Exam 3

Question 1

A complementation test is often used as a first step in determining whether
A. two mutant alleles are identical to one another
B. a chromosomal inversion is present
C. two mutants affect the same gene
D. one mutant can be rescued by a different allele at the same locus
E. a mutant individual’s vanity can get the best of it

Answer 1

C. two mutants affect the same gene

Question 2

Tardigrades are pretty remarkable animals in being resistant to environmental insults that are lethal
to humans and just about everything else. Genome sequencing and subsequent experimental
analyses revealed that tardigrades can survive exposure to ionizing radiation in part because they
have evolved a novel Damage suppressor (Dsup) protein. What evidence supports this conclusion?
A. Dsup expression reduces “comet tails” and recruitment of H2AX in irradiated cells
B. Dsup binds nucleosome monomers and recruits H2AX to damaged DNA
C. Dsup expression induces “comet tails” in irradiated cells and expression of H2AX in control
cells
D. Dsup localizes to the nucleus and binds both DNA and nucleosome monomers
E. Dsup localizes to the cytoplasm where it reduces formation of “comet tails” in control cells

Answer 2

A. Dsup expression reduces “comet tails” and recruitment of H2AX in irradiated cells

Question 3

Synteny refers to:
A. translocations that often occur between chromosomes during evolution
B. consequences of inversion for meiotic synapse formation
C. fusions that sometimes arise evolutionarily between non-homologous chromosomes
D. gene order that is shared on chromosomal regions of different species
E. synapses that can occur between non-homologous chromosomes during DNA repair

Answer 3

D. gene order that is shared on chromosomal regions of different species

Question 4

Which of the following statements about chromosomal inversions is true?
A. They are lethal when paracentric but viable when pericentric.
B. They often lead to cancer within generations but mechanisms that suppress cancer over
evolutionary time (e.g., tandem copies of tumor suppressor genes).
C. They often lead to reduced fertility within a generation but occasionally result in clusters of co-adapted and co-inherited alleles that evolve across generations.
D. They most often lead to thyroid cancer if followed by exposure to ionizing radiation.
E. All of the above

Answer 4

C. They often lead to reduced fertility within a generation but occasionally result in clusters of co-adapted and co-inherited alleles that evolve across generations.

Question 5

What sequence of events is shared among DNA mismatch repair, nucleotide excision and base
excision pathways?
A. recognition, endonuclease activity, DNA polymerization
B. recognition, DNA polymerization, exonuclease activity
C. recognition, exonuclease activity, DNA polymerization
D. exonuclease activity, recognition, DNA polymerization
E. DNA polymerization, endonuclease activity, recognition

Answer 5

C. recognition, exonuclease activity, DNA polymerization

Question 6

Fanconi anemia exemplifies:
A. The risk that ionizing radiation will lead to point mutations that affect mismatch repair and
chromosomal integrity, leading to chromosomal deletions and depletion of blood and other
stem cells.
B. The importance of telomere maintenance and rejuvenation in blood (and other) stem cell
derived lineages that have substantial requirements for on-going proliferation.
C. The heightened risk of single nucleotide mutations associated with long-term meiotic-arrest of
gametes as male humans age.
D. The role of intra-strand crosslinks in stalling replication fork advancement and the
consequences of defective repair mechanisms for accumulating cancer-causing point
mutations.
E. The importance of inter-strand cross link repair in maintaining chromosome integrity and the consequences of chromosomal rearrangements and deletions for development and cancer susceptibility.

Answer 6

E. The importance of inter-strand cross link repair in maintaining chromosome integrity and the consequences of chromosomal rearrangements and deletions for development and cancer susceptibility.

Question 7

Many events in animal development involve genes that function in just a few major signaling
pathways (Wnt, Hedgehog, RTK, Notch, BMP). Several of these genes and pathways were first
identified through analyses of Drosophila mutants and only later shown to be present, and functioning
similarly, in vertebrates. One example is hedgehog (hh; named after the first instar larva phenotype of
extra bristles). When a similar gene was found in vertebrates it was named Sonic hedgehog (Shh; yes,
after a cartoon character). Later, when another similar gene was found in vertebrates, it was named
Indian hedgehog (Ihh; after an actual species of hedgehog). Based on your knowledge of genome
evolution, which statement likely best describes the evolutionary relationships of these genes?
A. all are paralogues of one another, Shh and Ihh are ancient orthologues of each other, and hh is
paralogous to both Shh and Ihh
B. all are orthologues of one another, Shh and Ihh are ancient homologues of each another, and
hh is orthologous to both Shh and Ihh
C. all are homologues of one another, Shh and Ihh are ancient ohnologs of each another, and hh is orthologous to both Shh and Ihh
D. all are homologues of one another, hh and Shh are ohnologs of one another whereas hh and
Ihh are paralogues of one another
E. all are ohnologs of one another, hh and Shh are paralogues of one another whereas hh and Ihh
are paralogues of one another

