Sickle-cell anemia is a recessive trait in humans. In order to express the disease, an affected individual must inherit two recessive alleles (ss). What is the probability that the offspring of two heterozygous parents will also be heterozygous?

1/2

Explanation:

This is a cross involving a single gene coding for for possession or not of sickle cell anemia in humans. The 'S' allele encodes a normal cell while the 's' allele encodes a sickle cell. The allele for normal cell (S) is dominant over the allele for sickle cell (s) i.e. S allele will mask the phenotypic expression of s allele in a heterozygous state.

An heterozygous individual contains both dominant (S) and recessive (s) alleles. Hence, will have a genotype: Ss.

Using a punnet square, a cross between two heterozygous individuals i.e. Ss × Ss will produce four possible offsprings with genotypes: SS, Ss, Ss, ss

SS (1/4) is homozygous dominant i.e. normal offspring with same dominant alleles

Ss (2/4) is heterozygous dominant i.e. normal offspring with different alleles.

ss (1/4) is homozygous recessive i.e. sickle cell offspring with same recessive alleles.

Therefore, the probability that an offspring of the cross between two heterozygous parents will also be heterozygous is 2/4 = 1/2.

S. s

S SS Ss

s Ss ss

answer; atp and nadph;

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