1. 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

2. 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

3. Figure 22.3 DERIVING THE HARDY-WEINBERG PRINCIPLE-A NUMERICAL EXAMPLE
P1 = frequency of allele A1 = 0.7
1. Suppose that the allele
frequencies in the parental
generation were 0.7 and 0.3.
P2 = frequency of allele A2 = 0.3
Gametes from parent generation
2. 70% of the gametes in the
gene pool carry allele A1 and
30% carry allele A2 .
3. Pick two gametes at
random from the gene pool
to form offspring. Three
genotypes are possible. A1 A2 A2 A1 A1 A1 A2 A2
.07 x 0.3=0.21
.03 x 0.7=0.21
0.7 x 0.7
= 0.49
0.3 x 0.3
= 0.09
= 0.42
Homozygous
Heterozygous
Homozygous
4. Calculate the frequencies
of these three combinations
of alleles.
Gametes from offspring generation
5. When the offspring breed,
imagine that their gametes
go into a gene pool.
Figure: 22-03
Caption:
Deriving the Hardy-Weinberg Principle—A Numerical Example
6. Calculate the frequencies
of the two alleles in this
gene pool.
42% of the gametes
are from A1A2 parents.
Half of these carry A1 and half carry A2
49% of the gametes
are from A1A1 parents.
All of these carry A1
9% of the gametes
are from A2A2 parents.
All of these carry A2
BEHOLD! The allele frequencies of A1 and A2 have not changed from parent generation to offspring generation. Evolution has not occurred.
P1 = frequency of allele A1 = ( /2(0.42)) = ( ) = 0.7
P2 = frequency of allele A2 = (1/2(0.42) ) = ( ) = 0.3
Genotype frequencies will be given by: p12 : 2p1p2 : p22 as long as all Hardy-Weinberg assumptions are met

4. 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

5. Albinism in Hopi Native Americans
The incidence of albinism is remarkably common ( or 13 in every Hopis)
Assuming Hardy-Weinberg equilibrium, we can calculate q as the square root of = 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

6. 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

7. 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

8. 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

9. 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 = = 0.047
Next we determine the expected Hardy-Weinberg frequencies of genotypes (p2:2pq: q2)
That ratio is BB: Bb: bb

10. Now we can apportion the 497 individuals to their expected genotypes
Expected BB = x 497 = 451
Expected Bb = x 497 =45
Expected bb = x 497 = 1
Compare the observed genotypes in the field: 453, 42 and 2
These moths clearly conform to the Hardy-Weinberg expectations