Question

A fruit fly population has a gene with two alleles, A1 and A2. Tests show that 60% of the gametes produced in the population contain the A1 allele. If the population is in Hardy-Weinberg equilibrium, what proportion of the flies carry both A1 and A2?

1. 0.70                                         d. 0.48
2. 0.36                                         e. 0.24
3. 0.09

Answer 1

The correct answer is option d (0.48)

When a population is in H-W equilibrium, following equations can be used to determine allele/genotype frequencies

p + q = 1

and

p2 + 2pq + q2 = 1

Where

• p represents frequency of allele "A1",
• q represents frequency of allele "A2",
• p2 represents frequency of genotype "A1A1",
• q2 represents frequency of genotype "A2A2"
• 2pq represents frequency of genotype "A1A2" .

If 60 % of the gametes produced contain A1 allele in the polulation, its frequency can be measured as:

p = 60 % =0.6

Then, q = (1 - p) = (1 - 0.6) = 0.4

The proportion of the flies carrying both A1 and A2 allele will be 2pq

hence 2 pq = 2 (0.6 x 0.4)

= 0.48.

Hence the option d is correct.

Options a, b, c and e are incorrect because these values do not match with the A1A2 genotype frequency obtained from the above equation.