Question

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?

Answer 1

Answer:

Total $320$ 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

$= \frac{4}{100} * 1000\\= 40$

Out of $1000$, nearly $40$ 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 $0.04$

$q^2 = 0.04$

Frequency of recessive "s" allele will be

$q = 0.2$

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

$p + q = 1$

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

$p = 1-0.2\\p = 0.8$

Frequency of dominant genotype would be

$p^2 = 0.8^2\\p^2 = 0.64$

Hardy Weinberg's second equilibrium equation is

$p^2 + q^2 +2pq = 1$

Substituting the available values in above equation we get -

$2pq = 1 - 0.04 -0.64\\2pq = 0.32$

Hence, $32$ % of $1000$ babies are heterozygous (Ss) for the sickle-cell gene

Thus, total $320$ babies are heterozygous (Ss) for the sickle-cell gene