# PQ+1AB+answers - ABd(0.82(0.30(0.5 = 0.123...

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Biological Sciences 101 Dr. Mark Sanders Summer Session I 2011 Pop Quiz 1 Answers Each pair of adjacent genes is either recombinant or non-recombinant. The probability of recombination is equal to the recombination frequency and the probability of no recombination is equal to 1 – the recombination frequency. There are two equally likely recombinants and two equally likely non-recombinants for each gene pair, i.e. 0.5 each. Quiz 1A - genes A, B and D Recombination between A and B = 0.18, no recombination = 1 – 0.18 = 0.82 Recombination between B and D = 0.30, no recombination = 1 – 0.30 = 0.70 The answers to the quiz questions are shown in bold . The expected frequencies of all eight gametes are given assuming that interference is zero to simplify the calculation of double recombinants. Gamete Expected Frequency ABD (0.82)(0.70)(0.5) = 0.287 abd (0.82)(0.70)(0.5) = 0.287
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Unformatted text preview: ABd (0.82)(0.30)(0.5) = 0.123 abD (0.82)(0.30)(0.5) = 0.123 aBD (0.18)(0.70)(0.5) = 0.063 Abd (0.18)(0.70)(0.5) = 0.063 AbD (0.18)(0.30)(0.5) = 0.027 aBd (0.18)(0.30)(0.5) = 0.027 1.000 Quiz 1B – genes E, F and G Recombination between E and F = 0.10, no recombination = 1 – 0.10 = 0.90 Recombination between F and G = 0.26, no recombination = 1 – 0.26 = 0.74 The answers to the quiz questions are shown in bold . The expected frequencies of all eight gametes are given assuming that interference is zero to simplify the calculation of double recombinants. Gamete Expected Frequency efg (0.90)(0.74)(0.5) = 0.333 EFG (0.90)(0.74)(0.5) = 0.333 efG (0.90)(0.26)(0.5) = 0.117 EFg (0.90)(0.26)(0.5) = 0.117 eFG (0.10)(0.74)(0.5) = 0.037 Efg (0.10)(0.74)(0.5) = 0.037 EfG (0.10)(0.26)(0.5) = 0.013 eFg (0.10)(0.26)(0.5) = 0.013 1.000...
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## This note was uploaded on 09/20/2011 for the course BIS 101 taught by Professor Simonchan during the Summer '08 term at UC Davis.

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