Prelim+3+F23+Answer+Key

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BioMG2800 © Department of Molecular Biology and Genetics Cornell University 1 PBIOMG 2800 Lectures in Genetics and Genomics Third Preliminary Exam November 14, 2023 There are 14 pages to this exam, including this page. Be sure you have a complete set before you begin. Pages 13 and 14 are blank pages at the end. You may remove these and use them for scratch paper (you do not need to turn them in). Put all answers in the boxes or on the lines provided. Only answers in the boxes or lines will be graded. Please turn off all cell phones and other electronic devices. You may use a non-internet capable calculator. __ANSWER KEY_________________ _______________ Name: LAST, FIRST Net ID 1. ________/11 2. ________/15 3. ________/14 4. ________/16 5. ________/12 6. ________/16 7. ________/16 Total ________/100

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 2 1. (11 pts total) Short Answer questions a. (8 pts) For the following statements, indicate whether they are true or false. i. The heritability of the height of a field of sunflowers in a season with lots of variable hot and cold days would be higher compared to those in a season with consistently warm days. F ii. The Linkage Disequilibrium score between the genetic marker SNP007 and the SSR variant 003 is close to 0 in a population. We would expect to see random segregation of the alleles for SNP007 and SSR003 in this population. T iii. To package the long DNA molecules in the small space of the nucleus, eukaryotic cells rely on histone proteins, which are the major protein constituent of the chromatin and can effectively compact the DNA with little help from other factors. F iv. Deleterious alleles that have a severe consequence would be strongly selected against and rapidly disappear from a population over a few generations, regardless of whether they are dominant or recessive in nature. F b. (3 pts) Having a proper set of chromosomes undergo meiosis is critical for the development of normal gametes. Watermelon and bananas are exploited in agriculture to produce seedless fruits. Indicate all possible ploidies for those seedless fruits' somatic cells by placing an X in the box(es) to the left of your answer choice(s). Diploid (2x) X Triploid (3x) Tetraploid (4x) X Pentaploid (5x) +1.5 pts per correct answer. -1.5 pts per missing/incorrect answer.

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 3 2. (15 pts total) A bacterial species is discovered that can tolerate high levels of toxic petroleum hydrocarbons such as benzene. Realizing the commercial importance of these bacteria in cleaning up chemical spills, you decide to characterize the pathway that allows them to break down benzene into non-toxic molecules. Two enzymes D and E are expressed at high levels when the bacterial cells encounter benzene. You discover that the genes encoding these enzymes are organized into an operon. A diagram of the operon and its regulatory gene are shown below. The regulatory gene is an activator that, when bound to benzene, binds to the J region to activate the expression of the D and E genes. State the enzyme activities, + or - , of the enzymes encoded by genes D and E in the below genotypes in the presence of benzene. • The mutants reg-, D-, and E- are null alleles. • The J- mutant is a deletion of the entire J sequence. • The Reg 141 allele produces an activator that cannot bind benzene. • The Reg 537 allele produces an activator that cannot bind to the J region. Genotype D E Reg+ ; P+ J+ D+ E+ / reg- ; P+ J+ D+ E+ + + Reg+ ; P+ J- D+ E+ / reg- ; P+ J+ D- E+ - + reg- ; P+ J- D+ E+ / reg 537 ; P+ J+ D- E+ - - reg 141 ; P+ J- D+ E+ / reg- ; P+ J+ D- E+ - - reg 537 ; P+ J+ D+ E- / Reg+ ; P+ J- D+ E+ + - +1.5 pts per box.

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BioMG2800 © Department of Molecular Biology and Genetics Cornell University 4 3. (14 pts total) You identified a SNP that correlates with dark brown vs blond hair in a population. You hypothesize that the SNP affects an enhancer that regulates the expression of the gene KITLG , which encodes a melanin receptor. KITLG is expressed at higher levels in individuals with brown hair compared to those with blond hair. a. (8 pts) You developed a series of reporters to test the function of this SNP. In column 1 of the table on the next page, indicate in the table on the next page, the expected GFP expression levels (“high” or “low”) for each of the reporter constructs. b. (6 pts) This enhancer is regulated by the MITF transcription factor that binds to and acts on the enhancer sequence (containing either SNP) you have discovered. In columns 2 and 3 of the table on the next page, indicate the expected GFP expression levels upon RNAi knockdown of MITF (“Higher than no RNAi”, “Lower than no RNAi”, or “Same as no RNAi”). For column 2 assume MITF is a transcriptional activator. For column 3 assume MITF is a transcriptional repressor. promoter GFP C A CTAAAG promoter GFP C G CTAAAG 5kb 1kb promoter GFP C G CTAAAG 10kb promoter GFP GAAATC A C 4.5kb 1 2 3 4

