Question

The red-green color vision deficiency gene, a recessive trait, is on the X chromosome. Now, A...

The red-green color vision deficiency gene, a recessive trait, is on the X chromosome. Now, A male with normal vision and a female whose father could not see red or green are mated;

What is the probability that the child will not see red or green, if its a boy? what if it's a girl? Explain your answer.

what I think is that as the female's father gave her X chromosome and a female has XX chromosome (one active and one inactive), if the child is a boy he got the active X from mother and Y from father meaning he is at a higher risk of having the defect but if its a girl she will get one X from dad and one X from mom and there are chances that the X chromosome that carries color blindness might be inactive. Am I making it too complicated? Help Please

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Answer #1

Red-green color blindness is X-linked recessive trait.

Let, X = normal X chromosome, Xc = X chromosome containing color blind allele

Now, normal male will have genotype XY & affected men will have genotype XcY. Further, female who doesn't show the trait can have genotype XX (Normal genotype) or XXc (Carrier).

As father of the female show the trait, his genotype will be XcY. Further, he will transfer this defected X chromosome to the female. As a result, female will be carrier, i.e., XXc. Now, as male is normal, his genotype will be XY. The cross between male & female is shown in below Punnett square.

Gametes X Y
X XX (normal female) XY (Normal male)
Xc XXc (Carrier female; normal phenotype) XcY (Affected male)

From the above Punnett square, we find that probability that the child will not see red or green, if it's a boy is 1/2 & probability that the child will not see red or green, if it's a girl is zero.

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