4.population genetics2

8/7/2019 4.POPULATION GENETICS2

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Define gene pool and populationalong with the relationship withthe Hardy-Weinberg equation.

Use the Hardy-Weinbergequation to calculate the allele

frequency in a population.


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A group of individuals of the

same species living in thesame area that can freelyinterbreed and produce

fertile offspring naturally


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The study of genetic changes

in population Considers all of the allelesin a population

Potential for changes depends

on the gene pool and genefrequency


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All the genes and their alleles that are present inthe population of the particular species of organism

Includes all possible alleles of all genes present inthe population

Will be in genetic equilibrium from one generationto another as long as there is no factors disturbing

or changing the frequency of allele


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Example :

Tall trait controlled by 2 alleles; dominant A and rescessive a.Assuming that genotype of the individuals in the populationcan be TT,Tt and tt

Population : 100, 000 individuals

If the number of individuals with genotype TT : 50, 000Tt : 40, 000tt : 10,000

Gene Pool : 2 (50,000 + 40,000 + 10,000)= 200,000 alleles


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y The ratio of the total number of a given alleleto the total number of all alleles of that gene inthe pool or population

y Can be expressed as fraction / percentage /

decimal

Frequency of dominant allele (p)= No. of dominant alleles

total alleles in population

Frequency of recessive allele (q)= No. of recessive alleles

total alleles in population


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Frequency of allele A (p)

Frequency of allele a (q)

p + q = 1


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Example :

Population consist of 40 Individual AA, 47 Aa and 13 aa

AA Aa aa Total

No.Individual

40 47 13 100

No. Alelle,A 80 47 0 127

No. Alelle, a 0 47 26 73

Total allele 200


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Allele Frequencies (p) ,A

= 127/200= 0.635

Allele Frequencies (q) , a

= 73/200= 0.365

Total allele frequencies (p+q) = 0.635 + 0.365

= 1


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Example :

Population : 100, 000 individuals

If the number of individuals with genotypeTT : 50, 000,Tt : 40, 000, tt : 10,000

TT TtFrequency of T allele = (50,000 x 2 ) + (40,000 x 1)

100,000 x 2= 0.7

Frequency of t allele = (40,000 X 1) + (10,000 X 2)100,000 X 2

= 0.3


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Example :

Individual genotypes = 40 AA, 47 Aa & 13 aa

Genotype Frequencies :

- Homozygous dominant : 40/100 = 0.40- Heterozygous : 47/100 = 0.47- Homozygous recessive : 13/100 = 0.13


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A population

whose allele and genotype

frequencies do not

change from generation

to generation


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In 1908- G. H. Hardy (English mathemathician)- W. Weinberg (German physician)

identified a mathematical relationship betweenalleles and genotypes in populations.

This relationship has been called the

Hardy-Weinberg Equilibrium


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Hardy-Weinberg Equilibrium :

Frequency of alleles and genotypes in a

population will remains constant fromgeneration to generation provided certain

conditions are met.

In other words, it means that the

population is in genetic equilibrium


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Genetic drift : a change of gene frequencies

within population over time


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2. No migration that cause gene flow due

to immigration into or emigration out

from the population


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3. No net mutation becauseby changing one allele into

another, mutation alters

the gene pool


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4. Must have random mating/fertilization

because if individuals pick mates with

certain genotypes, the random mixingof gametes required for Hardy-

Weinberg equilibrium does not occur


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5. Selection

All genotypes are equally fertile so that no

natural selection is taking place

mechanism in which members of population are

more successfully adapted

to environment


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p2 + q2 + 2pq = 1

p = frequency of the dominant allele

q = frequency of the recessive allele


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OVUM SPERM ZYGOTE PROBABILITY

A A AA p xp = p2A a Aa p x q = pq

a A aA q xp = qp

a a aa qxq = q2

Therefore, zygotes are produced in the following

proportions :

P2 = frequency of homozygous dominant individuals (AA)

2pq = frequency of heterozygous individuals (Aa)

q2 = frequency of homozygous recessive (aa)


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1. In a population of 500 wildflowers, 20 are whiteflowers or having recessive phenotype and therest have dominant phenotype.

(a) What are the frequencies of :

(i) homozygous dominant individuals (p2)

(ii) heterozygous individuals (2pq)

(iii) homozygous recessive individuals (q2)

EXAMPLE :EXAMPLE :


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(i) Genotype frequency of recessive individuals

q2 = 20/ 500

q2 = 0.04 ( homozygous recessive individuals )

q = 0.04

= 0.2 ( recessive allele )

since p + q = 1

p = 1 0.2

p = 0.8 ( dominant allele )

Therefore p2 = (0.8 ) 2

p2

= 0.64 (homozygous dominant individuals)

therefore 2pq = 2 ( 0.8) (0.2)

= 0.32 (heterozygous individuals)

4.population genetics2

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