Download BIO152 Hardy Weinberg

BIO152 Hardy Weinberg
Tutorial 7
November 3
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Tutorial 7 Outline
„
Hardy-Weinberg model—and lab 4 (bring a
calculator)
„
Finish sex determination lecture
Reminder-Facilitated Study Groups
Tuesday 12-1 CCT2160
‰ Wednesday 1-2 CCT2110
‰ Thursday 1-2 SE1143
http://zoology.okstate.edu/zoo_lrc/biol1114/tutorials/Flash/life
4e_15-6-OSU.swf
‰
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1
1 minute write about:
Facilitated Study Groups
TOP: I would go to FSG go if they…
BOTTOM:I would find FSG more useful if I
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Evolution?
Evolution is the change in allele frequency over
time.
How do we tell if evolution has happened?
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If evolution has NOT happened…
What should we expect to see in the allele
frequency?
Answer = No change
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Hardy Weinberg principle
allele frequencies & genotype ratios in a
randomly-breeding population remain
constant from generation to generation
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gene pool
Individual members
of a population
contribute their
alleles to a
common pool of
genes.
www.brooklyn.cuny.edu/bc/
ahp/LAD/C21/C21_Gen
ePool.html
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Hardy-Weinberg model does not apply if
1.
2.
3.
4.
5.
Natural selection is acting on the gene
Genetic drift has affected the allele
frequencies
Individuals (thus alleles) move between
populations—gene flow from migration
Mutation has affected the allele(s)
Individuals are not mating randomly
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4
Random vs non random mating
Are all pairings equally likely in a population?
If not = assortative mating
e.g., humans seldom mate at random preferring
phenotypes like themselves (e.g., size, age,
ethnicity). This is called assortative mating
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Assortative mating
Marriage between close relatives is a special case of
assortative mating. The closer the kinship, the more
alleles shared and the greater the degree of
inbreeding.
►Inbreeding can alter the gene pool.
„ inbreeding populations predisposed to
homozygosity. (See Fig 24.13)
How can potentially harmful recessive alleles become
exposed in the children?
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Hardy-Weinberg model
=the NULL hypothesis for evolution
Because the HW model does not work if any
agent of evolution is operating
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Hardy-Weinberg equilibrium
(= no evolution is happening)
the distribution of genotype frequencies
depends ONLY on the allele frequencies.
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Example of stable allele frequencies
„
In general, populations with
the same allele frequency
can have different genotypic
frequencies.
„
However, when random
mating occurs, there is a
simple relationship between
the two.
„
With only random mating,
allele frequencies do not
change.
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Hardy-Weinberg (1908) theorem
If only random mating occurs, then
Allele frequencies remain unchanged over time
After one generation of random-mating, genotypic
frequencies are given by
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Importance of Hardy-Weinberg
equilibrium
„
„
Allele frequencies remain unchanged and
genotype frequencies (after the first
generation) are constant.
Thus, Mendelian genetics implies that genetic
variability can persist indefinitely, unless
other evolutionary forces act to remove it.
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Flash simulation
http://zoology.okstate.edu/zoo_lrc/biol1114/tuto
rials/Flash/life4e_15-6-OSU.swf
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Genetic composition of a population
= three aspects of variation within that
population:
„
The number of alleles at a locus.
„
The frequency of alleles at the locus.
„
The frequency of genotypes at the locus.
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example
AA Aa aa
Population 1 50 0 50
Population 2 25 50 25
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Frequency of A is 0.5
in both populations,
but the genotype
frequencies are very
different.
Sometime there are
differences among
genotypes due to the
probability of
survival.
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9
Lab 4: Hardy-Weinberg equilibrium
Page 4-3 Example- 25 diploid organisms
Typo
Allele frequency p+q=1.0
40 red
allele frequency of R = 40/50 = 0.8
(p)
10 white allele frequency of r = 10/50 = 0.2
(q)
--50 total ‘alleles’ for 25 diploid individuals
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Expected genotype frequency
p2 + 2pq + q2 = 1.0
Allele frequencies
p = 0.8
q = 0.2
Genotype frequencies
(0.8) 2 + 2(0.8x0.2) + (0.2) 2 = 1
0.64 [RR] + 0.32 [Rr] + 0.04 [rr] = 1
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Expected phenotype frequency
in population of 25
If red is dominant to white, both homozygous
RR and heterozygous Rr will look red
“red” 0.64 + 0.32 = 0.96 or 24 individuals
White 0.04
or 1 individual
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Another example
Example:
40 individuals which are AA
47 individuals which are Aa
13 individuals which are aa
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AA
Aa
aa
Total
#
individuals
40
47
13
100
#A
alleles
#a
Alleles
Total #
alleles
80
47
0
127
0
47
26
73
200
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Allele frequency
„
„
„
Allele frequency of A = 127/200 = 0.635
Frequency of A = 0.635 (p)
Frequency of a = 73/200 = 0.365 (q)
( also = 1- pA)
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Genotype frequency
40 AA 47 Aa 13 aa = 100 Total individuals
[p is the frequency of a genotype]
pAA = 40/100 = 0.4
pAa = 47/100 = 0.47** incomplete dominance
paa = 13/100 = 0.13
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Assumptions
(1) Organism is diploid
(2) Reproduction is sexual
(3) Generations are non-overlapping
(4) Mating occurs at random
(5) Population size is very large
(6) Migration is zero
(7) Mutation is zero
(8) Natural selection does not affect the gene in
question
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Freeman Figure 24.1
Good example of how to calculate allele and
genotype frequencies
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