Red an a normal in all respects, but her father is color blind and polydactylous

Question

Red an a normal in all respects, but her father is color blind and polydactylous. Bill is color blind and polydactylous. His mother has normal color vision and normal fingers and toes. color blindness is an X-linked recessive trait in humans. Polydactyly (extra fingers and toes) is al dominant trait. Martha has normal fingers and toes and nórmal cólor vision. Her mother is If Bill and Martha marry, what proportions of children with specific phenotypes would they be expected to are also expected to occur but are not included. produce? The answers only include the proportions of some of the possible phenotypes: other phenotypes O O O O 1/8 color-blind girls with normal fingers, 1/4 boys with normal vision and polydactyly 1/8 color-blind girls with polydactyly, 1/8 boys with normal vision and normal fingers 1/4 color-blind girls with normal fingers, 1/4 boys with normal vision and polydactyly 1/4 girls with normal vision and polydactyly. 1/8 boys with normal vision and polydactyly

Explanation / Answer

Correct option is option#2. 1/8 color-blind girls with polydactyly(PD), 1/8 boys with normal vision normal finger.

Explanation:

Martha is definitely a carrier for PD, so the probability of Martha giving a PD gene is (1/2). Bill is PD so the probability of bill giving a PD gene is also (1/2). Now , as Martha is heterozygous for color blind, the probability of giving a color blind (CB) gene will be 1/2. But for the daughter to be CB Bill has only one X chromosome, so the probability of him giving that X chromosome is 1.

Now multiply every probability for total probability that the daughter will be CB as well as PD

= 1/2 x 1/2 x 1/2 x 1 = 1/8

Similarly, for the boy to be normal vision(NV) the probability that he gets normal vision gene from gene is 1 (as Bill can only give Y, for it to be a son) and the probabilty that the son receives a normal gene for the vision from mother is eual to 1/2. Moreover, for the son to havenormal finger (NF) the probability that he gets good autosomal gene from both the parents is 1/2 ans 1/2 respectively.

Now multiply every probability for total probability that the son will be NV as well as NF

= 1 x 1/2 x 1/2 x 1/2 = 1/8

Hence option #2 is correct..

NOTE: Please feel free to comment in the answer below if you couldn't understant the analysis. I will be happy to help. Thank you.

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