BIOL 3010 Unit 3 MC Study Questions (Midterm 3)

Lectures 19-24 multiple choice study questions with answers

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

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

(Select all) Which would be useful to verify the correspondence of a candidate gene and mutant phenotype isolated by forward genetics? a. CRISPR/Cas9 targeting of the candidate gene phenocopies the original mutant b. isolation of independent alleles having the same or similar phenotype c. "rescue" of the mutant phenotype by providing the candidate gene product (e.g., transgenesis) d. isolation of a "revertant" allele at the same locus that restores the original phenotype e. crossing the mutant to a wild-type strain to verify recessivity

a, b, c, d — phenocopy by CRISPR knockout, independent alleles (complementation tests), rescue, and revertants all link gene to phenotype. (e only tests dominance.)

When oxidation of guanosine (G) occurs at a particular site, progeny of the cell will most often have at that site: a. G b. GO c. A d. C e. T

a. G — most of the time the lesion is repaired (and the C-template strand still makes G). If unrepaired, 8-oxoG pairs with A → G-to-T transversion (see the GO question).

The likelihood of aneuploidy increases with: a. reduced telomerase activity in older individuals b. the longer it takes to complete meiosis c. increased CENP-A synthesis in younger individuals d. the faster the rate of DNA replication e. the smaller the size of the chromosome being considered

b. the longer it takes to complete meiosis (oocytes arrested for decades → maternal age effect)

(Select all) Hox genes exemplify: a. genes required for body plan specification b. regions of synteny maintained over long periods of evolutionary time c. tumor suppressors amplified as larger bodies and life spans arose d. roles for tandem duplication in increasing gene numbers e. roles for whole genome duplication in increasing gene numbers

a, b, d, e — body plan genes; conserved clusters (synteny); the cluster arose by tandem duplication; multiple clusters (4 in vertebrates) came from whole genome duplication.

(Select all) Which are concerns about therapeutic applications of CRISPR/Cas9 gene editing? a. risk of off-target effects b. the relative efficiencies of NHEJ and homology directed repair c. the extreme inefficiency of Cas9 enzymatic activity in cells d. the requirement to first transplant edited cells into a foster mother e. possibility of pleiotropy having unexpected effects on other traits

a, b, e

In genetic mapping, a "marker" locus can be defined as: a. any restriction site closely linked to the induced mutation b. any microsatellite on the same chromosome as the mutant locus c. any site having variants that can be identified in a mapping cross d. any invariant site that reliably marks where the mutation occurred e. any gene that corresponds to the mutant phenotype

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

Drosophila Antennapedia, yellow peas, "double muscle" cows and "white" axolotls are all examples of: a. spontaneous mutants b. induced mutants c. copy number variants d. hypomorphic alleles e. chromosomal aneuploidies

a. spontaneous mutants

In a genetic screen, "saturation" most often refers to: a. the optimal dose of alkylating agent b. the plateau at which new mutant alleles most often represent loci already found c. gene products abundant enough that new alleles are recessive d. the minimal number of F2s needed to map at a set resolution e. the dosage of CRISPR/Cas9 reagents to knock out both alleles

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

What sequence of events is shared among mismatch repair, nucleotide excision and base excision repair? 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

a. recognition → endonuclease activity (nick/cut out the region) → DNA polymerization

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 an inversion d. crossing over has failed e. independent assortment has occurred

a. crossing over has occurred between them

The pioneering Drosophila early development screen revealed a hierarchy of gene action from early to late, including: a. gap → segment polarity → pair rule b. homeotic → cytoplasmic polarity → gap c. cytoplasmic polarity → gap → segment polarity d. segment polarity → pair rule → homeotic e. Hunchback gradient → cytoplasmic polarity → homeotic

c. cytoplasmic polarity genes → gap genes → segment polarity genes (full order: maternal/polarity → gap → pair rule → segment polarity; homeotic specify identity)

