answer all highlighted questions Normal No Spacing Heading1 Heading 2 NAME: BIOL
ID: 201376 • Letter: A
Question
answer all highlighted questionsNormal No Spacing Heading1 Heading 2 NAME: BIOL 102, CONCEPTS Breeding Bunnies Pre-Lab: Allele Frequencies and Hardy-Weinberg Equilibriunm The Hardy-Weinberg Principle was developed to describe the genetic characteristics of populations with no evolutionary forces acting on it (mating is random, there is no natural selection, no gene flow, no mutations, population size is extremely large). As a result, for populations in Hardy-Weinberg equilibrium the frequency of alileles and genotypes does not change over time (across generations). This also means that for any locus with two alleles, A (dominant allele) and a (recessive allele), if we know their frequencies in the population lfrealA)-p, freda)·al, the diploid genotype frequencies can be predicted from the allele frequencies as follows: freq(AA)-p2 the AA homorygotes in the population fregloo)- q the aa homozygotes in the population frealAa)- 2pa the heteroxypotes in the population Consider a gene with two alleles, A and a, with frequencies P·0.7 and q = 03, respectively. 1. Using this information answer the following questions a. Fil In the following table of expected genotype frequencies in this populations Table 1 Genotypes Aa Expected Frequency a. Assume the population we are studying is a species of plant and we wish to determine allele ampled 100 individual's from the population, how many would we expect to have the A allele? Now assume that we actually obtain the genotype for 100 members of the population and find the following number 2. Aa: 70 aa: 10 a. What are the observed genotype frequencles? AA: 02 Aa 0.7 aa: 0.1 b. Do you think the observed genotype numbers deviate significantly from those expected under Hardy Weinberg? if so, what might be the cause? Explain your answer
Explanation / Answer
1. aa - 0.09
AA - 0.49
Aa - 0.42
1A. AA and Aa = 0.49 + 0.42 = 0.91 = 91%
So 91/100 X 100 = 91 individual
2. A. observed genotype frequency -
AA-0.2 Aa-0.7 aa-0.1
Expected frequency
aa = 0.1 = 10
a = 0.31
A+a=1
S0, A=1-0.31 = 0.69
2Aa = 2X0.69X0.31 = 0.42 = 42
AA = 0.47 = 47
O E O-E (O-E)2 (O-E)2/ E
AA 20 47 -27 729 15.5
Aa 70 42 28 784 18.66
aa 10 10 0 0 0
Chi square = 34.16
Chi square 34.16 > than table value 3.14
Reject null hypothesis
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