SME
Evolutionary Biology
Natural selection
Use this for “p = 0.5 and aa survives half as well as AA and Aa. What is p after one generation?”
Diagram
Fitness: w(AA) = 1, w(Aa) = 1, w(aa) = 0.5; w̄ = 0.875. p = 0.5 → p′ = 0.5714: Δp = +0.07143, A rises.
50 → 57 of 100 alleles are A.
This is an example. Type your own values to start a new problem.
p
Frequency of A
example
q
Frequency of a
calculated
wAA
Fitness of AA
example
wAa
Fitness of Aa
example
wₐₐ
Fitness of aa
example
w̄
Mean fitness
worked out (not typed)
p′
Next generation’s p
worked out (not typed)
Δp
Change in p
worked out (not typed)
Formulas
p
=
1
−
q
0.5 = 1 − 0.5
w̄
=
p
2
×
wAA
+
2
×
p
×
q
×
wAa
+
q
2
×
wₐₐ
0.875
=
0.5
2
×
1
+
2
×
0.5
×
0.5
×
1
+
0.5
2
×
0.5
p
′
=
p
2
× wAA +
p
×
q
× wAa
w̄
0.5714
=
0.5
2
× 1 + 0.5 × 0.5 × 1
0.875
Δp
=
p
′
−
p
+0.07143 = 0.5714 − 0.5
Assumptions
•
Random mating in a large population, with no mutation, migration or drift: only selection acts.
•
Fitnesses are relative survival from 0 to 1; w̄ is the population’s mean fitness.
•
Newborns are in Hardy–Weinberg shares p², 2pq and q² before selection acts.
•
p′ counts the A alleles among survivors: AA carry two, Aa one.
Step-by-step
WE KNOW
p = 0.5, wAA = 1, wAa = 1, wₐₐ = 0.5
FIND
frequency of a (q), mean fitness (w̄), next generation’s p (p′), change in p (Δp)
Step 1 · Find frequency of a (q)
Use p = 1 − q
The two alleles’ frequencies add to 1.
q
=
1
−
p
q
=
1
−
0.5
q
=
0.5
Step 2 · Find mean fitness (w̄)
Use w̄ = p² × wAA + 2 × p × q × wAa + q² × wₐₐ
Each genotype’s Hardy–Weinberg share times its fitness, added.
w̄
=
0.5
2
×
1
+
2
×
0.5
×
0.5
×
1
+
0.5
2
×
0.5
w̄
=
0.25
×
1
+
0.5
×
1
+
0.25
×
0.5
w̄
=
0.25
+
0.5
+
0.125
w̄
=
0.875
Step 3 · Find next generation’s p (p′)
Use p′ = (p² × wAA + p × q × wAa) ÷ w̄
The A alleles among survivors (all of AA’s, half of Aa’s), out of all survivors.
p
′
=
0.5
2
× 1 + 0.5 × 0.5 × 1
0.875
p
′
=
0.25 × 1 + 0.25 × 1
0.875
p
′
=
0.5
0.875
p
′
=
0.5714
Step 4 · Find change in p (Δp)
Use Δp = p′ − p
The change in one generation: positive when selection favors A.
Δp = 0.5714 − 0.5
Δp = +0.07143
Check
Put the numbers back in. Both sides should match.
0.5 = 1 − 0.5 ✓
0.875
=
0.5
2
×
1
+
2
×
0.5
×
0.5
×
1
+
0.5
2
×
0.5
✓
0.5714
=
0.5
2
× 1 + 0.5 × 0.5 × 1
0.875
✓
+0.07143 = 0.5714 − 0.5 ✓
Topic check
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