Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: scientizscht on 26/06/2020 13:41:34
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Hello
I do not understand what specificity constant kcat/KM represents.
Its units are e.g. 10/M*s
I cannot make sense of units like 1/M*s
Can anyone explain what that really represents please?
Is it 10 substrate 'conversions' per second when the substrate is at concentration of 1M ?
Or what exactly?
Thanks!
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Units for rate constants can be funny. It's because you have to multiply by concentrations of reactants (depending on the molecularity of the reaction) to get actual rates (d(concentration of product)/dt or –d(concentration of reactant)/dt).
https://en.wikipedia.org/wiki/Rate_equation
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But if you multiple a constant of 1/M*s with a molarity, you get units of 1/s.
What would these units represent?
E.g. if an enzyme has a specificity constant of 1000/M*s and the substrate available in the solution is 5mM, what will the 5/s mean?
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It might mean that, on average, one molecule of the enzyme produces 5 molecules of the product per second. (in this case, the turnover frequency, or TOF, is 5 s–1)
Or it means that the rate constant must be multiplied by a term (or expression) that doesn't just have units of molarity, in order to provide the rate.
From what you have described, I can't tell which scenario applies (or maybe a third option).
Again, I urge you to read the wiki about rate laws I posted in my first response to this thread.
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Thanks, I read it but not sure how it answers my question.
What are the inputs needed to be able to convert the specificity constant into moles per second converted?
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What are the inputs needed to be able to convert the specificity constant into moles per second converted?
That depends on the rate law.
In some cases, rate = k[A]
In some cases, rate = k[A][ B]
In some cases, rate = k[A]2
In some cases, rate = k[A]1/2
(this is not an exhaustive list)
There isn't a way to derive this unless you know the actual mechanism of the rate-limiting step of the reaction (and sometimes information about bits that come before the rate limiting step is also needed). Typically these rate laws are determined experimentally. If the rate law has already been worked out by someone else, and you want to know the rate law, the units of k will tell you which form the rate law has.