The first row of the table below shows the number of individuals of each of thre
ID: 185361 • Letter: T
Question
The first row of the table below shows the number of individuals of each of three genotypes in a given population in the summer of 2015.
The second row shows the relative probability that each of these genotypes survives over the winter (the relative fitness based on survival).
If you resurvey the population in the spring of 2016, how many SS individuals would you expect to find?
Genotype
TT
TS
SS
Number of individuals (Fall)
582
400
1645
Probability of surviving the winter
.90
.85
.70
Genotype
TT
TS
SS
Number of individuals (Fall)
582
400
1645
Probability of surviving the winter
.90
.85
.70
Explanation / Answer
The number of SS individuals expected to find is 1182.
First thing to do is to find the allelic frequency i.e., the frequency of T allele and S allele. The allelic frequency can be calculated from the genotypic frequency.
The genotypic frequency is calculated as follows;
. f(TT) = Number of individuals of TT/ Total number of individuals in the population
= 582/2627 = 0.22
f(TS) = Number of individuals of TS/ Total number of individuals in the population
= 400/2627 = 0.15
f(SS) = Number of individuals of SS/ Total number of individuals in the population
= 1645/2627 = 0.63
Frequencies of alleles are calculated as follows:
f(T) = f(TT) + (1/2 x f(TS))
= 0.22 + 0.075 = 0.295
f(S) = f(SS) + (1/2 x f(TS))
= 0.63 + 0.075 = 0.705
To calculate the number of SS individuals in Spring 2016 after surviving in winter, we have to calculate the average population fitness.
Average population fitness = [(TT probability of surviving in winter) x (f(T))2 ] + [(TS probability of surviving in winter) x 2 x f(T) x f(S)] + [(SS probability of surviving in winter) x (f(S))2 ]
= [0.9 x (0.295)2] + [0.85 x 2 x 0.295 x 0.705] + [0.7 x (0.705)2]
= 0.780
Now, frequency of SS (f(SS)) after winter survival = [(SS probability of surviving in winter) x (f(S))2 ]
= 0.348/0.780 = 0.45
Hence, number of SS individuals expected to find in spring 2016 or after winter survival = [frequency of SS (f(SS)) after winter survival] x [Total individuals in the population] = 0.45 x 2627
= 1182 (approx.)
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