Hardy Weinberg law Hardy Weinberg Equilibrium with Solved Questions|CSIR NET|Life Sciences|GATE|JRF|ICMR| Video link: https://youtu.be/CUKGoxpptM8 Hardy Weinberg Law along with the assumptions the law is based on. Calculation of allelic and genotypic frequencies. Application of Hardy Weinberg law to different cases viz Multiple alleles, Polyploidy, Inbreeding and X-linked + Questions are discussed.

1. HARDY WEINBERG EQUILIBRIUM
R
r

2. 2
RR Rr rr
R- Dominant allele → p r- Recessive allele → q

3. Hardy Weinberg Law
Allelic and genotypic frequencies of an natural,
interbreeding mendelian population sharing a common gene
pool remains constant over generations provided no
assumption is violated
p+ q=1
(p+q)2
=p2
+2pq+q2
3

4. ASSUMPTIONS
4
Random
Mating
No natural
selection
No
mutation
No
migration
Large
population
size
Equal opportunity of mating with any other individual of the population- ‘panmictic population’
No individual will have a reproductive advantage over another because of its genotype
Random fluctuation in
allele
frequency-Genetic drift

5. Hardy Weinberg Law
Relates allelic and genotypic frequencies on an natural,
interbreeding mendelian population sharing a common gene
pool
Rule has 3 aspects
5
Allelic
Frequency
Equilibrium
Genotypic
Frequency
Equilibrium
Neutral
Equilibrium

6. 6
RR Rr rr

7. Calculating frequencies
Allelic frequencies
p= f(R)
= 2 x number of RR+ number of Rr
2 X Total number
= f(RR)+ ½ f(Rr)
q= f(rr)+ ½ f(Rr)
Genotypic
frequencies(Individuals)
RR: p2
Rr: 2pq
rr: q2
(p+q)2
=p2
+2pq+q2
=1
p+q=1
7
Homozygous or Heterozygous

8. EXAMPLE
In a population frequency of a homozygous recessive
disease is 16% then the frequency of dominant allele
would be?
a. 0.84 b. 0.6 c. 0.1 d. 0.4
8
q2
= 16%= 16/100=0.16
q= 0.4
p=1-q=1-0.4=0.6

9. EXAMPLE
In a population obeying Hardy-Weinberg equilibrium, the
frequency of recessive allele is 0.88, while of dominant
allele is 0.12. The frequency of heterozygotes in
population will be?
a. 11.1% b. 21.1% c. 79.9% d. 14.4%
9
q=0.88 p=0.12
2pq= 2 x 0.88 x 0.12 =21.12

10. Question (June 2006)
If the frequency of recessive allele causing disease in homozygous
recessive condition in a population of 10,000 is 0.04, then the number
of people affected by disease will be?
a. 16 b. 400 c. 3600 d. 496
10
q= 0.04 q2
= 0.0016
Affected= q = 0.0016 x 10000= 16

11. MULTIPLE
ALLELES
INBREEDING
SEX LINKED
POLYPLOIDY

12. CASES
Multiple alleles
(p+q+r)2
=p2
+q2
+r2
+2pq+2qr+2p
r
Eg. Blood groups
12
Blood types Genotype Frequency
A IA
IA
+ IA
i p2
+ 2pr
B IB
IB
+ IB
i q2
+ 2qr
AB IA
IB
2pq
O ii r2

13. EXAMPLE
In a large randomly mating human population, the frequencies of the IA
,IB
and i alleles are 0.7, 0.2 and 0.1, respectively. Calculate the expected
frequencies for each blood type
Let the Three alleles be p, q and r
Since population is in Hardy Weinberg population, genotypic frequencies
can be calculated as:
(p+q+r)2
= p2
+q2
+2pq+2pr+2qr
Let, f(IA
)=p=0.7 f(IB
)=q=0.2 f(i)=r=0.1

