If 4% of 1,000 babies in an African population are born with a severe form of sickle-cell anemia (ss), how many babies will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

Total babies are heterozygous (Ss) for the sickle-cell gene

Explanation:

It is given that babies with sickle cell anemia disease have genotype "ss"

The number of babies with sickle cell anemia disease

Out of , nearly species have ss genotype

Let us take "s" be recessive to "S"

Also let us assume that the given population is in Hardy Weinberg's equation-

Then frequency of recessive genotype would be

Frequency of recessive "s" allele will be

As per Hardy Weinberg's first equilibrium equation, we have -

Substituting the values of "q" in above equation, we get -

Frequency of dominant genotype would be

Hardy Weinberg's second equilibrium equation is

Substituting the available values in above equation we get -

Hence, % of babies are heterozygous (Ss) for the sickle-cell gene

Thus, total babies are heterozygous (Ss) for the sickle-cell gene

answer; i believe that the correct answer is (light);

answer: 45

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