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PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Figure 3–11 (a) Graph for converting 2 values to p values. (b) Table of 2 values for selected values of df and p.Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 381Biology 240 - General GeneticsSpring 2015Dr. Joe WilliamsContact: 974-6202, [email protected], January 22ndUnless otherwise indicated, photo credits are from:A.Klug WS, Cummings MR, Spencer CA, Palladino MA, Killian D. 2014. Concepts of Genetics. Boston: Pearson.B.Griffiths, S. Wessler, R. Lewontin, and S. Carroll. 2008. Introduction to Genetic Analysis, Eighth Edition. W. H. Freeman & Company.2Quiz todayReadings for today and next week are:Chapter 3.9-endAll of Chapter 4 Chapter 9 – generally, these deal with our next topic, “inheritance patterns”3XPure parentsX5852110 totalResultsIs 58:52 ratio consistent with expected 55:55 ratio?What is the probability of getting 58/110 heads when the coin is fair?4χ2 = Σ (observed – expected)2/expectedHow many degrees of freedom (df) are there?Degrees of freedom = number of independent variablesroll the dice 110 times:heads (y) = 52 then you know tails (Y) has to be 58 (since sum to 110)So, there is only one independent variable:You could express your results as P {y} = 52/110Same as P {Y} = 58/110Only 1 variable: P of getting y5χ2 = Σ (observed – expected)2/expectedClass Obs. Exp. (Obs.-Exp.)2(O-E)2/Ey 58 55 32 = 9 9/55= 0.16Y 52 55 -32 = 9 9/55= 0.16Total χ2 = 0.32Expect a 1:1 ratio if independent events110 events (flips, or meioses)expect ½ to be heads = 55expect ½ to be tails = 55how muchyou deviatedScaled to howlarge yourexpected is6χ2 = Σ (observed – expected)2/expectedHow many degrees of freedom (df) are there?Degrees of freedom = number of independent variablesIn our case, 2 classes minus 1 = df = 1If you know 1 class you can get the second, so there is only one independent variableClass Obs. Exp. (Obs.-Exp.)2(O-E)2/Ey 58 55 32 = 9 9/55= 0.16Y 52 55 -32 = 9 9/55= 0.16Total χ2 = 0.32General rule in probability tables: df = # classes – 1, or df = k-17χ2 = 0.32 for 1 df*The probability of getting 52 y alleles, if you expect to get 55(fair coin, equal segregation), is between 50% and 90%In other words, in about 6 of every 10 trials you woulddeviate from 55 by 3 or less just “by laws of chance”8“<5% is significant rule”It is not that unusual to get 52 “heads” by chance,when the dice (meiosis) is fairχ2 = 0.32 for 1 df*The null hypothesis of equal probability is not rejected hereScientists choose an arbitrary standard for rejecting the nullProbability of less than 1 in 20 of getting a particular result9Phenotypic ratio of9:3:3:1Dihybrid crossWe can use same approach for cases where results are not all of equal probabilityRoll a 4-sided dice 556 timesexpect 9 of every 16 yellow, round10Mendel’s example of independent assortmentClass O E (O-E)2(O-E)2/Eround, yellow 315 313 (2)2 = 4 4/313 = 0.013wrinkled, yellow 108 104 16 0.154round, green 101 104 9 0.087wrinkled, green 32 35 9 0.257Total χ2 = 0.510Expect a 9:3:3:1 ratio if independent assortment“If 556 seeds, then 9:3:3:1 ratio would be: 313:104:104:35”Ratio x # seeds = Expected #s9/16 x 556 = 312.75 = 3133/16 x 556 = 104.25 = 1043/16 x 556 = 104.25 = 1041/16 x 556 = 34.75 = 3511χ2 = Σ (observed – expected)2/expectedClass O E (O-E)2(O-E)2/Eround, yellow 315 313 4 0.013wrinkled, yellow 108 104 16 0.154round, green 101 104 9 0.087wrinkled, green 32 35 9 0.257556 556 Total χ2 = 0.510How many degrees of freedom (df) are there?Degree of freedom = number of independent variablesIn our case, 4 classes minus 1 = df = 3If you know 3 classes you can get the fourth, so there are 3 independent variables12χ2 = 0.510P = 1/20 or 0.05 is considered pretty improbable and is the generalscientific standard for saying something is “significantly different” from expectation.“significant”*We would expect a deviation from the expected ratioof this magnitude over 90 % of the time – i.e. this isvery close to the expected ratio13How to state a statistical conclusion:Mendel’s data is consistent with a 9:3:3:1 ratio, or Mendel’s data is not significantly different from a 9:3:3:1 ratio14How to state a statistical conclusion:Mendel’s data is consistent with a 9:3:3:1 ratio, or Mendel’s data is not significantly different from a 9:3:3:1 ratioIn this example, we can say that “there is no evidence that the color gene and the wrinkled gene are associated with each other.” They appear to be inherited independently.151913 – Elinor Carothers experiment in grasshoppers showing independent assortment of chromosomes“Purple” chromosome moved to the same pole as the short-armed chromosome exactly half of the time.Meiosis Iis fair!Figure 3–11 (a) Graph for converting 2 values to p values. (b) Table of 2 values for selected values of df and p. Notice that a chi-square value of just under 4 is never significant17Chapter 4 how phenotypes are inheritedInheritance patternsAutosomal inheritance pattern: either way you do the cross, you get same inheritance pattern1:1 ratio here 1:1 ratio hereSingle gene inheritance patterns – Dom/RecAutosomalAA1/4Aa1/4Aa1/4aa1/4Aa x AaA aAaPhenotypic ratio = 3:1Genotypic ratio = 1:2:1F1F2Monohybrid crossP AA x aaBecause of dominance or recessiveness, only two phenotypes can be detected20Conventions for abbreviating gene/allele namesWhen there is dominanceA dominant allele A/- dominant phenotype (AA or Aa)a recessive allele aa recessive phenotypeSingle gene inheritance patterns – Incomplete, or partial dominanceAutosomalAA1/4Aa1/4Aa1/4aa1/4Aa x AaA aAaPhenotypic ratio = 1:2:1Genotypic ratio = 1:2:1F1F2Monohybrid crossP AA x aaPartial dominancepattern22Conventions for abbreviating gene/allele namesCodominance(both alleles expressed)partialdominance© 2015 Pearson Education, Inc.Incomplete (partial)dominancepatternNeither allele isrecessive,but…The R1 allele determines the phenotype1 copy has lessexpression (pink)than 2 copies (red)atggctacctggtacctgcctgat………..cctacgctgcgtaatggactaccttagRemember, at the DNA sequence level, inheritance is always CODOMINANT in diploid individuals – two alleles are presentIt is the expression of those alleles,

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