1. Models of selection Photo: C. Heiser Capsicum annuum
How can we predict the speed of evolutionary change due to natural selection?
- genotype and allele frequencies
null models (no selection)
Photo: C. Heiser
Capsicum annuum

2. Foré genetics A population living in Papua New Guinea
Young women and older women surveyed for genotypes at one locus (2 alleles: M or V).
Genotype MM MV VV
juvenile
adult

3. Genotype frequencies
Consider a diploid population with two alleles at one locus, A and a
[NOTE: for this class, large letters DO NOT indicate dominance, just another allele!]
At time t,
C(t) = frequency of AA individuals
H(t) = frequency of Aa individuals
S(t) = frequency of aa individuals
C(t) + H(t) + S(t) = 1
(sum of the frequencies of all possibilities will always equal 1)

4. Genotype frequencies
The frequency of the A allele in this population, p(t), is given by the frequency of AA plus half of the frequency of Aa (since only half of the alleles in the heterozygote are A):
Similarly, the frequency of the a allele in this population, q(t) is given by the frequency of aa plus half of the frequency of Aa.
p(t) + q(t) = C(t) + H(t) + S(t) = 1
The sum of the frequencies of the two possible alleles equals 1.

5. Find the genotype and allele frequencies for this example.
Number AA = Frequency =
Number Aa = Frequency =
Number aa = Frequency =
 Total = Total frequency = 1
Number A = Frequency =
Number a = Frequency =
Total = Total frequency = 1

6. What happens to these frequencies over one generation?
We will assume none of the forces of evolution are operating:

7. Haploid stage (gametes)
Model without selection
Haploid stage (gametes)
frequency:
f(A) = p(t) = C(t) + 1/2 H(t)
f(a) = q(t) = 1 - p(t)
Meiosis
Random
union
Meiosis:
AA adults produce 100% A gametes
Aa adults produce 50% A gametes
aa adults produce 0% A gametes
Diploid stage
frequency: AA Aa aa
C(t) H(t) S(t)

8. Haploid stage (gametes)
Model without selection
Haploid stage (gametes)
frequency:
f(A) = p(t)
f(a) = q(t)
Meiosis
Random
union
If adults are equally fertile, then gamete frequencies equal the allele frequencies in the adults that produce them.
Diploid stage
frequency: A a
p(t) q(t)

9. Random mating AA (prob p2) Allele A p Aa (prob pq) aA (prob qp) q
Meiosis
Random
union
Random mating
Egg is drawn
Sperm is drawn
Offspring p q
Allele A
Allele a
AA (prob p2)
Aa (prob pq)
aA (prob qp)
aa (prob q2)

10. Random mating AA (prob p2) Allele A p Aa (prob pq) aA (prob qp) q
Meiosis
Random
union
Random mating
Egg is drawn
Sperm is drawn
Offspring p q
Allele A
Allele a
AA (prob p2)
Aa (prob pq)
aA (prob qp)
At time t+1
C(t+1) = freq(AA) =p2
H(t+1) = freq(Aa) = 2pq
S(t+1) = freq(aa) = q2
aa (prob q2)

11. Effect of random mating
At time t+1
C(t+1) = freq(AA) =p2
H(t+1) = freq(Aa) = 2pq
S(t+1) = freq(aa) = q2

12. Allele frequencies in next generation
At time t+1
C(t+1) = freq(AA) =p2
H(t+1) = freq(Aa) = 2pq
S(t+1) = freq(aa) = q2
so, p(t+1) = C(t+1) + 1/2 H(t+1)
= p2 + 2pq / 2
= p(p + q)
= p
Likewise, q(t+1) = q

13. Allele frequencies do not change in the absence of selection and with random mating.
What if one allele were dominant?

14. Using Hardy-Weinberg to detect selection
frequency:
f(A) = p(t)
f(a) = q(t)
Meiosis
Random
union
Adults
Selection
Selection affects aa genotype
f(AA) = p2
f(Aa) = 2pq
f(aa) < q2

15. H-W in practice We can use this to detect natural selection
In practice, we do not know p and q
We can calculate p and q for the adults.
Call these p' and q' (since they are after selection)
p' = f(AA) + 1/2 f(Aa)
q' = f(aa) + 1/2 f(Aa)
After selection p ≠ p‘ and q ≠ q'.
and f(AA) ≠ p'2
f(Aa) ≠ 2p'q'
f(aa) ≠ q'2
We can use this to detect natural selection

16. H-W example p = f(A) = 0.5, q = f(a) = 0.5 If we have 100 zygotes:
p2 * 100 = 25 AA; f(AA) = 0.25
2pq * 100 = 50 Aa; f(Aa) = 0.5
q2 * 100 = 25 aa; f(aa) = 0.25
10 aa die.
Adult population:
25 AA; f’(AA) = 25 / 90 = 0.28
50 Aa; f’(Aa) = 50 / 90 = 0.56
15 aa; f’(aa) = 15 / 90 = 0.17
p’ = f’(A) = (0.56 / 2) = 0.56
q’ = (0.56 / 2) = 0.45

17. H-W example p = f(A) = 0.5, q = f(a) = 0.5 If we have 100 zygotes:
p2 * 100 = 25 AA; f(AA) = 0.25
2pq * 100 = 50 Aa; f(Aa) = 0.5
q2 * 100 = 25 aa; f(aa) = 0.25
20 aa die.
Adult population:
25 AA; f’(AA) = 25 / 80 = 0.313
50 Aa; f’(Aa) = 50 / 80 = 0.625
5 aa; f’(aa) = 15 / 80 = 0.063
p’ = f’(A) = (0.556 / 2) = 0.625
q’ = (0.556 / 2) = 0.375
Expected numbers of each genotype if population is in H-W?
exp(AA) = p’2 * 90 = 31.3
exp(Aa) = 2p’q’ * 90 = 37.5
exp(aa) = 2q’2 * 90 = 11.3
Use 2 test to check:
2 =  (obs – exp)2 / exp
= (25 – 31.3)2 / 27.8
+ (50 – 37.5)2 / 44.5
+ (15 – 11.3)2 / 18.2 =
Critical value: 3.84

18. H-W problem

19. References, readings, and study questions
See sections 6.1 to 6.3 in the text, (3rd edition: 5.1 to 5.3) and answer question 1.