Final Answer:
for a human blood, there are two alleles (called s and s) and three distinct phenotypes that can be identified by means of the appropriate reagents. the following data was taken from people in britain. among the 1000 people sampled, the following genotype
(a) Yes, reject Hardy-Weinberg proportions
Step-by-step explanation:
To determine the frequency of alleles (s and s) in the population, start by calculating allele frequencies. The total number of alleles in the population is twice the number of individuals (1000 * 2 = 2000 alleles).
From the genotypes given, count the number of alleles:
- For ss (99 individuals), there are 2 * 99 = 198 alleles of s.
- For ss (418 individuals), there are 2 * 418 = 836 alleles of s.
- For ss (483 individuals), there are 2 * 483 = 966 alleles of s.
The total number of s alleles = 198 + 836 + 966 = 2000 alleles.
The total number of s alleles = 2000 alleles.
Calculate the frequencies:
Frequency of s allele = Total s alleles / Total alleles = 2000 / 4000 = 0.5
Frequency of s allele = Total s alleles / Total alleles = 2000 / 4000 = 0.5
The expected genotype frequencies under Hardy-Weinberg equilibrium for alleles s and s can be calculated from allele frequencies: s^2 = p^2 = 0.5^2 = 0.25 (for ss), 2pq = 2 * 0.5 * 0.5 = 0.5 (for ss), s^2 = q^2 = 0.5^2 = 0.25 (for ss).
Compare the observed genotype frequencies with the expected frequencies using the chi-square (\(\chi^2\)) test. Calculate the chi-square statistic and compare it to the critical value at a chosen significance level with 2 degrees of freedom. If the calculated \(\chi^2\) statistic exceeds the critical value, there is reason to reject the Hardy-Weinberg equilibrium assumption. Therefore, the correct answer is a) Yes, reject Hardy-Weinberg proportions.