Four percent of 1000 babies in an African population are bor

Four percent of 1000 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? Show your work.

Solution

Four percent of 1000 babies in an African population are born with a severe form of sickle-cell anemia (ss) where ss is the recessive gene.

ss= q²= 0.04

Therefore q= 0.2

From the Hardy Weinberg equation - :

p+q =1 where p represents SS dominant.

p= 1- q =1- 0.2 = 0.8

Therefore p=0.8

For the heterozygous genotype for the population n=1000

Ss = n2pq

= 1000(2(0.8) (0.2))

Ss= 1000x (0.32)

Ss= 320 babies

In the given population 320 babies out of 1000 will be more resistant to malaria.