POPULATION GENETICS
- INTRODUCTION TO POPULATION GENETICS
- Population
- Localized group of individuals capable of interbreeding
- Only members of the same species can produce viable offspring
- Often separated from others of the same species
- For example,
- Lake
- Island
- Roads
- Water hole
- Mountain
- Only occasional outbreeding
Genetic Structure
Allelic frequency
- Alleles
- Different forms of a gene
- For example; A, B, and O blood types
- Gene pool
- Total aggregate of all alleles in a population
- All alleles for all gene loci
Fixed genes
All individuals in a population are homozygous for a gene
Invariable within a population
Human gene pool
100,000 genes
22% are polymorphic
200,000 alleles in the human population
Example of allelic frequency of flower color
Population of 500 flowers
- 20 homozygous recessive (aa)
- 320 homozygous dominant (AA)
- 160 heterozygous (Aa)
1000 total genes for flower color in this population (diploids)
- a = 20 + 20 + 160 = 200
- A = 320 + 320 + 160 = 800
'a' has an allelic frequency of 20% (200 of 1000)
'A' has an allelic frequency of 80% (800 of 1000)
Genotype frequency
- Frequency of homozygous dominant, homozygous recessive, and heterozygous individuals in a population
- From above example:
Aa = 160 / 500 = 32%
AA = 320 / 500 = 64%
aa = 20 / 500 = 4%
Examples when allelic frequency is 50:50
100% Aa
50% aa & 50% AA
50% Aa, 25% aa & 25% AA
HARDY-WEINBERG EQUATION
- Introduction
Mathematical description of allelic frequencies
Works only with non-changing populations
- No mutations
- No outbreeding
- No environmental change
Assumes independent assortment of alleles and random fertilization
Two alleles
p + q = 1
- p represents one allele for a gene, q represents a different allele
- This formula gives you allelic frequencies
(p + q)2 = 1 = p2 + 2pq + q2
- p2 and q2 would indicate the percentages of homozygotes for those alleles, while 2pq indicates the number of heterozygotes
- This formula gives you genotypic frequencies
Three alleles
(p + q + r)2 = 1 = p2 + q2 + r2 + 2pq + 2pr + 2qr
You could use this formula for ABO blood types in humans
Example 1
Frequency for the dominant allele for red flower color is 0.8, and 0.2 for the recessive white allele
The next generation should have 64% AA, 32% Aa, and 4% aa
- 0.82 + 2(0.8)(0.2) + 0.22 = 1
Example 2
Phenylketonuria affect 1 in every 10,000 newborn caucasians. Determine the number of carriers.
- q2 = 0.0001
- 1 in 10,000 show disease, and therefore must be homozygous recessive
- q = 1%
- q = square root of 0.0001 = 0.01 = 1%
p = 99%
p + q = 1
Number of carriers is 1.98%
0.992 + 2 pq + 0.012 = 1
NON-EQUILIBRIUM
- Hardy-Weinberg equation works if:
Dealing with large population
Population is isolated
There are no mutations
Mating is random
There is no selection pressure
Non-equilibrium occurs when:
There is a small population
There is gene flow
There are mutations
There is selective mating
There is selective pressure