Question

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 ?

• A. 50%
• B. 75%
• C. 0%
• D. 25%

Hint:

Logic Tip: A mating between a heterozygous A and a heterozygous B is the only cross that can produce offspring of {all four} possible ABO blood types! Each type has a perfect 25% probability.

Answer 1

The Correct Option is D

Concept:
The ABO blood group system in humans is determined by a single gene ($I$) with three multiple alleles: $I^A$, $I^B$, and $i$. Alleles $I^A$ and $I^B$ are completely dominant over $i$, and they are co-dominant with each other. Blood group 'O' is the recessive phenotype, which only expresses when the genotype is homozygous recessive ($ii$).
Step 1:

• The mother is heterozygous for blood group 'A'. Therefore, her genotype must be $I^A i$.
• The father is heterozygous for blood group 'B'. Therefore, his genotype must be $I^B i$.
Step 2:

• Mother ($I^A i$) produces two types of ova: $I^A$ and $i$.
• Father ($I^B i$) produces two types of sperms: $I^B$ and $i$.
Step 3:
Cross: $I^A i \times I^B i$ [h] {|c|c|c|} Gametes & $I^B$ & $i$
$I^A$ & $I^A I^B$ (Type AB) & $I^A i$ (Type A)
$i$ & $I^B i$ (Type B) & $ii$ (Type O)
Step 4:
From the Punnett square, there are 4 possible genotype combinations, each with an equal 1/4 (25%) chance of occurring:
• 25% chance of $I^A I^B$ (Blood Group AB)
• 25% chance of $I^A i$ (Blood Group A)
• 25% chance of $I^B i$ (Blood Group B)
• 25% chance of $ii$ (Blood Group O)
Step 5:
The probability of having a child with the 'O' blood group is exactly 25%. Option (4) is correct.