Question

How to do part C?

B. Now we are going to further test the hypothesis that natural selection is at work with a different population, favoring the retention of the debilitating sickle cell disease (heterozygote dyantape) in a population (Yoruba tribe) exposed to malaria by determining whether or not the population is in Handy Weinberg (HW) equilibrium. The total Number of individuals surveyed in the Yoruba tribe was 12,387. They had the following genotypic breakdown genotype ΗΔΗ HH 9365 (2 dominant alleles cach!) 2993 29 (2 Recessive alleles each!) Total 12387 We must first determine exact allele frequencies from that information we must use the HW equation to predict the expected phenotypic frequencies if the population really was in HW equilibrium, and then finally we must compare observed and expected frequencies. We will compare the various genotypes to determine cach genotype's relative fitness. Teams fill in table below and answer questions on the board as indicated. M716 a) First we must determine the exact allele frequencies p and q. We are given the genotypes and the numbers of people with each genotype and so we have all the information we need. We must add up all the alleles of each type and divide by the total number of alleles in the population which will be 2 x total people. 2424 p- Number of non sickle cell alleles (2 x 9365 + 2993y Total Number of alleles (2 x 12387) e t 0.876 0.88 go 0.12 b) Now we need to determine the expected frequencies, fe, and then e) the expected number of people, N., that would be predicted if the population were in HW equilibrium. This is where we use the HW equation and substitute in a value for p and q. Write the HW equation out: (Note: Remember what each of the terms represents.) Expected frequencies (f.) c) Expected number of people (No=fox Nel 7 + 272 + Q?-1 p = 0.77 2pq = 0.21 & = 0.014

Answer 1

p= 0.88. p^2 = 0.77

q= 0.12 q^2= 0.0144

2pq= 0.21

C- expected number of people Ne = fe × Nt

Now Nt= 12387

And fe for different cases we jave already calculated.

( i) So expected number of people Ne = p^2 × 12387

0.77× 12387 = 9538

(ii). q^2 × 12387

= 0.0144 × 12387

= 178

(iii) 2pq × 12387

0.21 × 12387

= 2601

So the expected number of people with dominant H^A H^A = 9538

So the expected number of people with H^A H^ S condition= 2601

So the expected number of people with recessive condition = 178