Answers to Population Genetics Questions 4-7
4. p = 0.1
q = 0.9
w HH = 1.0
w Hh 0 1.0
w hh = 0.25
s Hh = 0
s hh = 0.75
s=1-w
s HH = 0
This simply tells us that there is strong selection against the hh.
To calculate allele frequency the next generation, you need to recognize that this is a case of
selection against the recessive allele, and use the formula from box 6.7 in your text.
q'

q(1 sq)
1 sq 2
where q’ = q the next generation (the frequency of the h allele in the next generation).
Conveniently, this is the answer to 4b.
Plugging in the values for q (.9) and s (.75), you should get
q’ = .7452
That’s the answer to question b
Notice how sharply the allele frequency has dropped in one generation as a result of strong
selection against the hh genotype.
a) - You can now answer question a very easily, if you recognize that the frequency of hh
individuals the next generation = q2 = (.7452)(.7452) = .5553
c) The frequency of the h allele will continue to decline under strong selection against the hh
genotype. You can calculate this decline over the next two generations by pplugging in the value
you get for h for each generation into the formula
q'

q(1 sq)
1 sq 2
This gives you
q2 = .5633
and q3 = .4268
There will be some round-off error
5. Light survival = 54/393 = .1374
dark survival = 19/406 = .0468
relative fitness (w for survival)
Light = .1374/.1374 = 1
Selection coefficient = 1-w
Light = 0 (no selection)
Dark = .0468/.1374 = .34
Dark = 1-.34 = .66
6. There are at least two ways of doing this: Here’s one way.
A. calculate genoty[pe frequencies:
N = total population size = 29+2993+9365 = 12387
HbS/HbS = 29/N = 29/12387 = .0023
HbS/HbA = 2993/12387 = .2416
HbA/HbA = 9365/12387 = .7560
B calculate allele frequencies
HbS = .0023 + ½(.2416) = .1231
HbA = 1 - HbS = .8769
C. Calculate expected genotype frequencies under H.W. equilibrium
HbS/HbS = (.1231)2 = .0151
HbA/HbS = 2pq = 2(.1231)(.8769) = .2159
HbA/HbA = (.8769)2 = .7690
(note at this point I always add my genotype frequencies to make sure they are ca. = to 1)
D. multiply by the total population size to estimate expected frequencies if under H.W.
equilibrium
Expected number of HbS/HbS = .0151(12387) = 187
Expected number of HbA/HbS = 2674
Expected number of HbA/HbA = 9525
I expect some roundoff error
7. I’ve defined survival as the (observed number)/(expected number).
HbS/HbS = 29/187 = .1551
HbA/HbS = 2993/2674 = 1.119
HbA/HbA = 9365/9525 = .9832
The heterozygotes have the highest survival rate, so we divide all genotype survival rates by
1.119 to get w.
HbS/HbS = .1386
HbA/HbS = 1
HbA/HbA = .8786
s=1-w
HbS/HbS = .8614
HbA/HbS = 0
HbA/HbA = .1214
Again I expect some round-off error