Quantitative Genetics
Hardy Weinberg Equilibrium
Dr. S.A. Mohammadi
Department of Plant Breeding & Biotechnology, Faculty of
Agriculture,
University of Tabriz, Tabriz, Iran
mohammadi@tabrizu.ac.ir
The Hardy-Weinberg Principle (1908)
Five H-W Equilibrium assumptions: If:
1.
The population size is very large
2.
Random mating is occurring
3.
No mutation occurs
4.
No selection occurs
5.
No alleles transfer in or out of the population (no migration occurs)
➢
Then allele frequencies in the population will remain constant through
future generations
➢ Genotype frequencies in progeny can be predicted from gene
frequencies of the parents
➢ Equilibrium attained after one generation of random mating
Simplifying Assumptions for Hardy-Weinberg Principle
1) diploid organisms
2) sexual reproducing organisms
3) generations are non-overlapping
4) all genotypes are equally viable
➢ If these simplifying assumptions are not met, it complicates the
mathematics for the analyses
➢ Whether or not these assumptions are all met, biologists can use
Mendelian ratios and Hardy-Weinberg analysis to measure rates of
evolution
H-W Equilibrium: Example
✓
✓
✓
✓
49 red-flowered RR
42 pink-flowered Rr
9 white-flowered rr
[100 diploid individuals carry 200
alleles]
49 RR and 42 Rr offspring have
(49 + 49 + 42 = ) 140 R alleles
42 Rr and 9 rr have
(42 + 9 + 9 = ) 60 r alleles
Freq of R = 140/200 = 0.7 and freq.
of r = 60/200 = 0.3
No allele freq. change in the F1
The Hardy-Weinberg Principle (1908)
Requirements of HW
Evolution
Violation
Large population size
Genetic drift
Random Mating
Inbreeding & other
No Mutations
Mutations
No Natural Selection
Natural Selection
No Migration
Migration
An evolving population is one that violates
Hardy-Weinberg Assumptions
Steps of deduction in the
proof of the H-W law, and
the conditions that must
hold
• Based on stable allele
frequencies and random
mating
• p2 +2pq + q2
Hardy-Weinberg Equilibrium (1908)
Genes in parents
A1
A2
A1A1
A1A2
A2A2
p
q
P = p2
H = 2pq
Q = q2
0.4
0.6
0.16
0.48
0.36
Frequencies
Example
Genotypes in progeny
➢ Note: for multiple alleles, expected genotype frequencies can be found
by expanding the binomial (p1 + p2)2
➢ For example, for three alleles:
2
2
2
p
+
p
+
p
=
p
+
2
p
p
+
2
p
p
+
p
+
2
p
p
+
p
( 1 2 3)
1
1 2
1 3
2
2 3
3
2
Hardy-Weinberg Equilibrium (1908)
Relationship between gene and genotype frequencies
1
➢ f(A1A2) has a maximum of 0.5,
➢ Most rare alleles occur in
heterozygotes
➢ Implications for
➢ F1?
➢ F2?
➢ Any BC?
0.8
Genotype frequency
which occurs when p=q=0.5
0.9
A1A1
0.7
A2A2
A1A2
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Frequency of A2
0.8
0.9
1
Applications of the Hardy-Weinberg Law
➢ Use frequency of recessive genotypes to estimate the frequency of
a recessive allele in a population
✓ Example: assume that the incidence of individuals homozygous
for a recessive allele is about 1/11,000.
q2 = 1/11,000
q  0.0095
➢ Estimate frequency of individuals that are carriers for a recessive
allele
p = 1 - 0.0095 = 0.9905
2pq = 0.0188  2%
Number of carrier = 206.8  207
Applications of the Hardy-Weinberg Law
✓
Frequency of the recessive: q2 = 1/20,000 = 0.00005
✓
Calculate the q value: √q2 = √ 0.00005 = 0.007
✓
Use the second equation: p + q = 1 to solve for p, p + 0.007 = 1, p =
0.993
Use p value to solve p2 and 2pq
p2 = (0.993) 2
p2 = 0.986
2pq = 0.014,
Number of carrier = 280
2pq = 2(0.993)(.007)
Testing for Hardy-Weinberg Equilibrium
All genotypes must be distinguishable
Genotypes
Gene frequencies
A1A1
A1A2
A2A2
A1
A2
Observed
233
385
129
0.5696
0.4304
Expected
242.36
366.26
138.38
N = N11+ N12+ N22= 233 + 385 + 129 = 747
pˆ1 =
N11 + 0.5 * N12
 1
= (233 +  385 ) / 747 = 0.5696
N
2
2
E (N11 ) = pˆ1 * N = (0.5696 ) * 747 = 242 .36
2
Chi-square test for Hardy-Weinberg Equilibrium
χ =
2
(Obs - Exp )2 = 1.96
Exp
2
critical χ1df
= 3.84
O-E
-9.36
18.74
-9.38
(O-E)2
87.61
351.19
87.98
(O-E)2/E
0.36
0.96
0.64
Σ((O-E)2/E)
1.96
c2, 1df
3.84146
2
Prob>c
calculated value of chi-square
critical value
0.16193
only 1 df because gene frequencies are estimated from the progeny data
➢ Accept H0: no reason to think that assumptions for Hardy-Weinberg
equilibrium have been violated
✓ does not tell you anything about the fertility of the parents
➢ When you reject H0, there is an indication that one or more of the
assumptions is not valid
✓ does not tell you which assumption is not valid
Chi-square test for H-W Equilibrium: Example 1
In a population with 200 genotypes, for a locus with an A/G SNP polymorphism,
the number of individuals with AA, AG and GG are 40, 70 and 90, respectively. Is
the population in H-W equilibrium for this locus?
✓ The proportional deficiency of heterozygotes (F) can be calculated based on
observed heterozygosity (0.350) and expected heterozygosity (2pq = (2)
(0.375)(0.625) = 0.469.
F = (0.350 - 0.469) / 0.469 = 0.253.
✓ A deficiency of heterozygotes is also called the Inbreeding Coefficient (F), if it is
attributable to selective union of similar gametes, and (or) selective mating of
similar genotypes.
Chi-square test for H-W Equilibrium: Example 2
For a locus with three alleles A, C and T, the number individuals for six possible
genotypes are as table. Is the population in H-W equilibrium for this locus?
✓ The same principles of two alleles locus are applied to calculations for a threealleles locus.
Exact Test for Hardy-Weinberg Equilibrium
➢ Chi-square is only appropriate for large sample sizes
➢ If sample sizes are small or some alleles are rare, Fisher’s Exact test is a
better alternative
N
N! n A ! na !2 Aa
Pr(N AA ,N Aa ,Naa n A , na ) =
N AA ! N Aa ! Naa ! (2N )!
✓ Calculate the probability of all possible arrays of genotypes for the
observed numbers of alleles
✓ Rank outcomes in order of increasing probability
✓ Reject those that constitute a cumulative probability of <5%
Exact Test for Hardy-Weinberg Equilibrium
Observed: 9 AA, 1 Aa, and 30 aa genotypes
N
N! nA ! na !2 Aa
Pr(N AA ,N Aa ,Naa nA , na ) =
N AA ! N Aa ! Naa ! (2N )!
N
40
40
40
40
40
40
40
40
40
40
NAA
9
8
7
6
5
0
4
1
3
2
NAa
1
3
5
7
9
19
11
17
13
15
Naa
30
29
28
27
26
21
25
22
24
23
nA
19
19
19
19
19
19
19
19
19
19
na
61
61
61
61
61
61
61
61
61
61
Probability
0.0000
0.0000
0.0001
0.0023
0.0205
0.0594
0.0970
0.2308
0.2488
0.3411
Cumulative
Probability
0.0000
0.0000
0.0001
0.0024
0.0229
0.0823
0.1793
0.4101
0.6589
1.0000
Reject H0
Reject H0
Reject H0
Reject H0
Reject H0
Accept H0
Accept H0
Accept H0
Accept H0
Accept H0
Likelihood Ratio Test
( )
( )

