Unformatted text preview: Bio Sci 97, Lecture Section C Exam A Final exam with explanations. Where appropriate, a discussion of the answer and any partial credit given is included. No questions required explanations; these are included here in parenthesis for your benefit. 1. In a certain population at Hardy-Weinberg equilibrium, one of every thirty individuals is affected by an autosomal recessive condition called "heavyeyelids" (he/he) which causes them to fall asleep during important lectures. a. What is the frequency of heterozygous carriers of "heavyeyelids" in this population (he/HE)? .298 (q2 = 1/30: q = .182, 1-.182 = p = .818, heterozygotes are 2pq) also accepted .3, .295 or .299 and any of these as percentages. Full credit given for setting up and not solving (i.e. no calculator). No credit given if you just wrote 2pq and didn't set up how to solve for p and q. b. What is the allele frequency of the dominant HE allele? HE = .818 (p) also accepted .82 and .83 and any of these as percentages. c. Would the expected frequency of "heavyeyelids" (he/he) among the offspring of first-cousin matings in this population be higher or lower than 1/30? higher 2. A man with no family history of cancer-prone genetic diseases dies of a brain tumor. The tumor does not metastasize, but grows so large that he dies from it. a. Would the recipient of his transplanted liver have a high risk of liver cancer? NO (his cancer was only in his brain and did not metastasize) b. Suppose his wife wants to use his sperm for in vitro fertilization--would the resulting children have a high risk of cancer? NO (cancer is NOT a genetic disease, but a disease of somatic cells--also, the man had no genetic predisposition as there was no family history) 3. Mutations in the K-ras gene are frequently found in colon cancer. These mutations are always missense mutations, primarily in codons 12 or 61. Mutations in the APC gene are also associated with colon cancer. In this case, however, nonsense and frameshift mutations are frequent and are scattered throughout most of the coding region. From this information, answer the following: a. What sort of gene is K-ras? Oncogene or proto-oncogene (because of gof mutations) For credit you must have written oncogene or protooncogene (or described it REALLY well)--question was what sort of GENE... b. What sort of gene is APC? Tumor suppressor gene (because of lof mutations) For credit you must have written tumor suppressor gene (or described it REALLY well)--question was what sort of GENE... 4. Some estimated broad-sense heritabilities include: stature, 0.85; systolic blood pressure, 0.57; twinning, 0.5. Which of these has the greatest genetic component? stature, 0.85 (both accepted) 5. Two highly inbred populations of tomato plants grow in adjacent fields. One field is watered regularly and the other is not. Seeds from the plants are taken to a greenhouse 1 Bio Sci 97, Lecture Section C Exam A where they can be grown under uniform conditions. The mean height of greenhouse grown representative plants at the time of flowering is measured: Watered field: Not watered: F1 F2 mean height = 62 cm, variance = 40 mean height = 136 cm, variance = 102 mean height = 102 cm, variance = 126 mean height = 104 cm, variance = 203 a. What is the environmental variance? 126 (this is exactly as the problem done in class. The F1 variance is entirely due to environment since all F1 are genetically identical) b. What is the broad-sense heritability for height (H2)? The formula you will need is: H2 = s2 /s2 ANSWER: 0.38 = (203-126)/203 g p (203 is the total variance and subtracting the environmental variance gives the genotypic variance. The broad-sense heritability is found by dividing the genotypic variance by the total variance it is the proportion of the total variance that is due to genetic effects) 6. How does the recessive allele that causes Tay Sachs disease survive in a population if all affected people with the disease die before they can reproduce? Recessive alleles are maintained in heterozygous individuals. Credit given for any answer with heterozygous or carrier in it. 7. A prominent local oncologist, RKB, is consulting with a family in which Li-Fraumeni syndrome is segregating. Li-Fraumeni syndrome is associated with loss of function mutation in p53. All family members that develop cancer are heterozygous for the loss of function p53 mutation. normal cancer RKB is studying a woman with ovarian cancer (Pt). Look Dad Mom cell cell at the abbreviated pedigree shown--gray individuals have developed cancer. RKB Mom Dad has isolated DNA from Mom, Dad, cancer cells from Pt and her normal somatic cells. RKB Pt is investigating the event leading to loss of heterozygosity (LOH) in the ovarian cancer. Using a probe specific for an SSR that is tightly linked to p53, RKB sees the bands shown: What is the nature of the mutation that led to LOH at the p53 locus in the cancer cell? deletion (the cancer cell lost an SSR relative to the normal cell from Pt. Thus the somatic mutation that led to LOH was due to a deletion of the p53 locus and the SSR) 8. Red-green colorblindness is inherited as an X-linked recessive (cb). A colorblind woman who has a normal number of chromosomes (XX) marries a man with normal color vision and a normal number of chromosomes. One of their children has an unusual chromosome complement: XXY. This child has normal vision. Assume that only one case of nondisjunction occurred to produce the XXY child. 