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 5 Table for question 3 Reporter construct GFP Levels (high or low) If MITF is an activator, expected GFP expression upon RNAi. (“Higher than no RNAi”, “Lower than no RNAi”, or “Same as no RNAi”) If MITF is a repressor, expected GFP expression upon RNAi. (“Higher than no RNAi”, “Lower than no RNAi”, or “Same as no RNAi”) 1 high Lower than no RNAi Higher than no RNAi 2 low Lower than no RNAi Higher than no RNAi 3 low Lower than no RNAi Higher than no RNAi 4 Low Same as no RNAi Same as no RNAi +2 pts per box for “GFP levels” +0.75 pts per box for “expected GFP expression upon RNAi”

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 6 4. (16 pts total) In a remote island population of the mythical Kudlebird, there are two feather colors controlled by one single locus with two alleles. The dominant allele (G) codes for the green feather color, while the recessive allele (g) results in the blue feather color. The different feather colors don’t confer any selection advantage or disadvantage. You are a biologist in a conservation group studying the endangered Kudlebird population, and you are interested in the alleles for this beautiful feather coloration. To do this, you conduct a thorough census of the Kudlebirds and find that out of 500 individuals, 50 exhibit blue color, 100 are carriers of the blue color allele. a. (4 pts) Calculate the frequency of the G allele in this Kudlebird population using 1 decimal place. 0.8 b. (4 pts) Is this Kudlebird population in Hardy-Weinberg equilibrium? Explain your reasoning in 1 concise sentence. No (+2). The predicted genotypes using HWE, 320 GG, 160 Gg, and 20 gg, do not match the observed genotypes 350 GG, 100 Gg, and 50 gg. For the reasoning part, correct calculations for the expected genotype frequency (+1), correct calculations for observed frequency (+1). c. (8 pts) As an effort to conserve this endangered Kudlebird population on this remote island, your group decided to introduce an artificially propagated Kudlebird population to the original 500 Kudlebirds population on the island. The introduced population is composed of 200 Kudlebirds, and the allele frequency for the dominant G allele ( G ) in this introduced population is 0.5 . Your group ensured that the merged Kudlebird population on the island is under Hardy- Weinberg Equilibrium population upon reintroduction. What is the expected frequency of blue feather Kudlebirds after 1 generation in this merged population (to 2 decimal places)? Show your work to get full or partial credit. 0.08. The merged population consists of 500 + 200 = 700 Kudlebirds (+2). Of the 700 × 2 = 1400 alleles, the number of g allele is 1000 × 0.2 + 400 × 0.5 = 400, thus the allele frequency of g allele in the merged population is 400 / 1400 = 0.286 (+2). As the merged population is in HWE, after 1 generation, the expected frequency of blue feather Kudlebirds (genotype frequency of gg) is 0.286 × 0.286 = 0.08 (+4). If not correct answer but attempt to merge the two populations: partial credit of +2. If not correct answer but merged the two populations and calculated the allele frequency correctly : +4.

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BioMG2800 © Department of Molecular Biology and Genetics Cornell University 7 5. (12 pts total) The Soay sheep of St. Kilda are a feral population of primitive domestic sheep, which have an unusual inherited polymorphism for horn type (long and short horns). Horns are a form of sexual weaponry present in all wild species of sheep and play an important role in competition between males for access to mates. For Soay sheep, both sexes have an inherited polymorphism for horn morphology. Your collaborators are interested in the genetics genetic architecture of this long/short horn morphs in the Soay sheep population. They conducted a genome-wide association study (GWAS) of horn length using ~36,000 single nucleotide polymorphisms (SNPs). This is the Manhattan plot showing the results of the GWAS study in male sheep and female sheep separately. Your collaborators want your insights on the interpretation of the results. a. (6 pts) What will be a plausible interpretation to the pattern you see above? (Select all that apply by placing an X in the box(es) next to your answer choice(s). X It’s possible that the same set of genetic elements determines horn lengths for both sexes of Soay sheep. The highest dot in Chromosome 10 in the female panel represents the SNP that is most likely to be causal to horn length morphology in female sheep. It’s possible that there is only 1 region related to the horn length, because the two peaks in the Manhattan plot could be caused by linkage disequilibrium. +6 correct answer. -3 for each incorrect answer.

