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MIT 7 03 - Study Guide

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ExamsFall 1993Name:Recitation day , time ,7.03 Hour Exam 1October 1, 1993Write your name on all five pages.Indicate your recitationsection on this pageWrite all answers on this handout onlyExam begins at 11:05 and ends at 11:55Time willbe announcedwhen 5 and 1 minutesremain,Problem 1 30 pointsProblem 2 30 pointsProblem3 40 oointsTotal 100 pointsName"1 In the following humanpedigreeindividualsexhibitinga commoninheritedallergy tomilk are shown by solid figures and unaffected individuals are shown by open figures.IO- 1 2femaler--] = maleIIIII1 6IV1 2 3 4 5 6 7'7(a 5pts.) Assumingcompletepenetrance,whatisthe probablemodeof inheritanceofthemilk allergy?(b 10pts.) Givethe genotypesof the followingindividualsusing + to indicatethe normalalleleand m to indicatethe allelespecifyingthe milkallergy. Inambiguouscases indicateallpossiblegenotypes.!1-1 111-211-2 111-511-3(¢ 15pts.) If cousinsIV-4 and IV-5 havea childtogetherwhatis the probabilitythatthechildwillhavethe milkallergy?Name:2 Consider two different antigen molecules produced on the blood cells of wild typemice according to the biosynthetic pathway below.enzymeB_,_,_ antigen Ienzyme/-A _ intermediatev enzyme"_lk antigen2Mice homozygousfor recessive alleles that blockthe productionof enzyme A (genotype a/a)do not make either antigen 1 orantigen2. Mice homozygousfor recessivedefects in enzymeB (genotypeb/b) do not make antigen1. Mice homozygousfor recessive defects in enzymeC (genotypec/c) do not make antigen2.(a 8pts.) Two differenttrue breedingstrainsof mice have been isolatedthat do not makeeither antigen 1 orantigen 2. When an individualfromone strainis crossedwith anindividualfrom the otherstrainall of the F1 miceproduceboth antigens. Write outthegenotypesfor bothstrains.(UseA, B, and C to designatethe wildtype and a, b, and c todesignate the defective alleles of the three enzymes).(b 12pts.) Two of the F1 mice are crossedto one another. The possible phenotypes forthe F2 progenyare shownbelow. What proportionof the F2 will be representedby eachphenotype on average?antigen 1+, antigen2+ 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 arephenotypically antigen 1- and antigen 2-. Suppose you wanted to test whether a given F2mouse that does not express either antigen is defective in production of enzyme A. Whatgenotype 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 inheritedanemia 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 populationis a heterozygote? Use the approximation that is valid for small q.Name:(¢ 10pts.) In this population marriages between a niece and her biological uncle occurquite 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 herhusband (and uncle) is also heterozygous and that the child of two heterozygotes ishomozygous.(d 10pts.) Given that uncle-niece marriages occurat a frequency of 0.008, use theanswers derived above to calculate the frequency within the population with which childrenwith the anemia are produced by uncle-niece marriages. Express your answer as a functionof q.(e 1Opts.) If half of the children with the anemia come from uncle-niece marriages and halfcome from marriageswith noobviousinbreeding,what is q? If helpful, you may use theapproximationthat the frequencyof randommarriagesis about one.EXAM#1 SOLUTIONSa) The mode of inheritance must be autosomal recessive. X-linked recessive is not apossibility, since male IV,5 is not affected.b) I1,1:m/+ 111,2:m/+11,2:m/m 111,5:m/m11,3:m/+ or+/+c) The probabilitythat the childhas the milkallergyisthe joint probabilitythat both parentsare heterozygous multipliedby the probabilitythat the childwillreceivetwo copiesof themutant gene-onefrom eachparent.p(mother heterozygous)=2/3 (mother iseither +/+ or m/+)p(father heterozygous)=lp(childwill inheritboth mutant copies)=1/4total probability= 2/3(1 )(1/4)=1/6Problem 2:a) One strain must be of genotype _ whilethe other must be AAbbc¢. This isthe onlyway the F1 couldproduceboth antigens(AaBbCc).b) antiaen 1+.antigen 2÷ must have the genotype A-B-C-. The proportionwillbe3/4*3/4*3/4 = 27/64 from the crossAaBbCcxAaBbCc.antiaen 1+.antigen 2- must havethe genotype A-B-cc. The proportionwill be3/4"3/4"1/4 = 9/64antigen 1-,antigen2+ must havethe genotype A-bbC-. The proportionwill be3/4"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 wouldgive antigen 1-,antigen 2- phenotypes are:A-bbccaaB-C-aabbC-aaB-ccaabl0cc In orderto distinguishthe genotypes that are deficient forenzymeA, the propertestcross wouldbe to an aaBBCCmouse. Any mouse that isaa willproduceallantigen 1-,antigen Z- offspringwhen crossedto the abovetestcross mouse. Anymouse that has at least one good copy of enzyme A will produceantigen 1+,antigen2+ offspringManystudents choseAAbbccasthe genotype for their testcross. This willnot allowyou todistinguishthe aabbcc mice from A-bbccmice. Usingthis testcross, one wouldconcludetheaabbccmousewasnot deficientinenzymeA, sinceit producesallantigen1-,antigen2-offspdnglOtherstudentschosea mousethat washeterozygousforA (Aa) fortheirtestcross.Thiswouldnotallowyouto drawa distinctconclusionbetweenaa andAamice. Intheory,ifthemouseto betestedwasaa, 1/2 of theprogenyfromthetestcrosswouldbeantigen1-,antigen2-• If the mouseto betestedwasAa, 1/4 of the progenywouldbeantigen1-, antigen2-. Giventhat micedonothavelargelitters,it maybeverydifficultto distinguishbetweenthesetwopossibilities.Finally,somestudentsoutlineda seriesof crossesfortheiranswer.Thehallmarkof a testcrossisthat it isonlyonecrossthatwillallowyouto determinethe genotypein questionlProblema Random mating means the population is in Hardy-Weinberg equilibrium.Thus the frequency of children born with the disease is simply o,2.b With a population in H-W equilibrium the frequency of heterozygotes issimply 2pq. Since q


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MIT 7 03 - Study Guide

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