A pure-breeding fruit fly with the recessive mutation cut wing, caused by the homozygous cc genotype, is crossed to a pure-breeding fly with normal wings, genotype CC. Their F1 progeny all have normal wings. F1 flies are crossed, and the F2 progeny have a 3:1 ratio of normal wing to cut wing. One male F2 fly with normal wings is selected at random and mated to an F2 female with normal wings. Using all possible genotypes of the F2 flies selected for this cross determine the probability of each cross. 1. what is the probability of the cross male CC x female CC?2. what is the probability of the cross male CC x female Cc?3. what is the probability of the cross male Cc x female CC?4. what is the probability of the cross male Cc x female Cc?

Question

Biology

Luke

28 October, 18:32

A pure-breeding fruit fly with the recessive mutation cut wing, caused by the homozygous cc genotype, is crossed to a pure-breeding fly with normal wings, genotype CC. Their F1 progeny all have normal wings. F1 flies are crossed, and the F2 progeny have a 3:1 ratio of normal wing to cut wing. One male F2 fly with normal wings is selected at random and mated to an F2 female with normal wings. Using all possible genotypes of the F2 flies selected for this cross determine the probability of each cross. 1. what is the probability of the cross male CC x female CC?2. what is the probability of the cross male CC x female Cc?3. what is the probability of the cross male Cc x female CC?4. what is the probability of the cross male Cc x female Cc?

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Answers (1)

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Yaretzi Harrison · Answer 1

1. 0% of probability, 2. 50% of probability, 3. 75% of probability.

Explanation:

In order to find the probabilities we have to draw Punnet squares. The mutated gene will be symbolized with a *.

1. Male CC with female CC, there is no mutation, the probability will be 0%.

X Y

X ║ XX ║ XY

X ║ XX ║ XY

2. The male has not got a mutation and the female has. When we draw the Punnet square, with one female X mutated, we find that there is a 50% of probability of mutation.

X Y

X * ║ X*X ║ X*Y

X ║ XX ║ XY

3. Now both, the female and the male, has the mutation. When we draw the Punnet square with mutations in the female and in the male, we find that there is a 75% of probability of mutation.

X * Y

X * ║ X*X * ║ X*Y

X ║ XX * ║ XY