1 answer

A geneticist has found 2 independent true breeding strains of mice that are completely hairless. Both...

Question:

A geneticist has found 2 independent true breeding strains of mice that are completely hairless. Both mutations are recessive to wild-type. When mice of the 2 diff hairless strains are crossed to each other, all the mice have hair, ie, they have the wild type phenotype. Based on this evidence, what is the minimum number of genes that can mtate to give hairless phentoype? when wildtype F1 mice of the cross bt the 2 hairless strains are crossed to generate an F2gen, what fraction of the offspring would you expect to be hairless? The answer is 2 and 7/16 but need to know HOW to get the answers. Thanks!


Answers

Suppose one mice with naked strain having allele n and normal allele N

Another mice hair less having h for hairless H normal allele.Cross between will normal hair bearing phenotype.

nnH . X NN hh

F . NnHh which has normal wild type phenotype I.e having hair.

.

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