In a population there are 400 AA, 400 Aa, and 200 aa individuals. (a) Calculate

Question

In a population there are 400 AA, 400 Aa, and 200 aa individuals. (a) Calculate allele frequencies. f(A) - f(a) (b) Calculate predicted HWE genotype frequencies. f(AA)- f(Aa) f(aa) (c) Is this population in Hardy-Weinberg equilibrium? Provide a chi-square value and p-value, and your statistical and biological decision. chi-square value (round up your answer nearest 100th) > p-value > Statistical decision: (reject fail to reject) Biological decision: This population (is / is not) in HWE. df 0.995 0.975 0.9 0.005 000 .000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 386 4.605 5.991 7.3789.210 10.597 3 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357 7.779 9.488 11.143 13.277 14.860 5 0.412 0.831 1.610 4.351 9.236 11.070 12.832 15.08616.750 6 0.676 1.237 2.204 5.348 10.645 12.592 14.449 16.812 18.548 0.5 0.1 0.05 0.025 0.01

Explanation / Answer

Answer:

AA+Aa+aa= 400+400+200= 1000

a) Allele frequencies:

AA=p2=400/1000 = square root of 0.4. therefore p= 0.6

f(A)= 0.6

aa=q2=200/1000 = square root of 0.2. therefore q= 0.4

f(a)= 0.4

b) expected HWE genotype frequencies:

AA=p2= 0.6*0.6= 0.36

Aa=2pq= 2*0.6*0.4= 0.48

aa=q2= 0.4*0.4= 0.16

c) Null hypothesis: Population is in Hardy weinberg equilibrium

Alternate Hypothesis : Population is not in Hardy weinberg equilibrium

AA Aa aa

Observed 400 400 200

Expected 360 480 160

(O-E)2/E 4.4 0.83 10

4.4+0.83+10= 15.23

Df= (c-1)*(r-1) r-row; c-column; Df= 2

Comparing the value of 27.7715.23 to the chi-square distribution for 21 degree of freedom, we estimate that the probability of getting this value or higher of the statistic is less than 1%. Therefore we reject null hypothesis.

Statistical decision: Reject null hypothesis

Biological decision: Population is not in Hardy Weinberg equilibrium

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