generally lethal, unless the individual is inbred.
Answer: A
Section: 6.2
Bloom's Taxonomy: Remembering/Understanding
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7) What is the leading cause of Down syndrome?
A) In older females, chromosome 21 is duplicated leading to abnormal egg formation.
B) In older parents, there is egg/sperm incompatibility leading to duplication of chromosome 21.
C) Aberrant implantation in the uterus leads to developmental problems in the fetus.
D) The nondisjunctional event that produces Down syndrome occurs more frequently during
oogenesis in women older than age 35.
E) In men over 25, sperm formation is impaired and produces monosomic children.
Answer: D
Section: 6.2
Bloom's Taxonomy: Remembering/Understanding
8) What experiment led to the identification of the DSCR (Down Syndrome Critical Region)?
A) transgenic humans were developed that had a duplicated DSCR
B) an extra DSCR was injected into yeast cells
C) a mouse model was developed trisomic for the DSCR
D) a mouse model was developed lacking the DSCR
E) cancer cells were sequenced to identify the VEGF gene
Answer: C
Section: 6.2
Bloom's Taxonomy: Remembering/Understanding
9) Name two methods used in genetic prenatal diagnostic testing in humans.
A) amniocentesis and DSCR
B) whole genome sequencing and PCR
C) chorionic villus sampling and cellular mitotic analysis
D) amniocentesis and NIPGD
E) amniocentesis and fetal uterine physiology compatibility test
Answer: D
Section: 6.2
Bloom's Taxonomy: Remembering/Understanding
10) Haploinsufficiency refers to ________.
A) the condition whereby a single cell is insufficient to divide to cause cancer
B) a state of being whereby a single gene is sufficient to cause several phenotypes
C) the condition whereby a single chromosome is insufficient to sustain life
D) the process by which a single gene will cause a cascading effect on a genome's phenotype
E) the genetic predisposition for some genes to come in only one copy in the genome
Answer: C
Section: 6.2
Bloom's Taxonomy: Remembering/Understanding
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11) 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)
A) 47, 47
B) 46, 47
C) 45, 47
D) 45, 46
E) 47, 45
Answer: C
Section: 6.2
Bloom's Taxonomy: Applying/Analyzing
12) Given that a human normally contains 46 chromosomes, give the chromosome number for
each of the following conditions:
Triploid
Trisomy 13
A) 69, 47
B) 69, 45
C) 138, 47
D) 138, 45
E) 96, 47
Answer: A
Section: 6.2
Bloom's Taxonomy: Applying/Analyzing
13) Assume that a species has a diploid chromosome number of 24. The term applied to an
individual with 25 chromosomes would be ________.
A) aneuploidy
B) euploid
C) triploid
D) trisomy
E) aneuploidy and trisomy
Answer: E
Section: 6.2
Bloom's Taxonomy: Remembering/Understanding
14) An individual with Patau syndrome would be ________.
A) termed a triploid
B) said to have a monosomy
C) said to have a trisomy
D) termed a euploid
E) termed a haploid
Answer: C
Section: 6.2
Bloom's Taxonomy: Remembering/Understanding
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15) Noninvasive prenatal genetic diagnosis is sometimes preferred to amniocentesis because
________.
A) it gives quicker results
B) it gives more precise results
C) it requires only a blood draw from the mother
D) it requires less technical time of a research lab
E) the mother does not have to be present
Answer: C
Section: 6.2
Bloom's Taxonomy: Remembering/Understanding
16) A child is born with Turner's syndrome and she is red green color blind (recessive X-linked).
Her father is red green color blind as well and her mother is homozygous dominant for color
sight. What had to happen in order for this child to be born with this chromosomal complement?
A) nondisjunction in meiosis I of the father
B) nondisjunction in meiosis II of the father
C) nondisjunction in either meiosis I or meiosis II of the mother
D) nondisjunction in either meiosis I or meiosis II of the father
E) nondisjunction in meiosis II of the mother
Answer: C
Section: 6.2
Bloom's Taxonomy: Evaluating/Creating
17) A son is born with Kleinfelter's syndrome and hemophilia. His father was normal and his
mother was a carrier for the recessive X-linked blood clotting disorder. What occurred in
meiosis to produce this genetic outcome?
A) nondisjunction in meiosis I of the father
B) nondisjunction in meiosis II of the father
C) nondisjunction in either meiosis I or meiosis II of the mother
D) nondisjunction in meiosis I of the mother
E) nondisjunction in meiosis II of the mother
Answer: E
Section: 6.3
Bloom's Taxonomy: Evaluating/Creating
18) What is the outcome of nondisjunction in meiosis I?
A) a resultant gamete that is triploid
B) a resultant gamete that may harbor from one chromosome both homologs from one parent or
none at all
C) a resultant gamete that may harbor from one chromosome both sister chromatids or none at all
D) a resultant gamete that is devoid of all chromosomes
E) four gametes that are all trisomic
Answer: C
Section: 6.3
Bloom's Taxonomy: Evaluating/Creating
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19) An expected meiotic pairing configuration in a ________ would be a trivalent.
A) monoploid
B) trisomic individual
C) monosomic individual
D) diploid
E) triploid
Answer: E
Section: 6.3
Bloom's Taxonomy: Applying/Analyzing
20) Name the polyploid condition that is formed from the addition of an extra set of
chromosomes identical to the normal diploid complement of the same species.
A) autotetraploidy, assuming the normal chromosome complement is haploid
B) autotetraploidy, assuming the normal chromosome complement is diploid
C) allotetraploidy, assuming the normal chromosome complement is diploid
D) alloeuploidy, assuming the normal chromosome complement is diploid
E) autooctoploidyploidy, assuming the normal chromosome complement is haploid
Answer: B
Section: 6.3
Bloom's Taxonomy: Remembering/Understanding
21) Colchicine is an alkaloid derived from plants. What is its effect on chromosome behavior?
A) By interfering with spindle formation, replicated chromosomes fail to migrate to the poles at
anaphase; thus, sister chromatids end up in the same nucleus.
B) By interfering with spindle formation, replicated chromosomes fail to migrate to the poles at
anaphase; thus, homologous chromosomes end up in the same nucleus.
C) By blocking DNA replication, chromosomes fail to migrate to the poles at anaphase; thus,
homologous chromosomes end up in the same nucleus.
D) By blocking DNA replication, chromosomes do not undergo meiosis I and instead all gametes
suffer from aneuploidy.
E) By blocking DNA replication, chromosomes do not undergo meiosis II and instead half the
gametes are empty.
Answer: A
Section: 6.3
Bloom's Taxonomy: Remembering/Understanding
22) Doubling the chromosomes of a sterile species hybrid with colchicine or cold shock is a
method used to produce a fertile species hybrid called a ________.
A) allopolyploid
B) autoploid
C) (amphidiploid)
D) triploid
E) autoallopolyploid
Answer: C
Section: 6.3
Bloom's Taxonomy: Remembering/Understanding
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23) Assume that an organism has a diploid chromosome number of 14. There would be
________ chromosomes in a tetraploid.
A) 14
B) 28
C) 42
D) 56
E) 104
Answer: B
Section: 6.3
Bloom's Taxonomy: Applying/Analyzing
24) An ________ may arise when three sperm cells are involved in fertilization of a single egg.
A) autotetraploid
B) autotriploid
C) allotriploid
D) allotetraploid
E) aneuploidy
Answer: A
Section: 6.3
Bloom's Taxonomy: Applying/Analyzing
25) Assume that an organism has a haploid chromosome number of 7. There would be ________
chromosomes in a monoploid individual of that species.
A) 14
B) 28
C) 42
D) 7
E) 3.5
Answer: D
Section: 6.3
Bloom's Taxonomy: Applying/Analyzing
26) Assume that a species has a diploid chromosome number of 24. The term applied to an
individual with 36 chromosomes would be ________.
A) triploid
B) monoploid
C) allopolyploid
D) aneuploidy
E) tetraploid
Answer: A
Section: 6.3
Bloom's Taxonomy: Remembering/Understanding
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27) Which of the following is an example of an endopolyploid?
A) two diploid mate and produce a tetraploid
B) an egg has complete nondisjunction and the resultant offspring is triploid
C) one chromosome is lost during cell division
D) cells lining the gut of mosquito larvae attain 16n ploidy
E) an individual suffers from a trisomy
Answer: D
Section: 6.3
Bloom's Taxonomy: Remembering/Understanding
28) How do deletions and translocations in chromosomes often occur?
A) chromosomes will fuse telomere to telomere
B) chromosomes will break and the sticky ends will rejoin
C) during meiosis I, sister chromatid exchange leads to abnormalities
D) when cells undergo meiosis II, chromosomes naturally break when sister chromatids are
being pulled apart
E) colchicine treatment causes chromosomal rearrangements
Answer: B
Section: 6.4
Bloom's Taxonomy: Remembering/Understanding
29) 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
30) What is formed when meiosis occurs in an individual who is heterozygous for an intercalary
deletion?
A) a translocation
B) a compensation loop
C) a terminal loop
D) a trifecta chromosomal cross
E) a translocation cross
Answer: B
Section: 6.5
Bloom's Taxonomy: Remembering/Understanding
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31) The chromosomal aberration that causes cri-du-chat syndrome can be referred to as a(n)
________.
A) duplication
B) inversion
C) reciprocal translocation
D) simple translocation
E) segmental deletion
Answer: E
Section: 6.5
Bloom's Taxonomy: Remembering/Understanding
32) Provide an example of gene redundancy that occurs in both eukaryotes and prokaryotes.
A) rDNA
B) CNV
C) centromeres
D) spindle fiber protein genes
E) cell wall proteins
Answer: A
Section: 6.6
Bloom's Taxonomy: Remembering/Understanding
33) Describe double Bar mutations in Drosophila melanogaster.
A) duplications in a portion of the X chromosome
B) large deletion in the X chromosome
C) an inversion heterozygote of the X chromosome
D) an X chromosome with two duplications of the 16A region
E) a female fly with only one X chromosome
Answer: D
Section: 6.6
Bloom's Taxonomy: Remembering/Understanding
34) Gene amplification is the basis for ________.
A) the nucleolar organizing region
B) a segmental deletion
C) bar-eyed flies
D) interstitial deletions
E) formation of the translocation cross
Answer: A
Section: 6.6
Bloom's Taxonomy: Remembering/Understanding
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35) What error of meiosis leads to both a duplication and a deletion?
A) replication errors
B) X-ray chromosomal breakage
C) unequal crossing over
D) D loop formation
E) replication cross formation
Answer: C
Section: 6.6
Bloom's Taxonomy: Remembering/Understanding
36) Which of the following is not a potential outcome of a gene duplication?
A) they may result in gene redundancy
B) they may result in providing the raw material for evolution
C) they may produce phenotypic variation
D) they may lead to the development of gene families
E) they may lead to translocation cross formation during synapsis
Answer: E
Section: 6.6
Bloom's Taxonomy: Remembering/Understanding
37) What is NOT a potential consequence of a copy number variation?
A) none
B) higher protein levels of specific genes
C) phenotypic consequences leading to enhanced ability to fight infection
D) loss of function at multiple loci
E) phenotypic consequences that track with the number of each gene
Answer: D
Section: 6.6
Bloom's Taxonomy: Applying/Analyzing
38) A chromosome without a centromere is ________.
A) paracentric
B) pericentric
C) acentric
D) dicentric
E) segmental
Answer: C
Section: 6.7
Bloom's Taxonomy: Remembering/Understanding
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39) A ________ inversion is one whose breakpoints do not flank the centromere.
A) paracentric
B) pericentric
C) acentric
D) dicentric
E) segmental
Answer: A
Section: 6.7
Bloom's Taxonomy: Remembering/Understanding
40) Inversion heterokaryotypes are often characterized as having reduced crossing over and
reduced fertility. Assume that you were examining a strain of organisms you knew to be
inversion heterokaryotypes and saw a relatively high number of double chromatid bridges
extending between anaphase I nuclei, as indicated in the following drawing. What is the product
of this type of inversion loop?
A) two acentric fragments and two dicentric chromosomes
B) one acentric fragment, one dicentric chromosome, and two wild-type chromosomes
C) four wild-type chromosomes
D) two chromosomes with deletions and two with duplications
E) two chromosomes with deletions and two acentric fragments
Answer: A
Section: 6.7
Bloom's Taxonomy: Evaluating/Creating
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41) What type of chromosomal configuration does the following diagram illustrate?
A) inversion (paricentric) heterozygote
B) reciprocal translocation heterozygote
C) inversion (paracentric) heterozygote
D) simple translocation heterozygote
E) inversion (paracentric) homozygote
Answer: C
Section: 6.7
Bloom's Taxonomy: Evaluating/Creating
42) What is a potentially evolutionary advantage of inversion heterozygosity?
A) there is none
B) enhanced genetic variation in the offspring
C) if a favorable complement of alleles occurs in an inversion heterozygote, they will not be
disrupted by crossing over
D) inversion loops stimulate favorable allele combinations
E) haploinsufficiency does not occur in inversion heterozygotes
Answer: C
Section: 6.7
Bloom's Taxonomy: Applying/Analyzing
43) 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
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44) Which of the following statements is TRUE?
A) Individuals with familial Down syndrome are trisomic and have 47 chromosomes.
B) In general, inversion and translocation heterozygotes are as fertile as organisms whose
chromosomes are in the standard arrangement.
C) Familial Down syndrome is caused by a translocation involving chromosome 21.
D) Translocations may be pericentric or paracentric.
E) Inversions and translocations are without evolutionary significance.
Answer: C
Section: 6.8
Bloom's Taxonomy: Remembering/Understanding
45) 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
46) Genetic anticipation is best described as ________.
A) the phenomenon by which synapsis is delayed due to translocation cross formation
B) trinucleotide repeats increase in future generations
C) trinucleotide repeats form fragile sites
D) the G2 gap before meiosis
E) cells stop during the cell cycle to check for DNA damage or acentric fragments
Answer: B
Section: 6.9
Bloom's Taxonomy: Remembering/Understanding
47) Recently, a gene located on chromosome 3 in humans, FHIT, has been shown to be
associated with the significant human malady known as ________.
A) 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
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48) What is a fragile site?
A) a structural component of a chromosome due to proteins bound to it that leads to a high
percentage of chromosomal breakage
B) a structural component of a chromosome due to the nucleotide sequence to it that leads to a
high percentage of chromosomal breakage
C) a site in a chromosome prone to duplication, such as the bar gene
D) an area around the centromere prone to inversion due to gene duplications
E) the telomere of a chromosome can fall off due to spindle malfunctions during meiosis
Answer: B
Section: 6.9
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 7 Chromosome Mapping in Eukaryotes
1) What term is applied when two genes fail to assort independently, that is, they tend to
segregate together during gamete formation?
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
2) Which of the following statements is true?
A) A centromere and its surrounding genes constitute a linkage group.
B) The chromosomal theory of inheritance denotes that linked alleles will never be separated.
C) Linkage without crossing over leads to all recombinant chromosomes.
D) The linkage ratio is best seen in genes on nonhomologous chromosomes.
E) If two gene loci are on nonhomologous chromosomes, genes at these loci are expected to
assort independently.
Answer: E
Section: 7.1
Bloom's Taxonomy: Remembering/Understanding
3) A ________ is all the genes on a single chromosome.
A) linkage group
B) recombination group
C) chromosomal conglomeration
D) genetic allele formation
E) gene loci line
Answer: A
Section: 7.1
Bloom's Taxonomy: Remembering/Understanding
4) If complete linkage occurs, we expect ________.
A) a 9:3:3:1 ratio in the F2
B) a 1:1:1:1 ratio in the test cross
C) to see only recombinant phenotypes in the F2 of the appropriate test cross
D) to see only parental phenotypes in the F2 of the appropriate test cross
E) to see all four types of offspring
Answer: D
Section: 7.1
Bloom's Taxonomy: Remembering/Understanding
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5) 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 which of the following?
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
6) Which of the following scenarios would erroneously lead you to believe two genes reside on
different chromosomes when they in fact are on the same chromosome?
A) a mating between consanguineous individuals
B) too large a sample size
C) the genes are very close together on the chromosome, such that there is never a crossover
between them
D) the genes are very far apart on the same chromosome, such that there is always a crossover
between them
E) the genes are very far apart on the same chromosome, such that there is never a crossover
between them
Answer: D
Section: 7.1
Bloom's Taxonomy: Applying/Analyzing
7) 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 of the following?
A) sex-linked inheritance with 30% crossing over
B) linkage with 50% crossing over
C) linkage with approximately 33 map units between the two gene loci
D) independent assortment
E) 100% recombination
Answer: C
Section: 7.2
Bloom's Taxonomy: Applying/Analyzing
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8) Who was the student in Morgan's laboratory who first realized that the sequence of genes
could be determined by Morgan's proposal that two genes located relatively close to each other
are less likely to form chiasma between them than if the two genes were relatively far apart?
A) Gregor Mendel
B) Barbara McClintock
C) A. H. Sturtevant
D) Thomas Hunt
E) Charles Darwin
Answer: C
Section: 7.2
Bloom's Taxonomy: Remembering/Understanding
9) At what stage of the meiotic cell cycle and during what chromosomal configuration does
crossing over occur?
A) at the four-strand stage of meiosis, after synapsis of homologous chromosomes, and before
the end of prophase I
B) in S phase of meiosis
C) during bivalent formation, after synapsis of homologous chromosomes, and before the end of
prophase I
D) during synapsis at the four-strand stage of meiosis
E) during synapsis in prometaphase
Answer: A
Section: 7.2
Bloom's Taxonomy: Remembering/Understanding
10) Below is a diagram during crossing over. Describe the outcome.
A) a 1:1 ratio
B) segregation of alleles (A and a) during meiosis II
C) segregation of alleles (A and a) during meiosis I
D) formation of parental alleles only
E) offspring will not look like either parent
Answer: B
Section: 7.2
Bloom's Taxonomy: Evaluating/Creating
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11) What is the relationship between the degree of crossing over and the distance between two
genes?
A) It is direct: as the distance increases, the frequency of recombination increases.
B) It is direct: as the distance decreases, the frequency of recombination increases.
C) It is indirect: as the distance increases, the frequency of recombination doubles.
D) There is no correlation.
E) It is indirect: as the distance decreases, the frequency of recombination doubles.
Answer: A
Section: 7.2
Bloom's Taxonomy: Remembering/Understanding
12) 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 mentioned
above?
A) 6
B) 12
C) 24
D) 48
E) 96
Answer: D
Section: 7.2
Bloom's Taxonomy: Evaluating/Creating
13) 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 ________.
A) the G and E loci show complete linkage
B) the G and E loci assort independently
C) the G and E loci are linked
D) the G and E loci reside on the same chromosome over 50 map units apart
E) the G and E loci are segregating independently
Answer: C
Section: 7.2
Bloom's Taxonomy: Evaluating/Creating
14) 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 ________.
A) the recombinant progeny are gE/ge and Ge/ge
B) the recombinant progeny are GE/ge and GE/ge
C) the recombinant progeny are GE/GE and ge/ge
D) the recombinant progeny are gE/GE and GE/ge
E) the recombinant progeny are Ge/Ge and gE/gE
Answer: A
Section: 7.2
Bloom's Taxonomy: Evaluating/Creating
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15) Choose the true statement below.
A) When two genes are over 50 map units apart, a crossover would theoretically occur 0% of the
time.
B) When two genes are over 50 map units apart, a crossover can theoretically occur 100% of the
time.
C) Linkage (viewed from results of typical crosses) always occurs when two loci are on the same
chromosome.
D) The percentage of tetrads involved in exchange between two genes is equal to the number of
recombinant progeny seen.
E) A single crossover on a chromosome will always yield the recombinant phenotype of the
genes you are interested in.
Answer: B
Section: 7.2
Bloom's Taxonomy: Applying/Analyzing
16) 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 there are ________ map units between the G
and E loci.
A) 5
B) 10
C) 15
D) 20
E) 25
Answer: D
Section: 7.3
Bloom's Taxonomy: Applying/Analyzing
17) The genes for purple eyes and curved wings are approximately 21 map units apart on
chromosome II in Drosophila. Assume that a purple-eyed female was mated to a curved wing
male and that the resulting F1 phenotypically wild-type females were mated to purple, curved
males. Of 1000 offspring, what would be the expected number of flies with purple eyes and
curved wings?
A) 395
B) 210
C) 105
D) 790
E) 540
Answer: A
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
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18) 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
light-straw males:
Phenotype
light-straw
wild-type
light
straw
Number
22
18
970
970
Total: 2000
Compute the map distance between the light and straw loci.
A) 2 map units
B) 98 map units
C) 4 map units
D) 12 map units
E) 8 map units
Answer: A
Section: 7.3
Bloom's Taxonomy: Applying/Analyzing
19) Assume that the genes for 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 mated to tan-bodied, barewinged males. Of 1000 offspring, what would be the expected of wild-type offspring, and in
what numbers would they be expected?
A) 8
B) 50
C) 75
D) 425
E) 350
Answer: D
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
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20) 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
Give the gene order of these three loci.
A) smooth, Lobe, straw
B) smooth, straw, Lobe
C) Lobe, smooth, straw
D) Lobe, straw, smooth
E) it is impossible to tell
Answer: A
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
21) 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
What is the coefficient of coincidence?
A) zero
B) 0.5
C) 0.25
D) 0.33
E) 0.75
Answer: A
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
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22) 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
Give the map units between Lobe and straw.
A) 5
B) 10
C) 7.5
D) 15
E) 20
Answer: D
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
23) In Drosophila, assume that the gene for scute bristles (s) is located at map position 0.0 and
that the gene for ruby eyes (r) is at position 15.0. Both genes are located on the X chromosome
and are recessive to their wild-type alleles. A cross is made between scute-bristled females and
ruby-eyed males. Phenotypically wild-type F1 females were then mated to homozygous double
mutant males, and 1000 offspring were produced. Give the frequency of scute expected.
A) 75
B) 425
C) 250
D) 360
E) 500
Answer: B
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
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24) 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 scenario described?
A) It is impossible to tell.
B) In the AaBb parent, each homolog has one dominant allele and one recessive allele.
C) In the AaBb parent, the dominant alleles are on one homolog and the recessive alleles are on
the other.
D) The genes are unlinked.
E) There is a very high coefficient of coincidence.
Answer: C
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
25) 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:
Phenotypes
spineless
wild
claret, spineless
claret
claret, hairless
hairless, claret, spineless
hairless
hairless, spineless
Number Observed
321
38
130
18
309
32
140
12
What is the gene order?
A) it is impossible to tell
B) claret, hairless, spineless
C) hairless, spineless, claret
D) spineless, claret, hairless
E) claret, spineless, hairless
Answer: B
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
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26) 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:
Phenotypes
spineless
wild
claret, spineless
claret
claret, hairless
hairless, claret, spineless
hairless
hairless, spineless
Number Observed
321
38
130
18
309
32
140
12
What are the map distances between hairless and spineless?
A) 10
B) 30
C) 40
D) 45
E) 50
Answer: A
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
27) Three loci, mitochondrial malate dehydrogenase that forms a and b (MDHa, MDHb),
glucuronidase that forms 1 and 2 (GUS1, GUS2), and a histone gene that forms + and − (H+,
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 expected numbers of the MDHa GUS2 H− cells she would produce.
Assume no chromosomal interference.
A) 0
B) 15
C) 235
D) 135
E) 315
Answer: D
Section: 7.3
Bloom's Taxonomy: Evaluating/Creating
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28) In a three-point mapping experiment, which class do you expect to find the least?
A) noncrossover
B) parental
C) double crossover
D) recombinant
E) single crossover
Answer: C
Section: 7.3
Bloom's Taxonomy: Remembering/Understanding
29) In a three-point mapping experiment, how many different genotypic classes are expected?
A) 0
B) 1
C) 2
D) 4
E) 8
Answer: E
Section: 7.3
Bloom's Taxonomy: Remembering/Understanding
30) 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
31) 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) 5
B) 30
C) 50
D) 80
E) 100
Answer: C
Section: 7.4
Bloom's Taxonomy: Remembering/Understanding
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32) Which of the following is an example of positive interference?
A) A single crossover stimulates further recombination.
B) A crossover in one region inhibits crossovers in nearby regions.
C) Double crossovers do not occur.
D) Crossovers will not occur near centromeres.
E) An actively transcribed gene is less likely to undergo crossover than a quiescent one.
Answer: B
Section: 7.4
Bloom's Taxonomy: Remembering/Understanding
33) Which of the following best describes why mapping is most accurate when genes are close
together on a chromosome?
A) This is not true; mapping genes is most accurate when genes are far away from each other.
B) This is not true; relative distance of two loci on a chromosome has no effect on accuracy of
mapping.
C) The centromere gets in the way.
D) Double crossover events yield a result that looks the same as no crossover in a two gene
mapping experiment and this throws off the calculations.
E) Double crossover events yield a result that looks the same as one crossover in a two gene
mapping experiment and this throws off the calculations.
Answer: D
Section: 7.4
Bloom's Taxonomy: Applying/Analyzing
34) The coefficient of coincidence reflects the frequency of observed double crossovers
compared to the frequency of expected double crossovers. What is the relationship between the
coefficient of coincidence and interference?
A) Interference is recombinants minus nonrecombinants.
B) Interference is one minus the coefficient of coincidence.
C) Interference equals the coefficient of coincidence.
D) Interference equals the coefficient of coincidence divided by the map units calculated.
E) Interference is the inverse square of the coefficient of coincidence.
Answer: B
Section: 7.4
Bloom's Taxonomy: Remembering/Understanding
35) If interference is complete, what would be the frequency of double crossovers?
A) 0
B) 1
C) 2
D) 0.25
E) 0.75
Answer: A
Section: 7.4
Bloom's Taxonomy: Applying/Analyzing
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36) ________ occurs when a crossover in one region of a chromosome reduces crossovers in
nearby regions.
A) Negative interference
B) Positive interference
C) Triple crossover
D) Coefficient of coincidence
E) Negative supercoiling
Answer: B
Section: 7.4
Bloom's Taxonomy: Remembering/Understanding
37) What two terms apply to the fusion of cultured human and mouse cells that produces cell
lines that are useful in assigning a gene to a particular human chromosome?
A) forward and reverse
B) lod score and pod score
C) heterokaryon and synkaryon
D) single and double
E) positive and negative
Answer: C
Section: 7.6
Bloom's Taxonomy: Remembering/Understanding
38) What is a lod score?
A) determination of the probability that two traits are linked by analyzing pedigrees
B) a method using cell fusion to determine on which chromosome a gene resides
C) determination of how often double crossovers occur
D) a calculation to show how much interference is occurring in crossover formation
E) determination of how often a gene mapping experiment gives the correct results
Answer: A
Section: 7.6
Bloom's Taxonomy: Remembering/Understanding
39) Methods for determining the linkage group and genetic map in humans involve which of the
following?
A) DNA markers
B) twin spots and tetrad analysis
C) tetrad analysis and bromodeoxyuridine
D) zygotene and pachytene DNA syntheses
E) chiasma-type and classical analyses
Answer: A
Section: 7.7
Bloom's Taxonomy: Remembering/Understanding
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40) What relatively recent scientific advancement has made mapping by linkage or classical
genetic mapping approaches virtually obsolete?
A) the genome sequence of a species
B) the inclusion of the X and Y chromosomes in SNP experiments
C) the use of synteny
D) positive interference
E) negative interference
Answer: A
Section: 7.7
Bloom's Taxonomy: Remembering/Understanding
41) What is a physical map of a chromosome?
A) description of the base pair difference between two loci
B) determination of where the centromere lies
C) determination of how far apart genes are my recombination mapping
D) identification of the length of the chromosome by comparing it to known chromosomes
E) determination of the p and q arm length
Answer: A
Section: 7.7
Bloom's Taxonomy: Remembering/Understanding
42) Based on the classic experiments of Creighton and McClintock with maize, crossing over
involves a physical exchange between chromatids. What particular chromosomal characteristic
allowed their experiments to succeed?
A) colored bands along the lengths of a chromosome
B) an extra chromosome
C) a missing chromosome
D) three chromosomes with identical structure
E) a chromosome with a unique cytological marker
Answer: E
Section: 7.8
Bloom's Taxonomy: Remembering/Understanding
43) Sister chromatid exchanges increase in frequency in the presence of X-rays, certain viruses,
ultraviolet light, and certain chemical mutagens. In what autosomal recessive disorder is there
known to be an increase in sister chromatid exchanges?
A) cystic fibrosis
B) Huntington's disease
C) achondroplasia
D) Bloom syndrome
E) Harlequin syndrome
Answer: E
Section: 7.9
Bloom's Taxonomy: Applying/Analyzing
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44) Which of the following statements is true?
A) Harlequin chromosomes are a result of homologous recombination.
B) Crossing over has a cytological basis.
C) Exchanges occur between homologous chromosomes, but never between sister chromatids.
D) Sister chromatid exchange would lead to a reshuffling of alleles.
E) Sister chromatid exchange would mutate one strand of DNA.
Answer: B
Section: 7.8, 7.9
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 8 Genetic Analysis and Mapping in Bacteria and Bacteriophages
1) Max Delbrück presented the first evidence that bacteria are capable of spontaneous mutation.
What is the name of the test that Delbrück used to demonstrate this phenomenon?
A) fluctuation test
B) logarithmic test
C) lag test
D) prototrophic test
E) auxotrophic test
Answer: A
Section: 8.1
Bloom's Taxonomy: Remembering/Understanding
2) ________ are bacteria that can grow on minimal medium and are assumed to be wild type.
A) Auxotrophs
B) Prototrophs
C) Mutants
D) F+ cells
E) Hfr
Answer: B
Section: 8.1
Bloom's Taxonomy: Remembering/Understanding
3) ________ have lost, through ________, the ability to grow on minimal medium.
A) prototrophs; mutation
B) auxotrophs; adaptation
C) auxotrophs; mutation
D) prototrophs; adaptation
E) mutants; adaptation
Answer: C
Section: 8.1
Bloom's Taxonomy: Remembering/Understanding
4) What is the adaptation hypothesis?
A) The phage stimulates the bacteria to mutate to become immune to the phage.
B) The phage mutates to infect bacteria more effectively.
C) The phage and bacteria both mutate so the phage becomes less virulent.
D) The bacteria will mutate so it can become an auxotroph.
E) The phage mutates so it can't infect bacteria any longer.
Answer: A
Section: 8.1
Bloom's Taxonomy: Remembering/Understanding
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5) In which phase of the bacterial cell cycle do the cells grow most quickly?
A) lag
B) log
C) stationary
D) adaptation
E) fluctuation
Answer: B
Section: 8.1
Bloom's Taxonomy: Remembering/Understanding
6) Analyze the following growth of a bacterial colony.
Medium
Complete with glucose
Complete with lactose (no glucose)
Minimal
Minimal media plus biotin, threonine, leucine, methionine, and
lactose
Minimal media plus biotin, leucine, and glucose
Minimal media plus threonine, leucine, and lactose
Minimal media plus leucine, methionine, and glucose
Growth
Yes
Yes
No
Yes
No
No
Yes
What is the genotype of the bacterium?
A) bio+, thr+, lac−
B) leu+, met+, lac+
C) leu+, met+, lac−
D) bio+, thr+, leu+, met+
E) bio+, leu+, lac−
Answer: B
Section: 8.1
Bloom's Taxonomy: Evaluating/Creating
7) Which of the following is true about the interrupted mating technique?
A) Time of contact is not applicable to this protocol.
B) Donor will pass only beneficial genes to the recipient.
C) Genes will be passed from the centromere.
D) Genes are transferred from the donor to recipient in a linear fashion based upon time of
contact.
E) The recipient bacterium will choose the best genes.
Answer: D
Section: 8.2
Bloom's Taxonomy: Applying/Analyzing
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8) 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?
A) adaptation
B) conjugation
C) transformation
D) transduction
E) time of transfer
Answer: B
Section: 8.2
Bloom's Taxonomy: Applying/Analyzing
9) Assume that the gene trpA in an auxotrophic strain of E. coli is located at 27 minutes, whereas
the gene pyrE is located at 81 minutes.
Present an experimental scheme that would allow you to convert one of the auxotrophic strains
to a prototrophic strain.
A) F+ (wild type) × F− (auxotroph)
B) F+ (auxotroph) × F− (wild type)
C) Hfr (auxotroph) × F− (auxotroph)
D) Hfr (auxotroph) × F− (wild type)
E) Hfr (wild type) × F− (auxotroph)
Answer: E
Section: 8.2
Bloom's Taxonomy: Evaluating/Creating
10) ________ bacteria contain an F factor or plasmid that is capable of initiating conjugation.
________ bacteria contain an F plasmid that possesses a portion of the bacterial chromosome.
A) Merozygotes; F−
B) F+; Hfr
C) F+; Merozygotes
D) F−; F+
E) F−; Hfr
Answer: C
Section: 8.2
Bloom's Taxonomy: Remembering/Understanding
11) Describing bacterial conjugation, which of the following matings is most likely to result in a
change from F- to F+?
A) donor F− recipient F+
B) donor F− recipient Hfr
C) donor Hfr recipient FD) donor Hfr recipient F+
E) donor F+ recipient F−
Answer: E
Section: 8.2
Bloom's Taxonomy: Applying/Analyzing
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12) What is a form of recombination in bacteria that involves the F plasmid?
A) conjugation
B) transduction
C) transformation
D) adaptation
E) fluctuation
Answer: A
Section: 8.2
Bloom's Taxonomy: Remembering/Understanding
13) How do different strains of E. coli can reveal different linkage arrangements of genes in Hfr
crosses?
A) They do not. All Hfr strains are identical.
B) Hfr strains will demonstrate different areas of recombination, due to dominance and
recessiveness of genes.
C) Different strains may have different F factors and therefore different initiation points for
chromosome transfer.
D) In different Hfr strains, genes can be in different order on the chromosome.
E) In different Hfr strains, different genes will be transferred based on where the centromere
resides.
Answer: C
Section: 8.2
Bloom's Taxonomy: Applying/Analyzing
14) Which of the following is true about a merozygote?
A) It is formed from an Hfr to F+ cross.
B) It can spontaneously form from an Hfr cell.
C) A merozygote can form if an F factor carries chromosomal DNA and is injected into an F−
cell.
D) A conjugation with an Hfr donor will never produce a merozygote.
E) Merozygotes were crucial to our understanding of the formation of the sex pilus.
Answer: C
Section: 8.2
Bloom's Taxonomy: Applying/Analyzing
15) In a bacterial cross in which the donor (Hfr) is a+b+ and the recipient strain (F−) is a−b−, it
is expected that recombinant bacteria ________.
A) will all be a+b+
B) will all remain a−b−
C) will be a mixture of a+b+, a−b−, and a−b+
D) will be a mixture of a+b+ and a−b−
E) will be a mixture of a+b+, a−b−, a+b−, and a−b+
Answer: C
Section: 8.2
Bloom's Taxonomy: Evaluating/Creating
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16) ________ can initiate chromosome transfer from one E. coli to another.
A) An F+ cell
B) An F− cell
C) A temperate phage
D) A plasmid
E) An Hfr cell
Answer: E
Section: 8.2
Bloom's Taxonomy: Remembering/Understanding
17) The interrupted mating technique was used with a series of Hfr strains and the following
results were obtained.
1) LQRYX
2) QLPZ
3) BTZP
4) LPZTB
5) PLQRY
What is the order of the genes?
A) BTPZLQRYX
B) XYRPZTBQL
C) PZTBXYRQL
D) XYRQLPZTB
E) RLQZTBXYP
Answer: D
Section: 8.2
Bloom's Taxonomy: Evaluating/Creating
18) Once an exogenous (foreign) piece of DNA is in the bacterial cell, it is incorporated into the
bacterial genome through ________.
A) DNA replication
B) crossing-over (recombination)
C) transformation
D) transduction
E) conjugation
Answer: B
Section: 8.2
Bloom's Taxonomy: Remembering/Understanding
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19) 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
20) What is an R plasmid?
A) They function in bacterial recombination.
B) They are the plasmids that carry the F factor.
C) They are a plasmid that can exist in the chromosome or outside the chromosome.
D) They carry a gene for resistance to an environmental attribute, such as heavy metals, and
genes for the transfer of the plasmid.
E) It is an F factor that has picked up part of a bacterial genome.
Answer: D
Section: 8.3
Bloom's Taxonomy: Remembering/Understanding
21) ________ can exist integrated into the bacterial chromosome or as an autonomous plasmid
unit in the cytosol.
A) Episomes
B) F′ plasmid
C) R factor
D) Temperate phages
E) Col plasmid
Answer: A
Section: 8.3
Bloom's Taxonomy: Remembering/Understanding
22) What are the general structural features of a prokaryotic plasmid?
A) linear double-stranded DNA integrated into the chromosome
B) made of protein and existing in the cytosol
C) circular double-stranded DNA in the cytosol
D) proteins on the bacterial chromosome
E) circular double-stranded DNA in the nucleus
Answer: C
Section: 8.3
Bloom's Taxonomy: Remembering/Understanding
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23) Bacteria that are in a particular physiological state to become transformed are called
________.
A) episomal
B) resistant
C) temperate
D) competent
E) transformable
Answer: D
Section: 8.4
Bloom's Taxonomy: Remembering/Understanding
24) After transformation, heteroduplex formation, and resolution, the resulting daughter cells are
________.
A) both mutant
B) one transformed and one nontransformed
C) both transformed
D) both wild type
E) both have plasmids
Answer: C
Section: 8.4
Bloom's Taxonomy: Applying/Analyzing
25) What is meant by the term cotransformation?
A) Several pieces of DNA can form heteroduplexes in a cell.
B) A cell can pick up more than one F factor.
C) More than one phage infects a bacterial cell.
D) An F− cell can become F+ several times in a life cycle.
E) Cotransformation occurs when several linked genes are transformed simultaneously.
Answer: E
Section: 8.4
Bloom's Taxonomy: Remembering/Understanding
26) Bacteriophages engage in two interactive cycles with bacteria. What are these cycles?
