Calculating velocity in enzyme kinetics

[Zahra B]

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 November 3, 2020 by Zahra B
Misspelling

[studiot]

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 K_m and V_max 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 < K_m

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

Substituting   = K_m yields a point on the linear section where v = V_max/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 =  K_m/2 into that linear simplification.

[studiot]

I apologise for the unwanted strikethrough.

It shouldn't be there and I can't remove it, though I have asked the mods to do so.

So please just carry on as though it were not there.

[Zahra B]

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 November 4, 2020 by Zahra B

[Zahra B]

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

[studiot]

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) =V_max/2 and the concentration (y axis) = K_M

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 November 4, 2020 by studiot

[CharonY]

!

Moderator Note

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

 

[studiot]

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.

[Zahra B]

@studiot Thank you so much for explaining! After reading it 10 times I actually think I got it now 😁

[BabcockHall]

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

Edited November 14, 2020 by BabcockHall