7. A recessive inherited disease is expressed only in individuals of blood group O and not expressed in blood groups A, B or AB. Alleles controlling the disease and blood group are independently inherited. A normal woman with blood group A and her normal husband with blood group B already had one child with the disease. The woman is pregnant for second time. What is the probability that the second child will also have the disease?
(1) 1/2 (2) 1/4
(3) 1/16 (4) 1/64
Step-by-Step Genetic Analysis
Step 1: ABO Genotypes of Parents
Parents: mother A, father B, already have O child.
Blood group O = ii → both parents must contribute i [web:122].
- Mother (A): IAi (AO)
- Father (B): IBi (BO)
From AO × BO: IAIB (AB), IAi (A), IBi (B), ii (O) – each 1/4 [web:123].
de>P(O child) = 1/4
Step 2: Disease Genetics
Disease: autosomal recessive, expressed only in group O individuals [web:155].
They have one affected child (O + dd):
- Both parents are carriers Dd at disease locus
- Dd × Dd → P(dd) = 1/4 [web:156][web:157]
Step 3: Probability Second Child Affected
Child must be:
- Blood group O (ii): 1/4
- Disease genotype dd: 1/4
Genes independent:
de>P(diseased) = P(O) × P(dd) = 1/4 × 1/4 = 1/16
Independent of first child outcome [web:158].
Option Review
- 1/2: carrier chance from carrier × normal (too high)
- 1/4: either O alone or dd alone (not both)
- 1/16: ✅ CORRECT – joint probability [web:155]
- 1/64: three independent 1/4 events (too low)
Key Insight
The disease requires TWO rare events:
- 1/4: child inherits i from both AO×BO parents
- 1/4: child inherits d from both Dd×Dd parents
Total: 1/16 per pregnancy
Final Answer
The probability second child will also have the disease is 1/16 [web:156].
