Exam 4 Genetics Review

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Apr 3, 2024

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Genes can switch “on” & “off” • differential expression of genes is essential (problematic to have all genes “on” at all times!) • some always “off” in certain tissues, or at particular developmental stages, others “on” but dynamic… • need to dramatically increase or decrease amount of gene product (# copies / cell), and do it fast Gene regulation • respond to changes in the external environment • also maintain normal on-going cellular functions • in bacteria, regulation of expression usually occurs at the level of transcription (i.e., DNA → mRNA) Prokaryotic gene regulation Always “on” (… no matter what!) • constitutive: a gene product produced constantly, regardless of external environmental conditions Sometimes “on”, but other times “off” • inducible: a gene product whose synthesis occurs only under certain conditions (envtl. cue is present) • repressible: a gene product whose synthesis is shut down when large amounts of ‘itself’ are present Wait… what? Inducible. gene “on” = enabled by presence of a protein (or AA) that is not the same as the protein (or AA) produced by that gene e.g., lac operon Repressible. gene “on” = enabled by absence of a protein (or AA) that is the same as the protein (or AA) produced by that gene e.g., trp operon Prokaryotic gene regulation • negative control: ‘normal’ state is “on”, transcription must be shut off by a regulator molecule • positive control: ‘normal’ state is “off”, transcription must be enabled by a regulator molecule Inducible. gene “on” = enabled by presence of a protein (or AA) that is not the same as the protein (or AA) produced by that gene Lactose metabolism Inducible. gene “on” = enabled by presence of a protein (or AA) that is not the same as the protein (or AA) produced by that gene

• environmental cue: lactose present = expression induced (lactose absent = expression repressed) • operon: genes with related function are side-by-side, & transcription is controlled by same regulatory region • cis- acting site: regulatory DNA region, adjacent to gene whose expression it controls • trans-acting molecule: a free molecule that binds to cis-acting sites to regulate expression • 3 structural genes, convert lactose → energy • lac Z: make glucose & galactose from lactose • lac Y: import lactose • lac A: export by-products … transcribed & translated together This figure represents the lac operon. Five features are labeled (letters), but two features are unlabeled (numbers). In the presence of lactose and the absence of glucose, what will be strongly bound to "2"? Selected Answer: D. nothing Answers: A. CAP B. lactose C. repressor D. nothing E. RNA polymerase Question 2 0.1 out of 0.1 points

This figure represents the lac operon. Five features are labeled (letters), but two features are unlabeled (numbers). In the absence of lactose and the presence of glucose, what will be strongly bound to "1"? Selected Answer: A. nothing Answers: A. nothing B. RNA Polymerase C. repressor D. CAP E. glucose Question 3 0.1 out of 0.1 points This figure represents the lac operon. Five features are labeled (letters), but two features are unlabeled (numbers). In the absence of lactose and glucose, what will be strongly bound to P? Selected Answer: B. RNA Polymerase

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Answers: A. repressor B. RNA Polymerase C. nothing D. lactose E. CAP Question 4 0.1 out of 0.1 points This figure represents the lac operon. Five features are labeled (letters), but two features are unlabeled (numbers). What is labeled "1", and what is labeled "2"? Selected Answer: D. CAP binding site; operator Answers: A. operator; structural gene B. operator; CAP binding site C. operator; repressor gene D. CAP binding site; operator E. CAP binding site; promotor

Sickle cell anemia in humans provides an example of the relationship between ______. Selected Answer: 5. genotype and fitness Answers: 1. DNA, RNA and protein 2. genotype and phenotype 3. phenotype and fitness 4. all of these are correct 5. genotype and fitness Question 2 0.05 out of 0.05 points What term would be applied to a regulatory condition that occurs when a non-enzymatic protein associates with a particular section of DNA and prevents transcription? Selected Answer: D. negative control Answers: A. positive control B. gene silencing C. inhibition D. negative control E. loss-of-function Question 3

0.05 out of 0.05 points Methylation is a mode of genetic regulation in eukaryotes, and it refers to ______. Selected Answer: C. addition of methyl groups to cytosine Answers: A. denaturation of RNA polymerase B. altering translational activity of charged tRNAs C. addition of methyl groups to cytosine D. alteration of DNA polymerase activity E. changes in DNA-RNA hydrogen binding Question 4 0.05 out of 0.05 points A similarity between the mechanism for controlling gene expression in the lac and trp operon is ______. Selected Answer: D. all of these are similarities

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Answers: A. the operator does not encode a gene product B. transcription generates a polycistronic mRNA strand C. the repressor protein is allosteric D. all of these are similarities E. the promoter does not encode a gene product Below is a set of contigs (6-bp each) from a linear chromosome that has been shotgun seqeunced. They have been aligned, but two contigs (labeled [1] and [2]) have not yet been added. AGGAAT AATCGC [1] ______ [2] ______ GCGTTT TTTTCC TCCGGA What is the identity of the 8th and 9th nucleotide (in order) of the consensus seqeunce? Selected Answer: B. GC Answers: A. CG B. GC C. AA D. GT E.

