Population genetics focuses on genetic changes within an interbreeding population Gene pool Allele Frequency all the genes in a population Shows frequency of ID: 918802
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Slide1
4-Population Genetics Notes
Slide2Population genetics
: focuses on genetic changes within an interbreeding population.
Gene pool = _______________________________
Allele Frequency =
all the genes in a population
Shows frequency of each allele in the gene pool (0.625 B, 0.375 b)
Slide3Changing the Gene Pool
Microevolution:
a change in a population’s gene pool over a succession of generations (a brief period of geologic time). Microevolution may be due to natural selection or other factors such as genetic drift.
Genetic Drift: random change of allele frequencies in a small population due to
chance , not natural selection.
Slide4Founder Effect
:
______________________
___________________________Example: ___________________________
migration of a small group to a new area can lead to genetic drift.
Amish in America (extra digits)
Slide5Bottleneck Effect:
________ ___________
Example: _________________
an event that drastically lowers the population numbers
endangered species
Slide6Slide7Genetic Equilibrium: Allele frequencies are not changing =
________________
no evolution
Slide8Slide9The Hardy Weinberg principle describes a hypothetical situation in which there is no change in the gene pool (frequencies of alleles), hence no evolution.
The frequencies
of
alleles will remain unchanged generation after generation (= NO EVOLUTION) if the following conditions are met:
Large populationRandom mating No mutation
No migrationNo natural selection
Slide10The Hardy-Weinberg Equation:
To estimate the frequency of alleles in a population, we can use the Hardy-Weinberg equation.
p
2
+ 2pq + q
2 = 1and p + q = 1Using the equation:
p
= the
dominant
allele
frequency
(represented here by
A
).
q
= the
recessive
allele
frequency
(
represented here by
a
)
For a population in genetic equilibrium:
p
+
q
= 1.0
(The sum of the frequencies of both alleles is
1.)
Slide11(
p
+
q
)2 = 1
so p2 + 2pq + q2 = 1
The three terms of this binomial expansion
indicate the frequencies of the three genotypes:
p
2
=
genotype frequency of
AA
(homozygous
dominant)
2pq
=
genotype frequency of
Aa
(heterozygous
)
q
2
=
genotype frequency of
aa
(homozygous recessive)
The Hardy-Weinberg Equation:
Slide12SAMPLE PROBLEM #1:
In pigs, black coat is recessive to white.
What is the
percentage
of
heterozygotes in this population?
Calculate
q
2
: Count the individuals that are
homozygous recessive in the illustration to the
left. Calculate the
frequency
of the total
population they
represent. This is
q
2
.
q
2
= 4/16 = 0.25
Slide13SAMPLE PROBLEM #1:
In pigs, black coat is recessive to white.
What is the
percentage
of
heterozygotes in this population? 2. Find q
: Take the square root of
q
2
to obtain
q
,
the frequency of the recessive allele.
q
= √0.25 = 0.5
Slide14SAMPLE PROBLEM #1:
In pigs, black coat is recessive to white.
What is the
percentage
of
heterozygotes in this population? 3. Find p
: The sum of the frequencies of both
alleles = 100%,
p
+
q
= 1. You know
q
, so what
is
p
, the frequency of the dominant allele?
p = 1 – q
p = 1 - 0.5 = 0.5
Slide15SAMPLE PROBLEM #1:
In pigs, black coat is recessive to white.
What is the
percentage
of
heterozygotes in this population? 4. Find
2
pq
: The frequency of the
heterozygotes
is represented
by 2
pq
.
Multiply by 100 to
calculate the percent of
the population that is
heterozygous
for white
coat
:
2pq = 2(0.5)(0.5)
2pq =
0.5
0.5 * 100=
50% Heterozygous
Slide16