Molecular variation
Joe Felsenstein
GENOME 453, Autumn 2013
Molecular variation – p.1/33
Views of genetic variation before 1966
The Classical view The Balancing Selection view
Hermann Joseph Muller Theodosius Dobzhansky
Most loci will be homozygous Most loci will be polymorphicfor the “wild-type allele" due to balancing selection
but a few mutants will exist with strong selection
Molecular variation – p.2/33
Gel electrophoresis
slots
gel of potato starch orpolyacrilamide
tank of buffer solution
wick
power supply
+
−
Molecular variation – p.3/33
After running the current
slots
gel of potato starch orpolyacrilamide
tank of buffer solution
wick
power supply
+
−
Molecular variation – p.4/33
Making one locus visible by staining
slots
gel of potato starch orpolyacrilamide
tank of buffer solution
wick
power supply
+
−
AA AA Aa aa Aa AA
stained
Molecular variation – p.5/33
A monomeric enzyme
AA
Aa
On gel:
Molecular variation – p.6/33
A dimeric enzyme
AA
Aa
On gel:
Molecular variation – p.7/33
Polymorphism on a gel
Molecular variation – p.8/33
Lewontin and Hubby’s 1966 work
Richard Lewontin, about 1980
Lewontin, R. C. and J. L. Hubby. 1966. A molecular approach to the studyof genic heterozygosity in natural populations. II. Amount of variation anddegree of heterozygosity in natural populations of Drosophilapseudoobscura. Genetics 54: 595-609.
Molecular variation – p.9/33
Measures of variability with multiple loci
Lewontin and Hubby (Genetics, 1966) suggested two measures ofvariability: polymorphism and heterozygosity.
Polymorphism is the fraction of all loci that have the most common alleleless than 0.95 in frequency (i.e. all the rarer alleles together add up to lessthan 0.05.
Heterozygosity is the estimated fraction of all individuals who areheterozygous at a random locus.
Molecular variation – p.10/33
Computing the average heterozygosity
If pi is the frequency in the sample of allele i at a locus, then theheterozygosity for that locus is estimated by taking the sum of squares ofthe gene frequencies (thus estimating the homozygosity) and subtractingfrom 1:
H = 1 −
alleles∑
i
p2
i
Molecular variation – p.11/33
An example:
locus 1 1locus 2 1locus 3 0.8 0.2locus 4 0.94 0.04 0.02
The heterozygosities are calculated as:locus 1 1 − 1
2 = 0
locus 2 1 − 12 = 0
locus 3 1 −
(
0.82 + 0.22)
= 0.32
locus 4 1 −
(
0.942 + 0.04
2 + 0.022)
= 0.1144
The average heterozygosity in this example is
H = (0 + 0 + 0.32 + 0.1144) / 4 = 0.1086
Molecular variation – p.12/33
Amounts of heterozygosity
Molecular variation – p.13/33
Kimura’s neutral mutation theory
Motoo Kimura and family, 1966 Tomoko Ohta, recently
Kimura, M. 1968. Evolutionary rate at the molecular level. Nature 217:624-626.
Kimura, M., and T. Ohta. 1971. Protein polymorphism as a phase ofmolecular evolution. Nature 229: 467-469.
Molecular variation – p.14/33
Neutral mutation theory
0
1
2
3
4
5
6
7
8
10
11
14
15
Heterozygosity
expected to be
4Nu
4Nu + 1at any point is
13
assume: population size N, rate u of neutral mutations, all different
9
Crow and Kimura, 1964; Lewontin and Hubby, 1966;Kimura, 1968; King and Jukes, 1969; Kimura and Ohta, 1971
16
12
Molecular variation – p.15/33
Crow and Kimura’s theoretical calculation
James F. Crow, about 1990
Kimura, M., and J. F. Crow. 1964. The number of alleles that can bemaintained in a finite population. Genetics 49: 725-738.