A) lytic and lysogenic
B) insertion and replication
C) auxotrophic and prototrophic
D) heteroduplex and homoduplex
E) negative and positive
Answer: A
Section: 8.5
Bloom's Taxonomy: Remembering/Understanding
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27) A bacteriophage that is capable of entering either a lytic or a 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
28) The clearing made by bacteriophages in a "lawn" of bacteria on an agar plate is called a
________.
A) clear zone
B) lysogenic zone
C) prophage
D) plaque
E) host range
Answer: D
Section: 8.5
Bloom's Taxonomy: Remembering/Understanding
29) Temperate phages are those that can enter either the ________ or the ________ cycle.
A) lytic; lysogenic
B) virulent; avirulent
C) functional; nonfunctional
D) former; nonformer
E) complementing; competing
Answer: A
Section: 8.5
Bloom's Taxonomy: Remembering/Understanding
30) What is the process by which a temperate bacteriophage infects a bacterial cell and
subsequently integrates its chromosome into the bacterial chromosome?
A) lysis
B) lysogeny
C) temperate
D) transformation
E) transduction
Answer: B
Section: 8.5
Bloom's Taxonomy: Remembering/Understanding
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31) Which of the following is a possible outcome of the lysogenic cycle?
A) bacterial cell lysis
B) prophage formation
C) bacterial fission
D) prophage formation and propagation
E) bacterial cell lysis, prophage formation and propagation, and bacterial fission are all possible
outcomes of the lysogenic cycle
Answer: E
Section: 8.5
Bloom's Taxonomy: Applying/Analyzing
32) Which of the following occurs during cell lysis?
A) prophage formation
B) binary fission
C) lysogeny
D) plaque formation
E) episome propagation
Answer: D
Section: 8.5
Bloom's Taxonomy: Remembering/Understanding
33) Which of the following is an episome?
A) F factor
B) temperate phage
C) lytic phage
D) temperate and lytic phages
E) F factor and temperate phage
Answer: E
Section: 8.5
Bloom's Taxonomy: Applying/Analyzing
34) What is the first step in phage formation inside the bacterial cell?
A) adherence to the pilus
B) tail formation
C) DNA packaging as the viral heads assemble
D) lysis
E) tail fibers are added
Answer: C
Section: 8.5
Bloom's Taxonomy: Remembering/Understanding
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35) When a bacteriophage genome incorporates itself into the chromosome of the host, that
phage genome is referred to as a(n) ________.
A) lytic phage
B) T4 phage
C) temperate phage
D) prophage
E) episome
Answer: D
Section: 8.5
Bloom's Taxonomy: Remembering/Understanding
36) 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?
A) 6.7 × 107 pfu/ml (pfu = plaque-forming units)
B) 67 pfu/ml
C) 6.7 × 105 pfu/ml
D) 6.7 × 10−5 pfu/ml
E) 0.1 × 105 pfu/ml
Answer: A
Section: 8.5
Bloom's Taxonomy: Evaluating/Creating
37) What is a bacteriophage?
A) a virus that exclusively undergoes the lytic cycle
B) a virus that exclusively undergoes the lysogenic cycle
C) a bacterium that infects other bacteria
D) a virus that has a bacterium as its host
E) a bacterium that infects a virus
Answer: D
Section: 8.5
Bloom's Taxonomy: Remembering/Understanding
38) What is a plaque composed of?
A) lysed bacteria
B) live bacteria and bacteriophage
C) bacteriophage and dead bacteria
D) bacteriophage and agar
E) lysed bacteria, phage, and agar
Answer: E
Section: 8.5
Bloom's Taxonomy: Applying/Analyzing
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39) Which of the following statements is false?
A) A symbiotic relationship between a phage and a bacterium apparently occurs in the process of
lysogeny.
B) Lysogeny is a process that occurs during transformation and conjugation.
C) A prophage can exist in a quiescent state in the bacterial chromosome.
D) A bacteriophage that can only undergo lysis is called a virulent phage.
E) Viral genomes comprise only genes for viral structure and assembly.
Answer: B
Section: 8.5
Bloom's Taxonomy: Remembering/Understanding
40) Transduction is a form of recombination in bacteria that involves ________.
A) 5-bromouracil
B) plasmids
C) bacteriophages
D) fertility factors
E) physical contact between the bacteria involved
Answer: C
Section: 8.6
Bloom's Taxonomy: Remembering/Understanding
41) Name two forms of recombination in bacteria.
A) lytic and lysogenic
B) auxotrophic and prototrophic
C) conjugation and transduction
D) mixed and generalized
E) insertion and replication
Answer: C
Section: 8.6
Bloom's Taxonomy: Remembering/Understanding
42) If two different auxotrophic strains are placed in a liquid medium culture tube, prototrophic
strains can sometimes be subsequently recovered. Choose the mechanism below that would not
allow this genetic change.
A) reverse mutation
B) genetic suppression
C) conjugation
D) transformation
E) transduction
Answer: E
Section: 8.6
Bloom's Taxonomy: Evaluating/Creating
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43) A form of bacterial recombination that involves a viral intermediate is called ________.
A) transduction
B) conjugation
C) transformation
D) plaque assay
E) competency
Answer: A
Section: 8.6
Bloom's Taxonomy: Remembering/Understanding
44) ________ is most likely associated with transduction.
A) Transformation
B) Conjugation
C) Lysogeny
D) factor
E) Plasmid ColE
Answer: C
Section: 8.6
Bloom's Taxonomy: Remembering/Understanding
45) ________ of genes is an indication that the genes are linked.
A) Reversion
B) Recombination
C) Transformation
D) Cotransduction
E) Conjugation
Answer: D
Section: 8.6
Bloom's Taxonomy: Remembering/Understanding
46) Viral mutations and variants are often categorized by changes in ________.
A) viral size
B) capsid shape
C) host range
D) plaque morphology
E) host range and/or plaque morphology
Answer: E
Section: 8.7
Bloom's Taxonomy: Remembering/Understanding
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47) In a mixed infection experiment, you study two plaque phenotypes of large (l+) versus small
(l−) and bumpy (b+) versus smooth (b−). You mix small (l− b+) with smooth (l+ b−) and
identify 50 small, smooth and 47 large, bumpy. How far apart are the genes?
A) 49
B) 99
C) 490
D) 1000
E) There is not enough information to calculate the distance.
Answer: A
Section: 8.7
Bloom's Taxonomy: Evaluating/Creating
48) In a mixed infection experiment, the results of an h+r and hr+ cross are as follows.
Genotype
h+r
hr+
h+r+
hr
Number of plaques
42
34
12
12
How far apart are these two genes?
A) 12
B) 24
C) 48
D) 2
E) 8
Answer: A
Section: 8.7
Bloom's Taxonomy: Evaluating/Creating
49) To produce recombinants in phage ________.
A) one crossover is required
B) two crossovers are required
C) phage recombinants do not occur
D) a bacterial cell must be competent
E) the F factor must be present
Answer: C
Section: 8.8
Bloom's Taxonomy: Applying/Analyzing
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50) In Seymour Benzer's studies, if he mixed two RII phage together (co-infection studies in
K12) and the resulting phage always failed to lyse K12 ________.
A) the two mutants were in the same cistron
B) the two mutants complemented each other
C) the two mutants were in different cistrons
D) the two mutants failed to infect bacteria
E) the two mutants represented mutations in both A and B
Answer: A
Section: 8.8
Bloom's Taxonomy: Applying/Analyzing
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 9 Extranuclear Inheritance
1) What is the name of a form of extrachromosomal inheritance that is due to transmission of
symbiotic or parasitic microorganisms?
A) organellar inheritance
B) microbial genomic transfer
C) infectious heredity
D) pathogenic paternal transfer
E) uniparental genome transfer
Answer: C
Section: Introduction
Bloom's Taxonomy: Remembering/Understanding
2) Forms of inheritance that do not follow typical Mendelian patterns and that appear to be more
influenced by the parent contributing the most cytoplasm to the embryo are grouped under the
general heading of ________.
A) sex-linked inheritance
B) neo-Mendelian inheritance
C) extrachromosomal inheritance
D) suppressive inheritance
E) dominance and/or recessiveness
Answer: C
Section: 9.1
Bloom's Taxonomy: Remembering/Understanding
3) Which of the following organelles are involved in the general category of organelle heredity?
A) mitochondria and chloroplasts
B) Golgi apparatus and nuclei
C) lysosomes and peroxisomes
D) factors and episomes
E) Golgi apparatus and rough endoplasmic reticulum
Answer: A
Section: 9.1
Bloom's Taxonomy: Remembering/Understanding
4) To which of the following does the term heteroplasmy refer?
A) cells with a variable mixture of normal and abnormal organellar genomes
B) heterozygous individuals with more than one gene pair involved
C) conditions in which the germ plasm is a mixture of dominant and recessive genes
D) a circumstance that is homologous to incomplete dominance
E) various stages of development of mitochondria and chloroplasts
Answer: A
Section: 9.1
Bloom's Taxonomy: Remembering/Understanding
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5) The plant coleus has three distinct leaf color phenotypes: green, white, and variegated (white
and green mixed) the crosses are made as follows:
female variegated leaves × male green leaves gives offspring with all variegated leaves
female green leaves × male with white leaves gives offspring with all green leaves
female with white leaves × male with variegated leaves gives offspring with all white leaves
This is an example of ________.
A) cytoplasmic inheritance
B) genomic imprinting
C) X-linked recessive disorder
D) Mendelian inheritance
E) maternal effect
Answer: A
Section: 9.1
Bloom's Taxonomy: Evaluating/Creating
6) ________ phenotypes are passed equally to offspring by both males and females.
A) Mitochondrial
B) Autosomal dominant
C) Autosomal recessive
D) X-linked recessive
E) X-linked dominant
Answer: B
Section: 9.1
Bloom's Taxonomy: Applying/Analyzing
7) Which of the following matings indicates that a Chlamydomonas mutant with photosynthetic
defects (that resulted from a deficiency in chloroplast translation) is located on the chloroplast
genome?
A) mt+ photo− × mt− photo+ yields all str−
B) mt+ photo+ × mt− photo− yields all str−
C) a petite × alpha wild type yields all petite
D) mt+ photo− × mt− photo+ yields mt+ photo+ and mt− photo−
E) mt+ photo− × mt− photo+ yields mt+ photo− and mt− photo+
Answer: A
Section: 9.1
Bloom's Taxonomy: Evaluating/Creating
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8) Why is variegation in four o'clock plants determined by the phenotype of the ovule source?
A) Dominant genes demonstrating variegation come from the ovule.
B) Recessive genes demonstrating variegation come from the ovule.
C) The chloroplasts determine the leaf coloration, and the chloroplasts are inherited through the
ovule.
D) The ovule nuclear genome contains the genes for leaf coloration.
E) The mitochondria determine the leaf coloration, and the mitochondria are inherited through
the ovule.
Answer: C
Section: 9.2
Bloom's Taxonomy: Remembering/Understanding
9) In what way do segregational petite mutations differ from neutral petites?
A) Neutral petites follow Mendelian inheritance as a result of the influence of nuclear genes;
segregational petites, when crossed to wild type, produce only wild-type, normal-sized colonies.
B) When a neutral petite is crossed to a segregational petite, all offspring are neutral petites.
C) Neutral petites are dominant and segregational petites are recessive mitochondrial genes.
D) Segregational petites follow Mendelian inheritance as a result of the influence of nuclear
genes; neutral petites, when crossed to wild type, produce only wild-type, normal-sized colonies.
E) When a segregational petite is crossed to a wild-type cell, all offspring are petite. However,
when a neutral petite is crossed to a wild-type cell, all offspring are wild type.
Answer: D
Section: 9.1
Bloom's Taxonomy: Remembering/Understanding
10) What is the expression pattern of dominant-negative mutations in petite strains of yeast?
A) Expression of wild-type mitochondria is enhanced.
B) Expression of mutant mitochondria resembles expression of wild-type mitochondria.
C) Mitochondria show enhanced capacity of oxidative phosphorylation.
D) The function of wild-type mitochondria is suppressed.
E) Mitochondrial membranes become hyperpolarized.
Answer: D
Section: 9.1
Bloom's Taxonomy: Applying/Analyzing
11) In Neurospora crassa, a mutation called poky (phenotype is slow growth) is passed down
cytoplasmically. Which cross below is not consistent this type of inheritance?
A) female poky × male poky yields all poky
B) female wild type × male wild type yields half poky
C) female wild type × male wild type yields all wild type
D) female wild type × male poky yields all wild type
E) female poky × male wild type yields all poky
Answer: B
Section: 9.1
Bloom's Taxonomy: Applying/Analyzing
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12) The inheritance of the petite phenotype in Saccharomyces cerevisiae is complicated by an
interaction of mitochondrial and nuclear genes. Which of the following is a false statement about
petites in S. cerevisiae?
A) Both nuclear and cytoplasmic (mitochondrial) genes contribute to the petite phenotype in
some cases.
B) The three categories of petites are segregational, neutral, and suppressive.
C) Neutral petites, when crossed to wild type, yield wild-type mitochondrial function.
D) Segregational petites are characterized by mutations in the nuclear genome.
E) Suppressive petites are characterized by mitochondria lacking most of its DNA.
Answer: E
Section: 9.1
Bloom's Taxonomy: Remembering/Understanding
13) You discover a new alien race that you decide to call the PAMMARIANS. They are
different from us in that they have three sexes, called Q, R, and S. All three sexes must fuse their
gametes to form a new PAMMARIAN (and you thought dating on the Earth was hard!). They
have nuclear DNA and extrachromosomal organelles, similar to mitochondria, called
MITOCHLORIANS, which have their own DNA. The R gender is like human females (they
donate the cytoplasm during the mating). Q and S do NOT donate cytoplasm during mating, like
human males. Which of the following is TRUE?
A) In a mating, if only the R gender individual has a mitochlorian DNA mutation that causes a
phenotype, the offspring will have NO mitochlorian mutant phenotype.
B) In a mating, if only the Q gender individual has a mitochlorian DNA mutation that causes a
phenotype, the offspring will have NO mitochlorian mutant phenotype.
C) In a mating, if only the S gender individual has a mitochlorian DNA mutation that causes a
phenotype, the offspring WILL HAVE a mitochlorian mutant phenotype.
D) In a mating, if only the Q gender individual has a mitochlorian DNA mutation that causes a
phenotype, the offspring WILL HAVE a mitochlorian mutant phenotype.
E) In a mating, if the S and Q genders have a mitochlorian DNA mutation, the offspring will be
mutant as well.
Answer: B
Section: 9.1
Bloom's Taxonomy: Evaluating/Creating
14) In organelle heredity, which of the following is TRUE?
A) In a reciprocal cross, the offspring will have the phenotype of the mother's genotype.
B) In a reciprocal cross, the offspring will have different phenotypes depending on which
phenotype the father has.
C) In a reciprocal cross, the imprinted gene will always come from the mother.
D) In a reciprocal cross, the offspring will always have the phenotype that the mother (egg) has.
E) In a reciprocal cross, the offspring will have varied phenotypes depending on which X
chromosome is converted into a Barr body.
Answer: D
Section: 9.1
Bloom's Taxonomy: Applying/Analyzing
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15) Inheritance of the green and white patches phenotype in Mirabilis jalapa (four o'clock) is an
example of ________.
A) maternal influence
B) organelle heredity
C) infectious heredity
D) maternal effect
E) reciprocal cytoplasmic inheritance
Answer: B
Section: 9.1
Bloom's Taxonomy: Remembering/Understanding
16) An example of ________ involves pigmentation in Mirabilis jalapa.
A) nucleolar inheritance
B) gene dosage effect
C) organelle heredity
D) maternal effect
E) X-linked recessive
Answer: C
Section: 9.1
Bloom's Taxonomy: Remembering/Understanding
17) In aerobically cultured yeast, you isolate a petite mutant. You believe it is a suppressive
mutant. Which cross would you do below to determine that it is indeed suppressive?
A) petite × wild type yields all petites
B) petite × wild type yields half petite and half wild type
C) petite × petite yields all petite
D) petite × petite yields half petite and half wild type
E) wild type × wild type yields petite
Answer: A
Section: 9.1
Bloom's Taxonomy: Evaluating/Creating
18) In what way do mitochondrial mutations influence phenotype?
A) They affect the production of ribosomes and protein synthesis.
B) They cause the nuclear DNA to segregate ineffectively.
C) Cells with mutated mitochondrial DNA cannot undergo anaphase.
D) They affect the production of ATP generated through cellular respiration.
E) Mutated mitochondrial DNA has no effect on phenotypes.
Answer: D
Section: 9.2
Bloom's Taxonomy: Remembering/Understanding
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19) Which three classes of macromolecules are encoded by mtDNA?
A) proteins, transfer RNAs, and ribosomal RNA
B) photosystem I, photosystem II, and tRNA
C) carbohydrates, enzymes, and proteins
D) transfer RNA, ribosomal RNA, and nuclear RNA
E) lipids, carbohydrates, and nucleic acids
Answer: A
Section: 9.2
Bloom's Taxonomy: Remembering/Understanding
20) The endosymbiont theory is support by several lines of evidence one of which is ________.
A) both mitochondria and chloroplasts generate ATP
B) both mitochondria and peroxisomes have circular chromosomes
C) both bacteria and lysosomes have degradative enzymes
D) the chloroplast genome resembles the genome of the mitochondria
E) mitochondria, chloroplast, and bacteria all have circular chromosomes
Answer: E
Section: 9.2
Bloom's Taxonomy: Applying/Analyzing
21) One explanation for organelle inheritance of disease phenotypes is that ________.
A) mitochondria and chloroplasts lack DNA and are therefore dependent on the maternal
cytoplasmic contributions
B) mitochondria and chloroplasts have DNA that is subject to mutation
C) organelles such as mitochondria are always wild type
D) chloroplasts, for example, are completely dependent on the nuclear genome for components
E) None of the answers listed are correct.
Answer: B
Section: 9.2
Bloom's Taxonomy: Applying/Analyzing
22) Approximately one in 5000 humans either have a mitochondrial DNA disorder or are at risk
for developing such a disorder. What future approach involving nuclear transplantation might be
available to treat mtDNA-based human disorders?
A) nuclear disintegration
B) mitochondrial replacement therapy
C) nuclear activation
D) mitochondrial activation
E) mitochondrial suppression
Answer: B
Section: 9.3
Bloom's Taxonomy: Remembering/Understanding
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23) Through the study of mitochondrial disorders, scientists have suggested a link between the
decline of mitochondrial function and aging. What process appears to be involved in this
suspected phenomenon?
A) accumulation of sporadic mutations in the nuclear genome
B) decline of mutations in the nuclear genome
C) accumulation of mutations in the mtDNA
D) recombination of mtDNA and nuclear DNA
E) recombination among mtDNAs of different mitochondria
Answer: C
Section: 9.3
Bloom's Taxonomy: Remembering/Understanding
24) Which criterion below does not indicate that a human disorder may be attributable to
genetically altered mitochondria?
A) maternal rather than Mendelian inheritance pattern
B) deficiency in some bioenergetic functions of mitochondria
C) documentation of a specific mitochondrial genetic mutation
D) deficiency in ability to generate ATP
E) passed down more readily to male offspring
Answer: E
Section: 9.3
Bloom's Taxonomy: Applying/Analyzing
25) Describe the molecular and transmission characteristics of Leber's hereditary optic
neuropathy (LHON).
A) cytoplasmic inheritance, which includes mtDNA lesions
B) cytoplasmic inheritance, which includes cpDNA lesions
C) maternal inheritance, which includes mtDNA lesions
D) X-linked recessive
E) dominant maternal inheritance
Answer: A
Section: 9.3
Bloom's Taxonomy: Remembering/Understanding
26) Name two human disorders that appear to be transmitted extrachromosomally.
A) poky chromosome syndrome and kidney cancer
B) myoclonic epilepsy and ragged red fiber disease and Leber's hereditary optic neuropathy
C) Duchenne muscular dystrophy and Down syndrome
D) cri-du-chat syndrome and Leber's hereditary optic neuropathy
E) reactive oxygen syndrome and myoclonic epilepsy and ragged red fiber disease
Answer: B
Section: 9.3
Bloom's Taxonomy: Remembering/Understanding
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27) Kearns-Sayre syndrome (KSS) results in progressive loss of vision and hearing, as well as
heart defects. What is the genetic basis for this disease?
A) Reactive oxygen species mutate DNA.
B) Too much sunlight damages mtDNA.
C) Deletions occur in mtDNA, and the proportion of defective mtDNAs increases as the severity
of the symptoms increases.
D) ATP builds up and runs the electron transport chain backwards.
E) Ageing shortens the telomeres of the mitochondrial DNA.
Answer: C
Section: 9.3
Bloom's Taxonomy: Remembering/Understanding
28) The genes for light eyes (lt; light) and straw bristles (stw; straw) are tightly linked on
chromosome 2 in Drosophila melanogaster. The Malpighian tubes of lt larvae and adults are
maternally affected in that Malpighian tubes of lt/lt organisms, whose mothers were lt+/lt and
have more yellow pigment than those from lt/lt mothers. There is no maternal effect associated
with the straw locus. Give the phenotypes of the offspring from the following crosses.
Female
Male
Cross #1:
lt stw/lt stw
lt+stw+/lt stw
lt stw+/lt stw
Cross #2:
lt+ stw/lt stw
lt stw/lt stw
Cross #3:
lt+ stw+/lt stw
Answer: Cross #1: 1/2 light Malpighian tubes, straw bristles, 1/2 wild type
Cross #2:
more pigmented Malpighian tubes, 1/2 straw bristles
Cross #3:
more pigmented Malpighian tubes, 1/2 straw bristles
Section: 9.4
Bloom's Taxonomy: Evaluating/Creating
29) Direction of shell coiling in Lymnaea peregra is strongly and most directly influenced by
________.
A) the genotype of the mother
B) the genotype of the father
C) the phenotype of the mother
D) the phenotype of the father
E) the genotype of the embryo itself
Answer: A
Section: 9.4
Bloom's Taxonomy: Remembering/Understanding
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30) Direction of shell coiling in the snail Lymnaea peregra is conditioned by a form of
extrachromosomal inheritance known as ________.
A) sex-linked inheritance
B) heteroplasmy
C) maternal effect
D) independent assortment
E) epistasis
Answer: C
Section: 9.4
Bloom's Taxonomy: Remembering/Understanding
31) The maternal effect in Lymnaea is such that the genotype of the egg determines the direction
of shell coiling regardless of the genotype of the offspring. Apparently, the cause of this
spectacular maternal effect results from ________.
A) orientation of the spindle apparatus in early cleavage
B) genophores present in the egg cytoplasm
C) the F factor exerting its influence on the centrosome
D) colicins "poisoning" one of the cleavage centers
E) allelic substitution as demonstrated from RNA injection experiments
Answer: A
Section: 9.4
Bloom's Taxonomy: Remembering/Understanding
32) Molecular/structural orientations (gradients) in an egg are thought to play a significant role in
development. What is the origin of such gradients?
A) Where the sperm penetrated the egg is the area of highest concentration.
B) During formation of the egg, nutritional as well as informational molecules (RNAs) are
placed in appropriate positions for the development of the embryo.
C) During early development, the nucleus is placed into position of the highest concentration.
D) The side facing up has the lowest concentration of molecules.
E) When an egg is fertilized, the gradient is set up based on the side of the egg on which the most
mitochondria reside.
Answer: B
Section: 9.4
Bloom's Taxonomy: Applying/Analyzing
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33) Molecular/structural orientations (gradients) in an egg are thought to play a significant role in
development. What evidence indicates that the maternal genotype is involved in providing such
gradients?
A) Results (non-Mendelian) indicate that the direction of shell coiling in snails is determined by
the genotype of the mother, not the genotype of the zygote.
B) Results indicate that the snail shell coiling is passed exclusively from the mother.
C) The snail shell coiling depends on the genotype of the mother and the offspring.
D) Snail shell's coil based on environmental conditions.
E) The snail's father's genotype dictates the next generation's phenotype.
Answer: A
Section: 9.4
Bloom's Taxonomy: Applying/Analyzing
34) Axis development is determined by a MATERNAL EFFECT gene in Drosophila. A
dominant Bicoid allele is required to set up the correct axis development in flies. bb (two loss of
function alleles) is LETHAL in this MATERNAL EFFECT gene.
A female Drosophila named Eustacia had all dead offspring. What is Eustacia's GENOTYPE?
A) BB
B) Bb
C) bb
D) Bb or bb
E) BB or Bb
Answer: C
Section: 9.4
Bloom's Taxonomy: Evaluating/Creating
35) In MATERNAL EFFECT, which of the following is TRUE?
A) The mother's genotype is the offspring's phenotype.
B) The mother's phenotype is the offspring's phenotype.
C) In a reciprocal cross, the offspring's genotype and phenotype are the same as the mother's.
D) In a reciprocal cross, only the father's alleles are expressed.
E) In a reciprocal cross, the grandfather's alleles have the most influence.
Answer: A
Section: 9.4
Bloom's Taxonomy: Applying/Analyzing
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36) Which of the following is false?
A) The direction of shell coiling in Lymnaea peregra is influenced by the orientation of the first
cleavage division.
B) It is safe to say that a maternal effect is caused by the genotype, not the phenotype, of the
parent producing the egg.
C) In maternal effect, an offspring will have the genotype of their mother.
D) In a reciprocal cross, if an offspring has the mother's phenotype for both crosses, cytoplasmic
inheritance is responsible.
E) When a maternal effect gene is seen, the offspring will demonstrate the phenotype of the
mother's genotype.
Answer: C
Section: 9.4
Bloom's Taxonomy: Remembering/Understanding
37) Snail shell coiling is determined by a MATERNAL EFFECT gene. Snail shells can have
either a right-handed (dextral) coiling or a left-handed (sinistral) coiling. Dextral (D) is
completely dominant to sinistral (d). Answer the following questions based upon this
information. A sinistral female snail named Eustacia has offspring that are all dextral.
What is the Eustacia's GENOTYPE?
A) DD
B) Dd
C) dd
D) Dd or dd
E) DD or Dd
Answer: B
Section: 9.4
Bloom's Taxonomy: Evaluating/Creating
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 10 DNA Structure and Analysis
1) The Central Dogma specifies that the information in ________.
A) protein is copied into an RNA molecule during translation, and the information in that RNA
molecule is used to make DNA during transcription
B) DNA is copied directly into a protein
C) DNA is copied into an RNA molecule during transcription, and the information in that RNA
molecule is used to make a protein during translation
D) RNA is copied into a DNA molecule during transcription, and the information in that DNA
molecule is used to make a protein during translation
E) DNA is copied into an RNA molecule during translation, and the information in that RNA
molecule is used to make a protein during transcription
Answer: C
Section: 10.1
Bloom's Taxonomy: Remembering/Understanding
2) A particular ________ carry the information for making a particular protein, but ________
can be used to make any protein.
A) gene and mRNA; a ribosome and a tRNA
B) gene and tRNA; a ribosome and an mRNA
C) tRNA and ribosome; a gene and an mRNA
D) gene and ribosome; a tRNA and an mRNA
E) ribosome and mRNA; a gene and a tRNA
Answer: A
Section: 10.1
Bloom's Taxonomy: Applying/Analyzing
3) Match the function of DNA on the left with the appropriate description on the right.
i. expression
ii. variation
iii. replication
iv. storage
a.
b.
c.
d.
stable maintenance and passage of information
duplication of genetic material
potential for alteration
production of a phenotype
Which of the choices below is correct?
A) i = d; ii = c; iii = b; iv = a
B) i = c; ii = a; iii = b; iv = d
C) i = d; ii = b; iii = c; iv = a
D) i = b; ii = d; iii = a; iv = c
E) i = d; ii = b; iii = a; iv = c
Answer: E
Section: 10.1
Bloom's Taxonomy: Remembering/Understanding
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4) In the mid-1940s, genetic information was known to consist of ________.
A) protein and carbohydrate
B) nucleic acid and lipid
C) carbohydrate and nucleic acid
D) lipid and protein
E) nucleic acid and protein
Answer: E
Section: 10.2
Bloom's Taxonomy: Remembering/Understanding
5) Who discovered that DNA was the genetic material (or "transforming principle") that could
convert nonvirulent R-type Diplococcus pneumoniae bacterium to the virulent S-type?
A) Frederick Griffith
B) Avery, MacLeod, and McCarty
C) Hershey and Chase
D) Erwin Chargaff
E) Watson, Crick, Wilkins, and Franklin
Answer: B
Section: 10.3
Bloom's Taxonomy: Remembering/Understanding
6) 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) oxygen and phosphorus
Answer: A
Section: 10.3
Bloom's Taxonomy: Remembering/Understanding
7) 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, and many 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: 10.3
Bloom's Taxonomy: Remembering/Understanding
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8) When a protoplast is infected by only the nucleic acid component of a virus, the infection is
referred to as ________.
A) transformation
B) lysis
C) transfection
D) recombination
E) transduction
Answer: C
Section: 10.3
Bloom's Taxonomy: Remembering/Understanding
9) Genetic functions in a eukaryote can be observed in ________.
A) chloroplasts
B) the nucleus
C) mitchondria
D) mitchondria and chloroplasts
E) chloroplasts, mitchondria, and the nucleus
Answer: E
Section: 10.4
Bloom's Taxonomy: Remembering/Understanding
10) If protein were the genetic principle and not nucleic acid, significant mutagenic effects
would be detected at ________.
A) 240 nm
B) 260 nm
C) 280 nm
D) 300 nm
E) 320 nm
Answer: C
Section: 10.4
Bloom's Taxonomy: Applying/Analyzing
11) 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) most lysozymes
Answer: A
Section: 10.5
Bloom's Taxonomy: Remembering/Understanding
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12) What are the two major components of the tobacco mosaic virus?
A) RNA and DNA
B) DNA and protein
C) RNA and protein
D) lipids and nucleic acids
E) carbohydrates and nucleic acids
Answer: C
Section: 10.5
Bloom's Taxonomy: Remembering/Understanding
13) The figure below shows a representative dinucleotide:
This figure represents ________, and the arrow is closest to the ________ end.
A) DNA; 3'
B) DNA; 5'
C) RNA; 3'
D) RNA; 5'
Answer: A
Section: 10.5
Bloom's Taxonomy: Applying/Analyzing
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14) The figure below shows a representative dinucleotide:
Spleen diesterase is an enzyme that breaks the covalent bond that connects the phosphate to the
5' carbon. If the dinucleotide is digested with spleen diesterase, to which base and to which
carbon on the sugar is the phosphate now attached?
A) A; 3'
B) A; 5'
C) T; 3'
D) T; 5'
Answer: A
Section: 10.5
Bloom's Taxonomy: Applying/Analyzing
15) The covalent linkage between the monomers in a dinucleotide is a(n) ________.
A) ester bond
B) peptide bond
C) disulfide bond
D) phosphodiester bond
E) ionic bond
Answer: D
Section: 10.6
Bloom's Taxonomy: Remembering/Understanding
16) The basic structure of a nucleotide includes ________.
A) amino acids
B) tryptophan and leucine
C) base, sugar, and phosphate
D) mRNA, rRNA, and tRNA
E) phosphorus and sulfur
Answer: C
Section: 10.6
Bloom's Taxonomy: Remembering/Understanding
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17) 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) nucleoside monophosphate
D) oligonucleotide
E) nucleoside
Answer: E
Section: 10.6
Bloom's Taxonomy: Remembering/Understanding
18) The figure below is a nucleotide:
If you were told that this molecule could be found in either DNA or RNA, depending on the
identity of the chemical groups labeled (X) and (Y), the nitrogen-containing ring structure would
have to be ________.
A) thymine
B) uracil
C) guanine
D) cytosine
E) adenine
Answer: D
Section: 10.6
Bloom's Taxonomy: Evaluating/Creating
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19) The figure below is a nucleotide:
If this molecule were found in a DNA chain, the chemical group labeled (X) would be ________
and the chemical group labeled (Y) would be ________.
A) H; H
B) H; OH
C) OH; H
D) OH; OH
Answer: C
Section: 10.6
Bloom's Taxonomy: Applying/Analyzing
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20) The figure below is a nucleotide:
If this molecule were found in an RNA chain, the chemical group labeled (X) would be
________ and the chemical group labeled (Y) would be ________.
A) H; H
B) H; OH
C) OH; H
D) OH; OH
Answer: D
Section: 10.6
Bloom's Taxonomy: Applying/Analyzing
21) Any two nucleic acid fragments will spontaneously hybridize if they encounter each other as
long as they are ________ and ________.
A) complementary; antiparallel
B) identical; compatible
C) DNA; RNA
D) primed; replicated
E) transcribed; translated
Answer: A
Section: 10.7
Bloom's Taxonomy: Remembering/Understanding
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22) 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) disulfide
Answer: D
Section: 10.7
Bloom's Taxonomy: Remembering/Understanding
23) The base content of a sample of nucleic acid is as follows: A = 31%, G = 31%, T = 19%, C =
19%. What conclusion should be drawn from this information?
A) This nucleic acid is made of RNA.
B) This nucleic acid is double-stranded DNA.
C) This nucleic acid is single-stranded DNA.
D) The strands in this molecule are parallel rather than antiparallel.
E) The purine/pyrimidine ratio in this molecule fits Chargaff's data.
Answer: C
Section: 10.7
Bloom's Taxonomy: Applying/Analyzing
24) The two DNA chains in a double helix ________.
A) wind around each other in such a way that purines are always opposite pyrimidines and vice
versa
B) are parallel to each other in 5' to 3' directionality
C) have identical base sequences
D) have an overall diameter of 2 μm
E) are held together due to the charges of their phosphate groups
Answer: A
Section: 10.7
Bloom's Taxonomy: Remembering/Understanding
25) If 15% of the nitrogenous bases in a sample of DNA from a particular organism are thymine,
what percentage should be cytosine?
A) 15%
B) 30%
C) 35%
D) 40%
E) 70%
Answer: C
Section: 10.7
Bloom's Taxonomy: Applying/Analyzing
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26) 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) G = T
Answer: C
Section: 10.7
Bloom's Taxonomy: Applying/Analyzing
27) In base pair between a guanine and a cytosine, ________ hydrogen bonds form between a
partially positive hydrogen on the guanine and a more electronegative atom on the cytosine.
A) 0
B) 1
C) 2
D) 3
E) 4
Answer: C
Section: 10.7
Bloom's Taxonomy: Applying/Analyzing
28) The most specific name for an incoming monomer about to be incorporated into a DNA
chain is ________.
A) deoxyribonucleoside
B) nucleotide
C) nucleoside monophosphate
D) ribonucleoside diphosphate
E) deoxyribonucleoside triphosphate
Answer: E
Section: 10.7
Bloom's Taxonomy: Applying/Analyzing
29) 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: 10.7
Bloom's Taxonomy: Applying/Analyzing
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30) The form of DNA that is observed to adopt a left-handed helix is ________.
A) A-DNA
B) C-DNA
C) D-DNA
D) P-DNA
E) Z-DNA
Answer: E
Section: 10.8
Bloom's Taxonomy: Remembering/Understanding
31) The form of DNA that is believed to be most biologically significant is ________.
A) A-DNA
B) B-DNA
C) D-DNA
D) E-DNA
E) Z-DNA
Answer: B
Section: 10.8
Bloom's Taxonomy: Remembering/Understanding
32) Compared with B-DNA, A-DNA is ________ compact, has ________ base pairs per turn,
and is a ________ helix.
A) more; fewer; right-handed
B) more; more; right-handed
C) less; fewer; left-handed
D) less; more; right-handed
E) more; fewer; left-handed
Answer: A
Section: 10.8
Bloom's Taxonomy: Applying/Analyzing
33) Arrange the principal forms of RNA in order of abundance from LEAST abundant to MOST
abundant.
A) tRNA, rRNA, mRNA
B) mRNA, rRNA, tRNA
C) tRNA, mRNA, rRNA
D) mRNA, tRNA, rRNA
E) rRNA, tRNA, mRNA
Answer: D
Section: 10.9
Bloom's Taxonomy: Evaluating/Creating
11
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34) When calculating a molecule's Svedberg coefficient, the rate of sedimentation is affected by
the molecule's ________.
A) mass
B) size
C) shape
D) size and shape
E) mass, size, and shape
Answer: D
Section: 10.9
Bloom's Taxonomy: Remembering/Understanding
35) Which type of rRNA can be found in both prokaryotes and eukaryotes?
A) 5S
B) 5.8S
C) 16S
D) 18S
E) 23S
Answer: A
Section: 10.9
Bloom's Taxonomy: Remembering/Understanding
36) Some forms of genetic regulation can be facilitated by ________.
A) miRNA
B) siRNA
C) lncRNA
D) lncRNA and siRNA
E) miRNA, siRNA, and lncRNA
Answer: E
Section: 10.9
Bloom's Taxonomy: Remembering/Understanding
37) FISH can be used to determine the ________.
A) melting temperature of a fragment of DNA
B) chromosomal location of specific genetic information
C) difference between A-form and B-form DNAs
D) locations of major and minor grooves in a double helix
E) difference in lengths between two fragments of DNA
Answer: B
Section: 10.10
Bloom's Taxonomy: Applying/Analyzing
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38) The figure below represents an aqueous solution of four identical DNA double helices in
pure water:
Which one of the following figures represents the state of the DNA molecules at the melting
temperature (Tm)?
A)
B)
C)
13
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D)
E)
Answer: A
Section: 10.10
Bloom's Taxonomy: Applying/Analyzing
39) Consider the following two statements about DNA melting temperature:
1) The melting temperature of a sample of DNA is the temperature at which 50% of the double
helices in the sample have completely denatured.
2) A molecule with 53% G—C base pairs would have a higher melting temperature than a
molecule of equivalent length with 53% A—T base pairs.
Determine which one of the statements below is CORRECT.
A) Statement 1) and statement 2) are true.
B) Statement 1) and statement 2) are false.
C) Statement 1) is true; statement 2) is false.
D) Statement 1) is false; statement 2) is true.
Answer: D
Section: 10.10
Bloom's Taxonomy: Evaluating/Creating
14
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40) Explain the tetranucleotide hypothesis and its relationship to the substance referred to as
nuclein.
