1. In a species of desert rodent coat color is affected by the T locus. TT indiv
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Question
1. In a species of desert rodent coat color is affected by the T locus. TT individuals have black coat, Tt are brown, and tt are tan. You sample 500 individuals from a desert population and observe 125 black, 250 brown, and 125 tan individuals.
Calculate the genotype frequencies and allele frequencies in this sample.
2. a. If you represent the allele frequencies as f(T)=p and f(t)=q, what does the Hardy-Weinberg Equilibrium Model predict the genotype frequencies will be in the next generation (show the calculations)?
b. What are the assumptions of the Hardy-Weinberg Equilibrium Model?
c. What does the model predict about the genotype and allele frequencies at generations 10, 100, 1000? (Note: This is where the
Explanation / Answer
From the given data,
TT – black – 125
Tt – brown – 250
Tt – tan – 125
Total = 125+250+125 = 500
a.
Genotypic frequency of black rodents (TT) = 125/500 = 0.25
Genotypic frequency of brown rodents (Tt) = 250/500 = 0.5
Genotypic frequency of tan rodents (tt) = 125/500 = 0.25
The sum of all the genotypic frequencies should be 1. 0.25+0.5+0.25 = 1.
b.
The population has two alleles. Thus, this entire population consists of 500 x 2 = 1,000 alleles.
Count the number of T or t alleles and divide by the total number of alleles for allelic frequency of each allele.
So, the allelic frequency for the T allele will be:
f(T)= [(2 x 125)+250]/1000
= [250+250]/1000
= [500]/1000
= 0.5
And the frequency of t allele will be:
f(t)= [(2 x 125)+250]/1000
= [250+250]/1000
= [500]/1000
= 0.5
Convert them into allele p and q for our convenience to use in Hardy Weinberg equilibrium. Thus, the result is p=0.5 and q=0.5. The total sum p+q should be 1.0.
Thus, 0.5 + 0.5 = 1
c.
The population has two alleles. Thus, this entire population consists of (80+240180=500) 500 x 2 = 1,000 alleles.
So, the allelic frequency for the T allele after 10 generations will be:
f(T)= [(2 x 80)+240]/1000
= [160+240]/1000
= [400]/1000
= 0.4
And the frequency of t allele will be:
f(t)= [(2 x 180)+240]/1000
= [360+240]/1000
= [600]/1000
= 0.6
Convert them into allele p and q for our convenience to use in Hardy Weinberg equilibrium. Thus, the result is p=0.4 and q=0.6. The total sum p+q should be 1.0.
Thus, 0.6 + 0.4 = 1
From the data, the total number of individuals (n) = 80+240+180 = 500
So, the Hardy–Weinberg expectation is:
Therefore, Chi-square value is:
Therefore, the chi-square value is 40.8
The degree of freedom = 1 (no. of genotypes – no. of alleles)
Since the p2+2pq+q2 is equal to 1, the given population is in Hardy Weinberg equilibrium.
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