Question

a male human is heterozygous for autosomal genes A and B and is also hemizygous for hemophillic gene h. what proportion of his genes will b abh?
1/8
1/16
1/4
1/32

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,,,,,,

The following genotype is formed in the human male where the autosomal genes are "A" and "B" respectively for Hemizygous genes it's given as a autosomal haemophilic gene that is, "h".

This sets a genotypic character of $\bf{Aa \: Bb \: X^h \: Y}$, since, as per the original question of a sufferer from Haemophilia will contain the Haemophilic gene it'll be a se.x characterised trait linked to the genes that are attached to the X-Chtomosomes.

Therefore, the total number and types of gametes going to be formed are:

$\bf{ 2 \times 2 \times 2 = 8 \: gametes}$

Cross linking them simultaneously will give us the following combinations:

$\bf{AB \: X^h, \: ABY, \: aB \: X^h, \: aBY, \: AbY, \: ab \: X^h \: and \: abY}$

$\bf{ab X^h}$ is the gene found to be the proportional gene pit of those 8 gametes.

So, total $\bf{\frac{1}{8}}$ of his sp.erms are going to be in a combination of "abh".

HOPE THIS DETAILED ANSWER HELPS YOU AND CLEARS THE DOUBTS FOR GENOTYPIC INHERITANCE!!!!!!!!