Question

i wouldnt mind if someone could check my work on the first page. but whats confusing me is section B. it starts on the bottom of the first page and carries onto the second page.

A. Estimation of allele frequency for a specific trait Fill in the tables below for taster and allele frequencies in your class section. Taster/Non-taster Frequency=Genotype frequency (1 pt) # of students #Tasters Frequency of # Non-tasters in Lab Section Tasters (p+2pg) 16 1 = 5125 Frequency of Non-tasters (g I = .6875 Now use the Hardy-Weinberg equation to determine the two allele frequencies. You should first determine the frequency of g by taking the square root of the non-taster frequency. Then the frequency of p can be determined by knowing that p+q-1. Fill in the table below with your calculations. The frequencies of tasters and non-lasters in North America are 0.55 and 0.45, respectively. Calculate the p and q allele frequencies for the North American population as well and fill in the appropriate spaces in the table below. Allele Frequency (2 pts) Frequency of p in Frequency of q in Frequency of p in Frequency of in your lab section your lab section North America North America 189: 121 h = ,829 11.671 = 329 45 , 671 Show your calculations to receive full credit (0.5 pt for calculation, 0.5 pt for correct answer) Question: What is the frequency of heterozygous tasters in your lab section? .284. 1 pt heterozygous = 2.p.97 21.121.829) = 283518 How does this compare to the percentage of heterozygous tasters in North America? 1 pt hetrozygous tutes 1942 onrclays freams of heterozygous tastors is gigantly less than of the North amech. by a fusor of abnost 35% kas common B. An idealized Hardy-Weinberg community Assume a population is having all heterozygous individuals. The genotype of all individuals will then be Aa. The initial allele frequencies will be 0.5 for both the dominant A and recessive a alleles and genotype frequencies will be 0 A4, 1.0 Aa and 0 a. So, we have a population of randomly mating heterozygous individuals. After random mating for 5 generations, the number of individuals with all three genotypes are given in the table below.

Answer 1

Section A is correct.

B.The total number of people are 200. Out of which 48 are AA, 95 are Aa and 57 are aa. Then genotypic frequencies are 48/200, 95/200 and 57/200 ie. 0.24, 0.475 and 0.285 for AA, Aa and aa respectively.

So here q^{​​​​​​2}​​​ is 0.285 , so q is √.285 ie. 0.53

So p is 1-.53 ie. 0.47

If the population is in equilibrium then the sum of allele frequencies p and q will be equal to 1.

Yes the results agree, the allele frequencies did not change much after 5 generation due to absence of any external pressures like migration, mutation, natural selection and no random mating.