POPULATION GENETICS

(2 Hour)

At the end of this topic, student should be able to:

* explain gene pool and allele frequency
* explain population genetics and its relation to allele frequency
* explain allele frequency and genetic equilibrium
* state the Hardy-Weinberg Law
* explain five assumptions of Hardy-Weinberg Law
* calculate allele and genotype frequencies

GENE POOL CONCEPT

• population genetics involves the study of genes in a population
• population is a group of individuals of the same species occupying a given area that can freely interbreed and produce fertile offspring in nature.
• All individuals do not necessarily have the same alleles for all the genes.
• The sum of all various alleles of all the genes in all of individuals of population is called the population`s gene pool.

• Allele frequency is a frequency of each allele in variation in a gene pool
• The gene pool consists of the total number of genes and allele at all gene loci in all individuals of the population.
• The gene pool always change from one generation to other generation. A changing is mean that population going to evolution and may change over time due to external factors.
• Evolution can be defined as the change allele frequencies in a population
• various genetics in gene pool organism diploid propagation via sex occur in by mutation and gene recombination
• In population genetics have 3 type population

           i)  nature population - population going to evolution
           ii) control population – population under control (e.g. in the lab)
          iii)  mathematics population – hypothesis population (probabilities of inheriting)

• Population genetics and its relation to 3 frequency

          i) allele frequency
          ii) genotype frequency
          iii) phenotype frequency

allele frequency

• Defined as the ratio of the total number of a particular allele to the total number of all the alleles in the gene pool

Example:

10 000 individuals make up a population
In this population, a particular character is controlled by a certain gene
The two alleles involved are taged as ‘A’ and ‘a’.
Let us assume that among the 10,000 individual is 6,000 have genotype AA, 2,000 have genotype Aa and 2000 have aa
--> The gene pool = 2(6,000 + 2,000 + 2,000)
= 20 000 alleles

or

-Total number of allele ‘A’= (6000 x 2) + 2000 = 14 000
-Total number of allele ‘a’ = (2000 x 2) + 2000 = 6000
-Total number of alleles in the gene pool = 10 000 x 2
= 20 000 alleles
- Frequency of allele ‘A’ = 14 000 ÷ 20 000 = 0.7
- Frequency of allele ‘a’ = 6000 ÷ 20 000 = 0.3

Genotype frequency

• Defined as the ratio of the total number of individuals with a particular genotype to the total number of all the individuals in the population

Example :
Frequency for genotype AA = 6000 ÷10 000 = 0.6
Frequency for genotype Aa = 2000 ÷10 000 = 0.2
Frequency for genotype aa = 2000 ÷10 000 = 0.2

Phenotype frequency

• A stating the number of times a specific phenotype occurs in a population in a single generation
• Phenotypes are the outcome of the genes expression.
• A pool of genetic resources is shared by all members of a population and passed on to the next generation.

HARDY-WEINBERG LAW

• In 1908, G. H. Hardy (an English mathemathician) and W. Weinberg (a German physician) independently identified a mathematical relationship between alleles and genotypes in populations.
• This relationship has been called the Hardy-Weinberg Law and it concerns allele frequency
• The Hardy-Weinberg Law states that :-

The frequency of alleles and genotypes in a population will remain constant from generation to generation and can breading via sex and very close genetics relationship in other words it means that the population is in genetic equilibrium.

• If a population is not evolving, it is in genetic equilibrium and the allele frequency do not change.
• When a population evolves, the allele frequency in the population will change
• Assumption of Hardy-Weinberg Law for a population , it must satisfy five main conditions:

i)very large population size -- so that genetic drift can be avoided (chance fluctuation in the gene that can cause phenotype frequencies to change over time)
ii)no migration -- that cause gene flow due to immigration into or emigration out from the population
iii)no net mutations -- because by changing one allele into another, mutation alter the gene pool
iv)random fertilization -- because if individuals pick mates with certain genotypes, the random mixing of gametes required for Hardy-Weinberg equilibrium does not occur all genotypes are equally fertile so that
v)no natural selection – individual can bread with same generation only no overlap of generation is taking place.

