Assume that there are only two alleles at a given locus in a population, and the population is in h-w equilibrium. if the frequency of one allele is 0.6, what is the frequency of heterozygotes in the populationa. -0.6.
b. 0
c. 0.4
d. 0.6
e. 1

Frequency of heterozygotes in the population is 0.4.

C. 0.4

Explanation:

If a population is in Hardy-Weinberg equilibrium, the sum of the frequencies of the two alleles will be equal to 1.  The letter p denotes the frequency of the first allele and the letter q denotes the frequency of the second allele.  

p+q=1

In the problem it is given that p=0.6

p+q=1

0.6+q =1  

q= 1-0.6

=0.4

We have to calculate the frequency of heterozygotes in the population. it is given by the equation

Number of heterozygous individuals =2pq

=

=0.48~0.4

answer; i believe the correct answer is (replication is not conservative);

9.9 years

a = p(1 + rt) . . account balance after time t at rate r starting with principal p

3000 = 2500(1 + 0.0202t) . . filling in the given numbers

1.2 = 1 + 0.0202t . . divide by 2500

0.2 = 0.0202t . . . subtract 1

0.2/0.0202 = t ≈ 9.901

it will take about 9.9 years for the account balance to reach $3000.