PowerPoint PresentationSlide 2Slide 3Slide 4ExampleSlide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Heritability1.Phenotypic variation2.Broad sense heritability3.Narrow sense heritability4.Artificial selection Please check announcements!!! I have better explained what you need to do for the lab assignment!Top 4 for Wednesday, Sept. 11, 2013© 2012 Pearson Education, Inc.Quiz Sept. 11, 2013Define the following:VP, VG, VE, VGxE•Heritability describes the proportion of total phenotypic variation in a population due to genetic factors•Many quantitative (and some other) characters, phenotypes are result of environment and genetic interaction•It does not indicate how much of a trait is genetically determined or the extent to which an individual's phenotype is due to genotype•A heritability estimate tells the proportion of phenotypic variation that can be attributed to genetic variation within a certain population in a particular environment•High heritability in multifactorial traits – much variation in pop attributed to variation in genetic factors (low environmental influence).•Low heritability – environment has high impact on phenotypic variation in pop.Example•Chickens in egg processing facility – environment, food controlled. –Egg production may have high heritability (differences in egg output among birds likely due to genetic differences)•Chickens in free range environment – –Egg production probably lower heritability – egg production may depend more on food acquisition, roosting spots etc. (environment more important).•Wording!!! Imagine human height heritability 0.65•Interpretation: in the population sampled, on average, 65% of overall variation in height could be explained by genotypic differences among individuals.•Does NOT mean that 65% of height is due to genes!•Phenotypic variance has components of–genotypic variance–environmental variance–genotype-by-environment interaction varianceVP = VG + VE + VGxEOr simplify as VP = VG + VE•Estimating heritabilities –Use inbred strains with highly homozygous genotypes–Test in range of environmental conditions•Variation between inbred strains in constant environment – predominantly genetic factors•Variation among members of same strain in different conditions – predominantly environmental factors•Broad-sense heritability – an estimate of how much of the observed (phenotypic) variation in the population can be attributed to genotypic variance•VG (genotypic variance) component includes all types of genetic variation (additive, epistatic and dominance effects). •Assumes VGxE is negligible.H2 = VG / VP = VG / (VE + VG)•High broad-sense heritability–Most of VARIATION in phenotype can be accounted for by genotypic differences among individuals in a population–Close to 1 – environment has little impact on phenotype–Close to 0 – most of phenotypic variation accounted for by environmental effects–Chickens:•Body weight 50%•Egg production 20%•Egg hatchability 15%•Narrow-sense heritability measure of genetic factors that contribute to variation that respond to selection - additive genotypic variance alone•Additive genotypic variance is partitioned into–additive variance–dominance variance–interactive varianceh2 = VA / VP VP = VG + VEVG = VA + VD + VI (interactive effects often negligible)h2 = VA / VP VP = VG + VEVG = VA + VD + VI (interactive effects often negligible)VA represents additive genetic variance that results from average effect of additive components of genesVD is the deviation from additive components that results when phenotypic expression in heterozygotes is not precisely intermediate between 2 homozygotesVI deviation from additive components that occurs when 2 or more loci behave epistaticallySO:h2 = VA / (VA + VD + VE) The higher the h2 the greater the ability to predict the phenotypes of offspring during selectionA hypothetical study investigated vitamin A content and the cholesterol content of eggs from a large population of chickens. The variances were calculated, as shown here:Variance Vitamin A CholesterolVP 123.5 862.0VE 96.2 484.6VA 12.0 192.1VD 15.3 185.3Calculate the narrow-sense heritability (h2) for both traits. Which trait, if either, is likely to respond to selection? Why?•Artificial selection is the process of choosing specific individuals with preferred phenotypes from an initially heterogeneous population for future breeding•The purpose is to develop a population containing a high frequency of individuals with the desired traits•Narrow-sense heritability estimates are valuable to plant and animal breeders because they estimate the proportion of total phenotype variance for that trait that is due to additive genetic varianceh2 can be measured with an artificial selection experimentFigure 23.7© 2012 Pearson Education, Inc.Table 23.4•Any single heritability estimate can only provide information about one population in a specific environment•Narrow-sense heritability is a more valuable predictor of response to selection when estimates are calculated for many populations and environments and show the presence of a clear trendh2 = M2-M / M1-M = R/SM2 = mean score for trait of offspring of selected individualsM = original population mean scoreM1 = mean score of selected individuals (parents)R = selection response – degree of response to mating the selected parentsS = selection differential – difference between mean for whole pop and mean for selected popFrom an artificial selection experiment can estimate REALIZED