The elementary evolutionary operator

1. Hardy-Weinberg Law

Allele frequencies are easily estimated from genotype frequencies How were allele frequencies estimated ? For Eskimo - Freq.of M = ( (0.5 x )) =.913 Freq. of N = ( (0.5 x )) =.087 Freq. of N = ( (0.5 x )) = ‘M’ alleles 1 ‘M’ + 1’N’ 2 ‘N’ alleles

Why Hardy-Weinberg Law? Hardy-Weinberg Law is a dynamical effect evolutionary operator

Scheme of genotypes locus gene Alleles of the gene : A B genotype One-locus genotypes Multilocus genotypes

One-locus population state N=N 1 +N 2 +N 3 - Size of population x AA = N 1 /Nx BB = N 2 /Nx AB = N 3 /N x AA,x BB,x AB - frequencies genotypes AA,BB,AB (x AA, x BB, x AB ) –state of the one-locus population x AA  0, x BB  0, x AB  0; x AA +x BB +x AB =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. 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. 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.

Pairs genotypes Frequencies AA,AA - x AA x AA AA,BB - x AA x BB AA,AB - x AA x AB BB,BB - x BB x BB BB,AB - x BB x AB AB,AB - x AB x AB Let state of population is x AA,x BB,x AB

f aaAAAa m AA Aa aa ½AA+½Aa ½aa+½Aa ¼AA+¼aa+½Aa AA aa Aa Mendelian First Law ½AA+½Aa

Evolutionary operator of the population f BBAAAB m AA AB BB ½AA+½AB ½BB+½AB ¼AA+¼BB+½AB AA aa Aa½AA+½AB Let state of population is x AA,x BB,x AB (AA,AA) - x AA x AA ; (AA,BB) – x AA x BB; (AA,AB) - x AA x AB; (BB,BB) - x BB x BB; (BB,AB) - x BB x AB; (AB,AB) - x AB x AB (x AA ) ´ = (x AA ) 2 + x AA x AB + ¼(x AB ) 2 (x BB ) ´ = (x BB ) 2 + x BB x AB + ¼(x BB ) 2 (x AB ) ´ = 2x AA x BB + x AA x AB + x BB x AB + ½(x AB ) 2

(x AA )´ = (x AA ) 2 + x AA x AB + ¼(x AB ) 2 = (x AA + ½x AB ) 2 (x BB )´ = (x BB ) 2 + x BB x AB + ¼(x BB ) 2 = (x BB + ½x AB ) 2 (x AB )´ = 2x AA x BB + x AA x AB + x BB x AB + ½(x AB ) 2 = 2(x AA + ½x AB )(x BB + ½x AB ) p = (x AA + ½x AB ); q = (x BB + ½x AB ); p+q=1 p and q is the frequencies of alleles A and B in the population. (x AA )´ = p 2 ; (x BB ) ´ = q 2 ; (x AB )´ = 2pq; p ´ = p 2 + ½ 2pq=p(p+q)=p q ´ = q 2 + ½ 2pq=q(p+q)=q Gene conservation Law Hardy-Weinberg Law p`=(x AA )´ + 1/2 (x AB )´ = p 2 + pq=p(p+q)=p q`=(x BB )´ + 1/2 (x AB )´ = q 2 + pq=q(p+q)=q

Current state (point) Next state (point)

P=p 2 Q=q 2 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. Mating pair Male offspring Thus, any state (x A, x a ) is equilibrious: (x A, x a )’ = (x A, x a ) State of generation for male part of population: (x A, x a ) and x A + x a =1. Evolutionary operator x A ’ = x A ; x a ’ = x a

II. X-linkage } } F M In this case there are two male genotypes A 1, A 2 and three female ones A 1 A 1, A 2 A 2, A 1 A 2. FM

f A 2 A1A1A1A1 A 1 A 2 m A1A1 A1A1 A2A2 A2A2 ½A 1 +½ A 2 A1A1 A2A2 f A 2 A1A1A1A1 A 1 A 2 m A1A1A1A1 A1 A2A1 A2 A1 A2A1 A2 A2 A2A2 A2 ½A 1 A 1 +½A 1 A 2 ½ A 2 A 2 +½A 1 A 2 A1A1 A2A2 The formation of male offspring The formation of female offspring II. X-linkage