TT Question 2 0 out of 0.05 points Below is a set of contigs (6-bp each) from a linear chromosome that has been shotgun sequenced. They have been aligned, but two contigs (labeled [1] and [2]), and underlined) have not yet been added. TTCCGG TTTTTC GCGTTT [1] ______ [2] ______ AATTCC AGGAAT Which of these are the most likely sequences of the two missing contigs? Selected Answer: B. [1] TCCTTT; [2] TTTCGC Answers: A. [1] GGAGCG; [2] TCCGGA B. [1] TCCTTT; [2] TTTCGC C. [1] AGGATC; [2] ATCCGC D. [1] CGCGCG; [2] TCCAAT E. [1] AAACGC; [2] AGGTTT Question 3 0.05 out of 0.05 points

Below is a set of contigs (6-bp each) from a linear chromosome that has been shotgun seqeunced. They have been aligned, but two contigs (labeled [1] and [2]) have not yet been added. AGGAAT AATCGC [1] ______ [2] ______ GCGTTT TTTTCC TCCGGA What is the expected sequence of the two missing contigs? Selected Answer: E. [1] CGCAAA; [2] AAAGCG Answers: A. [1] CGCAAA; [2] GCGTTT B. [1] GCGTTT; [2] GCGCGC C. [1] CGCAAA; [2] TTTGCG D. [1] GCGTTT; [2] AAACGC E. [1] CGCAAA; [2] AAAGCG Question 4 0.05 out of 0.05 points Genome projects that use shotgun sequencing involve the following four steps (in this order): Selected Answer: E. [1] fragment chromosomes, [2] sequence DNA, [3] align and assemble, [4] annotate Answers: A.

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[1] sequence DNA, [2] align and assemble, [3] annotate, [4] fragment chromosomes B. [1] fragment chromosomes, [2] annotate, [3] sequence DNA, [4] align and assemble C. [1] sequence DNA, [2] fragment chromosomes, [3] annotate, [4] align and assemble D. [1] annotate, [2] align and assemble, [3] fragment chromosomes, [4] sequence DNA E. [1] fragment chromosomes, [2] sequence DNA, [3] align and assemble, [4] annotate The main distinction between a benign tumor versus a malignant tumor is _______. Selected Answer: B. metastatic growth Answers: A. duration of exposure to a carcinogen B. metastatic growth C. a Philadelphia chromosome D. DNA methylation E. a mutant p53 gene Question 2 0.1 out of 0.1 points

What gene normally causes programmed death of a cell that has sustained severe DNA damage? Selected Answer: D. p53 Answers: A. cyclin B. BRCA1 C. proto-oncogene D. p53 E. oncogene Question 3 0.1 out of 0.1 points Cancer is associated with loss of control over growth and division of ______ cells, and it is usually caused by _______. Selected Answer: D. somatic; accumulation of several mutations Answers: A. germ-line; a single mutation B. none of these are true C. germ-line; accumulation of several mutations

D. somatic; accumulation of several mutations E. somatic; a single mutation Question 4 0.1 out of 0.1 points Mutant versions of genes that are normally involved in regulating cell growth and division are _______. Selected Answer: D. oncogenes Answers: A. tumor suppressors B. malignant genes C. proto-oncogenes D. oncogenes E. attenuators What three transitions in the cell cycle to serve as checkpoints? Answers: A. quiescence; senescence; and replication B. G0/G1; G1/G2; and G2/G3 C. arrest; apoptosis; and attenuation

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D. S-phase; mitosis; and meiosis E. G1/S; G2/M; and M Question 2 0.05 out of 0.05 points The protein p53 ________. Answers: A. stimulates both apoptosis and the arrest of the cell cycle B. is absent or nonfunctional in less than 30% of all cancers C. stimulates apoptosis, but not the arrest of the cell cycle D. stimulates the arrest of the cell cycle, but not apoptosis E. is found at high levels in all cells Question 3 0.05 out of 0.05 points Oncogenes are ________ for the cancer phenotype, whereas mutated tumor-suppressor genes are ________ for the cancer phenotype.

Answers: A. dominant; recessive B. recessive; incompletely dominant C. dominant; co-dominant D. co-dominant; incompletely dominant E. benign; malignant Question 4 0.05 out of 0.05 points Which phase of the cell cycle are cancer cells unable to enter? Answers: A. mitosis B. G2 C. G0 D. S-phase E. G1 Imagine that the fur color of squirrels is controlled by a single gene with two alleles ( A , and a ). In a population sample of 225, 90 individuals have a fancy pink fur color and they are either AA or Aa ; all other squirrels are green and they are aa . If you assume that this population is in Hardy–Weinberg equilibrium, what is the expected frequency of the " A " allele [1] and the " a " [2] allele?

