If the frequency of a recessive allele is 0.4. What will be the frequency of heterozygote carriers in a population of 2000?

The correct option is A 960
Assuming that the population is in the Hardy - Weinberg equilibrium, let’s suppose the frequency of the recessive allele is represented by q.
Then q = 0.4
We know p + q = 1
p + 0.4 = 1
p = 1 - 0.4
p = 0.6

To calculate the heterozygote frequency 2pq
2pq = $2\times 0.6\times 0.4$
2pq = 0.48

To calculate the number of heterozygous individual:
heterozygote frequency is multiplied by the total number of individuals
$0.4\times 2000$ = 960
So there are 960 carriers in a population of 2000 for the given gene.