Answer: In the 1860s, Friedrich Miescher isolated an acidic substance (called nuclein) from cell
nuclei, which turned out to be deoxyribonucleic acid. Although DNA was found to be present in
chromosomes, it fell out of favor as the presumptive genetic material because of the incorrect
hypothesis that DNA was a molecule of repeating units of four nucleotides, called the
tetranucleotide hypothesis.
Section: 10.2
Bloom's Taxonomy: Applying/Analyzing
41) 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. Since DNA can carry heritable "traits" from one organism to
another, it must be the genetic material.
Section: 10.3
Bloom's Taxonomy: Applying/Analyzing
42) Explain why the distribution of protein and nucleic acid in a eukaryotic cell favors nucleic
acid as the genetic material.
Answer: DNA is found where primary genetic functions occur; protein is found everywhere.
Section: 10.4
Bloom's Taxonomy: Applying/Analyzing
43) Identify the nitrogenous bases and pentose sugars in DNA and RNA.
Answer: DNA has deoxyribose as the sugar in its nucleotides. The purines that are found in
DNA are adenine and guanine, and the pyrimidines found in DNA are thymine and cytosine.
RNA has ribose as the sugar in its nucleotides. The purines that are found in RNA are adenine
and guanine, and the pyrimidines found in RNA are uracil and cytosine.
Section: 10.6
Bloom's Taxonomy: Remembering/Understanding
44) What is the difference between a polynucleotide and an oligonucleotide?
Answer: Polynucleotides are polymers longer than 30 nucleotides; oligonucleotides are
approximately 4-30 nucleotides in length.
Section: 10.6
Bloom's Taxonomy: Remembering/Understanding
45) What does it mean for a double helix of DNA to be antiparallel and complementary?
Answer: When the two strands in double-stranded DNA associate with each other, one strand
runs 5' to 3' in one direction, the other strand runs in the same plane, but with a 3' to 5'
orientation (antiparallel). Hydrogen bonding between strands occurs between adenines and
thymines, and between cytosines and guanines, which are in turn called complementary base
pairs.
Section: 10.7
Bloom's Taxonomy: Applying/Analyzing
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46) Why would D-DNA and E-DNA be unlikely to be found in the genomes of living
organisms?
Answer: They only occur in helices lacking guanine, and it is exceptionally unlikely to find an
organism that has no guanine.
Section: 10.8
Bloom's Taxonomy: Evaluating/Creating
47) Compare and contrast the principal types of RNA in terms of their cellular roles, relative
abundance, and variation in length.
Answer: rRNA is the most abundant, serves a structural and functional role in a ribosome, and
can be found in six widely different lengths. tRNAs are the next most abundant, carry amino
acids to ribosomes for incorporation into proteins, and exhibit a narrow range of sizes. mRNAs
are the least abundant, provide a blueprint for translation, and have the widest range of lengths
observed.
Section: 10.9
Bloom's Taxonomy: Evaluating/Creating
48) How do urea and NaCl influence temperature-induced DNA melting?
Answer: Urea competes for hydrogen bonds; thus, the bases pair with the urea rather than with
each other. This weakens the complementary associations that are required to hold the DNA
helix together; thus, less heat is required for melting. The sodium of sodium chloride associates
with and neutralizes the strong negative charges on the phosphates. The phosphates do not repel
each other with the sodium ion present; thus, the double-stranded structure requires more energy
to melt.
Section: 10.10
Bloom's Taxonomy: Applying/Analyzing
49) Explain the technique of electrophoretic separation of DNA fragments.
Answer: A mixture of different sized DNA molecules are subjected to an electric field, and
migrate through a semisolid matrix such that smaller fragments travel further in the matrix. Also
see Figure 10.18.
Section: 10.10
Bloom's Taxonomy: Applying/Analyzing
50) You determine the sequence of an isolated gene, and wish to find its location in the genome.
Propose a procedure using fluorescence to locate your gene of interest.
Answer: A single-stranded probe molecule could be synthesized, which is complementary to
one strand of your gene and labeled with a fluorescent marker. Mitotic cells could then be
immobilized and their chromosomes denatured, followed by introduction of labeled probe. After
hybridization, the fluorescent label is detected, identifying the chromosome and location of the
gene of interest.
Section: 10.10
Bloom's Taxonomy: Evaluating/Creating
16
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 11 DNA Replication and Recombination
1) Which of the following terms accurately describes the replication of DNA in vivo?
A) conservative
B) dispersive
C) semidiscontinuous
D) nonlinear
E) nonreciprocal
Answer: C
Section: 11.1
Bloom's Taxonomy: Remembering/Understanding
2) After ONE round of DNA replication in Meselson and Stahl's classic experiment, ________.
A) all three models of replication were still possibilities
B) the semiconservative model was conclusively proven to be the accurate model
C) the dispersive model was conclusively proven to be an inaccurate model
D) the conservative model was conclusively proven to be an inaccurate model
E) both the conservative and the semiconservative models were still possibilities
Answer: D
Section: 11.1
Bloom's Taxonomy: Applying/Analyzing
3) DNA replication proceeds ________.
A) unidirectionally
B) semiconservatively
C) progressively
D) discontinuously
E) dispersively
Answer: B
Section: 11.1
Bloom's Taxonomy: Remembering/Understanding
4) If DNA replication were fully conservative, how many intact parental double helices would
have been detected in Meselson and Stahl's experiment after three rounds of replication?
A) 0
B) 1
C) 3
D) 7
E) 8
Answer: B
Section: 11.1
Bloom's Taxonomy: Applying/Analyzing
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5) The amount of DNA that is replicated from a single origin of replication is referred to as a(n)
________.
A) primer
B) fork
C) ori
D) replicon
E) Okazaki fragment
Answer: D
Section: 11.1
Bloom's Taxonomy: Remembering/Understanding
6) Which DNA Polymerase III subunit is not part of the clamp loading assembly?
A) γ
B) δ
C) χ
D) θ
E) ν
Answer: D
Section: 11.2
Bloom's Taxonomy: Remembering/Understanding
7) DNA replication occurs by adding ________.
A) dNTPs to the 3′ end of the daughter strand
B) NTPs to the 5′ end of the daughter strand
C) dNTPs to the 3′ end of the template strand
D) NTPs to the 3′ end of the daughter strand
E) dNTPs to the 5′ end of the template strand
Answer: A
Section: 11.2
Bloom's Taxonomy: Remembering/Understanding
8) DNA polymerase III adds nucleotides ________.
A) to the 3′ end of the RNA primer
B) to the 5′ end of the RNA primer
C) in the place of the primer RNA after it is removed
D) to both ends of the RNA primer
E) to internal sites in the DNA template
Answer: A
Section: 11.2
Bloom's Taxonomy: Remembering/Understanding
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9) 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 RNA after it is removed
D) on single-stranded templates without need for an RNA primer
E) in a 5′ to 5′ direction
Answer: C
Section: 11.2
Bloom's Taxonomy: Remembering/Understanding
10) 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: 11.2
Bloom's Taxonomy: Remembering/Understanding
11) 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: 11.2
Bloom's Taxonomy: Remembering/Understanding
12) Which of the following statements about the oriC region in E. coli is incorrect?
A) It is directly recognized and bound by DNA helicase.
B) It is 245 base pairs in length.
C) It is rich in AT base pairs.
D) It includes five repeating sequences of 9 base pairs.
E) It includes three repeating sequences of 13 base pairs.
Answer: A
Section: 11.3
Bloom's Taxonomy: Remembering/Understanding
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13) The protein responsible for the initial step in unwinding the DNA helix during replication of
the bacterial chromosome is coded for by the ________ gene.
A) dnaA
B) dnaE
C) polA
D) polB
E) gyrA
Answer: A
Section: 11.3
Bloom's Taxonomy: Applying/Analyzing
14) If E. coli DNA polymerase III was able to initiate DNA synthesis opposite a single-stranded
template, which molecule(s) would no longer be essential?
A) single-stranded binding proteins
B) DNA gyrase
C) DNA polymerase I
D) helicase
E) DNA ligase
Answer: C
Section: 11.3
Bloom's Taxonomy: Applying/Analyzing
15) 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: 11.3
Bloom's Taxonomy: Remembering/Understanding
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16) Given the accompanying diagram, 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) dispersive
C) semiconservative
D) conservative and dispersive
E) semiconservative and dispersive
Answer: A
Section: 11.3
Bloom's Taxonomy: Applying/Analyzing
17) During DNA replication, ________ are 1000-2000 nucleotide long strands synthesized on
the ________ strand to maintain the ________ of replication.
A) Okazaki fragments; leading, accuracy
B) primers, lagging; continuity
C) Okazaki fragments; leading; semidiscontinuity
D) primers; leading; discontinuity
E) Okazaki fragments; lagging; bidirectionality
Answer: E
Section: 11.3
Bloom's Taxonomy: Remembering/Understanding
18) As unwinding of the helix occurs during DNA replication, tension referred to as ________ is
created ahead of the replication fork. This tension is relieved by the action of ________.
A) supercoiling; DNA gyrase
B) supercoiling; DNA helicase
C) supercoiling; single-stranded binding proteins
D) processivity; DNA gyrase
E) processivity; primase
Answer: A
Section: 11.4
Bloom's Taxonomy: Remembering/Understanding
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19) Which of the following statement about DNA replication in E. coli is correct?
A) Only the lagging strand needs to be primed.
B) There are two molecules of DNA polymerase III per replication fork.
C) There are two molecules of DNA helicase per replication fork.
D) Single-stranded binding proteins are only required on the leading strand.
E) DNA polymerase I follows closely behind DNA helicase in the replication fork.
Answer: B
Section: 11.4
Bloom's Taxonomy: Remembering/Understanding
20) The protein product of the E. coli dnaE gene is a component of ________.
A) DNA polymerase I
B) DNA polymerase III
C) primase
D) DNA ligase
E) DNA helicase
Answer: B
Section: 11.5
Bloom's Taxonomy: Remembering/Understanding
21) The activity of ________ would be quickly undone in the absence of single-stranded binding
protein.
A) DNA polymerase I
B) DNA polymerase III
C) primase
D) DNA ligase
E) DNA helicase
Answer: E
Section: 11.5
Bloom's Taxonomy: Applying/Analyzing
22) DNA replication in eukaryotes ________.
A) initiates at multiple origins
B) occurs without the need of a primer
C) synthesizes DNA approximately 25 times faster than in prokaryotes
D) takes place multiple times per cell cycle
E) utilizes a single type of DNA polymerase
Answer: A
Section: 11.6
Bloom's Taxonomy: Applying/Analyzing
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23) Chromatin is defined as ________.
A) DNA complexed with protein
B) hybrid molecules of DNA and RNA
C) genetic material found only in mitochondria
D) a catalytic molecule with a protein component and an RNA component
E) supercoiled loops of bacterial DNA
Answer: A
Section: 11.6
Bloom's Taxonomy: Remembering/Understanding
24) Assembly of new nucleosomes is carried out by complexes abbreviated ________.
A) CAF
B) DSB
C) TLS
D) TERC
E) SSB
Answer: A
Section: 11.6
Bloom's Taxonomy: Remembering/Understanding
25) Which eukaryotic DNA polymerase would least likely be involved in repair?
A) γ
B) δ
C) α
D) β
E) ε
Answer: C
Section: 11.6
Bloom's Taxonomy: Applying/Analyzing
26) Structures located at the ends of eukaryotic chromosomes are called ________.
A) centromeres
B) telomerases
C) recessive mutations
D) telomeres
E) permissive mutations
Answer: D
Section: 11.7
Bloom's Taxonomy: Remembering/Understanding
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27) Telomerase reads ________ as a template and synthesizes ________ as a product.
A) RNA; DNA
B) RNA; RNA
C) DNA; DNA
D) DNA; RNA
E) DNA; protein
Answer: A
Section: 11.7
Bloom's Taxonomy: Remembering/Understanding
28) The process of making DNA from an RNA template is referred to as ________.
A) reverse transcription
B) translation
C) transversion
D) transcription
E) recombination
Answer: A
Section: 11.7
Bloom's Taxonomy: Remembering/Understanding
29) Telomerase adds nucleotides to the ________.
A) 3′ end of the parental strand
B) 3′ end of the daughter strand
C) 5′ end of the parental strand
D) 5′ end of the daughter strand
Answer: A
Section: 11.7
Bloom's Taxonomy: Applying/Analyzing
30) A molecule that possesses both a protein component and an RNA component is referred to as
a ________.
A) ribonucleoprotein
B) dimer
C) polymerase
D) holoenzyme
E) tetramer
Answer: A
Section: 11.7
Bloom's Taxonomy: Remembering/Understanding
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31) Which of the following is not an essential step in homologous recombination?
A) exonuclease degradation
B) endonuclease nicking
C) branch migration
D) duplex separation
E) strand displacement
Answer: A
Section: 11.7
Bloom's Taxonomy: Remembering/Understanding
32) The Meselson and Stahl's 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: 11.1
Bloom's Taxonomy: Evaluating/Creating
33) 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: 11.1
Bloom's Taxonomy: Remembering/Understanding
34) Compare and contrast DNA Polymerases I, II, and III in E. coli. What capabilities are shared,
and which are exclusive to only one type of polymerase?
Answer: None of the polymerases are able to initiate chain synthesis, but all are able to extend
an existing fragment in the 5′ to 3′ direction. All three possess proofreading ability (3′ to 5′
exonuclease), but only DNA polymerase I possesses primer removal ability (5′ to 3′
exonuclease).
Section: 11.2
Bloom's Taxonomy: Applying/Analyzing
35) Explain why DNA primers can be used in in vitro DNA synthesis reactions, but not in vivo.
Answer: The need for the primer is to provide a 3′ OH group to which DNA-synthesizing
enzymes can add new nucleotides. In an in vitro reaction, provided the template sequence is
known, complementary single-stranded DNA primers can be synthesized and included in the
reaction.
Section: 11.3
Bloom's Taxonomy: Evaluating/Creating
9
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36) 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: 11.3
Bloom's Taxonomy: Remembering/Understanding
37) Why does DNA replication have to be semidiscontinuous in order to be bidirectional?
Answer: The requirement of synthesis to be 5′ to 3′, and the antiparallel nature of doublestranded DNA, means that in both replication forks one of the stands must be made in fragments
directed back toward the origin of replication.
Section: 11.4
Bloom's Taxonomy: Applying/Analyzing
38) Explain the need for, and action of, DNA gyrase in bacterial DNA replication.
Answer: DNA helicase induces supercoiling (tension) in the parental DNA as it processes.
DNA gyrase relieves this tension so that replication fork progression can continue without
grinding to a halt.
Section: 11.4
Bloom's Taxonomy: Remembering/Understanding
39) A drug is added to a culture of E. coli cells that inhibit DNA gyrase. Predict the effect of this
drug on DNA replication in the affected cells.
Answer: Without DNA gyrase to relieve the supercoiling of the DNA induced by DNA helicase,
the tension in the closed circle of DNA would cause replication machinery to stall, preventing
replication from completing and preventing the cells from dividing.
Section: 11.4
Bloom's Taxonomy: Applying/Analyzing
40) Describe the benefit to researchers of conditional mutants.
Answer: Many mutations that would be interesting to study are lethal. Conditional mutations
allow researchers to observe wild-type behavior under permissive conditions, and loss-offunction effects under restrictive conditions.
Section: 11.5
Bloom's Taxonomy: Remembering/Understanding
41) What is meant by a gene knockout?
Answer: The designed loss of a specific gene in order to study its function. This is often
accomplished in the context of a conditional knockout, where the gene is affected in some tissue
types but not others, allowing genes to be studied for which complete loss would be lethal.
Section: 11.5
Bloom's Taxonomy: Remembering/Understanding
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42) Compare and contrast DNA replication in prokaryotes and eukaryotes.
Answer: 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.
However, in eukaryotes, 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: 11.6
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: 11.6
Bloom's Taxonomy: Remembering/Understanding
44) What is meant by processivity of a DNA-synthesizing enzyme?
Answer: The strength of the association between the enzyme and its substrate, and thus the
length of DNA that is synthesized before the enzyme dissociates from the template.
Section: 11.6
Bloom's Taxonomy: Remembering/Understanding
45) 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: 11.6
Bloom's Taxonomy: Remembering/Understanding
46) 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: 11.7
Bloom's Taxonomy: Remembering/Understanding
47) Describe a somewhat extraordinary finding related to the Tetrahymena telomerase enzyme.
Answer: The enzyme contains a short piece of RNA that is essential for its catalytic activity.
Section: 11.7
Bloom's Taxonomy: Remembering/Understanding
48) Why are bacteria not dependent on telomerase for complete DNA replication?
Answer: Telomerase allows for the accurate replication of linear DNA molecules, and bacteria
typically possess circular chromosomes.
Section: 11.7
Bloom's Taxonomy: Applying/Analyzing
11
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49) What is meant by homologous recombination?
Answer: Genetic exchange at equivalent positions along two chromosomes with substantial
DNA sequence homology.
Section: 11.8
Bloom's Taxonomy: Remembering/Understanding
50) Describe two processes for which recombination is essential.
Answer: The exchange of genetic information during meiosis, and the repair of double-stranded
breaks in DNA molecules.
Section: 11.8
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 12 DNA Organization in Chromosomes
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: 12.1
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: 12.1
Bloom's Taxonomy: Remembering/Understanding
3) This figure represents supercoiled, circular, double-stranded DNA:
If this molecule of DNA originated as a linear molecule with a linking number (L) of 30, which
was then circularized and unwound, the L for the unwound circular molecule would be
________.
A) 28
B) 20
C) 18
D) 30
E) 32
Answer: A
Section: 12.2
Bloom's Taxonomy: Applying/Analyzing
1
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4) Type I topoisomerases differ from type II topoisomerases in that the type I enzymes
________.
A) cleave one strand of a double helix
B) cleave both strands of a double helix
C) induce negative supercoiling
D) decrease the linking number (L) of a double helix
E) ligate fragments of DNA that have been hydrolyzed
Answer: B
Section: 12.2
Bloom's Taxonomy: Applying/Analyzing
5) If relaxed circular DNA is designated (R), supercoiled circular DNA is designated (S), and
linear DNA is designated (L), arrange these conformations in order of increasing rate of
sedimentation during centrifugation.
A) (L), (R), (S)
B) (L), (S), (R)
C) (R), (S), (L)
D) (S), (R), (L)
E) (S), (L), (R)
Answer: A
Section: 12.2
Bloom's Taxonomy: Evaluating/Creating
6) In which of the following ways do polytene chromosomes differ from other chromosomes?
A) Polytene chromosomes are replicated but not separated.
B) Polytene chromosomes can only be found in bacteria.
C) Polytene chromosomes are separated but not replicated.
D) Polytene chromosomes are multiple copies of identical single-stranded DNA.
E) Polytene chromosomes represent uncoiled versions of meiotic chromosomes.
Answer: A
Section: 12.3
Bloom's Taxonomy: Applying/Analyzing
7) The condensed areas in polytene and lampbrush chromosomes are referred to as ________.
A) chromomeres
B) puffs
C) chromatids
D) centromeres
E) chiasmata
Answer: A
Section: 12.3
Bloom's Taxonomy: Remembering/Understanding
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8) Which statement about lampbrush chromosomes is incorrect?
A) The loops consist of two DNA helices, whereas the axis consists of a single DNA helix.
B) They are meiotic chromosomes.
C) Instead of being condensed, they are extended.
D) Looped out regions are active in RNA synthesis.
E) More DNA is looped out from the axis than is required for a single gene.
Answer: A
Section: 12.3
Bloom's Taxonomy: Applying/Analyzing
9) 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: 12.4
Bloom's Taxonomy: Remembering/Understanding
10) 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: 12.4
Bloom's Taxonomy: Remembering/Understanding
11) The condition that some organisms contain much larger amounts of DNA than are apparently
"needed" and that some relatively closely related organisms may have vastly different amounts
of DNA is more typical in ________.
A) viruses than in bacteria
B) RNA viruses than in DNA viruses
C) eukaryotes than in prokaryotes
D) the alphoid rather than the diploid family
E) prokaryotes than in eukaryotes
Answer: C
Section: 12.5
Bloom's Taxonomy: Remembering/Understanding
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12) 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 families
D) euchromatic regions
E) telomere-associated sequences
Answer: C
Section: 12.6
Bloom's Taxonomy: Remembering/Understanding
13) In addition to highly repetitive and unique DNA sequences, a third category of DNA
sequences exists. What is it called, and what types of elements are involved in it?
A) composite DNA; telomeres and heterochromatin
B) dominant DNA; euchromatin and heterochromatin
C) multiple gene family DNA; hemoglobin and 5.0S RNA
D) moderately repetitive DNA; SINEs, LINEs, and VNTRs
E) permissive DNA; centromeres and heterochromatin
Answer: D
Section: 12.6
Bloom's Taxonomy: Remembering/Understanding
14) Chromosomal regions that represent evolutionary vestiges of duplicated copies of genes that
have underdone sufficient mutations to render them untranscribable are called ________.
A) pseudogenes
B) alleles
C) transposons
D) satellites
E) LINEs
Answer: A
Section: 12.7
Bloom's Taxonomy: Remembering/Understanding
15) A particular variant of the lambda bacteriophage has a DNA double-stranded genome of
51,365 base pairs. How long would this DNA be?
Answer: One base pair is 0.34 nm; therefore, 51,365 bp × 0.34 nm/bp = 17,464 nm or 17.46 μm.
Section: 12.1
Bloom's Taxonomy: Applying/Analyzing
16) List several configurations that characterize different viral chromosomes.
Answer: DNA (single- and double-stranded), RNA (single- and double-stranded), linear,
circular
Section: 12.1
Bloom's Taxonomy: Remembering/Understanding
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17) Describe the chromosomal conformations of φX174 and polyomaviruses.
Answer: single-stranded DNA, circular; double-stranded DNA, circular, respectively
Section: 12.1
Bloom's Taxonomy: Remembering/Understanding
18) Explain how linear DNA can be supercoiled.
Answer: In eukaryotes with linear chromosomes, regions of DNA can be supercoiled if their
ends are anchored to chromatin proteins that serve as a structural framework, or lattice.
Section: 12.2
Bloom's Taxonomy: Remembering/Understanding
19) What is the relationship between topoisomers?
Answer: They are molecules of DNA that differ in their linking numbers (L) but are otherwise
identical.
Section: 12.2
Bloom's Taxonomy: Remembering/Understanding
20) How was it experimentally determined that "puffs" are transcriptionally active?
Answer: Puffs actively incorporate radioactively labeled RNA precursors during
autoradiography assays. Bands that are not extended into puffs incorporate fewer radioactive
precursors or none at all.
Section: 12.3
Bloom's Taxonomy: Applying/Analyzing
21) How does a polytene chromosome differ from a typical eukaryotic chromosome?
Answer: Polytene chromosomes are found in a variety of tissues in the larvae of some flies and
several species of protozoans and plants. A polytene chromosome contains banding patterns and
is large because of repeated replications of DNA without nuclear division.
Section: 12.3
Bloom's Taxonomy: Remembering/Understanding
22) 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. As 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: 12.4
Bloom's Taxonomy: Evaluating/Creating
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23) 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: 12.4
Bloom's Taxonomy: Applying/Analyzing
24) Describe the basic structure of a nucleosome. What is the role of histone H1?
Answer: Nucleosomes are composed of four different histone molecules, each of which exists
twice, thus forming an octamer. Histone H1 is between nucleosomes and is associated with
linker DNA.
Section: 12.4
Bloom's Taxonomy: Remembering/Understanding
25) Compare and contrast the chromosome structure of viruses, bacteria, and eukaryotes.
Answer: The amount of DNA per structure (virus particle, bacterium, or 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 DNAs,
the regularly spaced histones of eukaryotic chromosomes are unique.
Section: 12.4
Bloom's Taxonomy: Applying/Analyzing
26) List the components of a nucleosome.
Answer: Histones H2A, H2B, H3, and H4 exist as two types of tetramers: (H2A)2 + (H2B)2 and
(H3)2 + (H4)2.
Section: 12.4
Bloom's Taxonomy: Remembering/Understanding
27) What similarities do bacterial chromosomes have with eukaryotic chromosomes?
Answer: Both have double-stranded DNA and several types of proteins associated with that
DNA (nucleosomes in eukaryotes and HU and H1 proteins in bacteria).
Section: 12.4
Bloom's Taxonomy: Remembering/Understanding
28) What are histones, and how are they arranged in nucleosomes?
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.
Section: 12.4
Bloom's Taxonomy: Remembering/Understanding
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29) When native chromatin is digested with micrococcal nuclease, what significant result
occurs?
Answer: DNA fragments of approximately 200 base pairs in length are formed.
Section: 12.4
Bloom's Taxonomy: Remembering/Understanding
30) In the formation of nucleosomes, one histone class, H1, is not directly involved, yet it does
associate with DNA to form higher level chromosomal structures. Where does this histone (H1)
associate?
Answer: in the spaces between nucleosome and DNA complexes
Section: 12.4
Bloom's Taxonomy: Remembering/Understanding
31) What are minisatellites and microsatellites?
Answer: Both are highly repetitive, relatively short DNA sequences.
Section: 12.6
Bloom's Taxonomy: Remembering/Understanding
32) Briefly state what is meant by repetitive DNA.
Answer: DNA is present in repeated sequences—(GACAT)n, for example.
Section: 12.6
Bloom's Taxonomy: Remembering/Understanding
33) What is meant by SINE in terms of chromosome structure? By LINE? Why are they called
"repetitive"?
Answer: SINE = short interspersed elements, a moderately repetitive sequence class; LINE =
long interspersed elements. Multiple copies exist–up to 900,000 Alu, sequences, for example.
Section: 12.6
Bloom's Taxonomy: Remembering/Understanding
34) How do VNTRs relate to DNA fingerprinting?
Answer: VNTRs are variable number tandem repeats of 15-100 base pairs long that vary among
individuals. Because each non-twin individual has a different VNTR pattern, identity of
individuals is unique.
Section: 12.6
Bloom's Taxonomy: Remembering/Understanding
35) Approximately how much of the mammalian genome is composed of repetitive DNA?
Answer: About 5-10% of a mammalian genome is highly repetitive; about 30% is moderately
repetitive.
Section: 12.6
Bloom's Taxonomy: Remembering/Understanding
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36) Briefly describe the makeup of VNTRs (variable number tandem repeats).
Answer: VNTRs are repeating DNA sequences of about 15-100 base pairs long that are found
within and between genes. These sequences may be repeated to give regions 1000-5000 bp in
length. They are dispersed throughout the genome.
Section: 12.6
Bloom's Taxonomy: Remembering/Understanding
37) In instances in the eukaryotic genome, DNA sequences represent evolutionary vestiges of
duplicated copies of genes. What are such regions called and what are their characteristics?
Answer: Pseudogenes are duplicated copies of genes that have undergone considerable mutation
and share some homology to the original gene.
Section: 12.7
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 13 The Genetic Code and Transcription
1) Which of the following is true regarding the genetic code?
A) The genetic code is overlapping.
B) The genetic code is degenerate.
C) The genetic code is considered to not be universal.
D) The genetic code is ambiguous.
E) The genetic code has three start codons and one stop codon.
Answer: B
Section: 13.1
Bloom's Taxonomy: Remembering/Understanding
2) If there were 75 naturally occurring amino acids then what is the smallest codon size?
A) 1
B) 2
C) 3
D) 4
E) 5
Answer: D
Section: 13.2
Bloom's Taxonomy: Applying/Analyzing
3) 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) the code is triplet
B) the code contains internal punctuation
C) AUG is the initiating triplet
D) the code is overlapping
E) there are three amino acids per base
Answer: A
Section: 13.2
Bloom's Taxonomy: Remembering/Understanding
1
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4) The experiments that deciphered the genetic code used an enzyme called polynucleotide
phosphorylase. If researchers added this enzyme to a large quantity of only guanine (G)
ribonucleoside diphosphates then a RNA molecule would be produced that would code for which
amino acid?
A) phenylalanine (phe)
B) proline (pro)
C) glycine (gly)
D) lysine (lys)
E) arginine (arg)
Answer: C
Section: 13.3
Bloom's Taxonomy: Applying/Analyzing
5) 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: 13.3
Bloom's Taxonomy: Remembering/Understanding
6) 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) wobble
B) procodon
C) cistron
D) rho
E) codon
Answer: E
Section: 13.3
Bloom's Taxonomy: Remembering/Understanding
7) What is the initiator triplet in both bacteria and eukaryotes? What amino acid is recruited by
this triplet?
A) UAA, UGA, or UAG; no amino acid called in
B) UAA, UGA, or UAG; arginine
C) AUG; arginine
D) AUG; methionine
E) UAA, UGA, or UAG; methionine
Answer: D
Section: 13.4
Bloom's Taxonomy: Remembering/Understanding
2
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8) Due to the wobble hypothesis, which position in the codon, if changed to a different
nucleotide, would be least likely to cause a change in the amino acid encoded?
A) first nucleotide of the codon
B) second nucleotide of the codon
C) third nucleotide of the codon
D) any nucleotide of the codon.
E) Either the first or second nucleotide of the codon.
Answer: C
Section: 13.4
Bloom's Taxonomy: Applying/Analyzing
9) A particular mRNA is 300 nucleotides long. If a mutation in the middle of the sequence
changed a codon from a AAA to a UAA then what would be a reasonable prediction?
A) The protein coded by this mRNA would be shorter.
B) The protein coded by this mRNA would be longer.
C) The protein coded by this mRNA would be the same size.
D) The protein coded by this mRNA would not form due to a failure in initiation.
E) The protein coded by this mRNA would kill the cell.
Answer: A
Section: 13.4
Bloom's Taxonomy: Evaluating/Creating
10) 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 above
Answer: A
Section: 13.6
Bloom's Taxonomy: Remembering/Understanding
11) The nuclear genetic code is considered universal with a few exceptions. Some of these
exceptions are found when comparing the nuclear genetic code of humans to what?
A) acterial genetic code
B) phage M2 genetic code
C) angiosperm's genetic code
D) human mitochondrial genetic code
E) chimpanzee's genetic code
Answer: D
Section: 13.5, 13.6
Bloom's Taxonomy: Applying/Analyzing
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12) The finding that some phage genes have multiple initiation sites is evidence against what
characteristic of the genetic code?
A) The genetic code is universal.
B) The genetic code is commaless.
C) The genetic code is nonoverlapping.
D) The genetic code is unambiguous.
E) The genetic code is a triplet code.
Answer: C
Section: 13.7
Bloom's Taxonomy: Remembering/Understanding
13) The relationship among a gene, a messenger RNA, and a protein is that ________.
A) mRNAs make genes, which then make proteins
B) genes make mRNA, which then make proteins
C) proteins make mRNA, which then make genes
D) mRNAs make proteins, which then make genes
E) genes make proteins, which then make mRNA
Answer: B
Section: 13.8
Bloom's Taxonomy: Remembering/Understanding
14) Which of the following bacterial RNA polymerase subunits is found in the holoenzyme, but
not the core enzyme?
A) α (alpha)
B) β (beta)
C) β′ (beta prime)
D) ω (omega)
E) σ (sigma)
Answer: E
Section: 13.9
Bloom's Taxonomy: Remembering/Understanding
15) An mRNA that is being produced comes off of the ________.
A) coding strand in both bacteria and eukaryotes
B) template strand in both bacteria and eukaryotes
C) coding strand in bacteria and the template strand in eukaryotes
D) template strand in bacteria and the coding strand in eukaryotes
E) coding or template strand in both bacteria and eukaryotes depending upon the species
Answer: B
Section: 13.9
Bloom's Taxonomy: Remembering/Understanding
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16) 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 repeats
E) satellite DNAs
Answer: A
Section: 13.9
Bloom's Taxonomy: Remembering/Understanding
17) A mutation in a particular bacterial gene prevents the formation of a hairpin loop. This is
most likely to affect which part of transcription?
A) This mutation will affect initiation.
B) This mutation will affect elongation.
C) This mutation will affect termination.
D) This mutation will not affect any aspect of transcription in bacteria.
E) This mutation will affect initiation, elongation, and termination.
Answer: B
Section: 13.9
Bloom's Taxonomy: Applying/Analyzing
18) Eukaryotic transcription is different than bacterial transcription because eukaryotic
transcription ________.
A) occurs in the cytoplasm
B) only requires one RNA polymerase
C) requires the uncoiling of the chromatin fiber
D) does not require general transcription factors to initiate transcription
E) does not require the mRNA to be modified
Answer: C
Section: 13.10
Bloom's Taxonomy: Remembering/Understanding
19) Which transcription factor binds directly to the TATA-box sequence?
A) TFIIA
B) TFIIB
C) TFIIC
D) TFIID
E) enhancers
Answer: D
Section: 13.10
Bloom's Taxonomy: Remembering/Understanding
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20) A mutation of what type of sequence is predicted to have the greatest effect on the rate of
transcription?
A) core promoter
B) proximal-promoter element
C) enhancer
D) silencer
E) termination sequence
Answer: A
Section: 13.10
Bloom's Taxonomy: Applying/Analyzing
21) Which mRNA modification is likely absent if the mRNA is degrading prematurely from the
5′ end of the mRNA?
A) addition of the 3′ polyadenylated tail
B) addition of the 7-methylguanosine cap to the 5′ end
C) removal of introns
D) splicing together of exons
E) RNA editing
Answer: B
Section: 13.10
Bloom's Taxonomy: Applying/Analyzing
22) 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) There are more amino acids 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: 13.11
Bloom's Taxonomy: Applying/Analyzing
23) 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) the transfer RNA (tRNA) that binds to mRNA codon
E) the carbohydrate that serves as a signal for RNA transport
Answer: B
Section: 13.11
Bloom's Taxonomy: Remembering/Understanding
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24) 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: 13.11
Bloom's Taxonomy: Applying/Analyzing
25) 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: 13.11
Bloom's Taxonomy: Remembering/Understanding
26) Which of the following intron groups is matched appropriately with a true statement?
A) Group I introns are removed from rRNAs that are found in mitochondria and chloroplasts.
B) Group I introns are removed from rRNAs using large protein-based enzyme complexes.
C) Group II introns are removed from mRNA in mammals.
D) Spliceosomal introns are removed using the catalytic ability of the intron itself.
E) Spliceosomal introns are removed using large protein-based enzyme complexes.
Answer: E
Section: 13.11
Bloom's Taxonomy: Remembering/Understanding
27) The type of RNA modification that requires guide RNAs is ________.
A) addition of the 3′ polyadenylated tail
B) addition of the 7-methylguanosine cap to the 5′ end
C) removal of introns
D) splicing together of exons
E) RNA editing
Answer: E
Section: 13.12
Bloom's Taxonomy: Remembering/Understanding
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28) Describe the direction of information flow in living systems. Use appropriate, scientific
terms in your description.
Answer: DNA is replicated and passed to offspring through a variety of reproductive processes.
Information contained in the base sequences of DNA is transcribed into a variety of RNAs.
Certain RNAs (tRNA) carry amino acids to the site of translation where proteins are assembled.
Other RNAs (mRNA and rRNA) provide a mechanism for ordering the sequence of amino acids
in proteins.
Section: 13.2
Bloom's Taxonomy: Remembering/Understanding
29) Sidney Brenner argued that the code was nonoverlapping because he considered that coding
restrictions would occur if it were overlapping. A second major argument against an overlapping
code involved the effect of a single nucleotide change. In an overlapping code, how many
adjacent amino acids would be affected by a single nucleotide change? In a nonoverlapping
code, how many amino acid(s) would be affected?
Answer: two; one
Section: 13.2
Bloom's Taxonomy: Evaluating/Creating
30) 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 explain how the use of
polynucleotide phosphorylase and the triplet binding assay were used to decipher the genetic
code and why the triplet binding assay was more precise.
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 were used to verify information
provided earlier and to establish the ordered codon assignments.
Section: 13.3
Bloom's Taxonomy: Remembering/Understanding
31) 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 = 1/216
Section: 13.3
Bloom's Taxonomy: Applying/Analyzing
32) "Breaking the genetic code" has been referred to as one of the most significant scientific
achievements in modern times. Describe (in outline or brief statement form) the procedures used
to break the code.
Answer: (a) use of polynucleotide phosphorylase for the production of synthetic "mRNAs"
(b)introduction of synthetic mRNAs into the cell-free protein-synthesizing system
(c) frameshift mutations in the rII region of T4 phage to show code is triplet and degenerate
(d)triplet binding assay to produce mRNA–tRNA-ribosome complex
(e) development of regular copolymers for use in the cell-free protein-synthesizing system
Section: 13.3
Bloom's Taxonomy: Remembering/Understanding
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33) Reading the RNA sequences in a 5′ to 3′ direction, 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: 13.4
Bloom's Taxonomy: Remembering/Understanding
34) Explain the "wobble hypothesis" and why it allows a single tRNA to recognize more than
one codon.
Answer: The relaxed pairing at the third position of the codon and the first position of the
anticodon are less constrained and do not need to follow the traditional base pair rules. This
allows for a single tRNA to recognize multiple codons.
Section: 13.4
Bloom's Taxonomy: Remembering/Understanding
35) A particular mRNA makes a protein that has a unformylated methionine as its first amino
acid. Did this mRNA come from a bacterial or a eukaryotic cell? Explain.
Answer: Eukaryotic cell. Eukaryotic mRNA code for proteins starts with an unformylated
methionine whereas bacterial mRNA code for proteins starts with a formylated methionine.
Section: 13.4
Bloom's Taxonomy: Applying/Analyzing
36) Write a sequence of mRNA that codes for five amino acids and the mRNA should contain
proper punctuation and polarity.
Answer: 5′ AUG (any four amino acid coding codons) and (UAA, UGA, or UAG) 3′
Section: 13.4
Bloom's Taxonomy: Evaluating/Creating
37) There is some indication that the code is in some way ordered; a certain pattern exists.
Describe an observation that supports this view.
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: 13.4
Bloom's Taxonomy: Remembering/Understanding
38) 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: mitchondrial DNA; Mycoplasma capricolum; some protozoans
Section: 13.6
Bloom's Taxonomy: Remembering/Understanding
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39) 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: 13.7
Bloom's Taxonomy: Remembering/Understanding
40) What are three observations that suggested eukaryotic RNA was an intermediate between
DNA and protein.