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 8 Your collaborators further investigated the region that contains the highest peak in the Manhattan plot above for female sheep on Chromosome 10. Here is the same Manhattan plot but zoomed in only to the regions of interest. They also provide you with the linkage mapping and gene annotations in this region. b. (6 pts) Your collaborators want to know which genes could possibly contribute to the horn length morphology in this region. Please select list all candidate genes in this region shown above by placing an X in the box(es) to the left of your answer choice(s). X STARD13 RFC3 KL X RXFP2 X B3GLCT +2 per correct answer. -2 for each incorrect answer.

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 9 6. (16 pts total) Three selectively neutral SNPs on human chromosome 3 are followed in a population. It is known that their order is SNP1 SNP2 SNP3, and that there is 20 kb between SNP1 and SNP2 and also 20 kb between SNP2 and SNP3. The allele frequencies in the population are as follows: The frequencies of the various pairwise haplotypes observed in the population is as follows: a. (4 pts) Which locus pairs are in linkage disequilibrium? Indicate your answer by placing an X in the box to the left. SNP1 and SNP2 SNP1 and SNP3 X SNP2 and SNP3 All three loci are in linkage disequilibrium. All three loci are in linkage equilibrium.

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BioMG2800 © Department of Molecular Biology and Genetics Cornell University 10 b. (4 pts) Do the data suggest the presence of a recombination hotspot, and if so, where is it located? Indicate your answer by placing an X in the box to the left. No recombination hotspot is suggested by the data. A hotspot is to the left of all 3 SNPs. X A hotspot is between SNP1 and SNP2. A hotspot is between SNP2 and SNP3. A hotspot is to the right of all 3 SNPs. c. (8 pts) Which of the following are the most likely haplotypes present in ancient ancestors of the population under study? Select all that apply by placing an X in the box(es) to the left. SNP1: A SNP2: C SNP3: T X SNP1: A SNP2: G SNP3: T X SNP1: A SNP2: C SNP3: C SNP1: A SNP2: G SNP3: C SNP1: T SNP2: C SNP3: T X SNP1: T SNP2: G SNP3: T X SNP1: T SNP2: C SNP3: C SNP1: T SNP2: G SNP3: C +2 per correct answer. -2 for each incorrect answer.

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 11 7. (16 pts total) In a lab accident, a cage of WT male mice were exposed to a high dose of X-ray for a short time. These male mice were mated with true-breeding females with the recessive trait wavy hair, encoded by the gene Wy , which is located on chromosome 11. Most of the female offspring show wild-type phenotypes (straight hair), but one showed wavy hair. You mated this wavy hair female with a wild-type male. You then carried out a FISH experiment using DNA probe spanning the entire wavy hair gene using cells obtained from ~20 offspring. FISH is an assay where a fluorescently labeled DNA probe will base-pair and fluoresce if it can bind to any part of the gene. From the FISH assay, you saw two fluorescent spots for cells from about half of the offspring, but only one fluorescent spot for cells from the other offspring. a. (4 pts) What is the most likely chromosomal abnormality that accounts for the above data: X Deletion Duplication Inversion Translocation The known genetic distances between several markers and Wy in wild-type mice are shown below: To further characterize this chromosomal defect, you used genetic markers flanking the Wy gene and conducted mapping experiments. You obtained the following results for genetic distances (in m.u.): a-b 2 b-c 0.6 c-Wy <0.1 Wy-d <0.1 d-e 1.5 e-f 0.8 f-g 2 g-h 1

BioMG2800 © Department of Molecular Biology and Genetics Cornell University 12 b. (4 pts) Based on the additional mapping data, what chromosomal abnormality/abnormalities do you expect chromosome 11 of the X-rayed mice to have? Select all that apply by placing an X in the box(es) to the left of your answer choice(s). X (+1) Deletion Duplication X (+3) Inversion Translocation c. (8 pts) The locations of 7 PCR primers are shown above on the WT version of Chr 11. Indicate (yes or no) if the indicated PCR primers would make a PCR product on the WT chromosome or the rearranged chromosome 11. PCR primer pair WT chr 11 PCR product? Yes or No Rearranged chr 11 PCR product? Yes or No 1 + 2 Yes No 3 + 4 Yes No 5 + 6 Yes Yes 2 + 7 No Yes +1 per box.

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