quantitative genetics of natural variation: some questions do most adaptations involve the fixation...
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Quantitative Genetics of Natural Variation: some questions
Do most adaptations involve the fixation of major genes?
micromutationist view: adaptations arise by allelic substitution of slight effectat many (innumerable) loci, and no single substitution constitutes a majorportion of an adaptation (Darwin, Fisher)
macromutationist views:
1. single “systemic” mutations produce complex adaptations in essentially perfect form (Goldschmidt)
2. adaptation often involves one or a few alleles having large effects
• Of 8 studies, only 3 consistent with changes involving > 5 loci (Orr and Coyne 1992)
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Quantitative Genetics of Natural Variation: some questions
• How many loci contribute to naturally occurring phenotypic variation, and what are the magnitudes of their effects?
• What sorts of genes —and changes in these genes—are responsible for trait variation within populations (e.g., transcription factors, structural genes, metabolic genes)
• Do the same genes that contribute to variation within species also contribute to variation between species?
• What genes underlie evolutionary novelties?
• What are the genetic bases for evolutionary novelties?
• How do pleiotropic effects of genes evolve?
Answers require a mechanistic approach towards identifying the relevant loci and how genetic differences are translated into phenotypic differences
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Quantitative traits depend on multiple underlying loci
one locusone locus + environment
two loci + environment
four loci + environment
many loci + environment
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– a d + agenotypic value 0
A2A2 A1A2 A1A1genotype
Phenotypic Value and Population Means (Falconer and Mackay Ch. 7)
Phenotypic value = Genotypic value + Environmental Deviation
P = G + E
Genotype Freq Value Freq x ValA1A1 p2 +a p2aA1A2 2pq d 2pqdA2A2 q2 -a -q2a
Sum = Pop Mean = a(p-q) + 2dpq
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PredictableLarval Habitat
PredictableEphemeral Pond
Time
Hatching Metamorphosis
Timing of Metamorphosis
The majority of organisms on planet earth have complex life cycles
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T3
Hypothalamus
TRH
TSH
TRs
transcription
Target cells
TH
Pituitary
Thyroid
T4deiodionation
Thyroid Hormone Receptors as Candidate Genes forVariation in Metamorphic Timing
An extreme difference in metamorphic timing
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Thyroid Hormone ReceptorAlpha Genotype
Timing ofMetamorphosis
(Days)
A1A1 A1A2 A2A2
200 160 150
a -a
d
0Homozygote
Midpoint(175)
-15
-2525
Thyroid Hormone Receptors : A Hypothetical Example
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p = f(A1) q = f(A2)
0.0
0.3
0.5
0.7
1.0
1.0
0.7
0.5
0.3
0.0
A1A1 A1A2 A2A2
Genotype Freq Value Freq x ValA1A1 p2 25 p2(25)A1A2 2pq -15 2pq(-15)A2A2 q2 -25 -q2(25)
Sum = Pop Mean = 25(p-q) + 2(-15)pq
0 0 -25
2.25 -6.3 -12.25
6.25 -7.5 -6.25
12.25 -6.3 -2.25
25 0 0
Mean
-25 (150)
-16.3 (158.7)
-7.5 (167.5)
3.7 (178.7)
25 (200)
(reduces time)(adds time)
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Let’s Consider a Second Locus
Thyroid Hormone ReceptorAlpha Genotype
Timing ofMetamorphosis
(Days)
A1A1 A1A2 A2A2
200 160 150
a -a0
HomozygoteMidpoint
(170)
-3030
A1A1 A1A2 A2A2
200 140
Thyroid Hormone ReceptorBeta Genotype
Timing ofMetamorphosis
(Days)
0
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P = f(A1) Q = f(A2)
0.0
0.3
0.5
0.7
1.0
1.0
0.7
0.5
0.3
0.0
A1A1 A1A2 A2A2
0 0 -30
2.7 0 -14.7
0 0 0
14.7 0 -2.7
30 0 0
Mean
-30 (140)
-12 (158)
0 (170)
12 (182)
30 (200)
(reduces time)(adds time)
Genotype Freq Value Freq x ValA1A1 p2 30 p2(30)A1A2 2pq 0 2pq(0)A2A2 q2 -30 -q2(30)
Sum = Pop Mean = 30(p-q) + 2(0)pq
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a -a0
AverageHomozygote
Midpoint(172.5)
5555
227.5 117.5
Timing ofMetamorphosis
(Days)
Total Range = a=110
Consider the joint effect of both TH Loci
Th A1A1Th A1A1
Th A2A2Th A2A2
Overall Mean
= a(p-q) + 2dpq
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Genotypic value is not transferred from parent to offspring; genes are.
Need a value that reflects the genes that an individual carries and passes on to it’s offspring
Empirically: An individual’s value based on the mean deviation of its progenyfrom the population mean.
Theoretically: An individual’s value based on the sum of the average effectsof the alleles/genes it carries.
Breeding Value
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average effect of An:
n = mean deviation from the population mean of individuals that received An from one parent, if the other parent’s allele chosen randomly
1 = q [ a + d (q – p)]
2 = –p [ a + d (q – p)]
1 = pa + qd - [ a (p – q) + 2dpq ]
population mean
.
f (A1) f (A2)
Average Effect of an Allele
Type of Values and Freq Mean value Population Average gamete of gametes of genotypes mean effect of
geneA1A1 A1A2 A2A2
a d -a
A1 p q pa + qd -a(p-q) + 2dpq q[a+d(q-p)] A2 p q -qa + pd -a(p-q) + 2dpq -p[a+d(q-p)]
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When there are only two alleles at a locus
A1A1 A1A2
+a d
A2A2-a
Average effect of a gene substitution
(a - d) (d + a)
p(a - d) + q(d + a)
= a + d(q - p)
p