0.5 What makes the hardy-weinberg equation so useful?  (Page 3/4)

To answer this question, we need to determine the total number of alleles in the population and what fraction of that total is due to the CCR-5 and ccr-5 alleles respectively.

Since each of the 704 individuals in this sample has two alleles for this locus, one per chromosome, we are working with a total of 1408 alleles.

Of these 704 individuals, 582 have two copies of the CCR-5 allele and another 114 have one copy for a total of 1278 (2 x 582 + 114) CCR-5 alleles. Therefore, 1278 of 1408 alleles are CCR-5 alleles, a value equal to a frequency of 0.91 or 91% (1278/1408 x 100).

Now that we have calculated the frequency of the CCR-5 allele, there are two ways to confirm that the ccr-5 allele occurs with a frequency equal to 0.09. The first is to use the relationship p + q = 1. If we set p equal to 0.91, the frequency of the CCR-5 allele, then q, the frequency of the ccr-5 , must equal 0.09 because q = 1 - p.

We can also confirm this calculation using the genotype data from Table 1. If 114 individuals have one copy of the ccr-5 allele and another 8 have two copies, there are 130 copies (2 x 8 +114) of the ccr-5 allele in this population of 1408 alleles for a frequency of 130/1408 or 0.09, equivalent to 9% of the allele population.

To answer this, we must calculate the genotype frequencies we would see if this population is not evolving with respect to these alleles, the actual genotype frequencies observed in this population and finally, compare these two frequencies and draw a conclusion.

If the population is not evolving with respect to these alleles, the population of parents that produced these 704 offspring and the population of offspring themselves must have the same allele frequencies. Therefore, we can use the allele frequencies we calculated above and the Hardy-Weinberg equation to determine the genotype frequencies we would see in this offspring population under the assumption of no evolution.

According to the Hardy-Weinberg equation, the frequency of the

• CCR-5/CCR-5 genotype is equal to the square of the frequency with which this allele occurs or p2 = 0.91 x 0.91 = 0.83.
• ccr-5 / ccr-5 genotype is equal to the square of the frequency with which this allele occurs or q2 = 0.09 x 0.09 = 0.008.
• CCR-5/ ccr-5 genotype is equal to the two times the product of frequency with which of these two alleles occur or 2pq = 2(0.91 x 0.09) = 0.16.

Thus, if this population is not evolving, 83%, 0.8% and 16% of these 704 offspring should exhibit the CCR-5/CCR-5, ccr-5 / ccr-5 , and CCR-5/ ccr-5 genotypes respectively. We can confirm our mathematics and that we haven't missed any potential offspring genotypes by checking that these three frequencies sum to 1 as they do (0.83 + 0.16 = 0.008 = 1.0).

We now need to calculate the actual frequency with which these genotypes occur in this population of 704 offspring (and therefore 704 genotypes). From Table 1:

• 582 of 704 genotypes are CCR-5/CCR-5 for a frequency of 582/704=0.83.
• 8 of 704 genotypes are ccr-5 / ccr-5 for a frequency of 8/704=0.011
• 114 of 704 genotypes are CCR-5/ ccr-5 for a frequency of 114/704=0.16

Thus, 83%, 1.1% and 16% of this population of 704 offspring exhibit the CCR-5/CCR-5, ccr-5 / ccr-5 , and CCR-5/ ccr-5 genotypes respectively. We can confirm our mathematics by checking that these three frequencies sum to 1 as they do (0.83 + 0.16 + 0.011 = 1.0).

We are now prepared to compare the genotype frequencies we actually see in this population of 704 individuals to those we would see if it is not evolving with respect to these alleles. Reviewing the summarized data below, we can see that both sets of genotype frequencies are nearly identical confirming Samson et al.'s (1996) conclusion.

• 83% of the genotypes are CRR-5/CRR-5 in both data sets
• 16% of the genotypes are CRR-5/ crr-5 in both data sets.
• 0.08 and 1.1% , or ~0.1%, of the genotypes are crr-5 / crr-5 in the observed and expected data sets respectively.