Given the following reaction and equation for the initial velocity of the reaction: k₁ k3 E+SES E + P V=Keat [ES] = k3 [ES] k₂ where keat is the rate constant for the reaction which forms the product from the ES complex. Explain in words why the velocity is directly proportional to theamount of enzyme added in the presence of saturating substrate levels.
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![Given the following reaction and equation for the initial velocity of the reaction:
k₁ k3
E+S ES E + P
V=Keat [ES] = k3 [ES]
k₂
where keat is the rate constant for the reaction which forms the product from the
ES complex. Explain in words why the velocity is directly proportional to theamount of
enzyme added in the presence of saturating substrate levels.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ed96629-5430-48c1-a397-e5c0ff3303b9%2F7ff65a38-428f-444d-9483-bfaefb2c0dd2%2F3lj9d2p_processed.png&w=3840&q=75)
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- when saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.calculate the reaction velocity at saturating substrate concentrations. Your numerical answer is assumed to be in units of M sec-1. [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMAt what substrate concentration would an enzyme with a kcat of 33.0 s-1 and a Km of 0.0046 M operate at one-tenth of its maximum rate? Assume the enzyme obeys Michaeli s-Menten kinetics.
- 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-1An enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is 1 ×10-6 M?For a Michaelis-Menten reaction, k₁=5 x 107/M-s, k-1-2 x 104/s, and k2=4 x 10²/s. Calculate the Ks and KM for this reaction. Does substrate binding achieve equilibrium or steady state?
- In 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. b) By showing appropriate reaction mechanisms and rate equations, explain how enzyme catalytic reactions may be affected by competitive inhibitionAssume 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?In 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
- at what substrate concentration is v= 5.0 mM s^-1 for the enzyme catalyzed hydrolysis of trehalose?Given the following information, calculate the catalytic efficiency of the enzyme. Step by step please [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMa particular enzyme catalyzes a single reactant S to a single product P, following michaelis-menten kinetics rp=(VmaxCs) / (Km + Cs) 1. A reaction with this enzyme is carried out at very low substrate concentrations. Draw and label a curve on the plot that describes the reaction kinetics under those conditions.