7.03 Exam #1 Solutions

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, so the frequency of heterozygotes is 2q. c Since we are told that the niece is a heterozygote we do not use p or q at all in this part of the problem. If the niece is a heterozygote the chance that the uncle's sibling has the disease allele is 1/2. If the uncle's sibling is a heterozygote the chance the uncle has the disease allele is 1/2. If both the uncle and the niece are heterozygous for the disease allele the chance that they will have an affected child is 1/4. Thus p(affected child) = (1/2)(1/2)(1/4) = 1/16. d The chance that any woman is a heterozygote is 2q (see part b). The chance that this woman marries her uncle is .008. The chance that this uncle/niece couple have an affected child is 1/16 (seeabove). Thus the frequency within the population with which children with the anemia are produced by uncle-niece marriages is: (2q)(.008)(1/16) = o./1000. e If half the affected children come from uncle-niece marriages and half come from marriages with no inbreeding we simply set the answer to part d equal to the answer in part a. q2 = q/lO00. Solving for q we get q = 1/1000.