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Determine the type of inhibition of an enzymatic reaction from the following data collected in the presence and absence of the inhibitor.

[S] (mM)

v鈧 (mM min鈦宦)

v鈧 with I present (mM min鈦宦)

1

1.3

0.8

2

2.0

1.2

4

2.8

1.7

8

3.6

2.2

12

4.0

2.4

Short Answer

Expert verified

The inhibitor is a mixed inhibitor.

Step by step solution

01

Biological protein catalysts

Enzymes are biological protein catalysts that change the rate of a process in a living organism. The molecule that binds to the enzyme's substrate binding site is referred to as the substrate. Enzyme inhibitors stop enzymes from functioning.

02

Inhibition of an enzymatic reaction

Both free enzyme and enzyme-substrate complexes bind to mixed inhibitors. The following data is used to create a Lineweaver-Burk graph. Extrapolation is carried out on the acquired lines.

A factor was added to theMichaelis-Menton equation. 鈭 is a function of the concentration of a competitive inhibitor and the inhibitor's affinity for the enzyme. The concentration of an uncompetitive inhibitor and the inhibitor's affinity for the enzyme is used to calculate 鈭澦. If there is an inhibitor, 鈭 and 鈭澦 cannot be smaller than 1. Both with and without inhibitor are displayed on the graph. To the left of the 1/v axis, the lines cross.

03

Tabulation for plotting the graph

1/[S] (mM)

1/v鈧 (mM min鈦宦)

1/v鈧 with I present (mM min鈦宦)

1.0

0.7

1.2

0.5

0.5

0.8

0.3

0.4

0.6

0.1

0.3

0.5

0.08

0.2

0.4

04

Graphical representation of the Mixed inhibitor

05

Conclusion

Thus, the type of inhibition of an enzymatic reaction from the given data is collected in the presence and the absence of an inhibitor is mixed inhibition.

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Most popular questions from this chapter

You are attempting to determine KM by measuring the reaction velocity at different substrate concentrations, but you do not realize thatthe substrate tends to precipitate under the experimental conditions youhave chosen. How would this affect your measurement of KM?

Why are uncompetitive and mixed inhibitors generally considered to be more effective in vivo than competitive inhibitors?

The KM for the reaction chymotrypsin with N-acetylvaline ethyl ester is 8.8102M, and the KM for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6104M. (a) Which substrate has the higher apparent affinity for the enzyme? (b) Which substrate is likely to give a higher value for Vmax?

Calculate KMand Vmax from the following data:

[S](渭惭)

v0(mMs1)

0.1

0.34

0.2

0.53

0.4

0.74

0.8

0.91

1.6

1.04

Sphingosine-1-phosphate (SIP) is important for cell survival. The synthesis of SIP from sphingosine and ATP is catalyzed by the enzyme sphingosine kinase. An understanding of the kinetics of the sphingosine kinase reaction may be important in the development of drugs to treat cancer. The velocity of the sphingosine kinase reaction was measured in the presence and absence of threo-sphingosine, a stereoisomer of sphingosine that inhibits the enzyme. The results are shown below.

[Sphingosine]

(饾泹惭)

v鈧 (mg min鈦宦)

(no inhibitor)

v鈧 (mg min鈦宦)

(with threo-sphingosine)

2.5

32.3

8.5

3.5

40

11.5

5

50.8

14.6

10

72

25.4

20

87.7

43.9

50

115.4

70.8

Construct a Lineweaver-Burk plot to answer the following questions:

(a) What are the apparent KM and Vmax values in the presence and absence of the inhibitor?

(b) What kind of an inhibitor is threo-sphingosine? Explain.

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