Discussion of all aspects of biological molecules, biochemical processes and laboratory procedures in the field.

rewriter
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Joined: Fri Oct 27, 2006 5:34 am

i've been working on this problem for days ,it tortures me so.

here is the thing :

a 50 kDa enzyme can catalyse two different substrates A and B through the same sort of reaction

Km(A)=0.1mM Vmax(A)=100U/mg (1U can catalyse 1uM substrate in one minute)

Km(B)=5μM Kcat(B)=50/min

if we put A B both in a reaction system that catalysed by this enzyme

and make them the same cocentration:[A[=[B] then V(A)/V(B)=?

sdekivit
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rewriter
Garter
Posts: 2
Joined: Fri Oct 27, 2006 5:34 am
i've tried but really nothing worthy

i lay out the fomula of V(A)andV(B)

V(A)= Vmax[A]/{Km(A)+[A]}

V(B)= Kcat[Et][B]/{Km(B)+[B]}

as you see though [A]=[B] when i put them together it won't get me any constant ,neither to say that [Et] is unknown

it seems that the solution to this problem request some new method which i know nothing about

i cited this problem from a chinese postgraduate entrence exam paper.and now i doubt if there is a certain answer to this problem

sdekivit
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The maximal rate, V(max), reveals the turnover number when the enzyme is fully saturated with substrate. It's equal to k(cat)

druid
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Each of two substances is a competitive inhibitor of other ( A competes with B and B competes with A ).
We know that V=Vmax*[S]/Km(app)+[S], where Km(app) = Km+[I]/Ki, where Ki is a dissociation constant for inhibitor ( ie eq. constant for IE->I+E ).
So one needs to derive this Ki from the given data and i dont know how to do this ;(

I advise you to look at derivation of formula of reaction rate in presence of inhibitor. If you know how to derive V with apparent Km then you'll be able to use this method assuming A ( or B ) being an inhibitor.

P.S. If you found this problem in a usual biochem. book, there possibly must be some smart way to solve it without diving into kinetics details. But beautiful solutions are not for stupid me ;(

smrlmiao
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Joined: Thu Dec 21, 2006 5:20 am
[ES1]/[ES2]=Km2/Km1 (1)
V1/V2=Kcat1*[ES1]/Kcat2*[ES2] (2)
put (1) to (2)

V1/V2=Kcat1*Km2/Kcat2*Km1

Kcat1=Vmax1=100U/mg
Kcat2=50/min=50*6*107/60S
Km1=0.1mM=100μM
Km2=5μM

so ,V1/V2=107=10000000

sachin
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Any one studied co-enzyme kinetics.....

If present ...... pls tell me.....
Senior Education Officer, BNHS, India. www.bnhs.org

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