Question 1 (8 pts). In a population of wild hogs, fur color is determined by all

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Question 1 (8 pts). In a population of wild hogs, fur color is determined by alleles at one locus; allele A for black, and allele a for white. The alleles are co-dominant; so, the heterozygotes have a mixed (ie.,roan) phenotype. A sample 500 hogs is evaluated for fur color; 33 white, 172 black, and 295 roan.

a. What are the frequencies of the A and a alleles?

b. Assuming that the population is in Hardy-Weinberg Equilibrium, what is the expected frequency of roan hogs?

        

c. Assuming that the population is in H-W E, what is the expected number of white hogs?

        

d. How many degrees of freedom would you use in a Chi-Square test to assess if that population was in Hardy-Weinberg Equilibrium?

e. (2pts) what is the Chi-square value for testing if this population is in Hardy-Weinberg equilibrium?

        

f. What is the approximate probability (ie., the “P value”) that those observed data do not deviate from the frequencies expected of a population that is in Hardy-Weinberg equilibrium?

g. Is that population in Hardy-Weinberg equilibrium? Choose either “yes” or “no” and briefly explain your choice:

Explanation / Answer

a) Genotype for white fur is aa , black fur is AA and roan fur is Aa.

Frequency of A allele:

There are two A alleles in AA (black) = 172* 2= 344

There are one A allele in heterozygote Aa= 295

Frequency of allele A= p= (344+295)/ 500+500= 0.639

Frequency of allele a= q= 1-p= 1-0.639=0.361

b) The Hardy Weinberg equation is given as:

p2+2pq+q2=1

p2 = frequency of the black genotype AA, q2 = the frequency of the white genotype aa, and 2pq = frequency of the heterozygous genotype Aa.

Expected frequency of AA= p2= 0.639*0.639 = 0.408

Expected frequency of Aa= 2pq= 2X 0.639 X 0.361= 0.461

Expected frequency of aa= q2= 0.361 X 0.361= 0.130

Expected frequency of Roan hogs= 0.461

c) To obtain expected number of white hogs, multiply the expected frequency of white hogs with total population.

Expected number of white hogs= 0.130*500= 65

Expected number of roan hogs= 0.461*500=230.5

Expected number of black hogs= 0.408*500=204

d) Degree of freedom= Number of genotypes-1= 3-1=2

e) Ch-square test:

Null Hypothesis: there is no difference between the observed and expected values. Population is in HWE.

Oberved

Expected

(O-E)

(O-E)2

(O-E)2/E

AA

172

204

-32

1024

5.019608

Aa

295

230.5

64.5

4160.25

18.04881

aa

33

65

-32

1024

15.75385

Total

1

1

5.00E-01

6208.25

38.82226

The chi-square value is 38.822.

f) P value for chi square value of 38.822 at 2 degree of freedom= less than 0.00001

We can consider the value to be 0.00001.

g) No, the population is not in Hardy Weinberg equilibrium.

A Chi Square value between 1%-5% should allow you to reject the null hypothesis. 0.00001*100= 0.001% is a small P Value. Hence, the null hypothesis is rejected. The population is not in Hardy-Weinberg equilibrium.

Oberved

Expected

(O-E)

(O-E)2

(O-E)2/E

AA

172

204

-32

1024

5.019608

Aa

295

230.5

64.5

4160.25

18.04881

aa

33

65

-32

1024

15.75385

Total

1

1

5.00E-01

6208.25

38.82226

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