Question

Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some penicillin-resistant bacteria. The mass of this enzyme is 29.5 kD. The amount of penicillin hydrolyzed in 1 minute in a 10 mL solution containing 109 g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change appreciably during the assay. (a) Plot v versus. [Penicillin] and 1/ v versus 1/[Penicillin], (b) Determine the kinetic parameters, Km & Vmax. and (c) calculate k2. Velocity (v) Amount hydrolyzed (nM/min) у-аxis 1/v х-ахis Substrate = Pennicillin [Penicillin ИМ 1.0 1/Pennicillin] 0.11 0.25 3.0 0.34 5.0 0.45 10.0 0.58 30.0 0.61 50.0

Answer 1

a) Graph

b) From the equation of straight line, y = mx + c

y = 7.6225x + 1.4602 1/[Pencillin] 1/Velocity [Penicillin) ix-axis) velocity liy-axis) 0.11 9.090909091 3 0.333333333 0.25 5 0 .21 0.34 2.941176471 10 0.1 0.45 2.222222222 30 0.033333333 0.58 1.724137931 50 0.02 0.61 1.639344262 1/Velocity 0.6 1/(Penicillin]

Slope = 7.6225 = Km/Vmax

Intercept = 1.4602 = 1/Vmax

Vmax = 1/1.4602 = 0.6848 nM/min.

Km/Vmax = 7.6225

Km = 7.6225*Vmax = 7.6225*0.6848 = 5.220 uM of Penicillin.

c) k2 can be calculated as

Vmax = k2(total concentration of enzyme)

Concentration of enzyme = moles of enzyme/volume of solution

Molar mass of enzyme = 29.5 kDa = 29500 g/mol

Volume of solution = 10 ml = 0.010 L

Moles of enzyme = (10^-9 g)/(29500 g/mol)

= 3.390 x 10^-16 mol

Concentration = 3.390 x 10^-16 mol/0.01 L

= 3.390 x 10^-14 M or 3.390 x 10^-5 nM

k₂ = Vmax/(enzyme concentration)

= (0.6848 nM/min.)/(3.390 x 10^-5 nM)

= 1.912 x 10⁴ min^-1