Problem. Determine KM and Vmax for an enzyme from the following data : S(mM) vo(HM.s²) 1/S (mM) 1/v.(µM.s*) 1 2.5 1.00 0.40 2 4 0.50 0.25 5 6.3 0.20 0.16 10 7.6 0.10 0.13 20 9 0.05 0.11
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- Enzyme Kinetics question Enzyme used is 10uL of a 10 ng/uL solution to a reaction mix in a final volume of 2.0 mL. (Enzyme used is 20kDa monomeric enzyme if matters) Based off lineweaver burk where 300 uM of inhibitor is used noninhibited formula is y= 4x + 0.1 (x axis is 1/S 1/mM) (y axis is 1/Vo sec/mM) inhibited formula is y = 4x + 1 I found Km as 1 mM for inhibited Vmax as 1mM/sec for inhibited How would I find Kcat? How would I find Ki?If an enzyme catalyzed reaction has a KM of 5mM and a Vmax of 60 nm/sec, the substrate concentration at 30 nM/sec is? Thank you.Calculate α' for an inhibitor with KI' = 10.0 nmol L-1 when 100 nmol L-1 of inhibitor is present.
- Question: Determine the Km and Vmax for this enzyme/substrate combination. [Substrate] (mM) V0 (mM/min) 0.25 0.183 0.50 0.356 1.00 0.665 2.50 1.45 5.00 2.35 What is the concentration of substrate necessary to achieve a turnover rate of 1.00 mM/min?Initial rate data for an enzyme that obeys Michaelis–Menten kinetics areshown in the following table. When the enzyme concentration is 3 nmolml-1, a Lineweaver–Burk plot of this data gives a line with a y-intercept of0.00426 (μmol-1 ml s). (a) Calculate kcat for the reaction.(b) Calculate KM for the enzyme.(c) When the reactions in part (b) are repeated in the presence of 12 μM ofan uncompetitive inhibitor, the y-intercept of the Lineweaver–Burk plotis 0.352 (μmol-1 ml s). Calculate K′I for this inhibitor.An enzyme has a KM = 10 mM and Vmax = 100 mmol/min. Identify the substrate concentration (in mM) in which the velocity will near Vmax when there is a 10-fold decrease in KM.
- Modified TRUE or FALSE. Write the word TRUE if the statement is correct. If the statement is false, write the incorrect underlined word/s and indicate the correct word/s to make the statement true. The Michaelis-Menten Constant (Km) of an enzyme is equal to the enzyme concentration at which the initial velocity of the reaction is one half of maximum velocity (Vmax).For an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K MProblem: Alcohol Dehydrogenase catalyzes the conversion of ethanol to acetaldehyde. This enzyme, in its active state, consists of a protein molecule and a zinc ion. On the basis of this information, identify the following for this chemical system. Substrate Apoenzyme Cofactor Holoenzyme
- Chemical scheme for enzyme catalysis (a) Write the chemical equations for enzyme,substrate, enzyme•substrate complex, and product as for a typical Michaelis-Mentenenzyme (b) At what condition is half of the enzyme expected to be saturated withsubstrate? (c) Plot the rate of product formation as a function of substrate concentration.(d) Indicate the KM parameter on your rate vs substrate plot.An enzyme-catalyzes the isomerization of substrate S to product P. The enzyme has a molecular weight of 120,000 g/mol. In assays using 1 μg of enzyme per assay the Km was 3 x 10^-3M and the Vmax was 2.75 μmole per minute. What would be the Kcat (turnover number or molecular activity) of the enzyme under these conditions? 2.75 min^-1? 3,300,000 min^-1? 330,000 s^-1? 19,800,000 min^-1? 5,500 s^-1?MATHEMATICAL You do an enzyme kinetic experiment and calculate a Vmax of 100 mol of product per minute. If each assay used 0.1 mL of an enzyme solution that had a concentration of 0.2 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol?