1. POPULATION
GENETICS
G. H. Hardy
Wilhelm Weinberg

2. Population Genetics = the study of heredity within a given population
Population = a group of individuals of the same species occupying a given area at a certain time
Genetics = the study of heredity
Population Genetics = the study of heredity within a given population

3. Species
Species = a group of populations whose individuals have the potential to interbreed and produce fertile offspring in nature
Members of a population are more likely to breed within the population, so genes tend to stay in the same population for generations

4. Gene pool- all the genes in all the members of a population at one time. Immigration increases the gene pool and emigration decreases it.

5. Natural Selection is the basis for… Evolution
Evolution is the gradual change in the frequency of genes in a gene pool
Natural Selection is the basis for… Evolution

6. Each beak specialized for unique food source
Natural Selection
Natural selection  some traits are better equipped for survival in a given environment
They are “naturally selected” for through survival of the fittest
Example of natural selection:
Darwin’s finches
Different beak shapes for different food sources.
Each beak specialized for unique food source

7. The Importance of Variation
Variation among organisms may be in
Physical appearance
Metabolism
Fertility
Mode of reproduction
Behavior
Variation amongst organisms is dependent on variation in genes.
Albatross’ elaborate and unique courtship behaviour

8. Variation as a Means for Survival
Certain genotypes are better equipped than others for survival
Sexual reproduction ensures variations get passed on to offspring
Leads to natural selection for individuals in given conditions.

9. Studying Human Genetic Traits
Studying humans is problematic
Few offspring
Slow reproductive time
Environment affects phenotype
How do we study human populations then?
POPULATION SAMPLING!!

10. Electro-shocking for population sample
Population Sampling
Select a small sample of individuals from the population
Find the gene frequencies for a particular genetic trait in that sample
Apply gene frequencies to the whole population
This allows scientists to analyze
trends over time
Electro-shocking for population sample

11. Gene Frequencies
Sometimes, gene frequencies are associated with certain populations
Swedish people mostly blond haired and blue eyed
We say frequency of blond hair and blue eyes is HIGH in this population

12. Gene Frequencies
The recessive allele that causes Sickle cell anemia has different frequencies throughout Africa. The heterozygous condition provides immunity to malaria.

13. HARDY WEINBERG EQUILIBRIUM!!!
When studying GENE FREQUENCIES…
Must use…
HARDY WEINBERG EQUILIBRIUM!!!

14. Hardy Weinberg Equilibrium
Under specific conditions, after one generation of random mating, gene and allele frequencies (gene pool) will be stable from generation to generation.

15. Conditions necessary for HW equilibrium:
Large population
Random mating
No genetic drift
No gene flow - migration
No natural selection
No mutations
Animation of gene frequencies changing

17. Hardy-Weinberg Equilibrium Formulae
Alleles: p + q = 1
Genotypes: p2 +
2pq + q2 = 1 A a AA Aa aa
p = frequency of dominant allele in a population. (A)
q = frequency of recessive allele in a population. (a)
p2 = frequency of homozygous dominant genotype. (AA)
2pq = frequency of heterozygous genotype. (Aa)
q2 = frequency of homozygous recessive genotype. (aa)

18. p + q = 1 p2 + 2pq + q2 = 1 A a AA Aa aa Alleles: Genotypes:
These formulae allow scientists to determine whether evolution has occurred.
Changes in the gene frequencies over time indicates evolution.
No change in the gene frequencies indicates stability (no evolution).
The law states that if no evolution is occurring, then an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals.

