(background information) Imagine that this is a plot of Velocity (µM/min) vs Sub

Question

(background information)

Imagine that this is a plot of Velocity (µM/min) vs Subsrate Concentration (µM) with "Thrombin" which had a stock solution of 7µM but then you diluted it to 0.028µL and you know that the kcat is 3/sec and the eaction rate is 5µM/min

by sort of eyeballing it you can say that Vmax is about 5 µM/min and that means that Km is about 25µM just by looking at the graph.

Now here is the question.

c.) Suppose that you want to set up a reaction using twice the enzyme concentration, and a substrate concentration that will yield a reaction velocity of 4µM/min. What would the appropriate substrate concentration be?

d.) At what substrate concentration would you expect 30% of the enzyme molecules to have substrate bound?

Explanation / Answer

At very low [S], substrate concentration,

velocity of rxn=vo=vmax[S]/km+[S] becomes vo=vmax[S]/km=kcat[E][S]/km

or, vo=kcat[E][S]/km……

or, [E][S]=vo*km/kcat

for new substrate concentration [S]’ and enzyme conc [E]’

[E]'[S]'=vo'*km/kcat

Which gives

[E][S]/[E]’[S]’=vo/vo’………………(1)

Or,[ E][S]/[E]’[S]’=5 µM/4 µM

Or, ,[ E][S]/[E]’[S]’=5/4

As [E]’=2[E]

Or, ,[ E][S]/2[E][S]’=5/4

Or, [S]/2[S]’=5/4

Or, [S]/[S]’=5*2/4=5/2

[S]’=2/5*[S]=2/5*0.028=0.0112 µM

d) equation (1),

[E][S]/[E]’[S]’=vo/vo’

If [E]’=0.30[E]

[E][S]/0.30[E][S]’=5/4

[S]/0.3[S]’=5/4

[S]/[S]’=5/4*0.30=0.375

[S]’=1/0.375[S]=2.67*[S]

[S]’=2.67*[S]=2.67*0.028=0.0747 µM

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