1. 4-Population Genetics Notes

2. Population genetics: focuses on genetic changes within an interbreeding population.
Gene pool = _______________________________
Allele Frequency =
all the genes in a population
Shows frequency of each allele in the gene pool (0.625 B, b)

3. Changing the Gene Pool
Microevolution: a change in a population’s gene pool over a succession of generations (a brief period of geologic time). Microevolution may be due to natural selection or other factors such as genetic drift.
Genetic Drift: random change of allele frequencies in a small population due to chance , not natural selection.

4. Founder Effect:______________________ ___________________________
Example: ___________________________
migration of a small group to a new area can lead to genetic drift.
Amish in America (extra digits)

5. Bottleneck Effect: ________ ___________
Example: _________________
an event that drastically lowers the population numbers
endangered species

7. Genetic Equilibrium: Allele frequencies are not changing = ________________
no evolution

9. The Hardy Weinberg principle describes a hypothetical situation in which there is no change in the gene pool (frequencies of alleles), hence no evolution.
The frequencies of alleles will remain unchanged generation after generation (= NO EVOLUTION) if the following conditions are met:
Large population
Random mating
No mutation
No migration
No natural selection

10. The Hardy-Weinberg Equation:
To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation.
p2 + 2pq + q2 = 1
and
p + q = 1
Using the equation:
p = the dominant allele frequency
(represented here by A).
q = the recessive allele frequency (represented here by a)
For a population in genetic equilibrium: p + q = 1.0
(The sum of the frequencies of both alleles is 1.)

11. The Hardy-Weinberg Equation:
(p + q)2 = 1
so p2 + 2pq + q2 = 1
The three terms of this binomial expansion
indicate the frequencies of the three genotypes:
p2 = genotype frequency of AA (homozygous dominant)
2pq = genotype frequency of Aa(heterozygous)
q2 = genotype frequency of aa (homozygous recessive)

12. SAMPLE PROBLEM #1:
In pigs, black coat is recessive to white.
What is the percentage of heterozygotes in this
population?
Calculate q2: Count the individuals that are
homozygous recessive in the illustration to the
left. Calculate the frequency of the total
population they represent. This is q2.
q2= 4/16 = 0.25

13. SAMPLE PROBLEM #1:
In pigs, black coat is recessive to white.
What is the percentage of heterozygotes in this
population?
2. Find q: Take the square root of q2 to obtain q,
the frequency of the recessive allele.
q = √0.25 = 0.5

14. SAMPLE PROBLEM #1:
In pigs, black coat is recessive to white.
What is the percentage of heterozygotes in this
population?
3. Find p: The sum of the frequencies of both
alleles = 100%, p + q = 1. You know q, so what
is p, the frequency of the dominant allele?
p = 1 – q
p = = 0.5

15. SAMPLE PROBLEM #1:
In pigs, black coat is recessive to white.
What is the percentage of heterozygotes in this
population?
4. Find 2pq: The frequency of the heterozygotes
is represented by 2pq. Multiply by 100 to
calculate the percent of the population that is
heterozygous for white coat:
2pq = 2(0.5)(0.5)
2pq = 0.5
0.5 * 100=
50% Heterozygous