Michaelis and Menten used the simplified scheme for an enzyme catalyzed reaction: keat k₁ E+S=ES+E+P k-₁
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![Michaelis and Menten used the simplified scheme for an enzyme catalyzed reaction:
kat
k₁
E+S=ES¬E+P
k-₁
Using this reaction, the observed rate of product formation can be expressed as:
OkES
O [ES/e-kut
Okeat (EES)[S]
O keat ES
Ok₁([E] - ES)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5db6eb6d-ca25-4adc-b224-9a5c28243d09%2F2d4aad09-086c-4726-af47-8310836a876e%2F2wdsnmj_processed.jpeg&w=3840&q=75)
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- You will perform the protocol below for the calf intestinal alkaline phosphatase (CIP) provided. For each reaction, your final enzyme concentration should be 10 nM CIP. Note: Enzymes purchased are typically labelled with their “units of activity” (U), as this relates to how much enzyme is needed to catalyze a reaction. The 100 nM CIP provided has approximately 3 U/mL and was diluted 1 in 1,000 from a 500 U/mL purchased enzyme. 1) Create a table (similar to the one below) to help you determine and keep track of what to add to each of the cuvettes in which your reactions will be measured. The five different concentrations of PNPP should be: 25, 50, 100, 200, 300 μM. Each reaction will be in a final volume of 1 mL and contain 10 nM alkaline phosphatase. Concentrations of stock solutions: 1.0 mM PNPP, 100 nM calf intestinal phosphataseGiven the following reaction, identify the class and subclass of the enzyme involved. H. CH2OH H-C-OH C=0 но-с-н но-с-н H-C-OH H-C-OH H-C-OH H-C-OH CH2OPO,2 CH2OPO,2- Class: [ Select] [ Select ] Ligase Hydrolase Lyase Subcla Transferase Isomerase OxidoreductaseIn the scheme for enzymatic catalysed reaction proposed by Michaelis and Menten, the steps involve reversible formation of enzyme-substrate (ES) complex followed by conversion of the complex to the product (P). a) Derive the rate equation for enzymatic process. State an assumption made in this derivation. Terbitkan persamaan kadar tindak balas bagi proses enzim. b) By showing appropriate reaction mechanisms and rate equations, explain how enzyme catalytic reactions may be affected by competitive inhibition
- Given the following enzyme-catalyzed reaction, identify the class and subclass of the enzyme involved. HO Class: [Select] COO™ NH3* Subclass: [Select] H₂O H₂C= COO™ NH3*Beginning with the 1st tetrahedral intermediate, show the complete steps in chymotrypsin mechanism that occurs to form the 2nd chymotrypsin intermediate in the chymotrypsin active site. The substrate for chymotrypsin to be used is Ala-Tyr-Gly. Further, name the amino acid(s) that would be released as a result of the reactions you'd illustrated above.The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1
- Although graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction, sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained:A generalized enzyme active site is shaped like a hemisphere with a radius of 45Å. The active site holds the following amino acids in a homeostatic solution (pH = 7.38): -HAVARILKHAVARILKHAVARILK- Assuming the charge is distributed uniformly along the hemisphere, determine the force at which this active site acts upon a single ATP molecule at the center of the hemisphere.The following reactions were catalyzed by an enzyme that follows the Michaelis-Menten mechanism in the absence and presence of inhibitor (10mM). Assume [E]T is the same for each reaction. Determine K, and/or K'. [S] (mM) Uninhibited v (um/s) Inhibited v. (um/s) 1 3 1.67 2 4.5 2.6 6.8 4.5 10 8.1 6.2 20 9.5 7.7
- The KMof the enzyme for the substrate adenosine is 3 × 10ꟷ5M. The product inosine acts as an inhibitor of the reaction, with an inhibition constant (KI, the dissociation constant for enzyme-inhibitor binding) of 3 × 10ꟷ4M. However, a transition state analog,Inhibits the reaction with KIof 1.5 × 10ꟷ13M. Explain why 1,6-dihydroinosine serves as a better inhibitor of adenosine deaminase than inosine. Elaborate on your answeAssume that an enzyme-catalyzed reaction follows the scheme shown: E + S ES E + P k₁ = 1 x 10%/M-s k-1 = 2.5 x 10%/s k2= 3.4 x 107s What is the dissociation constant for the enzyme-substrate, Ks? What is the Michaelis constant, Km, for this enzyme? What is the turnover number, Kcat, for this enzyme? What is the catalytic efficiency for the enzyme? If the initial Et concentration is 0.25mM, what is Vmax?For an enzyme catalyzed reaction of the form: S + E → P + E, the rate of product formation, [P], is given by: d[P]/dt = k2[E}total [S] /(Km + [S]) = For the enzymatically catalyzed hydrolysis of ATP at 25 °C and pH 7.0, the Michaelis Menten constant, Km was found to be equal to 16.8 μmol L 1 and the value of k2[E]total was found to be 0.220 μmol L-¹s¹. Find the initial rate at an initial ATP concentration of 30.0 μmol L-1
