The heterozygous condition of sickle cell anemia results in resistance to malaria. If a homozygous normal father and a heterozygous mother have a child, what are the chances of that child being resistant to malaria but not having sickle cell anemia

1/2

Explanation:

This is a cross involving a single gene coding for sickle cell resistance (S) where the dominant allele is denoted by S and the recessive allele by s. Hence, an individual heterozygous will have a Ss genotype. According to the question, an individual with Ss genotype (heterozygous) will be phenotypically normal and resistant to malaria attack.

A cross between a homozygous normal male (SS) and heterozygous female (Ss), using a punnet square, will give rise to four possible offsprings with genotypes: SS, SS, Ss and Ss.

The SS genotype is phenotypically normal but not resistant to malaria while the Ss genotype is also phenotypically normal but resistant to malaria. Hence, the probability of their single child being phenotypically normal (not having sickle cell) and resistant to malaria is 2 out of 4 possible offsprings i.e. 2/4 = 1/2.

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the answer would be   10%