Transcript Slide 1

The elementary evolutionary
operator
1. Hardy-Weinberg Law
Allele frequencies are easily
estimated from genotype frequencies
2 ‘M’ alleles
1 ‘M’ + 1’N’ 2 ‘N’ alleles
0.913
0.913
go
How were allele frequencies estimated ?
For Eskimo - Freq .of M = (0.835 + (0.5 x 0.156 )) = .913
Freq. of N = (0.009 + .(0.5 x 0.156 )) = .087
Why Hardy-Weinberg Law?
Hardy-Weinberg Law is a dynamical effect
evolutionary operator
Scheme of genotypes
locus
gene
Alleles of
the gene :
One-locus
genotypes
AB
genotype
genotype
Multilocus genotypes
genotype
One-locus population state
N=N1+N2+N3 - Size of population
xAA= N1/N xBB= N2/N xAB= N3/N
xAA,xBB,xAB-frequencies genotypes AA,BB,AB
(xAA, xBB, xAB ) –state of the one-locus population
xAA  0, xBB  0, xAB  0;
xAA+xBB+xAB=1.
Random mating
Pairs genotype
AA,AA - 1 pair
AA,BB - 1 pair
AA,AB - 2 pairs
BB,BB - 2 pairs
BB,AB - 2 pairs
AB,AB - 0
What is meant by random mating?
Random mating means that, for any locus,
mating takes place at random with respect to
the genotypes in the population.
 Another way of saying this is that the
chance of an individual mating with another
of a particular genotype is equal to the
frequency of that genotype in the
population.

Let state of population is xAA,xBB,xAB
Pairs genotypes
Frequencies
AA,AA -
xAA xAA
AA,BB -
xAA xBB
AA,AB -
xAA xAB
BB,BB -
xBB xBB
BB,AB -
xBB xAB
AB,AB -
xAB xAB
Mendelian First Law
m
f
AA
aa
AA
AA
Aa
aa
Aa
aa
Aa
½AA+½Aa
½aa+½Aa
Aa
½AA+½Aa
½aa+½Aa
¼AA+¼aa+½Aa
Evolutionary operator of the population
m
AA
aa
Aa
AA
f
BB
AB
AA
AB
½AA+½AB
AB
BB
½BB+½AB
½AA+½AB
½BB+½AB
¼AA+¼BB+½AB
Let state of population is xAA,xBB,xAB
(AA,AA) - xAAxAA ; (AA,BB) – xAAxBB; (AA,AB) - xAA xAB; (BB,BB) - xBB xBB; (BB,AB) - xBB xAB; (AB,AB) - xAB xAB
(xAA ) ´ = (xAA)2 + xAAxAB + ¼(xAB ) 2
(xBB ) ´ = (xBB)2 + xBBxAB + ¼(xBB ) 2
(xAB ) ´ = 2xAAxBB + xAAxAB + xBBxAB + ½(xAB ) 2
(xAA )´ = (xAA)2 + xAAxAB + ¼(xAB ) 2 = (xAA+ ½xAB )2
(xBB )´ = (xBB)2 + xBBxAB + ¼(xBB ) 2 = (xBB+ ½xAB )2
(xAB )´ = 2xAAxBB + xAAxAB + xBBxAB + ½(xAB ) 2 = 2(xAA+ ½xAB )(xBB+ ½xAB )
p = (xAA+ ½xAB ); q = (xBB+ ½xAB );
p+q=1
p and q is the frequencies of alleles A and B in the population.
p`=(xAA )´ + 1/2(xAB )´ = p2 +pq=p(p+q)=p
q`=(xBB )´ + 1/2(xAB )´ = q2 +pq=q(p+q)=q
Hardy-Weinberg Law
(xAA )´ = p2; (xBB ) ´ = q2 ;
(xAB )´ = 2pq;
Gene conservation Law
p´ = p2 + ½ 2pq=p(p+q)=p
q´ = q2 + ½ 2pq=q(p+q)=q
Trajectory calculator
X11
0.2
X12
0.3
X22
.5
Current state (point)
generation
Go
0
1
2
3
reset
Next state (point)
Q=q2
P=p2
H=pq=p(1-p)
Hardy-Weinberg Law
The dynamics effects of sex linkage
We consider a bisexual population whose differentiation is
determined by two alleles A, a of some sex-linked locus. There
are three possibilities to consider
I. Y-linkage. Such locus is on the Y-chromosome and hence of the male only.
A male offspring inherits the father’s Y-chromosome.
State of generation for male part of population:
(xA, xa) and xA+ xa =1.
