1. 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.

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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.

8. Genetic Linkage

10. Two pops may have the same
allele frequencies but different
chromosome frequencies.

11. Conditions for Linkage Equilibrium
(Two locus case)
Frequency of B is the same on chromosome A and a
Frequency of haplotypes AB, ab, Ab, aB can be calculated
from allele frequencies A=p, a=q, B=t, and b=s.
f(AB) = pt f(ab) = qs f(Ab) = ps f(aB) = pt
3) Coefficient of linkage disequilibrium (D) = 0
f(AB) x f(ab) - f(Ab) x f(aB) = D
ps x qt - pt x qs = 0
(ranges from to +0.25)

12. Hardy Weinberg Principle for Two Loci
Chromosome frequencies remain unchanged across generations
if loci are in linkage equilibria.
If loci are in linkage disequilibria, the chromosome frequencies
will move closer to linkage equilibrium each generation.
What Creates Linkage Disequilibrium in a Population?
selection, drift, and population admixture

13. Effects of Selection on Chromosome Frequencies
ab/ab = body size of value 10
each additional A or B adds 1 value
predators eat all individuals of size 12 or less
65.28%
survive

14. Locus A and B are in disequilibrium among the survivors
Freq a = ( ) / .6528)/2
= 0.24
Freq b = ( ) / .6528)/2
= 0.09
Freq ab = .24 x .09 = .02
however!
65.28%
survive
Locus A and B are in disequilibrium
among the survivors