ExamsFall 1994Recitation day , time , TA7.03 Hour Exam 1October5, 1994Write your name on all six pages.Indicate your recitation section on this pageWrite all answers on this handout onlyExam begins at 11:05 and ends at 11:55Time will be announced when 5 and 1 minutes remain.Problem 1 30 pointsProblem 2 30 pointsProblem3 40 oointsTotal 100 pointsName:1 The traits in the pedigrees below are comoletely oenetrant.7"1 male 0 female z= I affected male 0 affected female(a 5 pts) The following pedigree shows inheritance of an X-linked recessive trait.X"¥9QWhat isthe probability that a son from the first-cousinmarriage shown will exhibit the trait?(b 10 pts) The following pedigree shows inheritance of the short-eared trait in mice.4f_.C_SS;V¢,, i_i_zr_f_aut "Zi) What is the probabilitythat mouse 1 is a heterozygote?+/+ii) What isthe probabilitythat mouse2 (the progenyfrom the brother-sistermating) willexhibit the short-eared phenotype?i IName:(c 15 #ts) Huntington's disease is caused by an autosomal dominant allele. This pedigreeshows inheritance of Huntington's disease in a family. Also present in this family is aparticular blood antigen which is a recessive trait. The Huntington's disease locus and thelocus fot the blood antigen are on the same chromosome and are 5 cM apart._ affected with Huntington's disease exhibits recessive blood antigenn(_EgP__¢of the individualsshownarebothaffectedwithHuntington'sandexhibitthe bloodantigen)4- & 14 +4. 0, "1- 4-or"14H 4. 4. _.+ o. 4. &What is the probability that the indicated grandchild will both exhibit the recessive bloodantigen and develop Huntington's disease? Please show your work.Lo.o ] ('/,): z.s7,Name:. Wild-type humbugs have brown bodies and brown eyes. You have isolated mutationsin three new humbug genes. The mutation sp is dominant and gives humbugs with spottedbodies. The mutation gr is recessive and gives humbugs with green bodies. The mutationbl is recessive and gives humbugs with black eyes.You cross two true-breeding mutant strains to produce F1 females heterozygous for sp, gr,and bl. These F1 females are then test-crossed to true-breeding black-eyed, green-bodiedmales. The phenotypes of 3000 progeny are scored as shown below:sp bl gr 53 L_>4. + 5r r_ sp + _r+ + + 61sp bi + 1390 _ p_r,,_{,l_+ + gr 1347 or,:l_r ;_ :+ bl + 4 _ c{o_l_l_ b_ _p ,_rsp + gr 2 ur_cv_r_sp + + 70+ bl gr 74(a 5 pts) What are the genotypesof the two true-breedingparentsof the F1 females?sip bi + _r 4- + _c(b 15 pts) Draw a map showing the order and distances between the sp, gr, and bl genes.i_l _p _rI I5,0 _.o(c 10 pts) A dominant mutation eyeless (ey) is identified. You want to map ey relative to blbut your lab partnerclaims this can1 be done sinceyou obviouslycan't score the presence ofblack or browneyes in an eyeless bug. You don'tagree and you cross a true-breedingeyeless bugto a true-breeding black-eyed bug. An F1 female is crossedto a true-breedingblack-eyed male. The followingphenotypesare observed in 100 progeny:eyeless 51 _ ,,,Io/st _r_ r_c_b;._fblack-eyed 39 '_ p_r_,_t,I ~ Ltl/_I _¢ p:,r_.,'l'_.lbrown-eyed 10 -_ rt_c_b;_.hf._What is the map distance between bl and ey?z( o)x leo : ZO _51 + .3_I _" lOName:, You have isolated two different yeast strains defective in methionine synthesis(met-).These strains are unable to grow on medium that does not contain methionine. When onestrain is"mated to the other the cliploids can grow on medium that does not containmethionine. From this you conclude that the mutations arein different genes which you callmet1 and met2. You sporulate the diploids and score 100 tetrads of three types.numberoftetrads4met- 75 PD3 met':1met+ 20 -r2 met-:2 met+ 5 _rj(a 10 pt8) Calculate the distance between met1 and met2.T -, r,,,'vFO zc -,3cdl._.= xlO0 = - -- Xto0 = Z5 cN_2-(fot_l ) Z _ ICC'(b 5 pie) YOU are interested in determining the genotypes of the individual spores from oneof the tetrads which has four met" spores. The genotype of each spore clone is tested bymating to each of four test strains that are of either mating type a or mating type a and areeither met1" or met2-. The results of these mating tests are shown below where + indicatesthat the mating produced diploids that can grow without methionine in the medium. Based onthe behavior of spores !-3, fill in the expected results for spore 4. Remember that in a tetradthere will be two spores of mating type a and two spores of mating type c_. Also rememberthat cells of the same mating type can't mate.o_met1- a met1- o:met2- _ met2-spore 1 4- "-- E --spore 2 --" m 4- mspore 3 "-- "- "-- 4-spore 4 _ 4- -- --(c 5 pts) A met"sporefrom one of the 2 met-:2 met+ tetrad types is mated to a wild typestrain. When the diploidissporulated,whichof the followingclasses doyou expect to occurleastfrequentlyand why?/ZzLt4 met- 3 met':1 met+ 2 met:2 met+i'fp_ 1 -- N?DName: _-k/<__ ;_& K_(Recitationday , time , TA7.03 Exam 2October 28, 1994Write your name on all seven pages.Indicate your recitation section on this pageWrite all answers on this handout onlyExam begins at 11:05 and ends at 11:55Time will be announced when 5 and 1 minutes remain.Problem 1 30 pointsProblem 2 30 pointsProblem3 40 pointsTotal 1O0pointsName:1 Bacteriophage 13makes plaques on bacterial host strains X14 and G13. Fivemutant strains a, b, ¢, and d of 13are able to make plaques on host X14 but not onG13. All five mutations are recessive and fail to complement one another andtherefore are in the same gene designated HI.In order to map the mutations with respect to one another, you coinfect X14 bacteriawith pairwise combinations of the mutants and titer the resulting phage on both X14and G13 hosts. The total number of plaques formed on X14 for each cross is~1x10 4,and the number of plaques formed on G13 for each cross is given in the table below:a b c d o,-b : z,,sO/Lo,_ = I cM_-C : z.,_l"tS/io4 -_ 3.5 cAAa 50 175 125o,-_ : 2Xl25/104 = 2.S _Mb 125 75c 50 b-_: }._IS/lo_ : 1.5,Mc-J: =,So/io_._ l_t_d(a 10 pts) Draw a map showing the order of the a, b, ¢, and d mutations and thedistances between them in map units.I I I to. b _ cThe protein product of the wild-type H1 gene produced when phage are grown oneither X14 or G13 has a molecular weight of 55,000 Da. When phage with the bmutation are grown on X14 the H1 protein produced is 34,000 Da. When phage w,ththe d mutation are grown on X14 the H1 protein produced is 45,000 Da. The mutantstrains b and d can make plaques on a derivative of G13 that contains the
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