Allele Frequency and The HardyWeinberg Equation

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Allele Frequency and The Hardy-Weinberg Equation
p q ? p22pqq2
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Genetic Variation Is Exhibited By All Organisms
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Objectives

• Allele Frequency
• - Random Mating
• - Hardy-Weinberg Rule/Law
• Calculating Allele Frequency
• Factors that Affect Allele Frequency

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Population

• A group of animals that ______ a common gene
  pool.
• e.g. - Herd of Cattle
• - Flock of Sheep at the OSU - Sheep Farm
• - Steelhead in ? River

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Allele Frequency
Relative _______ of a allele in a population
compared to the abundance of all the alleles of
that locus...
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Random Mating
an animal in a population has the ____
probability of mating with any animal of the
opposite sex in that population.
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Selective or mating, preference
exists in mate selection. Example Men with
brn. hair choosing women with blonde hair as
mates.
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• Calculating
• Allelic Frequency

• Genotypic Frequency

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• Example Two alleles at one locus are responsible
  for a trait. The frequency of the 1st allele
  in the population could vary, 0 100.The 2nd
  allele, 100 - the of the 1st allele.
• The frequency of the 2 sums to 100 or 1.
• Knowing the frequency of a recessive gene
  predicts the likelihood of its re-appearing
  in the next generation, and
• how many carriers exist.

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Number of copies of a specific allele in the
population
Allele frequency
Total number of alleles at that specific locus in
the population
Number of individuals with a specific genotype in
that population
Genotype frequency
Total number of individuals in that population
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• Consider a population of 100 hypothetical cows
  - tail length, normal vs short
• 64 normal tails with the genotype TT
• 32 normal tails with the genotype Tt
• 4 short tails with the genotype tt

Frequency of allele t
Homozygotes have two copies of allele t
Heterozygotes have only one
(2)(4) 32
Frequency of allele t
(2)(64) (2)(32) (2)(4)
All individuals have two copies of each allele
Frequency of allele t
0.2, or 20
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Hardy-Weinberg Rule / Equation

• A mathematical expression developed
  independently
• ________, a British mathematician
• and _________, a German physician.
• Explains how alleles are distributed in a
  randomly mated population.

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Hardy-Weinberg Rule / Equation

• To apply, there must be
• mutation,
• migration,
• nor selection at the locus involved.
• Population has to be large
• p2 2pq q2 1

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The Hardy-Weinberg Equation in Action!

• A polymorphic gene exists with two alleles,
  A and a
• The frequency of allele A is represented by p
• The frequency of allele a is represented by q
• p q 1
• Hardy-Weinberg equation states that
• (p q)2 1
• p2 2pq q2 1

Genotype frequency of AA
Genotype frequency of Aa
Genotype frequency of aa
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• Allele frequencies p 0.8 and q 0.2, and if
  the population is in Hardy-Weinberg equilibrium,
  then Genotype frequencies
• AA p2 (0.8)2 0.64
• Aa 2pq 2(0.8)(0.2) 0.32
• aa q2 (0.2)2 0.04

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Genotype Freq
X

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Calculating the Allele Frequency
Count the genes (genes exist in the pairs)
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Example of the the H-W Rule
Shorthorn cattle with Red (RR), Roan (Rr), and
White (rr) colors. Alleles are R r,
co-dominance . Frequencies of the alleles
present is unknown. Let p and q, frequencies
for alleles, R and r. Frequency of the RR
genotype probability (p) that an ovum contains
the R allele and (p) the sperm also contains R is
p x p p 2
Each genotype has its own phenotype
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Frequency of genotype rr probability (q) that
an ovum contains the r allele and the (q) sperm
also contains r is q x q q2
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Frequency of the genotype Rr - probability (p)
ovum contains allele R and the (q) sperm contains
allele r p x q pq The probability of the
opposite case (rR) is the same --- total
probability of the Rr genotype
2 x pq or 2pq Thus, p2 2pq q2
1
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After 1 generation of random mating, genes R and
r with frequencies p and q produce offspring as
follows RR p2 Rr pq or Rr
2pq rR qp rr q2 The sum of genotype
frequencies in the new generation is RR p2 2
pq Rr rr q2
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Applying Allelic Frequencies to the Equation
If p and q .7 and .3 (allele freqs.), p2
2pq q2 (genotype freqs.) p2 (.7) 2
.49 2pq .42 q2 (.3) 2 .09
1.
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H-W for Genes Expressing Inc-dominance (ea
genoty. differ. phenoty.)
Example Black(BB), blue (Bb) and white (bb) in
the chicken Total of chickens in the
population 150 ( of alleles 300
autosomes BB 95 birds ( of B alleles 190) Bb
50 birds ( of B alleles 50, of b alleles
50) bb 5 birds ( of b alleles
10) Calculate p (freq.) B Total of B alleles
in the population 190 50 240 Total of
all alleles in the population 190 50 50
10 300 or 150 x 2 300 p (B)
240/300 0.8 then q (b) 1 - 0.8 0.2
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H-W for Alleles Expressing Complete Dominance
(only 2 phenoty.)
Example The presence of the white belt in
cattle, Beltgtnon-belt Total of cattle 230
of belted cattle 147 (BB and Bb are
phenotypically identical) of non belted cattle
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How many Belted and nonBelted alleles are in
this herd?
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The phenotype frequency (non belted cattle)
its genotype frequency, bb frequency of belted
cattle q2 Calculation of q 2 83 /
230 q 2 0.361 square root of q 2 q
.601
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What is the Frequency of p ?
The phenotype frequency (belted cattle) the sum
of BB and Bb genotype frequencies , frequency of
belted cattle p 2 2pq
From H-W Law p2 2pq q 2 1, q2
.361 p2 2pq 1 -
.361 p2 2pq .639 ......use q2 0.361
(easy way) q .601 p q 1 p 1- q, p
1 - .601 or .399
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H-W for a Sex-Linked Trait Example color
blindness in humans A normal vision a
red-green color blindness XAXA norm female-not
carrier XAXa norm- carrier XaXa
color blind female XAY normal male XaY
color-blind male 1000 males and 1000 females
screened 61 males are color-blind 4 females
are color-blind
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61 males are color-blind
4 females are color-blind
One allele is present in the heterogametic
organism (male in mammals), the allele frequency
genotype frequency. The frequency of the male
genotype is 61/1000 0.061 q 61 / 1000
0.061 and p 1 - q, p 1 - 0.061 0.939
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Assume the same q as in the males, what are the
gene frequencies in 1000 females? A - norm
vision, a - red-green color blindness XAXA -
norm female-not carrier, XAXa - norm female,
carrier XaXa - color blind female XAY
- normal male XaY - color-blind male q
0.061 p 0.939
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Assuming the same q and p as in male group, how
many females (1000) are carriers ? 1000 x 2pq
1000 x 2 x 0.939 x 0.061 114 How many are
color-blind? 1000 x q2 1000 x (0.061)2 3.7
(4)
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Relationship between allele frequencies and
genotype frequencies according to the
Hardy-Weinberg equilibrium
This genotype predominates when the frequency of
allele a is intermediate
This genotype predominates when the frequency of
allele a is low
This genotype predominates when the frequency of
allele a is high
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Factors that Affect Allele Frequency