* actual reality this condition not occur, but from that law we can guess any changing in population genetics

Allele frequency

• equilibrium allele frequency, p+q=1
• Hardy-Weinberg equation code p for allele A and q for allele a (recessive)
• number 1 = 100% frequencies for some allele state

example calculation

normal human has dominant allele (N) can produce pigment of skin, hair and eye, but albino (n, recessive allele) not occurs. Assumption in population 10,000 people have one person albino, determine allele recessive of albino in this population.

calculation
All calculation must start from q2 (homozygous recessive) easy calculated cause we know total number that allele

q2 = 1/10,000 , q2 cause homozygous recessive genotype

q = √ 1/10,000 = 0.01

p + q = 1

p + 0.01 = 1

p = 1 – 0.01 = 0.99

dominant allele frequency (N) + recessive allele frequency (n)
0.99 + 0.01 = 1


Exercise,

the ability to taste the chemical phenylthiocarbamite (PTC) is controlled primarily by a single pair of autosomal alleles with dominance. Some individual will find that it tastes bitter (controlled by dominant allele, T), other individuals will find the chemical to be tasteless (controlled by recessive allele, t), assumption of 84% Malaysian citizen found PTC had taste and 16% found had no taste, determine dominant and recessive allele frequency

(calculation must begin from recessive allele first)

q2 = 0.16
q = 0.4

p = 1 – q = 1 – 0.4 = 0.6

allele frequency, T (p) = 0.6
allele frequency, t (q) = 0.4

Genotype frequency

• equilibrium genotype frequency, p2 +2pq + q2=1

(p + q)2 = p2 +2pq + q2

equilibrium verification

breeding between 2 homozygous,

P :               AA        x          aa

G :               A                       a

F1 :                        Aa (normal)


breeding between F1

To summarize
p = dominant allele frequency
q = recessive allele frequency
p2 = homozygous dominant genotype
2pq = heterozygous genotype
q2 = homozygous recessive genotype

It is possible to calculate all allele, genotype, dominant and recessive phenotype frequencies using the expressions :
allele frequency p + q = 1,
genotype frequency p2 + 2pq + q2 = 1,
dominant phenotype frequency = p2 + 2pq , and
recessive phenotype frequency = q2

Example calculation
In a population of 500 wildflowers, 20 are white flowers or having recessive phenotype and the rest have dominant phenotype. If the alleles involved are A and a and there are 320 AA plants and 160 Aa plants, what are the frequencies of Aa genotype individuals, dominant and recessive phenotype individuals ?

(*calculation must start from q first)

q2 = 20/500
q = 0.2
p = 1 – q = 1 – 0.2 = 0.8

Genotype frequency of AA (p2)
= 0.8 2 = 0.64

Genotype frequency of Aa (2pq)
= 2 x 0.8 x 0.2 = 0.32
Genotype frequency of aa = recessive phenotype frequency (q2)
= 0.2 2 = 0.04

Dominant phenotype frequency (p2 + 2pq)
= 0.64 + 0.32 = 0.96

This population is obey in Hardy-Weinberg equilibrium,
p2 + 2pq + q2 = 1
0.64 + 0.32 + 0.04 = 1

Exercise,

In a human population, a baby from 25,000 babies is born with a sickle-cell anemia . The patient is recessive homozygous (ss). Calculate
a) the frequency of recessive allele
b) the frequency of dominant allele
c) the number of carrier in 25,000 babies

(*calculation must start from q first)

a) ss → q2 = 1/ 25 000 = 0.00004
so, frequency allele q = 0.0063
b) p + q = 1
so, p = 1 – q
p = 1- 0.0063 = 0.9937
c) Frequency of carrier → 2pq = 2 (0.9937) (0.0063)
= 0.0125
so, 2pq x 25 000 = 0.0125 x 25 000 = 313 babies

Exercise,

Resistance to a poison to insects Plutella controlled by the dominant allele (L). 64% of the population of insects in the study area was found to be resistant to pesticides. Search
1) L and l allele frequency in the population
2) calculate the expected number of insects that have a genotypes LL, Ll and ll in a population consisting of 200 flys of Plutella
3) If all the insects that have no resistance in the population was killed, what is the genotype frequency of the insects that do not have the resilience to be delivered in the next generation

p2 + 2pq = 0.64 (dominant phenotype frequency)

q2 = 0.36 (1- 0.64)

q = 0.6

p = 1 – 0.6 = 0.4

so, allele frequency of L is a 0.4 and l is a 0.6

number of individuals with genotype LL = p2 x 200

= (0.4)2 x 200 = 32

number of individuals with genotype Ll = 2pq x 200

= 2 x 0.4 x 0.6 = 96

number of individuals with genotype ll = q2 x 200

= (0.6)2 x 200 = 72

individuals with genotype ll was killed, so the remaining 200 – 72 = 128

so, sum of allele still remain = 128 x 2

from gynotype of L1 = 96

allele frequency l (q) = 96 / 256 = 0.375

frequency of insects that have the genotype II which will be born

q2 = 0.3752 = 0.14

Retold by;

Amran Md Said
Matriculation College of Pahang


THE END