Problem SetsFall 1996(c) What is the maximum number of genes required for methionine synthesisrepresented by your twelve mutants?A colleague of yours isolates twelve more Met- mutants in his own lab. He performscomplementation tests among his mutants just as you did. The results of the tests onhis mutants are given in the table below.strains of mating type o_13 14 15 16 17 18 wildtype19 - - - + - - +2O -- -- -- + -- -- +strains of 21 - - - + - - +mating type a22 -- -- -- + -- -- +23 + + + + -- + +24 -- -- -- + -- -- +wildtype + + + + - + +(d) How do you explain the behavior of mutant 17?(e) Which of his mutations do you know to be in the same gene?(f) What is the minimum number of genes required for methionine synthesisrepresented by your colleagues twelve mutants?(g) What is the maximum number of genes required for methionine synthesisrepresented by your colleagues twelve mutants?2. (a) You are given a male and a female mouse that have chocolate brown fur.When you cross these mice all of the F1 progeny have chocolate brown fur. Similarly,crosses among the F1 mice give F2 mice that all have chocolate brown fur. What canyou say about the alleles governing coat color in the two mice that you were given?(b) Wild mice have gray fur. When you cross one of the chocolate brown mice to awild mouse all of the F1 progeny look like wild mice. What can you say about therelationship between the coat-color alleles in the chocolate brown mice relative to wildmice?(c) You produce a large number of F1 progeny by crossing chocolate brown miceto wild mice. Crosses among these F1 mice give 500 F2 mice. About half of the F2mice look like wild mice, many have coats with intermediate properties, and 5J lookjust like the chocolate brown grandparents. What is your best guess for the number ofcoat-color genes that differ between the chocolate brown mice and wild mice.(d) Evaluate your answer in part c by performing three different Chi-square tests.Forthe first test, use your answer for c to derive the expected number of chocolatebrownF2 mice. For the other tests, use one more and one less than your answer for cto derive the expected number of chocolate brown mice. Give the p values for each ofthe tests.:3. Each of the families below exhibits a different very rare genetic disorder whereindividuals expressing the disorder are shown by solid symbols. Assume completepenetrance and also assume that no new mutations have arisen in these families.Give the most likely mode of inheritance for each pedigree (autosomal recessive, X-linked recessive, or autosomal dominant). Also indicate the predicted genotypes ofeach individual in the pedigree using A for dominant alleles and a for recessivealleles (in ambiguous cases give all possible genotypes). Finally, calculate theprobability that the next child indicated by a (?) will be affected.(a)(b)(c)Problem Set #1 AnswersBiology 7.03I. Tn thisproblem we are assl_ming that each mutant has a single genemutated.(a) Mutants 2, 4 and 10 are in the same gene and mutants 6 and 8 are inthe same gene because they fail to complement which is the h_nrnark of twomutations being in the same gene.(b) The rninirullm number of genes is 4. Mutants 2, 4 and 10 representone gene. Mutants 6 and 8 represent one gene. Mutants 7, 9, 11 and 12 couldbe in the same gene and mutants 1, 3, and 5 could he m the same gene.(c) The mA_mnm mlmber of genes is 9. Mutants 2, 4 and 10 representone gene and mutants 6 and 8 represent one gene by their failure tocomplement. However, mutants 7, 9, 11 and 12 although they have the samecomplementation pattern in the complementation grid may all be in differentgenes as may mutants 1, 3 and 5. To confirm that they were mutations in thesame gene we would need to show that they all fail to complement the samemutation. Only failure to complement is taken as proof of the mutations beingin the same gene, not an identical pattern of complementation on thecomplementation grid.(d) Mutant 17 is a dominant mutation because it does not complementthe wild type str_ in.(e) Mutants 13, 14, 15, 18, 19, 20, 21, 22 and 24 are in the same genebecause they fail to complement each other.(f) The minimum m_mber of genes represented by these 12 mutants is 3.We Know mutants 13, 14, 15, 18, 19, 20, 21, 22 and 24 to be in the same geneand we also have mutants 16 and 23 which are in different genes. Mutant 17 isa dominant mutation so we cannot tell what gene it is in, so it could be amutation in one of those 3 genes.(g) The ma_rm_m number of genes represented by these 12 mutants is4 if mutant 17 where to be in a c]ifferent gene then all the other mutants._- (a) Since a different coat color does not appearinthe progeny, the line is1.iketyto be true breeding, that is homozygous for the chocolate brown furallele.(b) The grey coat color allele (wild type) is dominant to chocolate brown,(i.e., brown is recessive to grey).(c) In the F2 generation we have 5 out of 500 mice that express therecessive trait, that is 1/100. If the number of coat color genes involved is 1,we would expect 1/4 of the F2 generation to be chocolate brown; if there are twogenes, one would expect 1/16; if three genes are involved: 1/64; and if thereare four, 1/256. 1/100 is closest to 1/64; therefore, the best guess for thenumber genes involved is three.(d) Since we do not know the exact number of wild type mice and micewith intermediate properties, assume two phenotypic classes: chocolatebrown and non-brown.Assuming that three genes are involved, the expected number ofchocolate brown mice is 500 x 1/64 = 8. Hence,observed expectedn0n-br0wn 495 492brown 5 8%2= (O - E)2 / E = (495 - 492)2 / 492+ (5 - 8)2 / 8 = 1.14degrees of freedom = # classes -t = 2 - 1 =tUsing the %2table, we get: 0.1<p<0.5. That is, the three genes model is wellwithin the lirr_its.Similarly, for two genes:observed expectednon-brown 495 469brown 5 31--500 x 1/ 16%2= (O - E)2 / E = (495 - 469)2/ 469 + (5 -31)2/31 = 23df=-IThe p value is much lower than 0.005; hence the two gene hypothesis isrejected.Finally, for four genes:observed expectednon-brown 495 498brown 5 2 = 500 x 1/256%2= (O - E)2 / E = (495 - 498)2 / 498 + (5 - 2)2/2 = 4.52df-1" The p value is less than 0.05 (0.025<p<0.05); hence the four gene hypothesis isalsorejected.3. (a) The most likely mode of inheritance is autosomal dominant. Themother has a genotype aa, the father Aa, the male child aa and the femalechild Aa. The probability of the next child being affected is .5, the probabilitythat
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