1. Linkage and Linkage Disequilibrium
AB = 25%
Ab = 25%
aB = 25%
ab = 25%
f(Ab) = f(A) x f(b) A B B b a b A a
AB = 50%
Ab = 0%
aB = 0%
ab = 50% A B
f(Ab) ≠ f(A) x f(b) B b a 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)

3. Are alleles at separate loci paired at random?
Linkage equilibrium:
Are alleles at separate loci paired at random?
D = x11 − p1q1
D = x22 − p2q2

4. 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

5. LD as a two-locus Hardy Weinberg problem

6. Linkage disequilibrium (LD) decays with distance and time
AB = (1-r)/2
Ab = r/2
aB = r/2
ab = (1-r)/2 A B a b
r =
Rate of
recombination

7. Empirical demonstration of the
Decay of LD over time

11. Epistasis

15. QTL for flower traits in Mimulus (monkey flowers)
Different
pollinators
M. lewisii F1 M. cardinalis
F2’s

16. Genetic map of monkey flower

17. 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
Small
Large
Association between
Molecular marker (M)
and QTL(Q)
M, Q
m, q

18. detecting an association between a genetic marker (M)
QTL Mapping:
detecting an association between a genetic marker (M)
and a gene affecting a quantitative trait (Q).
QTL here
Marker here
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

19. 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

20. Genome wide association studies: GWAS
Mutation “causing”
variation in height
Tall A
Tall A
Tall A
Tall A
Short G
Short G
Short G
Short G
Adjacent SNPs are linked
Distant sites show no genotype-phenotype association
Problem: how do we find the causal SNPs? Needle in a haystack

21. 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
Small
Large
Association between
Molecular marker (M)
and QTL(Q)
M, Q
m, q

22. Human Chimp How does DNA evolve?
1 ATGCCCCAACTAAATACTACCGTATGGCCCACCATAATTACCCCCATACT 50
||||||||||||||||| ||||||| |||||||||||||||||||||||
1 atgccccaactaaataccgccgtatgacccaccataattacccccatact 50
51 CCTTACACTATTCCTCATCACCCAACTAAAAATATTAAACACAAACTACC 100
||| |||||||| ||| |||||||||||||||||||||| |||| ||||
51 cctgacactatttctcgtcacccaactaaaaatattaaattcaaattacc 100
101 ACCTACCTCCCTCACCAAAGCCCATAAAAATAAAAAATTATAACAAACCC 150
| ||||| ||||||||||| ||||||||||||||||| || || ||||||
101 atctacccccctcaccaaaacccataaaaataaaaaactacaataaaccc 150
151 TGAGAACCAAAATGAACGAAAATCTGTTCGCTTCATTCATTGCCCCCACA 200
||||||||||||||||||||||||| |||||||||||| ||||||||||
151 tgagaaccaaaatgaacgaaaatctattcgcttcattcgctgcccccaca 200
201 ATCC 204
||||
201 atcc 204

23. 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

24. Jukes Cantor One parameter model
= rate of substitution
PA(t) = ¼ + ¾ e-4at = probability that A remains A at time t
PNN = ¼ + ¾ e-8at = 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)

25. Kimura two-parameter modelb
a = rate of transition substitution
b = rate of transversion substitution
PAA(t) = ¼ + ¾ e-4bt + ½ e-2(a+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

27. Comparison of models
P-distance
Jukes Cantor
Kimura 2-parameter
Tamura-Nei
Etc…

28. 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

29. Anatomy of a phylogenetic tree
Terminal (external) nodes
Taxa =
OTUs =
Operational
taxonomic units
Taxon1
Taxon2
Taxon3
Taxon4
Taxon5
Taxon6
Polytomy
Non-dichotomous
splitting
External branch
Internal branch
Internal nodes
Root

30. 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

32. DNA test of neutrality Antigen binding sites: dN/dS > 1
“positive” selection
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 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
Rest of molecule: dN/dS < 1
Negative (purifying) selection

33. Maximum likelihood Likelihood of observing the data set
OTU1
OTU1
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
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
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
HTU1
HTU1
OTU1
OTU1