Rock pocket mice live in the desert of New Mexico and can be either tan or black

Question

Rock pocket mice live in the desert of New Mexico and can be either tan or black. Tan mice tend to be more common in areas where the desert floor is covered with sand and gravel. Black mice are more common in areas covered with dark lava flows. Scientists believe that predation pressure results in this pattern. Fur color is determined by a dominant black allele B and a recessive tan allele b. Suppose a population of mice with a B allele frequency of 0.5 is relocated to a basalt flow. Selection occurs due to predation, and the resulting breeding population has now produced on average 23 from tan mice, 122 from heterozygous black mice, and 58 from homozygous black mice. Calculate the selection coefficient, s, of tan mice after selection has occurred. Express your answer to at least two decimal places.

HINT:Calculating s of a genotype involves two steps. First, calculate the fitness, w, of the genotype by dividing the number of offspring from the genotype by the number of offspring from the most reproductively successful genotype. Next, calculate s with the equation s = 1 – w.

I did this and got .89 but it was the wrong answer. Can someone correctly answer this and please explain this step by step?

Explanation / Answer

In the original population, the frequency of dominant allele B, let’s call it p = 0.5

Since the sum of frequencies of all alleles =1, the frequency of recessive allele b let’s call it q = 1-0.5 = 0.5

According to Hardy-Weinberg equation,

P2+2pq+q2=1

Where p2 = the frequency of genotype BB = 0.25 [(0.5)2]

2pq is the frequency of the genotype Bb= 0.5 (2*0.5*0.5)

q2 = frequency of the genotype bb= 0.25  [(0.5)2]

When this population is subjected to selection, the new population will have a different frequency of the genotypes. The frequency of genotypes can be calculated from the data given

Frequency of genotype BB after selection = number of individuals with BB genotype/ Total number of individuals = 58/203 = 0.29

Likewise Frequency of Genotype Bb = 122/203 = 0.60 and

Frequency of Genotype bb = 23/203 = 0.11

[Total number of individuals = 23+58+122 = 203]

If you compare the frequencies of different genotypes it’s obvious that there is a decrease in frequency of the bb genotype (the tan color phenotype). This decrease in frequency can be considered due to decreased fitness of the bb genotype, w, where

w= 1-s , where s is the selection coefficient. If s=0 then w=1 which means there is no selection pressure on the genotype. If s=1 then the selection pressure is absolute and the genotype will vanish in the next generation. The value of s could be anywhere from 0 to 1.

The frequency of a genotype after selection is calculated from the initial frequency using the equation

Ffinal = Finitial * w

Solving for w in for the bb genotype

w = Ffinal/Finitial = 0.11/0.25 = 0.44

From the value of w (i.e fitness) the value of selection coefficient can be calculated

s =1-w = 1-0.44 = 0.56

Hence the value of selection coefficient s for the bb genotype is 0.56

If this answer is not correct, I might have made a mistake somewhere. An answer to a similar question is available here…

http://www.chegg.com/homework-help/questions-and-answers/rock-pocket-mice-live-desert-new-mexico-either-tan-black-tan-mice-tend-common-areas-desert-q12070549

I have access only to the question not the answer in the above link so you’ll need to verify if the answer is correct. Please let me know if my answer is correct or not

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