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boringboring

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Your sample's molecular monoisotopic mass is 944.5.
(M+H)+, A0: 945.5 (with zero C being C13)
A1:  946.5  (with 1 C being C13)
A2:  947.5  (with 2 C being C13), intensity ratio of A0:A1:A2 can yield number of carbons in your peptide.
(M+2H)++, A0: (944.5+2)/2=473.25
A1: 473.8=473.25+0.5
A2:  474.3=473.8+0.5
945.5-657.4=loss of mass 288 fragment, which is correspondent to its doubly charge 473.2-329.2.
3.98e3 is the ion count of the most abundant peak at 473.2, 100%.
7楼2013-07-24 05:29:46
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