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 predominantly 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. Natural Selection in Humans

13. 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…

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

15. 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

16. Any Questions??

17. Original problem source unknown
A population of turtles (in Hardy Weinberg equilibrium) is variable for the length of their tail. Long tails are dominant to short tails. 25 have long tails and 75 have short tails. What is the frequency of the allele for long tails? If 400 baby turtles are born in the population, predict how many of the offspring will have long tails and how many will have short tails?
Original problem source unknown

18. Original problem source unknown
A population of 1000 individuals has 49 people that are left-handed. Assume left-handedness is homozygous recessive and that this population is in Hardy Weinberg equilibrium. Calculate p and q. What is the frequency of homozygous dominants, heterozygotes, and homozygous recessives? How many individuals in the population carry the allele for left-handedness, but are not left-handed?
Original problem source unknown

19. What is the frequency of left handedness in our classroom population?
What if you measured another population and there were 200 individuals out of 1000 who were left handed? What is the frequency of the left hand allele in this population?
What is the frequency of left handedness in our classroom population?
49/1000 = frequency of left handers (q2 )
q= √q2 = √ = 0.22
p= 1-q = = .78
Homozygous dominant = p2 = (.78)(.78)= 0.61
Heterozygous = 2pq = 2 (.78)(.22) = 0.34
Homozygous recessive = q2 = (.22)(.22)= 0.05
Need heterozygotes = (0.34)(1000)= 340

20. Assume 1 in 10,000 babies born in the US have Phenylketonuria or PKU which is a rare inherited disorder that causes an amino acid called phenylalanine to build up in the body. Individuals with the disorder inherit 2 mutated alleles for their phenylalanine hydroxylase enzyme.
What is the frequency of carriers for this allele?

21. PKU is recessive. q2 = 1/10,000 = 0.0001 q = 0.01 p= 0.99
Assume 1 in 10,000 babies born in the US have PKU. What is the frequency of carriers for this allele?
PKU is recessive.
q2 = 1/10,000 =
q = 0.01
p= 0.99
Carriers are 2pq
2 (.99)(.01) = .0198

22. 30,000 individuals in the United States have cystic fibrosis
30,000 individuals in the United States have cystic fibrosis. Assuming the population is in Hardy Weinberg equilibrium how many individuals are carrier of the disease? Assume the US population is 316,000,000.
q2 = 30,000/316,000,000 =
q = 0.009
p= 0.991
Carriers are heterozygous or 2 pq
2 (0.991)(0.009) = 0.018
(0.018)(316,000,000) =5,688,000