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#1
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| Hi! I wonder how to calculate the ratio of reactionrate, Va/ Vb, for which an enzyme shows to substrate a, and substrate b, in a mixture of the following substrates where [a]=[b]: Substrate a = Km = 4yM kcat = 1 s-1 Substrate b: Km = 40mm kcat = 10 s-1 What is the difference in activation energies for that the enzyme substrate shows to substrate a, and substratemb. How can we show this in a diagram whith engeri on one axis and reationproecess on the other. This is a Michaelis-Menten equationproblem. thanks |
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#2
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| Hello, Well, the first thing to notice is that the specificity constant (that is the ratio kcat/Km IS THE SAME for both substrates. I'll call the specificity constant for substrate A Sfa, that is Sfa = kcat1/Km1 Similarly Sfb = kcat2/km2. Sfa = Sfb = 1 There is a very valid argument that we should consider the specificity constant and kcat as the two fundamental kinetic constants, and that the Michaelis constant is the RATIO of kcat/Km. That is, it is much more meaningful to consider Km, rather than the specificity constant, as the ratio of two fundamental kinetic constants. This argument is made very cogently by Fersht (see below). So ... From v1 = kcat1 a / (km1 + a) and v2 = kcat2 b /(km2 +b) (the MM equation in 'old-fashioned' form) we come up with the following for v1/v2 when a = b = s (s = substrate concentration) v1/v2 = (40 + s)/ (40 + 10 s) Notice that when s is very low (both substrate concentrations well below Km), the ratio v1/v2 equals 1. To put it another way, at low concentrations of both substrates, v1/v2 equals the ratio of the specificity constants. Are you sure that the question you want answered is to calculate v1/v2 when a = km1 and b = km2? This is a very interesting case, and v1/v2 = Sfa/Sfb For the example you give, this ratio will equal 1!! See Alan Fersht "Structure and Mechanism in Protein Science", 1999, pp 110 - 111 (W.H. Freeman & Co. are the publishers), or any of Fersht's books on enzyme kinetics. See also Dixon & Webb "Enzymes" Third edition 1979 (Longmans) pp 72 - 75 for a discussion of the case when a = Km1 and b = Kmb (this is a slightly more advanced treatment). I do not understand the second part of your question. Perhaps you could post a more specific update? Good luck, tgd |
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