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(B) The genotype of the male is $AaBbX^hY$. Since the genes $A$ and $B$ are autosomal and heterozygous,and the hemophilia gene $h$ is $X$-linked,we de

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$1/8$

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(B) The genotype of the male is $AaBbX^hY$. Since the genes $A$ and $B$ are autosomal and heterozygous,and the hemophilia gene $h$ is $X$-linked,we determine the gamete formation using the principle of independent assortment. The possible gametes for the autosomal genes $AaBb$ are $AB, Ab, aB,$ and $ab$. Each occurs with a probability of $1/4$. The sex chromosomes are $X^h$ and $Y$. Each occurs with a probability of $1/2$. To get the combination $abh$,the sperm must carry the autosomal combination $ab$ and the $X^h$ chromosome. The probability of $ab$ is $1/4$. The probability of $X^h$ is $1/2$. Therefore,the probability of the sperm being $abh$ is $(1/4) 	imes (1/2) = 1/8$.

$A$ human male is heterozygous for autosomal genes $A$ and $B$ and is also hemophilic (gene $h$). What proportion of his sperm will contain the $abh$ gene combination?

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