Lecture Summary February 2 2001: Neutral Theory (II)
Neutral gene substitution rate
The probability of fixation P is 1/(2Ne), the number
of new mutants in a generation is 2 Ne u (where u is the mutation rate per gene per generation).
| |
2 Neu |
| K = 2 Ne P = |
---------- | = q |
| |
2 Ne |
The gene substitution rate is simply the same as the mutation rate
and is independent of the population size. This makes intuitively sense,
because in small populations we have a small number of new mutants but a high
fixation probability, whereas in large populations we have a large number
of mutants with a low fixation probability.
Example: With a mutation rate of 0.00001 we have rate of turn over of
0.00001 per generation, so that about every 100000 generation an allele
takes over.
Gene substitution rate, when there is selection
The probability of fixation P for a beneficial allele is 2s,
| K = 2 Ne P =
2 Ne 2s = 4Ne u s |
The substitution rate is dependent on the size, the mutation rate, and
the selection coefficient.
Tests of Neutrality
The neutral theory predicts several outcomes:
- Looking at two populations that doe not exchange migrants the
course of change is independent, Looking at the same gene we can expect
that that both populations have a similar amount of variability dependent
only on the population size, because the mutation rate for the gene
will be the same.
- A high mutation rate gene will have many alleles in the population,
whereas a low rate gene will have only a few if any.
Relative Rate Test
We have 3 species (a,b,c). We use c as a baseline for a and b
who are more closely related to each other than to c.
One can depict this in a structure like this ((a,b),c).
We have sequences for each of our species but do not know when a and b
were a single species in the past. If we assume neutral change then
on both lineages from the ancestor A to a and from A to b a similar amount
of mutations have accumulated. We can test this by calculating the
genetic distances d and then test if d(A, a) are d(A, b) the same.
For that we set up a system that d(a,c) = d(a,A) + d(A,c),
d(b,c) = d(b,A) + d(A,c), and d(a,b) = d(a,A) + d(A,b). This can be solved
for d(A,a), d(A,b), and d(A,c).
And then we test if d(A,a) - D(A,b) = 0.
HKA-test (Hudson-Kreitman-Aguade') and McDonald-Kreitman test
Both test use the assumption that under neutrality
the expected variation of a specific marker is the same within a species
and between species. The HKA test can use any measure of distance between
two individuals (e.g. counting the number of different sites, or a more
sophisticated measure), whereas the MK test uses the differences
(only looking the fixed differences between the species and the polymorphic
differences in a species).
- HKA: constructs a goodness of fit (Chi-Square) statistic, where we
use the observed differences within two species
(K1i, K2i; i stands for locus i) and Di
which measures the difference between the species. For each of these
quantities there is also an expected value that is derived using
neutral theory and is based on population size and mutation rate.
(Hudson, Kreitman, Aguade' 1987: A test of neutral molecular evolution
based on nucleotide data. Genetics 116: 153-159.)
- MK: the ratio between the neutral replacement and synonymous
differences within and between species should be
the same if neutrality holds.
We test if Wr/Ws = Br/Bs,
where W is the Within species number of either synomous (=does not
change protein) or replacement (=changes protein) substitution;
Bis the number substitutions between species. (McDonald and Kreitman 1991 Nature 351: 652-654).