Question

When a woman heterozygous for $A$ blood group and heterozygous non albin marries a man who is heterozygous non albin and heterozygous for $B$ blood group. The probable number of kinds of phenotypes in their progeny are

• A. 4
• B. 9
• C. 8
• D. 6

Answer 1

The correct option is C 8

Let the woman's blood group be represented by $I^A$i and the non-albino condition as Aa. The man's blood group is $I^B$i and non-albino condition is Aa.

The gametes and cross are given below:

Female/Male	A $I^B$	Ai	a$I^B$	ai
A$I^A$	AA$I^A$$I^B$	AA$I^A$i	Aa$I^A$$I^B$	Aa$I^A$i
Ai	AA$I^B$i	AAii	Aa$I^B$i	Aaii
a $I^A$	Aa$I^A$$I^B$	Aa$I^A$i	aa$I^A$$I^B$	aa$I^A$i
ai	Aa$I^B$i	Aaii	aa$I^B$i	aaii

Note: AA and Aa indicate non albin while aa indicates albino condition.

$I^A$$I^B$ indicates blood group AB.

$I^A$ indicates blood group A.

$I^B$ indicates blood group B.

A) We can see from the table above that there are eight phenotypes.

B) We can see from the table above that there are twelve genotypes.
C) The probable number of kinds of phenotypes in their progeny are 8.

D) The number of phenotypes is 8.

So the correct answer is '8'.