read chapter 6 of text we saw in chapter 5 that a cross between two individuals heterozygous for a...

Chapter 6The ways of change: drift and selection

Assigned reading

Read Chapter 6 of text

Chapter 6 Genetics of populations

We saw in chapter 5 that a cross between two individuals heterozygous for a dominant allele produces a 3:1 ratio of individuals expressing the dominant phenotype: to those expressing the recessive phenotype.

For example brachydachtyly (shortening of the digits) displays this pattern of inheritance.

Population genetics

In the early 1900’s when Mendel’s work was rediscovered there was confusion about how these simple patterns of inheritance affected populations.

Why, for example, was not 3 of every 4 people a person with brachdactyly?

Why did not dominant alleles replace recessive alleles?

Population genetics

The confusion stemmed from confusing what was happening at the level of the individual with what occurs at the population level.

Individual-level thinking enables us to figure out the result of particular crosses.

Population genetics

Population level thinking however is needed to figure out how the genetic characteristics of populations change over time.

It enables us to figure out quantitatively what is happening in a population as a result of evolution. Remember, evolution occurs when genotype frequencies change over time.

Population genetics

Study of the distribution of alleles in populations and causes of allele frequency changes

Key Concepts

Diploid individuals carry two alleles at every locus› Homozygous: alleles are the same› Heterozygous: alleles are different

Remember Evolution: change in allele frequencies from one generation to the next

Key Concept

Hardy-Weinberg serves as the fundamental null model in population genetics

Hardy-Weinberg Model: a null model for population genetics.

Null models provide us with a baseline. They tell us what we expect to be the case if certain forces are not operating.

The Hardy-Weinberg equilibrium tells us what we expect to happen to genotype frequencies when forces such as natural selection are not operating on a population.

Hardy-Weinberg Model: a null model for population genetics.

The Hardy-Weinberg model enables us to determine what allele and genotype frequencies we would expect to find in a population if all that is happening is alleles are being randomly assigned to gametes when gametes are made (during meiosis) and those gametes meet up at random.

Hardy-Weinberg Model

The Hardy-Weinberg model examines a situation in which there is one gene with two alleles A1 and A2.

Recall that alleles are different versions of a gene.

There are three possible genotypes A1A1,

A2 A2,and A1 A2


Hardy and Weinberg used their model to predict what would happen to allele frequencies and genotype frequencies in a population in the absence of any evolutionary forces.

Their model produced three important conclusions


The three conclusions of the H-W model. In the absence of evolutionary processes acting on them:

1. The frequencies of the alleles A1 and A2 do not change over time.

2. If we know the allele frequencies in a population we can predict the equilibrium genotype frequencies (frequencies of A1A1, A2 A2,and A1 A2).


3. A gene not initially at H-W equilibrium will reach H-W equilibrium in one generation.

Assumptions of Hardy-Weinberg

1. No selection.› If individuals with certain genotypes

survived better than others, allele frequencies would change from one generation to the next.


2. No mutation› If new alleles were produced by mutation

or alleles mutated at different rates, allele frequencies would change from one generation to the next.


3. No migration› Movement of individuals in or out of a

population would alter allele and genotype frequencies.


4. Large population size.› Population is large enough that chance plays

no role. Eggs and sperm collide at same frequencies as the actual frequencies of p and q.

› If assumption was violated and by chance some individuals contributed more alleles than others to next generation allele frequencies might change. This mechanism of allele frequency change is called Genetic Drift.


5. Individuals select mates at random.› Individuals do not prefer to mate with

individuals of a certain genotype. If this assumption is violated allele frequencies will not change, but genotype frequencies might.

Deriving the H-W model

Hardy-Weinberg Equilibrium

Assume two alleles A1 and A2 with known frequencies (e.g. A1 = 0.6, A2 = 0.4.)

Only two alleles in population so their allele frequencies add up to 1.


Can predict frequencies of genotypes in next generation using allele frequencies.

Possible genotypes: A1A1 , A1A2 and A2A2


Assume alleles A1 and A2 enter eggs and sperm in proportion to their frequency in population (i.e. 0.6 and 0.4)

Assume sperm and eggs meet at random (one big gene pool).


Then we can calculate genotype frequencies.

A1A1 : To produce an A1A1 individual, egg and sperm must each contain an A1 allele.

This probability is 0.6 x 0.6 or 0.36 (probability sperm contains A1 times probability egg contains A1).


Similarly, we can calculate frequency of A2A2.

0.4 x 04 = 0.16.


Probability of A1A2 is given by probability sperm contains A1 (0.6) times probability egg contains A2 (0.4). 0.6 x 04 = 0.24.


But, there’s a second way to produce an A1A2 individual (egg contains A1 and sperm contains A2). Same probability as before: 0.6 x 0.4= 0.24.

Overall probability of A1A2 = 0.24 + 0.24 = 0.48.


Genotypes in next generation: A1A1 = 0.36 A1A2 = 0.48 A2 A2= 0.16 Adds up to one.


General formula for Hardy-Weinberg. Let p= frequency of allele A1 and q =

frequency of allele A2.

p2 + 2pq + q2 = 1.

Hardy Weinberg Equilibrium with more than 2 alleles

If three alleles with frequencies P1, P2 and P3 such that P1 + P2 + P3 = 1

Then genotype frequencies given by: P1

2 + P22 + P3

2 + 2P1P2 + 2P1 P3 +

2P2P3

Conclusions from Hardy-Weinberg Equilibrium

Allele frequencies in a population will not change from one generation to the next just as a result of assortment of alleles and zygote formation.

If the allele frequencies in a gene pool with two alleles are given by p and q, the genotype frequencies will be given by p2, 2pq, and q2.

Conclusions from Hardy-Weinberg Equilibrium

The frequencies of the different genotypes are a function of the frequencies of the underlying alleles.

The closer the allele frequencies are to 0.5 the greater the frequency of heterozygotes.

Working with the H-W equation

You need to be able to work with the Hardy-Weinberg equation.

For example, if 9 of 100 individuals in a population suffer from a homozygous recessive disorder can you calculate the frequency of the disease causing allele? Can you calculate how many heterozygotes are in the population?


p2 + 2pq + q2 = 1. The terms in the equation represent the frequencies of individual genotypes.

P and q are allele frequencies. It is vital that you understand this difference.


9 of 100 (frequency = 0.09) of individuals are homozygotes. What term in the H-W equation is that equal to?


It’s q2.

If q2 = 0.09, what’s q? Get square root of q2, which is 0.3.

If q=0.3 then p=0.7. Now plug p and q into equation to calculate frequencies of other genotypes.


p2 = (0.7)(0.7) = 0.49

2pq = 2 (0.3)(0.7) = 0.42

Number of heterozygotes = 0.42 times population size = (0.42)(100) = 42.

Working with the H-W equation: 3 alleles

There are three alleles in a population A1, A2 and A3 whose frequencies respectively are 0.2, 0.2 and 0.6 and there are 100 individuals in the population.

How many A1A2 heterozygotes will there be in the population?

Working with the H-W equation: 3 alleles

Just use the formulae P1 + P2 + P3 = 1 and P1

2 + P22 + P3

2 + 2P1P2 + 2P1 P3 + 2P2P3 = 1

Then substitute in the appropriate values for the appropriate term

2P1P2 = 2(0.2)(0.2) = 0.08 or 8 people out of 100.


Hardy Weinberg equilibrium principle identifies the forces that can cause evolution.

If a population is not in H-W equilibrium then one or more of the five assumptions is being violated.

read chapter 6 of text we saw in chapter 5 that a cross between two individuals heterozygous for a...

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