GENETIC POPULATION

GENETIC POPULATION

GENE POOL CONCEPT

·        Gene pool concept

·        Population genetics and its relation to allele frequency.

·        Relation between population genetic and evolution (changes in allele frequency)

·        Allele frequency and genetic equilibrium.

 

HARDY WEINBERG LAW

 

POPULATION GENETICS

*      The study of population genetics help us to better understand the evolutionary mechanisms.

*      To study evolution, we must study the changes in a population since a population is the basic unit of study in evolution.

*      Any change in a population reflect on evolution.

*      Population is a group of individuals of the same species occupying a given area that can freely interbreed and produce fertile offspring in nature.

 

GENE POOL

 

ALLELES

 

 

 

 

PHENOTYPES

 

THE HARDY-WEINBERG EQUILIBRIUM

 

ALLELE FREQUENCY

 

HARDY-WEINBERG LAW

v     Very large population size so that genetic drift can be avoided ( change fluctuation in the gene that can cause phenotype frequencies to change overtime )

v     No migration that cause gene flow due to immigration into or emigration out from the population.

v     No net mutations because by changing one allele into another, mutation alter the gene pool.

v     Random mating 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 no natural selection is taking place.

 

HARDY-WEINBERG EQUATION.

Ø      The letter p to represent the frequency of the dominant allele.

Ø      And the letter q to represent the frequency of recessive allele.

Ø      Example1

            To help understand the principle, consider a gene which has a dominant allele A  and a recessive allele a.

            Let p = the frequency of allele A

                   q = the frequency of allele a

           

            In diploid individuals these alleles occur in the combination given on table (refer table  below).

 

1st  allele

2nd allele

Frequency

A

A

p x p = p2

A

a

p x q

a

A

q x p

a

a

q x q = q2

 

·        As the homozygous dominant (AA) combination ¼ of the total possible genotypes, there is a ¼ (25%) of a single individual is being of this type.

·        Similarly, the change of it being homozygous recessive (aa) is ¼ (25%).

·        Whereas there is a ½ (50%) change of it being heterozygous.

·         There is 1/1 (100%) change of it being of this type.

 

*      Homozygous dominant (1/4) + Heterozygous (1/2) + Homozygous recessive (1/4) = 1.0 (100%)

*      Thus AA + 2Aa + aa = 1.0 (100%)

*      p2 + 2pq + q2 = 1.0 (100%)

*      The Hardy-Weinberg principle is expressed as:

*      To summarize

v     p = dominant allele frequency

v     q = recessive allele frequency

v     p2 = homozygous dominant genotype

v     2pq = heterozygous genotype

v     q2 = homozygous recessive genotype

*      The formula can be used to calculate the frequency of any allele in the population.

*      Example: Imagine that a particular mental defect is the result of recessive allele.

*      If the number of babies born with the defect 1:25000, the frequency of the allele can be calculated as follows.

v     The defect will only express itself in individuals who are homozygous recessive.

v     Therefore the frequency of these individuals (q2) = 1/25000 or 0.00004

v     The frequency of the allele (q) is therefore square root of 0.00004 = 0.0063 approx

v     As the frequency of both alleles must be 1.0 p + q = 1.0, then the frequency of the dominant allele (p) can be calculated.

v     p + q =1

v     p = 1.0 – q

v     p = 1.0 – 0.0063

v     p = 0.9937

v     The frequency of heterozygotes can now be calculated.

v     From the Hardy-Weinberg formula, the frequency of heterozygotes is 2pq, 2 x 0.9937 x 0.0063 = 0.0125

v     In other words, 125 in 10000 (or 313 in 25000) are carriers (heterozygotes) of the allele.

v     This means that in a population of 25000 individuals, just one individual will suffer the defect but around 313 will carry the allele.

v     The heterozygotes are acting as a reservoir of the allele, maintaining it in the gene pool.

v     As the heterozygotes are normal, they are not specifically selected against, and so the allele remains.

v     Even if the defective individuals are selectively removed, the frequency of the allele will hardly be affected.

v     In our population of 25000, there is one individual who has two recessive allele - total of 315.

v     The removal of the defective individual will reduce the number of alleles in the population by just 2, to 313.

v     Even with removal of all defective individuals it would takes thousands of years just to halve the allele’s frequency.

*      Example1

Ø      A wild flower population with two varieties contrasting in flower colour. An allele for pink flowers which will be symbolized by A are completely dominant to an allele for white flowers, symbolized by a.

Ø      If the frequency of allele A is 0.8 or 80% the frequency of allele a must be 0.2 or 20%.

Ø      Note that p + q = 1; the combined frequencies of all possible alleles must account for100% of the genes for that locus in the population.

Ø      If p + q = 1, then p = 1- q or q = 1- p.

Ø       When gametes combine their alleles to form zygotes, the probability of generating an AA genotype or individuals of homozygous dominant is p2 and for aa genotype or individuals of homozygous recessive is q2.

Ø      There are two ways in which an Aa genotype can arise, depending on which parent contributes the dominant allele. Therefore, the frequency of heterozygous individuals in the population is 2pq.

Ø      If we have calculated the frequencies of all possible genotype correctly, they should add up to 1:

Ø      p2 + 2pq + q2 = 1

Ø      p2 = frequency of AA

Ø      2pq = frequency of Aa

Ø      q2 = frequency of aa

Ø      In mathematical terms p + q = 1is the mathematical equation of probability and p2 + 2pq + q2 =1 is the binomial expansion of that equation, in this case:

Ø      p + q = 1

Ø      (p+q)2 =1^2

Ø      p2 +2pq + q2 = 1

Ø      To summarize :

p

Dominant allele frequency

q

Recessive allele frequency

p2

Homozygous dominant genotype

2pq

Heterozygous genotype

q2

Homozygous recessive genotype

 

Ø      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

Recessive phenotype frequency

q2

 

 

Ø      For each individual of dominant phenotype, must have at least one dominant allele ( AA or Aa ) and for each individual of recessive phenotype, all of it alleles must be recessive or aa.

·        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 thee are 320 AA plants and 160 Aa plants, what are the frequencies of Aa genotype individuals, dominant and recessive phenotype individuals?

·        Heterozygous individual showing normal phenotypic characteristics but possessing a recessive gene capable of producing some form of metabolic disorder when present in homozygous recessives are described as carriers.

·        For a population to be in Hardy-Weinberg equilibrium, it must obey the principle of p2 + 2pq + q2 = 1.

 

 LAMAN UTAMA