Download Hardy Weinberg PPT File

AGENDA – 2/21/17
• Take out your composition notebook and
pick up handouts!
Bell-Ringer: Genetics Review
Hardy Weinberg Quick Notes w/ practice
Hardy Weinberg Partner Practice
Homework:
– Work on Exam Study Guide --- test is on Friday!
Bell-Ringer: 2/21/17
1. What is the difference between an allele, a
genotype, and a phenotype?
2. What do the following genotypes mean?
(B= Dark eyes, b= light eyes)
- BB
- Bb
- bb
GENETICS AND EVOLUTION
• A population in which the frequency of alleles
remains the same over generations as being in
genetic equilibrium.
• A population that is in genetic equilibrium is NOT
evolving or changing.
“Evolution is simply a change in frequencies of alleles in the gene
pool of a population.”
The definition of evolution was developed in the early
20th century by
Godfrey Hardy, an English mathematician, and
Wilhelm Weinberg, a German physician.
Hardy
Weinberg
Hardy-Weinberg came up with five basic reasons
why a population would stay at genetic equilibrium:
1. the population is large, and no catastrophes occur.
2. there is no migration in or out of the population
3. no mutations occur in any individuals DNA within this population.
4. all mating is totally random (No sexual selection)
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
five conditions, let alone all of them, will happen in the
real world, evolution is inevitable.
Reproduction will not be random,
organisms will choose mates based
on certain traits…..
Hardy and Weinberg developed an equation we
can use to calculate how many individuals in a
population are:
BB (Homozygous Dom.)
Bb (Heterozygous)
bb (Homozygous Rec.)
Or in other words, determine the allelic
frequencies and genotypic frequencies in
a particular gene pool.
This is used to track allelic frequencies from generation to
generation in a population to monitor evolution OR changes
in the gene pool.
This is the Hardy-Weinberg equilibrium equation.
p² + 2pq + q² = 1
p is defined as the frequency of the dominant
allele
p = B
q is defined as the frequency of the recessive
allele
q = b
Because there are only two alleles in this
case, “B” and “b” the frequency of one plus
the other must equal 100%, so…
p + q = 1 (or 100%)
p² + 2pq + q² = 1
p = B
q = b
In this equation:
p² = homozygous dominant (BB) organisms in a
population.
2pq = heterozygous (Bb) organisms
q² = homozygous recessive (bb) ones
Albinism is a rare genetically inherited trait that is only
expressed 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)
Knowing one of the two variables
(q) in the Hardy-Weinberg 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 = AA = .986 = 98.6%
2pq = Aa = .014 = 1.4%
q² = aa = .00005 = .005%
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:
1. The frequency of the "aa" genotype.
The frequency of the “aa” genotype is given in
the problem as 36%!
2. 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%.
3. 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%.
4. 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)
5. 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%.
PRACTICE
HARDY-WEINBERG!
Pick any 3 problems to complete in
your notebook.
Round to 3 decimal places (thousandths place).
Hardy-Weinberg Problem Set
In your notebook:
• You will pick 3 problems to work out from
the Hardy Weinberg Problem Set!
– (We already did #1 together so pick 4 others)
• When you are finished, raise your hand so
that I can quickly check them!
• Whatever you don’t finish is homework!
AGENDA – 2/22/17
• Take out your composition notebook and
pick up handouts!
• Fishy Frequencies Lab
– All Data and Analysis Questions MUST be
in your notebook for this lab.
• Homework: Work on Exam Study Guide --- test
is on Friday!