In a Drosophila cross,you mate a parental line displaying crossed eyes (recessive trait,autosomal) and corkscrew bristles (recessive trait,autosomal) by a wild-type fly.The sites of the two genes are known to be unlinked.The F₁ flies are wild type.

You self the F₁ flies in the hope of regenerating the cross-eyed/corkscrew phenotype.

How many individual F₂ flies would you need to analyze to have 95% confidence of finding at least one cross-eyed/corkscrew fly?

Question

Quizplus · Accepted Answer

To determine the number of flies needed to analyze with a 95% confidence level, we can use the formula:
n = log(1-0.95)/log(1-(1/total number of possible outcomes))
Total number of possible outcomes = 2 x 2 = 4 (since there are 2 possible outcomes for each gene)
So, n = log(1-0.95)/log(1-(1/4)) = log(0.05)/log(0.75) = 46.8
We round up to 47 flies to ensure a 95% confidence level of finding at least one cross-eyed/corkscrew fly. Therefore, the best choice is D) 47 flies.

In a Drosophila Cross,you Mate a Parental Line Displaying Crossed