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Linkage and Linkage Disequilibrium
AB = 25%
A
B
Ab = 25%
aB = 25%
a
b
ab = 25%
A
B
a
b
AB = 50%
Ab = 0%
aB = 0%
ab = 50%
f(Ab) = f(A) x f(b)
B b
A
a
f(Ab) ≠ f(A) x f(b)
B b
A
a
A locus and B locus are in Linkage Disequilibrium
D = f(AB) x f(ab) - f(Ab) x f(aB)
Maximum with no recombination
D = 0 with free recombination (linkage equilibrium)
Linkage equilibrium:
Are alleles at separate loci paired at random?
D = x11 − p1q1
D = x22 − p2q2
A locus and B locus are in Linkage Disequilibrium
D = f(AB) x f(ab) - f(Ab) x f(aB) Maximum with no recombination
D -> 0 with free recombination (linkage equilibrium)
When allele frequencies are intermediate: f(A) = f(a) = f(B) = f(b) = 0.5,
and maximal LD occurs so that no recombinants are present:
f(AB) = f(ab) = 0.5, so D = 0.5 x 0.5 – 0.0 x 0.0 = 0.25
When allele frequencies are skewed: f(A) = 0.9, f(a) = 0.1; f(B) = 0.9, f(b) = 0.1
and maximal LD occurs so that no recombinants are present, D is less than 0.25:
f(AB) = 0.9, and f(ab) = 0.1, so D = 0.9 x 0.1 – 0.0 x 0.0 = 0.09
LD as a two-locus Hardy Weinberg problem
Linkage disequilibrium (LD) decays with distance and time
A
B
a
b
r=
Rate of
recombination
AB = (1-r)/2
Ab = r/2
aB = r/2
ab = (1-r)/2
Empirical demonstration of the
Decay of LD over time
r

D
p1 p2 q1q2
Epistasis
QTL for flower traits in Mimulus (monkey flowers)
M. lewisii
F2’s
F1
M. cardinalis
Different
pollinators
Genetic map of monkey flower
http://www.genetics.org/cgi/content/full/159/4/1701/F1
Quantitative trait locus (QTL)mapping:
Screen for marker-trait associations in F2s or RILs
Parentals
F1
M, Q
M, Q
M, Q
M, Q
M, Q
F2
Inbreed to make
Recombinant inbred lines (RILs)
Scan genome for association
Between molecular marker and phenotype
Association between
Molecular marker (M)
and QTL(Q)
Small
m, q
Large
M, Q
QTL here
Marker here
http://isotope.bti.cornell.edu/img/intro/qtl_fig_2.gif
QTL Mapping:
detecting an association between a genetic marker (M)
and a gene affecting a quantitative trait (Q).
QTL mapping works because there is linkage disequilibrium
(LD) between the marker (M) and the QTL (Q):
mm marker genotypes are correlated with small size
MM marker genotypes are correlated with large size
Most traits in organisms
Show continuous variation
How do we find the genes
That affect these
“quantitative” traits
Scan the genome for
Nucleotide sites that
Co-vary with the
phenotype
Genome wide association studies: GWAS
Mutation “causing”
variation in height
Tall
Tall
Tall
Tall
Short
Short
Short
Short
A
A
A
A
G
G
G
G
Adjacent SNPs are linked
Distant sites show no genotype-phenotype association
Problem: how do we find the causal SNPs? Needle in a haystack
What is better: More recombination, more markers?
Parentals
F1
M, Q
M, Q
M, Q
M, Q
M, Q
F2
Inbreed to make
Recombinant inbred lines (RILs)
Scan genome for association
Between molecular marker and phenotype
Association between
Molecular marker (M)
and QTL(Q)
Small
m, q
Large
M, Q
How does DNA evolve?
Human 1
Chimp 1
51
51
101
101
151
151
ATGCCCCAACTAAATACTACCGTATGGCCCACCATAATTACCCCCATACT
||||||||||||||||| ||||||| |||||||||||||||||||||||
atgccccaactaaataccgccgtatgacccaccataattacccccatact
.
.
.
.
.
CCTTACACTATTCCTCATCACCCAACTAAAAATATTAAACACAAACTACC
||| |||||||| ||| |||||||||||||||||||||| |||| ||||
cctgacactatttctcgtcacccaactaaaaatattaaattcaaattacc
.
