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## EVOLUTION

by: Freeda Strosin

28

0

5

# EVOLUTION BIOL 4415

Freeda Strosin
UCA
GPA 3.77

Staff

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COURSE
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Staff
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PAGES
5
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KARMA
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## Popular in Biology

This 5 page Class Notes was uploaded by Freeda Strosin on Thursday October 22, 2015. The Class Notes belongs to BIOL 4415 at University of Central Arkansas taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/227202/biol-4415-university-of-central-arkansas in Biology at University of Central Arkansas.

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Date Created: 10/22/15
Ileview f Funnyticquot Genetlcs liqllannnls 1 HardyWeinberg Equation p2 qu 12 1 Derivation Take a gene with two alleles call them A and a Dominance doesn t matter for our purposes this works equally well with codominance or incomplete dominance In a population some members will have the AA genotype some will have the Aa genotype and some will have aa Now imagine that you can somehow take all the gametes produced by the members of the populationifor simplicity we ll assume that these are eggs and sperm Some gametes of course carry A and some a p freq A q freq a NOTICE that the frequency of an allele is equal to the probability that a randomly chosen gamete will be carrying that allele Also notice that pq1 What s the chance that an egg and sperm drawn randomly will both be calrying A Obviously it s p x p or p2 And the chance that both gametes will both bear a is q x q or q2 There are two other possibilities sperm with A and egg with a or sperm with a and egg with A The chance of either one happening is p x q and the total probability of producing a zygote with the Aa genotype is twice that 2pq All of these probabilities sum to 1 So p2 qu q2 1 Since pq1 pq2 p2 qu q2 12 1 WHO CARES It s like this this equation links allele frequency to genotype frequency assuming certain conditions are met This means that l Ifyou know allele frequencies you can predict the genotype frequencies and compare them with the actual frequencies If they don t match then one of your assumptions is Violatedimaybe there is natural selection going on or immigration or nonrandom mating 2 Ifyou know genotype frequencies you can predict allele frequencies and compare them with the actual frequencies Again if they don t match then one of your assumptions is Violated 3 Ifyou know phenotype frequencies then you can estimate genotype and allele frequenciesibut you can t test the underlying assumptions 2 Fitness pzwAA quwAa q2 waa W p2WAA W 2pqwAa W q2 wanm 1 Derivation w in general means fitness a measurement of the relative ability ofmdividuals with a certain genotype to reproduce successfully WAA for instance means the relative ability of individuals with the AA genotype to reproduce successfully w is always a number between 0 and 1 Adding ws to the HardyWeinberg equation allows you to predict the effect of selection on gene and allele frequencies in the next generation Take the HardyWeinberg equation and multiply each term the frequency of each genotype by the fitness of that genotype Add those up and you get the mean fitness W wbar Divide through by W and you get the second equation Here each term of the equation is multiplied by the fitness of a genotype divided by the mean fitness If a genotype is fitter than average this quotient is greater than 1 and that genotype will increase in frequency in the next generation If a genotype is less fit than average the quotient is less than 1 and that genotype will decrease in frequency in the next generation 3 Mutation q qu upo qu Aq up vq Derivation Imagine that in each generation allele A mutates to allele a with a frequency of 11 and that allele a backmutates to A with a frequency of V Then in each generation q the frequency of the a allele increases by a factor of up the rate of mutation of A to a times the frequency of A and decreases by a factor of vq These will eventually balance each other out so that Aq 0 ie allele frequencies don t change any further When Aq 0 it must be true that up vq From this with a little algebraic jugglery you can derive the formula q u u V A where q qha 1s the equ111br1um frequency Slmllar equatlons let you der1ve p This isn t all that useful an equation however In real life mutation rates are usually on the order of 10 5 per locus per generation For example in humans the Huntington s chorea mutation spontaneously appears about once in every 200000 gametes produced This means that mutation by itself has very little effect on allele frequencies 4 Inbreeding freq AA p2 qu freq Aa 2pq 2qu freq aa q2 qu Derivation F is the inbreeding coef cient and it is the probability that two alleles in a diploid zygote are identical by descentiin other words that they are both descended from the same recent ancestor within the population The effect of inbreeding is to increase the frequency of homozygotes and decrease the frequency of heterozygotes These equations show how this effect is