Hardy-Weinberg Equilibrium Model

Aim:

1.  Calculate mathematical models based on Hardy Weinberg equilibrium to analyze genetic drift and effects of natural selection on allele frequencies in a population.

2. Extrapolate the data to predict what the effects of genetic drift, migration, and artificial selection on genotype frequencies of a population will be.

Brain storm: Plicker thinker’s For natural selection to occur, which must be

true of a population of organisms? A. Phenotypic variations must be based on

genetic variations rather than on varying environmental conditions.

B. The population must be occupying the same niche as another local population

C. Organisms must be capable of sexual reproduction

D. Food resources must be limited.

Which of the following statements concerning genetic variation is most accurate?

•  A. The process of natural selection directly creates genetic variation.

•  B. Changes in the environment of a population stimulate mutation and genetic change in that population.

•  C. Genetic variation must first be present in a population before natural selection can act on that population.

•  D. The higher the average heterozygosity in a population, the less the genetic variation in a gene pool.

A. Gene flow B. Geographic isolation C. Founder effect D. Bottleneck

1.  Darwin’s finches 2.  The establishment of a genetically unique

population through genetic drift. 3.  Natural disasters reduce the size of a

population randomly, resulting in a loss of genetic variation.

4.  An increase in diversity from movement of alleles into or out of a population from migration or wind.

The Hardy-Weinberg model states that the genetic variation in a

population will remain constant from one generation to the next in the

absence of disturbing factors.

A population (or an allele) is not evolving if it is in Hardy-Weinberg equilibrium.

Not Evolving

Evolving

Conditions for an allele to be in Hardy-Weinberg equilibrium:

These conditions

are impossible in nature!

We can compare a naturally occurring population to a theoretical population in

Hardy-Weinberg equilibrium to see if the natural population is evolving.

Naturally occurring population

Theoretical population In Hardy-Weinberg equilibrium

A population is evolving if:

Allele frequency is the proportion of gametes in a population. To

calculate allele frequency: •  There are always twice the number of

alleles as there are individuals in the population (every individual has two alleles for a gene).

AA Aa aa Homozygous Dominant Heterozygous Homozygous Recessive

The two Hardy-Weinberg formulae allow us to calculate allele frequency:

p2 + 2pq + q2 = 1 p + q = 1

Solve for q first!

Modeling Hardy-Weinberg frequencies: key phrases!

Frequency of the allele •  p = frequency of the

dominant allele (A)

•  q = frequency of recessive allele (a)

•  Phenotype frequency!

Frequency of alleles (genotypes!) •  p2= frequency of

homozygous dominant (AA)

•  2pq = frequency of heterozygous (Aa)

•  q2 = frequency of homozygous recessive (aa)

•  Genotype frequency!

p + q = 1 p2 + 2pq + q2 = 1

In a population that is in Hardy Weinberg equilibrium, the frequency of a particular allele a = 0.4. What percentage of the population is heterozygous for this trait?

A.  4% B.  16% C. 32% D. 48%

In a population of 1000 people, 90 have blue eyes. What percent of the population has hybrid brown eyes?

A.  9% B.  42% C. 3% D. 70%

How to calculate phenotypic frequency in a population:

f(white) =

f(black) =

3 8

5 8

Phenotypic frequency: 3 white 5 black 8 total mice

How to calculate genotypic frequency in a population:

f(white) =

f(aa) =

3 8

3 8

The frequency of the recessive genotype (we will call it aa) is the same as the frequency of the recessive phenotype.

How to calculate genotypic frequency in a population:

f(black)=

f(Aa) = ? f(AA) = ?

5 8

The frequency of the heterozygous genotype

(Aa) and the homozygous dominant genotype (AA) is harder

to tell.

p + q = 1

p = f (Dominant) = f (A)

q = f (aa) = f (a)

f (A) f (a)

f = frequency

AA

Aa

or

Recessive only! Always find q first!

The Hardy-Weinberg Variables: f =frequency in a population p2 = f (AA) = f (homozygous dominant)

2pq = f (Aa) = f (heterozygote)

q2 = f (aa) = f (homozygous recessive)

p2 + 2pq + q2 = 1 Homozyous dominant

Heterozygote Homozygous recessive

f (AA) f (Aa)

f (aa)

f (q2) = f (white)= = .36 3 8

Step 1: Find q2

Step 2: Find q

f(q) = f (q2)= .36 = .6 Step 3: Find p

p + q = 1 f (p) = 1 - .6 p = 1 – q f (p) = .4

Step 4: Find f (aa) f (aa) = f (q2) = .36

Step 5: Find f (Aa)

f (Aa) = 2pq = 2(.4)(.6) = .48 Step 6: Find f (AA)

f (AA) = f (p2) = .42 = .16

Using the Hardy-Weinberg equation to find f (AA), f (Aa) and f (aa):

The frequency of the recessive phenotype is equal to q2.

The frequency of the dominant phenotype is equal to p2 + 2pq.

A population is evolving if:

•  Any one of the five Hardy-Weinberg conditions are not met.

•  The allele frequency changes. •  It is not in Hardy-Weinberg Equilibrium

Example: In a population of mice, there are 3 bb, 4 Bb, and 3 BB. Find the

allele frequencies: B allele:

3 BB 4 Bb 3 bb

There are 2 B alleles here. There is only 1 B allele here.

There are no B alleles here.

Example: In a population of mice, there are 3 bb, 4 Bb, and 3 BB. Find the

allele frequencies: B allele:

3 BB 4 Bb 3 bb

There are 2 B alleles here. There is only 1 B allele here.

There are no B alleles here.

b allele:

3 BB 4 Bb 3 bb

There are 2 b alleles here.

There is only 1 b

allele here.

Example: In a population of mice, there are 3 bb, 4 Bb, and 3 BB. Find the

allele frequencies: B allele:

3 BB = 3 · 2 = 6 4 Bb = 4 · 1 = 4 3 bb 10

+

10 20

= .5 Total # of alleles

b allele:

3 BB 4 Bb = 4 · 1 = 4 3 bb = 3 · 2 = 6 10

Example: In a population of mice, there are 3 bb, 4 Bb, and 3 BB. Find the

allele frequencies: B allele:

3 BB = 3 · 2 = 6 4 Bb = 4 · 1 = 4 3 bb 10

+

10 20

= .5 Total # of alleles

= .5 10 20

+