The document discusses the Hardy-Weinberg equation, which relates allele and genotype frequencies in a population at genetic equilibrium. It describes how the equation predicts that allele and genotype frequencies will remain constant from generation to generation if certain conditions are met, including no mutations, natural selection, migration, genetic drift or non-random mating. It provides examples of using the Hardy-Weinberg equation to calculate expected genotype and allele frequencies based on observed data from populations of horses and butterflies.

1. THE HARDY-WEINBERG
EQUATION RELATES ALLELE AND
GENOTYPE FREQUENCIES
In 1980, Godfrey Harold Hardy an English
Mathematician and Wilhelm Weinberg a German
Physician independently derived a simple mathematical
expression called the Hard-Weinberg equation that
describes the relationship between allele and genotype
frequencies when a population is not evolving
(Brooker,R.J 2014.Pg.492).
The Hardy-Weinberg equation predicts that allele
genotype frequencies will remain the same, generation
after generation , but only when a population is in
equilibrium. T o be in equilibrium, evolutionary
mechanisms that can change allele and genotype
frequencies are not acting on a population.

2. For this to occur the following
conditions must be meet;
 No new mutation occur.
 No natural selection occurs. That is, no survival or
reproductive advantage exist for any of the
genotypes.
 No migration occurs between different
populations.
 The population is so large that allele frequencies
do not change due to random chance.
 Random mating occurs. That is, the members of
the population mate with each other without
regard to their phenotypes and genotypes.

3. Limitation
No population satisfies the Hardy-Weinberg
equilibrium completely. The population is in
disequilibrium, because evolutionary mechanisms
are affecting the population. Hence evolution
occurring.

4. THE HARDY-WEINBERG
EQUILIBRIUM EQUATION.
1. In a stable population, a gene with two alleles where `A` is dominant
allele and `a` is recessive allele.
2. Three possible genotypes are may occur, these are AA, Aa and aa .
3. If the frequency of occurrence of A=P and a=q then, the frequency
of occurrence of AA+AA+aa can be expressed by BINOMIAL
EQUATION. 𝑝2+2pq+q2=1. or (p+q)2.
where,
p is the frequency of the dominant allele
q is the frequency of the recessive allele.
since, there are only two alleles in this case , the frequency of one plus the
frequency of the other must be equal to 100%.
p+q=1 Hence p=1-q
therefore (AA+Aa)+(Aa+aa)=1
AA+2Aa+aa=1

5. Hence,p2+2pq+q2=1
Where,
p2 is the predicted frequency of homozygous
dominant(AA) individuals.
2pq is the predicted frequency of heterozygous
(Aa) individuals.
q2 is the predicted frequency of homozygous
recessive (aa) individuals.

6. Hence,p2+2pq+q2=1
Where,
p2 is the predicted frequency of homozygous
dominant(AA) individuals.
2pq is the predicted frequency of heterozygous
(Aa) individuals.
q2 is the predicted frequency of homozygous
recessive (aa) individuals.

7. QUESTION NUMBER 1AA. If 98 out of 200 horses in the population express
recessive phenotype, what percentage of the
population would you predict to be heterozygotes?
solution
Number of homozygote recessive individual=98
Total number of individuals in the population=200
q² =98/200=0.49
q=0.7 from p+q =1 then, p=1-q=1-0.7=0.3
Frequency of heterozygote (2pq)=2×(0.7)×(0.3)=0.42
The percentage of heterozygote in the population is
42%.

8. QUESTION NUMBER 1B
Your original population of 200 was hit by a tidal wave and
100 organisms were wiped out, leaving 36 homozygous
recessive out of the 100 survivors. If we assume that all
individuals were equally likely to be wiped out, how did the
tidal wave affect the predicted frequencies of the alleles in
the population?
Note: Assume the new population is at equilibrium after
the event so you are comparing two populations what are
at equilibrium to look for changes in allele frequencies.

