The enzyme succinoxidase adds oxygen to succinate to give fumarate. The reaction follows Michaelis-Menton kinetics. The following rate data were determined for a solution in which the enzyme concentration is 10.0 uM:

[S](mM) ----------------- 0.33 10.00
vo (uM s-1) at 20 C ----- 0.50 1.17
vo (uM s-1) at 40 C ----- 2.41 5.65

(a) Determine Vmax and Km for the reaction at 20 C and 40 C.

For this question, I know that the Michaelis-Menton equation is:

vo = Vmax/1 + Km/[S]
- the equation then becomes:
vo = k2*[E]/ 1 + Km/[S]

We know what vo is for, [E],and [S], but k2 is not known. I'm not sure how to proceed with this question unless k2 is known.

(b) Compute the activation energy of this reaction (succinate + O2 --> fumarate).

For this question, would the equation K2/k1 = -Ea/R[1/T2 - 1/T1] be used. T2 and T1 is known, R is a constant (8.314) but how is k2 and k2 calculated. This part I do not understand.

(c) Will the activation energy be higher or lower in the absence of succinoxidase? Why?

Solution Preview

Another form of the Michaelis-Menten equation is: vo/Vmax = [S]/(Km + [S])

In the (a) part, since you know vo at two different substrate concentrations, you can substitute these into the MM equation. Then you ...

Solution Summary

This solution includes a rationale on how to answer all three parts of this question regarding enzyme kinetics. Based on what is known for parts a-c, guidance is given in terms of what equations to utilize to calculate certain values and how to interpret the question. All of this is done in about 250 words.

You characterize a new enzyme's kinetics. Using a saturating (very high so all E is ES) substrate concentration, you measure the initial rate of the reaction at various enzyme concentrations. Does this data fit the Michaelis Menton model/equation? If not, what may be happening?
E (nM), rate (mM,s)
5, 50
10, 225
15

1. A new enzyme you are studying has a Km of 7.0mM and a kcat of 126 per second. What will be the initial rate of reaction in units mM/min when 9.3 nanometer enzyme is reacted with 2.7 mM substrate. Report your answer in units if mM/min to nearest 1mM/min?
2. A new enzyme you are studying has a Km of 12.6 mM and a kcat of

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A) What is the initial velocity v0 called when the substrate concentration [S] >>>KM ([S] )?
Label on graph.
B) Simplify the above MM equation using the assumption that [S] >>>KM. From this simplified equation, what is the rate const

I've attached a table and my calculations to find Km and Vmax for an enzyme following Michaelis Menten Kinetics and well as identifying what kind of inhibitor I is. I think my proposed solution is too simple and I must be missing something. If so, what am I overlooking and how do I figure this out?

Question:
It is possible to distinguish between an irreversible inhibitor and a classical mixed type reversible inhibitor by measuring Vmax of the enzymatic reaction as a function of [E]t in the absence and presence of the inhibitor.
What does the slope represent in each case?
What does the X-axis intercept represent in

For an enzyme which obeys Henri-Michaelis-Menten kinetics,
(1) At what substrate concentration will an enzyme characterized by a kcat of 30 s-1 and Km of 0.005 M show one quarter of its maximal rate?
(2) Calculate the fraction of Vmax that would be found when [S]=0.5 Km, [S]=3Km, and [S]=10Km

I am uncertain how to approach this problem:
uptake of galactose into red blood cells was looked at: the data is as follows,
galactose concentration mmol/l ^-1 galactose uptake(j)/nmol mg ^-1 protein min
^-1

With the following info for the activity of an enzyme in the absence and presence of compound A, what is the: Km of the enzyme, the Vmax of the enzyme, and the action of compound A ( an inhibitor, stimulator, modulator, etc) and please describe why.
S (mM) Vobs control Vobs with cmpd A
0.1 0.68 0.91
0.2 1.25 1.67
0.5

For a MM reaction
k1=5x10^7
k-1=2X10^4
k2=4x10^2
i calculated Km and Ks to be equal at about 4 x10^-4M
Does substrate binding achieve equilibrium or the steady state?
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