Basic Principles of Population Genetics Lecture 4

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Basic Principles of Population GeneticsLecture 4
Background Readings Chapter 1, Mathematical and
statistical Methods for Genetic Analysis, 1997,
Kenneth Lang.
This slide show follows closely Chapter 1 of
Langs book. Prepared by Dan Geiger.
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Founders allele frequency
In order to write down the likelihood function of
a data given a pedigree structure and a
recombination value ?, one need to specify the
probability of the possible genotypes of each
founder. Assuming random mating we have,
Pr(G1,G2)Pr(A1/A2, B1/B2)
?Pr(A1/A2, B1/B2) The likelihood function
also consists of transmission matrices that
depend on ? and penetrances matrices to be
discussed later.
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Hardy-Weinberg and Linkage Equilibriums
The task at hand is to establish a theoretical
basis for specifying the probability Pr(A1/A2,
B1/B2) of a multilocus, from allele frequencies.
We will derive under various assumptions the
following two rules which are widely used in
genetic analysis (Linkage Association) and
which ease computations a great deal. Of course,
the assumptions are not satisfied for all genetic
analyses.
Hardy-Weinberg (HW) Equilibrium Pr(A1/A2)
PA1 PA2, namely, the probability of an ordered
genotype A1/A2 is the product of the frequencies
of the alleles constituting that genotype.
Linkage Equilibrium Pr(A1B1) PA1 PB1,
namely, the probability of a haplotype A1,B1 is
the product of the frequencies of the alleles
constituting that haplotype.
These rules imply Pr(A1/A2, B1/B2)PA1 PA2
PB1 PB2
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A simple setup to study HW equilibrium
Consider a bi-allelic locus A with alleles A1, A2
.
Let u,v, and w be the frequencies of unordered
genotypes A1/A1, A1/A2, A2/A2. Clearly, uvw1.
How are these frequencies related to allele
frequencies p1 and p2 of A1 and A2 ,respectively ?
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Assumptions made to Justify HW

1. Infinite population size
2. Discrete generations
3. Random mating
4. No selection
5. No migration
6. No mutation
7. Equal initial genotype frequencies in the two
   sexes

HW equilibrium can be shown to hold under more
relaxed sets of assumptions as well. These
assumption are clearly not universal.
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What happens after one generation ?
Mating Type- Unordered genotype Nature of Offspring and segregation ratios Frequency of mates
A1/A1 x A1/A1 A1/A1 u2
A1/A1 x A1/A2 ½ A1/A1 ½ A1/A2 2uv
A1/A1 x A2/A2 A1/A2 2uw
A1/A2 x A1/A2 ¼ A1/A1 ½ A1/A2 ¼ A2/A2 v2
A1/A2 x A2/A2 ½ A1/A2 ½ A2/A2 2vw
A2/A2 x A2/A2 A2/A2 w2
(uvw)21
Frequency of A1/A1 after one generation
uu2 ½(2uv)
¼v2
(u ½v)2 p12
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After one generation
So, after one generation the genotype frequencies
u,v,w change to u,v,w as follows (using the
previous table)
Frequency of A1/A1 uu2uv ¼v2 (u ½v)2
p12
Frequency of A1/A2 v
uv2uw ½v2 vw 2(u½v)(½vw) 2p1p2
Frequency of A2/A2 w¼v2 vw w2 (½vw)2
p22
Hardy-Weinberg seems to be established after one
generation, but
u,v,w are frequencies for the second
generation while p1 and p2 are defined as the
allele frequencies of the first generation. Are
these also the allele frequencies of the second
generation ?
Yes ! Because p1 u ½v p12p1p2p1 and
similarly p2 p2.
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After yet another generation
Have we reached equilibrium ? Lets look at one
more generation and see that genotype frequencies
are now fixed.
Frequency of A1/A1 u(u ½v)2
(p12p1p2)2 p12
Frequency of A2/A2 w(½vw)2 (p22
p1p2)2 p22
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Use of Assumptions in the derivation

1. Infinite population size
2. Discrete generations (mating amongst ith
   generation members only)
3. Random mating
4. No selection
5. No migration
6. No mutation
7. Equal initial genotype frequencies in the two
   sexes

Segregation ratios below assume 1,2,3,7
Mating Type- Unordered genotype Nature of Offspring and segregation ratios Frequency of mates
A1/A1 x A1/A2 ½ A1/A1 ½ A1/A2 2uv
Frequency formula of A1/A1 after one generation
u2 ½(2uv) ¼v2 assume 4,5,6.
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An alternative justification
Previously, we started with arbitrary genotype
frequencies u,v,w and showed that they are
modified after one generation to satisfy HW
equilibrium. Now we start with arbitrary allele
frequencies p1 and p2. Random mating is
equivalent to random pairing of alleles each
person contributes one allele with the prescribed
frequencies.
So the frequency of A1/A1 in the new generation
is p12 , that of A1/A2 is 2p1p2 , and that of
A2/A2 is p22. Argument completed ?
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HW equilibrium at X-linked loci
Consider an allele at an X-linked locus. At
generation n, let qn denote that alleles
frequency in females and rn denote that alleles
frequency in males. More explicitly,

• Questions
• What is the frequency pn of the allele in the
  population ?
• Does pn converge and to which value p ?
• Does qn and rn converge to the same value ?