Molecular variation – p.16/33
Expected heterozygosity with neutral mutation
In a random−mating population with neutral mutation,a fraction F of the pairs of copies will be homozygous.Suppose all mutations create completely new alleles,and the rate of these neutral mutations is
one generation ago
F
nowF’ F’
Ndiploid population of size
u
Molecular variation – p.17/33
Expected heterozygosity with neutral mutation
of the time1/(2N)
and the rate of these neutral mutations is
one generation ago
of the time
F
nowF’ F’
1−1/(2N)
Ndiploid population of size
u
In a random−mating population with neutralmutation, a fraction F of the pairs of copies
will be homozygous. Suppose all mutationscreate completely new alleles,
Molecular variation – p.18/33
Expected heterozygosity with neutral mutation
of the time1/(2N)
and the rate of these neutral mutations is
one generation ago
of the time
F
nowF’ F’
1−1/(2N)
Ndiploid population of size
To be identical, both copies must not be new mutants,
and the probability of this is
u
2(1−u)
In a random−mating population with neutralmutation, a fraction F of the pairs of copies
2 [ x ] + (1−1/(2N)) F 1(1−u) (1/ (2N)) will be homozygous. Suppose all mutationscreate completely new alleles, F’ =
Molecular variation – p.19/33
Expected heterozygosity with neutral mutation
of the time1/(2N)
and the rate of these neutral mutations is
one generation ago
of the time
F
nowF’ F’
1−1/(2N)
Ndiploid population of size
To be identical, both copies must not be new mutants,
and the probability of this is
So if we have settled down to an equilibrium level of
heterozygosity, F’ = F , so that
u
2(1−u)
F = 2 [ x ] + (1−1/(2N)) F 1(1−u) (1/ (2N))
In a random−mating population with neutralmutation, a fraction F of the pairs of copies
2 [ x ] + (1−1/(2N)) F 1(1−u) (1/ (2N)) will be homozygous. Suppose all mutationscreate completely new alleles, F’ =
Molecular variation – p.20/33
Expected heterozygosity with neutral mutation
of the time1/(2N)
and the rate of these neutral mutations is
one generation ago
of the time
F
nowF’ F’
1−1/(2N)
Ndiploid population of size
To be identical, both copies must not be new mutants,
and the probability of this is
So if we have settled down to an equilibrium level of
heterozygosity, F’ = F , so that
which is easily solved to give
F = 1 −
2
2
u
2(1−u)
(1−u)
(1−u)
F = 2 [ x ] + (1−1/(2N)) F 1(1−u) (1/ (2N))
(1/ (2N))
(1 − 1/ (2N))
In a random−mating population with neutralmutation, a fraction F of the pairs of copies
2 [ x ] + (1−1/(2N)) F 1(1−u) (1/ (2N)) will be homozygous. Suppose all mutationscreate completely new alleles, F’ =
Molecular variation – p.21/33
Expected heterozygosity with neutral mutation
of the time1/(2N)
and the rate of these neutral mutations is
one generation ago
of the time
F
nowF’ F’
1−1/(2N)
Ndiploid population of size
To be identical, both copies must not be new mutants,
and the probability of this is
So if we have settled down to an equilibrium level of
heterozygosity, F’ = F , so that
which is easily solved to give
F = 1
heterozygosity is:or to good approximation:
1 + 4N1−F =
4N
F = 1 −
2
2
u
2(1−u)
(1−u)
(1−u)
u
u
u
F = 2 [ x ] + (1−1/(2N)) F 1(1−u) (1/ (2N))
(1/ (2N))
(1 − 1/ (2N))
1 + 4 N
In a random−mating population with neutralmutation, a fraction F of the pairs of copies
2 [ x ] + (1−1/(2N)) F 1(1−u) (1/ (2N)) will be homozygous. Suppose all mutationscreate completely new alleles, F’ =
Molecular variation – p.22/33
Heterozygosity in marine invertebrates (Valentine, 1975)