2 Bio Sci 97, Lecture Section C Exam A a. Is the child a boy or girl? boy b. In whose genome did nondisjunction occur and at which meiosis (I, II or can't tell). dad in meiosis I (0.5 for dad, 0.5 for meiosis I) 9. In maize, the genes v, pr and bm are all on chromosome 5. v is 20 cM from pr and bm is 10 cM from pr (pr is between v and bm). Assume there is no chromosome interference in this region. Calculate how many phenotypically v+ pr+ bm plants would you expect out of 1000 progeny from a cross of: v+ pr bm+ X v pr bm v pr+ bm v pr bm .2 x .1 = .02 x 1000 = 20 doubles. This is a crossover between v and pr, so total crossovers equals .2 x 1000 = 200. 200-20 = singles = 180. Divide by 2 to get the number of the specific phenotypic class asked for: 90. Credit given as follows: 4 pts: 90 3 pts: 180 1 pt: 200 or 100 10. Two loci are 22 map units apart. How many centiMorgans apart are they? 22 (0.22 NOT accepted) 11. Two loci are 52 map units apart. What is the frequency of recombination between the two loci? 50 % or 0.5 (come on guys!!) 12. Flies of genotype e+ f+ g+ h+ j+/ e+ f+ g+ h+ j+ (homozygous wild type) were irradiated in an attempt to produce rearranged chromosomes. The genes on the original unrearranged normal chromosome map as follows: Map units: e+ 10 f+ 5 g+ 5 h+ 10 j+ A rearrangement chromosome was obtained. Flies heterozygous for the rearranged chromosome and a normally arranged chromosome carrying all recessive mutant alleles (efghj) show the f, g and h mutant phenotypes. a. What sort of rearrangement describes the recovered rearranged chromosome? deletion b. Based on your hypothesis, what would you expect if linkage analysis was performed on the e-j interval for the rearranged chromosome? Would you find that the recombination frequency would be (choose one): A. 30% B. <30% C. >30% B. much less than the 30% of the unrearranged chromosome 3 Bio Sci 97, Lecture Section C Exam A 13. What gametes are produced by an individual with the genotype bD/Bd if the genes are a. completely linked bD and Bd (credit given if you goofed and wrote a slash between these two gametes) b. located on different chromosomes bD, bd, BD, Bd (no credit given for drawing diploid genotypes) 14. The Drosophila genes for white eyes (w), cut wings (ct) and tan body (t) lie at map positions 1.5, 20.0 and 27.5, respectively. Out of 576 progeny, 6 are double crossovers. 1.5 20.0 27.5 w ct t What is the interference across this region? First, calculate the distance between the loci. Then calculate expected numbers of dco class (.185)(.075)(576)=7.992 or 8 dco progeny. Observed dco progeny is 6. Calculate the coeficient of coincidence: observed/expected: 6/8 = .75. Finally, calculate the interference as 1-cc: 1-.75=.25. ANSWER is 0.25 or 25% or .249 No partial credit given. 15. A human geneticist, Toby, is trying to identify the genes that are mutated in individuals with heart disease. Toby finds a family in which an SSR (SSR6) seems to segregate with the disease: all 6 individuals with heart disease share the same SSR. Toby expands his studies to include other families with heart disease and finds that SSR6 is usually, but not always linked to the disease (182/218). Toby states that he's identified a QTL for heart disease. a. What characteristic of heart disease is Toby using to account for the individuals that have SSR6 but don't have heart disease? (hint: what sort of trait is it?) Because it's a threshold trait (not everyone will be above the threshold and show the trait). 0.5 given for stating multifactorial or explaining multifactorial. No credit for quantitative b. Is Toby going out on a limb here? Yes or No, please. THROWN OUT (NOT ALL STUDENTS KNOW THE MEANING OF THIS PHRASE--I APOLOGIZE). No--his conclusion is backed up by the facts. 16. Esme' has just joined her first genetics lab. She has been given the project of determining whether the human Factor VIII gene functions in Drosophila by making a transgenic Factor VIII fly. But Esme' doesn't have a clue how to proceed. Use the given list and order the four steps Esme' should take.
