A disease is caused by a dominant gene,but it | vedclass

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(C) Let the dominant gene causing the disease be $A$ and the normal recessive allele be $a$.The man is heterozygous $(Aa)$ and the woman is homozygous

Vedclass · Accepted Answer

$1/10$

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(C) Let the dominant gene causing the disease be $A$ and the normal recessive allele be $a$.The man is heterozygous $(Aa)$ and the woman is homozygous normal $(aa)$.The cross is $Aa 	imes aa$.The offspring genotypes are $50\% Aa$ and $50\% aa$.Only the individuals with the genotype $Aa$ carry the dominant gene for the disease.The probability of a child having the genotype $Aa$ is $1/2$.Since the penetrance of the gene is $20\%$ (or $1/5$),only $20\%$ of the individuals with the genotype $Aa$ will express the disease.Therefore,the proportion of children expected to have the disease is $1/2 	imes 1/5 = 1/10$.

$A$ disease is caused by a dominant gene,but it shows only $20\%$ penetrance. $A$ heterozygous man marries a homozygous normal woman. What proportion of the children in the population is expected to have the disease?

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