What is Population Genetics?
Download
Report
Transcript What is Population Genetics?
Population Genetics
Dr Pupak Derakhshandeh-Peykar, PhD
Ass Prof of Medical Science of Tehran
University
Ref.: Population and Evolutionary
Genetics: A primer
1
What is Population Genetics?
The genetical study of the process of
evolution
(The study of the change of allele
frequencies, genotype frequencies, and
phenotype frequencies)
2
Population genetics:
One of the oldest and richest
examples of success of mathematical
theory in biology
Mendelian genetics and Darwinian
natural selection in the first part of
the 20th century
“modern synthesis”
3
Population Genetics is…
About microevolution (evolution within
species)
Strongly dependent on mathematical models
A relatively young science (most important
discoveries are from after 1930)
4
Factors causing genotype
frequency changes
Selection
Mutation
Random Drift
Migration
Recombination
Non-random Mating
5
What forces are responsible for
divergence among populations?
Mutation
genetic diversity
Selection
genetic diversity
Genetic drift
genetic diversity
Migration
genetic diversity
Non-random
mating
genetic diversity
6
What's the most important
factor in evolution?
SELECTION
Natural selection causes evolution:
There is variation in fitness (selection(
That variation can be passed from one
generation to the next (inheritance(
This is the central insight of Darwin
7
THEORIES of EVOLUTION
and the
DARWINIAN REVOLUTION
8
Darwin's Theory of Evolution
Four Basic Themes:
1.
2.
3.
4.
Descent with Modification from Common
Ancestor
Diversity is result of Differential Survival
and/or Differential Reproduction among
individuals
with different Heritable characteristics
= Process of Natural Selection
Law of Evolution by Natural Selection
9
Charles Darwin (1809-1882)
10
Theory of Evolution by Natural
Selection (1859)
Charles Darwin (1809-1882)
Inherited Variation among individuals
↓
Differential survival and/or reproduction
(“hard” inheritance)
↓
Change in genetic composition of population
↓↓↓↓
Evolution
11
Jean Baptiste Lamarck (1744-1829)
12
Theory of Evolution
by Inheritance of Acquired Characteristics
(1809)
Jean Baptiste Lamarck (1744-1829)
Environmental change
↓
Change in organismal form
↓
Inheritance of acquired characteristics
(“soft inheritance”)
↓
Change in composition of population
↓↓↓
Evolution
13
Lamarck’s vs. Darwin’s Theories
=انقراض
=اصالح نژاد هدفدار
14
Dates, Contributors to Evolutionary
Thinking - 1
15
Dates, Contributors to Evolutionary
Thinking - 2
16
Genes in Populations:
Hardy Weinberg
Equilibrium
17
Alleles
Alternative forms of a particular
sequence
Each allele has a frequency
18
Alleles
Yeast:
12 Mb ; 6,340 genes
Nematode
elegance: 97 Mb;
19,100 genes
Human:
3,700 Mb; 75,000 genes!
19
Methods used to measure genetic
variation:
Genetic variation contains information
about an organism’s ancestry
determines an organism’s potential for
evolutionary change, adaptation, and
survival
1960s-1970s: genetic variation was first
measured by protein electrophoresis (e.g.,
allozymes)
20
1980s-2008s: genetic variation
measured directly at the DNA level (1):
Restriction Fragement Length
Polymorphisms (RFLPs)
Minisatellites (VNTRs)
DNA sequence
DNA length polymorphisms
Single-stranded Conformation
Polymorphism (SSCP)
21
1980s-2008s: genetic variation
measured directly at the DNA level (2):
Microsatellites
(STRs)
Random Amplified Polymorphic
DNAs (RAPDs)
Amplified Fragment Length
Polymorphisms (AFLPs)
Single Nucleotide Polymorphisms
(SNPs)
22
Types of measures of genetic
variation (1):
Polymorphism = % of loci or nucleotide
positions showing more than one allele
or base pair.
Heterozygosity (H) = % of individuals
that are heterozygotes
Allele/haplotype diversity = measure of
diversity and different alleles/haplotypes
within a population.
23
Types of measures of genetic
variation (2):
Nucleotide diversity = measure of number and
diversity of variable nucleotide positions
within sequences of a population.
Genetic distance = measure of number of base
pair differences between two homologous
sequences.
Synonomous/nonsynonomous substitutions =
% of nucleotide substitutions that do not/do
result in amino acid replacement.
24
Hardy-Weinberg
equilibrium
Properties of alleles in a population
Allele frequencies
Genotypes frequencies
25
Allele Frequency
For two alleles
Usually labeled p and q = 1 – p
For more than 2 alleles
Usually labeled pA, pB, pC ...
