lec 2 Probability_posted

lec 2 Probability_posted - Phenotype cross genotypes and...

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Punnett Square in a monohybrid cross F1 F2 Yy YY x Yy yy Y y Y y or or Male gametes Female gametes Y Male gametes y Y y YY Yy Yy yy Female gametes 3 yellow : 1 green 2 phenotypes, 3 genotypes Genotype : genetic constitution of an organism Phenotype : observable characteristics of an organism I II III III 1 YY (I) : 2 Yy (II) : 1 yy (III) Phenotype Monohybrid cross genotypes and phenotypes
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All yellow, round F1 Yy Rr YR yr yR Yr gametes Punnett Square in a dihybrid cross 9 : 3 : 3 : 1 Yellow, round Green, round Yellow, wrinkled Green, wrinkled 4 phenotypes, 9 genotypes YR YR Yr Yr yR yR yr yr YYRR YYrr YYRr YyRr YyRR YYRr YyRr Yyrr YyRR YyRr yyRR yyRr YyRr Yyrr yyRr yyrr Male gametes Female gametes I II III IV V V VI VI VII VII VIII VII VII VIII IX IX Phenotypes: Genotypes: 1 YYRR (I) : 1 YYrr (II) : 1 yyRR (III) 1 yyrr (IV) : 2 YYRr (V) : 2 YyRR (VI) 4 YyRr (VII) : 2 Yyrr (VIII) : 2 yyRr (IX) Dihybrid cross genotypes and phenotypes
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# genes 1 2 3 . . . n # phenotypes 2 4 8 . . . 2 n # genotypes 3 9 27 . . . 3 n # boxes in a Punnett square 4 16 64 . . . 4 n Following multiple numbers of genes poses problems to represent numbers of potential combinations Monohybrid cross Dihybrid cross How do we follow crosses with more than two segregating genes? Phenotype
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The Probability Method Probability of event A = P(A) = fraction of times event A occurs in a large number of trials P(A) = m/n, where m is the number of times event A occurs among n trials e.g. in the self fertilization of of Yy pea plants, for genotypes: P(YY) = 1/4 P(Yy) = 1/2 P(yy) = 1/4 P(genotypes) = P(YY) + P(Yy) + P(yy) = 1 for phenotypes: P(yellow seed) = 3/4 P(green seed) = 1/4 P(phenotypes) = P(yellow seed) + P(green seed) = 1 Note that - P(A) will always be a fraction between 0 and 1, where a probability of 0 means that the event will never occur, a probability of 1 that the event will always occur, a probability of 1/2 (= 0.5) that (in a large number of trials), the event will occur in one out of two trials, etc. - the sum of the probabilities for each of all possible outcomes equals 1 Probability
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The Probability Method The Multiplicative rule: if events A and B are independent * , then the Probability that they occur together, either simultaneously or sequentially, is P(A and B) = P(A) x P(B) * Independence of events: two events are independent if the realization of one event does not affect the realization of the other. Other examples of independent events - simultaneous coin tosses with two coins - sequential coin tosses with one coin, - segregation of alleles from different genes in Mendel’s experiments - the genotype, phenotype or sex of sequential children in a family e.g. in a deck of cards, where there are 52 cards (13 cards for each of four classes) P(ace of hearts) = P(ace and heart) = P(ace) x P(heart) P(ace) = 4/52 (there are four aces in a deck) P(heart) = 13/52 (there are 13 hearts in a deck, (=1/4)) So P(ace and heart) = 4/52 x 1/4 = 1/52 Probability
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The Probability Method The Additive Rule: if events A and B are independent, then the probability that at least one of them occurs, either simultaneously or sequentially, is P(A or B) = P(A) + P(B) - [P(A and B)] where P(A and B) = P(A) x P(B)
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lec 2 Probability_posted - Phenotype cross genotypes and...

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