Question

In population of 250 individuals in Hardy Weinberg equilibrium recessive and homozygous dominant individuals are 30% and 20% respectively. The frequency of recessive allele isSir please explain this question​

Answer 1

Given as :

For  Hardy Weinberg equilibrium

Total number of population = 250

Percentage Equilibrium recessive = q² = 30%

Percentage Homozygous dominant individuals = p² = 20%

To Find :

The frequency of recessive allele = q

Solution :

The frequency of dominant allele = q

Percentage of Heterozygous individual = 2 p q

The basic formula is

p² + q² + 2 p q = 1              ...........1

p + q = 1                              ...........2

From eq 1

30 % + 20 % + 2 p q = 1

Or, 0.3 + 0.2 + 2 p q = 1

Or,  2 p q = 1 - 0.5

Or,  2 p q = 0.5

∴        p q = $\dfrac{0.5}{2}$ = 0.25

From eq 2

∵ p + q = 1

So,  ( p - q )² = ( p + q )² - 4 p q

Or,                = 1² - 4 × 0.25

Or,                = 1 - 1

Or,                = 0

i.e   ( p - q )² = 0

Or,   p - q = 0                ........3

Solving eq 2 and eq 3

  ( p + q ) + ( p - q ) = 1 + 0

Or,   ( p + p ) + ( q - q ) = 1

Or,    2 q + 0 = 1

Or,       q = $\dfrac{1}{2}$ = 0.5

∴   The frequency of recessive allele = q = 0.5

Hence, The frequency of recessive allele is 0.5   Answer

Answer 2

Answer:

p2 +q2+2pq =1 hence here

p2 or homozygous dominant  =30/100=0.3

q2or homozygous recessive =20/100=0.2

p2+q2+2pq =1

0.3+0.2+2pq =1

2pq =0.5

p q=0.25

p=0.25/q    (1)

p +q =1

substituting 1 in this equation it will give

q2-q+0.25=1

from this  we get q =0.5

Step-by-step explanation: