JAMB Biology 2022 Paper

Concept:A man with sickle cell trait ($\mathrm{HbA\,HbS}$) should marry a woman with normal hemoglobin genotype to avoid producing children with sickle cell disease.Explanation:The man's genotype is $\mathrm{HbA\,HbS}$, meaning he is a carrier but usually not severely affected.If he marries a woman with $\mathrm{HbA\,HbA}$, the possible offspring are $\mathrm{HbA\,HbA}$ and $\mathrm{HbA\,HbS}$.Thus, no child inherits double sickle genes ($\mathrm{HbS\,HbS}$), and the risk of sickle cell disease is eliminated.Marrying a woman with $\mathrm{HbS\,HbS}$ would give a 50% chance of having a child with sickle cell disease.Therefore, the most preferred choice is a woman with genotype $\mathrm{HbA\,HbA}$.Answer:A. $\mathrm{HbA\,HbA}$