Transcript Hardy Weinberg

a
A Aa
a
Aa
The Hardy-Weinberg
A
a
principle is like a
a aa aa
Punnett square for
A AA Aa
populations, instead of
a Aa aa
individuals.
A Punnett square can predict the probability of
offspring's genotype based on parents' genotype
or the offsprings' genotype can be used to reveal
the parents' genotype.
Likewise, the Hardy-Weinberg principle is a tool
we can use to calculate the frequency of
particular alleles in a population.
The Hardy-Weinberg model enables us to
compare a population's actual genetic structure
over time with the genetic structure we would
expect if the population were in Hardy-Weinberg
equilibrium (or not evolving).
If genotype
frequencies differ
from those we
would expect
under equilibrium,
we can assume
that evolution is
taking place.
Evolution is simply a change in frequencies of alleles in
the gene pool of a population.
B=brown
• Alleles:
One of two or more forms of a gene that
code for different versions of the same
trait.
•Gene Pool:
b=blonde
All possible genes and genetic
combinations in a population.
group of organisms from the same
• Population: Aspecies
and the geographic location.
Let us assume that there is a trait that is determined by
the inheritance of a gene with two alleles--B and b.
If the parent generation had 92% B and 8% b and their
offspring collectively had 90% B and 10% b, it would be
evident that evolution had occurred between the
generations.
This definition of evolution was
developed in the early 20th century by
Godfrey Hardy, an English
mathematician, and Wilhelm
Weinberg, a German physician.
Hardy
They concluded in 1908 that gene
pool frequencies are inherently stable
but that evolution should be expected
in all populations virtually all of the
time. They resolved this apparent
paradox by analyzing the net effects
of potential evolutionary mechanisms.
Weinberg
Hardy-Weinberg Equilibrium maintains five basic
assumptions:
1. the population is infinitely large, and that
genetic drift is not an issue within the population.
2. there is no gene flow, or migration in or out of
the population
3. mutation is not occurring
4. all mating is totally random
5. natural selection is not occurring
Under these conditions it is obvious that evolution
would not occur. There are no mechanisms of
evolution acting on the population, so the process
cannot happen--the gene pool frequencies will
remain unchanged.
However, since it is highly unlikely that any one of
these seven conditions, let alone all of them, will
happen in the real world, evolution is inevitable.
Hardy and Weinberg went on to develop a simple equation that
can be used to discover the probable genotype frequencies in a
population and to track their changes from one generation to
another.
This has become known as the Hardy-Weinberg equilibrium
equation.
p² + 2pq + q² = 1
p is defined as the frequency of the dominant allele
q is the frequency of the recessive allele for a trait controlled
by a pair of alleles (A and a)
In other words, p equals all of the alleles in individuals who are
homozygous dominant (AA) and half of the alleles in people who
are heterozygous (Aa) for this trait in a population. In
mathematical terms, this is
p = AA + ½Aa
Likewise, q equals all of the alleles in individuals who are
homozygous recessive (aa) and the other half of the alleles in
people who are heterozygous (Aa).
q = aa + ½Aa
Because there are only two alleles in this case,
the frequency of one plus the frequency of the
other must equal 100%, so…
p+q=1
Since p+q=1, then logically p=1-q
There were only a few short steps from this knowledge for Hardy
and Weinberg before they realized that the chances of all
possible combinations of alleles occurring randomly is
(p + q)² = 1
Or more simply:
p² + 2pq + q² = 1
In this equation, p² is the predicted frequency of homozygous
dominant (AA) organisms in a population, 2pq is the predicted
frequency of heterozygous (Aa) organisms, and q² is the
predicted frequency of homozygous recessive (aa) ones!
Tudaaaaaaah!
From observations of phenotypes, it is usually only possible to know the
frequency of homozygous recessive organisms, or q² in the equation, since
they will not have the dominant trait.
This Drosophila
The Hardy-Weinberg equation
melanogaster is
allows us to predict which ones
white-eyed. The
they are. Since p = 1 - q and q is
genotype is rr.
known, it is possible to calculate p
as well.
