36. You place 1 black-bodied male fruit fly (bb) and 1 homozygous wild-type fema

Question

36. You place 1 black-bodied male fruit fly (bb) and 1 homozygous wild-type female (BB) in to each of 100 vials. This is what the distribution of allele frequencies looks like right now: 6 50 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Frequency of b alelle, generation O You allow the flies to breed. Draw the distribution of allele frequencies in the next generation: n 100 50 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Frequency of b alelle generation 1 You allow flies in each vial to mate randomly for 50 generations, keeping the population size in each vial to 10 flies. The b allele has no effect on fitness. Draw the most likely distribution of allele frequencies after 50 generations: 100 50 0 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Frequency of b alelle generation 50 1

Explanation / Answer

1. The cross between the parent generation would be Bb X Bb

Hence the following genotype will exist for F1 { BB, Bb, bb}

Thus frequency of allele b = 3/6 = 0.5

Thus the graph will be same as shown for Gen 0 .

2. F1 generation has the genotype { BB, Bb, bb}

Now if we allowed to mate them, the F2 generation will have the genotype { BB, Bb, bb}, as no other possibility can exist. Thus the frequency of allele in this will be 0.5 only. Graph same as Gen 0.

B b B BB Bb b Bb bb

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