7.03_FA09_EX1_key7.03_FA09_EX2_key7.03_FA09_EX3_key7.03 Exam 1 2009 Name: KEY TA: Jason Funt Wendy Niedelman Sarah Pfau David Gifford Amuda Panneerselvam Yun Sung Section time: There are 6 pages including this cover page Please write your name on each page.1. (30 points) You have two haploid strains of yeast that have adenine mutations: one is adex trp1 and the other is adey Trp1+. The adenine and tryptophan pathways are controlled by different genes. You mate the strains and obtain a diploid that is Trp+ Ade- (requires only adenine). A. (5 points) Given these data, select all statements that are consistent with the data: a. adex and adey are mutations in different genes b. adex and adey are mutations in the same gene c. adex and adey are recessive mutations in different genes d. adex is a dominant mutation a, b, d B. (12 points) You examine 100 meiotic tetrads and obtain the following results: trp adex + trp adex + + + adey + + adey 70 10 20 Draw a map showing any linked genes and the map distances. C. (5 points) Given the new data which of these statements is consistent with the data? a. adex and adey are mutations in different genes b. adex and adey are mutations in the same gene c. adex and adey are recessive mutations in different genes d. adex is a dominant mutation a, d D. (4 points) What is the probability of all double crossovers? E. (4 points) What is the probability of finding a tetrad with the following constitution? trp adex adey + adex adey trp + + + + + trp adex + + adex adey trp + + + + adey trp adex + trp adex adey + + + + + adey P (adex trp) P(trp adey) = .05 x .10 = .005 NPD = 1/4 of all double crossovers. So, 1/4 x.005 adex adey trp 5 102. (18 points) In Neurospora three different genes are needed for production of black pigment in spores. Recessive alleles at any one of these result in white spores. Three crosses of wild type by a mutant containing a different blk mutation were analyzed (120 asci/cross). Remember that in Neurospora the position of the spores in the ascus is related to the position of chromatids at the first meiotic division. Cross 1 blk1 x wild type Cross 2 blk2 x wild type Cross 3 blk3 x wild type Asci observed from: Cross 1 Cross2 Cross3 1. 6 0 20 2 6 0 20 3. 48 60 20 4. 6 0 20 5. 6 0 20 6. 48 60 20 For each cross show a map indicating what you can deduce about the location of the segregating gene. Cross 1 (6 points) Blk1 10 cM from centromere blk2 Cross 2 (6 points) Blk2 100% linked to centromere blk3 Blk3 unlinked to Cross 3 (6 points) centromere blk13. (22) You cross singed X snX sn female fruit flies by rasberry males (X r Y). Rasberry and singed are on the X chromosome. This cross gave rise to F1 progeny that are all wild type. You notice that the F1 females have occasional spots on their bodies that reveal recessive markers . Some flies have twin spots where one tissue is singed and the other is rasberry. Others have only a single singed spot. You never find flies that have only a rasberry spot. a. (6) Draw the event that gave rise to the twin spot b. (6) Draw the event that gave rise to singed c. (10) Based on these data draw the order of these genes with respect to the centromere singed singed rasberry + sn r + + sn r + + sn + + r sn r + + sn r sn + r + + singed Wild type r + r + + sn r + + sn r + + sn + sn + sn + sn r r + + singed rasberry r sn cen4. (30) Please answer the following questions in the context of the pedigree provided. Assume that individuals that are affected with a dominant trait are heterozygous for the trait. We do not know if individuals that are marked with “?” are affected or not. a) (8 points) For each mode of inheritance, if the mode of inheritance is not possible given the pedigree, write “Not Possible” next to the mode of inheritance. Otherwise, provide the probabilities for Events A and B as defined. Event A – Individual 6 is a carrier given it is not affected Event B – Individual 7 is a carrier given it is not affected P(A) P(B) Autosomal Dominant Not Possible Not Possible Autosomal Recessive 2/3 1 X-Linked Recessive ½ 0 X-Linked Dominant Not Possible Not Possible b) (4 points) Give the probability Individual 9 is affected assuming an autosomal recessive trait Event C – 9 is affected assuming an autosomal recessive trait Event D – 6 is a carrier assuming an autosomal recessive trait Event E – 7 is a carrier assuming an autosomal recessive trait P(C) = P(D) x P(E) x ¼ = 2/3 x 1 x ¼ = 1/6c) (4 points) Assuming an autosomal recessive trait, what is the probability that Individual 6 is a carrier given that Individual 9 is not affected? P(D | not C) = P(not C | D) P(D) / P(not C) P(D | not C) = ¾ x 2/3 / (1 – P(C)) = ¾ x 2/3 / 5/6 = 3/5 d) (7 points) Assuming an autosomal recessive trait, what is the probability that Individual 10 is affected given that Individual 9 is not affected? Event F – 10 is affected assuming an autosomal recessive trait P(F | not C) =P(F | DE) P(D | not C)P(E | not C) = ¼ x 3/5 x 1 = 3/20 e) (7 points) Assuming an X-linked recessive trait, what is the probability that Individual 10 is a carrier given that Individual 9 is not affected? Event G – 6 is a carrier assuming an X-linked recessive trait Event H – 9 is not affected assuming an X-linked recessive trait Event J – 10 is a carrier assuming an X-linked recessive trait P(G) = ½ P(H) = 1 – (P(G) x ½) = 3/4 P(G | H) = P(H | G) P(G) / P(H) = ½ x ½ / ¾ = 1/3 P(J | H) = P(G | H) x ½ = 1/61 Question 1____pts Question 2____pts Question 3____pts 7.03 EXAM TWO FALL 2008 Name: TA (Circle One Section): David Gifford: M1 Amudha Panneerselvam: M2 Jason Funt: M3 M4 Wendy Niedelman T10 T11 Sarah Pfau T12 T1 Yun Song: T2 M2 There are 3 questions, 8 pages including this cover page Please write your name on each page. Please… • Look over the entire exam so you don’t spend too much time on hard parts of questions leaving easy parts of questions unanswered • Check your answers to make sure that they make sense • To help us give partial credit,
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