Let distributions genotypes A 1 A 1, A 2 A 2, A 1 A 2 in female part of current generation are (x 11,x 22,x 12 ) accordingly, and distributions genotypes A 1, A 2 in male part of current generation are (y 1,y 2 ). As usual x and y nonnegative and x 11 +x 22 +x 12 =1; y 1 +y 2 =1. Evolutionary equations of male part of population y 1 ’ =x 11 y 1 +x 11 y 2 + ½x 12 y 1 + ½x 12 y 2 y 2 ’ =x 22 y 1 +x 22 y 2 + ½x 12 y 1 + ½x 12 y 2

II. X-linkage Evolutionary equations of female part of population x 11 = x 11 y 1 + ½x 12 y 1 x 22 = x 22 y 2 + ½x 12 y 2 x 12 =x 11 y 2 +x 22 y 1 + ½ x 12 y 1 + ½ x 12 y 2

II. X-linkage Evolutionary operator of the population y 1 ’ =x 11 y 1 +x 11 y 2 + ½x 12 y 1 + ½x 12 y 2 ;y 2 ’ =x 22 y 1 +x 22 y 2 + ½x 12 y 1 + ½x 12 y 2 x 11 ’ = x 11 y 1 + ½x 12 y 1 ;x 22 ’ = x 22 y 2 + ½x 12 y 2 x 12 ’ =x 11 y 2 +x 22 y 1 + ½ x 12 y 1 + ½ x 12 y 2 Let p f = x 11 + ½x 12 ; q f = x 22 + ½x 12 ; p m =y 1 ; q m =y 2 Then y 1 ’ =p f, y 2 ’ =q f genotype-gene x 11 ’ =p f p m, x 22 ’ =q f q m, x 12 ’ =p f q m +p m q f connection p f ’ = x 11 ’ + ½x 12 ’ = p f p m + ½ (p f q m +p m q f )= ½ p f (p m +q m )+½ p m (p f +q f )=½ (p f + p m ); p f +q f =x 11 +x 22 +x 12 =1; p m +q m =y 1 +y 2 =1 p m ’ = y 1 ’ = p f. p f, q f -frequencies A 1 and A 2 in female part of population; p m, q m -frequencies A 1 and A 2 in male part of population

II. X-linkage Evolutionary operator of the population (on gene level) p f ’ = ½ (p f + p m ); q f ’ = ½ (q f + q m ); p m ’ = p f ; q m ’ = q f Gene Conservation Low Indeed (2/3) p f ’ + (1/3) p m ’ = (1/3) (p f + p m )+ (1/3) p f = (2/3) p f + (1/3) p m The coefficient 2/3 and 1/3 correspond to the ratio 2:1 of X-chromosomes in female and male zygotes

p f ’ = ½ (p f + p m ); q f ’ = ½ (q f + q m ); p m ’ = p f ; q m ’ = q f Limiting behavior: p f = p m Equilibrium point p f = ½ (p f + p m ); => p f = p m ; q f = q m 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 x 11 ’ =p f p m, x 22 ’ =q f q m, x 12 ’ =p f q m +p m q f In equilibria point: x 11 ’ =p f p f, x 22 ’ =q f q f, x 12 ’ =2p f q f The Hardy-Weinberg Law is true for the female sex in an equilibrium state of population.

II. X-linkage. Limiting behavior p f ’ - p m ’ = ½ (p f + p m ) – p f = - ½ (p f - p m ); p f ’ = ½ (p f + p m ); q f ’ = ½ (q f + q m ); p m ’ = p f ; q m ’ = q f (p f - p m ) p f (n) – p m (n) = (- ½) n ( p f (0) - p m (0) ); p f (n) = const 1 + (- ½) n ( p f (0) - p m (0) )(1/3) Population trajectory Limit value

II. X-linkage Limit value Population trajectory Since the condition p f = p m is nessesary and sufficiently for an equilibrium then difference (p f - p m ) 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 (p f - p m )  0 in start point then (p f - p m )  0 along the trajectory.Therefore, the population under consideration is non-stationary.

II. X-linkage