1) The document summarizes Hardy-Weinberg equilibrium, which states that gene and genotype frequencies in a population will remain constant from generation to generation if the population is large and there is no selection, migration, mutation, or non-random mating. 2) It describes the mathematical relationship between gene frequencies (p and q) and genotype frequencies (p^2, 2pq, q^2) known as the Hardy-Weinberg principle or the square law. 3) Several conditions must be met for a population to be in Hardy-Weinberg equilibrium including large population size, equal mating opportunities, no selection, migration, or mutation.

1. Dr. Anitta P L
MVSc Scholar
Indian veterinary research institute, izatnagar

2. Reported the genetic equilibrium for the special case of equal gene frequencies at a locus with two alleles p=q=0.5
YULE (1902) W.E. CASTLE (1903) KARL PEARSON (1904)
G. H. HARDY W. WEINBERG
Mathematician, Cambridge University Physician, Germany

3. “In a large random-mating population with no-section, mutation, or migration, the
gene frequencies are constant from generation to generation; and; furthermore,
is a simple relationship between the gene frequencies and the genotype
 Deductions about population genetics and quantitative genetics
 If the gene frequencies of two alleles among the parents are p and q, then the genotype frequencies are p², 2pq and q²

4. No change in genetic structure of the population
Gene and genotype frequencies remain same from one generation to the
next
SQUARE LAW
(p+q)² = p² + 2pq + q²………………………….. 1

5. 1. Population Size:
For example, 5 horned cows in different population sizes shall indicate
varying proportions.
5/10 = 50%
5/100 = 5%
5/1000 = 0.5%
5/10
5/100
5/1000

6. 2. Differences in fertility and viability:
For example, the cows A, B, C and D may be different in
producing the number of calves
A B C D
5 calves 2 calves 1 calf No calving
Calf Death Adult

7. 3. Migration:
movement of individuals ,
introduction of individuals into the population ( ) or
Elimination of individuals from the population ( )
The gene and genotype frequencies may be changed by both
immigration and or emigration
4. Mutation:
Sudden change in gene structure is mutation
Wild type Mutant type
Both forward and reverse mutations may alter the gene frequencies of population

8. 5. Selection:
Choosing the parents of next generation Natural or artificial selection
Any type of selection may alter the gene frequencies of the population
Selection favouring or un-favouring certain genes may increase or decrease the frequency of
genes
6. Mating Systems:
Genotype frequencies in the offspring is influenced by the genotypes of the pairs that mate in
parent generation
The non-random mating or selective mating may alter the frequencies whereas random mating or
Panmixia may keep the frequencies constant
Genotypes of the
progeny
zygotes Union of
gametes
Mating systems of
the parents

9. Step Deduction from: to Conditions
1a
1b
Gene frequency in parents
Gene Frequency in all gametes
Gene frequency in gametes forming
zygotes
(1) Normal gene segregation
(2) Equal fertility of parents
(3) Equal fertilizing capacity of gametes
(4) Large population
2
3
4
Genotype frequencies in zygote
Genotypes frequencies in progeny
Gene frequency in progeny
(5) Random mating
(6) Equal gene frequencies in male and
female parents
(7) Equal viability
3

10. 1) From gene frequency in parents to gene frequency in gametes
Genes Genotypes
Frequencies
A1 A2 A1A1 A1A2 A2A2
p q P H Q
Genotypes(parents) Frequencies of the genotypes Gametes produced by parents Frequency of gametes (genes)
A1A1 p² = P
A1 p
A1A2 2pq = H
A2A2 q² = Q A2 q
Under normal conditions of segregation, A1A2 produce equal number of A1 and A2 gametes

11. 2) From gene frequency in gametes to genotype frequencies in zygote
Random mating between individuals is equivalent to random union among their gametes
The genotype frequencies among the zygotes ( Fertilized eggs ) are then the products of the frequencies of the gametic types that unite to produce them
3) Mating Frequencies
gametes and
their frequencies
gametes and their frequencies
A1
p
A2
q
A1 A1A1 A1A2
p p² pq
A2 A1A2 A2A2
q pq q²
Genotype and
frequency of
parent
Genotype and frequency of parents
A1A1 A1A2 A2A2
P H Q
A1A1 P P² PH PQ
A1A2 H PH H² HQ
A2A2 Q PQ HQ Q²

12. Mating Genotype and frequency of progeny
Type Frequency A1A1 A1A2 A2A2
A1A1 x A1A1 P² P² - -
A1A1 x A1A2 2PH PH PH -
A1A1 x A2A2 2PQ - 2PQQ -
A1A2 x A1A2 H² ¼H² ½H² ¼H²
A1A2 x A2A2 2HQ - HQ HQ
A2A2 x A2A2 Q² - - Q²
Sums (P + ½H)² 2( P + ½H)(Q + ½H) (Q + ½H)²
= p² 2pq q²

13. 4) From zygotes to adults
Zygotes must survive equally well
5) From genotype frequencies to gene frequencies in progeny
The frequency of A1 = p² + ½ (2pq)
= p² + pq
= p ( p+ q)
= p
 Since the gene frequencies are same in parents and progeny, the relationship between gene frequencies and genotype frequencies applies to
single generation
 The genotype frequencies in the progeny depend only on the gene frequencies in the parent and not on their genotype frequencies.
Two important aspects

14. The frequency of heterozygotes cannot be greater than 50% and this occurs when p =q =
0.5
When the gene frequency of an allele is low, the rare allele occurs predominantly in
heterozygotes and there are very few homozygotes

