If both parents are heterozygous for sickle-cell anemia ( HbA HbS ), what percentage of their offspring will have a homozygous genotype?A. 25B. 50C. 75D. 0

Sickle cell anemia is a genetic disorder that is inherited in an autosomal recessive pattern. If both parents are heterozygous for the trait (HbA HbS), they each carry one normal allele (HbA) and one mutated allele (HbS).

To determine the percentage of their offspring that will have a homozygous genotype, we can use a Punnett square. A Punnett square is a diagram that is used to predict the genotypes of a particular cross or breeding experiment.

Here's how the Punnett square would look for this cross:

    HbA    HbS

HbA HbA HbA HbA HbS HbS HbA HbS HbS HbS

From this, we can see that:

• 1 out of 4 (or 25%) will be homozygous normal (HbA HbA)
• 2 out of 4 (or 50%) will be heterozygous (HbA HbS)
• 1 out of 4 (or 25%) will be homozygous for the sickle cell trait (HbS HbS)

So, the percentage of their offspring that will have a homozygous genotype (either HbA HbA or HbS HbS) is 25% + 25% = 50%. Therefore, the correct answer is B. 50.