… subscripts A, B and C
indicate allele name
26
Genotype
The pair of alleles carried by an individual
Homozygotes
If there are n alternative alleles …
… there will be n(n+1)/2 possible genotypes
The two alleles are in the same state
Heterozygotes
The two alleles are different
27
The simple part …
Genotype frequencies lead to allele
frequencies…
For example, for two alleles:
pA = pAA + ½ pAB (> p=P+1/2 H*)
pB = pBB + ½ pAB (> q=Q+1/2 H)
However, the reverse is also
possible!
*H=2pq
28
Hardy-Weinberg Equilibrium
Relationship described in 1908
Hardy, British mathematician
Weinberg, German physician
Random union of games
Shows n allele frequencies determine
n(n+1)/2 genotype frequencies
Large populations
29
Hardy-Weinberg Equilibrium
Explains how Mendelian
segregation influences allelic and
genotypic frequencies in a
population
30
Required Assumptions in
Hardy-Weinberg law (1):
Diploid, sexual organism (Parthenogenetic)
Non-overlapping generations
Autosomal locus
Large population
Random mating
Equal genotype frequencies among
sexes
31
Required Assumptions in
Hardy-Weinberg law (2):
Absence of natural selection
Population is infinitely large, to avoid
effects of genetic drift
No mutation
No migration
< If assumptions are met, population will be in
genetic equilibrium
32
Two expected predictions:
Allele frequencies do not change over generations
After one generation of random mating, genotypic
frequencies will remain in the following
proportions:
(frequency of AA)
p2
(frequency of Aa)
2pq
(frequency of aa)
q2
*p = allelic frequency of A
*q = allelic frequency of a
*p2 + 2pq + q2 = 1
33
population is at equilibrium
A(p)=0.5
a(q)=0.5
A(p)=0.5
AA(p2)
0.5 x 0.5 = 0.25
Aa(pq)
0.5 x 0.5 = 0.25
a(q)=0.5
Aa(pq)
0.5 x 0.5 = 0.25
aa(q2)
0.5 x 0.5 = 0.25
34
Random Mating:
Mating Type Frequencies
P2
2PH
2PQ
H2
2QH
Q2
35
Mendelian Segregation:
Offspring Genotype Frequencies
Total
P2
P2
2PH
2PQ
H2
2QH
Q2
PH
1
_
¼ H2
_
_
p2
_
PH
2PQ
½ H2
QH
_
2pq
_
_
_
¼ H2
QH
Q2
q2
36
Conclusion
Genotype frequencies are function of
allele frequencies
Equilibrium reached in one generation
Independent of initial genotype
frequencies
Random mating, etc. required
Conform to binomial expansion
37
Simple HWE Exercise
If the defective alleles of the cystic
fibrosis (CFTR) gene have cumulative
frequency of 1/50 what is:
The proportion of carriers in the
population? p=P+1/2H H=2pq=2(p-P)=0.04
p=0.98
P=0.96
q=0.02
Q=0.0004
The proportion of affected children at
birth?
38
Frequencies of genotypes AA, Aa, and aa relative to the
frequencies of alleles A and a in populations at HardyWeinberg equilibrium
Max. heterozygosity
p = q = 0.5
39
Hardy-Weinberg for loci with more
than two alleles:
For three alleles (A, B, and C) with frequencies p, q, and r:
Binomial expansion
(p + q + r)2 = p2(AA) + 2pq(AB) + q2(BB) + 2pr(AC) +
2qr(BC) + r2(CC)
For four alleles (A, B, C, and D) with frequencies p, q, r,
and s:
(p + q + r + s) 2 = p2(AA) + 2pq(AB) + q2(BB) + 2pr(AC) +
2qr(BC) + r2(CC) + 2ps(AD) + 2qs(BD) + 2rs(CD) + s2(DD)
40
Hardy-Weinberg for X-linked
alleles (1):
e.g., Humans and Drosophila (XX = female, XY =
male)
XA(p)
Xa(q)
Y
XA(p)
XAXA
p2
XAXa
pq
XAY
p
Xa(q)
XAXa
qp
XaXa
q2
XaY
q
41
Hardy-Weinberg for X-linked
alleles (2):
Females
Hardy-Weinberg frequencies are the same for
any other locus:
p2 + 2pq + q2 = 1
Males
Genotype frequencies are the same as allele
frequencies:
p+q=1
Recessive X-linked traits are more common
among males.
42
Checking Hardy-Weinberg
Equilibrium
A common first step in any genetic study is to
verify that the data conforms to Hardy-Weinberg
equilibrium
Deviations can occur due to:
Systematic errors in genotyping
Unexpected population structure
Presence of homologous regions in the
genome
43
Testing
Hardy Weinberg Equilibrium
Consider a sample of 2N alleles
nA alleles of type A
nB alleles of type B
nAA genotypes of type AA
nAB genotypes of type AB
nBB genotypes of type BB
44
nA= nAA + ½ nAB / N
nB= nBB + ½ nAB / N
45
Simple Approach
Calculate allele frequencies (o) and
expected counts (e)
Construct chi-squared test statistic
Convenient, but can be inaccurate:
especially when one allele is rare
46