Individual iv-3 is born as a male affected with condition b but not condition

a. his parents are bred again, and an ultrasound shows that the fetus is another male. the relevant portion of the pedigree is shown below. use this new information to determine the parents' genotypes (indicated by red arrows). then calculate the probabilities that the second male offspring will have each condition. drag one pink label to each pink target and one blue label to each blue target. then use the white labels to answer questions 1 and 2. labels can be used once, more than once, or not at all.

Question

Salma Shepard

Biology

27 February, 03:08

Individual iv-3 is born as a male affected with condition b but not condition

a. his parents are bred again, and an ultrasound shows that the fetus is another male. the relevant portion of the pedigree is shown below. use this new information to determine the parents' genotypes (indicated by red arrows). then calculate the probabilities that the second male offspring will have each condition. drag one pink label to each pink target and one blue label to each blue target. then use the white labels to answer questions 1 and 2. labels can be used once, more than once, or not at all.

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Answers (1)

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Hoffman · Answer 1

The probable genotype of the IV-3 with the perspective of autosomal inheritance will be A because he does not have condition A. Therefore both parents must be in condition A.

The probable genotype of IV-3 with the perspective of set=linked inheritance will be BbY this is because he has B condition. Therefore the mothers' genotype must be XBXb and the father's genotype must be XBY.

The probability of IV-4 having condition A is 1/36