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Selection

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Although mutationand migration allinfluence allelicfrequencies, theydo not of

necessity producepopulation ofindividuals thatare betteradapted to their

environment.Natural selection,however, tends tothat end

African wild dog Coyote

Fox Wolf

Ancestral Canine

Natural Selection

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Artificialselection, aspracticed byanimal andplantbreeders,follows thesame rules

Saint Bernard Bulldog

Yorkshire terrier German shepherd

Artificial selection

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Selection is a process whereby onephenotype and, therefore, onegenotype leaves relatively moreoffspring than another genotype,

measured both by reproduction andsurvival

Selection is a matter of reproductivesuccess, the relative contribution ofthat genotype to the next generation

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Fitness

A measure of reproductive success isthe fitness, or adaptive value, of

genotypeFitness (W) usually is computed to vary

from zero to one (0-1) and is alwaysrelative to a given population at a giventime

W = 1 the genotype leaves themost offspring

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Selection Coefficient A selection coefficient measures the sum

of forces acting to prevent reproductive

success. It is usually given the letter s or t and is

defined by the fitness equation

s = 1 - W

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Component of Fitness

Natural selection can act at any stage

of the life cycle of an organism1. Zigotic Selection

2. Gametic Selection

3. Sexual Selection

4. Fecundity Selection

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Effect of Selection

1. Directional Selection

2. Stabilizing Selection3. Disruptive Selection

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Formula to examine allele

frequencies after selection

(q2)(W22)

= C

(2pq)(W12)

= B

p2(W11)

= A

Genotypic

frequencies

after

selection

(q2)(W22)(2pq)(W12)p2(W11)Ratio after

selection

W22W12W11Fitness (W)

1q22pqp2Initial

genotypicfrequencies

A2A2A1A2A1A1

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Allele frequencies after selection :

p1 = f (A1)1 = A + B

q1

= f (A2)1

= C + B

Change in allele frequencies after selection:

? p = p1 p

? q = q1 - q

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Problem (1)

Calculate the allele frequencies after

selection if :a. the initial allele frequency for A1(p) is 0.6 and A2 (q) is 0.4,

b. W11=0, W12= 0.4, W22=1

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Co-dominant Selection

p2(1) + 2pq(1+s) + q2(1-2s) = 1

wwAA= 1wwAa= 1 - s

wwaa= 1 - 2s

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Dominant Selection

p2(1) + 2pq(1-s) + q2(1-s) = 1

wwAA= 1

wwAa= 1 - s

wwaa= 1 - s

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Recessive Selection

p2(1) + 2pq(1) + q2(1-s) = 1

wwAA= 1

wwAa= 1

wwaa= 1 - s

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The allele frequency of recessive

allele after selection is :

q ( 1s.q)

q1 =( 1s.q2)

q1 = recessive allele frequency after selection

q = initial recessive allele frequency

s = selection coefficient

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Example :

The frequency of slow-moving allele of an enzyme (q)was initially 0.7. If the selection coefficient is 0.6,what is the allele frequency one generation afterselection

Answer :

(0.7) [ 1(0.6) (0.7) ]

q1 = = 0.58[ 1(0.6) (0.7)2 ]

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If both mutation and selection are operating, the

equilibrium frequency of the recessive allele after

selection is approximately :

q1 =

s

q1 = recessive allele frequency after selection = forward mutation rate (A a)s = selection coefficient

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Example :

In corn, the mutation rate from the normal allele to the

recessive sugary allele is 2.5 x 10-6 and the selection

coefficient is 0.1. What is the expected equilibrium

frequency of the sugary allele

Answer :

2.5 x 10-6

q1 = = 0.005

0.1

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Formula to examine forward mutation

rate if fitness are known

2b =

1 - w

b = frequency at equilibrium

= forward mutation rate

W = fitness of a dominant trait

1.

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Example :

A dominant trait has a fitness of 0.6. The frequency in a

population of this trait is 1 in 8000. What is the

mutation rate

Answer:

b = 1 ; w = 0.6 ; = ?

8000

1 = 2 16.000 = 10.6

8000 1 0.6 = 2.5 x 10-5

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= ( 1f ) x2.

= mutation rate

f = the reproductive finess of the abnormalgene (the frequency that the abnormal

allele is passed to the next generation)

x = the frequency of the abnormality in theindividuals in one generation

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Example :

In a population, the relative fitness of the allele for

chondrodystrophy is 0.1963. The data of a hospital shows

that there were 10 dwarf out of 94.075 births. What is the

mutation rate

Answer :

10 dwarf

x = = 0.00012 ; f = 0.1963

94.075 births

= ( 10.1963 ) ( 0.00012 ) = 4.27 x 10-5