1. Population Genetics Hardy Weinberg

2. Population Genetics
Mendelian genetics predicts the outcome of specific matings between individuals
What about the genetics of an entire population?
Population = all individuals of one species living in a given area
Population genetics works with the entire gene pool
or all the alleles present in the whole population

3. How will alleles change in the population?
Among a population of 2000 people:
720 have blue eyes (recessive)
1280 have brown eyes (dominant)
DNA testing reveals:
320 are homozygous for Brown Eyes(BB)
960 are heterozygous for Brown Eyes (Bb)

4. Allele frequency f(B)= 960 + 320 + 320/ 4000 = 0.4
Allele frequency is the fraction:
no. of a particular allele
no. of all alleles in population
For these 2000 people, there are 4000 alleles in the gene pool:
720 bb
960 Bb
320 BB
how many B alleles?
how many b alleles?
f(B)= / 4000 = 0.4
F(b) = /4000 = 0.6

5. What happens in the next generation?
In all the matings for this generation, what is the chance that an egg with the B allele will be fertilized by a sperm with the b allele and create a person with Bb genotype?
Recall:
40% of all eggs will carry B
60% of all sperm will carry b
Recall the Rule of Multiplication:
(prob. of event a) (prob of event b)= probability of both events happening
0.4 x 0.6 = 0.24
24% of offspring Bb
*Assuming no difference between sexes
and no mating preferences!

6. What happens in the next generation?
In all the matings for this generation:
0.4 x 0.4 = 16% BB
0.4 x 0.6 x 2 = 48% Bb
Bb and bB
(rule of addition)
0.6 x 0.6 = 36% bb
40%
60% B b BB Bb bb

7. What happens in the next generation?
In all the matings for this generation:
if 4000 offspring are born:
0.4 x 0.4 = 16% BB BB
0.4 x 0.6 x 2 = 48% Bb Bb
0.6 x 0.6 = 36% bb bb
2560 Brown
1440 Blue

8. What happens in the next generation?
New allele frequencies:
If 4000 offspring
640 BB
1920 Bb
1440 bb
3200 B allele/8000 = 0.4
4800 b allele/8000 = 0.6
After 5 generations:
64,000 offspring
10,240 BB
30,720 Bb
23,040 bb
51,200 brown alleles / 128,000 = 0.4
76,800 blue alleles / 128,000 = 0.6

9. After 5 generations (or any number):
Allele frequencies do not change!
Recessive alleles are maintained in the population
*If some specific assumptions are made

10. Hardy-Weinberg equilibrium
Godfrey Hardy (mathematician) and Wilhelm Weinberg (physician) (early 1900s):
Given some assumptions, allele frequencies won’t change:
The population is large
Mating is random
No migration in or out
No mutation
No selection (no allele is advantageous)
How often in nature are ALL of these assumptions met?
Rarely, if ever. This is an “ideal” state.

11. Does Hardy-Weinberg work?
In large populations, the Hardy-Weinberg equations predict results quite well for many traits
If a population is not in equilibrium:
Allele frequencies are changing
Evolution is occurring!

12. Hardy-Weinberg equations
Allele frequency:
Let p = frequency of the dominant allele
Let q = frequency of the recessive allele
Then, p + q = 1
Genotype frequency:
p2 = frequency of homozygous dominant genotype
q2 = frequency of homozygous recessive genotype
2pq = frequency of heterozygous genotype
p2 + 2pq + q2 = 1

13. 4 Steps to solving H-W Problems
set recessives = q2
Take square root of q2
1-q = p
Plug into expanded equation
Example: 16% of the cat population is white:
q2= 0.16
square root = 0.4
= p p = 0.6
(.6) (.4) = 1
so, 36% of population is TT
48% of population is Tt

14. Another Example:
Fraggles are mythical, mouselike creatures that live beneath flower gardens.
Of the 100 fraggles in a population, 75 have green hair (FF or Ff) and 25 have grey hair (ff).
Assuming genetic equilibrium:
What are the gene frequencies of F and f?
What are the genotypic frequencies?

15. Answer to Fraggle Problems:
Gene frequencies:
q2= .25, so:
q= .5
p= .5
Genotypic frequencies
FF = .25
Ff = .5
f f = .25

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

17. Sickle cell frequency High frequency of heterozygotes
1 in 5 in Central Africans = Ss
unusual for allele with severe detrimental effects in homozygotes
1 in 100 =ss
usually die before reproductive age
Why is the s allele maintained at such high levels in African populations?
Suggests some selective advantage of being heterozygous…

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

19. Heterozygote Advantage
In tropical Africa, where malaria is common:
homozygous dominant (normal)
die or reduced reproduction from malaria: SS
homozygous recessive
die or reduced reproduction from sickle cell anemia: ss
heterozygote carriers are relatively free of both: Ss
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

20. Sickle Cell Example:
If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria?
f(ss)= .09 = q2
q= .3
P= .7
2pq= .42
so 42% of the population is resistant to malaria.

21. Using Hardy-Weinberg
Cystic Fibrosis: 1 in 1700 US Caucasian newborns have cystic fibrosis. Use an F for the normal allele and f for recessive:
What percent of the above population have cystic fibrosis?
What percent are healthy, non carriers?
What percent are carriers of cyctic fibrosis?
In a population of 1700 people, how many would you expect to be homozygous normal?
In a population of 1700 people, how many would you expect to be heterozygous?
= q2
q= .024
p= .976
P2= .9524
2pq= .0468
1700 x .9524= 1619
1700 x .0468= 80

22. Stem Cells
Nova