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# H4_sol - ECEN 689 Statistical Computation in Genomic Signal...

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ECEN 689 - Statistical Computation in Genomic Signal Processing Fall 2009 Homework 4 – Solutions 1. Let n be the number of carbons. The second peak in the isotope pattern, according to the binomial law, has height 0 . 0108 × n × 100 (where the base peak is normalized to 100). Therefore, n = 34 . 56 / 1 . 08 = 32. For the purpose of checking, we can verify that the second peak has height 0 . 5 × 32 × 31 × 0 . 0108 2 × 100 = 5 . 78, as it should. Subtracting from the base peak mass the masses of C 32 and NO 2 yields the number of hidrogens: 471 - 32 × 12 - 14 - 2 × 16 = 41 The formula of Terfedine is therefore C 32 H 41 NO 2 . 2. (a) Let the two elements be A and B, and let the compound be A n B m . Let X = mass of A n Y = mass of B m X + Y = mass of A n B m The masses X and Y (and thus X + Y ) are discrete random variables, due to the random occurrence of isotopic variants of A and B . The corresponding probability mass functions p X , p Y , and p X + Y are the mass spectra (normalized to sum up to 1) of X , Y , and X + Y , respectively. Note that p X and p Y are determined from the

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H4_sol - ECEN 689 Statistical Computation in Genomic Signal...

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