Answer 7

C. all are homologues of one another, Shh and Ihh are ancient ohnologs of each another, and hh is orthologous to both Shh and Ihh

Question 8

Oxidation of nucleotide bases can result in the conversion of a G to an oxidized variant (“GO”). In
the absence of base or nucleotide excision repair, or mismatch repair, this damage could ultimately
result in:
A. a G to C transition mutation
B. a C to G transversion mutation
C. a G to T transversion mutation
D. a GO to G revertant mutation
E. a T to A transition mutation

Answer 8

C. a G to T transversion mutation

Question 9

In a forward genetic screen, researchers will usually:
A. pick a gene of interest, induce mutations, screen for mutants with phenotypic defects, then
verify that the chosen gene is affected in the mutant
B. pick a trait or process of interest, bring wild populations into the lab, inbreed to generate
additional copies of mutant alleles, then screen for individuals with phenotypic defects
C. pick a trait or process of interest, induce new mutations by CRISPR/Cas9 mutagenesis, screen
for mutant phenotypes, then sequence the genome to find the affected gene
D. pick a trait or process of interest, induce a polyploid state by mitotic arrest, then screen for the
preservation of duplicate genes that have become subfunctionalized to control that particular
phenotype
E. pick a trait or process of interest, induce mutations, screen later generations for individuals having phenotypic defects, then determine which gene is affected in the mutant

Answer 9

E. pick a trait or process of interest, induce mutations, screen later generations for individuals having phenotypic defects, then determine which gene is affected in the mutant

Question 10

Studies of human genome variation by “trio sequencing” show that an increased incidence of point
mutations is associated with increasing paternal age, in contrast to the increased incidence of
aneuploidy that is associated with increasing maternal age. This sex-difference is likely a result of:
A. less efficient proofreading activity of DNA polymerase in males and less efficient Dsup activity
in females
B. continuous replication of DNA in the male germline and delayed completion of meiosis in the female germline
C. the presence of only one X chromosome in males and the presence of two X chromosomes in
females
D. genetic imprinting in the male germline and low telomerase activity in the female germline
E. greater exposure to point mutagens in males and greater exposure to ionizing radiation in
females

Answer 10

B. continuous replication of DNA in the male germline and delayed completion of meiosis in the female germline

Question 11

Copy number variants are a common feature of genomes (more common than even single
nucleotide polymorphisms in the human genome). A nice example of a CNV that appears to have
phenotypic effects is the “ridge” found on the coats of Rhodesian ridgeback dogs, which can occur
along with a congenital skin defect, “dermoid sinus.” These traits together have been associated with
a duplicated chromosomal segment that contains several genes including 3 genes that encode
Fibroblast Growth Factors (FGF3, FGF4, and FGF19). Now imagine you have just adopted a
Rhodesian ridgeback, but on taking your pup to the vet it is found to have both a ridge and a dermoid
sinus. Ignoring other types of FGF genes in the genome (e.g., FGF2), what is the most likely (albeit not
definitive) explanation for your pup’s phenotype?
A. increased FGF activity, due to a single mutant chromosome per cell
B. increased FGF activity, due to the presence of 6 extra FGF genes per cell
C. decreased FGF activity, due to a single wild-type chromosome per cell
D. decreased FGF activity due to having two copies of a wild-type chromosome
E. decreased FGF activity, due to the presence of only 3 FGF genes per cell

Answer 11

B. increased FGF activity, due to the presence of 6 extra FGF genes per cell

Question 12

When seeking to meiotically map a new mutant, it is often valuable to have a large mapping cross
principally because having more individuals…
A. will reveal the causative mutation using any of several different genotyping methods
B. helps to ensure that phenotypes are genetically based and not due to environmental or
epigenetic effects
C. guarantees that some individuals will not be recombinant for the mutant locus and marker loci
on either side of it
D. allows more sampling of cross-overs and a narrower critical interval if marker densities are great enough
E. protects against revertant alleles that can arise spontaneously in previously mutagenized
individuals

Answer 12

D. allows more sampling of cross-overs and a narrower critical interval if marker densities are great enough

Question 13

In the context of a genetic screen, “saturation” most often refers to:
A. the optimal dose of alkylating agent needed to mutagenize a certain proportion of genes in the
genome
B. the observation that gene products are often sufficiently abundant that many new alleles are
recessive
C. the plateau at which new mutant alleles will most often represent loci already found
D. the minimal number of F2 individuals needed to map a recovered mutant at a resolution that
has been defined a priori by genetic distance
E. the dosage of CRISPR/Cas9 reagents needed to yield knockouts of both alleles at a target
locus

