Lecture 5

Lecture 5 - Biology 313 Lecture 5 Sept 1 2010 Corn...

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Unformatted text preview: Biology 313 Lecture 5 Sept 1, 2010 Corn detasseling • http://www.livinghistoryfarm.org/farmingint he30s/crops_03.html Predicting the outcomes of genetic crosses Punnett square Another method for determining the outcome of a genetic cross is to use rules of probability probability Probability: likelihood of the occurrence Probability: of a certain event of Rules of Probability Rules If • Sum rule: If two events are mutually exclusive, the probability that either of two events will occur is the either sum of their separate probabilities p(A or B) = p(A) + p(B) p(A If • Product rule: If two events are independent, the probability that the two events will both occur is the product of their separate probabilities p(A and B) = p(A) * p(B) p(A In a family of four children, what is the probability that all children are females? The probability that a female child is born is independent of whether or not that child has any sisters, therefore the product law applies. In a family of four children, what is the probability that all children are females? The probability of each child being female is ½. Therefore the probability of four children being all female is (1/2)4 or 1/16 The branching diagram The Consider the cross: Tt x Tt Gametes from Parent 1 = T and t Gametes Gametes from Parent 2 = T and t What is the expected ratio of progeny genotypes? TT, Tt, tt genotypes? Can use the product rule Gametes parent 1 Gametes parent 2 Progeny genotypes 1/2 T = 1/4 Tt = 1/4 tT 1/2 t 1/2 T 1/2 t 1/2 T 1/2 = 1/4 TT = 1/4 t Conclusion: 1 TT : 2 Tt : 1 tt tt Rules of probability in Mendelian crosses: Rules • • Given the cross: Rr X Rr Given A. What is the probability the first three offspring are all A. R phenotype? B. • B. What is the probability the first three offspring are all rr genotype? Answer: Genotype Phenotype = 1/4 RR, 1/2 Rr, 1/4 rr R = 1/2 + 1/4 = 3/4, r = 1/4 • Rr x Rr • Each offspring is independent, so the product rule is used for combinations. • A. all R Phenotype: 3/4 x 3/4 x 3/4 = 27/64 • B. all rr Genotype: 1/4 x 1/4 x 1/4 = 1/64 • C. With five offspring, what is the probability that 3 have the R phenotype and 2 have the r phenotype? The Binomial Expansion • binomial = (p + q)n binomial • • • • p = probability of R q = probability of r p + q = 1 I.e. they are mutually exclusive I.e. n = number of trials (e.g., number of offspring) Expand • Expand the binomial: • • • (p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5 Each term is the probability of one combination of five offspring. The probability of 3R and 2r = 10 (3/4)3(1/4)2 = 0.264 0.264 • P = n! psqt s!t! P = probability, n = total #, s = # first class, t = # second class probability, P= 5! (3/4)3 (1/4)2 = 5x4x3x2x1 (3/4)3 (1/4)2 = 10 (3/4)3(1/4)2 = 0.264 5! 0.264 3!2! 3x2x1x2x1 3!2! Mendel’s Monastery Brno, Czech Republic F2 P F1 P Selfed F2 Dihybrid Crosses Reveal the Principle of Independent Assortment Discovery of the Principle of Independent Assortment came from Dihybrid Crosses (crosses that involve two characteristics) Mendel Dihybrid Cross Assumption #1: Traits are on different chromosomes Assumption #2: There is Independent Assortment of alleles at Anaphase I Assumption #2: There is Independent Assortment of alleles at Anaphase I Can diagram results using Punnett Square or via Branching (later) Thought Question Assume Mendel did NOT obtain a 9:3:3:1 ratio in the F2. Would you conclude that there is no such thing as the 2nd Law? An example of the 2nd Law from Chapter 2 (Figure 2-17) : random distribution of chromosomes at Anaphase I produces genetic variation m = maternal p = paternal How predict progeny? • Branching method: an easier, faster way to predict progeny of crosses than Punnett Square method, especially for more than two traits • Independent assortment applies so can apply the multiplication rule Branching method for Dihybrid Cross Consider each gene (chromosome) separately Yy x Yy Can predict genotypes and phenotypes phenotypes • Given the cross: Rr Yy Tt X rr Yy tt rr What • What is the probability of an offspring with an rr YY tt genotype? genotype? Given the cross: Rr Yy Tt X rr Yy tt Given rr What is the probability of an offspring with an rr YY tt genotype? Answer: Genotype • • • Rr x rr wrinkled Yy x Yy Tt x tt = 1/2 Rr, 1/2 rr = 1/4 YY, 1/2Yy, 1/4 yy = 1/2 Tt and 1/2 tt Phenotype = 1/2 round 1/2 = 3/4 yellow, 1/4 green = 1/2 tall, 1/2 short Each event is independent, so the product rule is used for combinations. Genotype probability is 1/2 rr x 1/4 YY x 1/2 tt = 1/16 How to work a genetics problem? How See p. 60 (Good tips) Goodness-of-fit chi-square test When two (or more!) genes segregate independently, do we expect to see exactly the ratios predicted by rules of probability? NO !!!!!!!! Why is there deviation between an observed outcome and the expected outcome? chi-square test 2 steps: 1. Determine χ2 value χ2 = Σ (observed-expected)2/expected 2. Use the calculation of χ2 to determine a probability (p-value) Degrees of freedom (df) = one less than number of independent variables (for us, this is one less than the number of progeny classes) Mendel’s testcross: Rr Yy x rr yy Mendel’s • Hypothesis: Mendel’s F1 progeny are expected to have a Hypothesis: 1:1:1:1 ratio. 1:1:1:1 (null hypothesis = no difference between observed and expected results) • χ2 test: genotype Rr Yy Rr yy Rr rr Yy rr yy Total observed observed 55 expected expected 51.75 (obs-exp)2/exp (obs-exp) 0.204 0.204 51 49 52 51.75 51.75 51.75 0.011 0.146 0.001 207 207.00 0.362 = χ 2 0.362 Determine p-value Determine Degrees of freedom (df) = 4 -1 = 3 Degrees df = one less than the number of progeny classes classes Go to χ 2 distribution table Go Accept null hypothesis: chance explains the deviation of the observed from the expected Reject null hypothesis: something else explains the deviation of the observed from the expected p-value > 0.1 (i.e., a deviation between observed and p-value deviation expected due to chance is likely, thus accept the null likely thus hypothesis hypothesis 0.362 ...
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