species which is a samples/locus no. loci Het.Asterias vulgaris Northern sea star 19-27 26 1.1Cancer magister Dungeness crab 54 29 1.4Asterias forbesi Common sea star 19-72 27 2.1Lyothyrella notorcadensis brachiopod 78 34 3.9Homarus americanus lobster 290 37 3.9Crangon negricata shrimp 30 30 4.9Limulus polyphemus horseshoe crab 64 25 5.7Euphausia superba Antarctic krill 124 36 5.7Upogebia pugettensis blue mud shrimp 40 34 6.5Callianassa californiensis ghost shrimp 35 38 8.2Phoronopsis viridis horseshoe worm 120 39 9.4Crassostrea virginica Eastern oyster 200 32 12.0Euphausia mucronata small krill 50 28 14.1Asteriodea (4 spp.) deep sea stars 31 24 16.4Frielea halli brachiopod 45 18 16.9Ophiomusium lymani large brittlestar 257 15 17.0Euphausia distinguenda tropical krill 110 30 21.5Tridacna maxima giant clam 120 37 21.6
Molecular variation – p.23/33
An interesting case: Limulus polyphemus
Carboniferous (300 mya) Jurassic (155 mya) today
Selander, R.K., S.Y. Yang, R.C. Lewontin, W.E. Johnson. 1970. Geneticvariation in the horseshoe crab (Limulus polyphemus), a phylogenetic“relic." Evolution 24:402-414.
Molecular variation – p.24/33
An interesting case
Northern elephant seal Southern elephant sealMirounga angustirostris Mirounga leonina
Northern elephant seal: Population in 1890’s: 2-10 ?Population today: 150,000 or so (“911? help! there’s a monster dying onmy beach")
Bonnell, M.L., and R.K. Selander. 1974. Elephant seals: genetic variationand near extinction. Science 184: 908-909.
Molecular variation – p.25/33
A “population cage" for Drosophila
Molecular variation – p.26/33
Yamazaki’s population cage experiment
Yamazaki, T. 1971. Measurement of fitness at the esterase-5 locus inDrosophila melanogaster. Genetics 67: 579-603.
Molecular variation – p.27/33
Explaining Electrophoretic Polymorphisms
Can do it either way:
By neutral mutation: If H = 0.15 then if Ne = 1, 000, 000 weneed 4Neµ = 0.176 to predict this, so that implies µ = 4.4 × 10
−8.So we can explain the level of variation by a neutral mechanism.
By selection: To be effective in a populationwith Ne = 1, 000, 000 selection would need to be big enoughthat 4Nes > 1 so s > 1/4, 000, 000 which is quite small, andimpossible to detect in laboratory settings.
Molecular variation – p.28/33
DNA sequencing reveals a similar picture
Marty Kreitman
Kreitman, M. 1983. Nucleotide polymorphism at thealcohol-dehydrogenase Locus of Drosophila melanogaster. Nature 304:412-417.
Molecular variation – p.29/33
Kreitman’s sample of 11 ADH gene sequences, front end
Molecular variation – p.30/33
Kreitman’s sample of 11 ADH gene sequences, tail end
Molecular variation – p.31/33
SeattleSNPs data (Nickerson lab)
Matrix Metalloproteinase 3 SNP data
Molecular variation – p.32/33
Variation from the 1000 Genomes project
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Div
ers
ity
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
-2.5 kb
m other intron
TSS
last
CDS5�UTR first intronm m
first
CDSmiddle
CDSt tt
-250 bp TES
3�UTR
-25 kb 2.5 kb 25 kb
●
●
SNP diversity ( x 1,000)
1 bp Indel diversity ( x 1,000)
Average SNP diversity ( x 1,000)
Average 1 bp Indel diversity ( x 10,000)
TF motif (m) or miRNA binding target (t)
Annotation
SNP diversity 95% confidence interval contour
Indel diversity 95% confidence interval contour
Supplementary Figure S7
From paper:Mu, X.J., Z.J. Lu, Y. Kong, H.Y. Lam, M.B. Gerstein. 2011. Analysis of genomic variation innon-coding elements using population-scale sequencing data from the 1000 GenomesProject. Nucleic Acids Research 39(16): 7058-7076.
Molecular variation – p.33/33