19. FINDING THE HARDY-WEINBERG EQUILIBRIUMrr Rr r RR R r R
(p)
(q)
(p)
(p2)
(pq)
R = Tongue rolling
r = non-tongue rolling
(q)
(pq)
(q2) p2 +
2pq + q2 = 1
p = dominant allele (R)
q = recessive allele (r)

20. rr (q2) Rr (pq) r (q) RR (p2) R (p) 0.6 0.4 0.6 0.36 0.24 0.24 0.16
Let p  DOMINANT ALLELE frequency= 0.6
Let q  RECESSIVE ALLELE frequency= 0.4 rr
(q2) Rr
(pq) r
(q) RR
(p2) R
(p)
0.6
0.4
0.6
0.36
0.24
0.24
0.16
0.4

21. p2 + 2pq + q2 = 1 rr (q2) 0.16 Rr (pq) 0.24 r (q) 0.4 RR (p2) 0.36 R
0.6
What do the numbers add up to?
= 1
If we express this in algebra form, we get…
p2 + 2pq + q2 = 1
Extra practice: Visit the following sites. Read the brief explanations and answer the questions. Answer keys are provided.

22. If we know the frequency of the __________allele (p) and the frequency of the ____________ allele (q), then we can calculate the frequency of the genotypes _____ (p2), _____ (2pq) and ____ (q2) and vice versa.
dominant
recessive AA Aa aa

23. Step 1: Write down your known and unknowns
Steps to Solving Hardy Weinberg Questions
Step 1: Write down your known and unknowns
Step 2: Follow these steps:
(Write this on the left side of your page.) aa q2
= q2 a q A p
= 1 - q p2
= p  p AA Aa
2pq
= 2  p  q

24. 7 out of 9 mice are brown. Brown is dominant to white.
Example: Find all the Hardy Weinberg frequencies for the following.
7 out of 9 mice are brown. Brown is dominant to white.
Solution:
Since brown is the dominant trait, start by subtracting 7 from 9 to get the number of white mice, or the recessive individuals, which is 2. aa q2
= 2/9 =
= q2 a q
= 0.2222 = A p
= 1 - q = = AA p2
= p  p
= x = Aa
2pq
= 2  p  q
= 2 x x =
= 0.50

25. Example 1
In a population of flying pigs, some pigs have stubby wings. If there were 50 pigs in total and 8 have the recessive trait of stubby wings, how many are homozygous for the dominant allele?
Find: AA
= p2
= 0.16
Given: aa
= q2
= 8_ 50
= q2 q
= 0.16
= 0.4
= 1 – 0.4 p
= 1 - q
= 0.6
= 0.36 p2
= 0.62
0.36 x 50
= 18 pigs

26. Example 2
If the frequency of the recessive allele for blue eyes in a population is 30% and there are 500 individuals in the population, how many blue-eyed people would there be?
Find: aa
= q2
Given: a
= q
= 0.30
q2= (0.30)2 = 0.09
0.09 x 500
= 45 blue-eyed people

27. 160 students are heterozygous
Example 3
The dominant allele T controls the ability to taste PTC. Individuals with the T allele find PTC bitter, while tt individuals find it tasteless. In a sample of 320 students, 240 were tasters. Determine the number of heterozygous individuals in this population.
Find:
Tt = 2pq
Given:
TT and Tt =
p2 + 2pq =
240 students
q2 = = 80 =
= 0.25
q =
√q2 =
√0.25 =
0.5
p =
1-q = =
0.5
2pq =
2 (0.5) (0.5) =
0.5
0.5 x 320 =
160 students are heterozygous

28. at least one recessive allele?
Example 4
In cats, yellow eyes are controlled by a dominant allele and green eyes are the recessive trait. If 90 cats out of 250 have green eyes, how many cats have
at least one recessive allele?
Aa = 2pq AND
aa = q2
Find:
Given:
90 250
q2 =
=0.36
q =
√q2 =
√0.36 =
0.6
p =
1-q =
1- 0.6
= 0.4
2pq =
2 (0.4) (0.6) =
0.48
2pq + q2 = =
0.84
0.84 x 250 =
210 cats

29. Try this link for an online quiz for Hardy-Weinberg equilibrium
5 questions
Another quiz

30. Bozeman Hardy Weinberg Equation 9:23 Why do we always start with q2?
Bozeman:
Solving Hardy Weinberg Problems 11:07