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One female human has blood type AB+. A male human has blood type O-. One of...

One female human has blood type AB+. A male human has blood type O-. One of their children is blood type A-. The female human above remarries and has children with a male that has blood type B+. Assuming that the male human is heterozygous at both relevant loci, what is the probability of having a child that has blood type B-. Express as a percentage.

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Answer #1

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Based on the information, the genotype of female would be IAIB; Rh+Rh-

IAIB; RH+Rh- (AB+) (female) x (male) IBIO; Rh+Rh (B+) ---Parents

IAIB x IBIO = IAIB (1/4), IAIO (1/4), IBIB (1/4) & IBIO (1/4)

It means that B type (IBIB + IBIO) = ¼ + ¼ = ½

Rh+Rh- x Rh+ Rh- = Rh+ _ (3/4) & Rh-Rh- (1/4)

The probability of having a child that has blood type B- = 1/2 * ¼ = 1/8 = 12.5%

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