HARDY WEINBERG LAW | Online Biotech Notes
HARDY WEINBERG LAW
• According to Hardy (England,1908) and Weinberg (Germany, 1909), gene and genotype frequency of a Mendelian population remain constant generation after generation unless there is selection, mutation, migration or random drift.
• Suppose in a population of plant species, a gene has two allele having frequency 'p’ and 'q' resp.
• The frequencies of their genotypes are p^2, 2pq and q^2.
• Sum of both frequencies are equal to 1.
• P^2 + 2pq +q^2 =1 , this is also known as Hardy- Weinberg law equation.
• Acc. to this law, the frequency of p and q in a population remains constant generation after generation unless there's any factor's influence.
Random union of Gametes
• Simple mathematical expression that relates genotype and allele frequencies.
• The equation is used to determine genotype frequencies based on allelic frequencies.
• In random mating among gametes ;Male and Female gametes unite at random to yield zygotes.
• Frequency/probability of male and female gametes will be product of their individual frequency in the population.
u Let the frequency of two alleles A , a are
F(A)=p; F(a)=q
And we know that p + q=1
The union of male and female will yield genotypes –
AA, Aa, aa
The frequency of these genotypes will be – p2 ,2pq,q2
Respectively.
i.e. F(AA)=p2
F(Aa)=2pq
F(aa)=q2
• All AA individual will produce only A gametes.
• All Aa individuals will produce half A &half a gametes.
• All aa individuals will produce only a gametes.
• Frequency of A gametes = all gametes from AA individual + half from Aa
= p2 + ½(2pq)
= p2 +p q
= p(p + q) as p+ q =1
= p
Now, frequency of a gametes will be
= all gametes from aa individual + 1/2 gametes from Aa
= q2 + ½(2pq)
= q2 + p q
= q(p + q) as p + q =1
= q
u Hence , frequency of A & a gametes produced by new generation will be same as those of parental generation.
Therefore ,Hardy – Wienberg law is proved.
Example:-
Q : Say, in a population of 100 individuals 40 have MM , 40 MN , 20 have NN blood groups.
(a). Calculate the allelic frequency of M & N?
(b). What are the genotype frequency after one random mating?
A : (a) Allelic frequency of M and N:
Genotypes : MM MN NN
Proportions : .40 .40 .20
Frequency of M = .40 + .20 = .60
Frequency of N = .20 + .20 = .40
(b). Genotype frequency after one random mating.
Frequency of MM = .6*.6 = .36
Frequency of Mn = 2(.6 * .4) =.48
Frequency of NN = .4*.4 =.16
The fonding population are not in genetic equilibrium as the genotypic frequency were .40 MM , .20 MN & .20 NN.
After one random mating these are .36 , .48 , .16 .
u Again after one random mating the frequencies will be .36, .48 , .16 so the population has reached equilibrium.
Example:-
Q : If p = 0.8 ( i.e. frequency of A is 80% )
q = 0.2 ( i.e. frequency of a is 20% )
µ = 10-5
Find out the frequency of A & a after mutation ?
A : ∆q = pµ = 0.8 * 10-5 = 0.000008
therefore ,in the next generation
qn+1 = 0.2 + 0.000008 = 0.200008
pn+1 = 0.8 – 0.000008 = 0. 799992
Q : Find out the frequency of A after 1000 generations ? If p= 0.8 , t = 1000 , µ = 10-5
A : (1-µ)t = pt / po
(1- 0.00001)1000 =pt / 0.8
pt = 0.792
therefore, after 1000 gen. frequency
will drop from 0.8 to 0.792
u It means mutation rate affects allele frequency very slowly.
Factors Affecting The Hardy Weinberg Equilibrium
u There
are 4 factors that affects Hardy – Weinberg equilibrium
(1).Mutation
(2).Migration
(3).Random genetic drift
(4).Natural selection
1.Mutation
• Usually
convert one functional allelic form of a gene into less functional allelic
form.
• Mutation
of allele is a source of all genetic variations.
• New
alleles only occurs as a result of mutations.
• Thus
, mutation play important role in evolution.
• The
term mutation is used to express the rate at which more functional allele is
mutated into less functional allele.
• Symbolised
by µ.
