A man and woman have a son who is color blind, a recessive sex-linked trait carried on the X-chromosome, but neither parent is color blind. The woman is now pregnant with their second child. Which of the following statements is true about the second child?

Question

geometry_hater · Answer

They must've been carriers or something so thats why the first child was color blind

So that would probably make the second child color blind as well

anonymous · Answer

I don't know your answering options, but...

if it's a son, he will be color blind with a chance of 50%. if it's a girl, it won't.

geometry_hater · Answer

the mother is a carrier of the recessive allele so its a small chance of the child being color blind

blues · Answer

The Dad has only one X chromosome, so if he carried the allele for color blindness he would be color blind. We know he isn't; therefore, we know he's wild type.

Mom has two X chromosomes, either or both of which could have the allele for color blindness. If both her X chromosomes had the allele, she'd be color blind but we're told she isn't; therefore, she has one wild type and one (+) allele.

The probability that Dad gives his X to the offspring is (1/2) and the probability that Mom also gives her wild type X to the offspring is also (1/2), therefore, there is a (1/2)(1/2) = (1/4) chance that the child will be a non-carrying daughter.

The probability that Dad gives his X to the offspring is (1/2) and the probability that Mom gives her mutant X to the offspring is (1/2), so there is a (1/2)(1/2) = (1/4) chance that the offspring is a carrying but not expressing daughter.

The probability that Dad gives his Y to the offspring is (1/2) and the probability that Mom gives her mutant X to the offspring is (1/2), so the probability that the offspring will be a color blind son is (1/2)(1/2) = (1/4).

The probability that Dad gives his Y to the offspring is (1/2) and the probability that Mom gives her wild type X to the offspring is (1/2), so the probability that the offspring will be a non-colorblind son is (1/4).

I don't know what your options were, but the probability that the child will be color blind son is (1/4). The probability that it will be a carrying but not color blind daughter is (1/4). The probability that it will not have the color blind allele at all is (1/2).

blues · Answer

The earlier answers were very unclear.

anonymous · Answer

the woman is the carrier the son takes the x chromose from her

anonymous · Answer

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