14. Blood type
A = IA
IA
+ IA
i p2
+ 2pr= 0.49+0.14=0.63
B = IB
IB
+ IB
i q2
+2qr =0.04+0.04=0.08
AB= IA
IB
2pq=2x 0.7 x 0.2=0.28
O= ii r2
=0.01
Let, f(IA
)=p=0.7 f(IB
)=q=0.2 f(i)=r=0.1

15. CASES
Polyploidy (p+q)n
n=polyploidy
15
Diploid (p+q)2
= p2
+2pq+q2
Triploid (p+q)3
= p3
+ q3
+ 3p2
q+3q2
p
Tetraploid(p+q)4
= p4
+4p3
q+6p2
q2
+4pq3
+q4

16. June 2007
If an organism is triploid, then Hardy-weinberg theorem
applicable will be?
a. (p+q)3
b. (p+q)2
c. (p+q+r)3
d. (p+q+r)2
16

17. 17
Fruit colour of wild Solanum nigrum is controlled by two alleles of a gene (A and
a). The frequency of A, p=0.8 and a, q=0.2. In a neighbouring field a tetraploid
genotype of S.nigrum was found. After critical examination five distinct
genotypes found; which are AAAA, AAAa, AAaa and aaaa. Following Hardy-
Weinberg principle and assuming the same allelic frequency as that of diploid
population the numbers of phenotypes calculated within a population of 1000
plants are close to one of the following:
AAAA: AAAa: AAaa: Aaaa: aaaa
a. 409:409:154:26.2 b. 420:420:140:18.2 c. 409:409:144:36.2 d.
409:420:144:25.2

18. 18
Fruit colour controlled by two alleles frequency of A, p=0.8 and a, q=0.2
Neighbouring field a tetraploid genotype was found having five distinct
genotypes AAAA, AAAa, AAaa and aaaa. Numbers of phenotypes
calculated within a population of 1000 plants are close to one of the
following:
AAAA: AAAa: AAaa: Aaaa: aaaa
Tetraploid(p+q)4
= p4
+4p3
q+6p2
q2
+4pq3
+q4
f(AAAA)= p4
=(0.8)4
=0.4096; out of 1000 plants- 409.6
f(AAAa)=4p3
q=4x(0.8)3
x0.2=0.4096
f(AAaa)=6p2
q2
= 6x(0.8)2
x(0.2)2
=0.1536
a. 409:409:154:26.2 b. 420:420:140:18.2 c. 409:409:144:36.2 d. 409:420:144:25.2

19. Sex linked genes
Male= Heterogametic
Thus will need only 1 copy of
recessive allele to have the
disease
Frequency will be equal to
19
Genotype Genotype frequency
XA
Y p
XB
Y q
Female= Homogametic
They will need both copies of the allele
to be affected
Frequency will be equal to
Genotype Genotype
frequency
XA
XA
p2
XA
Xa
2pq
Xa
Xa
q2

20. Example
Frequency of recessive allele for an X-linked recessive disease
in human population is 0.02. Calculate proportion of diseased
individuals
20
MALES: 1 copy needed
Frequency of diseased
male=frequency of
recessive allele
=0.02
FEMALES: Both
copies needed
Frequency of
diseased=q2
=(0.02)2
=0.0004
Overall frequency(if
equal males & females)
=½ x 0.0.2+½ x
0.0004
=0.0102

21. Inbreeding
● Mating between individuals that are closely
related through common ancestry
● Extent of inbreeding can be measured by
“inbreeding coefficient”
● Reduction in heterozygote frequencies
How?
21

22. 22

23. Inbreeding coefficient
Relative reduction in heterozygosity in the population
due to inbreeding
Given by,
F=2pq-H = 1- H
2pq 2pq
23

24. DEC 2011
The frequencies of two alleles p and q for a gene locus in a
population at Hardy Weinberg equilibrium are 0.3 and 0.7
respectively. After a few generations frequency was found
to be 0.28. The inbreeding coefficient in this case is:
a. 0.42 b. 0.28 c. 0.33 d. 0.67
24
F= 1 -(H/2pq)= 1-(0.28/ 2x0.3x0.7)=1-(0.28/0.42)=1-0.67=0.33