Question

What is the probability that a man having AO\textsuperscript{B (ABO and Rh blood group) genotype and a woman having BO\textsuperscript{D} genotype will have OO\textsuperscript{d} genotype child ?}

• A. \( \frac{1}{16} \)
• B. \( \frac{1}{8} \)
• C. \( \frac{3}{16} \)
• D. \( \frac{1}{4} \)

Hint:

For combined genetic probabilities, calculate each trait separately and then multiply the probabilities using the principle of independent assortment.

Answer 1

The Correct Option is A

Concept: This question combines inheritance of:
• ABO blood group system,
• Rh blood group system. To solve the problem correctly, we calculate:
• Probability of child having OO genotype,
• Probability of child having Rh negative genotype \(dd\),
• Multiply both probabilities using laws of independent assortment.

Step 1: Understanding ABO inheritance. The father's genotype is: \[ AO \] The mother's genotype is: \[ BO \] Gametes produced by father: \[ A, O \] Gametes produced by mother: \[ B, O \] Now constructing ABO combination: \[ \begin{array}{c|cc} & B & O \hline A & AB & AO O & BO & OO \end{array} \] Possible offspring:
• AB,
• AO,
• BO,
• OO. Thus: \[ P(OO) = \frac{1}{4} \]

Step 2: Understanding Rh inheritance. Both parents are heterozygous: \[ Dd \times Dd \] Gametes: \[ D, d \] Punnett square: \[ \begin{array}{c|cc} & D & d \hline D & DD & Dd d & Dd & dd \end{array} \] Thus: \[ P(dd) = \frac{1}{4} \]

Step 3: Combining both probabilities. We need probability of: \[ OO^{d} \] That means:
• OO in ABO system,
• \(dd\) in Rh system. Using independent assortment: \[ P(OO \text{ and } dd) = P(OO)\times P(dd) \] Substituting values: \[ = \frac{1}{4}\times \frac{1}{4} \] \[ = \frac{1}{16} \]

Step 4: Final conclusion. Therefore, the required probability is: \[ \boxed{\frac{1}{16}} \] Hence, the correct answer becomes: \[ \boxed{\text{(A) } \frac{1}{16}} \]