1. Hardy-Weinberg Equillibrium
Do Now:
Consider the following gene pool:
P = .50, Q=.50
What are f(AA), f(Aa), and f(aa) as predicted by the Hardy-Weinberg equilibrium?

2. Hardy-Weinberg: The assumptions
For the Hardy-Weinberg equation to represent a population perfectly, some very unlikely (or impossible) things would have to be true:
No mutation (DNA changes)
No natural selection
No migration into or out of the population
All members of the population breed randomly, and produce the same number of offspring.
The population is infinitely large

3. HW = no evolution
The HW equilibrium represents a theoretical state where evolution is not occurring.
That’s why allele and phenotype frequencies of a population in HW equilibrium remain constant over time
In nature however, all populations evolve.
Populations of organisms can never actually be in Hardy-Weinberg equilibrium.

4. Populations Evolve Populations of organisms change over time.
The rate of evolution is measured by the change in allele frequencies in a gene pool over time.

5. So Why Use HW?
By comparing the genotype frequencies a real-world population to the expected HW equilibrium population, we can determine what type of evolution is happening!

6. P = 0.50, Q = 0.50
What genotype is being selected for in the environment?

7. Modes of Selection

8. Mode 1: Stabilizing Selection
The moderate phenotype is selected for.
P and Q become more equal over time, since heterozygotes (2pq) are becoming more common

9. Mode 2: Disruptive Selection
The moderate phenotype is selected AGAINST.
f(Aa) will be less than predicted by the HW equation.

10. Mode 3: Directional Selection
One homozygous genotype or the other is selected for.
f(AA) or f(aa) become greater over time.

11. The Quiz The quiz is open note.
It only includes problem solving with the H-W equation.
You have the remainder of the period.