Need an expert in Genetics to help me with my population genetics course assignment. The assignment documents has been attached with 3 questions. Also, a few other documents about sample calculations
Worked example of calculating F-statistics from genotypic data:
(From University of Wyoming, Molecular Markers course).
|
| Genotype |
|
| AA | Aa | aa |
Population 1 | 125 | 250 | 125 |
Population 2 | 50 | 30 | 20 |
Population 3 | 100 | 500 | 400 |
N (number of individuals genotyped):
Population 1: 500
Population 2: 100
Population 3: 1,000
Remember that the number of alleles is TWICE the number of genotypes (every genotype has two alleles).
Gene frequencies: Each homozygote will have two alleles, each heterozygote will have one allele.
p1 (frequency of allele A in Pop. 1) = (2*125 + 250)/1,000 = 0.50
q1 = 0.5
p2 = (2*50 + 30)/200 = 0.65
q2 = 0.35
p3 = (2*100 + 500)/2,000 = 0.35
q3 = 0.65
Observed heterozygosity (here we are counting heterozygous genotypes):
Hobs1 = 250/500 = 0.5
Hobs2 = 30/100 = 0.3
Hobs3 = 500/1000 = 0.5
Expected heterozygosity (2pq):
Hexp1 = 2*0.5*0.5 = 0.5 [Observed = Expected]
Hexp2 = 2*0.65*0.35 = 0.46 [Observed < Expected]
Hexp3 =2*0.35*0.65 = 0.46 [Observed > Expected]
Population inbreeding coefficient [F = (Hexp - Hobs) /Hexp]
F1 = (0.5 — 0.5)/0.5 = 0
F2 = (0.46 — 0.3)/0.46 = 0.341 [fewer heterozygotes than expected indicates inbreeding]
F3 = (0.46 — 0.5)/0.46 = -0.099 [more heterozygotes than expected means excess outbreeding]
p-bar (frequency of allele A) over all populations
(2*125 + 250 + 2*50 + 30 + 2*100 + 500)/(1,000 + 200 + 2,000) = 1,330 / 3,200 = 0.4156
or
(0.5*1,000 + 0.65*200 + 0.35*2,000) / 3,200
q-bar (frequency of allele a) over all populations
(2*125 + 250 + 2*20 + 30 + 2*400 + 500) / (1,000 + 200 + 2,000) = 1,870 / 3,2000 = 0.5844
Check: p-bar + q-bar = 1.0 (correct!)
Heterozygosity indices (over individuals, subpopulations and total population)
HI based on observed heterozygosities in populations
[= (Hobs1*N1 + Hobs2*N2 + Hobs3*N3)/NTOTAL] = (0.5*500 + 0.3*100 + 0.5*1000) / 1,600 = 0.4875
HS based on expected heterozygosities in populations
[= (Hexp1*N1 + Hexp2*N2 + Hexp3*N3)/NTOTAL] = (0.5*500 + 0.46*100 + 0.46*1000) / 1,600 = 0.4691
HT based on expected heterozygosities overall:
[= 2*p-bar *q-bar] = 2 * 0.4156 * 0.5844 = 0.0.4858
FINALLY, THE F-STATISTICS:
FIS [= (HS - HI)/HS] = (0.4691 - 0.4875) / 0.4691 = -0.0393
FST [= (HT - HS)/HT] = (0.4858 - 0.4691) / 0.4858 = 0.0344
FIT [= (HT - HI)/HT] = (0.4858 - 0.4875)/0.4858 = -0.0036
All of which tells us that: Pop1 is consistent with HWE,
Pop2 is inbred, and
Pop3 may have disassortative mating or be experiencing a Wahlund effect (more heterozygotes than expected – not discussed yet in class).
Subdivision of populations, possibly due to genetic drift, accounts for approx. 3.4% of the total genetic variation,and the set of populations, as a whole, shows no signs of inbreeding (FIT is nearly zero).