In a population of wild hogs, fur color is determined by alleles at one locus; a
ID: 261076 • Letter: I
Question
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 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?
b. Is that population in Hardy-Weinberg equilibrium? Choose either “yes” or “no” and briefly explain your choice:
Explanation / Answer
a. 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?
To find the out the p value, we need degress of freedom and probability.
The DF = number of categories – no of alleles = 3-2 = 1
Probability here is = 0.5
So, the p values at 1 DF and 0.05 probability is 3.84.
b. Is that population in Hardy-Weinberg equilibrium? Choose either “yes” or “no” and briefly explain your choice:
The answer is No.
For explanation find the below calculation.
The steps involved here is
1. Allele frequency determination
2. Genotype frequency determination
3. Chisquare test
Step 1: Allele frequency determination
Phenotype
Genotype
Freequency
Allele B
Allele b
Total
BLACK
BB
172
344
0
344
ROAN
Bb
295
295
295
590
WHITE
bb
33
0
66
66
Total
Total
500
639
361
1000
Allele B
= 639/1000
0.64
Allele b
= 361/1000
0.36
Step 2. Genotype frequency determination
BLACK
BB
=0.64*0.64=0.408
*500=204
ROAN
Bb
=2*0.64*0.36=0.461
*500=231
WHITE
bb
=0.36*0.36=0.130
*500=65
3. Chisquare test
Null hypothesis: The observed values are not deviating from the expected values.
Category
BB
Bb
bb
Observed values
172
295
33
Exprected Values
204
231
65
Deviation
-32
64
-32
D^2
1034.298
4137.191
1034.298
D^2/E
5.07
17.93
15.87
38.87
X^2
38.87
Degrees of freedom
1
Inference: The calculated chisquare value i.e. 38.87 is greater than the table value i.e. 3.84 at 1 DF and 0.05 probability, hence the null hypothesis is rejected.
Phenotype
Genotype
Freequency
Allele B
Allele b
Total
BLACK
BB
172
344
0
344
ROAN
Bb
295
295
295
590
WHITE
bb
33
0
66
66
Total
Total
500
639
361
1000
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