When meiotically mapping a new mutant, a large mapping cross is valuable principally because more individuals: a. reveal the causative mutation by any genotyping method b. ensure phenotypes are genetic, not environmental c. guarantee some individuals are non-recombinant d. allow more sampling of cross-overs and a narrower critical interval if marker density is high enough e. protect against revertant alleles

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

A standard explanation for why aneuploidy is most often deleterious: a. it leads to cancer b. it alters gene dosage c. it causes chromosome breakage during anaphase d. it leads to premature telomere shortening e. it reduces genetic diversity

b. it alters gene dosage

hedgehog (hh, fly), Sonic hedgehog (Shh) and Indian hedgehog (Ihh) (vertebrates). Which best describes their evolutionary relationships? a. all homologues; Shh and Ihh are ancient paralogues; hh is orthologous to both b. all paralogues; Shh and Ihh are ancient orthologues; hh paralogous to both c. all orthologues; Shh and Ihh ancient homologues; hh orthologous to both d. all homologues; hh & Shh paralogues, hh & Ihh paralogues e. all homologues; hh & Shh paralogues, hh & Ihh paralogues

a. all homologues; Shh and Ihh are paralogues (duplicated in vertebrate WGD); hh is orthologous to both (split by speciation)

A complementation test is often used as a first step in determining whether: a. two mutant alleles are identical b. a chromosomal inversion is present c. two mutants affect the same gene d. one mutant is recessive and another dominant e. homologous strand annealing can repair the mutant

c. two mutants affect the same gene

After Chernobyl (1986), pediatric thyroid tumors from patients near Chernobyl were more likely to have: a. aneuploidies causing p53 deficiency b. translocations bringing enzyme genes under RTK regulatory elements c. double strand breaks causing acute radiation syndrome d. translocations with gene fusions involving receptor tyrosine kinase coding sequences e. duplications of genes encoding RTK second messengers

d. translocations with gene fusions involving receptor tyrosine kinase coding sequences (RET/PTC fusions)

In a forward genetic screen for dominant mutant alleles, the researcher would: a. only find phenotypes after screening multiple F3 families b. only find embryonic lethals propagated from heterozygotes c. only find adults that underwent gene conversion d. first identify mutant phenotypes in mutagen-exposed F0 individuals e. first identify mutant phenotypes in the progeny of mutagenized F0 individuals

e. in the progeny (F1) of mutagenized F0 individuals

"Trio" sequencing shows most new point mutations arise on paternally derived chromosomes. A plausible explanation: a. males have more mutagen exposure b. male germ cells are arrested in meiosis II c. sperm DNA polymerases lack proofreading d. male germ cells continue to proliferate throughout life, allowing mutations to accumulate e. the single Y makes new mutations dominant

d. male germ cells continue to proliferate throughout life allowing mutations to accumulate

Which statement is correct about human copy number variants? a. they generally lead to disease b. individual repeats often involve >20% of a chromosome c. they only affect non-coding sequence d. they are more likely to involve non-coding genes than coding genes e. they are most often 11 kb or larger

d. more likely to involve non-coding genes than coding genes (coding genes are under stronger selection against dosage change). CNVs cover ~4.8–9.5% of the genome.

Which statement about chromosomal inversions is true? a. most often lead to thyroid cancer after radiation b. invariably lethal c. often cause cancer, suppressed over evolutionary time d. often reduce fertility within a generation but occasionally produce clusters of co-adapted, co-inherited alleles that evolve across generations e. often due to point mutations that resist correction

d. reduced fertility within a generation, but occasionally clusters of co-adapted, co-inherited alleles (supergenes)

Increased point mutations with paternal age vs. increased aneuploidy with maternal age is likely due to: a. less efficient proofreading in males / less Dsup in females b. continuous replication of DNA in the male germline and delayed completion of meiosis in the female germline c. one X in males, two in females d. imprinting in males, low telomerase in females e. more point mutagens in males, more radiation in females

