EVE100_142-177_2pp

EVE100_142-177_2pp - 1/20/10 1 142 142 Phenotypic evolution...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1/20/10 1 142 142 Phenotypic evolution and quantitative genetics (Ch 13) •What are “quantitative traits”? 143 143 Phenotypic evolution and quantitative genetics •Phenotypic variance = V P •Genetic variance = V G •Environmental variance = V E • V P = V G + V E •V G has an additive component, V A, and non-additive components resulting from dominance and epistasis. Dominance = interactions between alleles at a locus Epistasis = interactions between loci •V A determines the genetic response to selection and can also be thought of the genetic basis for parent-offspring similarity 1/20/10 2 144 144 Phenotypic evolution and quantitative genetics More on the evolutionary importance of additive variation •Consider a trait such as beak size, with the following genetics, no environmental effects, and allele freq = 0.5 Beak size Genotype freq AA 1 cm 0.25 Aa 1 cm 0.5 aa 0.5 cm 0.25 Mean phenotype is (0.25)1 + (0.5)(1) + (0.25)(0.5) = 0.875 145 145 Phenotypic evolution and quantitative genetics Beak size Genotype freq AA 1 cm 0.25 Aa 1 cm 0.5 aa 0.5 cm 0.25 Mean phenotype is (0.25)1 + (0.5)(1) + (0.25)(0.5) = 0.875 cm •Now, suppose we mate an AA individual (1 cm) with another, randomly selected individual. Regardless of which random individual is picked, the offspring have beak = 1 cm •Now, consider an Aa individual randomly mated to another bird: Avg. phenotype of offspring (based on Mendelian seg) Prob mate with AA = 0.25 0.5(1) + 0.5(1) = 1 Aa = 0.5 0.25(1) + 0.5(1) + 0.25(0.5) = 0.875 aa = 0.25 0.5(1) + 0.5(0.5) = 0.75 The exp. mean phenotype of offspring of this bird = 0.25(1) + 0.5(0.875) + 0.25(0.75) = 0.875 1/20/10 3 146 146 Phenotypic evolution and quantitative genetics To summarize: Mean population phenotype is (0.25)1 + (0.5)(1) + (0.25)(0.5) = 0.875 cm •Offspring of AA individual (1 cm beak ) all have beak = 1 cm •Offspring of Aa individual (1 cm beak ) have mean beak = 0.875 Both the AA and Aa parent have 1 cm beak, which deviates from population mean by 0.125 cm....
View Full Document

This note was uploaded on 03/29/2010 for the course EVE EVE 100 taught by Professor Davidbegun during the Winter '10 term at UC Davis.

Page1 / 18

EVE100_142-177_2pp - 1/20/10 1 142 142 Phenotypic evolution...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online