Problem Sets Fall 2000 2 Consider the following mouse breeding experiment involving two different rare traits A male mouse with both traits is crossed to a normal female and all of the offspring appear normal A female offspring from this cross is mated multiple times to a normal male to produce several litters of offspring A total of 32 offspring are scored as having the following characteristics 16 normal females 6 normal males 2 males with trait 1 1 male with trait 2 7 males with both traits a What is the mode of inheritance of each of the two traits Explain your reasoning b Use the chi square test to determine whether the two traits appear to be linked For this test you are trying to determine whether or not an expectation that the two traits are unlinked differs significantly from the observed data Note that there are a number of different ways to set up this test but there is one best way to test for linkage Show your work and use the table below which gives p values as a function of chi square values and degrees of freedom For your final answer use a p value 0 05 as the cut off for significant deviation from expectation p value 995 975 0 9 0 5 0 1 0 05 0 025 0 01 0 005 df 1 000 000 016 46 2 7 3 8 5 0 6 6 7 9 df 2 01 05 21 1 4 4 6 6 0 7 4 9 2 10 6 df 3 07 22 58 2 4 6 3 7 8 9 3 11 3 12 8 of the distance between c Based on the data give your best estimate trait 2 the genes for trait 1 and i 3 The producers of a soap opera have hired you as a consultant The story line includes two families each with individuals that have a rare trait The families are diagramed below individuals are numbered The scriptwriters two families to figure and those expressing are contemplating Because a number they are concerned out what the offspring 1 the trait are represented of different couplings with the genetic from each possible mating between accuracy individuals in the of the story they want you might be like 2 6 i 4 a Assume that the rare trait is recessive by the filled symbols 7 8 Consider the possible matings given below For each calculate the probability that the child will have the rare trait Female 2 and Male 5 Female 6 and Male 4 Female 7 and Male 4 Female 3 and Mate 8 b Now assume that the rare trait is dominant below calculate the probability Again for each of the possible matings given that the child will have the rare trait Female 2 and Male 5 Female 6 and Male 4 Female 7 and Male 4 Female 3 and Male 8 c Finally assume that the rare trait is X linked below calculate the probability For each of the possible matings given that the child will have the rare trait Explicitly give each ability in the cases where the probabilities for a boy or a girl having the trait differ Female 2 and Male 5 Female 6 and Male 1 Female 7 and Male 4 Female 3 and Male 8 prob ProblemSet1 Solutions ir la Those mutants that make tan colonies when crossed to another mutant can be assumed to carry recessive mutations because the white phenotype is not present in a heterozygote Those mutants that never make tan colonies as a diploid likely carry dominant mutations although we can not completely rule out the possibility that these mutants carry mutations in multiple genes discussed in part C Thus mutations 1 2 4 5 6 7 8 9 1O and 12 are recessive Mutations 3 and 11 are most likely dominant lb Based on non complementation of the recessive mutations 4 5 9 and 10 form one complementation gene Likewise 2 and 7 fail to complement complementation complements we can conclude that 1 group Group A and are mutations in the same and are members of a second oup Group B representing mutations in a second gene Mutant 6 members of both Group A and Group B and therefore represents a third gene Mutants 8 and 12 both complement Group A and Group B mutants as well as mutant 6 and therefore represent at least one more gene and possibly two genes We have no data regarding the phenotype of the 8 x 12 diploid and can t determine whether they are mutations in the same or different genes Taken together we can say that the recessive mutations represent at least 4 genes i We are unable to make any conclusions regarding the number of new genes represented by the dominant mutants 3 and 8 It is possible that 3 and 8 are mutations in the same gene It is also possible that one or both of these mutations is in one of the genes that we recovered a recessive mutation in above It is also possible that the white phenotype in one or more of the mutants is caused by mutations in more the one gene Thus the best answer for the number of mutants represented is at least 4 lc The most obvious ambiguity is whether 8 and i2 are mutations in the same gene This could be resolved by generating the 8 x 12 diploid Note that to do this you would have to sporulate for example the 8 x wild type diploid and select for a haploid spore of the appropriate mating type that shows the white phenotype A second ambiguity is whether 3 and 11 represent unique genes One way to approach this question would be to generate the 3 x 11 diploid sporulate and look at the segregation pattern of the white phenotype in the resulting haploids If any of the haploids form tan colonies then 3 and 11 are likely to be mutations in different genes A third ambiguity is whether or not the white phenotype in each case is due to a mutation in a single gene To determine this you would cross each mutant to wild type I sporulate the resulting diploids and look for 2 tan 2 white segregation in the haploids Any other segregation pattern is inconsistent with the white phenotype being caused by mutation iJf a single gene An extreme example of this possibility would be if a single strain carried recessive mutations in genes representing all of the complementation oups identified Such a mutant would appear dominant by the type of crosses performed in the table because it would fail to complement every other mutant To avoid this possibility you would cross each mutant to a wild type haploid and look at the diploid phenotype 2a Both traits are X linked recessive They must be recessive because all F1 offspring appear normal We can deduce that both traits are X linked because of the 32 progeny shown all females are normal and only males are affected 2b To use the Chi Square test we first need …
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