Plant Breeding: Theory and Practice: V.L. Chopra (2012) Breeding Field Crops: David Allen Sleper and John Milton Poehlman Plant Breeding: Principles and Methods- B.D.Singh Principles and Practice of Plant Breeding- J.R. Sharma Principles of Plant Breeding: R.W. Allard Plant Breeding: N.W. Simmonds

Reference Books

HARDY-WEINBERG HARDY-WEINBERG LAWLAW

Random Mating Random Mating PopulationsPopulations

Each individual of the population has equal Each individual of the population has equal opportunity of mating with any other individual of that opportunity of mating with any other individual of that population such a population is called random mating population such a population is called random mating populations, Mendelian populations or panmictic populations, Mendelian populations or panmictic populations.populations.

Gene Gene PoolPool

Gene pool may be defined as the sum Gene pool may be defined as the sum total of all the genes present in the population total of all the genes present in the population

PopulatiPopulationon

It consists of all such individuals that share the same gene pool, i.e., It consists of all such individuals that share the same gene pool, i.e., have an opportunity to intermate with each other and contribute to have an opportunity to intermate with each other and contribute to next generation of the populationnext generation of the population

Population Population GeneticsGeneticsPopulation genetics is that branch of Population genetics is that branch of

genetics that is concerned with the genetics that is concerned with the evolutionary processes of natural evolutionary processes of natural selection, genetic drift, mutation, selection, genetic drift, mutation, migration, and random mating.migration, and random mating.Hardy-Weinberg Hardy-Weinberg

LawLaw• This law was independently developed This law was independently developed by Hardy, in 1908, in England and by Hardy, in 1908, in England and Weinberg, in 1909, in GermanyWeinberg, in 1909, in Germany• “ “Gene and genotype frequency in a Gene and genotype frequency in a Mendilian population remain constant Mendilian population remain constant generation after generation if there is no generation after generation if there is no selection, mutation, migration or random selection, mutation, migration or random drift” drift”

Frequencies of three genotypes for a locus with Frequencies of three genotypes for a locus with two alleles, A and atwo alleles, A and a

AAAA p2p2AaAa 2pq2pqaaaa q2q2

Where, p represents the frequency of A and q Where, p represents the frequency of A and q of a allele in the population and p+q=1of a allele in the population and p+q=1

Such a population is said to be at equilibrium Such a population is said to be at equilibrium since the genotype frequencies would be since the genotype frequencies would be stable. This equilibrium is known as Hardy-stable. This equilibrium is known as Hardy-Weinberg equilibrium. A population is said to Weinberg equilibrium. A population is said to be at equilibrium when the frequencies of the be at equilibrium when the frequencies of the three genotypes, AA, Aa and aa are p2, 2pq and three genotypes, AA, Aa and aa are p2, 2pq and q2. q2.

Single gene with two allele, A and a there would be three Single gene with two allele, A and a there would be three genotypes, AA, Aa and aagenotypes, AA, Aa and aaSuppose the population has N individualsSuppose the population has N individuals

D individuals are AAD individuals are AAH individuals are AaH individuals are AaR individuals are aaR individuals are aa

Hence,Hence,D+H+R=ND+H+R=N

Total no. of alleles at this locus in the population would be Total no. of alleles at this locus in the population would be 2N, since each individual has two alleles at single locus2N, since each individual has two alleles at single locusTotal no. of A alleles = 2D+H, the ratio (2D+H)/2NTotal no. of A alleles = 2D+H, the ratio (2D+H)/2Ntherefore,therefore,

p= (2D+H)/2N or (D+½H)/Np= (2D+H)/2N or (D+½H)/Nq= (2R+H)/2N or (R+½H)/Nq= (2R+H)/2N or (R+½H)/N

since,since, p+q=1 p+q=1Therefore p=1-q or q=1-p Therefore p=1-q or q=1-p

Frequencies of different genotypes Frequencies of different genotypes produced by random union produced by random union

between A and abetween A and ap Ap A q aq a

p Ap A p2 AAp2 AA pq Aapq Aa

q aq a pq Aapq Aa q2 aaq2 aa

♂♂♀♀

Therefore, genotype frequencies in the next generation would be p2 2pq q2

It may be noted that D=p2, H=2pq and R=q2, It may be noted that D=p2, H=2pq and R=q2, Further, N=1 since,Further, N=1 since,

p2+2pq+q2=(p+q)2p2+2pq+q2=(p+q)2andand p+q=1 p+q=1Hence, p2+2pq+q2=1Hence, p2+2pq+q2=1

GenotypeGenotype AAAA AaAa aaaaFrequencyFrequency p2 p2 2pq 2pq q2 q2

This population would produce two types of gametes A and a; their frequencies can be calculated in a

similar manner as described before.

The frequencies of A and a gametes produced The frequencies of A and a gametes produced by the population may be calculated as by the population may be calculated as follows;follows;

Frequencies of gametes containing A allele= Frequencies of gametes containing A allele= (D+½H)/N(D+½H)/N = (p2+pq)/1=p2+pq= (p2+pq)/1=p2+pq = p(p+q)=p (since p+q=1)= p(p+q)=p (since p+q=1)

Frequencies of gametes containing a allele = Frequencies of gametes containing a allele = (R+½H)/N(R+½H)/N = (q2+pq)/1=q2+pq= (q2+pq)/1=q2+pq = q(p+q)=q (since p+q=1)= q(p+q)=q (since p+q=1)

Consequences of random mating of Consequences of random mating of genotypes in a Mendelian populationgenotypes in a Mendelian population

MatingMating Freq. of MatingFreq. of Mating Freq. of progeny from Freq. of progeny from the matingthe mating

AAAA AaAa AaAaAA x AAAA x AA p2 x p2=p4p2 x p2=p4 p4p4AA x AaAA x Aa 2(p2x2pq)=4p3q2(p2x2pq)=4p3q 2p3q2p3q 2p3q2p3qAA x aaAA x aa 2(p2 x 2(p2 x

q2)=2p2q2q2)=2p2q22p2q22p2q2

Aa x AaAa x Aa (2pq x (2pq x 2pq)=4p2q22pq)=4p2q2

p2q2p2q2 2p2q22p2q2 P2q2P2q2

Aa x aaAa x aa 2(2pq x 2(2pq x q2)=4pq3q2)=4pq3

2pq32pq3 2pq32pq3

aa x aaaa x aa q2 x q2=q4q2 x q2=q4 q4q4

The frequency of progeny with AA genotype would The frequency of progeny with AA genotype would be,be,

=p4+2p3q+p2q2=p4+2p3q+p2q2=p2(p2+2pq+q2) (p2 is taken as =p2(p2+2pq+q2) (p2 is taken as

common)common)=p2=p2 ( since p2+2pq+q2=1) ( since p2+2pq+q2=1)

Similarly, the frequency of aa progeny would be,Similarly, the frequency of aa progeny would be,=p2q2+2pq3+q4=p2q2+2pq3+q4=q2(p2+2pq+q2) (q2 is taken as =q2(p2+2pq+q2) (q2 is taken as

common)common)=q2=q2 (since p2+2pq+q2=1) (since p2+2pq+q2=1)

And the frequency of Aa progeny would be,And the frequency of Aa progeny would be,=2p3q+2p2q2+2p2q2+2pq3=2p3q+2p2q2+2p2q2+2pq3=2p3q+4p2q2+2pq3=2p3q+4p2q2+2pq3=2pq (p2+2pq+q2) (2pq is taken as =2pq (p2+2pq+q2) (2pq is taken as

common)common) =2pq=2pq (since (since p2+2pq+q2=1)p2+2pq+q2=1)