quantified Given that F is the probability that two alleles are identical by descent what is the probability that a given genotype will be AA There are two ways in which a genotype can be AA First one of the parental gametes could be carrying A and the other could be identical by descent the probability of this happening is p x F The other possibility is that one gamete could have the A allele the other gamete could have the A allele and the two alleles are not identical by descent The probability of that being the case is p x p x 1 F since if F is the probability of two alleles being identical by descent 1 F is the probability of two alleles not being identical by descent So the total probability of a given genotype being AA is the sum of these two pF p21 F We can expand that to pF p2 p2F rearrange to p2 pF p2F and regroup to p2 pF1 p 1 p of course is q so the formula for the frequency of the AA genotype is p2 qu You can work out the other two formulas in much the same way One way to use this formula is to calculate F if you know p and q for a population and you know the actual frequency of say the AA genotype you can plug those numbers in and calculate F If there is no inbreeding F 0 and you have the basic HardyWeinberg equilibrium if F 1 you have a completely inbred population with no heterozygotes at all To give a concrete example if you had a population in which only full siblings mated with each other after ten generations F would be about 085 5 Effective Population Size Here s the general idea Factors like genetic drift see below depend on population size But the census size of a populationithe raw number of membersimay not tell you what s really going on In a population of living things some members may not be reproducing because they re too young or too old because there s an excess of males or females because a few privileged males mate with multiple females etc etc The effective population size is the size of a hypothetical ideal population all of whose members have an equal probability of breeding that has as much inbreeding and genetic drift as the real population Exactly how you calculate this depends a lot on the situation and can be a little tricky but here s a sample In a population of diploid individuals with separate sexes Ne the effective population size is equal to 4Nme N Nm where N is the number of females and Nm is the number of males If N and Nm are equal then Ne reduces to N Nm which is just N the actual population size But suppose that this population is dominated by a single alpha male who does all the breeding which is close to what happens in some animals such as elephant seals Then Nm l and the formula becomes N8 4N N 1 So if you have a population of 100 elephant seals 50 males and 50 females N 100 but Ne 20051 39 In other words a population of 100 elephant seals in which only one male mates will have as much genetic drift and inbreeding as a population of four elephant seals in which all members could mate This has a lot of implications for conservation and breeding programs 6 Genetic Drift Vq poqoZN where Vq is the variance in the allele frequencies after one generation N6 is the effective population size and p0 and qo are the allele frequencies that you started with The variance of any set of numbers is a measure of how spread out the numbers are to be more exact it s the sum of the di erences between each number and the mean of the set The square root of the variance is the standard deviation which you might be more familiar with 7 NarrowSense Heritability h2 slope of least squares regression of mean offspring phenotype on mean parental phenotype Derivation Use this when you re dealing with a continuous traitiheight weight color number of parts etc as long as it can be quanti ed by integers or real numbersirather than a simple Mendelian dominantrecessive trait For each pair of parents nd the mean of the trait for each set of offspring find the mean value of the trait Then graph parental means vs offspring means on a Cartesian graph and take the slope of the regression line through the set of points Ifthe slope is nearly 0 then parental traits have nothing to do with what s seen in the offspring if the slope is close to 1 then parental traits and offspring traits are tightly correlated BIG DISCLAIMER The trait doesn t have to be genetically controlled AT ALL for this to work Poverty for example would show a strong narrowsense heritability children from rich families are usually rich themselves children of poor families often stay poor So would religious af liation with some exceptions children usually adopt the religion of theirparents But there are no poverty genes or Baptist genes 8 Response to Selection for a Continuous Trait R h2S Derivation For a continuous trait calculate the mean phenotype of the rst generation t and the mean phenotype of the second generation after a round of selection t The difference between the two is S the selection differential for a continuous trait s t t S multiplied by the narrowsense heritability h2 is the response to selection R This is a prediction of how a population will respond to directional selection on a continuous trait

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