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## Hot answers tagged enzyme-kinetics

3

Let's do some unit analysis: $\pu t$ = time $\pu U$ = units = $\mathrm{mol_{product}/t}$ $\pu s=$ specific activity $\mathrm{= U/g_{protein}}$ Therefore: $\mathrm{s = mol_{product} / (g_{protien} \times{t})}$ Your sheet says: $\mathrm{k_{cat}= s \times MW_{protien}}$ And we know that: $\mathrm{MW_{protein} = g_{protein} / mol_{protein}}$ ...

2

$K_M$ is a constant and is not $V_{\mathrm{max}}/2$ as they have different dimensions. But by the Michaelis Menten equation, $V=\frac{V_{\mathrm{max}}[S]}{K_M+[S]}$. So, when $[S]=K_M$, we have $V=\frac{V_{\mathrm{max}}[S]}{[S]+[S]}=\frac{V_{\mathrm{max}}}{2}$. Hence, $K_M$ is the substrate concentration, at which we would have $V=V_{\mathrm{max}}/2$. ...

2

If you apply quasi-steady state approximation for the complexes then you will get three equations for each of the complexes. Then you have to use the conservation law for enzyme to obtain the equation with respect to just substrate and inhibitor. If you try solving these algebraic equations, it gets very complex and difficult to solve by hand. However, in ...

1

The situation you described is entirely normal. $V_{max}$ is the same, so when described with a graph of initial reaction rate vs. substrate concentration, both curves will asymptotically approach the same maximum level. $K_M$ is higher in one than the other, so the exact shape of the curve as it goes from 0 to $V_{max}$ is different. As you may know, you ...

1

The order of magnitude for your intercept value depends on the axis of your graph, which in turn depends on the values / units of your input values. As an example if your substrate concentrations are in the range of $\mu mol / l$ (which is micromolar, not mircomoles), then that is also the order of magnitue you'll get out of the calculation. $dm^{-3}$ ...

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