Genotype & Allele Frequency

Exam 2 lecture 7

UIC BioS 101 Nyberg

Evolutionary Processes affecting
Allele Frequency &
Allele Frequency Change

Change of allele abundance/frequency is the basic evolutionary change.

Speaker Notes:

What does frequency mean to you?

Exam 2 lecture 7

UIC BioS 101 Nyberg

Reading Assignment

Chapter 25 Evolutionary Processes is the reading assigned for this lecture.
Genetic Drift and the neutral theory of evolution will be in next lecture.
Frequency is used in the sense of a proportion, frequency takes values between 0 and 1.

Speaker Notes:

Exam 2 lecture 7

UIC BioS 101 Nyberg

Counts and Frequencies

The increase in abundance is a true measure of fitness (= winning).
Most of population models deal with frequency, ranging from 0 to 1, of two or more types, i.e., relative abundance.
- We will continue in that tradition.
- Frequency = 0 means type no longer exists

Speaker Notes:

The maximum frequency is 1. If a gene has a frequency of 1 there are no other types in the population and the gene is said to be ‘fixed’ or monomorphic.

Exam 2 lecture 7

UIC BioS 101 Nyberg

aa = 6

Aa = 7

AA = 4

N = 17

Genotype frequencies are determined by counting the number of each type and dividing by the total.

Speaker Notes:

Check my counts.

Exam 2 lecture 7

UIC BioS 101 Nyberg

Freq (AA) =# of AA/N = 4/17 = 0.24

Freq (Aa) = count of Aa/total =7/17 = 0.41

Freq (aa) = # of aa divided by total =6/17 = 0.35

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1.00

The SUM of the frequencies of all the genotypes should equal 1.

Genotype frequency definitions

Speaker Notes:

These are DEFINITIONS. If the data allows you to calculate the definition, always use it.

Exam 2 lecture 7

UIC BioS 101 Nyberg

Phenotype frequencies

If the A locus has classic dominance, there are only two phenotypes, which we can symbolize as D and R.
The frequency of the dominant phenotype is 0.24 + 0.41 = 0.65; the frequency of the homozygous recessive phenotype is the same as the homozygous recessive genotype = 0.35

Speaker Notes:

Again when you know the numbers of each genotype or the frequencies of the genotypes USE them, not a formiula.

Exam 2 lecture 7

UIC BioS 101 Nyberg

Allele (or Gene) Frequencies

If the genotype frequencies are known, the allele frequencies are defined as:

Freq of A allele = Fr(AA) + ½•Fr(Aa) =0.24 + 0.205 =0.445 homozygous + ½ of heterozygotes (with allele)

Freq a allele = Fr(aa) + ½•Fr(Aa) =0.35 +0.205 =0.555

Speaker Notes:

Know the definition. Use the definition if it is possible to use the definition.

Exam 2 lecture 7

UIC BioS 101 Nyberg

AA = 4 = 8 copies of A

Aa = 7 = 7 copies of A & 7 copies of a

aa = 6 = 12 copies of a

15 A s 19 a s

Instead of 17 individuals think of the population as 34 (2x17) genes

Speaker Notes:

Each individual is diploid so if N is the number of individuals then 2N is the number of genes.

Exam 2 lecture 7

UIC BioS 101 Nyberg

RANDOM MATING

Random mating means that mating pairs are drawn independently from the available pool of individuals.
Independent means

Pr(AA x aa mating) = Pr(AA)•Pr(aa), etc.

With some algebra I will not show, it can be shown that random mating is equal to a random pick of alleles from a gene pool.

Speaker Notes:

Given there are 3 genotypes, list ALL 9 possible matings.

Exam 2 lecture 7

UIC BioS 101 Nyberg

RANDOM MATING

Let p = freq (A)

Let q = freq (a)

If A & a are

the only alleles

Then p + q = 1.

Expected offspring frequencies

If the previous generation

mated randomly.

Freq(AA) = freq(A)•freq(A) =

(freq(A))2 = p•p = p2

Freq(Aa) = p•q

Freq(aA) = q•p

Freq(aa) = q•q = q2

Both heterozygotes (Aa and aA) are grouped together as Aa so freq(Aa) = p•q + q•p = 2pq

Speaker Notes:

To do the proof let P = freq of AA, let H = freq of Aa and Q = freq of aa.

Distinguish between P (capital p) and p (lower case p).