Selected Answer: C. [1] 0.23, [2] 0.77 Answers: A. [1] 0.40, [2] 0.60 B. [1] 0.90, [2] 0.10 C. [1] 0.23, [2] 0.77 D. [1] 0.60, [2] 0.40 E. [1] 0.25, [2] 0.75 Question 2 0.1 out of 0.1 points The term ______ pool refers to the total genetic information carried by all members of a population. Selected Answer: A. gene Answers: A. gene B. genotype C. chromosome D. phenotype E. genome Question 3 0.1 out of 0.1 points

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Imagine that the fur color of squirrels is controlled by a single gene with two alleles ( A , and a ). In a population sample of 225, 90 individuals have a fancy pink fur color and they are either AA or Aa ; all other squirrels are green and they are aa . What is the observed frequency of the dominant [1] and the recessive [2] phenotypes? Selected Answer: E. [1] 0.40, [2] 0.60 Answers: A. [1] 0.75, [2] 0.25 B. [1] 0.77, [2] 0.23 C. [1] 0.90, [2] 0.10 D. [1] 0.60, [2] 0.40 E. [1] 0.40, [2] 0.60 Question 4 0.1 out of 0.1 points Imagine that the fur color of squirrels is controlled by a single gene with two alleles ( A , and a ). In a population sample of 225, 90 individuals have a fancy pink fur color and they are either AA or Aa ; all other squirrels are green and they are aa . If you assume that this population is in Hardy–Weinberg equilibrium, what is the expected frequency of the " AA " genotype [1] and the " Aa " genotype [2]? Selected Answer: B. [1] 0.05, [2] 0.35 Answers: A.

[1] 0.40, [2] 0.60 B. [1] 0.05, [2] 0.35 C. [1] 0.45, [2] 0.45 D. [1] 0.25, [2] 0.50 E. [1] 0.05, [2] 0.20 In comparing the effects of stabilizing selection [1] vs. directional selection [2] on the frequency of a continuously varying phenotypic trait within a population, several differences exist. These differences include: Selected Answer: E. no change [1] vs. change [2] in the mean trait value Answers: A. change [1] vs. no change [2] in levels of immigration B. no change [1] vs. change [2] in the mean trait value range C. bimdodal [1] vs. unimodal [2] trait value distribution D. unimdodal [1] vs. bimodal [2] trait value distribution E. no change [1] vs. change [2] in the mean trait value Question 2 0.1 out of 0.1 points

Imagine that the frequency distribution of a continuously varying phenotypic trait is bimodal in one population (pop. 1), but unimodal in another (pop. 2). From this, you would hypothesize that pop. 1 had experienced ______, and pop. 2 ______. Selected Answer: C. disruptive selection; had not Answers: A. stabilizing selection; had not B. disruptive selection; directional selection C. disruptive selection; had not D. directional selection; had not E. disruptive selection; stabilizing selection Question 3 0.1 out of 0.1 points Which of the following is not expected to cause deviation from genotype frequencies predicted by the Hardy–Weinberg law? Selected Answer: B. large population size Answers: A. a severe bottleneck B. large population size C. advantageous alleles

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D. positive assortative mating E. deleterious mutations Question 4 0.1 out of 0.1 points The relationship observed between the beak size of ground finches and a drought condition is an example of ______. Selected Answer: B. directional selection Answers: A. negative assortative mating B. directional selection C. disruptive selection D. inbreeding E. stabilizing selection In the study by Banks et al. (2003), the results ______ support the idea that female-biased predation of wombats was responsible for the previously reported ______ population sex ratio. Selected Answer: D. did not; male-biased Answers: A. did; unbiased B. did; male-biased

C. did; female-biased D. did not; male-biased E. did not; female-biased Question 2 0.1 out of 0.1 points In population genetics, a microsatellite marker (or locus) is a ______ section of DNA. Selected Answer: A. repetitive and length-variable Answers: A. repetitive and length-variable B. repetitive and length-invariable C. non-repetitive and length-invariable D. highly conserved and monomorphic E. non-repetitive and length-variable Question 3 0.1 out of 0.1 points In the study by Banks et al. (2003), the DNA extracted from hairy-nosed wombat carcasses was informative about ______.

Selected Answer: E. all of the three answers are correct Answers: A. none of the three answers are correct B. the sex of deceased individuals C. the health history of deceased individuals D. the age of deceased individuals E. all of the three answers are correct Question 4 0.1 out of 0.1 points In the study by Banks et al. (2003), the DNA extracted from canid faces was used to identify ______. Selected Answer: C. which species of canid had been at the crime scene and what they had recently eaten Answers: A. what the canid at the crime scene had recently eaten B. the degree of relatedness between dingoes and feral dogs C. which species of canid had been at the crime scene and what they had recently eaten D.

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the degree of relatedness between dingoes, feral dogs, and hairy-nosed wombats E. which species of canid had been at the crime scene

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Exam 4 Genetics Review

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