4.3. seleksi
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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
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