Answer: (1) DNA is found in the nucleus and protein synthesis occurs in the cytoplasm, (2)
RNA is synthesized in the nucleus and has a chemical similarity to DNA, and (3) RNA is
transported out of the nucleus and into the cytoplasm where protein translation occurs.
Section: 13.8
Bloom's Taxonomy: Remembering/Understanding
41) Describe the effects of mutating a sigma factor gene in bacteria.
Answer: The core RNA polymerase enzyme will form. Since there are more than one sigma
factor genes, by mutating a single sigma factor gene there will be some holoenzymes that will
not form which will affect specific gene transcription.
Section: 13.9
Bloom's Taxonomy: Applying/Analyzing
42) In a particular bacterial mutant, a specific gene is not producing a mRNA or the
corresponding protein. Researchers sequence the gene and determine that there are no mutations
in the coding portion of the gene. What could explain these observations?
Answer: One of the two consensus sequences are likely mutated. Either the Pribnow box
(TATAAT) or the −35 sequence (TTGACA).
Section: 13.9
Bloom's Taxonomy: Applying/Analyzing
43) Describe how rho-dependent termination occurs in bacteria.
Answer: A bacterial protein called rho factor binds to an mRNA at the rut site. It moves along
the mRNA in a 5′-to-3′ direction chasing after the RNA polymerase. When it reaches the hairpin
loop it removes it and then proceeds to break through the hydrogen bonds holding the RNADNA together, which successfully removes the RNA polymerase.
Section: 13.9
Bloom's Taxonomy: Remembering/Understanding
44) Describe a difference between the RNA polymerases of eukaryotes and prokaryotes.
Answer: In eukaryotes, three polymerases (I, II, and III) have been identified; only one has been
described in prokaryotes.
Section: 13.10
Bloom's Taxonomy: Remembering/Understanding
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45) 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: 13.10
Bloom's Taxonomy: Remembering/Understanding
46) What are four ways that mRNA can be modified?
Answer: (1) Adding a modified guanine to the 5′ end, (2) adding a polyadenylation tail to the 3′
end, (3) splicing of introns, and (4) RNA editing.
Section: 13.10
Bloom's Taxonomy: Remembering/Understanding
47) List four possible reasons that explain the importance of introns.
Answer: (1) introns allow for alternative splicing, (2) exon shuffling, which can aid in evolution
of new genes, (3) the intron can be a microRNA itself that can regulated gene expression, and (4)
introns can contain enhancer and silencer sequences that aid in gene regulation.
Section: 13.11
Bloom's Taxonomy: Remembering/Understanding
48) What is meant by the term heterogeneous nuclear RNA (hnRNA)?
Answer: pre-mRNA, primary transcripts before processing in eukaryotes
Section: 13.11
Bloom's Taxonomy: Remembering/Understanding
49) A researcher is looking at a cellular sample using an electron microscope. They observe that
while mRNA is being transcribed from the DNA that ribosomes are binding to the mRNA to
make proteins. Is this sample from a bacterial or a eukaryotic cell?
Answer: Bacteria. Since bacteria lack a nucleus, the ribosomes can bind to the mRNA as it's
being synthesized and initiate translation.
Section: 13.13
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 14 Translation and Proteins
1) A bacterium's large subunit is 60S and its small subunit is 40S. What is the size of the
ribosome containing both subunits?
A) 70S
B) 80S
C) 20S
D) 100S
E) 120S
Answer: B
Section: 14.1
Bloom's Taxonomy: Remembering/Understanding
2) Which of the following is true regarding the tRNA structure?
A) An amino acid binds to the 5′ end of the tRNA molecule.
B) The anticodon is found at the 3′ end of the tRNA molecule.
C) A tRNA molecule that has an amino acid attached to it is called a charged tRNA.
D) The nucleotides found in a tRNA molecule can only be adenine, uracil, guanine, and cytosine.
E) The tRNA binds to an mRNA's codon at the variable loop.
Answer: C
Section: 14.1
Bloom's Taxonomy: Remembering/Understanding
3) If humans had 25 amino acids instead of 20 amino acids then how many aminoacyl tRNA
synthetases would humans have?
A) 20
B) 3
C) 4
D) 25
E) 75
Answer: D
Section: 14.1
Bloom's Taxonomy: Applying/Analyzing
4) During initiation in bacterial translation, a particular mutation causes the premature binding of
the large ribosome subunit to the small ribosome subunit. Which component of initiation is not
working properly?
A) Shine-Dalgarno sequence on the mRNA
B) IF1
C) IF2
D) IF3
E) Ribosome's E site
Answer: D
Section: 14.2
Bloom's Taxonomy: Applying/Analyzing
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5) A mutation in EF-Tu would have a DIRECT effect on which part of translation?
A) The small and large subunits would not bind.
B) The peptide bond formation would be blocked.
C) The next tRNA would not be able to enter the A site of the ribosome.
D) Translocation would be directly blocked.
E) The polypeptide would form, but translation termination would be blocked.
Answer: C
Section: 14.2
Bloom's Taxonomy: Applying/Analyzing
6) What signals termination of translation in bacteria?
A) a tRNA specific to the stop codon enters the ribosome's A site
B) the stop amino acid is attached to the growing peptide chain
C) RF1 and RF2
D) EF-Tu and EF-G
E) ribozymes
Answer: C
Section: 14.2
Bloom's Taxonomy: Remembering/Understanding
7) A protein is 300 amino acids long. Which of the following could be the number of total
nucleotides in the section of DNA that codes for this protein? (Remember: DNA is doublestranded.)
A) 3
B) 100
C) 300
D) 900
E) 1800
Answer: E
Section: 14.2
Bloom's Taxonomy: Evaluating/Creating
8) The term peptidyl transferase relates to ________.
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: 14.2
Bloom's Taxonomy: Remembering/Understanding
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9) X-ray crystallography has revealed and confirmed many details about the ribosome. All of the
following are such details, EXCEPT ________.
A) the ribosome is a dynamic structure that changes during its different functional states
B) rRNAs are more likely than proteins to be the catalytic molecule of the ribosome
C) rRNA in the ribosome is in direct contact with the different loops of the tRNA molecule
D) the distance the tRNA must travel between the E, P, and A sites is considered to be relatively
short
E) the wobble hypothesis is further supported
Answer: D
Section: 14.3
Bloom's Taxonomy: Remembering/Understanding
10) Translation in bacterial and eukaryotic cells has many similarities, but there are also several
key differences. Which of the following is one of those differences that is seen in eukaryotes?
A) Translation and transcription are coupled.
B) Eukaryotic ribosomes are smaller with fewer proteins and RNA molecules.
C) Eukaryotes use the 5′ G-cap and Poly-A-tail on their mRNAs to initiate translation.
D) Eukartyotic mRNA contains a Shine—Dalgarno sequence that increases the efficiency of
translation.
E) Eukaryotes only require one release factor that recognizes all three stop codons.
Answer: C
Section: 14.4
Bloom's Taxonomy: Remembering/Understanding
11) 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: 14.6
Bloom's Taxonomy: Remembering/Understanding
12) 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: 14.6
Bloom's Taxonomy: Remembering/Understanding
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13) A researcher is studying the synthesis of a specific amino acid found in Neurospora. She
knows that the pathway begins with a precursor that is converted into the amino acid with two
known intermediates (Substance Blue and Substance Green). She accurately predicts that this
amino acid synthesis pathway is catalyzed by three enzymes, (I, II, and III). She subsequently
identifies three mutants that she calls Mutant I, Mutant II, and Mutant III. With the information
about the mutants below place the enzymes in the order that the enzymes act in this pathway.
(Note: the numbers I, II, and III don't necessarily indicate the order in which the enzymes
appear.)
Mutant I (only Enzyme I is mutated) is unable to synthesize the amino acid even if she provided
the mutant with both Substance Blue and Substance Green.
Mutant II (only Enzyme II is mutated) is able to synthesize the amino acid if she provided the
mutant with Substance Blue or Substance Green.
Mutant III (only Enzyme III is mutated) is able to make the amino acid if she provided the
mutant with Substance Green, but not if she provided the mutant with only Substance Blue.
A) I, II, III
B) II, III, I
C) III, II, I
D) II, I, III
E) III, I, II
F) More than one codon can specify a given amino acid.
Answer: B
Section: 14.6
Bloom's Taxonomy: Evaluating/Creating
14) 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: 14.7
Bloom's Taxonomy: Remembering/Understanding
15) Often point mutations can cause a protein to be made that differs in just one amino acid.
Which of the following changes is most likely to cause the greatest change in protein function?
A) a mutation that replaces a valine with an isoleucine
B) a mutation that replaces a glutamic acid with an aspartic acid
C) a mutation that replaces an asparagine with a glutamine
D) a mutation that replaces a proline with an aspartic acid
E) a mutation that replaces a serine with a cysteine
Answer: D
Section: 14.8
Bloom's Taxonomy: Applying/Analyzing
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16) 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: 14.8
Bloom's Taxonomy: Remembering/Understanding
17) The protein shown below provides an example of which level of protein folding?
A) primary
B) secondary
C) tertiary
D) quaternary
E) pentenary
Answer: D
Section: 14.8
Bloom's Taxonomy: Applying/Analyzing
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18) 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: 14.8
Bloom's Taxonomy: Remembering/Understanding
19) All of the following are examples of posttranslational modifications, EXCEPT________.
A) The N-terminal amino acid is often removed or modified.
B) Individual amino acid residues are sometimes modified, such as adding a phosphates, acetyl,
or methyl groups can be added.
C) Polypeptides can be degraded and then reassembled to produce entirely different sequenced
polypeptides.
D) Polypeptide chains can be cleaved to produce a shorter and functional polypeptide.
E) Prosthetic groups, such as metals or vitamins, can be added.
Answer: C
Section: 14.9
Bloom's Taxonomy: Remembering/Understanding
20) What is the name of the protein that helps fold other proteins into their final and functional
form?
A) chaperone
B) proteasomes
C) flippase
D) collagen
E) ubiquitin
Answer: A
Section: 14.9
Bloom's Taxonomy: Remembering/Understanding
21) Creutzfeldt-Jakob, Huntington, Alzheimer, and Parkinson disease are all characterized with
which part of protein synthesis?
A) initiation
B) elongation
C) termination
D) protein folding
E) N-terminal modification
Answer: D
Section: 14.9
Bloom's Taxonomy: Remembering/Understanding
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22) Which protein class directly controls many of the metabolic reactions within a cell?
A) structural proteins
B) repressor proteins
C) operator proteins
D) enzymes
E) hydrophilic proteins
Answer: D
Section: 14.10
Bloom's Taxonomy: Remembering/Understanding
23) Modular portions of a protein that fold into stable conformations with specific functional
capabilities are referred to as ________.
A) protein domains
B) protein secondary structures
C) protein quaternary structures
D) protein chaperones
E) protein subunits
Answer: A
Section: 14.11
Bloom's Taxonomy: Remembering/Understanding
24) Describe the structure of a tRNA molecule. Your answer should include a list of five
important sites on the tRNA and two reasons why tRNA's modified bases are important.
Answer: The five sites are amino acid binding site, ΤψC stem, variable loop, anticodon stem, D
stem, and acceptor stem. The modified bases are important in conferring structural stability and
play a role in hydroben bonding between tRNA and mRNA, which helps explain the wobble
effect.
Section: 14.1
Bloom's Taxonomy: Remembering/Understanding
25) Assume that a base addition occurs early in the coding region of a gene. Is the protein
product of this gene expected to have MORE or FEWER altered amino acids compared with the
original gene with a base deletion late in the coding region? Explain.
Answer: Since an addition or deletion of a base pair would change every codon after the
addition or deletion, then adding a base early in the coding region would change more amino
acids than deleting a base later in the coding region.
Section: 14.2
Bloom's Taxonomy: Applying/Analyzing
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26) The accompanying drawing represents simultaneous transcription and translation in E. coli.
The direction of the RNA polymerase is given by the arrow.
(a) Is the letter A nearer the 5′ or the 3′ end of the molecule?
(b) Is the letter B nearer the 5′ or the 3′ end of the molecule?
(c) Is the letter C nearer the 5′ or the 3′ end of the tRNA molecule?
(d) What is the "S" value for the large rRNA that is closest to the letter D?
(e) Which terminus (N or C) of the growing polypeptide chain is nearer to the letter E?
Answer: (a) 3′, (b) 5′, (c) 3′, (d) 23S, (e) N
Section: 14.2
Bloom's Taxonomy: Evaluating/Creating
27) Consider the roles of IF1, IF2, and IF3 during translation. Predict what the effect would be if
IF1, IF2, and IF3 were mutated.
Answer: If IF1 was mutated then the second tRNA would bind prematurely to the A site. If IF2
was mutated then the fMET tRNA would not be transferred to the P site of the small ribosome
subunit. If IF3 was mutated then the large ribosomal subunit would prematurely bind to the small
subunit.
Section: 14.2
Bloom's Taxonomy: Applying/Analyzing
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28) 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 and the effects on individual proteins.
(a) An acridine dye—induced mutation (adds or deletes single bases in DNA) leads to an mRNA
for one protein-producing gene. The condition is heterozygous in the involved cell.
(b) A deletion (homozygous) that removes approximately half of the 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 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: 14.2
Bloom's Taxonomy: Evaluating/Creating
29) What are polyribosomes and why are they important?
Answer: Polyribosomes are clusters of ribosomes held together on an mRNA molecule. This
allows for multiple proteins to be synthesized from a single mRNA before it is degraded. This is
particularly important in bacteria since they don't protect their mRNA.
Section: 14.2
Bloom's Taxonomy: Remembering/Understanding
30) Provide five reasons why translation is more complex in eukaryotes compared to bacteria.
Answer: There are several possible answers, but could include the following reasons. (1) In
eukaryotes, mRNA needs to be transported out of the nucleus prior to it being translated. (2)
Eukaryotic ribosomes are larger and are composed of more proteins and rRNA molecules. (3)
Translation initiation in bacteria requires the mRNA contain the Shine—Dalgarno sequence.
However, in eukaryotes, a more complex initiation process is needed that requires the 5'-G cap
and the 3'-poly-A tail. (4) Translation initiation in eukaryotes requires many more initiation
factors. (5) There are two tRNAs for methionine in eukaryotes: one for the initiating AUG codon
and a different one used for all other AUG codons. (6) Termination in eukaryotes uses one of
three possible Release Factors (one for each stop codon) while bacteria have a total of two
release factors.
Section: 14.4
Bloom's Taxonomy: Remembering/Understanding
31) Explain why a eukaryotic mRNA that has begun to be degraded from the 3' end, but still has
its complete coding region intact is not efficiently translated.
Answer: Eukaryotic translation uses a closed-loop process where the 5' G cap and the 3' poly Atail of the mRNA loop around and interact with the ribosome to initiate translation. If the 3' polyA tail has begun to degrade then translation will not efficiently be initiated.
Section: 14.4
Bloom's Taxonomy: Applying/Analyzing
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32) Given the accompanying table:
Supplement
Strain A
Strain B
Strain C
Strain D
1
+
+
+
-
2
+
+
-
3
+
+
+
+
4
+
-
Where numbers 1 through 4 indicate four supplements that must be added to sustain growth of
the organism. Determine a possible metabolic pathway that would give the results seen for the
four mutant strains, A through D.
Answer:
X → +→ 4 → +→ 2 → +→1 → +→ 3
↑
↑
↑
↑
C
B
A
D
Section: 14.6
Bloom's Taxonomy: Evaluating/Creating
33) 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 Supplement with
ICP 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.
(c) In bacteria it is often possible to make partial diploid strains. Assume that a diploid strain
was made containing the complete genomes of the trp-2 and trp-1 strains. Would this diploid
strain be able to grow on the unsupplemented medium? Explain your answer.
Answer:
(a) →+→>AA→+→>IGP→+→>IN→+→?Try
→+→>AA→+→>IGP→+→IN→+→Try
↑
↑
↑
↑
(b) 8
2
3
1
(c) Yes, complementation will occur because the trp-2 strain is also trp-1+ and the trp-1 strain is
also trp-2+.
Section: 14.6
Bloom's Taxonomy: Evaluating/Creating
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34) 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.
Answer:
succinate →+→ fumarate →+→ malate →+→ oxaloacetate
↑
↑
↑
l41
l62
l36
Section: 14.6
Bloom's Taxonomy: Evaluating/Creating
35) 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: 14.6
Bloom's Taxonomy: Remembering/Understanding
36) 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: 14.6
Bloom's Taxonomy: Remembering/Understanding
37) Sickle-cell anemia is caused when one nucleotide is changed to another nucleotide in the
hemoglobin gene. How does this one small change cause the life-threatening disease of sicklecell anemia?
Answer: Substituting this one nucleotide changes the amino acid from glutamic acid codon to
valine. Glutamic acid is a polar negatively charged amino acid whereas valine is a nonpolar.
Even though it is only a single base change, the result is a major change in the amino acid that
allows the hemoglobin protein to stick to other hemoglobin proteins that results in the sickling of
the red blood cell.
Section: 14.7
Bloom's Taxonomy: Remembering/Understanding
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38) Sickle-cell anemia is caused by a mutation that changes the second nucleotide of the codon
from GAG to GUG. This results in a change of the amino acid from glutamic acid to valine. If
the mutation affected the third nucleotide of the codon and changed GAG to GAU then the
amino acid would change from a glutamic acid to aspartic acid. Would you predict this change to
have the same effect?
Answer: A change in one polar-charged amino acid to another polar-charged amino acid would
be predicted to have less of a phenotype compared to a mutation that resulted in a change from a
polar-charged amino acid to a nonpolar amino acid. Changes to the third nucleotide of a codon
are often less damaging than changes to the first or second nucleotide.
Section: 14.8
Bloom's Taxonomy: Applying/Analyzing
39) 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: 14.8
Bloom's Taxonomy: Remembering/Understanding
40) 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: 14.8
Bloom's Taxonomy: Remembering/Understanding
41) Assuming that an amino acid sequence is 250 amino acids long, how many different
molecules, each with a unique sequence, could be formed?
Answer: 20250
Section: 14.8
Bloom's Taxonomy: Applying/Analyzing
42) 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: 14.8
Bloom's Taxonomy: Remembering/Understanding
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43) 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: 14.8
Bloom's Taxonomy: Applying/Analyzing
44) List at least three posttranslational modifications known to occur as a newly synthesized
protein matures.
Answer: N-terminal removal or modification of an amino acid, modification of some amino
acids, addition of carbohydrate side chains, trimming of polypeptide chains, removal of signal
sequences, addition of metals
Section: 14.9
Bloom's Taxonomy: Remembering/Understanding
45) The diseases bovine spongiform encephalopathy and Creutzfeldt—Jakob disease are caused
by what kind of protein? How does this protein cause these diseases?
Answer: These diseases are caused by the prion protein. The healthy prion (PrPC) is
characterized by the secondary structure of alpha helices whereas the diseased prion (PrPSc) is
characterized by the secondary structure of beta sheets.
Section: 14.10
Bloom's Taxonomy: Remembering/Understanding
46) Which class of protein functions primarily by lowering the energy of activation during a
reaction?
Answer: Enzymes
Section: 14.10
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 15 Gene Mutation, DNA Repair, and Transposition
1) In bacteria and eukaryotes, a mutation is when ________.
A) the amino acid sequence in a protein molecule is directly changed
B) the nucleotide sequence in an mRNA molecule is directly changed
C) the nucleotide sequence in a DNA molecule is directly changed
D) the expression of a gene changes without the DNA sequence being changed
E) DNA is unable to be replicated
Answer: C
Section: 15.1
Bloom's Taxonomy: Remembering/Understanding
2) A mutation that changes an amino acid-coding codon to a stop codon is classified as what
kind of codon?
A) missense
B) nonsense
C) neutral
D) silent
E) conditional
Answer: B
Section: 15.1
Bloom's Taxonomy: Remembering/Understanding
3) Which of the following mutations would be classified as a transversion?
A) CGA to a CGT
B) CGA to a CGU
C) CGA to a GGA
D) CGA to a CCA
E) CGA to GCT
Answer: A
Section: 15.1
Bloom's Taxonomy: Applying/Analyzing
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4) A geneticist is studying a particular Drosophila mutant that produces a phenotype of "no
wing" due to a mutation in a gene called wingless. She takes the mutant wingless and
mutagenizes it. She hypothesizes that any fly that emerges with wings will be due to the original
mutated gene reverting back to the wild-type allele. From this study, she identifies a fly that now
has wings. When she sequences the wingless gene, she discovers that the original mutation is still
there, and there are no other mutations in the wingless gene. However, a second gene at a
separate locus is mutated. The second mutation is best classified as a(n) ________.
A) conditional mutation
B) intragenic suppressor mutation
C) haploinsufficiency mutation
D) intergenic suppressor mutation
E) null mutation
Answer: D
Section: 15.1
Bloom's Taxonomy: Evaluating/Creating
5) Germ cell mutations ________.
A) are of no consequence to the next generation
B) affect somatic cell function in the current generation
C) are usually found in every cell of the next generation
D) are the cause of testicular cancer
E) are only seen in females
Answer: C
Section: 15.1
Bloom's Taxonomy: Remembering/Understanding
6) 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: 15.1
Bloom's Taxonomy: Remembering/Understanding
7) Mutations that arise in nature, from no particular artificial agent, are called ________.
A) nutritional mutations
B) induced mutations
C) spontaneous mutations
D) chromosomal losses
E) cosmic mutations
Answer: C
Section: 15.2
Bloom's Taxonomy: Remembering/Understanding
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8) At what stage is a mutation predicted to have the effect?
A) after birth but before puberty
B) after puberty
C) after the age of 60
D) shortly after birth
E) during early embryonic development
Answer: E
Section: 15.2
Bloom's Taxonomy: Remembering/Understanding
9) From the Luria-Delbruck fluctuation test, when would mutations arise in yeast cells to allow
for survival in the presence of toxic levels of hydrogen peroxide?
A) Mutations are adaptive and would have appeared immediately in response to the hydrogen
peroxide.
B) Mutations are random and would have appeared prior to being exposed to hydrogen peroxide.
C) Mutations are adaptive and would have appeared after most of the cells died in the presence
of hydrogen peroxide.
D) Mutations are random but would not have formed until after the hydrogen peroxide was
introduced.
E) Mutations are adaptive and would have formed prior to being exposed to hydrogen peroxide.
Answer: B
Section: 15.2
Bloom's Taxonomy: Applying/Analyzing
10) Tautomeric shifts ________.
A) are known to cause insertion or deletion mutations
B) are likely to cause frameshifts
C) can cause missense mutations due to mispairings
D) allow purines to bind to purines
E) allow pyrimidines to bind to pyrimidines
Answer: C
Section: 15.3
Bloom's Taxonomy: Remembering/Understanding
11) Which of the following two mutagens that would be classified as base analogs?
A) acridine orange and proflavine
B) ethylmethane sulfonate and ethylmethylketone peroxide
C) ultraviolet light and cosmic radiation
D) 5-Bromouracil and 2-amino purine
E) hydroxyurea and peroxidase
Answer: D
Section: 15.4
Bloom's Taxonomy: Remembering/Understanding
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12) Which f the following mutagens is matched correctly with its action?
A) 5-Bromouracil is an alkylating agent
B) 2-Amino purine is an adduct-forming agent
C) mustard gasses are base analogs
D) ethidium bromide is an intercalating agent
E) acetaldehyde is a free radicle
Answer: D
Section: 15.4
Bloom's Taxonomy: Remembering/Understanding
13) Thymine dimers form when ________.
A) two thymines on separate strands of DNA bind to together
B) two thymines on the same strand but not adjacent to each other bind together
C) two thymines on the same strand and adjacent to each other bind together
D) two thymines adjacent to each other are deleted from the DNA sequence
E) two thymines are inserted into a DNA sequence
Answer: C
Section: 15.4
Bloom's Taxonomy: Remembering/Understanding
14) Which of the following diseases are caused by an expansion of trinucleotide repeats?
A) beta-thalassemia
B) fragile X syndrome
C) cystic fibrosis
D) achondroplasia
E) Marfan's syndrome
Answer: B
Section: 15.5
Bloom's Taxonomy: Remembering/Understanding
15) Adenine methylase is an enzyme used during mismatch repair (MMR). It helps determine
which strand needs to be repaired by discriminating between template and new DNA strand. If
this enzyme is mutated, how would MMR be affected?
A) MMR would be hindered because both strands would be unmethylated and both would appear
as template DNA strands.
B) MMR would be hindered because both strands would be overmethylated and both would
appear as template DNA strands and it would not be possible to determine which strand has the
mutation.
C) MMR would be hindered because both strands would be unmethylated and both would appear
as new DNA strands and it would not be able to determine which strand has the error.
D) MMR would be hindered because both strands would be overmethylated and both would
appear as new DNA strands and it would not be able to determine which strand has the error.
E) MMR would not be hindered because adenine methylase is not directly involved in MMR.
Answer: C
Section: 15.6
Bloom's Taxonomy: Applying/Analyzing
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16) One would predict that bacteria that lack the enzyme photolyase would be more sensitive to
what kind of mutagen?
A) UV light
B) base analog
C) intercalating agents
D) reactive oxygen species
E) alkylating agents
Answer: A
Section: 15.6
Bloom's Taxonomy: Applying/Analyzing
17) Which repair mechanism is most likely affected if the enzyme DNA glycosylase is not
functioning properly?
A) photoreactivation repair
B) base excision repair
C) nucleotide excision repair
D) SOS repair
E) double-strand break repair
Answer: B
Section: 15.6
Bloom's Taxonomy: Remembering/Understanding
18) 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: 15.6
Bloom's Taxonomy: Remembering/Understanding
19) During the Ames test, his− bacteria are exposed to a chemical. If the bacteria remains his+,
then what can we say about the chemical?
A) The chemical is safe to consume and is not likely a mutagen.
B) The chemical is safe to consume even though it is likely a mutagen.
C) The chemical is not safe to consume because it is likely a mutagen.
D) The chemical is not safe to consume even though it is not likely a mutagen.
E) The chemical is certainly a cancer-causing chemical.
Answer: C
Section: 15.7
Bloom's Taxonomy: Applying/Analyzing
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20) Transposons, or jumping genes, are DNA elements that move within the genome. In which
organismic groups are transposons found?
A) bacteria
B) eukaryotes
C) mammals
D) ancient bacteria
E) all organismic groups
Answer: E
Section: 15.8
Bloom's Taxonomy: Remembering/Understanding
21) All insertion sequences (IS elements) contain two structural elements 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: 15.8
Bloom's Taxonomy: Remembering/Understanding
22) Some bacterial transposons, known as Tn elements, are larger than insertion sequences (IS
elements) and contain protein-coding genes that have human health significance. What might
such a bacterial transposon contain?
A) drug resistance
B) oncogene
C) pseudogene
D) proto-oncogene
E) dissociation element
Answer: A
Section: 15.8
Bloom's Taxonomy: Remembering/Understanding
23) 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) While 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: 15.8
Bloom's Taxonomy: Remembering/Understanding
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24) Compare and contrast between the following types of mutations: missense, nonsense, silent,
and neutral.
Answer: All of these mutations base substitutions (or point mutation); however, their effects are
different. Missense mutations change one amino acid to another amino acid; nonsense mutations
change an amino acid-coding codon to a stop codon; silent mutations change an amino acidcoding codon to a codon that codes for the same amino acid (due to the degeneracy of the genetic
code); and a neutral mutation changes a codon to a different amino acid, but it has no effect on
the protein's function.
Section: 15.1
Bloom's Taxonomy: Remembering/Understanding
25) A heterozygote could have a variety of phenotypes depending on whether or not the mutation
is recessive, dominant, or haploinsufficient. In this example, the gene A is necessary to form a
tail in mice. If the mouse is a heterozygote (A/a), then what would you predict the phenotype to
be if the mutation was recessive, dominant, or haploinsufficient.
Answer: If the mutation is recessive, then the phenotype should be similar to wild type, and tail
development should be unaffected. If the mutation is dominant, then the A/a mouse would likely
have no tail. If the mutation is haploinsufficient, then one might predict an abnormal tail
(perhaps shorter or misshaped).
Section: 15.1
Bloom's Taxonomy: Applying/Analyzing
26) 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 the 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 of the proteins
would be made in their normal form, but at reduced levels.
Section: 15.1
Bloom's Taxonomy: Applying/Analyzing
27) 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 the new mutation is dominant and 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: 15.2
Bloom's Taxonomy: Applying/Analyzing
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28) How was it determined that X-rays are mutagenic?
Answer: H. J. Muller determined that a direct relationship occurred between X-linked recessive
lethals and X-ray dose in Drosophila.
Section: 15.2
Bloom's Taxonomy: Remembering/Understanding
29) Explain the effects of a somatic mutation on the individuals and their offspring if the
mutation is recessive or dominant.
Answer: Somatic cell mutations can only affect the individuals they occur in and will not affect
their offspring. Whether or not the mutation affects the individuals can vary depending upon
whether or not the mutation is recessive or dominant. If the mutation is recessive, then it is less
likely to affect the individuals unless the wild-type allele also becomes mutated or if the
recessive mutation occurs on a male's X-chromosome. If the mutation is dominant, then it is
more likely to affect the individuals. However, even dominant mutations' phenotypes may have
less impact due to the thousands of other cells in the tissue that remain unmutated. It should be
noted that an accumulation of several mutations in a single cell could lead to diseases such as
cancer.
Section: 15.2
Bloom's Taxonomy: Remembering/Understanding
30) In a survey of 240,000 human births, six achondroplastic births were recorded to parents who
were unaffected. Given that this form of dwarfism is caused by a fully penetrant, dominant,
autosomal gene, what is the mutation rate?
Answer: 6/(4.8 × 105)
Section: 15.2
Bloom's Taxonomy: Applying/Analyzing
31) 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: 15.3
Bloom's Taxonomy: Remembering/Understanding
32) Describe the mutagenic action of the following two mutagens: 5-bromouracil and ultraviolet
light.
Answer: The mutagen 5-bromouracil is an analog of thymine, which anomalously pairs with
guanine. Ultraviolet light causes thymine dimers.
Section: 15.4
Bloom's Taxonomy: Remembering/Understanding
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33) Some mutagens cause genetic changes that can be "corrected" by reexposing cells to the
same mutagen. Other mutagens do not behave in this way. Provide one example of each of these
two types of agents and describe the mutational changes caused in DNA. Explain why some
mutagens behave in one way, whereas others do not.
Answer: Mutagens that cause base substitutions are "corrected" by mutagens of the same class
(nitrous acid, 2-aminopurine, and 5-bromouracil). Frameshift mutations are "corrected" by the
same class of frameshift mutagens, but not by mutagens that cause base substitutions. X-rays
cause major structural changes in chromosomes (deletions, translocations, etc.) and are not
"corrected" by any mutagen, including X-rays.
Section: 15.4
Bloom's Taxonomy: Remembering/Understanding
34) Considering the electromagnetic spectrum, identify likely mutagens from the following list:
radio waves, microwaves, infrared, ultraviolet, X-rays, gamma rays, and cosmic rays.
Answer: ultraviolet, X-rays, gamma rays, and cosmic rays
Section: 15.4
Bloom's Taxonomy: Remembering/Understanding
35) In World War I, mustard gas was used as a chemical weapon, and at a high dose, it would
cause death. However, it is now known to be a chemical mutagen. Explain how those who came
in contact with mustard gas and survived have an increase chance of developing mutations.
Answer: Mustard gas is an alkylating agent that donates an alkyl group to an amino or keto
group on nucleotides. This causes mismatch pairing and increases transition mutations.
Section: 15.4
Bloom's Taxonomy: Remembering/Understanding
36) Three human disorders—fragile X syndrome, myotonic dystrophy, and Huntington's
disease–are conceptually linked by a common mode of molecular upset. Describe the phenomena
that link these disorders.
Answer: All three 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).
Section: 15.5
Bloom's Taxonomy: Remembering/Understanding
37) 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: 15.6
Bloom's Taxonomy: Remembering/Understanding
38) Describe the action of the enzyme transposase.
Answer: Transposase makes staggered cuts in chromosomal DNA, into or out of which an IS
element can insert.
Section: 15.8
Bloom's Taxonomy: Remembering/Understanding
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39) How were insertion sequences (IS elements) first discovered?
Answer: IS elements were first discovered during the analysis of mutations in the gal operon of
E. coli.
Section: 15.8
Bloom's Taxonomy: Remembering/Understanding
40) What are LINES?
Answer: LINES are long interspersed elements that are plentiful in the human genome.
Section: 15.8
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 16 Regulation of Gene Expression in Bacteria
1) What term would be applied to a regulatory condition that occurs when a protein is associated
with a particular section of DNA and greatly reduces transcription?
A) negative control
B) positive control
C) induction
D) activation
E) stimulation
Answer: A
Section: 16.1
Bloom's Taxonomy: Remembering/Understanding
2) Enzymes that are continuously produced are said to have a(n) ________ expression.
A) inducible
B) repressible
C) constitutive
D) negative
E) positive
Answer: C
Section: 16.1
Bloom's Taxonomy: Remembering/Understanding
3) What term refers to clusters of bacterial genes that are under the control of a single regulatory
region?
A) autotroph
B) prototroph
C) operon
D) allosteric
E) attenuation
Answer: C
Section: 16.2
Bloom's Taxonomy: Remembering/Understanding
4) In the lac operon, the product of structural gene lacZ is capable of ________.
A) binding to the operator of the lac operon
B) forming lactose from two glucose molecules
C) transporting lactose into the cell
D) splitting the linkage of lactose to produce glucose and galactose
E) forming ATP from pyruvate
Answer: D
Section: 16.2
Bloom's Taxonomy: Remembering/Understanding
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5) 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: 16.2
Bloom's Taxonomy: Applying/Analyzing
6) To translate the three enzymes of the lac operon, ________ mRNA molecule(s) is/are
transcribed.
A) 0
B) 1
C) 2
D) 3
E) 4
Answer: B
Section: 16.2
Bloom's Taxonomy: Remembering/Understanding
7) 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: 16.2
Bloom's Taxonomy: Applying/Analyzing
8) Which binds to the operator of the lac operon?
A) permease
B) activator
C) beta galactosidase
D) repressor
E) lactose
Answer: D
Section: 16.2
Bloom's Taxonomy: Remembering/Understanding
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9) A protein, like the lac repressor, is said to be ________ because when it binds lactose, it
structurally changes and its chemical activity also changes.
A) enzymatic
B) structural
C) allosteric
D) multimeric
E) digestive
Answer: C
Section: 16.2
Bloom's Taxonomy: Remembering/Understanding
10) 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: 16.2
Bloom's Taxonomy: Applying/Analyzing
11) 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: 16.3
Bloom's Taxonomy: Remembering/Understanding
12) What is the name of the molecule that binds to CAP to activate transcription of the lac
operon?
A) glucose
B) cAMP
C) galactose
D) lactose
E) ATP
Answer: B
Section: 16.3
Bloom's Taxonomy: Remembering/Understanding
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13) When both lactose and glucose are available to E. coli, which part of the lac operon
regulation assures that glucose will be metabolized first?
A) The repressor binds to the operator in the absence of lactose.
B) The repressor cannot bind to the operator in the presence of lactose.
C) Permease allows lactose to enter the cell when lactose enters the cell.
D) CAP binds to cAMP and then to the promoter only in the presence of low glucose.
E) RNA polymerase transcribes the lac operon to make a single RNA molecule.
Answer: D
Section: 16.3
Bloom's Taxonomy: Remembering/Understanding
14) Through genetic analysis, Jacob and Manod were able to make predictions about the lac
operon. However, which of the following techniques confirmed their predictions?
A) DNA sequencing
B) PCR
C) crystal structure analysis
D) recombinant DNA technology
E) CRISPR
Answer: C
Section: 16.4
Bloom's Taxonomy: Remembering/Understanding
15) 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: 16.5
Bloom's Taxonomy: Remembering/Understanding
16) Genetic regulation in bacteria can involve alterations in RNA secondary structure. What
phenomenon occurs in the trp operon that involves such alterations?
A) transcription
B) capping
C) polyadenylation of the 3′ end of the mRNAs
D) intron processing
E) attenuation
Answer: E
Section: 16.6
Bloom's Taxonomy: Remembering/Understanding
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17) 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: 16.6
Bloom's Taxonomy: Applying/Understanding
18) What two important domains within a riboswitch involve the ligand-binding site?
A) aptamer and expression platform
B) A site and B site
C) B site and C site
D) alpha and beta
E) conformation and anticonformation
Answer: A
Section: 16.6
Bloom's Taxonomy: Remembering/Understanding
19) Small noncoding RNAs regulate gene expression in bacteria by ________.
A) activating transcription
B) repressing transcription
C) inhibiting translation
D) activating translation
E) by both inhibiting and activating translation
Answer: E
Section: 16.6
Bloom's Taxonomy: Remembering/Understanding
20) Compare and contrast positive and negative controls 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: 16.1
Bloom's Taxonomy: Remembering/Understanding
21) Present an overview of bacterial gene regulation in terms of growth efficiency.
Answer: Genetic systems have evolved that allow for "in-house" production of growth
substances when not supplied in the environment or when in full supply. When needed
substances are in full supply, such genetic systems are repressed.
Section: 16.1
Bloom's Taxonomy: Remembering/Understanding
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22) In the accompanying diagram, what type of control, positive or negative, is operating? The
oval structure is RNA polymerase.
Answer: Positive
Section: 16.1
Bloom's Taxonomy: Remembering/Understanding
23) 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 so that it can't bind to
the operator (a mutation in the repressor that prevents lactose binding would have the opposite
effect), or the operator is mutated so that it will not interact with the repressor.
Section: 16.2
Bloom's Taxonomy: Applying/Analyzing
24) 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: 16.2
Bloom's Taxonomy: Remembering/Understanding
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25) (a) Describe by labeled diagram the structural components of the lac operon in E. coli.
(b) State the function of the lac regulator gene.
(c) State the function of β-galactosidase in the lac system.
(d) Show by diagram the manner in which lactose brings about transcription of the three
structural genes of the lac operon.
