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)

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

Quantitative traits depend on multiple underlying loci

one locus one locus + environment

two loci + environment

four loci + environment

many loci + environment

– a d + agenotypic value 0

A2A2 A1A2 A1A1genotype

Phenotypic Value and Population Means

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

PredictableLarval Habitat

PredictableEphemeral Pond

Time

Hatching Metamorphosis

Timing of Metamorphosis

The majority of organisms on planet earth have complex life cycles

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

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

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)

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

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

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

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

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)]

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)

paverageeffect of

A1

averageeffect of

A2