SME
Evolutionary Biology
Genetic drift
Use this for “A population of Nₑ = 50 starts at H₀ = 0.5. What heterozygosity is left after 100 generations?”
Diagram
12 populations of Nₑ = 50, each starting at p = 0.5: by generation 100, 3 fixed (p = 1), 3 lost (p = 0), 6 still varying.
Expected
H
(dashed):
0.5
×
(1
−
1
2 × 50
)¹⁰⁰
=
0.183,
0.366
of
H₀
kept.
This is an example. Type your own values to start a new problem.
Hₜ
=
H₀
1
−
1
2
×
k
Share of heterozygosity kept
calculated
Formulas
k
=
(1 −
1
2 × Nₑ
)
t
0.366
=
(1
−
1
2 × 50
)¹⁰⁰
Hₜ
=
H₀
×
k
0.183 = 0.5 × 0.366
Assumptions
•
An ideal population: Nₑ diploids pass on 2Nₑ gene copies drawn at random each generation.
•
No selection, mutation or migration: drift alone changes the allele frequency p.
•
Each generation keeps 1 − 1 ÷ (2Nₑ) of the heterozygosity on average, so Hₜ = H₀(1 − 1 ÷ 2Nₑ)ᵗ.
•
Two alleles, so H₀ = 2p(1 − p) is at most 0.5; the 12 paths are random runs.
Step-by-step
WE KNOW
Nₑ = 50, H₀ = 0.5, t = 100
FIND
share of heterozygosity kept (k), heterozygosity after t generations (Hₜ)
Step 1 · Find share of heterozygosity kept (k)
Use k = (1 − 1 ÷ (2 × Nₑ))ᵗ
Each
generation
keeps
1
−
1
2N
ₑ
of
the
heterozygosity,
so
t
generations
multiply
it
in
t
times.
k
=
(1 −
1
2 × Nₑ
)
t
k
=
(1
−
1
2 × 50
)¹⁰⁰
k
=
(1 −
1
100
)
100
k
=
(1 − 0.01)
100
k
=
0.99
100
k
=
0.366
Step 2 · Find heterozygosity after t generations (Hₜ)
Use Hₜ = H₀ × k
The heterozygosity left is the starting one times the share kept.
Hₜ = 0.5 × 0.366
Hₜ = 0.183
Check
Put the numbers back in. Both sides should match.
0.366
=
(1
−
1
2 × 50
)¹⁰⁰
✓
0.183 = 0.5 × 0.366 ✓
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