Sol.
Michaelis-Menten(M-M) equation :
V₀ = Vmax [S] / ( KM + [S] )
So , in M-M plot , we plot V₀ vs [S]
Lineweaver - Burk(L-B) equation :
1/V₀ = ( KM / Vmax )*1/[S] + 1/Vmax
So , in L-B plot , we plot 1/V₀ vs 1/[S]

In L-B plot , equation of line ,
y = 13.77x +3.025
So , 1/Vmax = 3.025
and , Vmax = 1/3.025 = 0.33057 mM / s
Also , from equation of line ,
KM / Vmax = 13.77
KM = 13.77*0.33057 = 4.55194 mM
Also , Kcat = Vmax / [E]
As total enzyme concentration = [E] = 5
M = 0.005
mM
So , Kcat = 0.33057 / 0.005 = 66.114 s-1
Kcat / KM = 66.114 / 4.55194 = 14.52435 mM-1 s-1
Michaelis-Menten plot and Lineweaver-Burk plot calculations- use provided data to generate both M-M and L-B plots....
Michaelis-Menten plot and Lineweaver-Burk plot calculations: Use provided data to generate both M-M and L-B plots. Use scatter plots with markers on Excel: On the M-M Plot: estimate Vmax, KM On the L-B Plot: determine Vmax, KM, keat, kcat/Km. The total enzyme concentration is 5 uM. Graphs can be 1/2 page. Must be computer generated with all axes labeled. Substrate (mM) V. (mM/s) | 1/[S] (mM1) 1/V. (s/mM) 10 | 0 2.73 5.45 8.17 10.9 40.4 0.124 0.181 0.212 0.228...
(I need help with part C, Drawing the expected Michaelis-Menten plot; Do NOT draw the Lineweaver-Burk plot. thanks!) 1. Michaelis-Menten kinetics- use the M-M equation to answer the following: a. An enzyme (5 µM) has a Vmax of 450 mM/min. What is kcat? b. When the substrate concentration is 50 mM, the initial velocity (V0) was measured to be 375 mM/min. Under the conditions described above, calculate the KM. c. Draw the expected Michaelis-Menten plot (label your axes and include...
Write the equations that describe the Michaelis-Menten and the Lineweaver-Burk double-reciprocal plots. Draw examples of each plot, demonstrating how Km and Vmax can be determined. On the same graphs, draw another plot where the same enzyme-catalyzed reaction is subjected to inhibition by a competitive inhibitor.