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Calculating velocity in enzyme kinetics


Zahra B

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The question says "write the formula for calculating velocity of enzymatic reactions. From this formula, calculate the velocity of a reaction with concentration of the substrate equal to 0.5 Km" 

Im totally stuck. I know the formula is Vo = Vmax (S)/Km + (S) but I cant solve it any further

Edited by Zahra B
Misspelling
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Is this homework/coursework ?

Please note we use square brackets to denote concentration;
Curved brackets have a different meaning.

The Michaelis-Menton equation can be written


[math]v = \frac{{{V_{\max }}\left[ S \right]}}{{\left( {{K_m} + \left[ S \right]} \right)}}[/math]


Where Km and Vmax are constants

So the equation has two constants to be determined.

A general plot of the equations shows two regions.

A linear region from the start with < Km

and a rounding off region where the graph becomes asymptotic to Vmax.

Substituting   = Km yields a point on the linear section where v = Vmax/2, as shown on the graph.
 

This give a value for Km in terms of the others

So the equation can be further simplified in this region.

Since this is homework

I will leave you to calculate what happens when you substitute =  Km/2 into that linear simplification.

M-Mreaction.jpg.9fa9023a3a941de1bd10b221a1ff1ec0.jpg

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Hello! Thank you for taking your time to answer. Just wanted to clarify, this is not a homework, its a question from an old test I found while preparing for my test next week. I believe if we substitute km/2, the linear graph moves to the left since less substrate is needed to half saturate the enzyme?

I was also thinking that since substrate concentration is much less than Km, the velocity of the reaction is proportional to the substrate concentration (first order), so maybe the answer should simply be Vo = substrate concentration?

Ps. I dont know why but everytime I put the S in brackets, it strikes through the text

Edited by Zahra B
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1 hour ago, Zahra B said:

I think i solved it. Don't know if it is correct but it seems reasonable.

 

Sorry you haven't got the idea.

But then this is algebra and I expect you are a bioscientist.

Remember that the M-M equation is not concentration v time whose slope is the rate.
It is a graph of rate v concentration so starts of at the origin when the rate must be zero if the concentration is zero.

Let is consider the M-M equation


[math]v = \frac{{{V_{\max }}\left[ S \right]}}{{{K_m} + \left[ S \right]}}[/math]


[math]When\quad {K_m} = \left[ S \right]\quad then[/math]


[math]v = \frac{{{V_{\max }}\left[ S \right]}}{{\left[ S \right] + \left[ S \right]}} = \frac{{{V_{\max }}\left[ S \right]}}{{2\left[ S \right]}} = \frac{{{V_{\max }}}}{2}[/math]


Now this gives us two points on the x, y graph of the M-M equation, in the linear region.

the point (0, 0) where  concentration (x axis) = zero and the rate (y axis) = 0

and the point where the concentration (x axis) =Vmax/2 and the concentration (y axis) = KM

Because we are in the linear region we can redefine the equation in that region as rate = slope x concentration
We do not need to 'shift it'

So at our two known points


[math]slope = \frac{{\left( {{y_2} - {y_1}} \right)}}{{\left( {{x_2} - {x_1}} \right)}} = \frac{{\left( {\frac{{{V_{\max }}}}{2} - 0} \right)}}{{\left( {{K_m} - 0} \right)}} = \frac{{{V_{\max }}}}{{2{K_m}}}[/math]

Which give the simplified equation of the linear region as


[math]v = \frac{{{V_{\max }}}}{{2{K_m}}}\left[ S \right][/math]

Now we substitute the value of concentration   [math]\left[ S \right] = 0.5{K_m}[/math]

 


[math]{v_{s = 0.5{K_m}}} = \frac{{{V_{\max }}}}{{2{K_m}}}\left[ S \right] = \frac{{{V_{\max }}}}{{2{K_m}}}\frac{{{K_m}}}{2} = \frac{{{V_{\max }}}}{4}[/math]

 

Sorry the strikethrough is still wonky.

Though I can conquewr it using MathML.

:)

 

 

 

 

Edited by studiot
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19 minutes ago, CharonY said:
!

Moderator Note

Edited the strikethroughs. Not sure what happened, but I had to use an editor and copy it back in. 

 

Thanks. +1

@Zahra B

Good luck with your exam.

If you are studying this you need to be familiar with the M-M equation so now is the time to ask anything else.

I'm sure CharonY can answer any bio questions about it better than I can.

:)

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  • 2 weeks later...

I would not assume any particular value for Vmax.  Instead I would rearrange the Michaelis-Menten equation into the form v/Vmax = (S)/{KM + (S)}.  This means that the velocity one calculates will be relative to Vmax.  I decided to avoid using concentration brackets in this comment, because I think that they might be causing the unwanted strikethroughs.

Edited by BabcockHall
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