What is the probability of having children with ’O’ blood group, where both mother and father are heterozygous for ’A’ and ’B’ blood group, respectively ?

The correct answer is 25%.

To determine the chance of a child having blood type O, we examine the possible alleles each parent can pass to the child.

Mother is heterozygous for blood group A, so her genotype is $AO$. She can contribute either the $A$ allele or the $O$ allele.

Father is heterozygous for blood group B, so his genotype is $BO$. He can contribute either the $B$ allele or the $O$ allele.

When the alleles combine, the possible genotypes of the child are:

$AO$ (type A) – mother gives $A$, father gives $O$

$BO$ (type B) – mother gives $O$, father gives $B$

$AB$ (type AB) – mother gives $A$, father gives $B$

$OO$ (type O) – mother gives $O$, father gives $O$

These four outcomes are equally likely because each parent has two equally probable alleles to pass on. Therefore, the probability of the $OO$ combination, which results in blood type O, is

$1 out of 4 = \backslash frac\{1\}\{4\} = 0.25 = 25\backslash \%$

Thus, there is a 25% chance that a child of an $AO$ mother and a $BO$ father will have blood group O.