09.20.11 - Lecture 5 Extranuclear inheritance sex...

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Lecture 5. Extranuclear inheritance, sex chromosomes, and mating systems Readings: Chapt 2: 50-54 Chapt 3: 103-109 http://www.poorstudentscookbook.com/ “…dedicated to teaching college students how to cook real food for the same price as a frozen dinner or less.”
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Laws of Probability (IV) Combinatorial numbers Number of different ways of flipping m heads from N coin flips # combinations = N ! m !( N - m )! = N m ÷ = 52! 5!(52 - 5)! = 52 5 ÷ = 52 × 51 × 50 × 49 × 48 5 × 4 × 3 × 2 × 1 # possible poker hands = 2.6 x 10 6
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Laws of Probability (IV) Binomial Theorem: m N m a a m N m N P - - - = ) 1 ( )! ( ! ! represents one term in expanded binomial What is probability of having exactly 2 boys and 2 girls? Binomial theorem: (a+b) n = 1
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Laws of Probability (IV) Binomial Theorem: applies when there are 2 mutually exclusive events a and b such that P( a ) = 1-P( b ) where a is the probability of one outcome and ( 1-a ) the probability of the other outcome and m = the number of occurrences of outcome a and N is the total number of trials Examples: flipping coins, sex of child m N m a a m N m N P - - - = ) 1 ( )! ( ! !
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p (B) = p (G) = 1 2 3 4 5 ordered events = P (BBBGG) = P (BBGBG) = = p (any ordered event (p(B) #boys x p(G) #girls ) Question: If two parents have 5 children what is the probability that they will have 3 boys and 2 girls? Part I. What is the probability of having 3 boys and then 2 girls?
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Part II. Now we just need to know how many ordered series of boys and girls will give 3 boys and 2 girls Question: If two parents have 5 children what is the probability that they will have 3 boys and 2 girls? B 1 B 2 B 3 G 4 G 5 number of possible rearrangements = But if we only care about the sex of them, then we want to eliminate redundant ones: = N ! b !( g )! BBBGG GGBBB BBGBG GBGBB BBGGB GBBGB BGBBG GBBBG BGGBB BGBGB P = N ! m !( N - m )! a m (1 - a ) N - m 5! Number of ways to arrange 3 boys and 2 girls. 5!/3!2! = 10 * (1/32)= 5/16
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Data collection: 120 mice Yy (Yellow) x Yy (Yellow) The larger size of the yellow gene is due to pleiotropic effects of the agouti Y allele.
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Do these data fit a Mendelian model? Yellow (Yy) x Yellow (Yy) YY = 1/4 (Yellow) Yy = 2/4 (Yellow) yy = 1/4 (brown) Expected: 3 yellow: 1 brown Working Model: single gene trait, dominant brown (Y) allele. Use Mendel’s laws to predict expected genotypes Observed: 78 yellow, 42 brown Accept or reject model? χ 2 = 6.4, df=1 p -value Obs. Exp. Obs-Exp (O-E) 2 /E 78 90 12 144/90= 1.6 42 30 12 144/130=4.8 = 6.4 (sum of two) Yellow Brown
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Do these data fit a Mendelian model?
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