.
.
.
.
ACCTACCTCCCTCACCAAAGCCCATAAAAATAAAAAATTATAACAAACCC
| ||||| ||||||||||| ||||||||||||||||| || || ||||||
atctacccccctcaccaaaacccataaaaataaaaaactacaataaaccc
.
.
.
.
.
TGAGAACCAAAATGAACGAAAATCTGTTCGCTTCATTCATTGCCCCCACA
||||||||||||||||||||||||| |||||||||||| ||||||||||
tgagaaccaaaatgaacgaaaatctattcgcttcattcgctgcccccaca
201 ATCC 204
||||
201 atcc 204
50
50
100
100
150
150
200
200
Measuring DNA Evolution
•
Align sequences between
species
• Determine length of sequences,
L
• Count number of differences
• Divergence = proportion of
differences
• D = p-distance = (number of
differences) / (length of
sequence)
• Rate of divergence
  = (sequence divergence) /
(age of common ancestor)
  = D / time
• Rate of substitution
  = D / 2 x time
time
Example: 5 differences in 100
D = 0.05, t = 6 million years
Divergence = 0.05/6x106
Divergence = 8.3 x 10-9
Jukes Cantor One parameter model
= rate of substitution
PA(t) = ¼ + ¾ e-4t = probability that A remains A at time t
PNN = ¼ + ¾ e-8t = probability that two sequences have the same nucleotide at N
D = proportion of different nucleotides = 1 - PNN
Dhat = 3/4(1-e-8t)
K = - ¾ ln (1-4/3p)
where p = proportion of nucleotide differences (# diffs./total bp)
Kimura two-parameter model
 = rate of transition substitution
b
b
b
b
b = rate of transversion substitution
PAA(t) = ¼ + ¾ e-4bt + ½ e-2(+b)t
= probability that A remains A at time t
K = ½ ln(1/[1- 2P-Q]) + ¼ ln(1/[1-2Q])
where P = proportion of transitional differences
Q = proportion of transversional differences
Comparison of models
•
•
•
•
P-distance
Jukes Cantor
Kimura 2-parameter
Tamura-Nei
• Etc…
Molecular clocks
Approximately constant
Divergence of proteins
K = •f0
Rate of substitution =
Mutation rate x proportion of
neutral mutations
“Saturation” due to multiple
Hits in DNA evolution
Anatomy of a phylogenetic tree
Terminal (external) nodes
Taxa =
Taxon1
OTUs =
Operational
taxonomic units
External branch
Internal branch
Taxon2
Taxon3 Taxon4
Taxon5
Taxon6
Polytomy
Non-dichotomous
splitting
Internal nodes
Root
Relative rate test
• KAC = KBC
• KOC is shared
• Tajima test
• (m1-m2)2 / (m1+m2)
• Chi square, df=1
Species O
m1
m2
Species A
Species B
Species C
DNA test of neutrality
•
•
•
Neutral prediction:
amino acid (nonsynonymous) substitution
rate (dN) should be lower than silent
(synonymous) substitution rate (dS)
True for most genes
–
•
–
•
Follows from functional constraint
argument
Different for Major Histocompatibility
Complec (MHC) loci
–
•
Antigen binding sites: dN/dS > 1
“positive” selection
Antigen recognition sequence shows dN >
dS
Rest of molecule shows dN > dS, as
expected
Amino acid mutations are favored in
antigen recognition region
Promotes diversity, better recognition of
foreign peptides
http://depts.washington.edu/rhwlab/dq/3structure.html
Rest of molecule: dN/dS < 1
Negative (purifying) selection
Maximum likelihood
• Likelihood of observing the data set
– Assuming a given tree
– Assuming a given model of DNA evolution
• L = P(data|tree)
• Consider 4-taxon cases within a tree
OTU1
HTU1
OTU1
– For each site, Identify nucleotides at each of the four taxa
– Assume all 16 pairs of nucleotides at internal nodes
– Likelihood of observed 4 terminal nucleotides = sum of 16
independent probabilities
– Repeat likelihoods for each position in alignment
– Likelihood of tree = product of individual likelihoods
OTU1
HTU1
OTU1
• L = PLi for i = 1 to n positions in alignment (or sum of
log likelihoods)
• Calculate likelihood for other trees; choose tree with
maximum likelihood