A woman of blood group AB marries a man of blood group A whose own father (the g
ID: 99949 • Letter: A
Question
A woman of blood group AB marries a man of blood group A whose own father (the grandfather) was group O. Using a Punnett square and the laws of probability, first determine the genotypes of both parents then calculate the probability that:
a) one child will have group B
b) one child will have group O
c) the first child will have blood group AB
d) the first child will be son
e) the first child will be a son with blood group AB
f) the second will be a son with blood group B
g) the first child will be a son with blood group AB and the second will also be a son with blood group AB
Explanation / Answer
Woman has AB blood group, that means she is heterozygous for A and B alleles.
Man has A blood group, so he may be homozygous for A or heterozygous for A and O alleles. But his father had O group, which is homozygous for O allele. So the man must be heterozygous for A and O alleles, as he inherits one O allele from his father.
Woman (A and B)
Man
(A and O)
Gametes
A
B
A
AA
AB
O
AO
BO
0.25 will be A blood group with homozygosity for A allele
0.25 will be AB blood group
0.25 will be A blood group with heterozygosity for A and O alleles
0.25 will be B blood group with heterozygosity for B and O alleles
Total probability of A blood group is 0.5 (homozygous+heterozygous)
a) one child will have group B = 0.25
b) one child will have group O = 0 (Nil)
As there is no possible homozygosity for O allele
c) the first child will have blood group AB = 0.25
The assortment of alleles is independent, so no question of first and second child.
d) the first child will be son = 0.5
This depends on Y chromosome and always 0.5 for son or daughter
e) the first child will be a son with blood group AB = 0.5 X 0.25 = 0.125
Probability is cumulative and should be multiplied
f) the second will be a son with blood group B = 0.5 X 0.25 = 0.125
No difference for first and second child, as it is independent assortment.
g) the first child will be a son with blood group AB and the second will also be a son with blood group AB
Probability for first child to be son and AB is 0.125 and it is same for second child too. For both to occur combined probability is = 0.125 X 0.125 = 0.015625
Woman (A and B)
Man
(A and O)
Gametes
A
B
A
AA
AB
O
AO
BO
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