(e) Explain why certain mutations in the regulator gene (I−) of the lac system result in maximal
synthesis of β-galactosidase, permease, and transacetylase even in the absence of the inducer
(lactose).
Answer: (a) See appropriate diagrams in the text. (b) The regulator gene produces a repressor
protein that interacts with the operator to shut off transcription. In the presence of lactose, the
repressor protein does not interact with the operator. (c) β-Galactosidase cleaves the lactose
sugar into its components, glucose and galactose. (d) See appropriate diagrams in the text. (e)
Such mutations provide modified proteins that are unable to associate with the operator to shut
off transcription.
Section: 16.2
Bloom's Taxonomy: Remembering/Understanding
26) 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 ) Is 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: 16.2
Bloom's Taxonomy: Applying/Analyzing
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27) The accompanying table 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.
Genotype
(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) Is Oc Z- /F'Is 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: 16.2
Bloom's Taxonomy: Applying/Analyzing
28) 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: 16.2
Bloom's Taxonomy: Remembering/Understanding
29) When lactose is first introduced into the medium, the lac operon is not induced and thus
permease isn't induced. How it is possible for lactose to enter the cell before the lac operon is
induced?
Answer: While the lac operon is not induced yet, there is low level expression most of the time
which allows a small amount of permease to be expressed. This allows lactose to enter the cell at
low levels until the lac operon is needed for full induction.
Section: 16.2
Bloom's Taxonomy: Evaluating/Creating
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30) A constitutive mutation in the lac operon may be of several types. Name two types of
constitutive mutations and explain why they are constitutive.
Answer: lacI− is a lac operon constitutive mutation because it is unable to bind to the operator
DNA. If it can't bind, then the operon will not be inhibited. Similarly, the lacOc mutation causes
a constitutive expression of the lac operon genes. This is because the lacOc mutation removes
the operator and without an operator the operon cannot be turned off.
Section: 16.2
Bloom's Taxonomy: Applying/Analyzing
31) 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: 16.2
Bloom's Taxonomy: Remembering/Understanding
32) Explain why lacOc mutations are cis-acting while lacI mutations can be trans-acting.
Answer: The operator region does not produce a diffusible product, whereas the lacI gene does.
Section: 16.2
Bloom's Taxonomy: Remembering/Understanding
33) 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: 16.2
Bloom's Taxonomy: Applying/Analyzing
34) The cAMP-CAP complex and RNA polymerase bind more efficiently to the lac operon
together than either does alone. What term is applied to this increased efficiency of binding?
Answer: Cooperative binding
Section: 16.3
Bloom's Taxonomy: Remembering/Understanding
35) 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: 16.3
Bloom's Taxonomy: Remembering/Understanding
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36) 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: 16.3
Bloom's Taxonomy: Applying/Analyzing
37) 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 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: 16.5
Bloom's Taxonomy: Applying/Analyzing
38) What is an allosteric molecule?
Answer: One that can change in shape and chemical activity while interacting with other
molecules.
Section: 16.5
Bloom's Taxonomy: Remembering/Understanding
39) State whether the following statement is true or false, then give your reasoning. The
terminating "hairpin" loop occurs in the trp operon when sufficient tryptophan is present.
Answer: True; the hairpin loop terminates transcription because when tryptophan is present, it is
energetically favorable to prevent more of its synthesis.
Section: 16.5
Bloom's Taxonomy: Remembering/Understanding
40) 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: Attenuation occurs in which transcription of the operon is terminated.
Section: 16.6
Bloom's Taxonomy: Remembering/Understanding
41) Explain how small noncoding RNAs (sRNAs) can function as both a negative regulator and
a positive regulator.
Answer: sRNAs can negatively regulate translation by binding to ribosome-binding sequences
of mRNA. This blocks the ability for ribosomes to bind to and initiate translation. sRNAs can
positively regulate translation by disrupting hairpin loops that usually cover and inhibit the
ribosome-binding sequence. However, unmasking this area, the ribosomes are free to bind and
activate translation.
Section: 16.6
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 17 Transcriptional Regulation in Eukaryotes
1) 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: 17.1
Bloom's Taxonomy: Remembering/Understanding
2) The nuclear regions that contain RNA polymerase and transcription regulatory molecules that
help compartmentalize eukaryotic regulation are called ________.
A) chromosome territories
B) interchromatin compartments
C) transcription factories
D) nucleosomes
E) histones
Answer: C
Section: 17.2
Bloom's Taxonomy: Remembering/Understanding
3) In a particular organism, there are two similar genes called YFG1 and YFG2. YFG1 is
expressed in the liver and not in the pancreas, and YFG2 is expressed in the pancreas but not the
liver. Neither YFG1 nor YFG2 is expressed in the heart. Based on these observations, which of
the following is most true?
A) The YFG1 gene isolated from the pancreas and the YFG2 gene isolated from the liver would
be DNase I sensitive.
B) The YFG1 gene isolated from the liver and YFG2 gene isolated from the pancreas would be
DNase I sensitive.
C) Both YFG1 and YFG2 genes isolated from the heart would be DNase I sensitive.
D) Both YFG1 and YFG2 genes isolated from any tissue (the liver, the pancreas, and the heart)
would be DNase I sensitive.
E) Neither YFG1 nor YFG2 isolated from any tissue (the liver, the pancreas, and the heart) would
be DNase I sensitive.
Answer: C
Section: 17.2
Bloom's Taxonomy: Applying/Analyzing
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4) 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: 17.2
Bloom's Taxonomy: Remembering/Understanding
5) DNA is methylated on which nucleotides?
A) guanine
B) cytosine
C) adenine
D) thymine
E) uracil
Answer: B
Section: 17.2
Bloom's Taxonomy: Remembering/Understanding
6) If a promoter region is mutated in such a way that it can no longer be methylated, what would
the most likely effect be?
A) The gene linked to that promoter would be over expressed.
B) The gene linked to that promoter would not undergo replication.
C) The gene linked to the promoter would be under expressed.
D) The gene linked to the promoter would be expressed at regular levels.
E) The gene linked to the promoter would still be expressed, but the protein should contain
different amino acids.
Answer: A
Section: 17.2
Bloom's Taxonomy: Applying/Analyzing
7) Considering the location of genes in the interphase nucleus, certain chromosomal territories
appear to exist. Specifically, ________.
A) each chromosome appears to occupy a discrete domain
B) gene-poor regions of chromosomes are located outside the nucleus, whereas gene-rich regions
are located inside the nucleus
C) even-numbered chromosomes are located in the interior of the nucleus, whereas oddnumbered chromosomes are located peripherally
D) large chromosomes are more likely to be located in the center of the nucleus
E) small chromosomes are more likely to be located in the center of the nucleus
Answer: A
Section: 17.2
Bloom's Taxonomy: Remembering/Understanding
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8) 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: 17.3
Bloom's Taxonomy: Remembering/Understanding
9) 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: 17.2
Bloom's Taxonomy: Applying/Analyzing
10) What is the general position of the consensus sequence called the GC box? What is its
sequence?
A) promoter; CAAT
B) promoter; GGGCGG
C) terminator; CAAT
D) terminator; GGGCGG
E) attenuator; GGGCGG
Answer: B
Section: 17.3
Bloom's Taxonomy: Remembering/Understanding
11) Which process seems to be the most similar between eukaryotic and prokaryotic genetic
regulations?
A) transcriptional regulation
B) RNA splicing regulation
C) intron/exon shuffling
D) 5'-capping regulation
E) poly(A) tail addition
Answer: A
Section: 17.3
Bloom's Taxonomy: Remembering/Understanding
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12) Which of the following clusters of terms apply when addressing enhancers or silencers 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: 17.3
Bloom's Taxonomy: Remembering/Understanding
13) 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: 17.3
Bloom's Taxonomy: Remembering/Understanding
14) Which of the following is a functional domain that is found in eukaryotic transcription
factors and bacteria?
A) basic leucine zipper motif
B) zinc finger motif
C) helix-turn-helix motif
D) basic leucine zipper and zinc finger motifs
E) basic leucine zipper, zinc finger, and helix-turn-helix motifs
Answer: C
Section: 17.4
Bloom's Taxonomy: Remembering/Understanding
15) What type of DNA sequence a repressor protein be expected to bind?
A) silencer
B) enhancer
C) insulator
D) TATA Box
E) GC Box
Answer: A
Section: 17.4
Bloom's Taxonomy: Remembering/Understanding
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16) The pre-initiation complex (PIC) contains several proteins. What would be 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: 17.5
Bloom's Taxonomy: Applying/Analyzing
17) In what way do upstream activator sequences (UASs), regulatory sequences in yeast, differ
from enhancers and silencers?
A) UASs function only downstream.
B) UASs function only upstream.
C) Enhancers function only downstream.
D) UASs function in the middle of transcription units.
E) UASs can function only in the 5'→ 3' direction.
Answer: B
Section: 17.6
Bloom's Taxonomy: Remembering/Understanding
18) The GAL gene system in yeast is used to metabolize galactose. In the presence of galactose,
the GAL genes are expressed; however, in the absence of galactose, the GAL genes are repressed.
A yeast mutant is discovered that continuously expresses the GAL genes in the presence or
absence of galactose. Which GAL gene is likely mutated?
A) GAL4
B) GAL180
C) GAL3
D) GAL1
E) GAL 10
Answer: B
Section: 17.6
Bloom's Taxonomy: Applying/Analyzing
19) UASs (upstream activating sequences) are DNase hypersensitive. This means that ________.
A) a UAS is constitutively open
B) more than one strand of DNA exists in each UAS
C) each UAS is likely to be single stranded
D) each UAS has more histone-binding sites than non-UAS sites
E) any given UAS is composed of a double-stranded site with a bound repressor
Answer: A
Section: 17.6
Bloom's Taxonomy: Applying/Analyzing
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20) The ENCODE Project's goal was to identify all functional DNA sequences within the human
genome. Which of the following is a finding from ENCODE?
A) ENCODE determined that approximately 15% of the human genome codes for a protein.
B) ENCODE determined that approximately 25% of the human genome has some sort of
biochemical function.
C) ENCODE was able to classify that a large part of the human genome is "Junk DNA."
D) ENCODE was able to determine that each enhancer exerts its regulatory function on only a
single gene.
E) ENCODE was able to determine that 75% of the genome is transcribed in at least one cell
type.
Answer: E
Section: 17.7
Bloom's Taxonomy: Remembering/Understanding
21) Explain five ways that eukaryotic gene regulation is more complex than bacterial gene
regulation?
Answer: (1) Eukaryotes have histones, whereas bacteria don't have histones and these histones
have to be removed to turn on gene expression. (2) Eukaryotes have a nucleus and therefore
transcription occurs in the nucleus and translation occurs in the cytoplasm, whereas bacterial
transcription and translation are coupled in the cytoplasm. (3) mRNAs in eukaryotes are
processed (splice, poly-adenylated, and addition of modified guanine) before they are translated,
whereas bacterial mRNA are not processed in a similar way. (4) mRNA in eukaryotes is
typically much longer lived, whereas bacterial mRNAs are degraded shortly after they are
transcribed. (5) Eukaryotes have a larger array of post-translational mechanisms than bacteria.
Section: 17.1
Bloom's Taxonomy: Remembering/Understanding
22) List at least three levels or types of genetic regulation in eukaryotes.
Answer: Transcriptional regulation, DNA methylation, posttranscriptional regulation, and
posttranslational regulation.
Section: 17.1
Bloom's Taxonomy: Remembering/Understanding
23) Describe how the presence or absence of nucleosomes influences gene transcription.
Answer: The binding of transcription factors requires accessing nucleosomal DNA. When
nucleosomes are tightly associated with the DNA transcription, proteins can't access the DNA;
however, when nucleosomes are loosely associated with the DNA, then the DNA transcription
proteins have easier access to the DNA. To turn genes on and off, the nucleosomes need to
modulate from being on or off the DNA.
Section: 17.2
Bloom's Taxonomy: Remembering/Understanding
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24) Approximately 5% 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: 17.2
Bloom's Taxonomy: Remembering/Understanding
25) In what way 5'-azacytidine can influence transcription?
Answer: Nitrogen is at the 5' position of 5'-azacytidine. The base can be incorporated into DNA
in place of cytidine, but it cannot be methylated. Under methylation of the CG dinucleotide
increases transcription.
Section: 17.2
Bloom's Taxonomy: Remembering/Understanding
26) In what way 5'-azacytidine is used (experimentally) in the treatment of sickle-cell anemia?
Answer: The use of 5'-azacytidine reduces the amount of methylation in certain hemoglobin
genes and enhances expression (transcription).
Section: 17.2
Bloom's Taxonomy: Applying/Analyzing
27) Some diseases are linked to the under or over expression of specific genes. A particular
cancer is characterized by the overexpression of a gene. Do you predict the histones associated
with this gene to be acetylated? Describe why this would lead to increased gene expression and
how histone acetyl transfer (HAT) and histone deacetylase (HDAC) could be involved.
Answer: One would predict that since this gene is overexpressed that the histones are acetylated.
Therefore, HAT is working or perhaps over active and HDAC is not working properly. When
histones are acetylated, the histones will relax their association with the DNA allowing the gene
to be more likely expressed.
Section: 17.2
Bloom's Taxonomy: Applying/Analyzing
28) Enhancers are said to be cis-acting. What is meant by cis-acting, and what are enhancers?
Answer: Cis-acting means that the genes under control must be in the same chromosome as the
cis-acting element. Enhancers are sections of DNA that activate transcription of other sections of
DNA on the chromosome.
Section: 17.3
Bloom's Taxonomy: Remembering/Understanding
29) Describe three characteristics of enhancers and silencers.
Answer: Position need not be fixed. Orientation may be inverted without significant effect. They
can regulate other genes when inserted. They can act at a great distance from the promoter.
Section: 17.3
Bloom's Taxonomy: Remembering/Understanding
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30) Describe the difference between a focused promoter and a dispersed promoter.
Answer: A focused promoter specifies transcription initiation at a single specific nucleotide,
whereas a dispersed promoter directs the initiation of transcription from a number of weak
transcription start sites spread over a fairly wide region (50-100 nucleotides).
Section: 17.3
Bloom's Taxonomy: Remembering/Understanding
31) Describe the function and general natures of promoters in eukaryotes. Include the three
different parts of the promoter and their positions from the 5' end of a eukaryotic gene.
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 (-25 to -30) and a CAAT box (-70 to -80).
Section: 17.3
Bloom's Taxonomy: Remembering/Understanding
32) Mutations in the promoter region of the β-globin gene indicate that some areas are more
sensitive than others. When mutations occur in consensus sequences (modular elements such as
GC box, CAAT box, and TATA box), does transcription usually increase or decrease? Explain.
Answer: Transcription should decrease because by changing these modular elements, the
general transcription factors cannot bind efficiently or recruit RNA polymerase.
Section: 17.3
Bloom's Taxonomy: Applying/Analyzing
33) On a specific chromosome, an enhancer helps activate the expression of Gene A, but not
Gene B, which is equally close to the enhancer. A mutant is identified where this enhancer is
now activating the expression of both Gene A and Gene B. Explain what has likely been
mutated.
Answer: Gene A is activated by the enhancer but not Gene B due to an insulator sequence that
allows the enhancer to only activate Gene A and prevents activation of Gene B. However, in this
mutant, the insulator is mutated and can no longer block the activation of Gene B by the
enhancer and thus both Genes A and B are activated by this enhancer.
Section: 17.3
Bloom's Taxonomy: Applying/Analyzing
34) Transcription factors are important molecules that regulate gene activity in eukaryotes. What
are the two general classes of transcription factors that exist in eukaryotes? Describe where they
bind and what their effects are.
Answer: General transcription factors assemble at promoter regions adjacent to the site of
transcription. They are typically important in basal levels of transcription. The other class
includes activators and repressors that bind at more distant regions called enhancer and repressor
sequences, respectively. Activators and repressors are responsible for increased regulation and
are often important in tissue specific and developmental regulation.
Section: 17.4
Bloom's Taxonomy: Remembering/Understanding
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35) Explain why a transcription factor must have at least two functional domains.
Answer: Transcription factors are modular proteins with at least two functional domains: one
binds to DNA in promoters and enhancers, and the other activates transcription.
Section: 17.4
Bloom's Taxonomy: Remembering/Understanding
36) The human metallothionein IIA gene (hMTIIA) is transcriptionally regulated through the
interplay of regulatory elements and transcription factors. What is the function of this gene, and
how is it regulated by environmental circumstances?
Answer: The gene hMTIIA produces a protein that binds to heavy metals and protects cells from
their toxicity. The gene is transcribed at low levels in all cells but is induced to express at high
levels in cells exposed to heavy metals.
Section: 17.4
Bloom's Taxonomy: Remembering/Understanding
37) What are zinc finger motifs, and why are they frequently encountered in descriptions of
genetic regulation in eukaryotes?
Answer: Zinc fingers consist of amino acid sequences containing two cysteine and two histidine
residues at repeating intervals. Interspersed cysteine and histidine residues covalently bind zinc
atoms, folding the amino acids into loops (zinc fingers). They are one of the major groups of
eukaryotic transcription factors.
Section: 17.4
Bloom's Taxonomy: Remembering/Understanding
38) What is meant by the term helix-turn-helix (HTH)?
Answer: A geometric conformation is formed by two adjacent alpha helices separated by a
"turn" of several amino acids. Such motifs bind to the major grooves of DNA and interact with
the DNA backbone.
Section: 17.4
Bloom's Taxonomy: Remembering/Understanding
39) Describe the ENCODE project and list five of their findings.
Answer: The ENCODE project stands for Encyclopedia of DNA Elements. The project set out
to identify all functional DNA sequences that are found in the human genome. They have made
several discoveries including: (1) < 2% of the human genome codes for proteins, (2) >80% of the
genome has some sort of biochemical activity, (3) enhancers can regulated several different
promoters for different genes, (4) about 75% of the human genome is transcribe in one cell or
another, and (5) identify many diseases that are linked to nonprotein coding portions of the gene.
Section: 17.7
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 18 Posttranscriptional Regulation in Eukaryotes
1) Which of the following is the most common type of alternative splicing in animals and
involves making mRNA molecules that lack a complete exon?
A) alternative splice site
B) intron retention
C) cassette exons
D) mutually exclusive exons
E) alternative promoters
Answer: C
Section: 18.1
Bloom's Taxonomy: Remembering/Understanding
2) In a eukaryote, what is the best way to compare the number of genes in a genome to the
number of proteins that the genome can produce and why?
A) The number of proteins is equivalent to the number of genes due to alternative splicing.
B) The number of proteins is usually fewer than the number of genes due to alternate splicing.
C) The number of proteins is usually greater than the number of genes due to alternate splicing.
D) The number of proteins is usually fewer than the number of genes due to histone
modification.
E) The number of proteins is usually greater than the number of genes due to histone
modification.
Answer: C
Section: 18.1
Bloom's Taxonomy: Remembering/Understanding
3) Mutations that affect SR proteins will have what kind of effect on gene expression?
A) General transcription factors will be unable to bind to promoters.
B) Polyadenylation at the 3' end will be blocked.
C) RNA polymerase will be unable to terminate transcription.
D) Ribosomes will not be able to bind to mRNA molecules in the cytoplasm.
E) Splicing of the mRNA will not be activated.
Answer: E
Section: 18.1
Bloom's Taxonomy: Applying/Analyzing
4) Having two copies of the X-chromosome gene, ________, initiates the female sexdetermining pathway by which gene regulatory mechanism?
A) SXL; alternative splicing
B) SXL; alternative polyadenylation
C) DSX; alternative splicing
D) DSX; alternative polyadenylation
E) TRA; alternative splicing
Answer: A
Section: 18.1
Bloom's Taxonomy: Remembering/Understanding
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5) Which of the following statements is predicted if a population of fruit flies were fed a
chemical that blocked the function of the SXL protein?
A) The progeny in the next generation would be predominantly female.
B) The transformer (tra) gene would be maximally expressed.
C) The mRNA produced from the doublesex (dsx) gene would not be spliced.
D) The progeny in the next generation would be predominantly male.
E) The progeny in the next generation would be approximately 50% male and 50% female.
Answer: D
Section: 18.1
Bloom's Taxonomy: Applying/Analyzing
6) The human disease spinomuscular atrophy is caused by a failure of which posttranscription
regulatory mechanism?
A) alternative polyadenylation
B) RNA interference
C) mRNA surveillance
D) alternative splicing
E) translation initiation
Answer: D
Section: 18.1
Bloom's Taxonomy: Remembering/Understanding
7) 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: 18.2
Bloom's Taxonomy: Remembering/Understanding
8) ________ are cytoplasmic structures that will aggregate mRNA molecules for either
degradation or later translation.
A) Ribosomes
B) Mitochondria
C) Golgi apparatus
D) Proteasomes
E) Processing bodies
Answer: E
Section: 18.2
Bloom's Taxonomy: Remembering/Understanding
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9) Adenosine—uridine-rich elements are ________.
A) trans-acting elements that bind to mRNA to either degrade or stabilize mRNA
B) trans-acting elements that bind to mRNA to recruit ribosomes to mRNA for translation
initiation
C) cis-acting elements that recruit RNA-binding proteins to mRNA to either degrade or stabilize
the mRNAs
D) cis-acting elements where ribosomes bind to in order to initiate translation
E) trans-acting elements that block mRNA splicing
Answer: C
Section: 18.2
Bloom's Taxonomy: Remembering/Understanding
10) mRNA surveillance mechanisms are necessary ________.
A) in determining which alternative splice site should be used
B) to measure the concentration of mRNAs in the cytoplasm
C) for polyadenylation at the 3' end of mRNA
D) to eliminate mRNAs with a premature stop codon
E) to recruit general transcription factors to the promoter
Answer: D
Section: 18.2
Bloom's Taxonomy: Remembering/Understanding
11) 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: 18.3
Bloom's Taxonomy: Applying/Analyzing
12) 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: 18.3
Bloom's Taxonomy: Remembering/Understanding
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13) 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: 18.3
Bloom's Taxonomy: Remembering/Understanding
14) The discovery of RNA interference (RNAi) led to its use in biotechnology and medicine. All
of the following are examples of how RNAi could be used in biotechnology or medicine,
EXCEPT ________.
A) to treat a disease characterized by the buildup of a specific protein
B) to treat a disease characterized by the normal expression of a mutated gene
C) to treat a disease characterized by the overexpression of a nonmutated gene
D) to treat a disease characterized by a gene that is not expressed
E) to study loss-of-function phenotypes of specific genes in a model research organism
Answer: D
Section: 18.3
Bloom's Taxonomy: Remembering/Understanding
15) Which of the following statements is most true regarding long noncoding RNAs (lncRNAs)?
A) lncRNAs are characteristically about the same size as sncRNAs.
B) lncRNAs do not contain a 5' guanine cap or a 3' poly-A-tail.
C) Some lncRNAs will decrease the activity of miRNAs.
D) lncRNAs act only as posttranscriptional regulators and never as pretranscription regulators.
E) Some lncRNAs regulate alternative splicing.
Answer: C
Section: 18.3
Bloom's Taxonomy: Remembering/Understanding
16) A long poly-A-tail is necessary to initiate translation. Which protein binds directly to eIF4E
to inhibit translation initiation?
A) PABP
B) maskin
C) PARN
D) poly-A polymerase
E) CPEB
Answer: B
Section: 18.4
Bloom's Taxonomy: Remembering/Understanding
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17) A particular fibroblast mutation fails 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) poly-A polymerase
Answer: B
Section: 18.4
Bloom's Taxonomy: Applying/Analyzing
18) What posttranslational modification occurs on the protein CPEB to initiate mRNA
translation?
A) acetylation
B) ubiquitination
C) methylation
D) phosphorylation
E) glycosylation
Answer: D
Section: 18.4, 18.5
Bloom's Taxonomy: Remembering/Understanding
19) What is the general name of enzymes that add phosphate groups to protein targets?
A) ligases
B) acetylases
C) phosphatases
D) proteases
E) kinases
Answer: E
Section: 18.5
Bloom's Taxonomy: Remembering/Understanding
20) Which protein modification is most closely linked to protein degradation?
A) phosphorylation
B) ubiquitination
C) acetylation
D) methylation
E) glycosylation
Answer: B
Section: 18.5
Bloom's Taxonomy: Remembering/Understanding
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21) What is the general gene regulation mechanism that "cassette exons" and "intron retention"
are involved with? Explain their differences in regards to which kinds of organism use each and
how each is different in what they do.
Answer: They are different types of alternative splicing. Cassette exons is a method used mainly
by animals, and intron retention is used mainly by plants, fungi, and protozoa. Cassette exons
achieves alternative splicing by excluding one or more exons from the mature mRNA. Intron
retention achieves alternative splicing by including introns into the mature mRNA (this can lead
to novel functions or no function in the mature protein).
Section: 18.1
Bloom's Taxonomy: Remembering/Understanding
22) Mutating the gene Transformer (tra) in Drosophila that is chromosomally XX leads to the
development of only males. Explain.
Answer: In nonmutated fertilized eggs that are supposed to develop into females (XX), the two
copies of SXL causes the expression of tra, which then leads to an alternative splicing pathway
of the dsx gene that induces female development. However, even though sxl is present at levels
to cause female development (because of the two X chromosomes), only males form because the
SXL target is tra. If tra is not expressed or mutated, then males will form due to the alternative
splicing of dsx that leads to male development.
Section: 18.1
Bloom's Taxonomy: Applying/Analyzing
23) List three general pathways in which eukaryotic mRNA is typically degraded in eukaryotes.
Answer: Deadenylation-dependent decay, decapping, and nonsense-mediated decay
Section: 18.2
Bloom's Taxonomy: Remembering/Understanding
24) What is nonsense-mediated decay, why is it important, and what is a proposed mechanism as
to how it works?
Answer: Nonsense-mediated decay is a type of RNA surveillance that identifies mRNA with
premature stop codons and eliminates them. Nonsense-mediated decays are important because
mRNAs with premature stop codons will usually make nonfunctional proteins that could be
harmful or it's just a waste of cellular energy to make a protein that is not functional. How the
cell identifies this nonsense-containing mRNAs is not yet known, but a model is that the cell can
recognize such mRNAs due to the large distance between the poly-A tail and the stop codon
(usually this distance is relatively short).
Section: 18.2
Bloom's Taxonomy: Remembering/Understanding
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25) Cancer is caused by many different types of gene mutations. Some mutations are in protooncogenes, which lead to overexpression of the genes, and other mutations are in tumor
suppressor genes, which lead to under expression or no expression in these genes. Which kinds
of gene mutations would RNA interference (RNAi) be better at treating? Explain.
Answer: RNAi would be better at treating proto-oncogene mutations as RNAi will bind to
overexpressed mRNA molecules and degrade them. Tumor suppressor gene mutations lead to
fewer mRNA molecules, so RNAi would just make the situation worse by degrading the few
tumor suppressor mRNAs that remain.
Section: 18.3
Bloom's Taxonomy: Applying/Analyzing
26) What are two features of circular RNAs (cirRNAs) that limit their ability to be degraded?
Answer: Circular RNAs are closed and thus lack a 5' end or 3' end, so they cannot be degraded
by nucleases from either end.
Section: 18.3
Bloom's Taxonomy: Remembering/Understanding
27) Describe how cells control the size of the poly-A tail on mRNA molecules in order to
regulate translation initiation. Next explain how a kinase inhibitor could decrease gene
expression in regards to polyadenylation? Your answer should include the following proteins in
your answer: Maskin, PARN, CPEB, eIF4E, eIF4G, and Poly-A polymerase.
Answer: Typically, an mRNA cannot initiate translation unless it has a sufficiently long poly-A
tail at the 3' end of the mRNA. To inhibit translation initiation, a protein called CPEB binds to
the CPE sequence on the mRNA. This will then bring in an enzyme called PARN. PARN is a
ribonuclease that shortens the poly-A tail. In addition, CPEB also recruits the protein called
Maskin. Maskin binds to the translation initiation factor called eIF4E. This prevents eIF4E from
binding to eIF4G, which now blocks initiation. Together this keeps the 3' poly-A tail short and
blocks the interaction of the two IF proteins. To initiate translation, the protein CPEB gets
phosphorylated by a specific kinase. This causes a conformational change that excludes both
Maskin and PARN. With these two proteins gone, now the poly-A tail can extend and eIF4E and
eIF4G can interact with each other to initiate translation.
If a kinase inhibitor was introduced, then CPEB would not be phosphorylated, so no
conformational change would occur, and the Maskin and PARN would be able to continue to
exert their inhibitory effect on translation initiation.
Section: 18.4
Bloom's Taxonomy: Applying/Analyzing
28) Compare and contrast kinases and phosphatases.
Answer: Both are enzymes involved in posttranslational modifications involving phosphate
groups. These modifications typically activate or deactivate enzyme functions. Kinases add a
phosphate group and phosphatases remove phosphate groups.
Section: 18.5
Bloom's Taxonomy: Remembering/Understanding
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29) A particular mutant was found that produced a protein that was longer lived than its normal
protein. Upon sequencing the mutant, the researcher discovered that specific lysine amino acids
had been changed to arginines. Propose a model of what this mutation caused to allow the
proteins to be longer lived.
Answer: Proteins are often targeted for degradation by ligating ubiquitin molecules to a protein
at their lysine amino acids. As this protein is longer lived and has fewer lysines, then it is
reasonable to predict that the mutation prevents ubiquitination at these specific sites that causes it
to evade being degraded.
Section: 18.5
Bloom's Taxonomy: Applying/Analyzing
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 19 Epigenetic Regulation of Gene Expression
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: 19.1
Bloom's Taxonomy: Remembering/Understanding
2) All of the following are epigenetic mechanisms, EXCEPT ________.
A) changing gene expression by adding and removing chemical groups to histone proteins
B) changing gene expression by adding and removing methyl groups to DNA
C) changing gene expression by the use of long noncoding RNAs
D) changing gene expression by the use of small noncoding RNAs
E) changing gene expression by changing the nucleotide sequence through mutation
Answer: E
Section: 19.1
Bloom's Taxonomy: Remembering/Understanding
3) Which of the following is the most common place to find DNA methylation?
A) on cytosines when adjacent to guanines
B) on cytosines when adjacent to other cytosines
C) on guanines when adjacent to cytosines
D) on guanines when adjacent to other guanines
E) on cytosines when they are base paired to guanines
Answer: A
Section: 19.1
Bloom's Taxonomy: Remembering/Understanding
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4) In terms of epigenetics, how the promoter would be described 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: 19.1
Bloom's Taxonomy: Applying/Analyzing
5) Which DNA methyltransferase (DNMT) is important in maintaining DNA methylation
patterns during DNA replication?
A) DNMT1
B) DNMT2
C) DNMT3a
D) DNMT3b
E) DNMT 1 and DNMT3a
Answer: A
Section: 19.1
Bloom's Taxonomy: Remembering/Understanding
6) 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: 19.1
Bloom's Taxonomy: Remembering/Understanding
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7) Different modification of the histone have different effects on gene expression. For instance,
the ninth amino acid of histone 3 is a lysine (H3K9) can be acetylated once or methylated up to
three times. Which of the following is true regarding H3K9 modification?
A) If H3K9 is acetylated, then the gene will be repressed.
B) If H3K9 has a single methyl group, then the gene will be repressed.
C) If H3K9 has two methyl groups, then the gene will be activated.
D) If H3K9 has three methyl groups, then the gene will be repressed.
E) If H3K9 has three methyl groups, it is maximally activated compared to only having one
methyl group.
Answer: D
Section: 19.1
Bloom's Taxonomy: Remembering/Understanding
8) Long noncoding RNAs that partially overlap protein-coding genes and are transcribed in the
opposite direction are best classified as ________.
A) intronic lncRNA genes
B) bidirectional lncRNA genes
C) antisense lncRNA genes
D) piwi interacting RNAs
E) microRNAs
Answer: C
Section: 19.1
Bloom's Taxonomy: Remembering/Understanding
9) 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: 19.1
Bloom's Taxonomy: Applying/Analyzing
10) Which of the following 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 sourced allele is expressed, and in others, only the
maternal sourced allele is expressed.
Answer: E
Section: 19.2
Bloom's Taxonomy: Remembering/Understanding
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11) Which of the following genotypes would most likely lead to a low birth weight and a small
placenta?
A) Igf2+/Igf2+ where neither maternal nor paternal allele is mutated.
B) Igf2+/Igf2- where the mutated Igf2- allele comes from the maternal source.
C) Igf2+/Igf2- where the mutated Igf2- allele comes from the paternal source.
D) Igf2+/Igf2- and it doesn't matter which parental source the mutated Igf2- comes from.
E) Igf2+/Igf2- but only in female progeny as the Igf2 gene is on the X-chromosome.
Answer: C
Section: 19.2
Bloom's Taxonomy: Applying/Analyzing
12) What is the name of the DNA region that controls the expression of a cluster of genes
through methylation?
A) promoters
B) insulators
C) epimutations
D) imprinting control regions
E) enhancers
Answer: D
Section: 19.2
Bloom's Taxonomy: Remembering/Understanding
13) Prader-Willi (PW) and Angelman syndromes (AS) are imprinting disorders of the same
region on chromosome 15. Normally, the genes associated with PW are ________ imprinted
(silenced), and the one gene associated with AS is ________ imprinted (silenced).
A) maternally; paternally
B) paternally; maternally
C) paternally; paternally
D) maternally; randomly
E) randomly; paternally
Answer: A
Section: 19.2
Bloom's Taxonomy: Remembering/Understanding
14) In humans, the number of genes that are imprinted is less than 1%. These imprinted genes
are usually associated with ________.
A) brain function
B) growth regulation
C) apoptosis
D) DNA replication
E) metabolic pathways
Answer: B
Section: 19.2
Bloom's Taxonomy: Remembering/Understanding
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15) Which of the following statements is most true regarding X-inactivation?
A) X-inactivation is more common in males than females.
B) X-inactivation results in repressing the expression of approximately 75% of the genes on the
X-chromosome.
C) When a cell divides, the inactive X-chromosome remains inactive in the newly formed cell.
D) X-inactivation is caused primarily by the expression of several small noncoding RNAs
(sncRNA).
E) X-chromosomes that remain active are called Barr bodies.
Answer: C
Section: 19.2
Bloom's Taxonomy: Remembering/Understanding
16) In terms of overall methylation, what is the most accurate way to describe the genomes of
cancer cells?
A) Cancer cells have an overall hypermethylated genome compared to healthy cells.
B) Cancer cells have an overall hypomethylated genome compared to healthy cells.
C) Cancer cells have equivalent methylated genomes compared to healthy cells.
D) A few types of cancers are have hypermethylated genomes, but it is not common in most
types of cancers.
E) A few types of cancers are have hypomethylated genomes, but it is not common in most types
of cancers.
Answer: B
Section: 19.3
Bloom's Taxonomy: Remembering/Understanding
17) About 5% of those with Rubinstein-Taybi syndrome develops cancer as a result to decreased
gene expression. This is directly linked to abnormal histone modification caused by a
dysfunctional enzyme. What enzyme is this?
A) histone acetylase (HAT)
B) histone deacetyltransferase (HDAC)
C) DNA methyltransferase (DMT)
D) BRCA1
E) No enzyme could cause this as cancer is caused only by direct DNA mutations.
Answer: A
Section: 19.3
Bloom's Taxonomy: Remembering/Understanding
18) Which of the following statements is the most accurate in regards to mice pups who receive
less 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: 19.4
Bloom's Taxonomy: Remembering/Understanding
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19) List the three major epigenetic mechanisms.
Answer: (1) Addition or removal of methyl groups from DNA, (2) chromatin remodeling by
adding and removing chemical groups from histone proteins, and (3) regulation of gene
expression by noncoding RNA molecules.
Section: 19.1
Bloom's Taxonomy: Remembering/Understanding
20) Compare and contrast DNMT1, DNMT3a, and DNMT3b in regards to what they do, when
they act, and how they affect gene transcription.
Answer: DNMT1, DNMT3a, and DNMT3b are DNA methyltransferases and all add methyl
groups to cytosines that are typically adjacent to guanines (CG). Promoters that get methylated
will usually decrease gene expression. DNMT1 is primarily responsible for maintaining
methylation patterns as DNA is being replicated and DNMT3a and DNMT3b are responsible for
establishing the original methylation patterns.
Section: 19.1
Bloom's Taxonomy: Remembering/Understanding
21) Using the histone code different histone modifications follow a particular nomenclature.
Describe the histone with this code: H3K27me3 and what is the effect of this histone
modification?
Answer: The lysine at position 27 on histone 3 has three methyl groups. The effect of this
modification is linked to gene repression.
Section: 19.1
Bloom's Taxonomy: Remembering/Understanding
22) Describe the four types of long noncoding RNAs (lncRNA).
Answer: (1) Antisense lncRNA partially overlap protein-coding genes and are transcribed in the
opposite direction of the protein-coding gene. (2) Intronic lncRNA genes are located within
introns, and their transcription does not overlap with the adjacent exon. (3) Bidirectional lncRNA
genes use the promoter of a protein-coding gene but are transcribed in the opposite direction. (4)
Intergenic lncRNA genes are discrete transcription units located outside the protein-coding
region.
Section: 19.1
Bloom's Taxonomy: Remembering/Understanding
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23) 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 and a smaller placenta. 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 of the fetus and
placenta.
Section: 19.2
Bloom's Taxonomy: Applying/Analyzing
24) 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 from the
X-chromosome to be inactivated. The XIST lncRNA will repress the genes on the inactivated Xchromosomes. The TSIX is expressed from the X-chromosome that is to remain active. When
TSIX is expressed, its lncRNA will repress the XIST.
Section: 19.2
Bloom's Taxonomy: Remembering/Understanding
25) Is there a risk for epigenetic linked diseases in using artificial reproductive technologies
(ART) such as in vitro fertilization?
Answer: Yes. There is an increase in imprinting errors, and maternal specific methylation is
reduced when using ART. Some examples are an increase in Beckwith—Wiedemann syndrome,
Angelman syndrome, and Prader-Willi syndrome.
Section: 19.2
Bloom's Taxonomy: Remembering/Understanding
26) Even though cancer genomes are overall hypomethylated, explain why some genes are
hypermethylated in cancer cells.
