EXPLAIN PLEASE 1 M = 10-6 M; 1 nM = 10-9 M The Lineweaver-Burk plot of an enzyme

Question

EXPLAIN PLEASE

1 M = 10-6 M;

1 nM = 10-9 M

The Lineweaver-Burk plot of an enzyme is generated with a total enzyme concentration of 1.0 M. The intercept at the (1/V0) axis is 0.1 M-1.s and the intercept at the (1/[S]) axis is (- 0.2 M-1).

25) The Vmax of the enzyme is ____.

A) 10 M/s

B) 1.0 M/s

C) 5.0 /s

D) 2.0 M/s

26) The kcat of the enzyme is ____.

A) 1x104 M/s

B) 1x104 /s

C) 1x107 /s

D) 1x107 M/s

27) The KM of the enzyme is _____.

A) 5.0 M

B) 0.2 M

C) 10 M

D) 2.0 M

28) In the presence of 0.9 nM of an inhibitor, the intercept at the (1/[S]) axis changes to (- 0.02 M-1) while the intercept at the 1/Vmax axis stays the same, this is _____.

A) a noncompetitive inhibitor

B) a mixed inhibitor

C) a competitive inhibitor

D) an uncompetitive inhibitor

29) The values for this inhibitor under the conditions in question #28 is ____

A) 10

B) -0.1

C) -10

D) 0.1

30) The KI of the inhibitor used in #28 is ____.

A) 1.0 nM

B) 0.1 nM

C) 9.0 nM

D) 0.1 M

Explanation / Answer

A Lineweaver-Burk plot is constructed between 1/V and 1/[S] for an enzyme catalyzed reaction where V is the reaction velocity and [S] is the substrate concentration. Accordingly, the intercept on the x-axis represents the numerical value of -1/[Km] and the intercept on y-axis represents 1/Vmax. Thus, taking reciprocals of these values will give the direct values of Km and Vmax for an enzyme catalyzed reactions.

Thus, based upon the above given background, the answers can be found as below:

Answer 25: Choice A

(As described above, the Vmax of the reaction will be given by reciprocal of the intercept on y-axis i.e. 1/0.1 or 10 Ms-1)

Answer 26: Choice C

(The turnover number or Kcat is defined as the number of substrate molecules converted into proteins per unit enzyme catalysis sites by an enzyme. It is represented by the ratio of substrate concentration to the concentration of enzyme catalysis sites. Here, the Kcat can be calculated as Km/E i.e. 10uM/s or 107-s.

Answer 27: Choice A

(As described above, Km of the reaction can be calculated by taking modulus of the reciprocal of intercept on x-axis i.e. +1/0.2 or 5M)

Answer 28: Choice C

(When the 1/Vmax of teh enzyme catalysed reaction remains unchanged, but the Km changes, it suggests that the inhibitor used in the reaction competes with the enzyme for binding to the active sites. This does not change the reaction velocity but definitely changes the Km of the reaction.)

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