1. PCR amplify hybridize probe UV irradiate transform bacteria/recover lots of DNA inject into Drosophila PCR amplify transcribe ligate into vector Southern blot ligate into P element vector find an SSR 2. ligate into P element vector 3. transform into bacteria/recover lots of DNA 4. inject into Drosophila Credit given as follows: 4 Bio Sci 97, Lecture Section C Exam A 2 pts for all 4 in correct order. 1 pt if everything correct except student chose "ligate into vector" instead of "ligate into P element vector". 1 pt if 3 steps were in correct order, regardless of what 4th step was. 0.5 pt if 2 steps were in correct order, regardless of what the other two steps were. The emphasis in this question is to test whether you understood the logical progression of steps. 17. Your best buddy needs to determine whether dp and pp are linked in Drosophila. To determine whether these two loci are linked, your best buddy testcrosses an F1 female heterozygous for both loci; recessive loci are in cis in this F1 female. dp = dumpy wings, recessive dp+ = normal wings pp = pink eyes, recessive pp+ = normal red eyes a. What is the genotype of the female in the testcross? dp+ pp+/ dp pp No credit given for incorrect notation. No credit given for answers that did not clearly show the recessive alleles in cis (i.e. dp+/pp dp/pp+ OR dp+/pp+ dp/pp NOT accepted). Credit given if your brain "burped" and you substituted "e" for "pp" in all of question 17. b. What is the genotype of the male in the testcross? dp pp/ dp pp No credit given for incorrect notation. c. Fill in the blanks on your answer sheet for the progeny your buddy observes: genotype of maternal gamete # of flies dp+ pp+ 2449 dp pp 2451 dp+ pp 49 dp pp+ 51 Grading was more lenient here: Credit given for showing the diploid genotype of each instead of the requested maternal gamete genotype. Remember, rows 1 and 2 can be swapped and rows 3 and 4. 1 pt given if all were right but you dropped a + in haste. NO credit for switching parentals and recombinants. (FYI: last year's exam had the parentals in trans--some students memorized this and got it wrong this year, tsk, tsk). No credit for writing out phenotypes--you should know the difference between phenotypes and genotypes. d. What is the frequency of recombination between the two loci? 100/5000 =.02 or 2% or 2mu or 2 cM or just 2. 18. A "new and improved" (N/I) banana variety has been identified. Usually, banana gametes contain 3 chromosomes, but the gametes of the N/I banana contain 6 chromosomes, the rest of the N/I banana cells contain 12 chromosomes. a. Is the N/I banana euploid or aneuploid? euploid b. What is the "ploidy" of the new banana? tetraploid or 4 c. Is the N/I banana fertile or sterile? fertile 5 Bio Sci 97, Lecture Section C
d. How many N/I banana chromosomes are found in N/I somatic cells? 12 Exam A 19. Inheritance of ABO blood groups follows regular Mendelian patterns. The gene locus for ABO is given the symbol I; the three common alleles are the codominants IA and IB and the recessive IO. A population of UCI students has the following allele frequencies: 0.7 IO, 0.2 IA, 0.1 IB. The genotypes for the phenotypic blood groups are: AB (IA IB); A (IA IA as well as IA IO); B (IB IB as well as IB IO); O (IO IO). At what frequency is the A blood type expected if the population is in Hardy-Weinberg equilibrium? (0.2)2 + 2(0.2)(0.7) = .32 (NO credit for 1/3, .33 ,.34) 20. Smurfs are normally blue (Bl), fuzzy (Fz) and industrious (In). Data from a three-point testcross are shown. Phenotype Normal White Bald Lazy white, bald white, lazy bald, lazy white, bald, lazy Genotype Bl-, Fz-, Inbl, Fz-, InBl-, fz, InBl-, Fz-, in bl, fz, Inbl, Fz-, in Bl-, fz, in bl, fz, in Number 324 49 3 42 32 5 51 309 On the genetic map, which gene lies in the middle? Fz, fuzzy (ok if you drew out the map) SECTION II. Choose the best answer (1) for the questions 22 and 23 based on the given pedigree. *********Questions 21 and 22 were thrown out because there was more than one interpretation of the pedigree. The pedigree shows the inheritance of a genetic disease in three generations of a family.
1 2 3 7 4 5 6 8 9 10 21. Based on the pedigree, the mode of inheritance of this disease could be a. Autosomal recessive b. X-linked recessive 6 Bio Sci 97, Lecture Section C
c. Autosomal dominant d. X-linked dominant e. C or D Exam A 22. Assume that #10 marries an unaffected woman, and they have an unaffected daughter. Given that assumption, then the mode of inheritance of this disease is most likely... a. Autosomal recessive b. X-linked recessive c. Autosomal dominant d. X-linked dominant e. Y linked dominant SECTION III. Answer True or False (T or F) on your answer sheet for each of the following questions (8 points total): 23. Failure of a checkpoint has no impact on the development of cancer. False 24. The mutation rate at all loci in the human genome is the same. False 25. Only rarely will a female human being be identified that is a genetic mosaic for an Xlinked gene. False 26. Recombination frequencies are the same for cis and trans heterozygotes. True 27. Digesting genomic DNA with a restriction enzyme will result in one or two fragments. False only if you use a Southern blot to detect...remember, digestion of genomic DNA with a RE gives a smear of thousands, perhaps millions of products. 28. GM crops are banned in the United States. False 29. Chemotherapy works by stopping all replicating cells from dividing. True 30. If the green hair trait is due to an X-linked recessive gene and 1/2500 males has green hair, then the value of q for the population is 1/2500. True 31. Cells from an XXX female have one Barr Body. False 32. Even though the rate of spontaneous mutation is high in our cells, very few proteins are affected because the protein translation machinery compensates for incorrect codons. False 33. Polymorphisms are found most frequently in gene-poor regions. True 34. A dicentric chromosome is genetically unstable. True 35. A karyotype is the best way to find a SNP. False 36. A person with a reciprocal translocation is not missing any genetic information. True 37. Reciprocal crosses in Drosophila identified the existence of a special chromosome, the ring chromosome. False 7 Bio Sci 97, Lecture Section C Exam A 38. Because mitochondrial DNA recombines so frequently, it is an unreliable indicator of relatedness. False F F F T F F T T F F T T F T F F F F T T F T F F F F T F F T T F F F F F T F F T T F T T F T F F 8 ...
View Full Document