9. QUESTION NUMBER 1B CONT……
Solution.
Number of homozygous recessive=36
Total number of individuals in the population=100
q²=36/100=0.36, q=0.6
from p+q=1 then, p=1-q
p=1-0.6=0.4
Frequency of heterozygote(2pq) = 2(0.4)(0.6)=0.48
Frequency homozygote dominant(p²)=(0.4)²=0.16
proof.p²+2pq+q²=1
0.16+0.48+0.36=1 hence proved

10. QUESTION 1B, SOLUTION
CONT……
The drastic reduction in size of population due to some
chance event called bottleneck event particularly when
the original gene pool(allele frequencies) is no longer
represented in the surviving population. Since there is
now small population, chances are likely that it will
subjected to genetic drift and continue to shift away from
the original allele frequencies(pre-tidal wave).

11. QUESTION No. 2
B. If 9% of an African population is born with severe
form of sickle - cell anemia (ss) ,what percentage
of the population will be more resistant to malaria
because they are heterozygous (Ss) for the sickle
-cell gene?
solution
(ss)= homozygous recessive gene for sickle-
anemia
(Ss)=heterozygous gene for sickle -cell anemia
( SS)=homozygous dominant gene normal person

12. Qn.2B solution Conti…..
.9%=percentage of recessive sickle -cell anemic
Let y=number of recessive sickle-cell anemic
individuals in the population
.100=total number of individuals
(y/100)1×00%=9%
Y=9
Number of dominant individuals in the
population=100-9=91

13. Qn. 2B solution cont’d
phenotype
genotype SS +
Ss
ss
Number of individuals in
the population
91 9
Frequency of genotype
of individuals in the
population
91/100=
0.91
9/100=
0.09

14. Qn. 2B solution cont’d
 Frequency of genotype of homozygous recessive
=q² but q=0.09
 Frequency of homozygous recessive
allele(q)=square root of frequency of homozygous
recessive genotype (q²)
 Frequency of homozygous recessive allele(q)=
0.3
P²=frequency of homozygous dominant genotype

15. Qn.2B solution cont’d
 P=frequency homozygous dominant allele
 p+q =1
 But q=0.3
 P=1-q=1-0.3=0.7
 Frequency of homozygous dominant
genotype=p²=0.49
 Let y=number of homozygous dominant
individuals

16. Qn. 2B solution cont’d
 0.49=y/100
 Y=49
 Number of heterozygous individuals
=91-49=42
 Percentage of heterozygous individuals in the
population =(42/100)×100%=42%

17. Qn. 2C
C. Coloration in tiger butterfly species behave as a
single locus, two allele system with incomplete
dominance .Data 1612 individuals are given
below
white –spotted(AA)=1469, intermediate (Aa)=138,
little spotted (aa)=5.
calculate the following frequencies
(i) A (ii) a (iii) AA (iv) Aa (v) aa

18. Qn.2C solution
phenotype White-spotted Intermediate Little spotted
genotype AA Aa aa
Number of
individuals in the
population
1469 138 5
Frequency of
genotype
1469/1612=0.91
p²=0.91
138/1612=0.085
6
2pq=0.0856
5/1612=0.0031
q²=0.0031
Number of
alleles in the
population(2 per
individual)
2938A 138A+138a 10a

19. Qn.2C Solution cont…….
Total number of homozygous dominant allele
frequencies=2938A+138A=3076A,
Total number of alleles in the
population=(2938+138+138+10)=3224
Frequency of homozygous dominant
allele(p)=3076/3224=0.95
Total number of homozygous recessive allele=148a ,
frequency homozygous recessive allele(q)
=148/3224=0.0459

20. Qn. 2C Solution cont…….
i. A=P= 0.95
ii. a=q= 0.0459
iii. AA= p²=0.91
iv. Aa= pq =0.044
v. aa = q²=0.0031

21. REFERENCIES
1. Brooker,R.J (2014).Biology. Second Ed.New
York, McGRaw Hill(pg.492-494)