Problem Sets Fall 1996 c What is the maximum number of genes required for methionine represented by your twelve mutants synthesis A colleague of yours isolates twelve more Met mutants in his own lab He performs complementation tests among his mutants just as you did The results of the tests on his mutants are given in the table below strains of mating type o strains of mating type a 13 14 15 16 17 18 19 2O 21 22 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 wildtype f What is the minimum number of genes required for methionine represented by your colleagues twelve mutants synthesis g What is the maximum number of genes required for methionine represented by your colleagues twelve mutants synthesis 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 can you 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 a wild mouse all of the F1 progeny look like wild mice What can you say about the relationship between the coat color alleles in the chocolate brown mice relative to wild mice c You produce a large number of F1 progeny by crossing chocolate brown mice to wild mice Crosses among these F1 mice give 500 F2 mice About half of the F2 mice look like wild mice many have coats with intermediate properties and 5J look just like the chocolate brown grandparents What is your best guess for the number of coat 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 For the first test use your answer for c to derive the expected number of chocolate brown F2 mice For the other tests use one more and one less than your answer for c to derive the expected number of chocolate brown mice Give the p values for each of the tests 3 Each of the families below exhibits a different very rare genetic disorder where individuals expressing the disorder are shown by solid symbols Assume complete penetrance and also assume that no new mutations have arisen in these families Give the most likely mode of inheritance for each pedigree autosomal recessive Xlinked recessive or autosomal dominant Also indicate the predicted genotypes of each individual in the pedigree using A for dominant alleles and a for recessive alleles in ambiguous cases give all possible genotypes Finally calculate the probability that the next child indicated by a will be affected a b c Problem Set 1 Answers Biology 7 03 I Tn thisproblem we are assl ming mutated a Mutants has a single gene 2 4 and 10 are in the same gene and mutants the same gene because mutations that each mutant they fail to complement 6 and 8 are in which is the h nrnark of two being in the same gene b The rninirullm number one gene Mutants of genes is 4 Mutants 6 and 8 represent be in the same gene and mutants c The mA mnm one gene and mutants complement complementation one gene Mutants of genes is 9 Mutants 6 and 8 represent mutants pattern genes as may mutants 7 9 11 and 12 could 1 3 and 5 could he m the same gene mlmber However 2 4 and 10 represent 2 4 and 10 represent one gene by their failure to 7 9 11 and 12 although in the complementation they have the same grid may all be in different 1 3 and 5 To confirm that they were mutations same gene we would need to show that they all fail to complement mutation Only failure to complement in the same gene not an identical complementation d Mutant the same is taken as proof of the mutations pattern of complementation in the being on the grid 17 is a dominant mutation because it does not complement the wild type str in e Mutants because 13 14 15 18 19 20 21 22 and 24 are in the same gene they fail to complement f The minimum We Know mutants m mber of genes represented mutation by these 12 mutants is 3 13 14 15 18 19 20 21 22 and 24 to be in the same gene and we also have mutants a dominant each other mutation 16 and 23 which are in different genes Mutant 17 is so we cannot tell what gene it is in so it could be a in one of those 3 genes g The ma rm m 4 if mutant number of genes represented by these 12 mutants 17 where to be in a c ifferent gene then all the other mutants is a Since a different coat color does not appearinthe progeny the line is 1 iketyto be true breeding that is homozygous for the chocolate brown fur allele 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 the recessive 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 two genes one would expect 1 16 if three genes are involved 1 64 and if there are four 1 256 1 100 is closest to 1 64 therefore the best guess for the number genes involved is three d Since we do not know the exact number of wild type mice and mice with intermediate properties assume two phenotypic classes chocolate brown and non brown Assuming that three genes are involved the expected number of chocolate brown mice is 500 x 1 64 8 Hence observed expected n0n br0wn 495 492 brown 5 8 2 O E 2 E 495 492 2 492 5 8 2 8 1 14 degrees of freedom classes t 2 1 t Using the 2 table we get 0 1 p 0 5 That is the three genes model is well within the lirr its Similarly for two genes observed non brown 495 brown 5 expected 469 31 500 x 1 16 2 O E 2 E 495 469 2 469 5 31 2 31 23 df I The p value is much lower than 0 005 hence the two gene hypothesis is rejected Finally for four genes observed non brown 495 brown 5 expected 498 2 500 x 1 256 2 O E 2 E 495 498 2 498 5 2 2 2 4 52 df 1 The p value is less than 0 05 0 025 p 0 05 hence the four gene hypothesis is alsorejected 3 a The most …
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