In humans, the allele for sickle cell trait is common in some Caribbean populations. A man who is heterozygous for sickle cell (HbA HbS) marries a woman with the same genotype. What percentage of their children would be expected to have sickle cell disease?

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Question

In humans, the allele for sickle cell trait is common in some Caribbean populations. A man who is heterozygous for sickle cell (HbA HbS) marries a woman with the same genotype. What percentage of their children would be expected to have sickle cell disease?

Accepted Answer

25% — Award 1 mark for correctly applying Mendelian inheritance. The cross HbA HbS × HbA HbS produces: HbA HbA (normal), HbA HbS (carrier), HbA HbS (carrier), HbS HbS (sickle cell disease). Only 1 in 4 (25%) would have sickle cell disease. A is incorrect — affected offspring are possible. C is incorrect — this represents carriers, not affected individuals. D is incorrect — this would require different parental genotypes.

Answer

75%

Answer

0%

Answer

25%

Answer

50%

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