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Week 4, Lecture 3, Mendel

Week 4, Lecture 3, Mendel - Gregor Mendel Solving Mendelian...

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1 26-Sep-09 Gardiner 1 Gregor Mendel Fig. 12-2, p. 236 B I O L O G Y 0 2 3 26-Sep-09 Gardiner 2 Mendel’s Law of Segregation From monohybrid cross work The two chromosomes of a homologous pair separate (are segregated) during metaphase I of meiosis I B I O L O G Y 0 2 3 26-Sep-09 Gardiner 3 Mendel’s Law of Independent Assortment Comes from the dihybrid cross work Independent assortment of alleles occurs as the various homologous pairs of chromosomes within a cell assort independently during anaphase I of meiosis I B I O L O G Y 0 2 3 26-Sep-09 Gardiner 4 Solving Mendelian Problems Determine the simple probability of gamete genotype A Aa P (A) = ½ AA P (A) = 1 B I O L O G Y 0 2 3 26-Sep-09 Gardiner 5 Simple Probability The probability of equally likely outcomes is 1 Number of possible outcomes E.g. A coin toss has 2 possible outcomes P (H) = P (T) = 1/2 B I O L O G Y 0 2 3 26-Sep-09 Gardiner 6 Solving Mendelian Problems Determine the probability of gamete genotype AB AaBb P (AB) = ¼ Rule of multiplication - “A and B”
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2 B I O L O G Y 0 2 3 26-Sep-09 Gardiner 7 Compound probability: multiplication rule The probability of two independent events having a specified outcome is P 1 x P 2 For example, a two-coin toss has four possible outcomes ½ tails x ½ tails = ¼ double tails Multiplication Rule : If events A and B are independent, then P(A and B) = P(A)P(B) Suppose a six sided dice is rolled twice and the rolls are independent. A = rolling a 1 on the first roll B = rolling a 1 on the second roll P(rolling a 1 on both rolls) = P(A and B) =P(A)P(B) = 1/6 x 1/6 = 1/36 Segregation of alleles & fertilization as chance events B I O L O G Y 0 2 3 26-Sep-09 Gardiner 10 Dihybrid Cross So: Parent YyRr probability gamete will carry Y and R alleles is ¼ In the F 2 we don’t have to construct a Punnett square – probability of YYRR is 1/16 ( ¼ chance ovum YR x ¼ chance YR sperm) B I O L O G Y 0 2 3 26-Sep-09 Gardiner 11 Compound probability : addition rule The probability of alternative outcomes is P 1 + P 2 For example, in a two coin toss, the probability that they will be the same is = ¼ double tails + ¼ double heads = 1 /2 B I O L O G Y 0 2 3 26-Sep-09 Gardiner 12 Solving Mendelian problems Determine the probability of gamete genotype Ab or aB AaBb P (Ab ) = 1/4 AaBb P (aB ) = 1/4 P (Ab or aB ) = 1/2 Rule of addition “Ab or aB
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3 PpYyRr x Ppyyrr What fraction of offspring will exhibit the recessive phenotype for at least two of the three traits? Pyr pyr PYR PPYrRr PpYyRr PYr PPYyrr PpYyrr PyR PPyyRr PpyyRr Pyr PPyyrr Ppyyrr pYR PpYyRr ppYyRr pYr PpYyrr ppYyrr pyR PpyyRr ppyyRr pyr Ppyyrr ppyyrr PpYyRr x Ppyyrr What fraction of offspring will exhibit the recessive phenotype for at least two of the three traits?
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