36. On a lineweaver-Burk plot of kinetic data for an enzyme in the presence 1 x 10 M of a classic noncompetitive inhibitor, the y-intercept = 4 x 10M min and the x-intercept = -5 x 10' M'. What is the true Km of the enzyme? a. 2.0 x 10 M b. 2.5 x 10 M c. 1.2 x 10 M d. 0.8 x 10 M e. 1.2 x 107 M

Question 15 Plots A and B show enzyme kinetics data in the absence and presence of inhibitor. Which response describes the effect of the inhibitor? B Vo (M¹) 2960 10 A [S] (mm) 6 8 10 Apparent Km > Km; Apparent Vmax > Vmax Apparent Km > Km; Apparent Vmax Vmax -1 1.2 1 0.8 0.6 0.4 0.2 0 -0.2 1 1/[S] (mM¹¹) Analyse the plot and calculate the values. This question is harder than it seems. 2

Enzyme activity Can you please briefly explain

Construct a figure of Lineweaver-Burk plot. Determine the K, and Vmx of the enzyme- substrate complex from this figure. Please make sure that your figure has all the necessary components such as axis title, axis scale, axis unit etc. concentration (in mM) of substrate (NPP): 0.005mM 0.01mM 0.025mM 0.5mM 0.075mM 0.1mM 0.125mM p-nitrophenol (PNP): 0.00021mMmin-1 0.00023mMmin-1 0.00029mMmin-1 0.00043mMmin -1 0.00069mMmin-1 0.00028mMmin -1 0.00055mMmin -1

Below which is false about the Michaelis constant Km in Michaelis-Menten kinetics? k₂ k₁ E+SESE + P k₁ k₂ A higher K indicates a higher affinity between the substrate and the enzyme. Km is effectively equal to the Dissociation Constant (k_₁/k₁) for ES. When [Substrate] = Km, Vo = 1/2 V max When [Substrate] = Km [ES] = [Efree] = [Etotal]

4. Graphing the results from kinetics experiments with enzyme inhibitors The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the presence of two different inhibitors (2) and (3) at 5 mM concentration. Assume [Er] is the same in each experiment. [S] (mM) 1 2 4 8 12 1 2 4 8 12 [S] (mm) a. Determine Vmax and Km for the enzyme. b. Determine the type of inhibition and the Kıfor each inhibitor. 1/[S] M-1 Plot these data as double-reciprocal Lineweaver-Burk plots and use your graph to answers a. and b. 1000 500 250 125 83.3 Answer: The data may be analyzed using double-reciprocal variables. For each [S] and corresponding u, we will calculate 1/[S] and 1/v. sed to get mo Helmien (1) v (μmol/mL sec 12 20 29 35 40 1/v (1) v (1) μmol/mL se mL-sec/umol C 12 20 29 35 40 Please explain how to get K 1/v 20 × 104 10 × 104 -400 8.33x104 5.00x104 3.45x104 2.86x104 2.50x104 Condition No inhibitor 0 5 mM inhibitor (2) 5 mM inhibitor (3) Plots of 1/v vs. 1/[S]…

The enzyme examase catalyses the following reaction, A - B where A is the reactant and B is the product A595 measurements were used to determine the catalytic activity of examase by measuring the rate at which B is produced. From this, the activity of the enzyme was determined to be 42 µmol/mL per minute. A standard albumin solution (concentration = 10 mg/mL) was used to prepare a standard curve of Abs595 vs amount of protein in mg. The protein assay was also performed using 100 uL of a 1 in 10 dilution of the original examase sample. For the absorbance measurements, it was determined that the sample used in the A595 measurement contains 1.2 mg of examase. Which of the following statements correctly gives the specific activity of examase? 1.0.35 U/mg O 2.0.035 U/mg 3. 5040 U/mg O 4.35 U/mg O 5.3.5 U/mg

Using this graph, what can be determined about the effect of enzyme concentration on the initial rate of the reaction and on the amount of product formed?

If inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM, calculate the new Vmax. Calculate the new Km of the above lactase reaction, with inhibitor X. Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. Inhibitor X exerts which of the following effects on the above enzyme (lactase)?

Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. (inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM)  0.20 per minute 0.50 per minute 1.0 per minute 2.0 per minute 5.0 per minute