L r z

=
L z
• Maximum of the likelihood function given the data (z)
when some parameters are assigned hypothesized
values
• Maximum of the likelihood function given the data (z)
when there are no restrictions
When the hypothesis is true:
(

) ( )

LR = −2 ln  = −2 L r z − L  z
c2 df=#parameters assigned values
Likelihood ratio tests for multinomial proportions are often
called G-tests (for goodness of fit)
Lynch and Walsh Appendix 4
Likelihood Ratio Test for HWE
 Nˆ ij
G = −2 Nij ln
 Nij
i =1 j  i

n
n




where N̂ ij is the expected number
and Nij is the observed number of the ijth genotype
Genotypes
A1A1 A1A2 A2A2
Observed 233 385 129
Expected 242.36 366.26 138.38
O*ln
(E/O)
9.1768
- 9.0546
8 19.211
5
G=
1.96
Prob>c 0.1615
2
3
Exercise
Consider a population of wildflowers that is incompletely dominant for color:
✓ 320 red flowers (CRCR)
✓ 160 pink flowers (CRCW)
✓ 20 white flowers (CWCW)
➢ Examine the H-W equilibrium of this locus in the population under study
HW: Does
it Matter
PlantBreeder?
HW: Does
it Mattertotothe
the Plant
Breeder?
➢ All selection theory that is the underpinning of plant breeding based on HW
populations
➢ One classic example of an HW population is the Aztec farmer’s field of open
pollinated (OP) maize
✓ Each plant has the opportunity to mate with any other plant
✓ OP varieties were commonly grown before hybrids took hold
✓ OP landraces are still common in developing
➢ The other classic example of an HW pop is the F2 generation of a self
pollinated crop like wheat or soybean
✓ In F2, the frequency of the A and the a alleles are 0.5 and the frequency
of AA, Aa and aa are 0.25, 0.5 and 0.25, respectively