1. Given,
S1=0.003, Km=0.06, Vo1=0.0015, Vo2=0.003, S2=?
Answer 1: We know that,
V0= Vmax [S1] / Km + [S1]
0.0015= Vmax * 0.03 / 0.06 + 0.03
0.015= Vmax * 0.03 / 0.09
0.015 * 0.09= Vmax * 0.03
0.00135 = Vmax * 0.03
Vmax = 0.00135 / 0.03
Vmax =0.045
0.03 = 0.045 * S2 / 0.06 + S2
S2= 0.12 mmol/L
2. Given,
S1= 0.03, Km= 0.06, Vo1= 1.5 * 10-3, S2= 0.12 mmol/L, V02=?
Answer 2:- We know that, V01= Vmax [S1] / Km + [S1]
1.5 * 10-3 = Vmax * 0.03 / 0.06 + 0.03
1.5 * 10-3 = Vmax * 0.03 / 0.09
0.000135 = Vmax * 0.03
Vmax = 0.000135 / 0.03
Vmax = 0.0045
V02 = 0.0045 * 0.12 / 0.06 + 0.12
V02 = 0.00054 / 0.18
V02 = 3 * 10-3 mmol/L.min-1
6. An enzyme with a km of 0.06mmol/L hydrolyzed a substrate of a concentration 0.03 mmol/L....
4. An enzyme hydrolyzed a substrate concentration of 0.03mmol/L, the initial velocity was 0.5 X 10-3 mmol/L.min' and the maximum velocity was 4.5 x 10-3 mmol/L.min.l. Calculate the Km value. 5. Urease hydrolyzed urea at [s]=0.03mmol/L with a km of 0.06 mmol/L. The initial velocity observed was 1.5X10-3 mmol/L.min-1 Calculate the maximum velocity of the enzyme reaction.
112 marks] 3. The relationship between initial velocity (V.) and substrate concentration of most of the enzyme- catalized reactions are explained by Michaelis-Menten equation. IMPORTANT: Show the calculations and indicate the units for all your answers. a. For an enzyme which follows the Michaelis-Menten enzyme kinetics, Km is 50 mmol L. Calculate the substrate concentration required to obtain the initial velocity (V.) equivalent to 90% of the maximum velocity (Vmax). b. The Vmax of the above reaction is 250 mmol...
At what substrate concentration would an enzyme with a kcat of 30.0 sec-1 and a Km of 0.0030 M have an initial velocity that is 25% of the maximum velocity? a. 0.0010M b 0.0090 M c. 0.0015 M d. 0.0060 M e.Cannot be determined
What is substrate concentration, expressed as a multiple of Km, when an enzyme reaction is observed to have an initial rate Vo = 0.75 Vmax. Select one: O a. [S] = 0.33 x km O b. [S] = 0.25 km O c. [S] = 0.75 x Km O d. [S] = 0.3 x km O e. [S] = 0.5 x km Check Next page ime Jump to...
Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L. Part A If the total enzyme concentration was 6 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute? Express your answer to three significant figures. kcat kcat = 2.73×104 min−1 Part B Calculate kcat/KM for the enzyme reaction.
(15 points) The following data is for a reaction catalyzed by tyrosine monoxygenase: Substrate Concentration (mol/L) Initial Velocity (mM/min) 1.5 0.66 1.2 0.65 0.81 0.45 0.65 0.39 0.49 0.32 0.27 0.21 a) Plot the velocity (y-axis) versus substrate concentration [S] (x-axis) curve and insert/draw the graph in the space below. What are the approximate KM and Vmax values? b) Construct a 1/v (y-axis) versus 1/[S] (x-axis) plot in the space below. Calculate the KM and Vmax values. c) Calculate the...
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where [S]=6mM, it was found that Vo=480nM/min. What was the enzyme concentration? Give your answer in nM. Using the same kcat and Km as the previous question, if [Et]=0.5uM gives a Vo=5 uM/min, what wat the [S]? Give your answer in uM. reaction is run with the kcat=20/min and the Km=4uM. Use the enzyme concentration from question 1 above. A very strong inhibitor is added creating...
1. An enzyme with a Km of 1x10 % M was assayed using an initial substrate concentration of 3x10-- M. After 2 min, 5 percent of the substrate was converted. How much substrate will be converted after 10 min. 30 min. 60 min? How long must the reaction be run to achieve 99% conversion? (Assume that the enzyme follows Michaelis-Menten kinetics.)
1) 2) If the total enzyme concentration was 9 nmol/Lnmol/L, how many molecules of substrate can a molecule of enzyme process in each minute? Express your answer to three significant figures. 3) The kcat for neuraminidase at pH=6.15pH=6.15, 37 ∘C∘C is 26.8 s−1s−1. Calculate KMKM for the hydrolysis of sialic acid. Express your answer with the appropriate units. An enzyme that follows Michaelis-Menten kinetics has a Ky value of 6.00 uM and a keat value of 176 s-1. At an...
a. An enzyme has a Vmax of 100 umol/min and a Km of 40 uM. When substrate concentration is 40 uM what is the initial reaction rate? b. An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted with a irreversible active site specific inhibitor. After reaction with the inhibitor, the enzyme was assayed using a 2 mM concentration of substrate, and it gave a reaction rate of 20 umol/min. What percentage of the...