When sickle cell carrier marries a normal woman, the probability of them having a normal offspring is

Analyze the question A sickle cell carrier has the genotype AS (heterozygous). A normal woman has the genotype AA. We need the probability that a child is "normal".

Concept

• AA = normal
• AS = carrier (sickle cell trait, usually healthy)
• SS = sickle cell anaemia

Punnett square

| | A (mother) | A (mother) | |---|---|---| | A (father) | AA | AA | | S (father) | AS | AS |

Offspring: 2 AA : 2 AS, so 50% AA and 50% AS.

Interpreting "normal" Here "normal" means the genotype AA, i.e. a child who is neither a carrier nor a sufferer. The probability is 2/4 = 0.5.

(If "normal" meant simply "not having sickle cell anaemia", every child would qualify, with a probability of 1. Option D, ".1", looks like a typo for 1. But the standard JAMB reading of "normal offspring" is the AA genotype, which gives 0.)

Answer: C (0.5)

Exam tip For a carrier (heterozygote) × normal (homozygous dominant) cross, remember the ratio 1 : 1. Half the children are normal and half are carriers. For carrier × carrier, the ratio is 1 : 2 : 1, which gives 0.25 for a normal child.