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

Which is true about non-homologous end-joining? a. repairs double strand breaks and often leads to small deletions or insertions b. repairs DSBs by crossing over without indels c. exploited in CRISPR for error-free editing d. allows targeted mutagenesis with viability selection in mouse ES cells e. the only mechanism used in CRISPR editing

a. repairs double strand breaks and often leads to small deletions or insertions (indels)

In a forward genetic screen, researchers will usually: a. pick a trait/process, induce mutations, screen later generations for phenotypic defects, then determine which gene is affected b. induce polyploidy and screen for subfunctionalized duplicates c. induce mutations by CRISPR, screen, then sequence d. inbreed wild populations and screen e. pick a gene, induce mutations, screen, verify the gene

a. phenotype first → induce mutations → screen later generations → identify the gene (e describes reverse genetics)

In studies of genome evolution the gar has been useful because: a. it shows chromosome diminution and DNA loss b. it is more closely related to teleosts than to tetrapods, yet its lineage underwent only two rounds of WGD, like tetrapods c. its prokaryotic genome identifies the minimal gene set d. same chromosome number as chicken and human e. retained ≥85% of genes after three WGD rounds

b. a ray-finned fish that branched off before the teleost-specific (3rd) WGD, so it has 2 rounds like tetrapods — a bridge for comparisons

(Select all) Potential consequences of single nucleotide alterations: a. truncated protein product b. amino acid replacement c. mis-splicing of RNA d. altered transcriptional efficiency e. silent substitution

a, b, c, d, e — all of them (nonsense, missense, splice-site, promoter/regulatory, synonymous)

Rhodesian ridgeback: ridge + dermoid sinus are associated with a duplicated segment containing FGF3, FGF4, FGF19. Your pup has BOTH a ridge and a dermoid sinus. Most likely explanation (ignoring other FGFs)? a. decreased FGF activity, only 3 FGF genes per cell b. decreased FGF activity, two wild-type chromosomes c. decreased FGF activity, a single wild-type chromosome d. increased FGF activity, due to the presence of 6 extra FGF genes per cell e. increased FGF activity, a single mutant chromosome per cell

d. increased FGF activity from 6 extra FGF genes per cell — duplication on BOTH chromosomes (homozygous: 3 genes duplicated × 2), which raises dermoid sinus risk

(Select all) Which would be considered "transition mutations"? a. A to G b. A to C c. G to T d. C to T e. C to G

a and d — transitions are purine↔purine (A↔G) or pyrimidine↔pyrimidine (C↔T); purine↔pyrimidine changes are transversions

(Select all) Which is true of Klinefelter syndrome? a. individuals are female because of an extra X b. individuals are male because both X's are inactivated c. tall because of three active SHOX copies in their germ line d. tall because of three active SHOX copies in their somatic cells e. tall because of twice the dosage of X genes outside PAR1/PAR2

d only — XXY males (Y makes them male); SHOX is in PAR1 and escapes X-inactivation, so 3 active copies in body (somatic) cells → tall. Genes outside the PARs are dosage-compensated by X-inactivation.

Sea cucumber cells are mostly 2N. Skin gland cells are 48N with one large nucleus; most epidermal cells are 4N with two nuclei. What can you infer? a. gland cells lack S; epidermal cells lack M b. gland cells lack an M phase, whereas epidermal cells have an abbreviated M phase c. gland cells extended M; epidermal abbreviated S d. gland cells extended G2; epidermal abbreviated G1 e. gland cells lack G1 and S; epidermal lack G2 and M

b. gland cells endoreplicate (S without M); epidermal cells do mitosis without cytokinesis (binucleate)

Oxidation of G to "GO": in the absence of BER, NER, or mismatch repair, this damage could ultimately result in: a. a T to A transition b. a GO to G revertant c. a G to T transversion mutation d. a C to G transversion e. a G to C transition

c. G to T transversion — GO pairs with A, and in the next round A pairs with T

Tardigrades survive ionizing radiation partly due to Damage suppressor (Dsup) protein. What evidence supports this? a. Dsup expression reduces "comet tails" and recruitment of H2AX in irradiated cells b. Dsup binds nucleosomes and recruits H2AX c. Dsup induces comet tails in irradiated cells d. Dsup localizes to the nucleus and binds DNA and nucleosomes e. Dsup in cytoplasm reduces comet tails in control cells

a. reduced comet tails (less DNA fragmentation) and less γH2AX (fewer DSBs) in irradiated cells expressing Dsup — functional evidence

