In certain African countries 4 of the newborn babies have si

In certain African countries 4% of the newborn babies have sickle-cell anemia, which is a recessive trait. Out of a random population of 1,000 newborn babies, how many would you expect of each of the three possible genotypes for this trait? Choose the letters you will use to represent the alleles & genotypes: Translate to known frequencies: Choose which Hardy-Weinberg formula you will primarily use: Solve. Neatly show work below. Write final answer in box.

Solution

Dominant allele

p

Recessive allele

q

In a Hardy weinber formula, dominant allele is represented as p and recessive allele is represented by q.

and if the population is in equilibrium, p+q = 1

it is given that in a population of 1000, 4% of newborn are having Sickle cell Anaemia disease.

we know that it is a trait inherited by homozygous recessive allele.

So if 4% of 1000 people have disease, it means number of homozygous recessive = ss = 4% of 1000 = 40

ss = q² = 40

so q = square root of 40 = 6.32

now p will be calculated as 1- q = -5.32

Homozygous dominant will be SS = p² = (-5.32)² = 28.30

Hardy weinberg formula will be calculated as

p² + 2pq + q² = 28.30 + (2 * 5.32 *6.32 ) + 40 = 135.44

Now follow the tables for the genotype:

P (Dominant allele)

5.32

q (Recessive allele)

6.32

Genotypes are :

Homozygous Dominant

SS

Heterozygous

Ss

Homozygous Recessive

ss

p²   Homozygous Dominant

28.30

2pq Heterozygous

67.244

q² Homozygous Recessive

40

Dominant allele	p
Recessive allele	q