Transcript GLYPHOSATE RESISTANCE Background / Problem

Lecture 3: Allele Frequencies
and Hardy-Weinberg
Equilibrium
August 27, 2012
Last Time
u Review of genetic variation and
Mendelian Genetics
u Methods for detecting variation
 Morphology
 Allozymes
 DNA Markers
ä Anonymous
ä Sequence-tagged
Today
Sequence probability calculation
Molecular markers: DNA sequencing
Introduction to statistical distributions
Estimating allele frequencies
Introduction to Hardy-Weinberg
Equilibrium
Using Hardy-Weinberg: Estimating allele
frequencies for dominant loci
If nucleotides occur randomly in a genome,
which sequence should occur more
frequently?
AGTTCAGAGT
AGTTCAGAGTAACTGATGCT
What is the expected probability of each
sequence to occur once?
How many times would each sequence be
expected to occur by chance in a 100 Mb
genome?
What is the expected probability of each
sequence to occur once?
AGTTCAGAGT
What is the sample space for the first position?
A
T
Probability of “A” at that position? 1
G
4
C
Probability of “A” at position 1, “G” at position 2, “T”
at position 3, etc.?
1 1 1 1 1 1 1 1 1 1
x x x x x x x x x  0.2510  9.54 x10 7
4 4 4 4 4 4 4 4 4 4
AGTTCAGAGTAACTGATGCT
0.2520  9.09x1013
How many times would each sequence be
expected to occur in a 100 Mb genome?
AGTTCAGAGT
9.54x10 10   95.4
7
8
AGTTCAGAGTAACTGATGCT
9.09x10 10   9.1x10
13
8
5
Why is this calculation wrong?
AGTTCAGAGTAACTGATGCT
AGT TCA GAG TAA CTG ATG CT
UCA AGU CUC AUU GAC UAC GA
Ser Cys
Phe Ile Asp Tyr
UGA AGU CUC AUU GAC UAG GA
Stop Cys Phe Ile Asp Stop
A
P( A  B)  P( A)  P( B)  P( A  B),
B
P( A  B)  P( A | B) P( B),
DNA Sequencing
Direct determination of
sequence of bases at a
location in the genome
 Shotgun versus PCR
sequencing
Dye terminators (Sanger)
and capillaries revolutionized
DNA sequencing
Modern sequencing methods
(sequencing by synthesis,
pyrosequencing) have
catapulted sequencing into
realm of population genetics
Human genome took 10 years
to sequence originally, and
hundreds of millions of
dollars
Now we can do it in a week
for <$2,000
SNPs
A Single Nucleotide Polymorphism
(SNP) is a single base mutation in
DNA.
The most common source of genetic
polymorphism (e.g., 90% of all
human DNA polymorphisms).
Identify SNP by screening a
sample of individuals from study
population: usually 16 to 48
Once identified, SNP are
assayed in populations using
high-throughput methods
Genotyping by Sequencing
New sequencing methods generate 10’s of millions of short sequences
per run
Combine restriction digests with sequencing and pooling to genotype
thousands of markers covering genome at very high density
Presence-Absence
Polymorphism
SNP
Generate 10’s of thousands of markers
for <$100 per sample
http://www.maizegenetics.net/images/stories/GBS_CSSA_101102sem.pdf
Genotyping by Sequencing Cost Example
http://www.maizegenetics.net/gbs-overview
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 A1 = p
Frequency of A2 = q
1
N11  N12
2 N11  N12
2
p

,
N
2N
1
N 22  N12
2 N 22  N12
2
q

,
N
2N
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:
SEp 
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 )
SEpi 
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)
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?
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
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