University:
Swansea University
Course:
PM-149 | Genetics, Genomics & Evolution
Academic year:

2023
Views:

56

Pages:

6
Author:

Amal Jith

Genetic duftneutral allele frea change in small propln purely by chancein called genetic duffmax in small poplnselectively neutralMignitude of Genetic Driftfounder effectMGD= PqBottle neck effect2NCreneticduittelfect max when P and q are equalN=5MGD= Pq = 0.5 x0.52N 2x5if P=0.5= 0.025= 25 magnitude of genetic diffif P= 0.20.2x0.810= 0.016 16 magnitode of geneticdrift. Popln size increase from550MGD = 0.25100= 0.0025x 1000 = 2.5 MGDif popln size in small impact in largeLitresSelectioncoefficient(s)= I-f$fruit fly lavval grown in rottenfroit that for contain alcohol, flypocas alcohol debydiogenase enzymeto detoxity alcohol andthis promote survival of lavval, 2alleles of Alcohol dehydiogenareobserved in the froit. fly populationAdhf/Adhf=120large no:masffitAdbf/Adbs=60Ads/AdD's =30f1 = 120 =1S1=1-1120= 0f2= 60.0.5S2 = 1-0.5120= 0.5f3== = 30 ==0.25S3 = 1-0.25120=0.75 Catcolate If a poptn has an intial freq of Adhf = 0.2,what will be the freqin nex genP=0.2 q=0.82pq.q?intial genotype freq = P20.04 0.32 0.164. = 1After one generation = intial freq x fitness.0.04x1 0.32x0.5 0.64x0.250.16 = 0.320.040.160.040.36divide with0.040.160.160.36.0.36 for0.360.36conveting to Q.10.H4 = I0.110.44genotype change afte onegenerationp2 0.11P= 0.11=0.33q=1-0.33= 0.670.44x0.25Afte 2 generation 0.11x1 0.44x0.50.110.220.11 = 0.110.22=0.440.440.44 0.110440.50.25= 0.25P=0.5q=0.5 Popln size increase from 550MGD = 0.25100= 0.0025 x 1000 = 2.5MGDif poplin size in small,impact in largefitnesSelectioncoefficient(s)= I-f$fruit fly lairal grown in rotten froit that Ber contain alcohol,thypocas alcohol debydiogenase enzymeto detoxity alcohol oodthis promote survival of lavval, 2alleles of Alcohol dehydiogenareobserved in the fruit fly populationAdhf/Adhf=120large no: masffitAdbf/Adbs=60Ads/AdD =30f1 = 120 =181=1-1120= Of2= 60 0.5S2 = 1-0.5120=0.5f3 30 = 0.25S3=1-0.25120=0.75 CalcolateIf a poptn has an intial freq of Adhf = 0.2, what will be the freqin nex genP= 0.2 q=0.8intial gonotype freq = p22pq. q?0.04 0.32 0.164 = = 1After one generation = intial freq X fitness0.04x10.32x0.50.64x0.250.16. = 0.320.040.160.040.36divide with0.040.160.160.36.0.36 for0.360.36conveting to Q.10.44 = I0.110.44genotype change afte onegenerationP2.0.11P= 0.11=0.33q=1-0.33= 0.670.11x10.44x0.50.44x0.25Afte 2 generation0.110.220.11 =0.110.0=0.440.440.44 0.11= 0.250.50.25P=0.5q=0.5 harmful dominant allele eliminate rapidly than harmful receisiveallele because receisive allele can mask their effect in the heterozygocscondition but dominont allele cannotChisquare analysisx2= E (0-E)2o observed=EE = expecteddegree of freedom = n-1n= no:of variablesBlack catDF= 2-1Grey cat= IBB6bf1BbF2 30black 20greyobserved freq.if it's a mendalion phenolype12.5xf2= 3:1 (phenolypic ratio).12.53753 x50= 37.5H1,x50Expected phenotypic frequency.4= 12.5x2= (30-37.5)2(20-12.5)+12.5E37.5= (-7.5)2(a 7.5) 2+37.512.5=6

Understanding concept of Genetic drift and the Important questions.pdf

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