Question

The normal chemokine receptor 5 gene (CCR5) allele “+” and a variant allele D was found in a population of people. The variant D has a deletion of 32 base pairs called “Delta 32”. A sample of people from this population were tested for the variation in the CCR5 gene. The researchers found

16 people were homozygous for Delta 32;

165 people were heterozygous for the Delta 32 deletion;

433 people did not have the Delta 32 deletion.

Choose all the correct answers to complete the sentence describing this population. The genotype frequency of the Delta 32 heterozygotes is ____, which would correspond to _____ if this population is at Hardy-Weinberg Equilibrium.

Options - select all possible answers

a. q= 0.16

b. 2pq=0.134

c. f(++)=433/614

d. 2pq= 0.269

e. f(D+ or DD)=181/614

f. f(D+)=165/614

g. f(DD)= 16/614

h. p = 0.84

Answer 1

Fill in the blank 1)

To find the frequency of heterozygous individuals, the formula is rather simple. We divide the number of heterozygotes with total number of people.

Now total number is = 16 + 165 + 433 = 614

Now the frequency of heterozygous is = 165/614 = 0.269.

So the right option is Option (d) 2pq = 0.269 and also option (f) f(D+) = 165/614 because this is the formula for calculating the heterozygous frequency.

Fill in the blank 2)

Now we are assuming that the population is in hardy weinberg equilibrium. Then we can assume that the frequency of Homozygous recessive (Delta 32) will be 16/614 = 0.026

(We use the same formula as we used for heterozygous in the first question but replace the numbers which correspond to whose frequency we are looking for)

And the frequency for Homozygous CCR5 allele (+) is = 433/614 = 0.705

Now we have the frequency of Homozygous population. But it is p2 for dominant and q2 for recessive population.

To find the value of p and q, we have to under root them.

So q = √0.026 = 0.16

And p = √0.705 = 0.84

Now if we look at the options given, then the options (c) and option (g) are right as they are the formula for calculations. And option (a) and option (h) are also right.

While p and q values correspond to allelic frequency. And they will be true if population is at hardy weinberg equilibrium.