Answer: Hypermethylation of promoters will lead to reduced expression of those genes. Some
genes such as cell cycle stop and DNA repair genes need to be expressed. When they aren't
expressed, this can lead to cancer. Hypermethylation will reduce their expression.
Section: 19.3
Bloom's Taxonomy: Remembering/Understanding
27) Explain how HATs and HDACs can lead to the formation of cancer.
Answer: HATs usually lead to gene expression, and HDACs usually lead to gene repression. In
cancer cells if HATs are mutated, then genes that are normally expressed to prevent cancer are
now repressed, which can lead to cancer. In addition, in cancer cells if HDACs are mutated, then
genes that are normally inactive to suppress cancer will now be expressed leading to cancer.
Section: 19.3
Bloom's Taxonomy: Remembering/Understanding
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28) 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: 19.3
Bloom's Taxonomy: Applying/Analyzing
29) What are the two 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: 19.5
Bloom's Taxonomy: Remembering/Understanding
30) The data from the NIH Roadmap Epigenomics Project are classified into which five
categories?
Answer: (1) Histone modifications, (2) DNA methylation, (3) open and closed chromatin
configurations, (4) expression levels of protein-coding genes, and (5) expression profiles for
small nuclear RNAs such as miRNA.
Section: 19.5
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 20 Recombinant DNA Technology
1) Recombinant DNA technologies are methods used to do which of the following?
A) manipulate DNA for the study of specific sequences
B) integrate DNA in computers to make them faster
C) join together DNA molecules from the same source
D) induce homologous recombination inside a cell
E) combine chromosomes to make them easier to study
Answer: A
Section: 20.0
Bloom's Taxonomy: Remembering/Understanding
2) Cloned DNA is useful in studying which area(s) of genetic manipulation?
A) learning about the structure and organization of DNA
B) learning about introns
C) learning about translation of a gene
D) determining size of an organism's genome
E) producing commercial products
Answer: E
Section: 20.0
Bloom's Taxonomy: Remembering/Understanding
3) A restriction enzyme that uses a six (6) base recognition sequence will cut DNA, on average,
every ________ bases, if all four nucleotides are present in equal proportions.
A) 256
B) 4096
C) 1296
D) 500
E) 5000
Answer: B
Section: 20.1
Bloom's Taxonomy: Evaluating/Creating
4) Recognition sequences for restriction enzymes possess the unique quality of being the same
when read 5′ to 3′ on either strand. What is this property called?
A) palindromic sequence
B) consensus sequence
C) gene sequence
D) origin sequence
E) recombination sequence
Answer: A
Section: 20.1
Bloom's Taxonomy: Remembering/Understanding
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5) When choosing a restriction enzyme for use in recombinant DNA technologies, it is often
preferred that the enzyme generate cohesive, or "sticky," ends. Why is this preferred?
A) The sticky ends do not have hydrogen bonds to help in re-annealing the cut DNA.
B) The sticky ends make the DNA bind tighter to any cut DNA.
C) The sticky ends prevent the DNA from re-annealing to any DNA.
D) The sticky ends have hydrogen bonds that help re-anneal the cut DNA.
E) The sticky ends stick to the purification medium making the fragments easier to purify.
Answer: D
Section: 20.1
Bloom's Taxonomy: Remembering/Understanding
6) Assume that a plasmid (circular) is 2800 base pairs in length and has restriction sites for
EcoRI at the following locations: 400, 700, 1400, and 2600. Give the expected sizes of the
restriction fragments following complete digestion with EcoRI.
A) 200, 400, 700, 1200
B) 400, 1200, 1600
C) 300, 700, 2200
D) 700, 400, 1400, 2600
E) 300, 600, 700, 1200
Answer: E
Section: 20.1
Bloom's Taxonomy: Applying/Analyzing
7) DNA ligase ________.
A) reconnects the bases together between the DNA strands
B) reconnects the phosphodiester linkage between bases on the same strand of DNA
C) cuts the DNA to produce sticky or blunt ends
D) adds bases into a growing DNA molecule
E) removes bases from a DNA molecule
Answer: B
Section: 20.1
Bloom's Taxonomy: Remembering/Understanding
8) Transformation of cells using the technique of heat shock requires which of the following?
A) brief electric shock
B) brief exposure to elevated temperature
C) the use of calcium ions
D) the use of calcium and brief exposure to elevated temperature
E) the use of zinc ions
Answer: D
Section: 20.1
Bloom's Taxonomy: Remembering/Understanding
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9) A plasmid contains a multiple cloning site inside the coding region of the lacZ gene and also
contains an ampicillin resistance gene at a separate locus. When cells are transformed with a
successfully recombinant plasmid containing a piece of DNA cloned into the multiple cloning
site, what color will the colonies be when grown on an agar plate containing ampicillin and Xgal?
A) blue
B) white
C) green
D) clear
E) colonies will not grow
Answer: B
Section: 20.1
Bloom's Taxonomy: Applying/Analyzing
10) There are multiple cloning vector types in modern recombinant DNA technology ranging
from plasmids to viral vectors. Which vector type is most useful when cloning an insert of
approximately 500kb?
A) bacterial artificial chromosome
B) bacterial plasmid
C) viral vector
D) human artificial chromosome
E) yeast artificial chromosome
Answer: E
Section: 20.1
Bloom's Taxonomy: Remembering/Understanding
11) Restriction mapping is used to characterize cloned DNA. What does a restriction map tell the
researcher about the cloned DNA?
A) The number of restriction sites for the specific restriction enzyme.
B) The size of the genome the cloned DNA was isolated from.
C) The distances between restriction sites for the specific restriction enzyme.
D) The restricted conditions under which the organism can grow.
E) The number of sites and distance between them for the specific restriction enzyme.
Answer: E
Section: 20.1
Bloom's Taxonomy: Remembering/Understanding
12) If one wishes to clone a gene using typical restriction endonucleases, how does the restriction
endonuclease identify the appropriate cut sites in the genome?
A) The endonuclease recognizes the gene of interest.
B) The endonuclease identifies its specific recognition sequence.
C) The endonuclease cannot identify the cut sites.
D) The endonuclease cuts randomly in the genome.
Answer: B
Section: 20.1
Bloom's Taxonomy: Applying/Analyzing
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13) The following are four processes common to most cloning experiments:
i) transforming bacteria
ii) plating bacteria on selective medium
iii)cutting DNA with restriction endonucleases
iv) ligating DNA fragments
Place the components of this list in the order in which they would most likely occur during a
cloning experiment.
A) i, ii, iii, iv
B) ii, i, iv, iii
C) iv, i, iii, ii
D) iii, iv, i, ii
E) ii, iii, i, iv
Answer: D
Section: 20.1
Bloom's Taxonomy: Evaluating/Creating
14) What is necessary to generate a genomic DNA library?
A) Many overlapping fragments of the genome with at least one copy of every DNA sequence in
the organism's genome.
B) Many overlapping fragments of the genome with at least one copy of every DNA sequence in
the organism.
C) A single fragment that contains at least one copy of the DNA sequence in the organism's
genome.
D) Many fragments of the genome with at least one copy of every DNA sequence in the
organism's genome.
E) Many fragments of the genome with at least one copy of every DNA sequence in the
organism.
Answer: A
Section: 20.2
Bloom's Taxonomy: Remembering/Understanding
15) When generating a cDNA library, one may get different sequences returned if samples are
taken at different times. Why is this the case?
Answer: The cDNA library uses reverse transcriptase to copy mRNA present in the cell to
DNA. If cellular conditions change over time the genes being expressed could change as well
resulting in a change in the mRNA present when samples are taken. This would result in a
different cDNA library being generated.
Section: 20.2
Bloom's Taxonomy: Evaluating/Creating
16) What is the most serious drawback to using library screening to study genetics?
Answer: The sequence of the gene must be known in order to design probes to locate the gene in
the library.
Section: 20.2
Bloom's Taxonomy: Remembering/Understanding
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17) Which of the following best describes a cloning vector?
A) The direction in which DNA is cloned.
B) A DNA molecule that accepts DNA fragments and replicates the fragment in a host.
C) A DNA molecule that accepts DNA fragments and degrades them in a host.
D) The fragment of DNA encoding a gene of interest.
Answer: B
Section: 20.1
Bloom's Taxonomy: Remembering/Understanding
18) A scientist is troubleshooting the synthesis of a cDNA library. The scientist performs both a
Northern and a Southern blot. The Northern blot demonstrated the presence of RNA while the
Southern blot indicated that no cDNA was present in the sample. What is likely to be the cause
of the failed synthesis of the cDNA library?
A) proper primers
B) defective reverse transcriptase
C) too many dNTPs
D) temperature too cold for annealing
Answer: B
Section: 20.2
Bloom's Taxonomy: Evaluating/Creating
19) The genomics era began with the development of which of the following?
A) whole-genome sequencing
B) whole-transcriptome sequencing
C) PCR
D) third-generation sequencing
Answer: A
Section: 20.2
Bloom's Taxonomy: Remembering/Understanding
20) A probe with the sequence 5'-A-T-G-C-C-A-G-T-3' will serve as a probe for which
sequence?
A) 3'-T-G-S-C-C-G-T-A-5'
B) 3'-A-T-G-C-C-A-G-T-5'
C) 3'-T-A-C-G-G-T-C-A-5'
D) 3'-A-C-T-G-G-C-A-T-3'
Answer: C
Section: 20.2
Bloom's Taxonomy: Applying/Analyzing
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21) Genomic libraries are constructed from the fragments generated from a ________ digest.
A) gene specific
B) random
C) complete
D) incomplete
Answer: B
Section: 20.2
Bloom's Taxonomy: Remembering/Understanding
22) The development of polymerase chain reaction (PCR) was a watershed event in recombinant
DNA technology. What are two characteristics of PCR that make it such an important
development?
Answer: One advantage of PCR over bacterial cloning is that exponentially more DNA clones
can be made in a shorter time frame. A second advantage is the cost reduction in generating
cloned DNA through the use of thermostable Taq polymerase.
Section: 20.3
Bloom's Taxonomy: Evaluating/Creating
23) During the hybridization/annealing phase of a PCR reaction, the primers bind to the target
DNA sequence at a specific temperature. What effect would an increase in the GC content of the
primer have on the optimal temperature for annealing the primer to a target DNA sequence?
A) the annealing temperature would decrease
B) the annealing temperature would increase
C) the annealing temperature would remain the same
Answer: B
Section: 20.3
Bloom's Taxonomy: Applying/Analyzing
24) Briefly explain the three major steps in the polymerase chain reaction.
Answer: The first step is denaturation of the target DNA molecule by exposure to high
temperature. The second step is annealing of primers to the target DNA. The third step is the
synthesis of the target DNA sequence by a thermostable DNA polymerase.
Section: 20.3
Bloom's Taxonomy: Remembering/Understanding
25) RT-PCR (reverse transcription-polymerase chain reaction) can be used to compare gene
expression levels during an organism's response to the environment. A scientist monitoring a
tetracycline resistance gene will see what response in the expression when the organism is
challenged?
A) increased gene expression
B) decreased gene expression
C) no change in gene expression
Answer: A
Section: 20.3
Bloom's Taxonomy: Applying/Analyzing
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26) Why are DNA binding dyes such as SYBR green integral to the technique of qPCR?
Answer: The SYBR green dye binds double-stranded DNA. As more DNA is synthesized in the
qPCR reaction, the light emitted by SYBR green increases in proportion to the amount of DNA
present. This is what allows real-time quantification of the DNA.
Section: 20.3
Bloom's Taxonomy: Remembering/Understanding
27) In a PCR, the polymerase adds nucleotides to the 3'-hydroxyl of the terminal nucleotide of
each primer. Briefly describe how this results in the amplification of the single region of interest
in the genome.
Answer: The primers are used to initiate DNA synthesis by the polymerase and the antiparallel
nature of the DNA ensures that the region between the primers is replicated while the region
outside the primers is not.
Section: 20.3
Bloom's Taxonomy: Applying/Analyzing
28) What is the order of the three main steps in a PCR?
A) denaturation, elongation, annealing primers
B) elongation, annealing primers, denaturation
C) annealing primers, denaturation, elongation
D) denaturation, annealing primers, elongation
Answer: D
Section: 20.3
Bloom's Taxonomy: Remembering/Understanding
29) A PCR was designed to clone a DNA sequence of 1.5 kb. When the products were run on an
agarose gel, there was a smear of bands ranging from 1.5 kb to 6 kb. Which of the following is a
reason for this result?
A) the polymerase overran the reverse primer
B) the polymerase denatured
C) one of the primers did not bind
D) the thermocycler malfunctioned
Answer: C
Section: 20.3
Bloom's Taxonomy: Evaluating/Creating
30) 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: Since the smaller fragments migrate toward the "+" pole, away from the origin, they
bind less stain than the larger fragments near the origin.
Section: 20.4
Bloom's Taxonomy: Evaluating/Creating
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31) Due to single nucleotide polymorphisms (SNPs), all humans have a slightly different
nucleotide sequence for their genome. Restriction mapping is used in forensic sciences to match
DNA evidence to a potential suspect of a crime. How does restriction mapping reduce the
likelihood of misidentifying a suspect?
A) SNPs add and remove restriction sites resulting in a unique "fingerprint" for each person
B) SNPs change the genes that are expressed
C) SNPs only add new restriction sites to the genome resulting in a unique "fingerprint" for each
person
D) SNPs only remove restriction sites from the genome resulting in a unique "fingerprint" for
each person
Answer: A
Section: 20.4
Bloom's Taxonomy: Evaluating/Creating
32) Nucleic acid blotting is commonly used in molecular biology. Northern blotting is used to
detect the presence of which type of nucleic acid?
A) ribonucleic acids
B) deoxyribonucleic acids
C) triphosphate nucleic acids
D) transfer ribonucleic acids
E) unnatural nucleic acids
Answer: A
Section: 20.4
Bloom's Taxonomy: Remembering/Understanding
33) Fluorescence in situ hybridization is another way to visualize the presence of a nucleotide
sequence. Which of the following is an advantage of FISH over Northern and Southern blots?
A) Probes are not needed in FISH.
B) Blotting is not used in FISH.
C) Fluorophores are not used in Northern or Southern blots.
D) There are no advantages to FISH over Northern and Southern blots.
Answer: B
Section: 20.4
Bloom's Taxonomy: Remembering/Understanding
34) When a 5-kb circular plasmid is digested with a restriction enzyme that has three recognition
sites on the plasmid, how many bands can be visualized on an agarose gel?
A) 1
B) 2
C) 3
D) 4
Answer: C
Section: 20.4
Bloom's Taxonomy: Applying/Analyzing
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35) Spectral karyotypes take advantage of which of the following?
A) Chromosomes can be labeled with different radioactive isotopes.
B) Probes can be labeled with different color fluorophores.
C) Probes can be made with different radioactive isotopes.
D) Chromosomes are different colors under a microscope.
Answer: B
Section: 20.4
Bloom's Taxonomy: Remembering/Understanding
36) Agarose gels separate DNA fragments based on what property?
A) amount of adenine bases in the sequence
B) charge of the DNA molecule
C) size of the fragment
D) amount of agarose in the DNA
Answer: C
Section: 20.4
Bloom's Taxonomy: Remembering/Understanding
37) What is the main purpose of a DNA probe?
A) extend the growing polynucleotide
B) hybridizes to a target sequence
C) binds to proteins
D) cuts DNA targets
Answer: B
Section: 20.4
Bloom's Taxonomy: Remembering/Understanding
38) Next-generation sequencing takes advantage of the concept of sequencing-by-synthesis.
What is sequence-by-synthesis?
A) The use of PCR to incorporate ddNTPs, followed by electrophoresis.
B) The use of a multiwell system and microfluidics to detect incorporated nucleotides at each
base.
C) The use of a single polymerase in a nanopore that reports on a single molecule of synthesized
DNA.
D) The use of firefly luciferase and pyrophosphate to report on nucleotide incorporation.
Answer: B
Section: 20.5
Bloom's Taxonomy: Applying/Analyzing
39) Third-generation sequencing strategies are based on which of the following?
A) sequencing a single strand of DNA
B) sequencing an ensemble of DNA strands with the same sequence
C) sequencing multiple strands of DNA with different sequences to generate a library
D) sequencing DNA using PCR
Answer: A
Section: 20.5
Bloom's Taxonomy: Remembering/Understanding
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40) Sanger sequencing is based on the order in which ddNTPs are added to a growing
polynucleotide. Why are ddNTPs integral to the Sanger sequencing method?
A) They have a 2′ hydroxyl that allows for extension of the polynucleotide.
B) They have a 3′ hydroxyl that allows for extension of the polynucleotide.
C) They do not have a 2′ hydroxyl, which does not allow the extension of the polynucleotide.
D) They do not have a 3′ hydroxyl, which does not allow the extension of the polynucleotide.
Answer: D
Section: 20.5
Bloom's Taxonomy: Remembering/Understanding
41) High throughput sequencing takes advantage of automated reading of sequencing data.
Which of the following assists in automating the sequencing readout?
A) fluorescently tagged ddNTPs
B) radioactive ddNTPs
C) fluorescently tagged dNTPs
D) radioactive dNTPs
Answer: A
Section: 20.5
Bloom's Taxonomy: Remembering/Understanding
42) When performing a sequencing reaction using Sanger sequencing, which ddNTPs must be
included in the reaction?
A) ddATP
B) ddTTP
C) ddCTP
D) ddGTP
E) all four ddNTPs must be present
Answer: E
Section: 20.5
Bloom's Taxonomy: Remembering/Understanding
43) ddATP, ddTTP, ddCTP, and ddGTP are labeled with red, green, yellow, and blue fluorescent
dyes, respectively. A five-base read from a sequencing reaction produced the following color
sequence, read by the computer: red, yellow, yellow, green, green. What is the sequence of the
template DNA?
A) ACCTT
B) AAGGT
C) TTCCA
D) TGGAA
Answer: A
Section: 20.5
Bloom's Taxonomy: Evaluating/Creating
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44) What is the function of a ddNTP in DNA sequencing?
A) methylation of guanine
B) termination of DNA synthesis
C) enhancing the processivity of the polymerase
D) provide a 3′ hydroxyl for continued elongation
Answer: B
Section: 20.5
Bloom's Taxonomy: Remembering/Understanding
45) Which of the following describes a "knock-out" organism?
A) An organism that loses consciousness randomly.
B) An organism that is capable of out-competing another organism.
C) An organism that is negative for alleles coding a gene of interest.
D) An organism that is positive for alleles coding a gene of interest.
Answer: C
Section: 20.6
Bloom's Taxonomy: Remembering/Understanding
46) A knock-out mouse is made with respect to the PFKL gene, and this mouse is now described
as which of the following?
A) a gain-of-function PFKL mouse
B) a loss-of-function PFKL mouse
C) a retention-of-function PFKL mouse
D) an enhancement-of-function PFKL mouse
Answer: B
Section: 20.6
Bloom's Taxonomy: Remembering/Understanding
47) The generation of a knock-out organism generally requires all the following EXCEPT
________.
A) embryonic stem cells
B) a targeting vector
C) knowledge of the target sequence
D) a selectable marker
E) knowledge of the sequence product
Answer: E
Section: 20.6
Bloom's Taxonomy: Remembering/Understanding
48) Conditional knock-out organisms provide which advantage over full knock-out organisms?
A) They allow scientists to turn the gene off at any time during the organism's development.
B) They allow scientists to turn on the gene at any time during the organism's development.
C) They allow scientists to render the organism unconscious at any time.
D) They allow scientists to study genome size.
Answer: A
Section: 20.6
Bloom's Taxonomy: Remembering/Understanding
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49) Transgenic animals are useful in studying the effects of a gene product on the organism's
phenotype. Briefly describe how you would study the effect of overexpression of a pancreatic
enzyme using a transgenic animal.
Answer: The gene of interest can be inserted into the organism's genome under the regulation of
a pancreas-specific promoter. This will cause the enzyme to be overproduced in the pancreatic
tissue, potentially resulting in a changed phenotype.
Section: 20.6
Bloom's Taxonomy: Evaluating/Creating
50) Protospacer adjacent motifs, PAMs, allow bacteria to do which of the following?
A) recognize and digest their own DNA
B) recognize and digest foreign DNA
C) recognize proper spacing for bacterial transcription initiation
D) recognize and integrate foreign DNA
Answer: B
Section: 20.6
Bloom's Taxonomy: Remembering/Understanding
51) Which of the following is not a common system for generating knock-out organisms?
A) cre-lox
B) CRSPR-CAS
C) homologous recombination
D) transformation
Answer: D
Section: 20.6
Bloom's Taxonomy: Remembering/Understanding
52) The CRISPR-Cas system of gene editing is based on what naturally occurring biological
process?
A) bacterial defense against viral infection
B) eukaryotic defense against viral infection
C) viral defense against bacterial defenses
D) viral defense against eukaryotic defenses
Answer: A
Section: 20.6
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 21 Genomic Analysis
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 ________.
A) incorrect sequencing
B) incorrect annotation
C) incorrect BLAST parameters
D) incorrect enzyme experiments
E) incorrect qPCR
Answer: B
Section: 21.3
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: 21.3
Bloom's Taxonomy: Evaluating/Creating
3) How many potential reading frames are there in a 1500-bp segment of double-stranded DNA?
A) 1
B) 2
C) 4
D) 5
E) 6
Answer: E
Section: 21.3
Bloom's Taxonomy: Remembering/Understanding
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: 21.4
1
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Bloom's Taxonomy: Evaluating/Creating
5) The Human Genome Project, which got under way in 1990, is an international effort to
________.
A) determine the base sequence of the human genome and to identify all the genes within
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: 21.4
Bloom's Taxonomy: Remembering/Understanding
6) The sequencing of the human genome lead to the realization that chromosome 19 contains
many genes while chromosomes 13 and Y contain relatively few. This finding implies what
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 large noncoding
regions.
D) Genes have no organization and randomly appear throughout the genome and chromosomes.
Answer: C
Section: 21.5
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 in formulating policy and legislation related
to the application of human genomic information. 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: 21.5
Bloom's Taxonomy: Remembering/Understanding
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: 21.5
Bloom's Taxonomy: Remembering/Understanding
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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: 21.5
Bloom's Taxonomy: Remembering/Understanding
10) Compared with eukaryotic chromosomes, bacterial chromosomes are ________.
A) large, mainly organized in single-gene transcription units without introns
B) small, mainly organized in single-gene transcription units with introns
C) large, mainly organized in polygenic transcription units without introns
D) small, with high gene density
E) large, triple-helix, Z-DNA, organized in single gene units with introns
Answer: D
Section: 21.6
Bloom's Taxonomy: Remembering/Understanding
11) Compared with prokaryotic chromosomes, eukaryotic chromosomes are ________.
A) large, mainly organized in single-gene transcription units without introns
B) small, mainly organized in single-gene transcription units with introns
C) large, mainly organized in polygenic transcription units without introns
D) small, mainly organized in polygenic transcription units without introns
E) large, linear, less densely packed with protein-coding genes, mainly organized in single-gene
units with introns
Answer: E
Section: 21.6
Bloom's Taxonomy: Remembering/Understanding
12) Most of the bacterial genomes described in the text have fewer than ________.
A) 10,000 genes
B) 5000 base pairs
C) 500 genes
D) 10,000 base pairs
E) 50 genes
Answer: A
Section: 21.6
Bloom's Taxonomy: Applying/Analyzing
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13) A bacterial polygenic transcription unit ________.
A) contains information for one protein product
B) contains information for more than one protein product
C) is capped at the 5′ end and carries a poly-A tail at the 3′ end
D) is void of start (AUG) and termination (UAA, UGA, UAG) triplets
E) None of the answers listed is correct.
Answer: B
Section: 21.6
Bloom's Taxonomy: Remembering/Understanding
14) What is the typical gene density per kilobase in prokaryotes?
A) 426,000
B) 3000
C) 1200
D) 1
E) 12
Answer: D
Section: 21.6
Bloom's Taxonomy: Applying/Analyzing
15) In general, the organization of genes in bacteria is different from that in eukaryotes. In E.
coli, approximately 27 percent of all genes are organized into contiguous, functionally related
units containing multiple genes under coordinated control that are transcribed as a single unit.
Such contiguous gene families are called ________.
A) transcriptomes
B) proteomes
C) contigs
D) operons
E) pseudogenes
Answer: D
Section: 21.6
Bloom's Taxonomy: Remembering/Understanding
16) Which of the following is not an emerging "-omics" field?
A) pharmacogenomics
B) metagenomics
C) glycomics
D) chromosomics
E) transcriptomics
Answer: D
Section: 21.9
Bloom's Taxonomy: Remembering/Understanding
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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: 21.6
Bloom's Taxonomy: Evaluating/Creating
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 gen location in the genome
D) differences in proteins expressed between cells of an organism
Answer: B
Section: 21.6
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: 21.6
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) LUMINA
C) NCBI
D) NIH
E) ENCODE
Answer: E
Section: 21.6
Bloom's Taxonomy: Remembering/Understanding
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: 21.6
Bloom's Taxonomy: Remembering/Understanding
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22) Several years ago, scientists compared the genome of a wooly 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: 21.6
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) splice-outs
D) inteins
E) exteins
Answer: A
Section: 21.7
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: 21.7
Bloom's Taxonomy: Remembering/Understanding
25) Aligning genome sequences from different organisms to determine their evolutionary history
is referred to as ________.
A) comparative genomics
B) metagenomics
C) evogenomics
D) stone-age genomics
Answer: A
Section: 21.7
Bloom's Taxonomy: Remembering/Understanding
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26) The advent of genomic sequencing and annotations using bioinformatics has led to the
realization that some of the long-held beliefs about prokaryotic genome structure were incorrect.
Which of the following has not been refuted by modern bioinformatic discoveries?
A) Bacterial genomes are always smaller than eukaryotic genomes.
B) All bacterial genomes are circular.
C) Prokaryotes have a single chromosome.
D) Prokaryote genomes do not contain introns.
Answer: D
Section: 21.7
Bloom's Taxonomy: Remembering/Understanding
27) Which of the following is a major contributor to variation in gene density in eukaryotes?
A) repetitive sequences
B) transposons
C) restriction sites
D) heterochromatin
Answer: A
Section: 21.7
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: 21.7
Bloom's Taxonomy: Applying/Analyzing
29) The genome of the domesticated dog, Canis familiaris, was sequenced in 2005. Using the
dog as a model system provides insight into which of the following human disease states?
A) aneuploidies
B) multifactorial diseases
C) behavioral conditions (i.e., obsessive-compulsive disorders)
D) aneuploidies, multifactorial diseases, and behavioral conditions
Answer: D
Section: 21.7
Bloom's Taxonomy: Remembering/Understanding
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 the 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 trends.
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: 21.7
Bloom's Taxonomy: Remembering/Understanding
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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 organisms 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: 21.8
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: 21.8
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: 21.8
Bloom's Taxonomy: Applying/Analyzing
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: 21.9
Bloom's Taxonomy: Remembering/Understanding
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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 transcriptase
B) protein quantification
C) heterochromatin
D) euchromatin
Answer: A
Section: 21.9
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: 21.9
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: 21.9
Bloom's Taxonomy: Remembering/Understanding
38) Using RNA-seq generates data in situ. What does RNA-seq not provide data for?
A) gene expression levels
B) sequence of the RNA
C) location of transcript in the genome
D) amount of protein produced by RNA
Answer: D
Section: 21.9
Bloom's Taxonomy: Applying/Analyzing
39) Proteomics is the "-omics" study of proteins, investigating the complete set of proteins in a
cell. What does proteomics provide data on?
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: 21.10
Bloom's Taxonomy: Remembering/Understanding
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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: 21.10
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: 21.10
Bloom's Taxonomy: Remembering/Understanding
42) MALDI-mass spectrometry was used to identify a protein from preserved T. rex samples.
Briefly describe how this was done.
Answer: The T. rex sample was believed to be collagen. A synthetic collagen was made and
examined using MALDI-mass spectrometry. This generated a "fingerprint" for collagen. The T.
rex sample was then examined with MALDI-mass spectrometry. The "fingerprints" matched,
confirming the T. rex sample was collagen.
Section: 21.10
Bloom's Taxonomy: Evaluating/Creating
43) Describe the similarities between sequencing a protein and sequencing a genome.
Answer: Similar to genome sequencing protein sequences are broken up into smaller strings
then overlapping regions are stitched together, similar to contigs until the entire sequence is
known.
Section: 21.10
Bloom's Taxonomy: Applying/Analyzing
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: 21.10
Bloom's Taxonomy: Remembering/Understanding
10
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 22 Applications of Genetic Engineering and Biotechnology
1) Biotechnology is commonly used to produce transgenic plants and animals that are then
harvested for useful products. This process is commonly referred to as which of the following?
A) animal farming
B) tuna farming
C) biopharming
D) culturing
E) genetic production
Answer: C
Section: 22.1
Bloom's Taxonomy: Remembering/Understanding
2) Mastitis in cows results in infected mammary glands. The use of biotechnologies to introduce
the antibiotic gene lysostaphin from Stpahylococcus simulans is an example of generating what
type of animal?
A) a knock-out animal
B) a transgenic animal
C) a conditional knock-out animal
D) a synthetic animal
Answer: B
Section: 22.3
Bloom's Taxonomy: Evaluating/Creating
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: 22.4
Bloom's Taxonomy: Evaluating/Creating
4) SNPs are characterized by being ________.
A) highly heterogeneous with variable amino acid substitutions
B) highly uniform in the population with variable numbers of tandem repeats
C) variable in the population with variable base sequences
D) homogeneous with various nucleotides transcribed repeatedly
E) lethal genes with small noteworthy transcribed regions
Answer: C
Section: 22.4
Bloom's Taxonomy: Remembering/Understanding
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5) A DNA microarray (also called a DNA chip) can be used to ________.
A) generate mutations in the gene of interest
B) detect mutations that may indicate a risk of disease
C) detect RFLPs
D) determine protein translation levels
Answer: B
Section: 22.4
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: 22.4
Bloom's Taxonomy: Remembering/Understanding
7) Which of the following best describes the term edible vaccine?
A) A recombinant organism that expresses the vaccine in a consumable portion of the organism.
B) Part of a recombinant organism that the vaccine can be purified from.
C) A vaccine that can be eaten by a human and alter the genome.
D) A DNA that a human can eat and relies on the human cells to generate the vaccine.
Answer: A
Section: 22.1
Bloom's Taxonomy: Remembering/Understanding
8) Generally, vaccines are used to stimulate the immune system by providing antigens of
potential pathogens. What is the typical composition of a vaccine?
A) naked DNA
B) RNA from bacteria
C) inactivated or attenuated viruses
D) genes encoding for virulence factors
Answer: C
Section: 22.1
Bloom's Taxonomy: Remembering/Understanding
9) A transgenic organism is one in which ________.
A) DNA from a different organism is introduced to produce a biopharmaceutical
B) DNA from a different organism is introduced to produce a more resilient bacterium
C) its genes have transferred to new chromosomes
D) a naturally occurring plasmid has been transferred through natural means
Answer: A
Section: 22.1
Bloom's Taxonomy: Remembering/Understanding
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10) Selective breeding has been used for centuries to ________.
A) ensure purity of genetic lines
B) remove favorable traits from plants and animals
C) enhance growth and yield from domesticated plants and animals
D) train crops to grow in more favorable environments
Answer: C
Section: 22.2
Bloom's Taxonomy: Remembering/Understanding
11) As proof of a functional synthetic genome the JCVI conducted ________ to demonstrate that
JCVI-syn1.0 was a functional genome.
A) gene editing
B) genome transplantation
C) whole genome sequencing
D) homologous recombination
Answer: B
Section: 22.8
Bloom's Taxonomy: Remembering/Understanding
12) Using single-cell RNA sequencing a clinician can now determine ________ in one
experiment instead of two?
A) heterozygosity
B) ratio of genes present to the expression levels of those genes
C) gene expression and protein levels associated with those genes
D) whether the organism is transgenic or not
Answer: B
Section: 22.5
Bloom's Taxonomy: Applying/Analyzing
13) Personal genomics has led to the discovery of many new genes associated with disease
states. Studies on autism spectrum disorder have revealed how many genes to have an impact on
the disease state?
A) 1
B) 5
C) 63
D) >100
Answer: D
Section: 22.5
Bloom's Taxonomy: Remembering/Understanding
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14) Which of the following is not an application of RFLPs?
A) forensic science
B) paternity testing
C) evolutionary biology
D) proteomics
Answer: D
Section: 22.4
Bloom's Taxonomy: Remembering/Understanding
15) 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: 22.7
Bloom's Taxonomy: Evaluating/Creating
16) One procedure that has greatly enhanced prenatal screening is ________.
A) chorionic villus sampling
B) fetal blood sampling
C) GWAS
D) RNA-seq
Answer: A
Section: 22.4
Bloom's Taxonomy: Remembering/Understanding
17) 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: 22.4
Bloom's Taxonomy: Evaluating/Creating
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18) The CJVI undertook a comparison between M. genitalium and Haemophilus influenzae in an
effort to determine what?
A) the core set of genes needed to sustain life
B) the core set of genes that cause H. influenzae virulence
C) the core set of genes needed for M. genitalium to infect humans
D) the core set of genes needed for understanding life
Answer: A
Section: 22.8
Bloom's Taxonomy: Remembering/Understanding
19) One possible use of synthetic biology is bioremediation; this application is best described as
________.
A) isolating bacteria from polluted areas and adjusting their genome to process the pollutant
B) transplanting a completely synthesized genome into a host that can then use genes to process
the pollutant
C) synthesizing an entire organism that can process the pollutant
D) collecting pollutants and subjecting naturally occurring organisms to extreme levels to force
evolution to process the pollutant
Answer: B
Section: 22.8
Bloom's Taxonomy: Evaluating/Creating
20) Synthetic biologists are creating gene clusters that have logic functions and memory called
________.
A) synthetic genes
B) synthetic gene circuits
C) genetic computers
D) organic computers
Answer: B
Section: 22.8
Bloom's Taxonomy: Remembering/Understanding
21) The organization founded to ensure privacy and fairness in the interpretation of genetic
information is called ________.
A) CJVI, Craig J. Venter Institute
B) ELISA, Ethical Listing in Sequence Association
C) ELSI, Ethical, Legal, and Social Implications
D) GWAS, genome-wide associated sequences
Answer: C
Section: 22.9
Bloom's Taxonomy: Remembering/Understanding
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22) 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: 22.9
Bloom's Taxonomy: Remembering/Understanding
23) 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: 22.9
Bloom's Taxonomy: Remembering/Understanding
24) There is a fear that preconception genetic testing may lead to "designer babies." What
historical practice comes to mind regarding the production of such babies?
A) eugenics
B) gene driving
C) gene editing
D) inbreeding
Answer: A
Section: 22.9
Bloom's Taxonomy: Remembering/Understanding
25) 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: 22.9
Bloom's Taxonomy: Remembering/Understanding
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26) Due to the relatively unregulated environment that genomic information is being collected in,
which of the following is not a concern?
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: 22.9
Bloom's Taxonomy: Remembering/Understanding
27) Currently, the most common method of vaccination is injection. Modern biotechnology is
likely to make injection less common through the development of ________.
A) edible vaccines
B) gene editing
C) biopharming
D) fusion proteins
Answer: A
Section: 22.1
Bloom's Taxonomy: Applying/Analyzing
28) There are several problems associated with the use of new GMOs. One of which is
________.
A) the ecological impact is well known
B) the yields from the genetic modification may not be useful in the field
C) the ecological impact is largely unknown
D) the GMO could prove dangerous for human consumption
Answer: C
Section: 22.1
Bloom's Taxonomy: Applying/Analyzing
29) To fight the recent outbreak of Zika virus in the Americas, scientists took advantage of
________ to force a gene to be passed on to progeny that resulted in short-lived offspring that
did not reach maturity.
A) gene drive
B) selective breeding
C) DNA vaccines
D) CRISPR-Cas
Answer: A
Section: 22.3
Bloom's Taxonomy: Remembering/Understanding
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30) The use of ASOs in the process of in vitro fertilization to determine the genetic risk factors
present is called ________.
A) next-generation sequencing
B) preimplantation genetic diagnosis
C) RFLP analysis
D) genome scanning
Answer: B
Section: 22.4
Bloom's Taxonomy: Remembering/Understanding
31) Evolving genome sequencing methods continue to be used in ever more time-sensitive
situations. Which of the following is an emerging situation in which the CDC is using WGS,
NGS, and TGS?
A) pathogen identification
B) vaccine generation
C) tracking impact of pathogens on host genomes
D) isolating pathogens after they have spread
Answer: A
Section: 22.4
Bloom's Taxonomy: Remembering/Understanding
32) The incorporation of a heavy metal inducible promoter in front of RFP in the transgenic
animal GloFish is an example of what?
A) a bioassay for water contamination by heavy metals
B) an in vitro assay for water contamination by heavy metals
C) a transgenic fish that glows under UV light when heavy metals are absent
D) a transgenic animal sold as a gimmick
Answer: A
Section: 22.3
Bloom's Taxonomy: Remembering/Understanding
33) What is the ultimate goal of the Human Genome Project?
A) to rid the human population of genetic disease
B) to better understand the genes and regulatory elements of the genome
C) to map all the genes in the genome so they can be edited with CRISPR-cas
D) to understand all of the proteins generated in a cell
Answer: B
Section: 22.4
Bloom's Taxonomy: Applying/Analyzing
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34) The incorporation of the lacZ gene into the JCVI-syn1.0 genome was used as an indicator for
successful transplantation because ________.
A) the M. capricolum genome did not include lacZ
B) tetracycline resistance is conferred by lacZ
C) the original genome of M. mycoides contained lacZ
D) tetracycline resistance is a selective marker
Answer: A
Section: 22.8
Bloom's Taxonomy: Remembering/Understanding
35) The CJVI has identified approximately 473 genes make up the core set of genes for
prokaryotes. What did their findings suggest is the core set of genes in the human genome?