15.  Maintenance or change in genetic structure
Genetic structure of a population can be conserved under random mating
Hence, random mating is advocated if the population has optimum fitness
 Estimation of genotypic frequencies
( p + q)² = p² + 2pq + q²
 Estimation of gene frequencies
When there is dominance relationship b/w the alleles, the estimation of gene frequencies is not possible either by gene counting or from genotypic frequencies.
Frequency of recessive allele may be estimated as q = 𝒔𝒒𝒖𝒂𝒓𝒆 𝒓𝒐𝒐𝒕 𝒐𝒇 𝒒²
 Frequency of carriers
The frequency of heterozygotes among all individuals =
The frequency of heterozygotes among normal individuals, H‘ = Ratio of genotype frequencies Aa/(AA + Aa)
H‘ = 2q (1-q) = 2q
(1-q)² + 2q(1-q) 1 + q

16.  Test of Hardy-Weinberg equilibrium
If population is in equilibrium All conditions are fulfilled
Chi-square test x² = (A - P²N)² + (B – pqN)² + (C - q²N)² = A² + B² + C² - N
p²N 2pqN q²N p²N 2pqN q²N
Where A, B, and C are the observed number of three genotypes
N = A + B + C
 To establish Mendelian hypothesis i.e., mode of inheritance
Wright (1917)- Genetic basis of coat colour in short horn breed of cattle controlled by a sinlge locus with 2 co-dominant alleles rather than two loci hypothesis
Bernstein (1925)- A, B, O blood antigens as three alleles of a single locus rather than two loci hypothesis

17. 1) Equal gene frequencies in two sexes:
If population is not in equilibrium, it will attain equilibrium in next generation if random mating is followed and gene frequencies are equal in both the sexes.
Follows square law
2) Unequal gene frequencies in two sexes
Requires 2 generation of random mating to attain genetic equilibrium
The gene frequencies in the first generation of random mating become equal in both sexes which are the average of the frequencies in the parents of two sexes
The genotype frequencies in the second generation of random mating attain equilibrium

18.  Also applicable to the case of multiple alleles
 The genotype frequencies for multiple alleles in a large random mating population are obtained from expansion of square of multinomial
Consider three alleles (A1, A2 and A3) at locus A with their frequencies as p, q and r.
Then p + q + r = 1
In this case there will be 6 types of genotypes with their proportions as under:
Alleles and their frequencies Genotype and their frequencies
A1 A2 A3 A1A1 A2A2 A3A3 A1A2 A1A3 A2A3
(p + q+ r)² = p² + q² + r² + 2pq + 2pr + 2qr = 1
Homozygotes = Sum of squared frequencies = p² + q² + r²
Heterozygotes = Twice the product of each pair of frequency = 2(pq + pr + qr)
Genotypes in equilibrium population =
Taking “k” as the number of multiple alleles at a locus, No. of total genotypes = k(k + 1) = k + k(k -1)
2 2
For example

19. Sex-linked genes
 The relationship between gene frequency and genotype frequency in the homogametic sex is same as the with an autosomal gene
 But in case of heterogametic individual, it carries only one gene instead of two
 Thus, two-thirds of the sex linked genes in the population are carried by the homogametic sex and one-third by the heterogametic
Frequency of A1 among females, pf = P + ½H
Frequency of A1 among males, pm = R
Frequency of A1 in the whole population = p’ =
2
3
pf +
1
3
pm =
1
3
(2𝑝𝑓 + pm ) =
1
3
( 2P + H + R)
Frequency
Females Males
A1A1 A1A2 A2A2 A1 A2
P H Q R S

20.  The gene frequency in the population as a whole does not change, but its distribution b/w the two sexes oscillates as population approaches equilibrium
 Males get their sex-linked genes only from their mothers.
p’m = pf
 Females get their sex-linked genes equally from both the parents.
p’f = ½ (pf + pm)
 The difference between the frequencies in the two sexes
p’f –p’m = ½ (pm +pf) –pf = -½ (pf-pm)
The distribution of the genes between the two sexes oscillates, but the
difference is halved in successive generations and population rapidly
approaches equilibrium in which the frequencies in the two sexes are equal

21.  With multiple genes, equilibrium is not established in one generation of random mating
 But approaches very rapidly in subsequent generations
 If there is linkage, the approach to equilibrium will be slower than if they are independent.
Linkage Rate of approach to equilibrium
 Disequilibrium with respect to two or more loci is called
 Intermixture of population with different gene frequencies
 Chance in small population
 Selection favouring one combination of alleles over another
 Disequilibrium can be expressed by reference to genotypes by
comparing the frequencies of coupling and repulsion of double heterozygote
A1B1/A2B2 Coupling heterozygote
A1B2/A2B1 Repulsion heterozygote
Genes A1 A2 B1 B2
Gene frequencies pA qA pB qB
Gametic types A1B1 A1B2 A2B1 A2B2
Frequencies, equilibrium pApB pApB qApB qAqB
Frequencies, actual r s t u
Differences from equilibrium +D -D -D +D
D = ru-st Half the difference b/w coupling and repulsion heterozygotes
After any number n of generation, Disequilibrium; Dn = D0(1-c)ˆn

22. References
• Falconer, D.S., Mackey, T.F.C., 1996. Introduction to quantitative genetics (4th Ed.). Pearson
Education Limited, United Kingdom, 1: 1-22
• Singh, R., Tomar, A.K., Tomar, S.S., 2006. Animal genetics and breeding (1st Ed.). Daya
Publishing House, New Delhi, 12: 188-202
• Kanakaraj, P
., 2001. A textbook on animal genetics (1st Ed.). International Book Distributing
Company, Uttar Pradesh, 25: 309-328
• Gillespie, J.S., 1998. Population genetics – A concise guide (1st Ed.). The John Hopkins
University Press, Baltimore and London, 1: 1-17