
Read Chapter 6 of text


Brachydachtyly displays the classic 3:1 pattern of inheritance (for a cross between heterozygotes) that mendel described.




In the early 1900’s there was confusion about how such simple patterns of inheritance affected populations.
Why, aren’t 3 of every 4 people brachdactylyous?
Why don’t dominant alleles replace recessive alleles?


Problem stems from confusing what happens at the individual level with what occurs at the population level .

Individual-level thinking enables us to figure out the result of particular crosses.


Population level thinking is used to figure out how the genetic characteristics of populations change quantitatively over time.


Population genetics is the study of the frequency distribution of alleles and genotypes in populations and the causes of allele frequency changes

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

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


The Hardy-Weinberg model examines a situation in which there is one gene with two alleles A
1 and A
2
.

A
2
There are three possible genotypes A
1
A
1,
A
2
,and A
1
A
2



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 acting on the population.
Their model produced three important conclusions




In the absence of evolutionary processes acting on them:
1. The frequencies of the alleles A
1
A
2 do not change over time.
and
2. If we know the allele frequencies in a population we can predict the equilibrium genotype frequencies
(frequencies of A
1
A
1,
A
2
A
2
,and A
1
A
2
).

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


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 does not affect allele frequencies.
› If assumption is 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.


Assume two alleles A
1 frequencies (e.g. A
1 and A
2
= 0.6, A
2 with known
= 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: A
1
A
1
, A
1
A
2
A
2
A
2 and


Assume alleles A
1 and A
2 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.
A
1
A
1
: To produce an A
1
A
1 egg and sperm must each contain an A allele.
individual,
1
This probability is 0.6 x 0.6 or 0.36
(probability sperm contains A
1 times probability egg contains A
1
).


Similarly, we can calculate frequency of
A
2
A
2
.

0.4 x 04 = 0.16.


Probability of A
1
A sperm contains A egg contains A
2
2
1 is given by probability
(0.6) times probability
(0.4). 0.6 x 04 = 0.24.


But, there’s a second way to produce an
A
1
A
2 individual (egg contains A sperm contains A
2 before: 0.6 x 0.4= 0.24.
1 and
). Same probability as

Overall probability of A
1
0.48.
A
2
= 0.24 + 0.24 =






Genotypes in next generation:
A
1
A
1
A
1
A
2
= 0.36
= 0.48
A
2
A
2
= 0.16
Adds up to one.



General formula for Hardy-Weinberg.
Let p= frequency of allele A frequency of allele A
2
.
1 and q =
 p 2 + 2pq + q 2 = 1.


If three alleles with frequencies P
1
P
3 such that P
1
+ P
2
+ P
3
= 1
, P
2 and


Then genotype frequencies given by:
P
1
2
2P
2
P
3
+ P
2
2 + P
3
2 + 2P
1
P
2
+ 2P
1
P
3
+


1. 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.

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



3. 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.


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?


 p 2 + 2pq + q 2 = 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 q 2 .

If q 2 = 0.09, what’s q? Get square root of q 2 , 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.

 p 2 = (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.


There are three alleles in a population A
1
,
A
2 and A
3 whose frequencies respectively are 0.2, 0.2 and 0.6 and there are 100 individuals in the population.

How many A
1
A
2 heterozygotes will there be in the population?


Just use the formulae P
1
P
1
2 + P
2
2 + P
3
2 + 2P
1
P
2
+ P
+ 2P
1
P
2
3
+ P
3
+ 2P
= 1 and
2
P
3
= 1
Then substitute in the appropriate values for the appropriate term
2P
1
P
2
100.
= 2(0.2)(0.2) = 0.08 or 8 people out of


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.