1 / 14

Quantitative Genetics of Natural Variation: some questions

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 effect at many (innumerable) loci, and no single substitution constitutes a major

ailish
Télécharger la présentation

Quantitative Genetics of Natural Variation: some questions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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 major portion 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)

  2. 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

  3. Quantitative traits depend on multiple underlying loci one locus + environment two loci + environment one locus four loci + environment many loci + environment

  4. A2A2 A1A2 A1A1 genotype – a 0 d + a genotypic value Phenotypic Value and Population Means P = G + E Phenotypic value = Genotypic value + Environmental Deviation Genotype Freq Value Freq x Val A1A1 p2 +a p2a A1A2 2pq d 2pqd A2A2 q2 -a -q2a Sum = Pop Mean = a(p-q) + 2dpq

  5. Timing of Metamorphosis The majority of organisms on planet earth have complex life cycles Predictable Larval Habitat Hatching Metamorphosis Predictable Ephemeral Pond Time

  6. Thyroid Hormone Receptors as Candidate Genes for Variation in Metamorphic Timing Hypothalamus TRH Pituitary TSH Thyroid TH Target cells T4 deiodionation T3 TRs transcription An extreme difference in metamorphic timing

  7. Thyroid Hormone Receptors : A Hypothetical Example Thyroid Hormone Receptor Alpha Genotype A1A1 A1A2 A2A2 Timing of Metamorphosis (Days) 200 160 150 d -15 -a a -25 25 0 Homozygote Midpoint (175)

  8. Genotype Freq Value Freq x Val A1A1 p2 25 p2(25) A1A2 2pq -15 2pq(-15) A2A2 q2 -25 -q2(25) Sum = Pop Mean = 25(p-q) + 2(-15)pq (adds time) (reduces time) p = f(A1) q = f(A2) A1A1 A1A2 A2A2 Mean 0 0 -25 2.25 -6.3 -12.25 6.25 -7.5 -6.25 12.25 -6.3 -2.25 25 0 0 0.0 0.3 0.5 0.7 1.0 1.0 0.7 0.5 0.3 0.0 -25 (150) -16.3 (158.7) -7.5 (167.5) 3.7 (178.7) 25 (200)

  9. Let’s Consider a Second Locus Thyroid Hormone Receptor Alpha Genotype A1A1 A1A2 A2A2 Timing of Metamorphosis (Days) 200 160 150 0 Thyroid Hormone Receptor Beta Genotype A1A1 A1A2 A2A2 Timing of Metamorphosis (Days) 200 140 0 -a a -30 30 Homozygote Midpoint (170)

  10. Genotype Freq Value Freq x Val A1A1 p2 30 p2(30) A1A2 2pq 0 2pq(0) A2A2 q2 -30 -q2(30) Sum = Pop Mean = 30(p-q) + 2(0)pq (adds time) (reduces time) P = f(A1) Q = f(A2) A1A1 A1A2 A2A2 Mean -30 (140) -12 (158) 0 (170) 12 (182) 30 (200) 0 0 -30 2.7 0 -14.7 0 0 0 14.7 0 -2.7 30 0 0 0.0 0.3 0.5 0.7 1.0 1.0 0.7 0.5 0.3 0.0

  11. Consider the joint effect of both TH Loci Total Range = 2Sa=110 Tha A1A1 Thb A1A1 Tha A2A2 Thb A2A2 Timing of Metamorphosis (Days) 227.5 117.5 0 -a a 55 55 Average Homozygote Midpoint (172.5) Overall Mean = Sa(p-q) + S2dpq

  12. 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 Breeding Value Empirically: An individual’s value based on the mean deviation of its progeny from the population mean. Theoretically: An individual’s value based on the sum of the average effects of the alleles/genes it carries.

  13. . a1 = 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 gene A1A1 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)] average effect of An: an = mean deviation from the population mean of individuals that received An from one parent, if the other parent’s allele chosen randomly a1 = q [ a + d (q – p)] a2 = –p [ a + d (q – p)]

  14. When there are only two alleles at a locus Average effect of a gene substitution A1A1 A1A2 A2A2 +a d -a (a - d) (d + a) p(a - d) + q(d + a) a = a + d(q - p) average effect of A1 average effect of A2 a pa

More Related