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A male human is heterozygous for autosomal genes $\mathrm{A}$ and $\mathrm{B}$ and is also hemizygous for hemophilic gene $h$. What proportion of his sperms will be $a b h$ ?
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The correct answer is:
$\frac{1}{8}$
Genotype of male $=\mathrm{AaBbX} \mathrm{X}^{\mathrm{h}} \mathrm{Y}$
Number of gametes $=2^n$
Where $\mathrm{n}=$ no. of heterozygous alleles (genes A, B, sex chromosome)
Thus, Number of gametes $=(2)^3=8$
The gametes produced wil be $=\mathrm{abX}^{\mathrm{h}}$, $a b Y, \mathrm{ABX}^{\mathrm{h}}, \mathrm{aBY}, \mathrm{aBX}{ }^{\mathrm{h}}, \mathrm{AbY}, \mathrm{ABY}$ and $\mathrm{AbX}^{\mathrm{h}}$.
Hence, number of sperm carrying $\mathrm{abh}=1 / 8$.
Number of gametes $=2^n$
Where $\mathrm{n}=$ no. of heterozygous alleles (genes A, B, sex chromosome)
Thus, Number of gametes $=(2)^3=8$
The gametes produced wil be $=\mathrm{abX}^{\mathrm{h}}$, $a b Y, \mathrm{ABX}^{\mathrm{h}}, \mathrm{aBY}, \mathrm{aBX}{ }^{\mathrm{h}}, \mathrm{AbY}, \mathrm{ABY}$ and $\mathrm{AbX}^{\mathrm{h}}$.
Hence, number of sperm carrying $\mathrm{abh}=1 / 8$.
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