A) about 500
B) about 1000
C) about 2000
D) about 20,000
Answer: C
Section: 22.8
Bloom's Taxonomy: Remembering/Understanding
36) 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: 22.8
Bloom's Taxonomy: Applying/Analyzing
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 23 Developmental Genetics
1) Which term refers to the regulatory events that establish a specific pattern of gene activity and
developmental fate for a given cell?
A) lysogen
B) differentiation
C) determination
D) gradient regulated
E) attenuation
Answer: C
Section: 20.0
Bloom's Taxonomy: Remembering/Understanding
2) Which general genetic process is believed to account for the variety of cellular structures and
functions in eukaryotic cells?
A) variable gene activity
B) negative control exclusively
C) maternal environmental activities
D) intron processing
E) RNA processing
Answer: A
Section: 23.1
Bloom's Taxonomy: Remembering/Understanding
3) Immediately after fertilization of a Drosophila egg, the zygote nucleus undergoes a series of
divisions. Subsequent nuclear migration generates a(n) ________.
A) syncytium
B) maternal effect
C) homeodomain
D) zygote
E) cleavage nucleus
Answer: A
Section: 23.3
Bloom's Taxonomy: Remembering/Understanding
4) In what order do the segmentation genes function in Drosophilia?
A) gap, segment-polarity, pair-rule
B) pair-rule, transdeterminal, gap
C) transdeterminal, gap, pair-rule
D) gap, pair-rule, segment-polarity
E) segmentational, helical, spherical
Answer: D
Section: 23.4
Bloom's Taxonomy: Remembering/Understanding
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5) Mutations that eliminate a contiguous region in the Drosophila embryo's segmentation pattern
are called ________.
A) homeodomains
B) gap genes
C) compartment genes
D) linkage genes
E) segment genes
Answer: B
Section: 23.4
Bloom's Taxonomy: Remembering/Understanding
6) Which of the following is a fundamental characteristic of the Notch pathway?
A) It works through nuclear signals involving the lac operon.
B) It works through direct cell-to-cell contact.
C) It allows mRNAs to accumulate in the posterior portion of the Drosophila embryo.
D) It directs mRNAs to antisense systems in the posterior portion of all organisms.
E) It provides symmetry to plant flowers.
Answer: B
Section: 23.7
Bloom's Taxonomy: Remembering/Understanding
7) Provide an explanation of the differences between differentiation and determination. Provide
examples of each process and indicate how each is involved in development.
Answer: Determination is the early commitment of a cell to an eventual developmental fate.
Differentiation is the set of functional and structural changes associated with the expression of
that developmental fate. When nuclei of the developing Drosophila embryo reach the blastoderm
stage, they become determined. Differentiation of the cells in Drosophila occurs during
metamorphosis. Development, therefore, probably depends first on the determination of cells,
then on the differentiation of cells.
Section: 23.0
Bloom's Taxonomy: Remembering/Understanding
8) How does determination relate temporally to differentiation?
Answer: Determination occurs when a cell's developmental fate is set, whereas differentiation is
the expression of that determined state. Determination occurs before differentiation.
Section: 23.0
Bloom's Taxonomy: Evaluating/Creating
9) Describe the difference between a totipotent cell and a pluripotent cell.
Answer: Totipotent cells are capable of becoming any type of cell in the mature organism.
Pluripotent cells are capable of becoming any cell in the organism with the exception of
placental and embryonic cell types.
Section: 23.1
Bloom's Taxonomy: Remembering/Understanding
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10) Provide a brief description of the variable gene activity hypothesis as it relates to
development. What information is often provided in support of this hypothesis?
Answer: Since all cells apparently have the same genetic information, yet may differ structurally
and functionally, the variable gene activity hypothesis suggests that differential gene
transcription (genetic regulation) accounts for such cellular diversity.
Section: 23.1
Bloom's Taxonomy: Remembering/Understanding
11) Design an experiment that would allow you to determine if a particular nucleus in a
Drosophila embryo is capable of directing development of an entire new fly.
Answer: Experiments in support of such genomic equivalence include the following: the
observation that chromosome number and structure do not consistently change in different cells
of an organism, nuclear transplantation in amphibians, and the presence of genes but no gene
products in some tissues (hemoglobin, for example). For the Drosophila experiment, one could
do nuclear transplantation similar to the classic experiments in amphibians and recent
experiments in mammals or transplant a cell from embryos of one genotype to the posterior end
of the early embryos of another genotype. The transplanted cells would be reprogrammed into
germ cells by the posterior P granules. The transplant can be detected in the next generation by
the recognizable different phenotypes due to different genotypes of the donor and recipient.
Section: 23.3
Bloom's Taxonomy: Evaluating/Creating
12) It is often said that development is a two-step process. What two steps are likely to be
referred to here?
Answer: determination and differentiation
Section: 23.0
Bloom's Taxonomy: Remembering/Understanding
13) What kinds of general observations cause one to conclude that development is the result of
variable gene activity?
Answer: Individual cells of multicellular organisms produce different gene products, and each
apparently has an equal genomic complement.
Section: 23.1
Bloom's Taxonomy: Applying/Analyzing
14) Three investigators–Nüsslein-Volhard, Wieschaus, and Lewis–won the Nobel Prize for
physiology and medicine in 1995 for work they did in 1970 with Drosophila. Briefly describe
their findings.
Answer: They identified and characterized a number of maternal-effect and zygotic genes that
control pattern formation in Drosophila.
Section: 23.3
Bloom's Taxonomy: Remembering/Understanding
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15) Maternal effects are cases of extrachromosomal inheritance in which the genotype of the
mother influences the phenotype of her immediate offspring in a non-Mendelian manner.
Provide a rationale for the molecular basis of a maternal effect.
Answer: During development of the egg, females provide numerous nutritional and
informational substances, which direct and support early embryonic development. These
substances are often in the form of transcription factors, receptors, mRNA, and proteins,
although other substances are also likely (substrates and products).
Section: 23.3
Bloom's Taxonomy: Applying/Analyzing
16) Experiments involving nuclear transplantation in amphibians indicate that nuclei derived
from blastula are more likely to support development of complete and normal adults compared
with those derived from later stages of development. What do these experiments tell us about the
process of development?
Answer: Development occurs as a series of cascades, with early genes influencing late genes, at
times with stability. Such progressive determination, if stable, may be irreversible and fail to
support development of an entire organism from a single cell. A number of epigenetic factors
such as chromatin remodeling and DNA methylation are also likely to come into play.
Section: 23.1
Bloom's Taxonomy: Evaluating/Creating
17) Which functions earlier in development: maternal-effect genes or zygotic genes?
Answer: maternal-effect genes
Section: 23.3
Bloom's Taxonomy: Remembering/Understanding
18) In Drosophila, which type of genes are controlled by maternal gene products laid down in
gradients in the egg?
A) pair-rule genes
B) segment polarity genes
C) gap genes
D) all genes are controlled by maternal gene products
Answer: C
Section: 23.4
Bloom's Taxonomy: Remembering/Understanding
19) In most species the expression of Hox genes is colinear with the anterior to posterior
organization of the organism. Describe what this means.
Answer: Hox genes expressed at the anterior of the organism are located at the 3′ end of the
gene cluster. Those expressed in the posterior are transcribed form the 5′ end of the cluster.
Section: 23.4
Bloom's Taxonomy: Remembering/Understanding
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20) Arabidopsis thaliana has homologous gene clusters to the Hox genes found in Drosophila.
What are these master regulatory clusters called?
A) MADS-box proteins
B) Hox genes
C) MRCs
D) CURLY LEAF genes
Answer: A
Section: 23.6
Bloom's Taxonomy: Remembering/Understanding
21) Flowers that develop in Arabidopsis originate from concentric rings of cells. What class of
gene(s) is responsible for the development of sepals?
A) A genes
B) A and B genes
C) C genes
D) B and C genes
Answer: A
Section: 23.6
Bloom's Taxonomy: Remembering/Understanding
22) C. elegans is used as a model organism in the study of the Notch signaling pathway. What is
the method communication between cells in the Notch signaling pathway?
A) release of extracellular signals to be detected by neighboring cells
B) electrical propagation of signals across small gaps
C) direct cell-to-cell contact through transmembrane proteins
D) cells do not communicate using the Notch signaling pathway
Answer: C
Section: 23.7
Bloom's Taxonomy: Remembering/Understanding
23) A gene that specifies the fate of a particular anatomical segment in Drosophila is called a(n)
________.
A) selector gene
B) silent gene
C) MADS genes
D) determinant genes
Answer: A
Section: 23.5
Bloom's Taxonomy: Remembering/Understanding
24) Which class of genes controls the developmental identity of segments along the anterior—
posterior axis?
Answer: Homeotic genes
Section: 23.3
Bloom's Taxonomy: Remembering/Understanding
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25) Mutations in the selector genes sometimes form structures in the wrong segment,
transforming the antenna of a fly into a leg, for example. What are such mutations called?
A) silent mutations
B) homeotic mutations
C) homologous mutations
D) lethal mutations
Answer: B
Section: 23.3
Bloom's Taxonomy: Remembering/Understanding
26) Over the course of evolution, humans have obtained four Hox gene clusters due to
duplication. The presence of multiple Hox gene clusters suggests what?
A) Three of the four clusters are silent and have no effect on development.
B) All four clusters are active and represent the added complexity of vertebrate development.
C) Only two of the clusters are active and represent the added complexity of vertebrate
development.
D) Only one of the clusters is silent since it would result in Drosophila development not
vertebrate.
Answer: B
Section: 23.5
Bloom's Taxonomy: Remembering/Understanding
27) It is presently believed that the asymmetric distribution of substances in the egg, positioned
by the mother, is responsible for establishing the determined state. Which general term describes
such molecular positioning?
Answer: maternally constructed molecular gradients
Section: 23.3
Bloom's Taxonomy: Remembering/Understanding
28) What is the significance of the homeodomain?
Answer: The homeodomain is a highly conserved protein of 60 amino acids found in a variety
of organisms, which, in conjunction with other factors, is thought to play a role in DNA binding
and transcriptional activation.
Section: 23.5
Bloom's Taxonomy: Applying/Analyzing
29) From a human developmental standpoint, what is the significance of Hox genes?
Answer: Hox genes are highly conserved among all organisms with bilateral symmetry. The
arrangement and expression pattern of Hox genes establish anterior/posterior development in
mammals. Some limb malformations in humans are caused by mutations in Hox genes.
Section: 23.5
Bloom's Taxonomy: Evaluating/Creating
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30) The bicoid gene of Drosophila generates embryos with two posterior regions. What is the
likely aspect of embryonic development that the bicoid gene probably controls?
Answer: Anterior—posterior polarity
Section: 23.3
Bloom's Taxonomy: Evaluating/Creating
31) Give a brief definition of a homeobox.
Answer: a highly conserved genetic element that encodes a 60 amino acid sequence with DNAbinding characteristics
Section: 23.5
Bloom's Taxonomy: Remembering/Understanding
32) During development, many genes act in such a way as to decrease the number of alternative
developmental pathways that a cell can follow. Usually there are two alternative developmental
paths for a cell to follow. What is the term given to genes that act in this fashion?
Answer: binary switch genes
Section: 23.8
Bloom's Taxonomy: Remembering/Understanding
33) What significant conclusion has the classic work on the eyeless and Pax6 genes provided in
terms of developmental mechanisms?
Answer: Some vertebrate versions of binary switch genes are capable of triggering
developmentally significant processes across broad evolutionary distances. Such genes are
therefore related evolutionarily.
Section: 23.8
Bloom's Taxonomy: Applying/Analyzing
34) The process whereby a cell's eventual developmental fate is set is called ________.
A) determination
B) differentiation
C) notching
D) embryogenisis
Answer: A
Section: 23.0
Bloom's Taxonomy: Remembering/Understanding
35) The process whereby a cell's determined state is expressed is called ________.
A) determination
B) differentiation
C) notching
D) embryogenisis
Answer: B
Section: 23.0
Bloom's Taxonomy: Remembering/Understanding
7
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36) A mutation seen in Drosophila is found to cause antennae to grow on the abdomen of the
flies. What type of mutation is this likely to represent?
A) homeotic mutation
B) lethal mutation
C) silent mutation
D) evolutionary mutation
Answer: A
Section: 23.5
Bloom's Taxonomy: Applying/Analyzing
37) Which of the following is not a set of master regulatory genes that control homeobox genes
during development?
A) CURLY LEAF
B) Polycomb
C) Hox
Answer: C
Section: 23.6
Bloom's Taxonomy: Remembering/Understanding
38) After fertilization of an egg the totipotent cell divides and the cells begin to specialize. These
30—40 cells are called ________ and are pluripotent.
A) embryonic stem cells
B) somatic cells
C) germ line cells
D) T cells
Answer: A
Section: 23.1
Bloom's Taxonomy: Remembering/Understanding
39) A homeobox gene is one that produces a ________.
A) repetitive nucleotide sequence such as VNTR
B) 60 amino acid DNA binding domain
C) maternal transcription factors
D) transcription factor that strips methylation from the entire genome
Answer: B
Section: 23.5
Bloom's Taxonomy: Remembering/Understanding
40) Development of the anchor cell in the vulva in the nematode C. elegans is the result of
________.
A) specific upregulation of the Delta protein
B) stochastically increased production of Delta protein
C) specific upregulation of the LIN-12 protein
D) stochastically increased production of the LIN-12 protein
Answer: B
Section: 23.7
Bloom's Taxonomy: Remembering/Understanding
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41) Caenorhabditis elegans is extremely useful as an experimental model system due to all of
the following EXCEPT ________.
A) its genetics are well known
B) its genome has been sequenced
C) adults contain a small, fixed number of cells
D) it is highly variable in size and shape making it good for evo-devo studies
Answer: D
Section: 23.7
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 24 Cancer Genetics
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: 24.1
Bloom's Taxonomy: Remembering/Understanding
2) Which of the following general mechanisms appear to be involved in the formation of cancer
cells?
A) genomic instability, DNA repair failure, chromatin modifications
B) inversions, operon formation, methylation
C) RNA failure, DNA phosphorylation, phosphorylation of adenyl cyclase
D) transdetermination, mutation, allosteric interactions
E) suppression, tabulation, projection
Answer: A
Section: 24.1
Bloom's Taxonomy: Remembering/Understanding
3) What is the name of the protein that 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: 24.3
Bloom's Taxonomy: Remembering/Understanding
4) Which of the following proteins function as a cell-cycle regulator and transcription factor that
can result in cell death (apoptosis) to a damaged cell?
A) p34
B) p102
C) cyclin
D) p53
E) phosphokinase
Answer: D
Section: 24.4
Bloom's Taxonomy: Remembering/Understanding
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5) Mutations in the ras gene family induce normally quiescent cells to proceed into the
replication cycle. This converts the ras gene from a ________ gene to a ________ gene.
A) proto-oncogene; oncogene
B) oncogene; proto-oncogene
C) mutant; oncogene
D) tumor suppressor; proto-oncogene
E) pseudooncogene; proto-oncogene
Answer: A
Section: 24.4
Bloom's Taxonomy: Applying/Analyzing
6) Mutant versions of genes that are normally 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: 24.4
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 forms 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: 24.6
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: 24.4
Bloom's Taxonomy: Remembering/Understanding
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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: 24.1
Bloom's Taxonomy: Remembering/Understanding
10) Briefly describe what a passenger mutation is with respect to cancer genetics.
Answer: A passenger mutation is a genetic alteration that does not result in a tumor. It is a
mutation that confers a selective advantage to tumorigenesis in the presence of an oncogene.
Section: 24.1
Bloom's Taxonomy: Remembering/Understanding
11) Chronic myelogenous leukemia appears to be associated with a chromosomal rearrangement.
How is a chromosomal rearrangement responsible for this disease?
Answer: Joining of chromosomes 9 and 22 through translocation generates a hybrid gene bcr/cabl that produces a protein that is inappropriately active and causes the disease.
Section: 24.2
Bloom's Taxonomy: Remembering/Understanding
12) Signal transduction is best described as ________.
A) communication between cells through physical interaction
B) transmission of signals external to the cell to the nucleus
C) transmission of signals from cancer cells to normal cells
D) communication between nuclei of two organisms
Answer: B
Section: 24.3
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: 24.3
Bloom's Taxonomy: Evaluating/Creating
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 either CDK or its partner cyclin could deregulate key
checkpoints resulting in cancer cell proliferation.
Section: 24.3
Bloom's Taxonomy: Evaluating/Creating
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15) A loss-of-function mutation in BAX proteins could eliminate apoptotic response in cell.
Describe why this type of mutation is a passenger mutation and not a driver mutation.
Answer: The loss of function of the apoptotic genes are a passenger mutation as they do not
contribute to tumorigenesis but confer a selective advantage by eliminating the apoptotic defense
mechanism.
Section: 24.3
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 apoptotic
response or necrosis. Normal cells are nonproliferating and have functional DNA repair
mechanisms that prevent their apoptotic response.
Section: 24.3
Bloom's Taxonomy: Applying/Analyzing
17) Which three stages or transitions in the cell cycle seem to serve as points of control
(checkpoints)?
Answer: G1/S, G2/M, M
Section: 24.3
Bloom's Taxonomy: Remembering/Understanding
18) Describe how tumor-suppressor genes keep cells in check.
Answer: Tumor-suppressor genes generate gene products that regulate cell cycle checkpoints
and may initiate apoptosis. When these genes suffer mutations their ability to regulate cell cycle
and apoptosis is lost resulting in tumrigenesis.
Section: 24.4
Bloom's Taxonomy: Remembering/Understanding
19) Name two of the classes of proteins that combine to directly control progression through the
cell cycle.
Answer: CDK and cyclins
Section: 24.4
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: 24.4
Bloom's Taxonomy: Applying/Analyzing
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21) Chronic myelogenous leukemia is a cancer found in white blood cells. What is its supposed
genetic basis of this disease?
Answer: CML is the result of a specific translocation between chromosome 9 and 22. The CABL gene from chromosome 9 is translocated next to the BCR gene on chromosome 22. This
results in a fusion protein being expressed. The new chromosomal construct is called the
Philadelphia chromosome.
Section: 24.2
Bloom's Taxonomy: Remembering/Understanding
22) Describe the normal function of a tumor-suppressor gene.
Answer: A tumor-suppressor gene is a gene whose normal function is to suppress cell division.
When mutant, cell division control is lost and a cancer may form.
Section: 24.4
Bloom's Taxonomy: Remembering/Understanding
23) Describe the cellular and molecular function of the ras gene family and the consequences of
mutations in ras.
Answer: The ras gene family encodes a protein that is involved with signal transduction in the
cell membrane. Point mutations may cause changes in function that promote abnormal signaling,
thus stimulating uncontrolled cell growth.
Section: 24.4
Bloom's Taxonomy: Remembering/Understanding
24) Describe three genetic mechanisms whereby proto-oncogenes can become overexpressed.
Answer: Mutations can occur which cause an enhancement of the promoter or other upstream
regulatory signals.
Section: 24.4
Bloom's Taxonomy: Applying/Analyzing
25) What is the name of a normal gene that serves to promote cellular division?
A) proto-oncogene
B) post-oncogene
C) oncogene
D) cancer gene
Answer: A
Section: 24.4
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: 24.4
Bloom's Taxonomy: Remembering/Understanding
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27) Describe the molecular nature of mutation, as related to cancer, in a ras gene.
Answer: The ras gene family encodes a protein of 189 amino acids, which is involved with
signal transduction in the cell membrane. Point mutations may cause changes in function to
allow signals to enter the cell abnormally, thus stimulating uncontrolled cell growth.
Section: 24.4
Bloom's Taxonomy: Remembering/Understanding
28) Briefly describe the following types of genes: tumor-suppressor gene, proto-oncogene, and
oncogene.
Answer: Tumor-suppressor genes normally function to inactivate or repress cell division. Protooncogenes normally function to promote cell division, whereas oncogenes are mutant forms of
proto-oncogenes.
Section: 24.4
Bloom's Taxonomy: Remembering/Understanding
29) Much has been written about p53 in terms of cancer biology. What is p53, and what is its
significance?
Answer: Mutations in the p53 gene are important in the development of a number of cancers. It
is a tumor-suppressor gene that normally functions to control the transition from late G1 to S
phase. The product of p53 has DNA-binding properties.
Section: 24.4
Bloom's Taxonomy: Remembering/Understanding
30) What are two properties that various types of cancer cells share?
Answer: uncontrolled proliferation and accumulation of somatic mutations
Section: 24.1
Bloom's Taxonomy: Remembering/Understanding
31) Most p53 mutations that result in cancer are a result of loss of function in p53. Describe one
method through which a gain-of-function mutation in p53 could cause cancer.
Answer: The gain-of-function mutation in p53 could result in the overproduction of chromatinremodeling enzymes. The improper expression of these enzymes would likely result in
misexpression of other genes and could lead to tumorigenesis.
Section: 24.4
Bloom's Taxonomy: Remembering/Understanding
32) Most cancers arise due to mutations in somatic cells, if someone has a predisposition to
cancer, what genetic circumstance likely exists?
Answer: a germ-line mutation in a proto-oncogene or tumor suppressor, which is inherited
Section: 24.6
Bloom's Taxonomy: Remembering/Understanding
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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: 24.6
Bloom's Taxonomy: Applying/Analyzing
34) The genetic difference between familial retinoblastoma and sporadic retinoblastoma appears
to be based on those with the familial form starting out being ________, whereas those with the
sporadic form start out being ________.
Answer: heterozygous for the retinoblastoma gene; homozygous normal
Section: 24.6
Bloom's Taxonomy: Evaluating/Creating
35) Most cancer-causing animal viruses are RNA viruses called ________.
A) neoviruses
B) retroviruses
C) classic viruses
D) DNA viruses
Answer: B
Section: 24.7
Bloom's Taxonomy: Remembering/Understanding
36) In what way might a virus contribute to cancer formation?
Answer: Proviral DNA may integrate near a proto-oncogene and alter its expression, thus
generating an oncogene, or it may bring in an oncogene during an infection. In either case, the
cell cycle can be upregulated by a virus.
Section: 24.7
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: 24.8
Bloom's Taxonomy: Remembering/Understanding
38) Provide a simple definition of a carcinogen.
Answer: a cancer-causing agent
Section: 24.8
Bloom's Taxonomy: Remembering/Understanding
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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: 24.1
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: 24.2
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: 24.1
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: 24.3
Bloom's Taxonomy: Remembering/Understanding
43) The gene p53 is called the ________ because it corrects mutations in the spindle apparatus
before nondisjunction can occur.
A) guardian of the genome
B) M phase checkpoint monitor
C) S phase checkpoint monitor
D) tumor-suppressor monitor
Answer: A
Section: 24.4
Bloom's Taxonomy: Remembering/Understanding
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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: 24.4
Bloom's Taxonomy: Remembering/Understanding
45) Which virus type results in the immediate onset of cancer phenotype upon infection of a
host?
A) retrovirus
B) acute transforming retrovirus
C) adenovirus
D) lentivirus
Answer: B
Section: 24.7
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: 24.8
Bloom's Taxonomy: Remembering/Understanding
47) Examined cancer cells from the lungs of smokers demonstrate which of the following?
A) normal mutation rates compared to non-smokers
B) unregulated 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: 24.8
Bloom's Taxonomy: Remembering/Understanding
48) A retrovirus uses ________ to make a DNA copy of RNA.
A) reverse transcriptase
B) ras proteins
C) cyclins
D) CDKs
Answer: A
Section: 24.7
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 25 Quantitative Genetics and Multifactorial Traits
1) Meristic traits are characterized by having phenotypes that are described by ________.
A) whole numbers
B) irrational numbers
C) imaginary numbers
D) prime numbers
E) complex numbers
Answer: A
Section: 25.1
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: 25.1
Bloom's Taxonomy: Remembering/Understanding
3) Assume that a cross is made between tall and dwarf tobacco plants. The F1 generation showed
intermediate height, while 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: 25.2
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: 25.3
Bloom's Taxonomy: Remembering/Understanding
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5) When multiple genes contribute to the expression of a phenotype they are called ________.
A) nonadditive alleles
B) heterozygous alleles
C) additive alleles
D) homozygous alleles
E) concordant alleles
Answer: C
Section: 25.2
Bloom's Taxonomy: Remembering/Understanding
6) Continuous inheritance is often related to the term quantitative inheritance. Why?
Answer: In continuous inheritance, we consider each involved locus as having a quantitative
input on the production of a single characteristic of the phenotype. In addition, although it may
not always be the case, we would consider each gene product to be qualitatively similar.
Section: 25.1
Bloom's Taxonomy: Remembering/Understanding
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?
Answer: 1:4:6:4:1
Section: 25.2
Bloom's Taxonomy: Applying/Analyzing
8) How many gene pairs are involved in generating a typical 1:4:6:4:1 ratio?
A) 2
B) 3
C) 4
D) 6
Answer: A
Section: 25.2
Bloom's Taxonomy: Remembering/Understanding
9) When a trait is polygenic, but has a small number of discrete classes, it is said to be a(n)
________ trait.
A) meristic
B) continuous
C) oncogenic
D) qualitative
E) threshold
Answer: E
Section: 25.1
Bloom's Taxonomy: Remembering/Understanding
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10) 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 traits
B) indeterminate traits
C) threshold traits
D) continuous traits
Answer: A
Section: 25.2
Bloom's Taxonomy: Remembering/Understanding
11) Assume that you are studying a trait determined by a number of polygenes. If there are seven
phenotypic categories, how many genes are probably involved?
A) 2
B) 3
C) 4
D) 7
Answer: B
Section: 25.2
Bloom's Taxonomy: Applying/Analyzing
12) Polygenic traits often demonstrate a ________ distribution when the phenotypes from a
proper sample are represented in a histogram.
A) normal
B) Poisson
C) binomial
D) uniform
Answer: A
Section: 25.2
Bloom's Taxonomy: Remembering/Understanding
13) If the proportion of F2 individuals resembling either of the two most extreme phenotypes
(the parental phenotypes) can be determined, what formula can be applied to determine the
number of gene pairs involved?
A) 1/4n (where n is the number of gene pairs)
B) 2n + 1 (where n is the number of gene pairs)
C) s =
where s is the number of gene pairs)
D)
=
where n is the number of gene pairs)
Answer: A
Section: 25.2
Bloom's Taxonomy: Remembering/Understanding
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14) What formula can be used to determine the number of categories (phenotypes) possible in
the F2 results of a polygenic system?
A) 1/4n (where n is the number of gene pairs)
B) 2n + 1 (where n is the number of gene pairs)
C) s =
where s is the number of gene pairs)
D)
=
where n is the number of gene pairs
Answer: B
Section: 25.2
Bloom's Taxonomy: Remembering/Understanding
15) 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?
A) 3
B) 6
C) 16
D) 32
E) 64
Answer: A
Section: 25.2
Bloom's Taxonomy: Applying/Analyzing
16) Assume that four polygenic gene pairs are involved in determining phenotypes of F2. How
many phenotypic classes are expected?
A) 4
B) 9
C) 8
D) 16
Answer: B
Section: 25.2
Bloom's Taxonomy: Applying/Analyzing
17) What is the term given to a random subset of individuals who are selected for measurement
in a particular study?
A) sample
B) representatives
C) controls
D) placebos
Answer: A
Section: 25.3
Bloom's Taxonomy: Remembering/Understanding
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18) Which of the following is not a statistical value often used to represent a sample when
evaluating QTLs?
A) mean
B) standard deviation
C) variance
D) mode
Answer: D
Section: 25.3
Bloom's Taxonomy: Remembering/Understanding
19) Given the following numbers, calculate the mean: 10, 12, 14, 16, 18.
A) 14
B) 10
C) 18
D) 13
Answer: A
Section: 25.3
Bloom's Taxonomy: Applying/Analyzing
20) s2 = S (Xi − X)2/n − 1 is the formula for calculating which statistical value?
A) mean
B) covariance
C) standard deviation
D) variance
Answer: D
Section: 25.3
Bloom's Taxonomy: Remembering/Understanding
21) To estimate how much the means of a variety of like samples drawn from the same
population might vary, what statistic is often used?
A) relative error of the mean
B) standard error of the mean
C) standard deviation of the mean
D) variance of the mean
Answer: B
Section: 25.3
Bloom's Taxonomy: Remembering/Understanding
22) Which statistical value is used to represent the linkage of the increase in one measurement to
the increase in a second measurement?
A) standard deviation
B) correlation coefficient
C) covariance
D) variance
Answer: B
Section: 25.3
Bloom's Taxonomy: Remembering/Understanding
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23) Identical twins are ________.
A) monozygotic
B) dizygotic
C) homozygotic
D) hemizygotic
Answer: A
Section: 25.5
Bloom's Taxonomy: Remembering/Understanding
24) Define the term broad-sense heritability (H2). What is implied by a relatively high value of
H2? Express aspects of broad-sense heritability in equation form.
Answer: Broad-sense heritability (H2) refers to the degree to which phenotypic variation is due
to genetic factors. A very high value of H2 indicates that the environment had a relatively low
impact on phenotypic variation. VP = VE + VG + VGE.
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
25) If one is attempting to determine the influence of genes or the environment on phenotypic
variation, inbred strains with individuals of a relatively homogeneous or constant genetic
background are often used. 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 various
environmental conditions?
Answer: Nongenetic factors generally categorized as "environmental" components
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
26) What is the formal expression used to examine the relative importance of genetic versus
environmental factors?
A) broad-sense heritability index (H2) = VG/VP
B) narrow-sense heritability index (h2) = VA/VP
C) genotype-by-environment interaction variance VP = VG + VE + VG*E
D) correlation coefficient s =
Answer: A
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
27) We often refer to the host of genetic factors in the genome that have an influence on a
particular gene's expression as its genetic background. What influence does genetic background
have over the phenotype of the organism?
Answer: Genetic background reflects all of the epigenetic influences exerted on a single or a set
of genes that directly impact the observed phenotype.
Section: 25.4
Bloom's Taxonomy: Applying/Analyzing
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28) Which of the following is not a component of phenotypic variance?
A) environmental variance
B) genetic variance
C) the interaction between genetics and environment
D) the covariance
Answer: D
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
29) Interpret the meaning of an H2 (broad-sense heritability) value that approaches 1.0.
Answer: Almost all the phenotypic variation is determined by genotype differences and
environmental factors have little to no effect.
Section: 25.4
Bloom's Taxonomy: Applying/Analyzing
30) 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: 25.4
Bloom's Taxonomy: Applying/Analyzing
31) Provide a brief definition of the term interactive variance.
Answer: Interactive variance occurs when two or more loci behave epistatically.
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
32) Which statement best describes h2 (narrow-sense heritability)?
A) It measures the contribution genotypic variance to total phenotypic variance.
B) The proportion of phenotypic variance due to additive genotypic variance alone.
C) The deviation from additive components when they behave epistatically.
D) The genotypic variance due to additive action of various QTLs.
Answer: B
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
33) Which value of narrow-sense heritability (h2) indicates a selection will have a significant
impact in altering the initial population?
A) 99
B) 0.99
C) 20
D) 0.20
Answer: B
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
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34) 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) artificial selection
C) diversity selection
D) exclusive selection
Answer: B
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
35) Which of the following mammalian traits exhibits a relatively low h2 value?
A) tail length in mice
B) birth weight in cattle
C) milk yield in cattle
D) litter size in mice
Answer: D
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
36) 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. 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: 25.4
Bloom's Taxonomy: Evaluating/Creating
37) 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: 25.5
Bloom's Taxonomy: Remembering/Understanding
38) 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: 25.5
Bloom's Taxonomy: Evaluating/Creating
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39) Which human trait would be expected to have high concordance?
A) eye color
B) weight
C) contracting measles
D) hair length
Answer: A
Section: 25.5
Bloom's Taxonomy: Remembering/Understanding
40) Individuals with the same genetic background and a high degree of homozygosity are said to
be isogenic. Of what value are isogenic strains in genetic studies?
Answer: Because the genetic background is the same (or at least very similar), phenotypic
variation must be due to nongenetic factors.
Section: 25.5
Bloom's Taxonomy: Remembering/Understanding
41) 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 being reared in the same household), one can again estimate the
influence of heredity on behavioral traits.
Section: 25.5
Bloom's Taxonomy: Evaluating/Creating
42) 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: 25.5
Bloom's Taxonomy: Remembering/Understanding
43) What are QTLs?
Answer: QTLs are quantitative trait loci and refer to genes controlling the expression of
quantitative traits.
Section: 25.6
Bloom's Taxonomy: Remembering/Understanding
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44) 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 trait loci
(QTLs). DNA polymorphisms generate molecular markers RFLPs, which can serve as reference
points in mapping QTLs.
Section: 25.6
Bloom's Taxonomy: Remembering/Understanding
45) A "marker" in a genetic sense usually represents a site along a chromosome where a specific
nucleotide sequence exists. What specific phrase is used when such markers are identified by
restriction endonucleases and a particular set of DNA fragments is generated?
A) QTL
B) RFLP
C) expression QTL
D) restriction site
Answer: B
Section: 25.6
Bloom's Taxonomy: Remembering/Understanding
46) Expression QTL mapping allows researchers to characterize ________.
A) loci that control gene expression
B) loci that encode phenotypes
C) loci that modify the rate of protein translation
D) loci that have high mutating rates
Answer: A
Section: 25.6
Bloom's Taxonomy: Remembering/Understanding
47) Polygenes are involved in determining ________.
A) continuously varying traits
B) traits that are strictly determined by genetic variance
C) traits that are strictly determined by environmental factors
D) traits that have a distinct all or none phenotype
Answer: A
Section: 25.1
Bloom's Taxonomy: Remembering/Understanding
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48) Which theory suggests that many factors or genes contribute to the phenotype in a
cumulative or quantitative manner?
A) additive gene theory
B) multiple factor hypothesis
C) heritability theory
D) threshold gene theory
Answer: B
Section: 25.1
Bloom's Taxonomy: Remembering/Understanding
49) ________ alleles are those that are not epistatic over nonallelic genes that influence the same
phenotypic characteristic.
A) Additive
B) Nonadditive
C) Threshold
D) Discontinuous
Answer: A
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
50) Heritability is a measure of the degree to which the phenotypic variation of a given trait is
due to ________.
A) genetic factors
B) environmental factors
C) epigenetic regulation
D) somatic mutations
Answer: A
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
51) Which trait is probably not caused primarily by genes that behave codominantly or
epistatically?
A) height
B) skin color
C) general body structure
D) eye color
Answer: D
Section: 25.4
Bloom's Taxonomy: Remembering/Understanding
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52) Concordance refers to the frequency with which members of a twin pair express a ________
trait.
A) different
B) similar
C) mutated
D) unrelated
Answer: B
Section: 25.5
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Chapter 26 Population and Evolutionary Genetics
1) What is the term given to a group of individuals belonging to the same species that live in a
defined geographic area that actually or potentially interbreed?
A) population
B) consanguinity
C) hybrid vigor
D) genetics
E) cytogenetics
Answer: A
Section: 26.1
Bloom's Taxonomy: Remembering/Understanding
2) What 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: 26.1
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: 26.1
Bloom's Taxonomy: Applying/Analyzing
4) In a population of 100 individuals, 49% are of the NN blood type. What percentage is
expected to be MN assuming Hardy-Weinberg equilibrium conditions?
A) 9%
B) 21%
C) 42%
D) 51%
E) There is insufficient information to answer this question.
Answer: C
Section: 26.2
Bloom's Taxonomy: Applying/Analyzing
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5) 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: 26.2
Bloom's Taxonomy: Applying/Analyzing
6) 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: 26.3
Bloom's Taxonomy: Remembering/Understanding
7) In small isolated populations, gene 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: 26.3
Bloom's Taxonomy: Remembering/Understanding
8) 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: 26.9
Bloom's Taxonomy: Remembering/Understanding
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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) 9%
B) 58%
C) 91%
D) 42%
Answer: D
Section: 26.2
Bloom's Taxonomy: Applying/Analyzing
10) Which assumption pertains to a population in a Hardy-Weinberg equilibrium?
A) a small population
B) a selectively mating population
C) a selective advantage of one genotype
D) no migration, mutation or, genetic drift
Answer: D
Section: 26.2
Bloom's Taxonomy: Remembering/Understanding
11) Assume that in a Hardy-Weinberg population, 9% of the individuals are of the homozygous
recessive phenotype. What percentage are homozygous dominant?
A) 49%
B) 91%
C) 51%
D) 9%
Answer: A
Section: 26.2
Bloom's Taxonomy: Applying/Analyzing
12) In a population that meets the Hardy-Weinberg equilibrium assumptions, 81% 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) 81%
B) 19%
C) 18%
D) 72%
Answer: C
Section: 26.2
Bloom's Taxonomy: Applying/Analyzing
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13) In a population of cattle, the following color distribution was noted: 36% red (RR), 48% roan
(Rr), and 16% white (rr). This population is in a Hardy-Weinberg equilibrium. What will be the
distribution of genotypes in the next generation if the Hardy-Weinberg assumptions are met?
A) 36% red; 48% roan; 16% white
B) 30% red; 50% roan; 20% white
C) 30% red; 60% roan; 10% white
D) 40% red; 40% roan; 20% white
Answer: A
Section: 26.2
Bloom's Taxonomy: Remembering/Understanding
14) In a population of 10,000 individuals, where 3600 are MM, 1600 are Mm, and 4800 are mm,
what are the frequencies of the M alleles and the m alleles?
Answer: M = 0.6; m = 0.4
Section: 26.2
Bloom's Taxonomy: Remembering/Understanding
15) What is meant by the equation p + q = 1.0?
Answer: The sum of the relevant individual alleles in a population is equal to 100% of those
alleles.
Section: 26.2
Bloom's Taxonomy: Remembering/Understanding
16) What does the variable 2pq represent?
Answer: the expected frequency of heterozygotes in a population in Hardy—Weinberg
equilibrium
Section: 26.2
Bloom's Taxonomy: Remembering/Understanding
17) Assuming that p = 0.3 for a population in Hardy-Weinberg equilibrium, what would be the
expected frequency of heterozygotes for the involved allelic pair?
Answer: 0.42
Section: 26.2
Bloom's Taxonomy: Applying/Analyzing
18) Give a brief definition of the term genetic equilibrium.
