A woman who is a heterozygous carrier of an X-linked recessive disease gene mates with a...

A woman who is a heterozygous carrier of an X-linked recessive disease gene mates with a phenotypically normal male. They have six sons and four daughters. This disease gene has a penetrance of 80%. How many children will be affected by this disease?

Expert Solution

In the case of X linked recessive only when both alleles is present in the recessive condition it shows the disease in both males and females. It is more prevalent in males than females because for males only one copy of a gene is required to cause the disease so the probability is more There are many X linked recessive diseases like Hemophilia, Colorblindness.

	X	Y
X	XX(Normal female)	XY(Normal male)
Xc	XcX (carrier female )	XcY (affected male

So in this case out of 4 children, only one is affected so the probability of affected child is 25 % But the penetrance is 80% so the probability reduces to 20 % as 25*80 = 20%

in this case, if the parent has 6 sons and 4 daughters, and the probability of affected children is asked not for the only son so the probability of children affected will be probability is 20 % so out of 10 children of only 2 is affected which are 2 sons. no daughter is affected

2 children are affected

gladiator answered 2 years ago

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