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MSU LBS 148 - Overview of Lecture: Microevolution II
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Overview of Lecture: Microevolution II Read: Text pgs 450-451, ch 22 & 23 More on microevolutionary patterns and processes with a careful look at the concept and use of heritability, the distinction between having particular alleles and using (expressing) them, and some ways parents control gene expression. Bullet Points: • phenotypic variation • selection within generations • heritability across generations • phenotypic response across generations • fluctuating selection on finch beaks • some complications when partitioning variance heritability of physical traits and IQ • having a gene vs expressing a gene • microarrays and the cellular “RNA world” • epigenetic effects on gene expression • methylation & imprinting • parental modulation of gene expression • loss of imprinting and diseasehttp://www.youtube.com/watch?v=FnzmxeZJeho&feature=relatedhttp://www.youtube.com/watch?v=faRlFsYmkeYAn organism exposes its phenotype … not its genotype to the environment. big benthic little limnetic sticklebacks2. Pile individuals into the bins, making a bar graph 3. and/or draw a line through the tops of the bars 5. Relabel the y axis to show the proportion in each bin; do this by dividing the # in each bin by the total n (ex: 50) 20/50 = 0.4 This is convenient for comparing different graphs; they all have the same y scale & sum up to 1 It is conventional & convenient to describe samples or pops with a graph called a frequency distribution We can describe a frequency distribution with a few descriptive statistics, like: ave = mean = 170 a. the mean (average) b. various measures of variation around the mean, ex: variance, std deviation, etc. 160 165 170 175 180 185 1. Divide the range of data into bins (x axis); ex: 5 cm student height (cm) 4. Label the y axis to show the # indiv’s in each bin 20 Ex: a quantitative trait: height of a sample of 50 female Vet students http://www.husdyr.kvl.dk/htm/kc/genetik/6/calc.htm20/50 = 0.4 160 165 170 175 180 185 At this point, we can see and describe the distribution of the phenotypic trait, offspring height (cm) but, the variation around the mean is just random, noise until we can relate deviations from the mean to some other, ‘explanatory’ variable: Well – parents and offspring have similar genes, so if genes have something to do with height, Hyp: taller than ave offspring from taller than ave parents? 160 165 170 175 180 185 mid-parent (ave) height (cm) 20/50 = 0.4 Does parental height ‘explain’ any of the var in offspring height? How can we see if offspring deviations from ave are related to (correlated with; ‘explained by’) parental deviations from ave? ex: can we ‘explain’ why this indiv is above ave?It is obvious from inspection that deviations of parents and offspring are correlated How can we get a more precise, quantitative measure of the extent to which offspring trait deviation from mean can be predicted from knowing parental trait deviation from mean? This is exactly why Galton invented the statistical procedure linear regression. 160 165 170 175 180 185 160 165 170 175 180 185 mid-parent (ave) height (cm) offspring height (cm) AVE x AVE y The procedure fits a line y = a·x + b through the data that pivots around the point (mean y, mean x) to find the slope a that best predicts how much you expect particular y’s will deviate from ave given the amount particular x’s deviate from ave. 0.75 if y = a·x + b , then (y - ave y) = a·(x - ave x) So, if we have a pair (or sample) of parents that are about 8cm below ave, how much would you predict/expect their offspring to deviate from ave ? ex: (y - ave y) = 0.75·(- 8 cm) = -6cm (minimizes the sum of vertical deviations (‘errors’) squared) In our student height ex, the slope is a ≅Note: if y = a·x + b , then (y - ave y) = a·(x - ave x) {just translating the origin} . If the slope = a, how much should we expect the the next generation (offspring) to respond R to this selection S? Let xs be the ave trait value of the breeding parents; while x0 is the ave trait value for all adults; call this diff the selection differential S = (xs - x0) 160 165 170 175 180 185 160 165 170 175 180 185 mid-parent (ave) height (cm) offspring height (cm) x0 y0 expect: R = a ·(xs - xo) = a·S, ex: R = 0.75 · -8 = -6cm* *technically, we would convert the units to standard deviations and not work in cm {What if parents and offspring have similar environments as well as similar genes?}The slope of the linear regression line is one estimate of heritability. 1. If knowing parents trait deviation tells nothing about offspring trait deviation (ex all var is env ‘noise’), dots fall in circular cloud, slope = 0, heritability h2 = 0, and the response R to any selection S is 0. mid-parent (ave) height offspring height (cm) 2. If knowing parents trait deviation tells everything about offspring trait deviation (ex all var is genetic), dots fall along a 45° line, slope = 1, heritability h2 = 1, and the response R = selection S. mid-parent (ave) height offspring height (cm) 3. Galton called this ‘regression’ because offspring are rarely as ‘deviant’ as their parents; they ‘regress toward the mean,’ slope < 1, heritability h2 < 1 mid-parent (ave) height offspring height (cm)Evolutionary beacon. Medium ground finch beaks wax and wane with climate shifts. CREDIT: JOSEPH W. DOUGHERTY … On page 707 of this issue the Grants review 30 years of evolution among Darwin's finches. … {this is the Science version of News & Views, called News of the Week} {Volume 296, Number 5568, 26 Apr 2002, pp. 633-635.}, within speciesFrom 1972 to 2001, Geospiza fortis (medium ground finch) and Geospiza scandens (cactus finch) changed several times in body size and two beak traits. Natural selection occurred frequently in both species and varied from unidirectional to oscillating, episodic to gradual.Figure 3. Predicted and observed evolutionary responses to natural selection on beak size ( ) and shape ( ) in G. fortis and beak size


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MSU LBS 148 - Overview of Lecture: Microevolution II

Course: Lbs 148-
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