Answer:
3/16
Explanation:
Both parents are both of IA IB Hh genotype (AB bloodtype). They were able to express the A and B antigens due to the presence of Hh genotype. This genotype always determines if the A and B antigens will be expressed on the surface of the RBCs, the hh shows lack of the antigen resulting in the OO phenotype.
To check for the probability whether their first child will express the B bloodtype, we will just do a crossing between the Hh genotypes of both parent. Thus we have:
Hh x Hh
HH Hh hh
3/4 1/4
HH and Hh produces the A or the B or AB bloodline while he produces O bloodline (lack of the A and B antigens)
Thus the probability is 3/4 of producer a child with B bloodline
Then a cross between the two heterozygous parent (AB)
AB x AB
AA AB BB
¼ ½ ¼
Prob. of B bloodtype is ¼
Probability of producing B bloodtype with expression of B antigen with Hh is thus:
3/4 x 1/4 = 3/16