BIOB 272 1st Edition Lecture 11Outline of Last Lecture Dominance Continued, Pedigrees, and LinkageI. Mutiple LociII. Genes & environment: Expression variation III. Pedigree Analysis1. Autosomal, recessive 2. Autosomal, dominant3. X-linked, recessive4. X-linked, dominantIV. Linkage & Mendel’s Principles1. Dominance2. Segregation3. Independent AssortmentV. LinkageOutline of Current Lecture I. Probability: a. Product Ruleb. Sum Rulec. Conditional ProbabiltiyII. Genes on Same Chromosome vs. Genes on Different Chromosomes after MeiosisIII. Hypothesis TestingThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Chi-Square goodness of fit testCurrent LectureStats, Hypothesis Testing, and LinkageI. Probability: - (P) of an event is number of times event will occur (a) divided by total number of possible events (n). P = a/n- Example: The probability of flipping a heads…o Pheads=1 heads/(1 heads + 1 tails)= ½ or 50%a. Product Rule: - If events A and B are independent, then the probability that they both occur is: P(A and B) = P(A) x P(B)- The probability of 2 or more independent events occurring simultaneously is equal to the product of their individual probabilities- Example: What’s the probability of flipping 2 heads in arow?P(A) = 0.5 , P(B) = 0.5P(A and B) = P(A) x P(B)0.5 x 0.5 = 0.25 = 25%- Example: Progressive Retinal Atrophy (PRA) -common recessive disease causingblindness in dogso PP = healthy/non-carrier, Pp = healthy/carrier, pp = affectedo Is a healthy dog a carrier?? PP x pp= all progeny Pp and healthy Pp x pp= 50% progeny Pp and healthyo We do the cross. We get 5 healthy puppies in a row. What’s the probability of that outcome if the test dogis a carrier (i.e., Pp)? Answer = (.5*.5*.5*.5*.5)= 3% =1/32b. Sum Rule: - Probability of one of two or more mutually exclusive events occurring is equal to the sum of their individual probabilities: P(A or B) = P(A) + P(B)- Example: Probablity of a tall plant from a Tt x Tt cross? Tall if TT, Tt, or tTo Punnett Squares results: TT, Tt, tT, tt P(tall)= ¼+ ¼ + ¼ = ¾ c. Conditional Probability: - Probability of an event occurring conditional upon a limited set of possibilities- Example: what is the probability of rolling a3 with a dice if we role an odd number?o P 3/odd= P3/Podd= (1/6)/(1/2)= 0.33- Example: What is the conditional probability that a tall plantfrom a cross between two heterozygotes (Tt x Tt)will be heterozygous?o Punnett Square Results: TT, Tt, Tt, tto PTt/tall= (0.25 + 0.25) the Tt heterozygous results from the Punnett Square/(0.25 +0.25 +0.25) all of the tall options in the Punnett Square = 0.67II. Genes on Same Chromosome vs. Genes on Different Chromosomes after Meiosis:a. Genes on different chromosomes at the start of Meiosis I will create equal amounts of each allele pair- Example: YY, RR, yy, rr loci all on different chromosomes at the start ofMeiosis=> after Meiosis II will have 25% Ry, 25% rY, 25%ry, 25%RYb. Genes on the same chromosomes at the start of Meiosis I will create:- Possible haploid cells without crossing over- Possible haploid cells with crossing over. Crossing over can reassort linked alleles.- They may not show independent assortmentIII. Hypothesis Testing- In science, when we test a hypothesis, our experimentis designed to falsify not “prove” our hypothesis.-Do the observed data fit the expected distributionthat we’d find if our null hypothesis (HO) is true?Chi-Square goodness of fit test:- X^2= ((O1-E1)^2)/ E1 + ((O2-E2)^2)/E2- O=observed data, E=expected dataExample:Null hypothesis (HO): Mendelian ratio of 1:2:1Do these values fit this hypothesis?F2 Phenotype Observed # Expected #Red Flower 62 62.5Pink Flower 131 125White Flower 57 62.5Total 250 250Chi-square Statistic: X^2= ((62-62.5)^2)/62.5 + ((131-125)^2)/125 + ((57-62.5)^2)/62.5= 0.776Find the degrees of freedom (df)= total number of classes (of flower color) minus one:df=n-1 = df= 3-1=2Use a chi-square table (**in lecture notes) to determine the critical value given the degrees of freedom (2) and using a P value of 0.05:For this example you get: 5.991The x^2 (chi-square statistic #) is 0.776 which is less than 5.991 so we accept the null hypothesis(Ho)-if it were to be larger than the 5.991 we would reject the null
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