1. Measuring Evolution of Populations

2. 5 Agents of evolutionary change
Mutation
Gene Flow
Non-random mating
Genetic Drift
Selection

3. Populations & gene pools
Concepts
a population is a localized group of interbreeding individuals
gene pool is collection of alleles in the population
remember difference between alleles & genes!
allele frequency is how common is that allele in the population
how many A vs. a in whole population

4. Evolution of populations
Evolution = change in allele frequencies in a population
hypothetical: what conditions would cause allele frequencies to not change?
non-evolving population
REMOVE all agents of evolutionary change
very large population size (no genetic drift)
no migration (no gene flow in or out)
no mutation (no genetic change)
random mating (no sexual selection)
no natural selection (everyone is equally fit)

5. Hardy-Weinberg equilibrium
Hypothetical, non-evolving population
preserves allele frequencies
Serves as a model (null hypothesis)
natural populations rarely in H-W equilibrium
useful model to measure if forces are acting on a population
measuring evolutionary change
G.H. Hardy (the English mathematician) and W. Weinberg (the German physician) independently worked out the mathematical basis of population genetics in Their formula predicts the expected genotype frequencies using the allele frequencies in a diploid Mendelian population. They were concerned with questions like "what happens to the frequencies of alleles in a population over time?" and "would you expect to see alleles disappear or become more frequent over time?"
G.H. Hardy
mathematician
W. Weinberg
physician

6. Hardy-Weinberg theorem
Counting Alleles
assume 2 alleles = B, b
frequency of dominant allele (B) = p
frequency of recessive allele (b) = q
frequencies must add to 1 (100%), so:
p + q = 1 BB Bb bb

7. Hardy-Weinberg theorem
Counting Individuals
frequency of homozygous dominant: p x p = p2
frequency of homozygous recessive: q x q = q2
frequency of heterozygotes: (p x q) + (q x p) = 2pq
frequencies of all individuals must add to 1 (100%), so:
p2 + 2pq + q2 = 1 BB Bb bb

8. H-W formulas Alleles: p + q = 1 Individuals: p2 + 2pq + q2 = 1 B b BB

9. Using Hardy-Weinberg equation
population: 100 cats
84 black, 16 white
How many of each genotype?
q2 (bb): 16/100 = .16
q (b): √.16 = 0.4
p (B): = 0.6
p2=.36
2pq=.48
q2=.16 BB Bb bb
Must assume population is in H-W equilibrium!
What are the genotype frequencies?

10. Using Hardy-Weinberg equation
p2=.36
2pq=.48
q2=.16
Assuming H-W equilibrium BB Bb bb
Null hypothesis
p2=.20
p2=.74
2pq=.10
2pq=.64
q2=.16
q2=.16
Sampled data 1:
Hybrids are in some way weaker.
Immigration in from an external population that is predomiantly homozygous B
Non-random mating... white cats tend to mate with white cats and black cats tend to mate with black cats.
Sampled data 2:
Heterozygote advantage.
What’s preventing this population from being in equilibrium. bb Bb BB
Sampled data
How do you explain the data?
How do you explain the data?

11. Application of H-W principle
Sickle cell anemia
inherit a mutation in gene coding for hemoglobin
oxygen-carrying blood protein
recessive allele = HsHs
normal allele = Hb
low oxygen levels causes RBC to sickle
breakdown of RBC
clogging small blood vessels
damage to organs
often lethal

12. Sickle cell frequency High frequency of heterozygotes
1 in 5 in Central Africans = HbHs
unusual for allele with severe detrimental effects in homozygotes
1 in 100 = HsHs
usually die before reproductive age
Sickle Cell:
In tropical Africa, where malaria is common, the sickle-cell allele is both an advantage & disadvantage. Reduces infection by malaria parasite.
Cystic fibrosis:
Cystic fibrosis carriers are thought to be more resistant to cholera:
1:25, or 4% of Caucasians are carriers Cc
Why is the Hs allele maintained at such high levels in African populations?
Suggests some selective advantage of being heterozygous…

13. Malaria
Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells 1 2 3

14. Heterozygote Advantage
In tropical Africa, where malaria is common:
homozygous dominant (normal)
die or reduced reproduction from malaria: HbHb
homozygous recessive
die or reduced reproduction from sickle cell anemia: HsHs
heterozygote carriers are relatively free of both: HbHs
survive & reproduce more, more common in population
Hypothesis:
In malaria-infected cells, the O2 level is lowered enough to cause sickling which kills the cell & destroys the parasite.
Frequency of sickle cell allele & distribution of malaria

15. Any Questions??

16. Hardy-Weinberg Lab Data
Mutation
Gene Flow
Genetic Drift
Selection
Non-random mating

17. Hardy Weinberg Lab: Equilibrium
Original population
Case #1 F5
18 individuals
36 alleles
p (A): 0.5
q (a): 0.5
total alleles = 36
p (A): (4+4+7)/36 = .42
q (a): (7+7+7)/36 = .58 AA 4 Aa 7 aa 7 AA
.25 Aa
.50 aa
.25 AA
.22 Aa
.39 aa
.39
How do you explain these data?

18. Hardy Weinberg Lab: Selection
Original population
Case #2 F5
15 individuals
30 alleles
p (A): 0.5
q (a): 0.5
total alleles = 30
p (A): (9+9+6)/30 = .80
q (a): (0+0+6)/30 = .20 AA 9 Aa 6 aa AA
.25 Aa
.50 aa
.25 AA
.60 Aa
.40 aa
How do you explain these data?

19. Hardy Weinberg Lab: Heterozygote Advantage Original population
Case #3 F5
15 individuals
30 alleles
p (A): 0.5
q (a): 0.5
total alleles = 30
p (A): (4+4+11)/30 = .63
q (a): (0+0+11)/30 = .37 AA 4 Aa 11 aa AA
.25 Aa
.50 aa
.25 AA
.27 Aa
.73 aa
How do you explain these data?

20. Hardy Weinberg Lab: Heterozygote Advantage Original population
Case #3 F10
15 individuals
30 alleles
p (A): 0.5
q (a): 0.5
total alleles = 30
p (A): (6+6+9)/30 = .70
q (a): (0+0+9)/30 = .30 AA 6 Aa 9 aa AA
.25 Aa
.50 aa
.25 AA .4 Aa .6 aa
How do you explain these data?

21. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5-1
6 individuals
12 alleles
p (A): 0.5
q (a): 0.5
total alleles = 12
p (A): (4+4+2)/12 = .83
q (a): (0+0+2)/12 = .17 AA 4 Aa 2 aa AA
.25 Aa
.50 aa
.25 AA
.67 Aa
.33 aa
How do you explain these data?

22. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5-2
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
total alleles = 10
p (A): (0+0+4)/10 = .4
q (a): (1+1+4)/10 = .6 AA Aa 4 aa 1 AA
.25 Aa
.50 aa
.25 AA Aa .8 aa .2
How do you explain these data?

23. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5-3
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
total alleles = 10
p (A): (2+2+2)/10 = .6
q (a): (1+1+2)/10 = .4 AA 2 Aa 2 aa 1 AA
.25 Aa
.50 aa
.25 AA .4 Aa .4 aa .2
How do you explain these data?

24. Hardy Weinberg Lab: Genetic Drift
Original population
Case #4 F5
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
AA Aa aa p q AA
.25 Aa
.50 aa
.25
How do you explain these data?

25. Any Questions??