Question

If a Lineweaver-Burk plot gave a line with an equation of y = 0.490 x + 0.059, what is the velocity at a substrate concentration of 5 mM? The original units for substrate were in mM and velocity in mM/s.

a) 0.122 mM/s
b) 0.341 mM/s
c) 0.490 mM/s
d) 0.549 mM/s

Answer 1

To find the velocity at a substrate concentration of 5 mM using the Lineweaver-Burk plot, we need to determine the reciprocal of the substrate concentration and velocity. By substituting the values in the Lineweaver-Burk equation, we find that the velocity at a substrate concentration of 5 mM is approximately 5.0072 mM/s.

Step-by-step explanation:

To determine the velocity at a substrate concentration of 5 mM using the Lineweaver-Burk plot, we first need to determine the reciprocal of the substrate concentration (1/[S]) and the reciprocal of the velocity (1/v). Using the equation of the Lineweaver-Burk plot (y = 0.490x + 0.059), we can substitute the values for x and y with the reciprocals of the substrate concentration and velocity respectively. Solving for x gives us the reciprocal of the substrate concentration (1/[S]). Once we have the reciprocal of the substrate concentration, we can substitute this value into the Lineweaver-Burk equation and solve for the reciprocal of the velocity (1/v). Finally, we can take the reciprocal of the velocity (1/v) to find the velocity at a substrate concentration of 5 mM in mM/s.

Using the equation y = 0.490x + 0.059, we have:

0.490x = 1/5

x = 1/(0.490*5)

x = 0.4082

Now, substituting x into the Lineweaver-Burk equation, we have:

y = 0.490(0.4082) + 0.059

y = 0.1996

Finally, taking the reciprocal of the velocity, we have:

1/v = 0.1996

v = 1/0.1996

v ≈ 5.0072 mM/s