Evolution PCB4674 HWE practice questions Hardy-Weinberg equilibrium practice que
ID: 196357 • Letter: E
Question
Evolution PCB4674 HWE practice questions Hardy-Weinberg equilibrium practice questions You observe a field of wildflowers with the following genotype frequencies: AA: 44, Aa: 66, aa: 22. What are the allele frequencies? Is this population in Hardy-Weinberg equilibrium? observed genotype # frequency frequency expectedexpected genotype allele genotype # trequency 44.917 64.167 22.917 0.583 0.333 0.5 0.167 0.340 0.486 0.174 0.417 132 Yes, observed genotype numbers are close to expected so population is in HWE 2 Given initial allele frequencies of A: 0.3 and a: 0.7, calculate the final allele frequencies for a population of 10000 individuals one generation later assuming HWE conditions are met. What are the genotype frequencies, and numbers? expected genotype frequency expected genotype frequency allele frequency 900 4200 4900 0.3 0.09 0.42 0.49 0.7 A: 0.3 a:0.7 Recall if HWE assumptions are met, allele frequencies do not change generation after generation 3 Repeat question 2 assuming that mutation converts A to a at a rate of 1 in 10,000 expected genotype expected allele genotype# frequency allele frequency allele mutation frequency frequency 0.09 0.42 0.49 0.3 0.3 0.0001) 0.3 0.29997 900 4200 4900 0.3 Aa 0.7+ 0.7 0.3(0.0001) 0.70003 0.7Explanation / Answer
1. The answer is
Frequency of alleles and genotypes estimation:
Allele frequency estimation:
Genotype
Freequency
Allele A
Allele a
Total
AA
44
88
0
88
Aa
66
66
66
132
aa
22
0
44
44
Total
132
154
110
264
Allele A
= 154 / 264
= 0.58
Allele a
= 110 / 264
= 0.42
Expected genotype frequencies:
AA
= 0.58*0.58 = 0.340
*132 = 45
Aa
= 2 *0.58 * 0.42 =0.486
*132 = 64
aa
= 0.42 *0.42 = 0.174
*132 = 23
Chisquare test:
Null hypothesis: The observed values are not deviating from the expected values.
Test static:
Category
AA
Aa
aa
Observed values
44
66
22
Exprected Values
45
64
23
Deviation
-1
2
-1
D^2
0.840278
3.361111
0.840278
D^2/E
0.02
0.05
0.04
0.11
X^2
0.11
Degrees of freedom
1
Inference: The calculated chisquare value i.e. 0.11 is less than the table value i.e. 3.84 at 1 DF and 0.05 probability, hence the null hypothesis is accepted. Which means the population is in HW equilibrium.
ACCORDING TO CHEGG GUIDELINES WE HAVE TO ANSWER ONE QUESTION AT A TIME. POST THE REST AS SPERATE QUESTIONS, THEN I CAN HELP YOU.
Genotype
Freequency
Allele A
Allele a
Total
AA
44
88
0
88
Aa
66
66
66
132
aa
22
0
44
44
Total
132
154
110
264
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