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Argument Outline
Assuming equal number of males and females, we
have pn 2/3 qn 1/3 rn for every n.
Let p p0 2/3 q0 1/3 r0. We will now show
that both qn and rn converge quickly to p (but
not in one generation as before).
Having shown this claim, the female genotype
frequency of A1/A1 must be p2 , that of A1/A2 is
2p(1-p) , and that of A2/A2 is (1-p)2, satisfying
HW equilibrium. For male, genotypes A1 and A2
have frequencies p and 1-p.
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The recursion equations
Because a male always gets his X chromosome from
his mother, and his mother precedes him by one
generation,
rn qn-1
(Eq. 1.1)
Similarly, females get half their X-chromosomes
from females and half from males,
qn ½ qn-1 ½ rn-1
(Eq. 1.2)
2/3(½ qn-1 ½ rn-1 ) 1/3 qn-1
2/3 qn-1 1/3 rn-1
It follows that the allele frequency pn 2/3 qn
1/3 rn never changes and remains equal to p0 p.
To see that qn converges to p, we need to relate
the difference qn-p with the difference qn-1-p.
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The fixed point solution
qn-p qn- 3/2 p ½ p
½ qn-1 ½ rn-1 -
3/2 (2/3 qn-1 1/3 rn-1) ½ p
- ½ qn-1 ½ p (just cancel terms)
- ½ (qn-1- p)
Continuing in this manner, qn-p - ½ (qn-1- p)
(- ½)2 (qn-2- p) (- ½)n (q0- p) ? 0
So in each step the difference diminishes by half
and qn approaches p in a zigzag manner. Hence, rn
qn-1 also converges to p. What does this mean ?
Having shown this claim, the female genotype
frequency of A1/A1 must be p2 , that of A1/A2 is
2p(1-p) , and that of A2/A2 is (1-p)2, satisfying
HW equilibrium. For male, genotypes A1 and A2
have frequencies p and 1-p. HW equilibrium is not
reached in one generation but gets there fast
(quite there in 5 generations).
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Linkage equilibrium
Let Ai be allele at locus A with frequency pi
Let Bj be allele at locus B with frequency qj
Denote the recombination between these loci by
?f and ?m for females and males,
respectively. Let ? (?f ?m )/2. Linkage
equilibrium means that Pr(Ai Bj) piqj
We use the same assumptions employed earlier to
demonstrate linkage equilibrium, namely, to show
that Pn(Ai Bj) converges to piqj at a rate that
is fastest when the recombination ? is the
largest.
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Convergence Proof
Pn(Ai Bj) ½ gamete from female ½ gamete
from male
½ gamete from male
½ (1-?f )Pn-1(Ai Bj) ?f piqj ½ (1-?m
)Pn-1(Ai Bj) ?m piqj
(1-? )Pn-1(Ai Bj) ?piqj
So, Pn(Ai Bj) - piqj (1-? ) Pn-1(Ai Bj)
piqj
(1-? )nP0(Ai Bj) piqj
Exercise Repeat this analysis for three loci
(Problem 7, with guidance, in Kenneth Langs
book).
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Ramifications for Association studies
Many diseases are thought to been caused by a
single random mutation that survived and
propagated to offspring, generation after
generation.
Would we see association at random population
samples?
If the mutation happened many generations ago, no
trace will be significant. Allele frequency will
reach linkage equilibrium ! We need a
combination of close markers and recentallele age
of the disease. Association studies like that are
also called linkage disequilibrium mapping or LD
mapping in short.
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Selection and Fitness
Fitness of a genotype is the expected genetic
contribution of that genotype to the next
generation, or to how many offspring it
contributes an allele. Let the fitness of the
three genotypes of an autosomal bi-allelic locus
be denoted by wA/A, wA/a and wa/a .
If pn and qn are the allele frequencies of A and
a, then the average fitness under HW equilibrium,
is wA/Apn2 wA/a 2pnqn wa/a qn2.
Interpretation When sr0, there is no
selection. When r is negative A/A has advantage
over A/a. Similarly with negative s. When r is
positive (must be fraction), A/A has a
disadvantage over A/a. When both s and r are
positive, there is a heterozygous advantage.
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Assuming selection exists
Our goal is to study the equilibrium of allele
frequencies under various selection possibilities
(namely, different values for r and s).
In our new notations the average fitness wn at
generation n is given by wn ? (1-r)pn2 2pnqn
(1-s)qn2 1-rpn2 -sqn2
a/a
A/a
A/A
To find equilibrium we study the difference ?pn ?
pn1 - pn
?pn ? pn1 - pn (1-r)pn2 pnqn / wn - pn
(1-r)pn2 pnqn- (1-rpn2 -sqn2)pn / wn
pnqn (s- (rs) pn) / wn
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Interpretation when rgt0 and s?0
We just derived ?pn pnqn (s- (rs) pn) / wn
Convergence occurs when ?pn0, namely, when pn0,
pn1 (i.e., qn0) or pns/(rs). Where should it
converge to ?
Claim When (rgt0 and s ? 0), pn ? 0, i.e., allele
A disappears. In the opposite case (r?0 and sgt0),
allele a should be driven to extinction. (Why is
this extinction process sometimes halted in real
life ? )
Proof When (rgt0 and s ? 0), the linear function
g(p)s-(rs) p satisfies g(0) ? 0 and g(1) lt 0,
hence it is negative at (0,1). Thus, ?pn
monotonically decreases at each step. So ?pn must
approach 0 at equilibrium. Similarly, with the
other case.
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when r and s have the same sign
Conclusion I (for negative sign) If r and s are
negative, ?(pn ) gt 1, so pn ? 0 for p0 above
s/(rs), and pn ? 1 for p0 below s/(rs). In
other words, s/(rs) is an unstable equilibrium.
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when r and s are both positive
If both r and s are positive (Heterozygous
advantage), then
Conclusion II If both r and s are positive, pn
? s/(rs) and this point is a stable equilibrium.
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Heterozygous advantage
If we observe a recessive disease that is
maintained in high frequency, how can we explain
it ? Intuition says that it should disappear.
However, if the A/a genotype has an advantage
over other genotypes, then the defective allele
would be kept around. Technically, if both r and
s are positive, then the A/a genotype has the
best fit.
The best evidence for such phenomena is the
sickle cell anemia. In some part of Africa, this
anemia, despite being a recessive disease, is
kept in high frequency. It turns out that the
A/a genotype appears to provide protection
against malaria ! (so it has high fit in
swamp-like areas).
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Sickle cell anemia????? ?????? -
Medical Encyclopedia Red blood cells, sickle cell