• Selection
• Mutation
• Genetic drift
• Migration
• Mixing Populations

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Selection
Process of determining which animals breed and
produce the next generation. Natural selection -
"biology" of the animal determines the potential
to breed. Artificial selection - potential to
breed is controlled (by humans).
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Example horned and polled cattle
H - polled p (H) 0.5 h - horned q (h)
0.5 Assume Random mating population, N 100
animals How many animals are HH ?
N.p2 100 x (.5)(.5) 25 Hh ?
N.2pq 100 x 2 (.5)(.5) 50 hh ?
N.q2 100 x (.5)(.5)
25 Completely dominant gene! Phenotypically,
HHHh HH 75 polled Hh
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Selection against horns hh animals are
culled Population has only p2 2pq
75 animals (Selection continued) How many
alleles in new population? 2 x 75 150 How
many heterozygous animals? no. of HH
25 75 no. of Hh 50 How many alleles in
heterozygous animals? 50
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What is the h allele frequency in the NEW
population? q 50/150 1/3 or .33 Frequency
of H allele in new population? p q 1 p 1
- q p 1 - .33 .67 Difference between new
and old populations Old New After 1
gen of Sel. p 0.5 p 0.67 q
0.5 q 0.33
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Selection Against a Recessive Allele
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• Mutation
• The allele undergoes a change. Mutation is
• the original source of genetic variation and
• causes small changes in allelic frequency per
  generation.
• Types of the mutations
• Spontaneous -- appears without apparent
  explanation
• Induced -- caused by mutagenic agents x-ray,
  chemicals, etc.

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Mutation The number of alleles in the population
can be altered by mutation. If p 0.5 q
0.5 Assume 50 of the dominant allele mutated
to a recessive form -------------- 50 of 0.5
0.25 In the new population p 0.5 -
0.25 0.25 q 0.5 0.25 0.75
1.00
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Change in gf, Mutation is the only Influence
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Migration
New genetic material, animals being introduced
into the mating population. Example A colony of
gray house bats exists in So. Oregon. Established
colony acquires 3 new gray bats from So.
Mississippi (someones pets escaped)
interbreeding occurs.
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Genetic Drift
Random fluctuations of gf due to sampling errors,
occurs in all populations -- in small populations
large changes in allelic frequency. Example
Population of 5 animals (10 alleles) All
alleles have the same chance of being in the
gametes, the proportion of one allele may be gt
than other due to a chance. Like flipping a
coin - the more flips the better chance that the
results are 11
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Example - Codominance
Cattle Roan (R and r) 40 Red, 58 Roan, 2 White
What are the gene frequencies? p and q can be
determined by adding all the alleles in the
population. (The three phenotypes are
distinguisable). How many R alleles are
present? What is the total number of alleles in
this population? p ? P 138 / 200 .69 q ?
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Is This Herd in H-W Equilibrium at the Roan Locus
?
Allele freqs .69 and q .31 the genotypic and
phenotypic frequencies should be (pq)2
P22pqq2 Expect 48 Red 42 Roan 10
White Observed 40 Red 58 Roan 2 White
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• Are the observed numbers in agreement with the
  expected?
• How do we evaluate this in an unbiased manner?
• Goodness of Fit Test
• Calc X2 with 2 df _at_ .05 13.8 What is your
  conclusion?

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Example - Complete Dominance
Combs in chicken Rose, R gt Single, r
gt
Flock contains 5 Rose combs , 95 Single
combs The allele frequencies are ? q2 (Single
comb) .95 q .975 p 1- q 1 -
.975 .025
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If this flock contains 135 birds,
How many are heterozygous Rose comb? _______
birds x 2pq (2x.975x.025) _____ How many are
homozygous for Rose comb? 135 birds x p2 135 x
(.025)2 .08