Hardy-Weinberg

Exam 2 lecture 7

UIC BioS 101 Nyberg

Speaker Notes:

Exam 2 lecture 7

UIC BioS 101 Nyberg

IF Mating is Random, then

Expected genotype frequencies can be predicted from allele frequencies.

e.g. freq(IBi) = 2•freq(IB)•freq(i)

If genotype frequencies can be predicted then phenotype frequencies are also predicted.

e.g. freq(B blood type) = 2•freq(IB)•freq(i)

+ freq(IB)• freq(IB)

Speaker Notes:

This is known as the Hardy-Weinberg equation and is sometimes referred to as an ‘equilibrium’, i.e. freqAA = p2, freq(Aa) =2pq and freq(aa) =q2.

Exam 2 lecture 7

UIC BioS 101 Nyberg

A worked example using ABO

Given freq(IA) = 0.3, freq(IB) = 0.1 so freq(i) = 1 -0.3 -0.1 = 0.6 (as they must add to 1)
THEN the expected frequency of the IBi genotype (using Hardy-Weinberg) is 2•.1•.6 = 0.12
The expected frequency of the B phenotype is expected freq(IBi) + freq(IBIB) = 2•.1•.6 + .1•.1 = 0.13 or 13%

Speaker Notes:

Calculate the expected frequencies of each genotype and check if they add up to 1.

Exam 2 lecture 7

UIC BioS 101 Nyberg

Allele frequencies can be estimated from phenotype frequencies (but only if one assumes mating is random).

If the frequency of the RH- phenotype is 0.4, then the allele frequency is estimated to be (0.4)½ = 0.63.

If the frequency of the RH- phenotype is 0.02, then the allele frequency is estimated to be (0.02)½ = 0.14.

Allele frequencies are fairly high even for very rare recessive phenotypes.

Speaker Notes:

X½ is the square root of x.

Exam 2 lecture 7

UIC BioS 101 Nyberg

Directional Selection of Alleles

If an allele increases in frequency no matter what its initial frequency, it is considered to be directionally selected.
The most clear examples of directional selection are examples of DDT resistance in mosquitoes and antibiotic resistance in human diseases.

Speaker Notes:

Exam 2 lecture 7

UIC BioS 101 Nyberg

Phenotype Selection Patterns

Directional selection at phenotypic level results when survival increases (or decreases) in increasing phenotypic value.
Stabilizing selection is the pattern where the middle phenotypic values have the greatest survival and the extremes have the least.

Speaker Notes:

Stabilizing selection of phenotype does NOT automatically mean stabilizing selection of alleles.

Can you create a counter example.

Directional Selection

Exam 2 lecture 7

UIC BioS 101 Nyberg

Speaker Notes:

An extreme weather event resulted in directional selection.

Gene Flow
Reduces differentiation due to drift

Exam 2 lecture 7

UIC BioS 101 Nyberg

Speaker Notes:

Mutation increases fitness over many generations in a bacterial population

Split a population into parts
Freeze most parts
Allow the unfrozen part to grow for many generations
Compare evolved populations to the original population

Exam 2 lecture 7

UIC BioS 101 Nyberg

Speaker Notes:

Exam 2 lecture 7

UIC BioS 101 Nyberg

Speaker Notes:

Increase in fitness
Each step is the result of a mutation that increases fitness

Exam 2 lecture 7

UIC BioS 101 Nyberg

Speaker Notes:

Inbreeding
Changes genotype frequency, NOT allele frequency

Exam 2 lecture 7

UIC BioS 101 Nyberg

Speaker Notes:

Selfing over four generations

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UIC BioS 101 Nyberg

Speaker Notes:

Regulating marriages

Exam 2 lecture 7

UIC BioS 101 Nyberg

Speaker Notes:

Exam 2 lecture 7

UIC BioS 101 Nyberg

BIOLOGICAL POPULATIONS ARE FINITE

The calculation of expected genotype frequencies given random mating assumes that p does not change with time. That is not possible with finite populations.
All real populations have a finite number of members, and even if mating is independent of genotype, the actual numbers change slightly from the expected due to sampling and the population size, N, becomes an important variable.

Speaker Notes:

Exam 2 lecture 7

UIC BioS 101 Nyberg

Vocabulary

Allele frequency
Genotype frequency
Phenotype frequency
Random mating
Expected genotype frequency

Hardy-Weinberg
Directional selection
Stabilizing selection
Gene flow
Inbreeding

Speaker Notes: