6. Consider the following population of really cool stream-dwelling salamanders:

Question

6. Consider the following population of really cool stream-dwelling salamanders: A1A1 A1A2 A2A2 Relative frequencies 0.60 0.30 0.10 Probability of survival 0.60 0.80 0.40 Number of offspring 15.0 5.0 10.0 a. Calculate the relative fitness of each genotype. b. Calculate the selection coefficient for each genotype. c. If the strength and direction of natural selection in this population remain the same, what will the frequency of the A1 and A2 alleles be when the population reaches Hardy-Weinberg equilibrium (just explain in words, you do not need math)? d. Calculate p for A1 assuming that the strength and direction of natural selection remain unchanged. e. Now suppose that the probability of survival for the A2A2 genotype increased to 0.90, but everything else remained the same. How would this change your answer to part c)?

Explanation / Answer

A) W11=(.6)(15)=9

W12=(.8)(5)=4

W22=(.4)(10)=4

w11=9/9=1

w12=4/9=0.44

w22=4/9=0.44

B) S11=1-1=0

S12=1-.44=0.56

S22=1-0.44=0.56

C) A1=p will go to 1; p^=1, fixation

A2=q will go to 0; q^=0, extinction

D) First calculate the allele frequencies: p=P+1/2H

=0.6+(1/2)*0.3

= 0.75q=1-0.75=0.25

Then, calculate w :

w=(0.75)2*(1)+2(0.75)(0.25)*(0.44)+(0.25)2*(0.44)

=0.563+0.165+0.0275

=0.76

After that, calculate p:

p=(0.75/0.76)*(0.75*1+0.25*0.44-0.76)

=0.99*0.1

=0.099

E) W22=9

w22=1

This is the same as A1A1. So, the selection would underdomanince. Since the relative frequency of p is greater than q then p^ would still equal 1 and go to fixation and q^ would still equal 0 and go to extintion. Therefore, this wouldn’t change our answer to part C.

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