Question

In a population of 5000 individuals is 450 individuals show a recessive trait.What will be no.Of heterozygous individuals in this population.

P.S. please provide the solution with satisfactory explanation not just the answer

Answer 1

Answer:

p2+q2+2pq=1

Explanation:

q= 450/5000 = 0.09

p= 1-0.09 =0.91

heterozygote=2pq=2*0.09*0.91

= 0.1638

Answer 2

This can be solved using Hardy-Weinberg Equation;

Explanation:

• p = AA (dominant)
• q = aa  (recessive)
• 2pq = Aa (Heterozygous)
• $p^{2} + 2pq + q^{2}$
• p  +  q = 1
• p = no. of dominant individuals/total population
• q = no. of recessive individuals/total population
• q = 450/5000 = 0.09
• p + q = 1
• p = 1 – q = 1 – 0.09 = 0.91
• 2pq = 2 x 0.91 x 0.09 = 0.1638
• Total no. of heterozygous individuals = 0.1638 x 5000
• = 819