Introductory Genetics Lecture 2 PDF

BIO207H5S Introductory Genetics Winter 2024 Lecture 2 Preeti Karwal, Ph.D. Forked line/Branch Diagram for Dihybrid Cross : applying Probability rules Forked line/Branch Diagram for Trihybrid Cross : applying Probability rules Round seeds, yellow endosperm, gray seed coat X Wrinkled seeds, gree...

BIO207H5S Introductory Genetics Winter 2024 Lecture 2 Preeti Karwal, Ph.D. Forked line/Branch Diagram for Dihybrid Cross : applying Probability rules Forked line/Branch Diagram for Trihybrid Cross : applying Probability rules Round seeds, yellow endosperm, gray seed coat X Wrinkled seeds, green endosperm, white seed coat RRYYCC X rryycc F1 : RrYyCc Selfing F2 : R_Y_C_ : R_Y_cc : R_yyC_ : rrY_C_ : R_yycc : rryyC_ : rrY_cc : rryycc Phenotype Ratio : Try calculating yourself!! Problem Solving for calculation of probability of obtaining offspring…… Aa Bb cc Dd Ee × Aa Bb Cc dd Ee Calculate the probability of obtaining offspring with the genotype aa bb cc dd ee The probability of an offspring from this cross having genotype aa bb cc dd ee : ¼ × ¼ × ½ × ½ × ¼ = 1/256 The Chi-Square (𝛘2) Goodness-of-Fit Test The ratios of genotypes and phenotypes actually observed among the progeny, however, may deviate by chance from the expected ratio as per Mendelian principles of segregation, independent assortment, and dominance. This test indicates the likelihood of chance that could produce the deviation between the expected and the observed values. Scientific Question: Is there a difference between observed and expected ratio? Null hypothesis: There is no significant difference between the observed and the expected ratio. Degrees of freedom (d. o. f) = No. of categories (n) - 1 TABLE 3.7 Critical values of the 2 distribution The critical value, P(0.05) at d.o.f = 1 is 3.841. P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 3 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357 7.779 9.488 11.143 13.277 14.860 5 0.412 0.831 1.610 4.351 9.236 11.070 12.832 15.086 16.750 6 0.676 1.237 2.204 5.348 10.645 12.592 14.449 16.812 18.548 7 0.989 1.690 2.833 6.346 12.017 14.067 16.013 18.475 20.278 8 1.344 2.180 3.490 7.344 13.362 15.507 17.535 20.090 21.955 9 1.735 2.700 4.168 8.343 14.684 16.919 19.023 21.666 23.589 10 2.156 3.247 4.865 9.342 15.987 18.307 20.483 23.209 25.188 11 2.603 3.816 5.578 10.341 17.275 19.675 21.920 24.725 26.757 12 3.074 4.404 6.304 11.340 18.549 21.026 23.337 26.217 28.300 13 3.565 5.009 7.042 12.340 19.812 22.362 24.736 27.688 29.819 14 4.075 5.629 7.790 13.339 21.064 23.685 26.119 29.141 31.319 15 4.601 6.262 8.547 14.339 22.307 24.996 27.488 30.578 32.801 P, probability; df, degrees of freedom. *Most scientists assume that, when P < 0.05, a significant difference exists between the observed and the expected values in a chi-square test. Since 𝛘2 = 0.46 is less than 3.841, the null hypothesis fails to be rejected or in other words, the null hypothesis is accepted. In other words, since the value of 𝛘2 = 0.46 lies between P(0.1) and P(0.5), the probability of deviation occurring by chance lies between 10% - 50% and hence null hypothesis is accepted. So, we conclude that there is no significant difference between the observed and the expected ratio. Chi Square Analysis Problem solving for a monohybrid cross…… Using Binomial Equation to calculate Probability Binomial expansion helps solve complex genetics problems related to probability. It helps calculate all possibilities for a given set of two unordered events. If there are 2 events with alternate independent events having probabilities p and q, then in n number of trials, the probabilities of various combinations of events is given by : = (p + q)n where p + q = 1 e.g., the two categories can be Normal and Diseased, WT and Mutant, Tall and Short. Binomial probability that exactly x events belonging to a category occur in n trials = n = no. of trials x = no. of events in one category n – x = no. of events in another category p = Probability of occurrence of x events out of n trials q = Probability of occurrence of other events (n-x) out of n trials where n! = 1 x 2 x 3 x 4 ….n Probability calculation using Binomial Expression….. In a family of six children, what is the probability that atleast four will be girls? G = Girl B = Boy Probability of having a G = ½ Probability of having a B = ½ Event Binomial Expression Probability 4G, 2B 6!/(4! 2!) x (½)4 X (½)2 15/64 5G, 1B Try calculating yourself!! 6G, 0B Total prob. = 22/64