1. Modern Evolutionary Biology I. Population Genetics
A. Overview
Sources of Variation Agents of Change
Mutation N.S.
Recombination mutation (polyploidy)
- crossing over
- independent assortment
VARIATION

2. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
G. Hardy and W. Weinberg
1. Definitions
- Evolution: a change in the genetic structure of a population
- Population: a group of interbreeding organisms that share a common gene pool; spatiotemporally and genetically defined
- Gene Pool: sum total of alleles held by individuals in a population
- Genetic structure: Gene array and Genotypic array
- Gene/Allele Frequency: % of alleles at a locus of a particular type
- Gene Array: % of all alleles at a locus: must sum to 1.
- Genotypic Frequency: % of individuals with a particular genotype
- Genotypic Array: % of all genotypes for loci considered; must = 1.

3. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
1. Definitions
2. Basic Computations AA Aa aa
Individuals 70 80 50
(200)

4. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
1. Definitions
2. Basic Computations AA Aa aa
Individuals 70 80 50
(200)
Genotypic Array
70/200 = 0.35
80/200 = .40
50/200 = 0.25
= 1

5. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
1. Definitions
2. Basic Computations AA Aa aa
Individuals 70 80 50
(200)
Genotypic Array
70/200 = 0.35
80/200 = .40
50/200 = 0.25
= 1
''A' alleles
140
220/400 = 0.55

6. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
1. Definitions
2. Basic Computations AA Aa aa
Individuals 70 80 50
(200)
Genotypic Array
70/200 = 0.35
80/200 = .40
50/200 = 0.25
= 1
''A' alleles
140
220/400 = 0.55
'a' alleles
100
180/400 = 0.45

7. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
1. Definitions
2. Basic Computations
- Determining the Gene Array from the Genotypic Array
a. f(A) = f(AA) + f(Aa)/2 = /2 = = .55
b. f(a) = f(aa) + f(Aa)/2 = /2 = = .45
KEY: The Gene Array CAN ALWAYS be computed from the genotypic array; the process just counts alleles instead of genotypes. No assumptions are made when you do this.

8. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
1. Goal:
Describe what the genetic structure of the population would be if there were NO evolutionary change – if the population was in equilibrium.

9. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
1. Goal:
Describe what the genetic structure of the population would be if there were NO evolutionary change – if the population was in equilibrium.
For a population’s genetic structure to remain static, the following must be true:
- random mating
- no selection
- no mutation
- no migration
- the population must be infinitely large

10. Initial genotypic freq.
Modern Evolutionary Biology
I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
2.Example: AA Aa aa
Initial genotypic freq.
0.4
0.2
1.0
Gene freq.
Genotypes, F1
Gene Freq's
Genotypes, F2

11. Initial genotypic freq.
Modern Evolutionary Biology
I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
2.Example: AA Aa aa
Initial genotypic freq.
0.4
0.2
1.0
Gene freq.
f(A) = p = /2 = 0.6
f(a) = q = /2 = 0.4
Genotypes, F1
Gene Freq's
Genotypes, F2

12. Initial genotypic freq.
Modern Evolutionary Biology
I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
2.Example: AA Aa aa
Initial genotypic freq.
0.4
0.2
1.0
Gene freq.
f(A) = p = /2 = 0.6
f(a) = q = /2 = 0.4
Genotypes, F1
p2 = .36
2pq = .48
q2 = .16
= 1.00
Gene Freq's
Genotypes, F2

13. Initial genotypic freq.
Modern Evolutionary Biology
I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
2.Example: AA Aa aa
Initial genotypic freq.
0.4
0.2
1.0
Gene freq.
f(A) = p = /2 = 0.6
f(a) = q = /2 = 0.4
Genotypes, F1
p2 = .36
2pq = .48
q2 = .16
= 1.00
Gene Freq's
f(A) = p = /2 = 0.6
f(a) = q = /2 = 0.4
Genotypes, F2
After one generation with these conditions, the population equilibrates

14. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
2.Example
3. Utility:
If no populations meets these conditions explicitly, how can it be useful?

15. Initial genotypic freq. CONCLUSION:The real population is NOT in HWE.
Modern Evolutionary Biology
I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
2.Example
3. Utility:
If no populations meets these conditions explicitly, how can it be useful?
For comparison, like a “perfectly balanced coin” AA Aa aa
Initial genotypic freq.
0.5
0.2
0.3
1.0
Gene freq.
f(A) = p = /2 = 0.6
f(a) = q = /2 = 0.4
HWE expections
p2 = .36
2pq = .48
q2 = .16
= 1.00
CONCLUSION:The real population is NOT in HWE.

16. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
3. Utility:
- if a population is NOT in HWE, then one of the assumptions must be violated.
Sources of Variation Agents of Change
Mutation N.S.
Recombination Drift
- crossing over Migration
- independent assortment Mutation
Non-random Mating
VARIATION
So, if NO AGENTS are acting on a population, then it will be in equilibrium and WON'T change.

17. Modern Evolutionary Biology I. Population Genetics
A. Overview
B. The Genetic Structure of a Population
C. The Hardy-Weinberg Equilibrium Model
3. Utility:
- if a population is NOT in HWE, then one of the assumptions must be violated.
-Also, If HWCE is assumed and the frequency of homozygous recessives can be measured, then the number of heterozygous carriers can be estimated.
For example:
If f(aa) = .01, then estimate f(a) = .1 and f(A) must be .9. f(Aa) = 2(.1)(.9) = 0.18.