BIO 325 1st Edition Lecture 6Outline of Last Lecture I. Alleles are carried in chromosomesII. Sex chromosome in DrosophilaIII. Sex-linked traitsIV. MeiosisOutline of Current Lecture II. Pedigree AnalysisIII. Conditional ProbabilityIV. Pedigree ProbabilityV. Linkage (Beginning of Lecture 6)VI. Chi Square TestVII. Recombination FrequencyCurrent LectureI. X and Y linked traits in humans are identified by pedigree analysisa. X-linked traits 1. Recessiveii. Trait appear in more males than femalesiii. Never pass from father to soniv. Affected male passes mutation to all daughters who are heterozygous carriers v. Often skips a generationvi. Examples: Colorblindness, hemophilia, Duchenne’s muscular dystrophy1. Dominantvii. Affected males never have affected sonsviii. All daughters will be affectedix. Examples: Defective tooth enamelII. Conditional Probabilitya. Ex: What proportion of round peas are homozygous? Rr x Rr = ¼ RR, ½ Rr, ¼ rrP(A given B)=P(A and B)/P(B)1/3=(1/4)/(3/4)i. A: Round peas that are homozygous (1/4)ii. B: The rest (3/4)These 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.III. Probability Recapa. “At least”: 1-P(A)b. Unordered Events: Binomial equation or can be multinomial [Mathisfun.com]c. Conditional probability- just be able to deduce probabilities from given pedigrees Linkage Part IIV. If genes are very close together on the same chromosome, there is dependent linkagea. i.e Two genes on same chromosome segregate togetherV. recombinant: combination of alleles in child that you did not see in parentVI. parental: same as parent (opposite of recombinant)*Relative terms; dependent on parents/what genes are present*Can only be 0 to 50% recombinant. a. Independent assortment results in 50% recombination and 50% parentalb. Dependent linkage deviates from the 1:1:1:1 ratio of genotype of F1 gametesc. Recombinant ratio is neither 0% nor 50% signaling that crossing over occurred between 2 linked genesd. When genes are linked, parental classes are always the majorityVII. Chi-square Test: pinpoints the probability that ratios are evidence of linkagea. Sample size must be largeb. Equation: X^2=(O1-E1)^2/E1+(O2-E2)^2/E2+…….(ON-EN)^2/ENc. O= observed #d. E= expected #e. Df= degrees of freedom; should be 3 when testing 1:1:1:1 ratioVIII. Meiosisa. Recombination occurs during crossing overb. Chiasmata: crossing over at the site of pairing between homologous chromosomes (homologues) that physically changes the chromosomeLocus=LocationRecombination frequency= # of recombinants/total x 1001 RF= 1 map unit (m.u.); Directly proportional to distance between two genes on a chromosomeMapping: Locating genes along a chromosome When the m.u is > 50%, the genes will behave independently even if they are linkedThree point crossesFigure out the order of the three genes; which one is in the middleby comparing the distances between each