Lecture 17- basic principles of heredity independent assortment

Lecture 17- basic principles of heredity independent assortment

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Unformatted text preview: Lecture 17 06-03-11 Chapter 3 Basic Principle of Heredity nd law: Independent Assortment 1. 2. Branch diagram to determine the phenotype and genotype from the cross 3. Goodness-of-fit chi-square test  Mendels DihybridCross Experiment Cross involving two pairs of contrasting traits F1:RrYy ConclusionfromDihybridCrossExperiment Independent assortment The inheritance of one trait does not influence the inheritance of another trait in a dihybrid cross Monohybrid cross: The F1 generation - dominant phenotype: Round or Yellow. Dihybrid cross: The F1 - dominant phenotype for each trait: Round and Yellow. Two independent monohybrid crosses During gamete formation, segregation of alleles at independent or unlinked loci is independent of the segregation of each other. - The independent allele segregation and subsequent random union of gametes at fertilization determine the phenotypes observed. BranchDiagramfor DihybridCross ThePunnettSquare canbeused,buta branchdiagram (Forked Line)is ofteneasier Dihybridcanbe understoodas twoindependent monohybrid crosses Multiply probabilitiesfor eachcombination BranchDiagram/ForkedLine You will get the same results if you first list color and then shape TrihybridCross Three pairs of contrasting traits: A/a, B/b, C/c Cross between AABBCC x aabbcc Gametes: ABC x abc F1 hybrid: AaBbCc F1 Gametes (8 types): 2n ABC, ABc, AbC, Abc, aBC, aBc, abC, abc Punnett square results in 64 boxes = cumbersome Use the Branch Diagram/Forked-Line Method Branch Diagram for Trihybrid F i g : 3.10, K & C Practice:DihybridTestcross Inthetestcross, oneparentishomozygousrecessiveatbothloci rryy(wrinkled,green). Theotherparenthasdominantphenotyperoundandyellow.The genotypecanbehomozygousdominantgeneonbothlocioronly oneloci,orheterozygousonbothloci. R_Y_=RRYY,RRYy,RrYY,RrYy(RoundandYellow) ry Practice:RrYyxrryy Rr x rr = Rr rr Yy x yy= Yy yy Yy yy 1/4 RrYy 1/4 Rryy 1/4 rrYy rryy 1/4 Practice:DihybridTestcross ObservedandPredictedRatioofProgeny Genetic ratios are prediction based on Mendilian principles. RrYy x rryy = ¼ RrYy: ¼rrYy: ¼ Rryy:1/4rryy Predicted outcome 1 (round yellow):1(wrinkled yellow):1(round green):1(wrinkle green). The actual phenotypes we observed when we look at the progenies from the cross may deviate from the expectation - RrYy x rryy = 1000 progeny - Expecting 250 round yellow, 250 wrinkle yellow, 250 round green, and 250 wrinkle green - In fact, we may see 243: 262: 254: 239 The deviation between the expectation and actual observation is due to chance Example: toss coin 1000 times 486 heads and 514 tails are within reason. Deviation is attributed to chance. EvaluatingFitnessofGeneticData If other than chance causes the deviation, how can we tell? Observed 25 round and 15 winkled seeds, would the ratio still be 1:1? How to evaluate the deviation between observation and expectation is due to chance or not? Maybe the inheritance of this character (seed shape) is more complicated than was assumed Chi-Square (X2) a statistic test to evaluate the fitness of observed results to theoretical expectations TheGoodness ­of ­fitChi ­SquareTest Chi-square test evaluates the probability that the deviation between the observed and the expected values is due to chance Establish null hypothesis (Ho): yes due to chance Ho: No real difference exists between the observed values and expected values. Calculate the Chi-Square value, which is measuring the deviation between the observed and expected value Used to find the probability that the deviation is due to chance The lower the Chi-Square value is, the higher the possibility that the deviation is due to chance, the Ho is truer The higher the Chi-Square value is, the higher the likelihood that the other factor other than chance alone causes the deviation. Ho is false. We need to reject the null hypothesis: Ho CalculateChi ­Square(X2)Value Chi-Square (X2) X2 ( O b s e r v e d E x p e c t e d )2 E xpected with degrees of freedom (df) = n-1, where n = the number of different phenotypic classes. 3:1 ratio, n = 2, so df = 2 1 = 1 9:3:3:1 ratio, n = 4, df = 3 Use the chi-square value and df to find the probability of fitness (p-value) using a chi-square table. Use the 5% level, p-value = 0.05, as a threshold for determining whether to reject (p < 0.05) or accept (p 0.05) the null hypothesis.   chi-square table Accept Ho Reject Ho Chi ­SquareAnalysis ­Example MonohybridCross,Purple&WhiteFlowers 3:1 E xpected R a t io O b se r v e d (o) E xpected (e) ⇠ 105 ⇠ (150) = 112.5 1 /4 45 (150) = 37.5 (o (o e) (o e )2 e )2 e 105-112.5 = -7.5 (-7.5)2 = 56.25 56.25/112.5 = 0.5 45-37.5 = 7.5 (7.5)2 = 56.25 56.25/37.5 = 1.5 X 2 = 2.0 150 .1 < P < .5 2 (observed - expected) 2 expected n=2,df=1 Conclude:p>0.05,FailtoReject,theobservedDatafitstheexpected3:1ratio. Steps to perform a Chi-square test 1. Find following experimental data 1. Total number of offspring analyzed 2. Number of different classes of offspring 3. Number of offspring observed in each class 2. Calculate how many offspring would be expected in each class if Ho is correct. 1. Multiply the expected ratio (portion) by the total number of offspring analyzed in the experiment. 3. Calculate the Chi-square value. 4. Compute the degrees of freedom 5. Use the chi-square value and df to find the probability of fitness (p-value) using a chi-square table. 1. Use the 5% level, p = 0.05, as a threshold for determining whether to reject (p < 0.05) or fail to reject (p 0.05) the null hypothesis. Chi ­SquareAnalysis ­Example n=4,df=3 DihybridCross 9:3:3:1 E xpected R a t io O b se r v e d (o) E xpected (e) 9/16 587 567 + 20 400 0.71 3 / 16 197 189 +8 64 0.34 3 / 16 168 189 -21 441 2.33 1 / 16 56 63 -7 49 0.78 1008 (o (o e) (o e )2 e )2 e X 2 = 4.16 .1< P <.5 Conclude:p>0.05,FailtoReject,theobservedDatafitstheexpected9:3:3:1ratio. Chi ­SquareAnalysis ­Practice Dihybrid test cross E x pected R a t io O b se r v e d (o) 1/2 1 262 1/2 258 1/2 E xpected (e) df=? 243 1/2 1:1:1:1 (o (o e) (o e )2 e )2 e 237 1000 X2 = <P< ...
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