Sickle cell anemia is a recessive trait in humans. In a cross between a father who has sickle cell anemia and a mother who is heterozygous for the gene, what is the probability that their first three children will have the normal phenotype

1/8

Explanation:

Assuming the allele for sickle cell is represented by s, the alternate non-sickling allele will be S.

A ss father marries a mother that is heterozygous for the trait:

ss   x   Ss = Ss, Ss, ss and ss.

Thus, 50% of their children will be heterozygous normal for the trait while 50% will be homozygous recessive (affected) for the trait.

The probability of producing normal children is 1/2. Hence, the probability of their first three children having a normal phenotype is equivalent to the probability of the first being normal, the second being normal and the third being normal. Mathematically;

    = 1/2 x 1/2 1/2 = 1/8

The probability that their first three children will have the normal phenotype is 1/8.

answer; i believe that the correct answer is ; males are hemizygous for the x chromosome;

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explanation: