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Lecture 3: Allele Frequencies
and Hardy-Weinberg
Equilibrium
August 24, 2015
u Review of genetic variation and
Mendelian Genetics
 Sample calculations for Mendelian expectations: see
solutions in excel file on website
u Methods for detecting variation
 Morphology
 Allozymes
 DNA Markers (deferred to Friday: Guest lecture)
ä Anonymous
ä Sequence-tagged
Today
Introduction to statistical distributions
Estimating allele frequencies
Introduction to Hardy-Weinberg
Equilibrium
Using Hardy-Weinberg: Estimating allele
frequencies for dominant loci
Statistical Distributions: Normal Distribution
Many types of estimates follow normal distribution
 Can be visualized as a frequency distribution (histogram)
 Can interpret as a probability density function
2 sd
1 sd
1 n
Expected Value (Mean): x   xi
n i 1
where n is the
number of samples
Variance (Vx): A measure of
the dispersion around the
mean:
1 n
Vx 
( xi  x ) 2

n  1 i 1
Standard Deviation (sd): A
measure of dispersion around
the mean that is on same scale
as mean
sd  Vx
Standard Error of Mean
Standard Deviation is a measure of how individual points
differ from the mean estimates in a single sample
Standard Error is a measure of how much the estimate
differs from the true parameter value (in the case of
means, μ)
 If you repeated the experiment, how close would you expect
the mean estimate to be to your previous estimate?
Standard Error of the Mean (se):
95% Confidence Interval:
Vx
se 
n
x  1.96( se)
Estimating Allele Frequencies, Codominant Loci
Measured allele frequency is maximum likelihood estimator
of the true frequency of the allele in the population (See
Hedrick, pp 82-83 for derivation)
p
1
N12
2
N
N11 
Expected number of observations of allele A1: E(Y)=np
 Where n is number of samples
 For diploid organisms, n = 2N , where N is number of
individuals sampled
Expected number of observations of allele A1 is analogous
to the mean of a sample from a normal distribution
Allele frequency can also be interpreted as an estimate of
the mean
Allele Frequency Example
Assume a population of Mountain Laurel
(Kalmia latifolia) at Cooper’s Rock, WV
Red buds: 5000
Pink buds: 3000
White buds: 2000
A1A1
A1A2
A2A2
Phenotype is determined by a single,
codominant locus: Anthocyanin
What is frequency of “red” alleles (A1), and “white”
alleles (A2)?
Frequency of A2 = q
Frequency of A1 = p
p
1
N12
2 N11  N12
2

,
N
2N
N11 
q
1
N12
2 N 22  N12
2

,
N
2N
N 22 
Allele Frequencies are Distributed as Binomials
Based on samples from a population
 For two-allele system, each sample is like a “trial”
 Does the individual contain Allele A1?
 Remember, q=1-p, so only one parameter is estimated
Binomials are variables that can be interpreted as the
number of successes and failures in a series of trials
 n  y n y
P(Y  y)   s f ,
 y
Number of ways
of observing y
positive results
n 
in n trials
 
where s is the probability
of a success, and
f is the probability of a
failure
Probability of
observing y positive
results in n trials once
n!
 y   C  y!(n  y )!
 