Evolutionary operator
xA’ = xA; xa’ = xa
Mating pair
Male offspring
Thus, any state (xA, xa) is equilibrious:
(xA, xa)’ = (xA, xa)
II. X-linkage
In this case there are two male genotypes A1, A2 and
three female ones A1A1, A2A2, A1A2.
}
M
}
F
M
F
II. X-linkage
The formation of male offspring
f
m
A1
A2
A1A1
A1
A1
A2 A2
A2
A2
A1 A2
½A1+½ A2
½A1+½ A2
The formation of female offspring
m
A1
A2
A1A1
A1A1
A1 A2
f
A2 A2
A1 A2
A2 A2
A1 A2
½A1A1+½A1 A2
½ A2 A2 +½A1 A2
II. X-linkage
Let distributions genotypes A1A1, A2A2, A1A2 in female part of
current generation are (x11,x22,x12) accordingly, and distributions
genotypes A1, A2 in male part of current generation are (y1,y2). As
usual x and y nonnegative and x11+x22+x12=1; y1+y2=1.
Evolutionary equations of male part of population
y1’=x11y1+x11y2+ ½x12y1+ ½x12y2
y2’=x22y1+x22y2+ ½x12y1+ ½x12y2
II. X-linkage
Evolutionary equations of female part of population
x11= x11y1+ ½x12y1
x22= x22y2+ ½x12y2
x12=x11y2+x22y1+ ½ x12y1+ ½ x12y2
II. X-linkage
Evolutionary operator of the population
y1’=x11y1+x11y2+ ½x12y1+ ½x12y2;
x11’= x11y1+ ½x12y1;
y2’=x22y1+x22y2+ ½x12y1+ ½x12y2
x22’= x22y2+ ½x12y2
x12’=x11y2+x22y1+ ½ x12y1+ ½ x12y2
Let pf= x11+ ½x12; qf= x22+ ½x12; pm=y1; qm=y2
pf+qf=x11+x22+x12=1; pm+qm=y1+y2=1
pf, qf -frequencies A1 and A2 in female part of population; pm, qm -frequencies A1 and A2 in male part of
population
Then
y1’=pf, y2’=qf
x11’=pfpm, x22’=qfqm, x12’=pfqm+pmqf
genotype-gene
connection
pf ’= x11’+ ½x12’ = pfpm+ ½ (pfqm+pmqf)=
½ pf (pm+qm)+½ pm (pf+qf)=½ (pf + pm);
pm’ = y1’= pf.
II. X-linkage
Evolutionary operator of the population (on gene level)
pf ’= ½ (pf + pm); qf ’= ½ (qf + qm); pm’ = pf; qm’ = qf
Gene Conservation Low
Indeed
(2/3) pf ’ + (1/3) pm’ = (1/3) (pf + pm)+ (1/3) pf = (2/3) pf + (1/3) pm
The coefficient 2/3 and 1/3 correspond to the ratio 2:1 of
X-chromosomes in female and male zygotes
pf’= ½ (pf + pm); qf’= ½ (qf + qm); pm’ = pf; qm’ = qf
Equilibrium point
pf= ½ (pf + pm); => pf = pm ;
Limiting behavior:
pf = pm
qf = qm
In a state of equilibrium (and in such a state only) the probability of every
gene in the male sex is equal to its probability in the female sex.
Evolutionary equation are x11’=pfpm, x22’=qfqm, x12’=pfqm+pmqf
In equilibria point: x11’=pfpf, x22’=qfqf, x12’=2pfqf
The Hardy-Weinberg Law is true for the female sex in an equilibrium state
of population.
II. X-linkage. Limiting behavior
pf ’ - pm’ = ½ (pf + pm) – pf = - ½ (pf - pm);
(pf - pm)
pf ’= ½ (pf + pm); qf ’= ½ (qf + qm); pm’ = pf; qm’ = qf
pf(n) – pm(n) = (- ½)n( pf(0) - pm (0));
pf(n) = const1 + (- ½)n( pf(0) - pm (0))(1/3)
Population trajectory
Limit value
II. X-linkage
Population trajectory
Limit value
Since the condition pf = pm is nessesary and sufficiently for an equilibrium then difference (pf - pm)
may be regarded as the measure of disequilibrium of state of the population. The modulus of
measure of disequilibrium is halved for one generation, and its sing alternates, I.e. an excess of
genes is pumped from one sex to another.
If (pf - pm)  0 in start point then (pf - pm)  0 along the trajectory.Therefore, the
population under consideration is non-stationary.
II. X-linkage