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

Step 1 : Understanding the Question:
This problem involves basic Mendelian genetics applied to the human ABO blood group system. We need to determine the mathematical probability of producing an offspring with type 'O' blood from parents with specific heterozygous genotypes.
Step 2 : Key Formulas and Approach:
1. Identify the alleles: $I^A$ and $I^B$ are codominant, while $i$ is recessive.
2. Determine parental genotypes: Heterozygous 'A' is $I^A i$; Heterozygous 'B' is $I^B i$.
3. Use a Punnett square to predict the genotypic and phenotypic ratios of the offspring.
Step 3 : Detailed Explanation:

Parental Genotypes: The mother's blood type is A but she is heterozygous, so her genotype is $I^A i$. The father's blood type is B but he is heterozygous, so his genotype is $I^B i$.

Gamete Production: The mother produces two types of ova: $I^A$ and $i$. The father produces two types of sperm: $I^B$ and $i$.

Punnett Square Analysis: \includegraphics[width=0.3\linewidth]{143_sol.png}

Calculating Ratios: Out of 4 possible genetic combinations:
1/4 is $I^A I^B$ (Type AB)

1/4 is $I^A i$ (Type A)

1/4 is $I^B i$ (Type B)

1/4 is $ii$ (Type O)


Step 4 : Final Answer:
The probability of a type 'O' offspring ($ii$) is 1 out of 4, which is 25%. This leads to the selection of option (D).