Brachydactyly and evolutionary change

We know the gene for brachydactyl
fingers is dominant to normal fingers

A man named Yule suggested that
short-fingered people should
become more common through time

Godfrey Hardy showed this
inference was wrong

Wilhelm Weinberg derived the same
solution to the problem
independently
The Hardy-Weinberg Law - the most important
principle in population genetics

The law is divided into three parts: a set of assumptions and two major
results
• In an infinitely large, randomly mating population, free from mutation,
migration and natural selection (note there are five assumptions here)
• the frequencies of the alleles do not change and
• as long as the mating is random, the genotypic frequencies will
remain in the proportions p2 (frequency of AA), 2pq (frequency of Aa)
and q2 (frequency of aa) where p is the frequency of A and q is the
frequency of a
• The sum of the genotypic frequencies should be:
p2 + 2pq + q2 = 1
Figure 22.3
DERIVING THE HARDY-WEINBERG PRINCIPLE-A NUMERICAL EXAMPLE
1. Suppose that the allele
frequencies in the parental
generation were 0.7 and 0.3.
P1 = frequency of allele A1 = 0.7
P2 = frequency of allele A2 = 0.3
Gametes
from parent
generation
A1 A1
2. 70% of the gametes in the
gene pool carry allele A1 and
30% carry allele A2 .
A1 A2
A2 A1
A2
.07 x 0.3=0.21 .03 x 0.7=0.21
A2
0.7 x 0.7
= 0.49
0.21 + 0.21 = 0.42
0.3 x 0.3
= 0.09
Homozygous
Heterozygous
Homozygous
Gametes
from
offspring
generation
49% of the gametes
are from A1A1 parents.
All of these carry A1
3. Pick two gametes at
random from the gene pool
to form offspring. Three
genotypes are possible.
4. Calculate the frequencies
of these three combinations
of alleles.
5. When the offspring breed,
imagine that their gametes
go into a gene pool.
42% of the gametes
are from A1A2 parents.
Half of these carry A1
and half carry A2
9% of the gametes
are from A2A2 parents.
All of these carry A2
P1 = frequency of allele A1 = (0.49 + 1/2(0.42)) = (0.49 + 0.21) = 0.7
P2 = frequency of allele A2 = (1/2(0.42) + 0.09) = (0.21 + 0.09) = 0.3
Genotype frequencies will be given by: p12 : 2p1p2 : p22 as long
as all Hardy-Weinberg assumptions are met
6. Calculate the frequencies
of the two alleles in this
gene pool.
BEHOLD! The allele frequencies of A1
and A2 have not changed from parent
generation to offspring generation.
Evolution has not occurred.
Genotypic frequencies under the Hardy-Weinberg Law

The Hardy-Weinberg Law
indicates:
• At equilibrium, genotypic
frequencies depend on the
frequencies of the alleles
• The maximum frequency for
heterozygotes is 0.5
• If allelic frequencies are
between 0.33 and 0.66, the
heterozygote is the most
common genotype
Albinism in Hopi Native Americans

The incidence of albinism is remarkably
common (0.0043 or 13 in every 3000
Hopis)

Assuming Hardy-Weinberg equilibrium, we
can calculate q as the square root of
0.0043 = 0.066

p is therefore equal to 0.934

The frequency of heterozygotes in the
population is 2pq = 0.123

In other words, 1 in 8 Hopis carries the
gene for albinism!

Take-home Lesson: For a rare allele,
heterozygotes can be relatively common
Extensions of the Hardy-Weinberg Law

X-linked traits (two alleles)
• Females have two copies of the gene but males are
hemizygous
• Female genotypes should be in the familiar p2: 2pq: q2 ratio
• Male genotypes should be in a p:q ratio
• This result shows why X-linked recessive traits are much more
frequently observed in males
Extensions of the Hardy-Weinberg Law

Multiple alleles
• Let p, q and r represent the frequencies of the three
alleles
• p + q + r = 1.0
• The Hardy-Weinberg model predicts the genotype
frequencies will be (p + q + r)2
• The result is a ratio of p2: 2pq: q2: 2pr: r2: 2pq
Testing for Hardy-Weinberg distributions


If we know the frequency of genotypes in a population, we
can test to see if those frequencies conform to the
expectations from the Hardy-Weinberg model
Let’s reconsider the data for scarlet tiger moths
• Number of BB = 452
• Number of Bb = 43
• Number of bb = 2

Now we need to determine the allele frequency of B and b

Let’s determine the frequency of allele B

p = f(B) = [(2 x 452) + 43]/994 = 0.953

We can determine the value of q by difference:
• q = 1 - p = 1 - 0.953 = 0.047

Next we determine the expected Hardy-Weinberg frequencies of
genotypes (p2:2pq: q2)
• That ratio is 0.908 BB: 0.090 Bb: 0.002 bb

Now we can apportion the 497 individuals to their expected
genotypes
• Expected BB = 0.908 x 497 = 451
• Expected Bb = 0.090 x 497 =45
• Expected bb = 0.002 x 497 = 1

Compare the observed genotypes in the field: 453, 42 and 2

These moths clearly conform to the Hardy-Weinberg
expectations