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When [S] = Km, the velocity of an enzyme catalyzed reaction is about? Select one:a. 0.1 Vmaxb. 0.9 Vmaxc. 0.4 Vmaxd. 0.7 Vmaxe. 0.5 Vmax |
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Answer» How to calculate the km and Vmax values of an enzyme when I have substrate/product INHIBITION? Hello, Could anyone please help me in calculating the Km and Vmax values of an enzyme (I am working on dihydrofolate reductase DHFR) when I have substrate/product inhibition? I SEE the inhibition on (specific activity nmol/min/mg to substrate CONCENTRATION µM) curve, with substrate concentrations near the theoretical Vmax. (I mean the calculated Vmax by ignoring the inhibition). (I attached a link of the curve) - More details: the Km of my enzyme is about 1µM. (in case I ignored the inhibition). The specific activity with 5µM substrate is 2µM/min/mg. (in my case this is Vmax when I ignore the inhibition). The specific activity with 10µ substrate is 1,7µM/min/mg. (it is lower here, that MEANS here we have inhibition). The specific activity with 20µM substrate is 1,5µM/min/mg. (again inhibition). I don't know what is the right method to calculate the Vmax and Km for that enzyme and substrate with that inhibition phenomenon, WAITING kindly for your guiding.Explanation: |
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