Knowing p and q, it is a simple
matter to plug these values into
This Drosophila
the Hardy-Weinberg equation
melanogaster is red(p² + 2pq + q² = 1). This then
eyed. The genotype
could be RR, or Rr,
provides the predicted
as red is a dominant
frequencies of all three genotypes
trait.
for the selected trait within the
population.
Those who express the trait in their phenotype could be either
homozygous dominant (p²) or heterozygous (2pq).
Albinism is a rare genetically inherited trait that is
only expressed in the phenotype of homozygous
recessive individuals (aa).
The most characteristic symptom is
a marked deficiency in the skin and
hair pigment melanin.
This condition can occur among
any human group as well as
among other animal species. The
average human frequency of
albinism in North America is only
about 1 in 20,000.
The Hardy-Weinberg equation (p² + 2pq + q² = 1),
and the frequency of homozygous recessive
individuals (aa) in a population is q². Therefore,
in North America the following must be true for
albinism: q² = 1/20,000 = .00005
By taking the square root of both sides of this equation, we get:
q = .007 (rounded)
In other words, the frequency of the
recessive albinism allele (a) is .007
or about 1 in 140. Knowing one of
the two variables (q) in the HardyWeinberg equation, it is easy to
solve for the other (p).
p=1–q
p = 1 - .007
p = .993
The frequency of the dominant, normal allele (A) is,
therefore, .99293 or about 99 in 100.
The next step is to plug the frequencies of p and q into
the Hardy-Weinberg equation:
p² + 2pq + q² = 1
(.993)² + 2 (.993)(.007) + (.007)² = 1
.986 + .014 + .00005 = 1
This gives us the frequencies for each of the three genotypes for
this trait in the population:
p² = predicted frequency
(AA)
of homozygous
dominant individuals = .986 = 98.6%
2pq =
predicted frequency (Aa)
of heterozygous
individuals = .014 = 1.4%
q² = predicted frequency
of homozygous
(aa)
recessive individuals
(the albinos) = .00005 = .005%
1. You have sampled a population in which you know that
the percentage of the homozygous recessive genotype
(aa) is 36%. Using that 36%, calculate the following:
A. The frequency of the "aa" genotype.
The frequency of the “aa” genotype is given in
the problem as 35%!
B. The frequency of the "a" allele.
The frequency of aa is 36%, which means that q2 = 0.36,
by definition.
If q2 = 0.36, then q = 0.6, again by definition.
Since q equals the frequency of the a allele, then the
frequency is 60%.
C. The frequency of the "A" allele.
Since q = 0.6, and p + q = 1, then p = 0.4;
the frequency of A is by definition equal to p, so the
answer is 40%.
D. The frequencies of the genotypes "AA" and "Aa."
The frequency of AA is equal to p2, and the frequency of
Aa is equal to 2pq.
So, using the information above, the frequency of AA is
16%
(p2 = 0.4 x 0.4 = 0.16)
and Aa is 48%
(2pq = 2 x 0.4 x 0.6 = 0.48)
E. The frequencies of the two possible phenotypes if "A"
is completely dominant over "a."
Because "A" is totally dominate over "a", the dominant phenotype will
show if either the homozygous "AA" or heterozygous "Aa"
genotypes occur.
The recessive phenotype is controlled by the homozygous aa
genotype. Therefore, the frequency of the dominant phenotype equals
the sum of the frequencies of AA and Aa, and the recessive phenotype
is simply the frequency of aa.
Therefore, the dominant frequency is 64% and, in the first part of this
question above, you have already shown that the recessive frequency
is 36%.
• http://anthro.palomar.edu/synthetic/synth_2.htm
• http://media.nasaexplores.com/lessons/04006/images/9-12_redeyefly.gif
• http://www.kstate.edu/parasitology/biology198/hardwein.html
• http://en.wikipedia.org/wiki/Hardy-Weinberg_principle