Sickle cell anemia is an inherited autosomal
recessive blood disease in which the red blood
cells produce abnormal pigment (hemoglobin). The
abnormal hemoglobin causes deformity of the red
blood cells into crescent or sickle-shapes, as
seen in this photomicrograph. The sickle cell
mutation is a single nucleotide substitution (A ?
T) at codon 6 in the beta-hemoglobin gene,
resulting in the following substitution of amino
acids GAG (Glu) ? GTG (Val).
Source (Edited) http//www.nlm.nih.gov/medlineplu
s/ency/imagepages/1212.htm
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Facts about Sickle cell Disease

• Sickle Cell Disease is much more common in
  certain ethnic groups affecting approximately one
  out of every 500 African Americans.
• Because people with sickle trait were more likely
  to survive malaria outbreaks in Africa than those
  with normal hemoglobin, it is believed that this
  genetically aberrant hemoglobin evolved as a
  protection against malaria.
• Although sickle cell disease is inherited and
  present at birth, symptoms usually don't occur
  until after 4 months of age.
• Sickle cell anemia may become life-threatening
  when damaged red blood cells break down (and
  other circumstances). Repeated crises can cause
  damage to the kidneys, lungs, bones, eyes, and
  central nervous system.
• Blocked blood vessels and damaged organs can
  cause acute painful episodes. These painful
  crises, which occur in almost all patients at
  some point in their lives. Some patients have one
  episode every few years, while others have many
  episodes per year. The crises can be severe
  enough to require admission to the hospital for
  pain control.

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Balance of Mutation and Selection
Most mutations are neutral or deleterious. We
discuss balance between deleterious mutations and
selection. Let ? denote the mutation rate from a
to A. Suppose the equilibrium frequency of allele
A is p and of a is q1-p. When is a balance
achieved between selection (say, preferring
allele a ) and mutation that changes allele a
back to allele A ? The frequencies p and q must
satisfy the equilibrium condition
This yields 1- rp2 1-? and thus p2 ?/r and a
balance is achieved that retains both alleles.
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Finite Population ?Genetic Drift
Alelle 10
Alelle 5
Source Gideon Greenspan
After 800 generations, by simulation, from the
ten alleles only two remain numbered 5 and
number 7.