POPULATION GENETICS
I.
INTRODUCTION TO POPULATION GENETICS
A. Population
1. Localized group of individuals capable of interbreeding
a) Only members of the same species can produce viable offspring
b) Often separated from others of the same species
(1) For example,
(a) Lake
(b) Island
(c) Roads
(d) Water hole
(e) Mountain
(2) Only occasional outbreeding
B. Genetic Structure
1. Allelic frequency
a) Alleles
(1) Different forms of a gene
(a) For example; A, B, and O blood types
b) Gene pool
(1) Total aggregate of all alleles in a population
(2) All alleles for all gene loci
c) Fixed genes
(1) All individuals in a population are homozygous for a gene
(2) Invariable within a population
d) Human gene pool
(1) 100,000 genes
(2) 22% are polymorphic
(3) 200,000 alleles in the human population
e) Example of allelic frequency of flower color
(1) Population of 500 flowers
(a) 20 homozygous recessive (aa)
(b) 320 homozygous dominant (AA)
(c) 160 heterozygous (Aa)
(2) 1000 total genes for flower color in this population (diploids)
(a) a = 20 + 20 + 160 = 200
(b) A = 320 + 320 + 160 = 800
(3) 'a' has an allelic frequency of 20% (200 of 1000)
(4) 'A' has an allelic frequency of 80% (800 of 1000)
2. Genotype frequency
a) Frequency of homozygous dominant, homozygous recessive, and heterozygous
individuals in a population
b) From above example:
(1) Aa = 160 / 500 = 32%
(2) AA = 320 / 500 = 64%
(3) aa = 20 / 500 = 4%
c) Examples when allelic frequency is 50:50
(1) 100% Aa
(2) 50% aa & 50% AA
(3) 50% Aa, 25% aa & 25% AA
II. HARDY-WEINBERG EQUATION
A. Introduction
1. Mathematical description of allelic frequencies
2. Works only with non-changing populations
a) No mutations
b) No outbreeding
c) No environmental change
3. Assumes independent assortment of alleles and random fertilization
B. Two alleles
1. p + q = 1
a) p represents one allele for a gene, q represents a different allele
b) This formula gives you allelic frequencies
2. (p + q)2 = 1 = p2 + 2pq + q2
a) p2 and q2 would indicate the percentages of homozygotes for those alleles, while
2pq indicates the number of heterozygotes
b) This formula gives you genotypic frequencies
C. Three alleles
1. (p + q + r)2 = 1 = p2 + q2 + r2 + 2pq + 2pr + 2qr
2. You could use this formula for ABO blood types in humans
D. Example 1
1. Frequency for the dominant allele for red flower color is 0.8, and 0.2 for the recessive
white allele
2. The next generation should have 64% AA, 32% Aa, and 4% aa
a) 0.82 + 2(0.8)(0.2) + 0.22 = 1
E. Example 2
1. Phenylketonuria affect 1 in every 10,000 newborn caucasians. Determine the number
of carriers.
a) q2 = 0.0001
(1) 1 in 10,000 show disease, and therefore must be homozygous recessive
b) q = 1%
(1) q = square root of 0.0001 = 0.01 = 1%
c) p = 99%
(1) p + q = 1
d) Number of carriers is 1.98%
(1) 0.992 + 2 pq + 0.012 = 1
III. NON-EQUILIBRIUM
A. Hardy-Weinberg equation works if:
1. Dealing with large population
2. Population is isolated
3. There are no mutations
4. Mating is random
5. There is no selection pressure
B. Non-equilibrium occurs when:
1. There is a small population
2. There is gene flow
3. There are mutations
4. There is selective mating
5. There is selective pressure