1. Polymorphism
Polymorphism: when two or more alleles at a locus exist in a population at the same time.
Nucleotide diversity:
P = xixjpij
considers # differences
and allele frequency ij
Freq
(x)
Seq 1 G A G G T G C A A C
Seq 2 G A G G A C C A A C
Seq 3 G A G C T G G A A G
p p13
p23
P = (0.4)(0.5)(0.2) + (0.4)(0.1)(0.3) + (0.5)(0.1)(0.5) = 0.077
p12
p13
p23

2. In Theory: Under infinite-sites model: Expectation (P 
4Nem = frequency of heterozygotes per nucleotide site

3. Nucleotide diversity is low in humans

4. Expectation (K
Polymorphism is also estimated by:
ATCCGGCTTTCGA
K = 3 for-->ATCCGAATTTCGA
ATTCGCCTTTCGA
K= Number of segregating (variable) sites in a sample of alleles.
In Theory:
Expectation (K
Where a = 1 + 1/2 + 1/3 +……..1/n-1

5. Coalescent Process t2 t3 t4 t5 Gene Tree tm is time for coalescence
from m to m-1 sequences t3 t4 t5
Gene Tree

6. Coalescent Process a b c d e f g h Gene Tree
The geneology of n sequences
has 2(n-1) branches.
n = number of external
branches. c d
n-2 are internal e f g h
Gene Tree

7. How long will the coalescence process take?
Simplest case: If pick two random gene copies, probability that
the second is the same as the first is 1 / (2N). This is the probability
that two alleles coalesce in previous generation.
It follows that / (2N) is the probability that two sequences
were derived from different sequences in the preceding generation.
Therefore, the probability that 2 sequences derived from the same
ancestor 2 generations ago (grandparent) is / (2N) x 1 / (2N).
It can be shown that the probability that two sequences were
derived from the same ancestor t generations ago is:
[1 - 1 / (2N)t x (1 / (2N)] ~ (1 / (2N) e(-t/(2N)

8. Consider probability of common ancestry for:
[1 - 1 / (2N)g-1 x (1 / (2N)]
Because N is in denominator, the probability will depend on sample size
Consider probability of common ancestry for:
Generations ago Prob(N=5) Prob(N=10)
It can be shown that the average time back to common ancestry
of a pair of genes in a diploid population is 2N, and the average
time back to common ancestry of all gene copies is 4N.

9. Large pop
Time back to common ancestor
Small pop

10. Coalescence with no mutation
The average degree of relationship increases with time.
All of the gene copies in a population can be traced back
to a single ancestral gene.
A population will eventually become monomorphic
for one allele or another, with this probability determined by initial allele frequencies.

11. Coalescence with mutation
If each lineage experiences m mutations per generation, then the number of base pair differences between them will be #dif = 2mtca.
If the average time to coalescence is 2N for two randomly chosen gene copies, then #dif = 2 m (2N).
Therefore, expect the
average number of base pair differences between gene
copies to be greater in a larger population.

12. Total length of branches of gene tree
I + L = J
Internal
branches
External
branches
Total time
length + =
Now consider mutation among branches
during the coalescent process.
i) + e) = 
Mutations
internal
branches
Mutations
external
branches
Total number
of mutations
in gene tree + =
In theory: total number of mutations equals the
number of segregating sites (K)

13. Testing for Selective Neutrality
Using the difference in
estimates of polymorphism
to detect deviation from
neutrality.
Tajima’ s Test (1989):
P- K / a
D =
V(P- K/a)
Normalizing factor
Rationale:
Pand K are differentially influenced by the
frequency of alleles.

14. P K/a
Few alleles at intermediate frequency
>
<
Many low frequency, variable alleles
D = 0 neutral prediction
D > 0 balancing selection
D < 0 directional selection

15. Fu and Li’s Test (1993): Using the difference in
# mutations in gene tree
to detect deviation from
neutrality.
i - e / (a - 1)
D =
V[i - e / (a - 1)
Rationale: An equivalent number of mutations is expected
between interior verses exterior branches of a neutral
gene tree.

16. i  e
Few alleles at intermediate frequency
>
<
Many low frequency, variable alleles
D = 0 neutral prediction
D > 0 balancing selection
D < 0 directional selection

17. Gene genealogies under no selection, positive
selection, balancing selection, and background selection.
No Selection : 7 neutral mutations accumulate since
the time of the last common
ancestor.
D = 0

18. Consider the Effects of Selection on Neutral Sites Linked to a Selected Site
Positive Selection : neutral variation at linked sites will be eliminated (swept away) as the advantageous allele quickly is fixed in the population. This process is
also called hitch-hiking.
Time
D < 0

19. Consider the Effects of Selection on Neutral Sites Linked to a Selected Site
Balancing Selection : neutral variation at linked sites accumulates during the long
period of time that both
allele lineages are maintained.
Time
D > 0

20. Consider the Effects of Selection on Neutral Sites Linked to a Selected Site
Background Selection : gene lineages become extinct not only by chance, but because
of deleterious mutations to which they are linked, which
eliminates some gene copies.
Time
D < 0

21. Problem: Background selection and hitchhiking are
contrasting processes that lead to the same pattern.
How to differentiate?
Dramatic examples of reduced polymorphism=hitchhiking.
Less dramatic examples=background selection.