Name: ___________KEY________________17.03 Final Exam -- 2005 KEYName: ____________KEY_______________The exam starts at 9 am and ends at 12 pm.There are 18 pages including this cover page.Please write your name on each page.Only writing on the front of every page will be graded.Question 1 24 pts________Question 2 26 pts________Question 3 20 pts________Question 4 24 pts________Question 5 24 pts________Question 6 22 pts________Question 7 34 pts________Question 8 26 pts________TOTAL out of 200_______Name: ___________KEY________________21. (24 pts) You are studying three autosomal mutations in flies. Each of these threemutations lies in a different gene. All three genes lie on the same autosome. The wn1–mutation is recessive and causes the phenotype of short wings (wild-type flies havelong wings). The wn2– mutation is recessive and also causes the phenotype of shortwings. The ey– mutation is dominant and causes the phenotype of small eyes (wild-type flies have big eyes). You cross true-breeding wn1– wn2– short-winged femalesto true-breeding ey– males to obtain an F1 generation. You then cross female F1 fliesto true-breeding wn1– wn2– big-eyed males. You analyze the resulting progeny, andfind that there are flies in the progeny from all four phenotypic classes: Short wings Small eyesLong wings Small eyesShort wings Large eyesLong wings Large eyesFor parts (a) – (c), write out complete genotype(s) and phenotype(s) of the flies we askfor. By complete genotype, we mean the genotype at all loci discussed in the problem.By complete phenotype, we mean the phenotype at all traits discussed in the problem.If there are multiple answers, write ALL POSSIBLE answers.Use “+” to indicate wild-type alleles.(a, 6pts) Write out complete genotype(s) and phenotype(s) of both parents.Phenotype GenotypeP generation motherShort wings, big eyeswn1– wn2– ey+wn1– wn2– ey+P generation fatherLong wings, small eyeswn1+ wn2+ ey–wn1+ wn2+ ey–(b, 6pts) Write out complete genotype(s) and phenotype(s) of F1 flies.Phenotype GenotypeF1 generation(mother of F2)Long wings, small eyeswn1+ wn2+ ey–wn1– wn2– ey+Father to whom youcross the F1 motherShort wings, big eyeswn1– wn2– ey+wn1– wn2– ey+Name: ___________KEY________________3(c, 8pts) Write out complete genotype(s) of the different F2 flies.Phenotype GenotypeShort wings Small eyeswn1– wn2– ey– OR wn1– wn2+ ey– OR wn1+ wn2– ey–wn1– wn2– ey+ wn1– wn2– ey+ wn1– wn2– ey+Long wings Small eyeswn1+ wn2+ ey–wn1– wn2– ey+Short wings Large eyeswn1– wn2– ey+ OR wn1– wn2+ ey+ OR wn1+ wn2– ey+wn1– wn2– ey+ wn1– wn2– ey+ wn1– wn2– ey+Long wings Large eyeswn1+ wn2+ ey+wn1– wn2– ey+(d, 4pts) Remember from class that, oddly enough, male flies do not undergorecombination during meiosis. You cross true-breeding wn1– wn2– short-wingedfemales to true-breeding ey– males to obtain an F1 generation. You then cross maleF1 flies to true-breeding wn1– wn2– big-eyed females. If you analyze 2000 resultingprogeny, predict the number of the following kinds of flies that you will get:Phenotype Number of fliesShort wings Small eyes 0Long wings Small eyes 1000Short wings Large eyes 1000Long wings Large eyes 0You cross an F1 male wn1+ wn2+ ey– to a female wn1– wn2– ey+ wn1– wn2– ey+ wn1– wn2– ey+There are no recombinants, so the male either gives the top chromosome or the bottomchromosome, and the female always gives wn1– wn2– ey+. Thus all F2 flies are eitherwn1+ wn2+ ey– OR wn1– wn2– ey+wn1– wn2– ey+ wn1– wn2– ey+Name: ___________KEY________________42. (26 pts) You are studying a new species of primate that is diploid, and has fourpairs of autosomes. You have found a rare autosomal recessive disease that is lethal inold age, and is prevalent in a primate family living in the wild. The mother (Individual 2)has already died from this disease. You want to find the genetic locus responsible forthis disease, and decide to use SSR mapping to do so. Your first step is to determinewhich chromosome the locus responsible for the disease is located on. You haveaccess to blood samples of all living members of the family, and you use these bloodsamples to genotype each living member of the family at four SSRs: SSR12, on chromosome 1 SSR13, on chromosome 2 SSR14, on chromosome 3 SSR17, on chromosome 4The pedigree of the primate family, and the SSRs possessed by each family member,are shown in the chart below. Assume complete penetrance and no new mutations.ABAAAAABBBABAABBAABBABABABBCACBBBBACACBBACBCBBAAABBBACABBCBCABABABBCBCABBCB?ABABABABABABBBBBABBBBBABB?BCIndividual 2Individual 1SSR12131417Name: ___________KEY________________5(a, 4pts) Fill in the empty column of the chart, which indicates the deceased mother’sgenotypes at each of the four SSRs. Indicate any ambiguous alleles with a question-mark (?).(b, 4pts) If you want to determine the LOD score for this family for the locusresponsible for the disease and one SSR, which parent(s) would be relevant (IndividualOne, Individual Two, or both)?Individual One, because he is heterozygous at all loci involved.(c, 4pts) Do you know the phase of the parent(s) you listed in part (b)?No, because you do not know his parents.For parts (d) and (e), calculate the LOD score at θ = 0.1 for this family for the locusresponsible for the disease and each of the following SSRs. For each LOD score,clearly write the expression you used to calculate the LOD score.(d, 7pts) the locus responsible for the disease and SSR12.log (1/2) (0.45)3 (0.05)4 + (1/2) (0.05)3 (0.45)4 = -1.33(0.25)7Since you do not know the genotypes of the parents of Individual One, there are twopossible phases for Individual One. We will call + the allele that does not confer the diease,and – the allele that does confer the disease.Phase One Phase Two–B from father = parental –B from father = recombinant+A from father = parental +A from father = recombinant–A from father = recombinant –A from father = parental+B from father = recombinant +B from father = parentalThe first child received +AThe second child received –AThe third child received – and ???The fourth child received +BThe fifth child received + and
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