Answer 13

C. the plateau at which new mutant alleles will most often represent loci already found

Question 14

Following the Chernobyl nuclear plant explosion and meltdown in 1986, there was an increased
risk of pediatric thyroid cancer. When primary tumors were analyzed, patients in the vicinity of
Chernobyl were more likely than other patients to have:
A. aneuploidies leading to deficiencies in p53 activity
B. translocations bringing transcription factor genes under the control of receptor tyrosine kinase
regulatory elements
C. double strand breaks leading to acute radiation syndrome
D. translocations with gene fusions involving receptor tyrosine kinase coding sequences
E. duplications of genes encoding receptor tyrosine kinase second messengers

Answer 14

D. translocations with gene fusions involving receptor tyrosine kinase coding sequences

Question 15

For genetic mapping, production of a 100-individual family using an incross design (heterozygous
F1 crossed to heterozygous F1) rather than a back-cross design (heterozygous F1 crossed to
homozygous mutant parent) would be expected to:
A. decrease the mapping resolution by 50%
B. increase the mapping resolution by 200%
C. double the number of informative meioses to 200
D. double the number of informative meiosis to 100
E. halve the number of informative meioses to 100

Answer 15

C. double the number of informative meioses to 200

Question 16

Drosophila Antennepedia, yellow peas, “double muscle” cows and “white” axolotls are all
examples of:
A. recessive mutants
B. hypomorphic alleles
C. copy number variants
D. induced mutants
E. spontaneous mutants

Answer 16

E. spontaneous mutants

Question 17

Which of the following are potential consequences of single nucleotide alterations?
A. truncated protein product
B. amino acid replacement
C. mis-splicing of RNA
D. altered transcriptional efficiency
E. all of the above

Answer 17

E. all of the above

Question 18

Imagine you are studying the development of sea cucumbers, the cells of which are predominantly
2N. You find that the typical skin gland cell is 48N and has a single large nucleus, whereas most
epidermal cells are 4N with two nuclei. What might you infer about cycling of these cells?
A. gland cells lack an M phase, whereas epidermal cells have an abbreviated M phase
B. gland cells lack an S phase, whereas epidermal cells lack an M phase
C. gland cells have an extended M phase, whereas epidermal cells have an abbreviated S phase
D. gland cells have an extended G2 phase, whereas epidermal cells have an abbreviated G1
phase
E. gland cells lack G1 and S phases, whereas epidermal cells lack G2 and M phases

Answer 18

A. gland cells lack an M phase, whereas epidermal cells have an abbreviated M phase

Question 19

In a typical “F3 screen” for recessive mutant alleles, the F2 generation is an important step
because it:
A. results in homozygous individuals that may have mutant phenotypes
B. selects mutant alleles present uniquely in the germ line of F1s (as opposed to somatic cells) for
subsequent propagation
C. generates additional copies of new mutant alleles, allowing for screening of homozygous individuals in the F3
D. allows dominant alleles to be discarded from the screen before F3 families are contaminated
with them
E. ensures induced aneuploidies are not propagated into the F3 generation, in which they could
lead to deviations from expected Mendelian ratios

Answer 19

C. generates additional copies of new mutant alleles, allowing for screening of homozygous individuals in the F3

Question 20

Which statement is true about forward genetic screens:
A. mutations are specifically targeted to genes that function in DNA repair
B. mutations are preferentially induced in genes of interest
C. mutations are often homozygous when they first arise
D. mutations are induced at sites throughout the genome
E. mutations are confined to coding regions

Answer 20

D. mutations are induced at sites throughout the genome

Question 21

In studies of genome evolution the gar has been a useful species for making comparisons
because:
A. it has retained at least 85% of the genes produced after each of two successive rounds of
whole genome duplication
B. it has exactly the same number of chromosomes as chicken and human
C. its prokaryotic genome is often informative when trying to identify the minimal gene set by
whole-genome alignment and synteny
D. it is more closely related to teleost fishes than to tetrapods, yet its phylogenetic lineage has undergone only two rounds of whole genome duplication, similar to tetrapods
E. it exhibits a phenomenon of chromosome diminution and DNA loss that serves as a strong
counterpoint to the notion that the genome remains intact during development

Answer 21

D. it is more closely related to teleost fishes than to tetrapods, yet its phylogenetic lineage has undergone only two rounds of whole genome duplication, similar to tetrapods

