1. Population Genetics: Selection and mutation as mechanisms of evolution
Population genetics: study of Mendelian genetics at the level of the whole population.

2. Hardy-Weinberg Equilibrium
The Hardy-Weinberg Equilibrium Principle allows us do this. It enables us to predict allele and genotype frequencies from one generation to the next in the absence of evolution.

3. Hardy-Weinberg Equilibrium
Assume a gene with two alleles A and a with known frequencies (e.g. A = 0.6, a = 0.4.)
There are only two alleles in the population so their frequencies must add up to 1.

4. Hardy-Weinberg Equilibrium
Using allele frequencies we can predict the expected frequencies of genotypes in the next generation.
With two alleles only three genotypes are possible: AA, Aa and aa

5. Hardy-Weinberg Equilibrium
Assume alleles A and a are present in eggs and sperm in proportion to their frequency in population (i.e. 0.6 and 0.4)
Also assume that sperm and eggs meet at random (one big gene pool).

6. Hardy-Weinberg Equilibrium
Then we can calculate expected genotype frequencies.
AA: To produce an AA individual, egg and sperm must each contain an A allele.
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7. Hardy-Weinberg Equilibrium
Similarly, we can calculate frequency of aa.
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8. Hardy-Weinberg Equilibrium
Probability of Aa is given by probability sperm contains A (0.6) times probability egg contains a ___________________________________________________

9. Hardy-Weinberg Equilibrium
But, there’s a second way to produce an Aa individual (egg contains A and sperm contains a). Same probability as before: ________________
Hence the overall probability of Aa = ____________________________

10. Hardy-Weinberg Equilibrium
Genotypes in next generation:
AA = _________
Aa = __________
aa= ____________
These frequencies add up to one.

11. Hardy-Weinberg Equilibrium
General formula for Hardy-Weinberg.
Let p= frequency of allele A and q = frequency of allele a.
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12. Conclusions from Hardy-Weinberg Equilibrium
If the allele frequencies in a gene pool with two alleles are given by p and q, the genotype frequencies will be given by p2, 2pq, and q2.

13. Working with the H-W equation
You need to be able to work with the Hardy-Weinberg equation.
For example, if 9 of 100 individuals in a population suffer from a homozygous recessive disorder can you calculate the frequency of the disease-causing allele? Can you calculate how many heterozygotes are in the population?

14. Working with the H-W equation
p2 + 2pq + q2 = 1. The terms in the equation represent the frequencies of individual genotypes. [A genotype is possessed by an individual organism so there are two alleles present in each case.]
P and q are allele frequencies. Allele frequencies are estimates of how common alleles are in the whole population.
It is vital that you understand the difference between allele and genotye frequencies.

15. Working with the H-W equation
9 of 100 (frequency = 0.09) of individuals are homozygous for the recessive allele. What term in the H-W equation is that equal to?

16. Working with the H-W equation
It’s q2.
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If _________________. Now plug p and q into equation to calculate frequencies of other genotypes.

17. Working with the H-W equation
_______________________________
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To calculate the actual number of heterozygotes simply multiply 0.42 by the population size = __________________.

18. Other examples of working with HW equilibrium: is a population in HW equilibrium?
In a population there are 100 birds with the following genotypes:
49 AA
32 Aa
24 aa
How would you demonstrate that this population is not in Hardy Weinberg equilibrium

19. Three steps Step 1: calculate the allele frequencies.
Step 2: Calculate expected numbers of each geneotype (i.e. figure out how many homozygotes and heterozygotes you would expect.)
Step 3: compare your expected and observed data.

20. Step 1 allele frequencies
Step 1. How many “A” alleles are there in total?
49 AA individuals = 98 “A” alleles (because each individual has two copies of the “A” allele)
32 Aa alleles = 32 “A” alleles
Total “A” alleles is =130.

21. Step 1 allele frequencies
Total number of “a” alleles is similarly calculated as 2* = 80
What are allele frequencies?
Total number of alleles in population is = 200 (or you could calculate it by multiplying the number of individuals in the population by two 100*2 =200)

22. Step 1 allele frequencies
Allele frequencies are:
A = 120/200= 0.6. Let p = 0.6
a = 80/200 = 0.4. Let q = 0.4

23. Step 2 Calculate expected number of each genotype
Use the Hardy_Weinberg equation
p2 + 2pq + q2 = 1 to calculate what expected genotypes we should have given these observed frequencies of “A” and “a”
Expected frequency of AA = p2 = 0.6 * 0.6 = 0.36
Expected frequency of aa = q2 = 0.4*0 .4 =0.16
Expected frequency of Aa = 2pq = 2*.6*.4 = 0.48

24. Step 2 Calculate expected number of each genotype
Convert genotype frequencies to actual numbers by multiplying by population size of 100
AA = 0.36*100 = 36
aa = 0.16*100 = 16
Aa = 0.48*100 = 48