1. The Hardy-Weinberg Theorem
Hardy WHO???

2. What the heck is the Hardy-Weinberg Theorem?
States that… frequencies of alleles & genotypes in a population’s gene pool remain constant between generations
Theorem shows how Mendelian inheritance can be preserved in populations that are NOT evolving BUT…
It also shows how evolutionary change can happen over time
What the heck is the Hardy-Weinberg Theorem?

3. In order for a stable H-W equilibrium to work…
You need a VERY large population
There can be NO gene flow
There can be NO mutations
There MUST BE random mating
There can be NO natural selection
In order for a stable H-W equilibrium to work…

4. Some questions to ponder…
How many populations remain in a stable H-W equilibrium for a long time?? Are the five conditions ever met long- term? What types of situations will throw a population OUT of H-W equilibrium?
Some questions to ponder…

5. If populations show the SAME allele frequencies generation to generation, they are said to be in a state of Hardy- Weinberg Equilibrium
Let’s take a look at the H-W Equation:
p2 + 2pq + q2 = 1

6. p2 + 2pq + q2 = 1 So… p2 = homozygous dominant frequency (AA)
p = dominant (A) allele frequency  q = recessive (a) allele frequency 
So… p2 = homozygous dominant frequency (AA) 
q2 = homozygous recessive frequency (aa) 2pq = heterozygous frequency (Aa)
p2 + 2pq + q2 = 1

7. If we know the frequency of one of the alleles, we can calculate the frequency of the other allele:
p + q = 1
p = 1 – q
q = 1 - p

8. Guess what we’re doing next… (da da da duuum!)
The hardest part of the H-W Theorem is applying it to mathematical problem sets.
Guess what we’re doing next… (da da da duuum!)
Learn how to weave baskets
Make daisy chains
Draw stick men
Do some H-W problems!

9. 1 in 1700 US Caucasian newborns have cystic fibrosis
1 in 1700 US Caucasian newborns have cystic fibrosis. C for normal is dominant over c for cystic fibrosis.
1. When counting the phenotypes in a population why is cc the most significant?
When dominance is the case. CC and Cc are both normal. The recessive genotype cc is the only one that will result in cystic fibrosis.
2. What percent of the above population have cystic fibrosis (cc or q2)?
1/1700 are cc. This is or 0.059% of the population.
Problem #1

10. ALLELE FREQUENCY CALCULATIONS Now we know q2 we can calculate q
ALLELE FREQUENCY CALCULATIONS Now we know q2 we can calculate q. q = square root of = or 2.4% We know that p = 1 – q so… p = 1 – = or 97.6%

11. GENOTYPE FREQUENCY CALCULATIONS CC homozygous dominant (normal) = p2 = (0.976)2 = or 95.3% Cc heterozygous dominant (carrier) = 2pq = 2(0.976)(0.024) = or 4.68% How many of the 1700 of the population are homozygous normal? (0.953)(1700) = ~1620 individuals How many of the 1700 in the population are heterozygous carrier? (0.0468)(1700) = ~79 individuals