n
y
Given the allele frequencies that you
calculated earlier for Cooper’s Rock
Kalmia latifolia, what is the probability
of observing two “white” alleles in a
sample of two plants?
Variation in Allele Frequencies, Codominant Loci
Binomial variance is pq or p(1-p)
Variance in number of observations of A1: V(Y) = np(1-p)
Variance in allele frequency estimates (codominant, diploid):
Vp 
p(1  p)
2N
Standard Error of allele frequency estimates:
SE p 
p(1  p)
2N
Notice that estimates get better as sample size increases
Notice also that variance is maximum at intermediate allele
frequencies
Maximum variance as a function of allele
frequency for a codominant locus
0.3
0.25
p (1-p )
0.2
0.15
0.1
0.05
0
0
0.1
0.2
0.3
0.4
0.5
p
0.6
0.7
0.8
0.9
1
Why is variance highest at intermediate
allele frequencies?
p = 0.5
p = 0.125
If this were a target, how variable would your outcome
be in each case (red versus white hits)?
Variance is constrained when value approaches limits (0 or 1)
What if there are more than 2 alleles?
General formula for calculating allele frequencies in
multiallelic system with codominant alleles:
1 n
N ii   N ij
2 j 1
pi 
, ji
N
Variance and Standard Error of allele frequency estimates
remain:
V pi 
pi (1  pi )
SE pi 
2N
pi (1  pi )
2N
How do we estimate allele frequencies for
dominant loci?
Codominant locus
-
+
A1A1
A1A2
A2A2
Dominant locus
A1A1
A1A2
A2A2
Hardy-Weinberg Law
After one generation of random mating,
single-locus genotype frequencies can be
represented by a binomial (with 2 alleles)
or a multinomial function of allele
frequencies
( p  q)  p  2 pq  q
2
Frequency of A1A1 (P)
2
Frequency of A1A2 (H)
2
Frequency of A2A2 (Q)
Hardy-Weinberg Law
Hardy and Weinberg came up with this
simultaneously in 1908
After one generation of random mating,
single-locus genotype frequencies can be
represented by a binomial (with 2 alleles)
or a multinomial function of allele
frequencies
( p  q)  p  2 pq  q
2
Frequency of A1A1 (P)
2
Frequency of A1A2 (H)
2
Frequency of A2A2 (Q)
Hardy-Weinberg Equilibrium
After one generation of random mating,
genotype frequencies remain constant, as
long as allele frequencies remain constant
Provides a convenient Neutral Model to
test for departures from assumptions
Allows genotype frequencies to be
represented by allele frequencies:
simplification of calculations
New Notation
Genotype
AA
Aa
aa
Frequency
P
H
Q
Allele
A
a
Frequency
p
q
Hardy-Weinberg Assumptions
Diploid
Large population
Random Mating: equal probability of
mating among genotypes
No mutation
No gene flow
Equal allele frequencies between sexes
Nonoverlapping generations
Graphical Representation of
Hardy-Weinberg Law
(p+q)2 = p2 + 2pq + q2 = 1
Relationship Between Allele
Frequencies and Genotype Frequencies
under Hardy-Weinberg
Hardy-Weinberg Law and Probability
A(p)
a(q)
A
(p)
AA (p2)
Aa (pq)
a
(q)
aA (qp)
aa (q2)
p2 + 2pq + q2 = 1
How does Hardy-Weinberg Work?
Reproduction is a sampling process
Example: Mountain Laurel at Cooper’s Rock
Red Flowers: 5000
Pink Flowers: 3000
White Flowers: 2000
Alleles:
: A2=14
: A1=26
A1A1
A1A2
A2A2
Frequency of A1 = p = 0.65
Frequency of A2 = q = 0.35
What are expected numbers of phenotypes and
genotypes in a sample of 20 trees?
What are expected frequencies of alleles in pollen and ovules?
Genotypes:
: 4
: 10
: 6
Phenotypes:
: 4
: 10
: 6
What will be the genotype and
phenotype frequencies in the next
generation?
What assumptions must we make?
What about a 3-Allele System?
Alleles occur in gamete pool at same frequency as in adults
Probability of two alleles coming together to form a zygote
is A B
U
A1 (p)
Pollen Gametes
A2 (q)
A3 (r)
A1A1 = p2
A1A2 = 2pq
A1
(p)
A1A3 = 2pr
A2A2 =
q2
A3A3 = r2
Ovule Gametes
A2A3 = 2qr
A2
(q)
A3
(r)
From Neal, D. 2004. Introduction to Population Biology.
Equilibrium
established with
ONE
GENERATION of
random mating
Genotype
frequencies remain
stable as long as
allele frequencies
remain stable
Remember
assumptions!