Question

In: Biology

how and dumpy interact such that flies who are homozygous for the dumpy LOF mutation (dp)...

how and dumpy interact such that flies who are homozygous for the dumpy LOF mutation (dp) and heterozygous for the complete LOF mutation of how (h-) have wild type wings (normal length).

Mutants when dealing with only one gene (wildtype/normal phenotypes are dominant to mutant phenotypes):

dp/dp = truncated (short) wings

h-/h- = dies when trying to exit pupae

Mutants with the two genes interacting (wildtype/normal phenotypes are dominant to mutant phenotypes; so if there is even one dp+ allele they will have wildtype/normal length wings):

dp h+/ dp h+ = truncated wings that can fold down

dp h+ / dp h- = wild type wings (can fold down and are normal length)

dp h- / dp h - = dies when trying to exit pupae

You mate two wild type flies who are both carriers for the dumpy LOF mutation (dp) and the complete LOF how mutation (h-). They both have the same genotype: h- dp / h+ dp+.

dumpy and how are 9 map units apart. If the cross produced 1000 larvae how many, on average, would grow up to have truncated wings? Remember, the complete loss of function mutant dies when coming out of its pupal case, the larval stage is before the pupal stage.

Select one:

a. 2

b. 90

c. 0

d. 4.5

e. 9

Expert Solution

Out of 1000 larvae a) 2 will have truncated wings.

Explanation is given below:

gladiator answered 4 months ago

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