Please explain the following problem clearly. A certain enzyme with alpha_4 quat
ID: 1048118 • Letter: P
Question
Please explain the following problem clearly.
A certain enzyme with alpha_4 quaternary structure contains one active site per 85 kDa subunit. A series of enzyme assays with (and without) an inhibitor are carried out. Each enzyme assay contained 3.3 mu g of enzyme. The enzyme assay data is plotted below. Answer the questions that follow. Clearly show oil of your work. (a) DEFINE k_cat then use the data above to CALCULATE the value of k_cat (min^-1) in the absence of inhibitor. (b) DEFINE K_m AND then calculate its value (in mM) using the data above.Explanation / Answer
Ans. Mass of 1 subunit = 85kDa = 85 x (1.66 x 10-15 ug) ; 1 kDa = 1.66 x 10-15 ug
= 1.411 x 10-13 ug
Mass of 1 enzyme molecules (tetramer, 4 active sites) = 4 x mass of 1 subunit
= 4 x 1.411 x 10-13 ug = 5.644 10-13 ug
No. of enzyme molecules in 3.3 ug enzyme = (1 / 5.644 10-13 ug) x 3.3 ug = 5.8469 x 1012 molecules
Moles of enzyme = No. of enzyme molecules / Avogadro No.
= (5.8469 x 1012 molecules) / 6.0221409 x 1023 = 9.709 x 10-12 moles
= 9.709 x 10-6 mmoles
Using, Kcat = Vmax / [ET] ; where, [ET] = concentration of enzyme.
From LB plot: Lineweaver-Burk plot equation in from of y = a x + c
where, y = 1/ V0, x = 1/ [S], c = 1/ Vmax , a = Km/ Vmax
So, from equation: y = 1.9x + 2.9
Vmax= (1/ 2.9) = 0.344827 umol min-1
Or, Kcat = (0.344827 umol min-1) / (9.709 x 10-6 umol) = 0.035516 x 106 min-1 = (3.55 x 104 min-1)*
Since, each molecule has 4* active sites, the actual Kcat is equal to 4 times calculated above.
Hence, Kcat = 4 x (3.55 x 10 min-1) = 14.206 x 104 min-1 =1.42 x 105 min-1
Ans. b. 1.9 = (Km/ Vmax) = (Km/ 0.344827)
Or, Km = 1.9 x 0.344827 = 0.655 mM
Note: While using the LB graph equation, the units of the calculated Km and Vmax is the same as those used to plot the graph.
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