Transcript Evolution 1/e - SUNY Plattsburgh


Read Chapter 6 of text
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.

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?


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 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.

Study of the distribution of alleles in
populations and causes of allele
frequency changes

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

Hardy-Weinberg serves as the
fundamental null model in 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.

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.
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.

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.

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.

If three alleles with frequencies P1, P2 and
P3 such that P1 + P2 + P3 = 1
Then genotype frequencies given by:
 P12 + P22 + P32 + 2P1P2 + 2P1 P3 +
2P2P3


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.
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?
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.

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?
Just use the formulae P1 + P2 + P3 = 1 and
P12 + P22 + P32 + 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.