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An enzyme has a $V_{max}$ of 50 $\mu$mol product formed $(min * \text{mg protein})^{-1}$ and a $K_m$ of 10 $\mu$M for the substrate. When a reaction mixture contains the enzyme and a 5$\mu$M substrate, which of the following percentages of $V_{max}$ will be closest to the initial reaction rate ($V_o$)

a) 5% b) 15% c) 33% d) 50% e) 66%

Can someone explain this to me; I know michaelis-menton but I don't understand the solution to this or how to find it.

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closed as off-topic by AliceD, March Ho, MattDMo, WYSIWYG Oct 19 '15 at 5:56

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You literally just have to plug in the numbers in the Michaelis Menten equation. $\endgroup$ – Inhibitor Oct 17 '15 at 11:00
  • $\begingroup$ Giving a value for Vmax is superfluous. The question may be answered without any knowledge of the Vmax value. If the enzyme obeys MM kinetics, the velocity will be one-third of Vmax at S=Km for all values of Vmax. $\endgroup$ – user1136 Mar 24 '18 at 10:17
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I don't think you understand MM kinetics.

In your case the $V_{max}$ value depends on the amount of the protein. If we assume 1mg protein, then $V_{max} = 50\frac{\mu{M}}{min}$.

According to MM kinetics:

$V = V_{max}.\frac{S}{K_M+S}$

so

$V_0 = V_{max}.\frac{S_0}{K_M+S_0} = V_{max}.\frac{5\mu{M}}{10\mu{M}+5\mu{M}} = V_{max}.\frac{1}{3}$

or in other terms

$\frac{V_0}{V_{max}} = \frac{1}{3} = 33.3\%$

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  • $\begingroup$ @ChrisCook If you consider that inf3rno answered your question, then you should check it to indicate future users that you received an answer that helped you. It also reward the one who answered with 15 points of reputation and therefore encourage users to answer your future posts. $\endgroup$ – Remi.b Oct 18 '15 at 22:51
  • $\begingroup$ sorry, new to the site; I checked the check under his answer, I assume this is what you meant? thanks $\endgroup$ – Chris Cook Oct 19 '15 at 6:47

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