Exams Fall 1993 Name Recitation day time 7 03 Hour Exam 1 October 1 1993 Write your name on all five pages Indicate your recitationsection on this page Write all answers on this handout only Exam begins at 11 05 and ends at 11 55 Time will be announcedwhen 5 and 1 minutes remain Problem 1 Problem 2 Problem3 Total 30 points 30 points 40 ooints 100 points Name 1 In the following humanpedigree individuals exhibitinga common inherited allergy to milk are shown by solid figures and unaffected individuals are shown by open figures I O female 1 r male 2 I III 1 6 IV 1 2 3 4 5 6 7 7 a 5pts milk allergy Assumingcompletepenetrance what isthe probablemode of inheritanceof the b 10pts Give the genotypesof the followingindividualsusing to indicatethe normal allele and m to indicatethe allelespecifyingthe milkallergy In ambiguouscases indicateall possiblegenotypes 1 1 111 2 11 2 111 5 11 3 15pts If cousinsIV 4 and IV 5 havea child together what is the probabilitythat the childwill have the milkallergy Name 2 Consider two different antigen molecules produced on the blood cells of wild type mice according to the biosynthetic pathway below enzyme B A antigen I enzyme v intermediate enzyme lk antigen2 Mice homozygousfor recessive alleles that block the production of enzyme A genotype a a do not make either antigen 1 or antigen 2 Mice homozygousfor recessive defects in enzyme B genotype b b do not make antigen 1 Mice homozygousfor recessive defects in enzyme C genotype c c do not make antigen 2 a 8pts Two differenttrue breedingstrainsof mice have been isolated that do not make either antigen 1 or antigen 2 When an individualfrom one strain is crossedwith an individualfrom the other strain all of the F1 mice produce both antigens Write out the genotypesfor both strains Use A B and C to designatethe wild type and a b and c to designate the defective alleles of the three enzymes b 12pts Two of the F1 mice are crossedto one another The possible phenotypes for the F2 progeny are shown below What proportionof the F2 will be represented by each phenotype on average antigen 1 antigen 2 antigen 1 antigen 2 antigen 1 antigen 2 antigen 1 antigen 2 Name 2 10pts Among the F2 there will be mice of several different genotypes that are phenotypically antigen 1 and antigen 2 Suppose you wanted to test whether a given F2 mouse that does not express either antigen is defective in production of enzyme A What genotype would you choose for a mouse used for such a test cross of the F2 mouse Describe the possible outcomes of this cross and how you would interpret them 3 a 5pts Suppose that in an isolated population there exists a very rare inherited anemia which is autosomal recessive Given the frequency of the allele for the anemia as q calculate the probability that a child will be born with the anemia assuming random mating Express your answer as a function of q b 5pts What is the probability as a function of q that a given individual in the population is a heterozygote Use the approximation that is valid for small q Name 10pts In this population marriages between a niece and her biological uncle occur quite often Given the niece in such a marriage is heterozygous for the allele for the anemia what is the probability that her child will have the anemia This is the joint probability that her husband and uncle is also heterozygous and that the child of two heterozygotes is homozygous d 10pts Given that uncle niece marriages occur at a frequency of 0 008 use the answers derived above to calculate the frequency within the population with which children with the anemia are produced by uncle niece marriages Express your answer as a function of q e 1Opts If half of the children with the anemia come from uncle niece marriages and half come from marriages with no obviousinbreeding what is q If helpful you may use the approximationthat the frequency of random marriagesis about one EXAM 1 SOLUTIONS a The mode of inheritance must be autosomal possibility recessive X linked recessive is not a since male IV 5 is not affected b I1 1 m 111 2 m 11 2 m m 111 5 m m 11 3 m or c The probabilitythat the childhas the milkallergyis the joint probabilitythat both parents are heterozygous multipliedby the probabilitythat the childwill receivetwo copiesof the mutant gene one from each parent p mother heterozygous 2 3 mother is either or m p father heterozygous l p childwill inherit both mutant copies 1 4 total probability 2 3 1 1 4 1 6 Problem 2 a One strain must be of genotype while the other must be AAbbc This is the only way the F1 could produceboth antigens AaBbCc b antiaen 1 antigen 2 must have the genotype A B C The proportionwill be 3 4 3 4 3 4 27 64 from the crossAaBbCcxAaBbCc antiaen 1 antigen 2 must have the genotype A B cc The proportionwill be 3 4 3 4 1 4 9 64 antigen 1 antigen 2 must have the genotype A bbC The proportionwill be 3 4 1 4 3 4 64 antiaen 1 antioen 2 will comprisethe rest of the population or 1 27 64 9 64 9 64 19 64 c The possiblegenotypes that would give antigen 1 antigen 2 phenotypes are A bbcc aaB CaabbCaaB cc aabl0cc In order to distinguishthe genotypes that are deficient for enzyme A the proper testcross would be to an aaBBCCmouse Any mouse that isaa will produceall antigen 1 antigen Z offspring when crossedto the above testcross mouse Any mouse that has at least one good copy of enzyme A will produceantigen 1 antigen2 offspring Manystudents choseAAbbcc as the genotype for their testcross This will not allow you to distinguishthe aabbcc mice from A bbcc mice Usingthis testcross one would concludethe aabbccmousewasnot deficientin enzymeA sinceit producesallantigen1 antigen2offspdnglOtherstudentschosea mousethat washeterozygousforA Aa fortheir testcross Thiswouldnot allowyouto drawa distinctconclusion betweenaa andAa mice In theory if the mouseto be testedwasaa 1 2 of the progenyfromthe testcrosswouldbe antigen1 antigen2 If the mouseto be testedwasAa 1 4 of the progenywouldbe antigen1 antigen2 Given that micedo not havelargelitters it maybeverydifficultto distinguish betweenthesetwo possibilities Finally somestudentsoutlineda seriesof crossesfortheiranswer The hallmarkof a testcrossisthat it is onlyonecrossthat willallowyouto determinethe genotype in questionl Problem a Random mating means the population is in Hardy Weinberg Thus the frequency of children born with the disease is simply o 2 equilibrium b With a population in H W equilibrium the frequency of heterozygotes is simply 2pq Since q is small p is about 1
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