Download Hardy Weinberg

Measuring
Evolution of Populations
Hardy Weinberg
AP Biology
2010-2011
5 Agents of evolutionary change
Mutation
Gene Flow
Non-random mating
Chemical
Changes to DNA
Migration
Sexual Selection
AP Biology
Genetic Drift
Selection
Small population
Natural Selection
Differential Survival
III. Genetic Equilibrium
 Researchers know whether or not a population is


evolving by tracking deviations from a baseline of
genetic equilibrium.
If you notice, they violate all the mechanisms of
microevolution.
Five conditions required for a stable gene pool:
1.
2.
3.
4.
5.
Mutations do not occur
Population is infinitely large (genetic drift eliminated)
No gene flow
Random mating (no sexual selection)
All individuals survive and reproduce equally (no NS)
 If these 5 things happen, new alleles are not introduced

to a population and therefore, the population isn’t
evolving.
This doesn’t happen in reality but the rate these change
has a direct affect on the rate of evolution.
AP Biology
Genotype Frequencies Vs. Allele Frequencies
Studying population genetics requires understanding alleles in a
gene pool.
Each individual has 2 alleles for each trait.
This is the genotype, the phenotype is dependent upon this.
The gene pool represents all available alleles in a population so
genotypic and allelic frequencies always add up to 1.
AP Biology
Genotype Frequency
(frequency of EE) +
(frequency of Ee) +
(frequency of ee) = 1
Allele Frequency:
(frequency of E) +
(frequency of e) = 1
The Hardy-Weinberg Formula. Rules
 Determining Genetic Equilibrium… the BASELINE
 The Hardy-Weinberg formula can be used to
determine if a population is in genetic equilibrium
p2 + 2pq +q2= 1.0
is essentially: (AA) + (Aa) + (aa) = 1.0
 It should be noted that H-W examines only traits that
have 2 alleles!

We couldn’t use it for our turtle population b/c there
were 6 phenotypes.
 The frequency of the dominant allele (A) plus the
recessive allele (a) equals 1.0
p + q = 1.0
AP Biology
The Hardy-Weinberg Principle
 The Hardy-Weinberg

principle describes a
population that is not
evolving
If a population does
not meet the criteria
of the HardyWeinberg principle, it
can be concluded
that the population is
evolving.
AP Biology
Finding out whether a population is evolving. The frequencies of
wing-color alleles among all of the individuals in this hypothetical
population of morpho butterflies are not changing; thus, the
population is not evolving.
AP Biology
Fig. 18-3a, p. 280
490 AA butterflies
dark-blue wings
490 AA butterflies
dark-blue wings
490 AA butterflies
dark-blue wings
420 Aa butterflies
medium-blue wings
420 Aa butterflies
medium-blue wings
420 Aa butterflies
medium-blue wings
90 aa butterflies
white wings
90 aa butterflies
white wings
90 aa butterflies
white wings
Starting Population
Next Generation
Next Generation
AP Biology
Fig. 18-3b, p. 280
Class Example 1
 Let’s see if we can figure out allele
frequencies for a population of pigs using
H-W.
 The goal is to find: q2, p2, p, q, 2pq,
AP Biology
Example:




White (pink) coats are dominant over black.
Can you identify who is dominant and recessive.
Yes.
Can you determine who has what alleles?
 Only for the
homozygous
recessive!
 You have to
calculate the
others….
AP Biology
Hardy-Weinberg Sample
 Pink is dominant over black.
 Calculate q2: Count the individuals that are
homozygous recessive in the illustration
above. Calculate the percent of the total
population they represent. This is q2.
q2 = 25%
AP Biology
 Find q. Take the square root of q2 to
obtain q, the frequency of the recessive
allele.
q = .5
AP Biology
 Find p. The sum of the frequencies of
both alleles = 100%, p + q = l. You know
q, so what is p, the frequency of the
dominant allele?
p=1–q
p = 1 - .5
p = .5
AP Biology
 Find 2pq. The frequency of the
heterozygotes is represented by 2pq.
This gives you the percent of the
population that is heterozygous for
white coat:
q = .5
p = .5
2pq = 2(.5)(.5)
2pq = .5
This represents the carriers of the recessive alleles
AP Biology
 This shows the allele frequency in
generation 1.
 One would study the allele frequencies
over several generations to see if the
population is evolving.
AP Biology
What is the Point of H-W?
 Why we use the H-W equation is to




determine if a population is evolving.
By calculating the equilibrium in the
beginning of a population we know the
distribution of alleles.
If we examine a population later in time we
can see if there is a change.
According to H-W, if there is no change
there is no evolution.
If there is change evolution has occurred.
AP Biology
Independent Practice






Complete Question #1 on the handout.
You may ask questions – share out answers.
The goal is to find: q2, p2, p, q, 2pq
You start by determining the % of
homozygous recessives in a population (this
is q2)
It is also important why you are doing this…
it is to determine the genetic equilibrium
baseline for the population.
Complete the rest of the worksheet for
classwork/homework.
AP Biology
Example Of Strong Selection Pressure
 Tay Sachs
Primarily in Ashkenazi Jews & Cajuns
 Recessive disease = aa
 lysosomal storage disease

 lack of one functional digestive enzyme in
lysosome
 build up undigested fat in
brain cells
 children die before they
are 5 years old
So where do new cases of
come from?
APTay-Sachs
Biology
Example of heterozygote advantage
 Sickle cell anemia

inherit a mutation in gene coding for one
of the subunits in hemoglobin





oxygen-carrying blood protein
normal allele = Hb
mutant allele = Hs
recessive trait = HsHs
low oxygen levels causes
RBC to sickle
 clogging small blood vessels
 damage to organs

AP Biology
often lethal
Single-celled eukaryote parasite
(Plasmodium) spends part of its
life cycle in red blood cells
Malaria
1
liver
2
AP Biology
3
Sickle cell frequency
 High frequency of heterozygotes
1 in 5 in Central Africans = HbHs
 unusual for allele with severe
detrimental effects in homozygotes

 1 in 100 = HsHs
 usually die before reproductive age
Why is the Hs allele maintained at such high
levels in African populations?
Suggests some selective advantage of
bHs
being
heterozygous…
H
AP Biology
Heterozygote Advantage
 In tropical Africa, where malaria is common:

homozygous dominant (normal)
 reduced survival or reproduction from malaria: HbHb

homozygous recessive
 reduced survival & reproduction from sickle cell anemia: HsHs

heterozygote carriers
 survival & reproductive advantage: HbHs
Hypothesis:
In malaria-infected
cells, the O2 level is
lowered enough to
cause sickling which
kills the cell & destroys
the
parasite.
AP Biology
Frequency of sickle cell allele
& distribution of malaria
Concept Check
 What is the usefulness of the H-W
Law?
 Homework
Complete the H-W Online Tutorial by
your next period.
 http://www.montereyinstitute.org/cou
rses/AP%20Biology%20I/course%20fi
les/multimedia/lesson23/lessonp.html
?showTopic=2

AP Biology
Any Questions??
AP Biology
2010-2011