• Range 10-5 - 10-6
per generation.
How mutation rate affects allele frequencies?
u First
case:
Suppose gene A is mutating to gene a at the rate of µ.
F(A) = po ; F(a)
= qo before mutation
F(A) = p ; F(a) = q
after mutation
Frequency of newly produce a allele = po µ
Total frequency of a allele after mutation = qo + po µ
Total frequency of A allele after mutation = po - qo µ
Hence, q = qo
+ po µ
∆q = po
µ
If po is replaced by p then
∆q = pµ
It should be noted that value of p will decrease with
every generation as
this is the case when reverse mutation is taken as
compared to forward mutation.
u Second
case:
When reverse mutation is taken into consideration. Let us
suppose that rate of mutations from A to a is “µ” & from a to A is “v”.
Forward mutation = A to a
Reverse mutation = a to A
F(A) = po ; F(a)
= qo
Thus , due to forward and reverse mutation po will decline po µ & increase by qo v.
so, the value of
p after mutation = p1
p1 = po - po µ +qo v
∆p = p1 - po = qo
v – p µ
This is the case of forward and reverse mutation thus
, p & q will change due to mutation
when
po µ ≠ qo v
But , when po µ = qo v ; the values of p & q will not be affected by mutation and they
will be
at equilibrium known as Mutational Equilibrium.
u At Mutational
equilibrium :
po µ = qo v
p0 µ = v(1- po )
po µ = v – po v
v =
po µ + po v
v =
po (µ + v)
po = v/( µ + v)
po = equilibrium frequency of A
qo = equilbrium frequency of A
2. Migration or Gene Flow
It is the movement of genes which takes place only when
organism or Gametes migrate and contribute their gene to the gene pool of the
recipient population.
Migration Ratio- M / (M+N)
‘M’ = No. of
migration individual.
‘N’= No. of
individual in the population before Migration.
‘M+N’ = total
no. of individuals (after migration).
Example:- A species of bird may occupy two geographic regions on rare occassion, the birds from western population by flying to eastern population. If two population have different allele frequencies and if migration occur in sufficient no. it may alter the allelic frequency of eastern population.
3. Random Genetic Drift
Random change in allelic frequency due to chance or sampling error is called random genetic drift. The effect of this is large on small population.
Example:-
u Suppose 10 individuals are inhabiting at an island. 5 of them have green eyes and 5 have brown eyes and green eye color is recessive to brown eye color.
Answer :-
F(B) = 0.6 ;
F(b) = 0.4;
One day a typhoon strikes and kills 50% population. And
all the killed ones were green eyed
color individuals.
Thus, evolution occurs in the population during which the
allele frequency of brown eyes is changed from 0.6 to 1.0, simply as a result
of chance.
So , in small populations one of the allele is eliminated
and other is fixed at 100%.and the allele become monomorphic & cannot
fluctuate any further.
We can determine the no. of generations required by an
allele to be fixed by the formula:
E = 4N
E = avg. no. of generation to achieve fixation
N = no. of
individuals in a population , assuming that male & female contribute equally in
each suiciding generation.
4. Natural Selection
u In
1850’s Charles Darwin and Alfreel Russell Wallace. Independently proposed the
theory of Natural Selection.
u According
to this theory, there is a “struggle for existence’’. The conditions found in
nature result in the selective survive and reproduction of individuals whose
characteristics makes them well adapted according to their environment. These
individual are more likely to reproduce and control offspring to next
generation. More recently it has been realized that natural selection can be
related not only to differential survival but also to mating efficiently and
fertility.
4(a). Directional Selection
u Favours the survival of one extreme of a phenotypic distribution, that is better adapted.
For
example:- In a dimly lit forest ,
directional selection world favors snail with darker shells ,because they world
be better camouflaged and less susceptible to predation.
4(b). Disruptive Selection
u Disruptive
selection favors the survival of two or more different phenotype classes of
individual.
u Example
– disruptive selection favors the survival of dark shelled and lighted snails
that occupy a heterogenous environment.
4(c). Stabilizing Selection
u It
favors the survival of individuals with intermediate phenotypes.
u For Example:- Moderately colored snails may survive better in an environment that has an intermediate level of life
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