Ranking DSB repair mechanisms from best outcome (no errors) to worst (large deletions/rearrangements): a. homologous recombination > non-homologous end joining > single strand annealing b. SSA > alt-EJ > NHEJ c. NHEJ > SSA > HR d. alt-EJ > SSA > HR e. alt-EJ > SSA > NHEJ

a. HR (error-free, uses template) > NHEJ (small indels) > SSA (deletes sequence between repeats)

Huntington disease exemplifies human disease arising from: a. trinucleotide repeat expansion affecting a coding sequence b. telomeric repeat expansion c. microsatellite expansion in a promoter d. centromeric repeat expansion e. chromosomal segment expansion

a. trinucleotide (CAG) repeat expansion in coding sequence → polyglutamine

(Select all) Long-term retention of gene copies following whole genome duplication can occur because: a. mutations arise in different cis-regulatory elements in each 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. copied genes within an inversion aren't subject to crossing over e. synteny normally prevents gene loss

a and c — subfunctionalization (a) and neofunctionalization (c). b = nonfunctionalization (copy is lost).

Which is true about forward genetic screens? a. mutations are preferentially induced in genes of interest b. mutations are induced at sites throughout the genome c. mutations target DNA repair genes d. mutations are often homozygous when they first arise e. mutations are confined to coding regions

b. mutations are induced at sites throughout the genome (random)

Fanconi anemia exemplifies: a. radiation-induced point mutations affecting mismatch repair b. telomere maintenance in stem cells c. single nucleotide mutations from meiotic arrest in aging males d. the importance of inter-strand cross link repair in maintaining chromosome integrity and the consequences of rearrangements/deletions for development and cancer susceptibility e. intra-strand crosslinks stalling forks and point mutations

d. inter-strand crosslink repair → chromosome integrity; defects cause rearrangements/deletions, developmental defects, cancer risk

In a typical "F3 screen" for recessive mutant alleles, the F2 generation is important because it: a. results in homozygous individuals with mutant phenotypes b. selects germline-only alleles from F1s c. lets dominant alleles be discarded d. keeps aneuploidies out of the F3 e. generates additional copies of new mutant alleles, allowing screening of homozygous individuals in the F3

e. F2 family has multiple heterozygous carriers (additional copies) so F2 × F2 intercrosses give homozygous mutants in the F3

In cervids (deer) and other ungulates, whole genome sequencing and synteny analysis revealed: a. genome duplication leading to elk and cow b. extensive gene loss leading to M. muntjak c. genome duplication leading to M. reevesi d. fusion of chromosomes in the lineage leading to M. muntjak e. aneuploidies in M. reevesi worsening in M. muntjak

d. chromosome fusions in the lineage leading to M. muntjak (Indian muntjac, very few chromosomes)

(Select all) In CRISPR/Cas9 mutagenesis, a synthetic guide RNA functions to: a. cut target RNA b. interact with Cas9 c. initiate non-homologous end joining d. recognize DNA-target sequence e. encode Cas9 for translation

b and d — the gRNA binds Cas9 and base-pairs with the target DNA; Cas9 does the cutting

Fragile X: a CGG repeat in the 5' UTR of FMR1 expands to several hundred copies (X-linked, ~2x more common in males). A plausible explanation for the effect on gene function: a. inversions between 5' UTR and 3' UTR promoter elements b. reduced proofreading by DNA pol alpha in CGG-rich domains c. enhanced recruitment of DNA methyltransferase leading to DNA methylation and transcriptional silencing d. a longer Arg-rich domain causing toxicity e. recessive in males, dominant in females

c. CpG-rich expansion gets methylated → FMR1 silenced (loss of functio