Question 22

Huntington disease exemplifies instances of human disease arising from:
A. trinucleotide repeat expansion affecting a coding sequence
B. telomeric repeat expansion affecting proliferative senescence
C. microsatellite repeat expansion affecting a proximal promoter
D. centromeric repeat expansion affecting chromosome segregation
E. chromosomal segment expansion resulting in localized aneuploidy

Answer 22

A. trinucleotide repeat expansion affecting a coding sequence

Question 23

Long-term retention of gene copies following whole genome duplication can occur because:
A. mutations arise in different cis regulatory elements in each gene copy
B. one copy acquires a premature termination codon and the other does not
C. one copy acquires a novel protein function and the other does not
D. both A and B
E. both A and C

Answer 23

E. both A and C

Question 24

Analyses of mutants isolated in the pioneering screen to identify genes required for early
Drosophila development revealed a hierarchy of genes actions from early to late, including:
A. gap genes ➝ segment polarity genes ➝ pair rule genes
B. cytoplasmic polarity genes ➝ gap genes ➝ segment polarity genes
C. homeotic genes ➝ cytoplasmic polarity genes ➝ gap genes
D. pair rule genes ➝ segment polarity genes ➝ homeotic genes
E. Hunchback protein gradient ➝ cytoplasmic polarity genes ➝ homeotic genes

Answer 24

B. cytoplasmic polarity genes ➝ gap genes ➝ segment polarity genes

Question 25

When genes are linked, the presence of some recombinant phenotypes in a cross indicates that:
A. crossing over has occurred between them
B. the genes reside on different chromosomes
C. the genes occur within a chromosomal inversion
D. both A and B
E. both A and C

Answer 25

A. crossing over has occurred between them

Question 26

In a forward genetic screen for dominant mutant alleles, the researcher would:
A. only find mutant phenotypes after screening as many as 6 F3 families
B. only find embryonic lethal phenotypes that must be propagated from wild-type heterozygotes
C. only find adult mutants that have undergone gene conversion during crossing over
D. first identify mutant phenotypes in F0 individuals that have been exposed the mutagen
E. first identify mutant phenotypes in the progeny of mutagenized F0 individuals

Answer 26

E. first identify mutant phenotypes in the progeny of mutagenized F0 individuals

Question 27

Which of the following would be useful to verify the correspondence of a candidate gene and
mutant phenotype?
A. isolation of independent alleles having the same or similar phenotype
B. “rescue” of the mutant phenotype by providing the candidate gene product to the affected
cells (e.g., by transgenesis)
C. isolation of a “revertant” allele at the same locus that restores the original phenotype in the
mutant background
D. all of the above
E. none of the above

Answer 27

D. all of the above

Question 28

In the context of genetic mapping, a “marker” locus can be defined as:
A. any gene that corresponds to the mutant phenotype
B. any invariant site that reliably marks where the mutation has occurred
C. any site having variants that can be identified in a mapping cross
D. any microsatellite on the same chromosome as the mutant locus
E. any restriction site that is closely linked to the induced mutation

Answer 28

C. any site having variants that can be identified in a mapping cross

Question 29

Imagine you are interested in how cells are organized into particular arrangements and you are
using pigment stripes of the genetic model organism zebrafish to understand such processes. You
have identified a spontaneous mutant, casper, that completely lacks stripes when homozygous
because of a 0.2 Mb deletion on chromosome 4 that removes the coding sequence and proximal
promoter of an essential receptor-encoding gene. Later, in a forward genetic screen you happen upon
a similar phenotype, which you christen casper-too, but it has about 20 residual pigment cells when
homozygous. You are suspicious the new allele might represent the same locus as your deletion allele,
so you perform a complementation test and find that the progeny have about 10 pigment cells each.
Which result and interpretation makes sense?
A. casper and casper-too are alleles of different genes and casper has residual activity
B. casper and casper-too are alleles of different genes and casper-too is a null allele
C. casper and casper-too are alleles of the same gene and casper has residual activity
D. casper and casper-too are alleles of the same gene and casper-too is a null allele
E. none of the above

Answer 29

E. none of the above

Question 30

When cells attempt to repair double strand breaks, different mechanisms lead to different
likelihoods of errors and mutations being introduced, when comparing the repaired product to the
original chromosome. A reasonable ranking of such mechanisms, from best outcome (no errors) to
worst (large chromosomal deletions or rearrangements) would be:
A. single strand annealing > alternative end joining > non-homologous end joining
B. non-homologous end joining > single strand annealing > homologous recombination
C. alternative end joining > single strand annealing > homologous recombination
D. homologous recombination > non-homologous end joining > single strand annealing
E. alternative end joining > single strand annealing > non-homologous end joining

Answer 30

D. homologous recombination > non-homologous end joining > single strand annealing