Answer: a condition for a population in which the frequency of given alleles remain constant
from generation to generation
Section: 26.2
Bloom's Taxonomy: Remembering/Understanding
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19) A certain form of albinism in humans is recessive and autosomal. Assume that 1% of the
individuals in a given population are albino. Assuming that the population is in Hardy-Weinberg
equilibrium, what percentage of the individuals in this population is expected to be
heterozygous?
A) 82%
B) 9%
C) 99%
D) 18%
Answer: D
Section: 26.2
Bloom's Taxonomy: Applying/Analyzing
20) Which of the following does not change gene frequencies in populations?
A) mutation
B) diet
C) migration
D) selection
Answer: B
Section: 26.3
Bloom's Taxonomy: Remembering/Understanding
21) Which factor contributes to the phenomenon of natural selection?
A) lack of competition in mating
B) genetic homogeneity in a population
C) competition for survival
D) underpopulation
Answer: C
Section: 26.4
Bloom's Taxonomy: Remembering/Understanding
22) Selection that favors the middle phenotype and selects against the extreme phenotypes is said
to be ________.
A) directional selection
B) stabilizing selection
C) diselection
D) disruptive selection
Answer: B
Section: 26.4
Bloom's Taxonomy: Remembering/Understanding
23) One of the Hardy-Weinberg assumptions states that all genotypes in the population are free
of selective advantage. What influence might a selective advantage of a genotype have on a
Hardy-Weinberg equilibrium?
Answer: Certain alleles will reach the next generation on a nonrandom basis, thus upsetting the
equilibrium.
Section: 26.4
Bloom's Taxonomy: Evaluating/Creating
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24) Mutation and migration introduce new alleles into populations. What is the most likely
principal force that will shift allelic frequencies within large populations?
Answer: natural selection
Section: 26.4
Bloom's Taxonomy: Remembering/Understanding
25) Which term is given to the measure of the proportion of offspring that a particular phenotype
will contribute to the next generation?
Answer: fitness
Section: 26.4
Bloom's Taxonomy: Remembering/Understanding
26) Briefly describe how the founder effect can impact the allelic frequencies in a population.
Answer: The founder effect generates an increase in recessive allelic frequencies by having an
individual carrier found an isolated population with minimal migration effects. This preserves
the frequency of recessive alleles.
Section: 26.7
Bloom's Taxonomy: Evaluating/Creating
27) Which would change allele frequencies more quickly: selection against a dominant allele or
selection against a recessive allele?
Answer: selection against a dominant allele
Section: 26.4
Bloom's Taxonomy: Remembering/Understanding
28) Which natural factors affect changes in genetic variation?
Answer: Chromosomal assortment and crossing over reshuffle combinations of mutations at
different loci; forces such as selection, genetic drift, and migration can increase or decrease the
frequencies of mutant alleles in populations.
Section: 26.5
Bloom's Taxonomy: Remembering/Understanding
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?
Answer: 1/100,000
Section: 26.5
Bloom's Taxonomy: Applying/Analyzing
30) Migration occurs when individuals move between populations. Considering a single pair of
alleles, A and a, what formula is used to indicate the new frequency of A in one generation of
migration?
Answer: pi' = (1 - m)pi + mpm
Section: 26.6
Bloom's Taxonomy: Remembering/Understanding
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31) One of the Hardy-Weinberg assumptions states that the population is infinitely large. What
influence might a small population size have on a Hardy-Weinberg equilibrium?
Answer: Sampling error would cause random and possibly significant fluctuations in
determination of gene frequencies.
Section: 26.7
Bloom's Taxonomy: Evaluating/Creating
32) What are the genetic consequences of inbreeding?
Answer: For a given allele, inbreeding increases the proportion of homozygotes in a population.
Section: 26.8
Bloom's Taxonomy: Remembering/Understanding
33) In zoo animals, inbreeding often occurs because of a lack of a sufficient pool of breeding
individuals. Under such conditions, what two characteristics are often exhibited among inbred
organisms?
Answer: higher frequency of aberrant phenotypes and higher mortality rates
Section: 26.8
Bloom's Taxonomy: Remembering/Understanding
34) Two species that have experienced several thousand years of isolation from one another are
now incapable of producing viable offspring despite mating efforts. What type of reproductive
isolating mechanism developed between the two species?
Answer: postzygotic mechanisms
Section: 26.9
Bloom's Taxonomy: Applying/Analyzing
35) Provide a general definition for 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: 26.9
Bloom's Taxonomy: Remembering/Understanding
36) What 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: 26.9
Bloom's Taxonomy: Remembering/Understanding
37) Define the term fitness.
Answer: Fitness is a measure of the proportion of offspring that a particular phenotype will
contribute to the next generation.
Section: 26.4
Bloom's Taxonomy: Remembering/Understanding
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38) Present a rationale for using DNA sequence polymorphisms as an index of genetic 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, while there may be a
general relationship between genetic diversity and evolutionary (phylogenetic) diversity, it is
likely that there are exceptions.
Section: 26.10
Bloom's Taxonomy: Evaluating/Creating
39) Why is mitochondrial DNA often used in construction of phylogenetic trees?
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: 26.10
Bloom's Taxonomy: Applying/Analyzing
40) What type of evolutionary information can be gained by examining the DNA sequence of a
gene for a protein such as cytochrome c?
Answer: Cytochrome c is a respiratory pigment found in the mitochondria of eukaryotes.
Because of its vital role in aerobic metabolism, it has evolved very slowly. Differences signify
vast evolutionary distances.
Section: 26.10
Bloom's Taxonomy: Evaluating/Creating
41) Cytochrome c is a respiratory pigment found in the mitochondria of eukaryotes. Compared
with some other proteins, it has changed very slowly over long periods of time. Why?
Answer: It serves a vital function; therefore, any changes in amino acid sequence are usually
strongly selected against.
Section: 26.10
Bloom's Taxonomy: Applying/Analyzing
42) What are phylogenetic trees?
Answer: diagrammatic representations of evolutionary relationships among organisms
Section: 26.10
Bloom's Taxonomy: Remembering/Understanding
43) 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: 26.10
Bloom's Taxonomy: Applying/Analyzing
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44) What information have the results of mitochondrial DNA analysis offered concerning the
phylogenetic relationships between Neanderthals and modern humans?
Answer: Although Neanderthals and humans share a common ancestor, Neanderthals were a
separate hominid line and probably did not contribute mitochondrial genes to H. sapiens.
Section: 26.10
Bloom's Taxonomy: Remembering/Understanding
45) In the 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: 26.10
Bloom's Taxonomy: Remembering/Understanding
46) In the case of complete dominance in a population in equilibrium, we cannot tell which
individuals are homozygous dominants and which are heterozygous, what information is needed
to determine the frequency of homozygous dominant and heterozygous genotypes?
A) the frequency of homozygous recessive genotype
B) the frequency of heterozygous recessive genotype
C) the size of the sample population
D) the mutation rate in the population
Answer: A
Section: 26.2
Bloom's Taxonomy: Remembering/Understanding
47) Given an inheritance pattern of incomplete dominance and 81 flowers that are red (R1R1), 18
flowers that are pink (R1R2), and 1 flower that is white (R2R2), what is the frequency of the R1
allele?
A) 0.1
B) 0.8
C) 0.9
D) 0.2
Answer: C
Section: 26.2
Bloom's Taxonomy: Applying/Analyzing
48) For a given locus, in a population with two alternative alleles, the allele frequencies p + q =
________.
A) 100
B) 1.0
C) p × q
D) 0
Answer: B
Section: 26.2
Bloom's Taxonomy: Remembering/Understanding
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49) If a gene has three alleles in a population, their frequencies must add up to ________.
A) 0
B) 1.0
C) 1.5
D) 3.0
Answer: B
Section: 26.3
Bloom's Taxonomy: Remembering/Understanding
50) Directional selection generates a phenotypic shift ________.
A) toward an extreme of the phenotype
B) away from either extreme of the phenotype
C) toward both extremes of the phenotype
D) toward the mean of the extremes of the phenotype
Answer: A
Section: 26.4
Bloom's Taxonomy: Remembering/Understanding
51) Natural selection occurs when there is nonrandom elimination of genotypes from a
population due to ________.
A) differences in viability or reproductive success
B) differences in reproductive isolation mechanisms
C) similarities in viability or reproductive success
D) decreased predation
Answer: A
Section: 26.4
Bloom's Taxonomy: Remembering/Understanding
52) What is the differential reproduction of genotypes, resulting from their variable fitness?
A) selection
B) discrimination
C) genetic drift
D) evolution
Answer: A
Section: 26.4
Bloom's Taxonomy: Remembering/Understanding
53) What is regarded as a strong evolutionary mechanism for changing allelic frequencies?
A) mutations
B) recombination
C) selection
D) migration
Answer: C
Section: 26.5
Bloom's Taxonomy: Remembering/Understanding
10
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54) Genetic drift is primarily associated with what size breeding populations?
A) very small
B) very large
C) mid-sized
D) Genetic drift has the same effect on all population sizes.
Answer: A
Section: 26.7
Bloom's Taxonomy: Applying/Analyzing
55) Genetic drift arises from all the following EXCEPT ________.
A) a small breeding population
B) a genetic bottleneck
C) a founder effect
D) mutations
Answer: D
Section: 26.7
Bloom's Taxonomy: Remembering/Understanding
56) Inbreeding will change ________.
A) gene frequencies
B) species
C) mating isolation mechanism
D) diet
Answer: A
Section: 26.8
Bloom's Taxonomy: Remembering/Understanding
57) Evolution is dependent on ________ in the evolving population.
A) genetic diversity
B) inbreeding
C) equilibrium
D) constant gene frequencies
Answer: A
Section: 26.10
Bloom's Taxonomy: Remembering/Understanding
58) Conservation of amino acid sequence among distantly related groups of organisms is
suggestive of a(n) ________.
A) important function for that sequence
B) lack of selective pressure on that sequence
C) lack of common ancestor
D) intrinsic resistance to mutations in that sequence
Answer: A
Section: 26.10
Bloom's Taxonomy: Applying/Analyzing
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Concepts of Genetics, 12e, Global Edition (Klug)
Special Topics 1 CRISPR-Cas and Genome Editing
1) CRISPR-Cas functions in bacteria to ________.
A) fight viruses
B) replicate viral DNA
C) edit genomes
D) code for restriction enzymes
Answer: A
Section: ST 1.1
Bloom's Taxonomy: Remembering/Understanding
2) Mechanisms of adaptive immunity include ________.
A) restriction enzymes
B) blocking phage adsorption
C) blocking phage DNA insertion
D) learning from past exposures
Answer: D
Section: ST 1.1
Bloom's Taxonomy: Remembering/Understanding
3) cas genes code for a variety of proteins, including ________.
A) RNases
B) polysaccharides
C) lysozymes
D) crRNAs
Answer: A
Section: ST 1.1
Bloom's Taxonomy: Remembering/Understanding
4) Which of the following is the correct order of steps for the CRISPR-Cas mechanism?
A) cleavage, target interference, biogenesis
B) spacer acquisition, biogenesis, target interference
C) target interference, transcription, biogenesis
D) transcription, cleavage, spacer acquisition
Answer: B
Section: ST 1.1
Bloom's Taxonomy: Remembering/Understanding
5) Type II CRISPR-Cas systems are the most widely used systems because ________.
A) they were discovered first
B) Streptococcus pyogenes is an important research organism
C) they use a nuclease complex
D) they only require a single protein, Cas9
Answer: D
Section: ST 1.1
Bloom's Taxonomy: Applying/Analyzing
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6) Which of the following is required to efficiently target DNA in vitro using CRISPR-Cas9?
A) Cas9
B) tRNA
C) target RNA
D) native crRNA
Answer: A
Section: ST 1.2
Bloom's Taxonomy: Remembering/Understanding
7) Genome editing is based on ________.
A) acetylating specific DNA sequences with specific nucleases
B) methylating specific DNA sequences with specific proteins
C) targeting and manipulating specific DNA sequences with specific nucleases
D) transcribing specific DNA sequences with specific proteins
Answer: C
Section: ST 1.2
Bloom's Taxonomy: Remembering/Understanding
8) Which of the following is a pathway available to cells to repair double-stranded DNA breaks?
A) homologous directed end joining
B) homologous end-joining
C) nonhomologous directed repair
D) nonhomologous end-joining
Answer: D
Section: ST 1.2
Bloom's Taxonomy: Remembering/Understanding
9) The promiscuity of the CRISPR-Cas9 system likely benefits bacteria by ________.
A) facilitating defense against new viral attacks
B) continuing the evolutionary process
C) enabling faster search of large genomes
D) increasing the safety of medical applications
Answer: A
Section: ST 1.2
Bloom's Taxonomy: Applying/Analyzing
10) The primary benefit of CRISPR-Cas over TALENs and ZFNs is ________.
A) the ability of Cas9 to bind engineered sgRNA
B) the specificity of the nuclease
C) the speed of the process
D) the cost
Answer: A
Section: ST 1.2
Bloom's Taxonomy: Evaluating/Creating
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11) A strategy to induce random indels in a target sequence might use ________.
A) HDR to match damaged and undamaged chromosomes
B) NHEJ to ligate fragments
C) HDR and a donor template
D) NHEJ and Cas9
Answer: B
Section: ST 1.2
Bloom's Taxonomy: Evaluating/Creating
12) Reverse genetics is a strategy with which researchers ________.
A) investigate the function of proteins
B) turn on a gene and monitor the effects
C) delete genes and monitor the effects
D) delete a genome and investigate the function of plasmids
Answer: C
Section: ST 1.3
Bloom's Taxonomy: Remembering/Understanding
13) dCas9 is a useful tool in the CRISPR-Cas toolkit because it can ________.
A) deactivate the transcription of a gene
B) fluoresce and help researchers locate target sequences
C) bind DNA, but it cannot travel down the template strand
D) bind DNA, but it cannot cut DNA
Answer: D
Section: ST 1.3
Bloom's Taxonomy: Remembering/Understanding
14) All of the following are challenges in gene therapy EXCEPT ________.
A) unknown specific mutations
B) inaccessible tissues
C) unknown cell types
D) designing a strategy
Answer: D
Section: ST 1.3
Bloom's Taxonomy: Remembering/Understanding
15) Examples of modifications to Cas9 include all of the following EXCEPT ________.
A) epigenetic modifiers
B) fluorescent proteins
C) promotor domains
D) repression domains
Answer: C
Section: ST 1.3
Bloom's Taxonomy: Remembering/Understanding
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16) The primary value of the CRISPR-Cas system in biotechnology is ________.
A) the removal of ineffective genes
B) simple modifications of genomes
C) rapid and cost-effective production of genetically modified organisms
D) addition of random indels
Answer: C
Section: ST 1.3
Bloom's Taxonomy: Remembering/Understanding
17) Applications of CRISPR-Cas in biotechnology include all of the following EXCEPT
________.
A) creating bioreactors to manufacture human proteins
B) modulating epigenetic regulation
C) increasing drought resistance of crops
D) engineering pigs that are resistance to PRRS
Answer: B
Section: ST 1.3
Bloom's Taxonomy: Applying/Analyzing
18) An advantage of base editor technology is that ________.
A) DNA is not cut, so there are no indels
B) the hybrid protein changes A to T for greater variability
C) uracil is more specific than cytosine and enables greater fidelity
D) specific basic sequences are edited, leading to a more specific product
Answer: A
Section: ST 1.3
Bloom's Taxonomy: Applying/Analyzing
19) A major limitation of CRISPR-Cas to treat human disease in vivo is modifying millions of
cells. What are the characteristics of successful delivery vehicles?
A) They are encapsulated in target cells (via endosomes).
B) They bind to target proteins.
C) They induce an immune response.
D) They are carried by erythrocytes.
Answer: A
Section: ST 1.3
Bloom's Taxonomy: Evaluating/Creating
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20) A strategy to reduce off target effects of the CRISPR-Cas system is to limit homology with
other regions of the genome. This principle capitalizes on the specificity of which step of the
mechanism?
A) spacer acquisition
B) target interference
C) crRNA biogenesis
D) leader sequence determination
Answer: B
Section: ST 1.3
Bloom's Taxonomy: Evaluating/Creating
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Concepts of Genetics, 12e, Global Edition (Klug)
Special Topics 2 DNA Forensics
1) VNTRs are useful in DNA forensics because ________.
A) the number of VNTRs varies between people
B) the VNTRs have exactly the same sequence between people
C) the VNTRs act as an identifier for a group of individuals
D) the VNTRs have very little variation in length
Answer: A
Section: ST 2.1
Bloom's Taxonomy: Remembering/Understanding
2) Justify the FBI's use of only 20 short tandem repeats (STRs) as their core set of STRs for
forensic analysis.
Answer: By characterizing 20 STRs, the analysis covers over two billion combinations even if
each STR only exhibits four alleles each. This is highly unlikely as STRs vary in repeats from 7
to 40 times.
Section: ST 2.1
Bloom's Taxonomy: Evaluating/Creating
3) The development of which biotechnology revolutionized the field of DNA forensics?
A) PCR
B) GWAS
C) Sanger sequencing
D) capillary electrophoresis
Answer: A
Section: ST 2.1
Bloom's Taxonomy: Remembering/Understanding
4) Present evidence supporting the argument that Y chromosome STR profiling is not sufficient
for proper DNA profiling.
Answer: Y chromosomal STR profiling is not sufficient for DNA profiling as it only focuses on
the Y chromosome of individuals. As Y chromosomes do not undergo recombination, they are
directly inherited from father to son and as such, all males of the same patrilineage will be
identified by the same STR pattern.
Section: ST 2.1
Bloom's Taxonomy: Applying/Analyzing
5) Mitochondrial DNA profiling is useful in developing DNA profiles from samples that are in
less than ideal condition. What is a major limitation of using mitochondrial DNA profiling?
A) It is present in high copy number.
B) It is useful in identifying victims of disasters when relatives are available for reference.
C) It is not possible to differentiate between maternal relatives.
D) It is possible to differentiate between maternal relatives.
Answer: C
Section: ST 2.1
Bloom's Taxonomy: Remembering/Understanding
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6) Single-nucleotide polymorphisms (SNPs) are being used more and more in forensic analysis
yet have not been fully embraced. However, SNPs have found an exemplary use in which of the
following fields?
A) evolution studies
B) protein stability studies
C) VNTR copy number studies
D) epigenetic regulation studies
Answer: A
Section: ST 2.1
Bloom's Taxonomy: Remembering/Understanding
7) DNA phenotyping represents an emerging technology that uses SNPs to determine physical
features. Describe why DNA phenotyping faces skepticism in court rooms and from other
scientists with regard to its accuracy.
Answer: DNA phenotyping faces scrutiny and skepticism due to its reliance on multiple genes
as well as multiple SNPs in those genes to provide a rather low probability of a correct
identification.
Section: ST 2.1
Bloom's Taxonomy: Applying/Analyzing
8) DNA phenotyping poses several concerns as an emerging technology. Which of the following
is not a concern with DNA phenotyping?
A) racial profiling
B) privacy violations
C) intellectual property of the company's methodologies
D) its ability to help identify missing persons
Answer: D
Section: ST 2.1
Bloom's Taxonomy: Remembering/Understanding
9) Provide support for using the product rule in generating a high confidence that an individual
has a unique DNA profile using multiple STRs.
Answer: The product rule states that the probability of an individual having certain alleles in a
population is the result of the product of each of the individual allelic frequencies in the
population. As such, one STR locus with two alleles at a frequency of 0.361 and 0.141 would
have a 10% chance of being unique in a population. However, by examining a second locus
containing one allele at a frequency of 0.243, we can demonstrate that the likelihood of an
individual having both STRs is ~0.6%.
Section: ST 2.2
Bloom's Taxonomy: Evaluating/Creating
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10) Which of the following contribute to an increased probability of a random match of a DNA
profile in a population?
A) a random population with many relatives
B) a population with inbreeding
C) a population with a small number of identical twins
D) a population containing limited VNTR variation
Answer: B
Section: ST 2.2
Bloom's Taxonomy: Remembering/Understanding
11) CODIS (Combined DNA Index System) is a database that contains DNA profiles from all of
the following EXCEPT ________.
A) people convicted of certain crimes
B) unidentified remains
C) crime scene evidence
D) public servants
Answer: D
Section: ST 2.2
Bloom's Taxonomy: Remembering/Understanding
12) There are several concerns with using DNA profiling in forensics with the advent of cheaper
synthesis methods for DNA. Propose a mechanism to determine if crime scene evidence is
natural or synthetic.
Answer: Testing evidence for methylation patterns and comparing those patterns to those seen in
natural DNA samples.
Section: ST 2.3
Bloom's Taxonomy: Evaluating/Creating
13) Why does DNA profiling pose potential ethical problems when a partial match occurs in
CODIS?
Answer: A partial match in CODIS could lead investigating agencies to focus on family
members of the partial match. The ethical issue arises when it is considered right to suspect
someone of a crime based on his or her DNA.
Section: ST 2.2
Bloom's Taxonomy: Applying/Analyzing
14) When using DNA profiling in prosecuting criminal cases, lawyers must be very thorough in
their investigation of the DNA evidence. Which of the following could cause issues with the
DNA evidence presented in a court case?
A) transference
B) quantity
C) deference
D) recombination
Answer: A
Section: ST 2.3
Bloom's Taxonomy: Remembering/Understanding
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15) Explain why Identical twins can have the same STRs in a DNA profile yet still exhibit
differing phenotypes.
Answer: While the DNA sequence of identical twins is the same, there is evidence that
environmental factors affect the epigenetic regulation of genes that could result in differing
phenotypes.
Section: ST 2.2
Bloom's Taxonomy: Evaluating/Creating
16) Using the values in TABLE ST 2.2; what is the expected genotype frequency for the two loci
profile consisting of D8S1179 and D5S818?
A) 0.001936
B) 0.000421
C) 0.102
D) 0.019
Answer: A
Section: ST 2.2
Bloom's Taxonomy: Applying/Analyzing
17) STRs are repeated elements in the genome. They have ________ nucleotides per repeat
compared to VNTRs.
A) more
B) fewer
C) the same number
D) precisely two times the number
Answer: B
Section: ST 2.1
Bloom's Taxonomy: Remembering/Understanding
18) One of the advantages of STR profiling over VNTR profiling is that ________.
A) it presents more variable sequences reducing false positive matches
B) it requires much less DNA because PCR can be used to amplify small samples
C) STRs are variable in the number of repeats, whereas VNTRs are not
D) STRs are not variable in the number of repeats, whereas VNTRs are
Answer: B
Section: ST 2.2
Bloom's Taxonomy: Remembering/Understanding
19) Mitochondrial DNA profiling is used to trace the maternal side of family trees. Why is this
done using mtDNA?
A) The mtDNA only comes from the father so all differences are from the mother.
B) The mtDNA is supplied to the zygote only from the egg.
C) The mtDNA undergoes recombination allowing allelic mixture.
D) The mtDNA under low selective pressure mutates readily.
Answer: B
Section: ST 2.1
Bloom's Taxonomy: Remembering/Understanding
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20) Mitochondrial DNA profiling is primarily used to differentiate ________.
A) mothers from daughters
B) siblings
C) mothers and sons
D) unrelated individuals
Answer: D
Section: ST 2.2
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Special Topics 3 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 3.1
Bloom's Taxonomy: Remembering/Understanding
2) Personalized medicine is ________.
A) a subset of precision medicine
B) based on an average of molecular profiles
C) used to design treatments for multiple patients
D) involves screening drugs on individual patients
Answer: A
Section: ST 3.1
Bloom's Taxonomy: Remembering/Understanding
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 3.1
Bloom's Taxonomy: Evaluating/Creating
4) Using "-omics" data prior to designing a treatment for patients is the goal of ________.
A) precision medicine
B) classic medicine
C) next generation medicine
D) neopharmaceuticals
Answer: A
Section: ST 3.1
Bloom's Taxonomy: Remembering/Understanding
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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 3.1
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 3.1
Bloom's Taxonomy: Applying/Analyzing
7) Immunotherapies seek to take advantage of ________ in treating cancers.
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 3.2
Bloom's Taxonomy: Remembering/Understanding
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 exome of the patient, mutations in coding regions can be
identified. These mutations can then be compared to databases that contain drugs and their
targets. If the mutated genes determined in the whole genome sequencing show up in the drug
database, a potential therapeutic has been found.
Section: ST 3.2
Bloom's Taxonomy: Evaluating/Creating
9) Neoantigens are ________.
A) mutations that result in cancer
B) mutated proteins expressed in the cytosol of cancer cells
C) mutated proteins expressed on the outer membrane of cancer cells
D) mutations that are silent
Answer: B
Section: ST 3.2
Bloom's Taxonomy: Remembering/Understanding
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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 3.2
Bloom's Taxonomy: Applying/Analyzing
11) In order to take advantage of adoptive cell therapies, the tumor must be biopsied and the
tumor infiltrating leukocytes are ________ for reactivity and then cultured ________ prior to
being reinfused into the patient.
A) screened; in vivo
B) screened; in vitro
C) mutated; in vivo
D) mutated; in vitro
Answer: B
Section: ST 3.2
Bloom's Taxonomy: Remembering/Understanding
12) One method to target the immune system of a patient to a cancer cell that it is nonresponsive
to is the use of ________.
A) CARs
B) WGS
C) SNPs
D) VNTRs
Answer: A
Section: ST 3.2
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 3.2
Bloom's Taxonomy: Evaluating/Creating
14) Which of the following is not a side effect of current CAR T cell therapies?
A) inflammatory response
B) neurotoxicity
C) eventual tumor resistance
D) headaches
Answer: D
Section: ST 3.2
Bloom's Taxonomy: Remembering/Understanding
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15) The age of precision medicine comes with many legal and ethical issues. Evaluate the current
hurdles to personalized medicine becoming common place.
Answer: There are several hurdles that precision medicine must overcome to be more
accessible. One 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 3.4
Bloom's Taxonomy: Evaluating/Creating
16) In future, medical schools will have to adapt their education to include ________ to better
equip physicians to interpret precision medicine data.
A) bioinformatics
B) anatomy
C) pharmacology
D) biochemistry
Answer: A
Section: ST 3.4
Bloom's Taxonomy: Applying/Analyzing
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 invariant domains as well as the variant domains
that bind antigens. CAR T cell receptors are much simpler; comprised of a signaling domain a
single transmembrane domain and two variable domains connected with a linker sequence.
Section: ST 3.2
Bloom's Taxonomy: Applying/Analyzing
18) All of the following are ways tumors avoid immune response EXCEPT ________.
A) the use of checkpoint molecules
B) the presence of regulatory T cells
C) abnormal expression of MHC
D) the use of endocytosis to destroy T cells
Answer: D
Section: ST 3.2
Bloom's Taxonomy: Remembering/Understanding
19) Adoptive cell transfer uses autologous cells. What does autologous cells?
A) automatically made
B) cells are taken from a donor other than the patient
C) cells are taken from the patient, augmented, and then reinfused to the patient
D) cells that are automatically expressing chimeric receptors
Answer: C
Section: ST 3.2
Bloom's Taxonomy: Remembering/Understanding
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20) Immune system cells that are responsible for the recognizing neoantigens and targeting
tumors are called ________.
A) macrophages
B) tumor-infiltrating lymphocytes
C) regulatory T cells
D) monocytes
Answer: B
Section: ST 3.2
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Special Topics 4 Genetically Modified Food
1) Describe the difference between genetically modified (GM) food and foods that are
selectively breed for 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 4.1
Bloom's Taxonomy: Applying/Analyzing
2) When GMOs are generated through the transfer of genetic material across species, the GMO
is said to be ________.
A) cisgenic
B) transgenic
C) isogenic
D) exogenic
Answer: B
Section: ST 4.1
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 4.1
Bloom's Taxonomy: Evaluating/Creating
4) What is a major concern surrounding GMOs that critics cite as an argument against GMOs?
A) They could escape containment and outcompete non-GMOs resulting in extinction of the
GMO.
B) They could escape containment and breed with non-GMOs and create unregulated hybrids.
C) They could escape containment and pollute other gene pools.
D) They could escape containment and produce even more useful genetic variants.
Answer: B
Section: ST 4.1
Bloom's Taxonomy: Remembering/Understanding
5) Bt crops are an insect-resistant GM plant. 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 4.1
Bloom's Taxonomy: Remembering/Understanding
1
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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 1590 million children, 19 million pregnant women, and almost 500,000 children. The
Golden Rice 2 GMO is grown on the same land as non-GMO rice and can be made sterile to
prevent the GMO from reproducing out of control.
Section: ST 4.1
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 4.2
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 4.2
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 4.2
Bloom's Taxonomy: Remembering/Understanding
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10) The use of beta-glucuronidase (GUS) in the generation of Roundup-Ready soybeans is an
example of a ________.
A) negative marker
B) positive marker
C) herbicide resistance
D) agrobacterium resistance
Answer: B
Section: ST 4.2
Bloom's Taxonomy: Remembering/Understanding
11) Predict the phenotype of the following gene construct if it is successfully transfected into
plant cells. Promoter- GUS- PFK1
A) blue precipitate
B) white precipitate
C) no precipitate
D) green precipitate
Answer: A
Section: ST 4.2
Bloom's Taxonomy: Applying/Analyzing
12) Which method would best meet the need in generating GM plants if food crops were
threatened by invasive insect species?
A) biolistic methods
B) agrobacterium-mediated technology
C) selective breeding
D) CRSPR-cas
Answer: D
Section: ST 4.2
Bloom's Taxonomy: Applying/Analyzing
13) Gene-edited organisms are seeing much less regulatory oversight than transgenic organisms.
Why is this the case?
A) The gene-edited organisms are safer.
B) The gene-edited organisms have no foreign DNA in them and do not fall under current
regulatory statutes.
C) Transgenic organisms are more complicated and require more testing.
D) Transgenic organisms are simpler than gene-edited organisms and have more regulations
because of it.
Answer: B
Section: ST 4.2
Bloom's Taxonomy: Applying/Analyzing
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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 4.3
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 4.3
Bloom's Taxonomy: Applying/Analyzing
16) GM foods are considered safe by advocates who cite ________.
A) 20 years of consumption with no reliable correlation to disease or 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 4.3
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 4.3
Bloom's Taxonomy: Remembering/Understanding
18) The future of GM organisms is encouraging. A project is underway to produce a GM chicken
that is incapable of ________.
A) spreading avian influenza
B) reproducing but containing twice as much breast meat
C) regulating the number of eggs laid
D) contracting avian influenza
Answer: A
Section: ST 4.4
Bloom's Taxonomy: Remembering/Understanding
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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 4.3
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 4.4
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Special Topics 5 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 5.1
Bloom's Taxonomy: Remembering/Understanding
2) Gene therapy is ________.
A) capable of treating diseases associated with an unknown gene
B) able to treat any type of cell no matter where it is in the body
C) an example of translational medicine
D) not a good candidate for cancer therapeutics
Answer: C
Section: ST 5.1
Bloom's Taxonomy: Remembering/Understanding
3) Minimizing off-target effects of gene therapy is a concern with ________.
A) ex vivo gene therapy
B) in vivo gene therapy
C) autologous gene therapy
D) the use of adenovirus therapy
Answer: B
Section: ST 5.2
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 5.2
Bloom's Taxonomy: Evaluating/Creating
1
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5) Lentiviral vectors are preferred over adenoviral 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 5.2
Bloom's Taxonomy: Applying/Analyzing
6) Viral vectors are not the only method being studied for gene therapy purposes. Which of the
following is a nonviral delivery method 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 5.2
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 5.2
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 5.3
Bloom's Taxonomy: Evaluating/Creating
2
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9) Despite the promise of gene therapy treating disease, there were two major setbacks to clinical
trials. This setbacks brought to light the disadvantages of using retroviral vectors for gene
therapy, which include ________.
A) controlled insertion of the therapeutic gene in the host genome
B) the unknown response of a patient to the retroviral vector
C) the viral vector will not impact the host
D) certain viral vectors do not insert in important genes of the host
Answer: B
Section: ST 5.4
Bloom's Taxonomy: Remembering/Understanding
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 do?
A) metabolizes rod cell waste
B) metabolizes cone cell waste
C) metabolizes retinol
D) metabolizes retinal
Answer: C
Section: ST 5.4
Bloom's Taxonomy: Remembering/Understanding
11) A serious concern about gene therapies and their commercialization is ________.
A) their cost to the consumer
B) their speedy development process
C) the lack of individuals treatable with the drug
D) the ease of achieving regulatory approval
Answer: A
Section: ST 5.5
Bloom's Taxonomy: Remembering/Understanding
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 5.6
Bloom's Taxonomy: Applying/Analyzing
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13) With gene therapy being at the cutting edge of medical technologies, it is not surprising that
serendipitous treatments are discovered. Explain how treating an HIV-positive patient with stem
cells from an HIV-resistant donor cured him or her of HIV while targeting his or her acute
myeloid leukemia.
Answer: Using 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 5.6
Bloom's Taxonomy: Evaluating/Creating
14) The use of CRISPR-cas has revolutionized the gene therapy field. One of the first uses of
CRISPR-cas to edit genes in vivo was done at MIT. What disease did they cure?
A) type 1 tyrosinemia
B) HIV
C) RDEB
D) B-thalassemia
Answer: A
Section: ST 5.6
Bloom's Taxonomy: Remembering/Understanding
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 5.6
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 5.6
Bloom's Taxonomy: Remembering/Understanding
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17) What is the target of the first CRISPR-cas gene editing in humans?
A) PD-1
B) CAR T cells
C) FUH
D) COL7A1
Answer: A
Section: ST 5.6
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 5.6
Bloom's Taxonomy: Remembering/Understanding
19) One gene therapy that does not rely on gene editing is RNA interference. This therapy uses
________ to interfere with expression of proteins at the translational level.
A) tRNA
B) miRNA
C) mRNA
D) mtRNA
Answer: B
Section: ST 5.6
Bloom's Taxonomy: Remembering/Understanding
20) Which gene therapy 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 5.7
Bloom's Taxonomy: Remembering/Understanding
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Concepts of Genetics, 12e, Global Edition (Klug)
Special Topics 6 Neurodegenerative Diseases
1) Neurogenetics is ________.
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 genes associated with the formation of neurons
Answer: B
Section: ST 6.1
Bloom's Taxonomy: Remembering/Understanding
2) Huntington's 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 6.1
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 6.1
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 wrote that "when either or both parents have shown manifestations of the
disease…the offspring almost invariably suffer from the disease." This describes 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 6.1
Bloom's Taxonomy: Evaluating/Creating
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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 6.1
Bloom's Taxonomy: Remembering/Understanding
6) After James Gusella's first attempt to link the RFLP marker to HD failed, he used a new 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 6.1
Bloom's Taxonomy: Remembering/Understanding
7) The second study by 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 6.1
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 6.1
Bloom's Taxonomy: Remembering/Understanding
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9) Explain how the use of sequencing technologies finally brought the search for the HD gene to
a close as reported by the Huntington's Disease Collaborative Research Group.
Answer: Sequencing of the contigs that covered the end of Chromosome 4 that had been
previously identified as the location of the HD locus revealed an "interesting contig" that
contained a CAG repeat sequence. The population of individuals without HD had 17 alleles with
CAG repeats varying from 11 to 34 copies. HD individuals had alleles with 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 6.1
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 minors should not be tested for
HD unless they demonstrate juvenile HD symptoms.
Answer: minors should not be tested for HD because there is currently not cure or preventative
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. In addition, minors may be denied health
coverage based on their genetic test results. This is a continuing ethical dilemma for genetic
testing.
Section: ST 6.1
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 6.2
Bloom's Taxonomy: Remembering/Understanding
12) Why does the mHTT gene have such wide-ranging effects in the disease state?
Answer: The mHTT causes a wide range of disease phenotypes because of the ubiquitous nature
of huntingtin. Huntingtin is pivotal in protein-protein interactions 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 6.2
Bloom's Taxonomy: Applying/Analyzing
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13) The favored model organism for studying the development of HD is ________.
A) C. elegans
B) D. melanogaster
C) H. sapiens
D) M. musculus
Answer: D
Section: ST 6.2
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 cloned 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 6.4
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 left the normal HTT
expression level unchanged
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 6.5
Bloom's Taxonomy: Remembering/Understanding
16) Researcher Xiao-Jiang Li has used ________ to edit the exon of hua mutant HTT allele in
mice. Three weeks after the injection of therapeutic into the brain the mice demonstrated a
reduction in protein aggregates.
A) CRISPR-cas
B) ZFNs
C) TALENs
D) lentiviral transfection
Answer: A
Section: ST 6.5
Bloom's Taxonomy: Remembering/Understanding
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17) HD heredity seems to increase if a father inherits the mutant allele from his mother. This is
most likely caused by ________.
A) recombination
B) imprinting
C) selection
D) segregation
Answer: B
Section: ST 6.6
Bloom's Taxonomy: Remembering/Understanding
18) Other neurodegenerative diseases also function through aggregation of misfolded proteins.
However, the initial protein aggregates are different in each disease. Work on HD may still be
vital to treating other diseases as ________.
A) other neurodegenerative diseases use axonal transport to spread to unaffected cells
B) the method of aggregation is the same in all neurodegenerative disease
C) the method of disaggregation is similar in all neurodegenerative disease
D) not all neurodegenerative disease spread
Answer: A
Section: ST 6.6
Bloom's Taxonomy: Remembering/Understanding
19) How do Parkinson's causing aggregates enter unaffected cells?
A) endocytosis
B) exocytosis
C) phagocytosis
D) alpha-synuclein aggregates cannot enter unaffected cells
Answer: A
Section: ST 6.6
Bloom's Taxonomy: Remembering/Understanding
20) How does the development of protein aggregates in Huntington's disease bring about
dementia in patients?
A) disruption of neuronal connection between the medulla and the caudate region
B) disruption of neuronal connection between the caudate region and the frontal lobe
C) disruption of the neuronal connection of the entire central nervous system
D) disruption of the neuronal connection to the frontal lobe resulting in brain weight gain of
~30%
Answer: B
Section: ST 6.6
Bloom's Taxonomy: Remembering/Understanding
5
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