In a 250ul reaction containing 0.3 nmol of a Michaelis enzyme with Km = 4.4 x 10- M and saturating substrate concentration, product was formed at an initial velocity of 6.3 x 10 M min-1. What is kcat for this enzyme? Give your answer in 3 sigfigs in scientific notation, for example 1.20 e
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- Instantaneous rates for the reaction of hydroxide ion with Cv+ can be determined from the slope of the curve in Figure 11.3 at various concentrations. They are (1) At 4.0 105 mol/L, rate = 12.3 107 mol L1 s1 (2) At 3.0 105 mol/L, rate = 9.25 107 mol L1 s1 (3) At 2.0 105 mol/L, rate = 6.16 107 mol L1 s1 (4) At 1.5 105 mol/L, rate = 4.60 107 mol L1 s1 (5) At 1.0 105 mol/L, rate = 3.09 107 mol L1 s1 (a) What is the relationship between the rates in (1) and (3)? Between (2) and (4)? Between (3) and (5)? (b) What is the relationship between the concentrations in each of these cases? (c) Is the rate of the reaction proportional to the concentration of Cv+? Explain your answer.The following initial rate data were obtained on the rate of binding of glucose with the enzyme hexokinase (from yeast) present at a concentration of 1.34 mmol/dm3. [C6H12O6] / (mmol dm-3) 1.00 1.54 3.12 4.02 Initial speed v0 / (mol dm-3 s-1) 5.00 7.60 15.5 20.0 use the data above to calculate the rate constant. k = ________ (mol/L/s, s-1, L/mol/s, or L2/mol2/s). 3 s.f. normal format.Reaction Velocity (mm/sec) Ma no !! L [1 1 L IM 3M FO Substrate Concentration (M) 1. IN Which statement is true? KM AN ALM 4V The graph above displays the kinetic profiles of the same enzyme catalyzing two different substrates. The red curve displays the higher turnover number (kcat). The red curve displays higher affinity between substrate and enzyme. The blue curve displays competitive inhibition. The red curve displays higher cooperativity,
- The following initial rate data were obtained on the rate of binding of glucose with the enzyme hexokinase (from yeast) present at a concentration of 1.34 mmol/dm3. [C6H12O6] / (mmol dm-3) 1.00 1.54 3.12 4.02 Initial speed v0 / (mol dm-3 s-1) 5.00 7.60 15.5 20.0 (a) What is the order of the reaction with respect to glucose? (b) use the data above to calculate the rate constant. k = ________ (mol/L/s, s-1, L/mol/s, or L2/mol2/s). 3 s.f. normal formatFor an enzyme -catalyzed reaction, KM=10.0 mmol dm-3, vmax=0.250 mmol dm²³ s-'; [E]o= 2.3 x 10-6 mmol dm-3. Calculate the catalytic efficiency (=kb/KM) of the enzyme. 0.920 dm³ mol-l s-1 3.1 x 107 dm³ mol-1 s-1 1.1 x107 dm³ mol-l s-1 0.250 dm³ mol-l s-1The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics. Product formed (mmol/min) Substrate added (mmol/L) 217 0.8 325 2 433 4 488 6 647 1000 The Km for this enzyme is approximately: 4 mM. 2 mM. 1 mM. 6 mM. 1000 mM.
- i need this reaction mechanism draw pls MeO Me Me SO₂Ph TMS₂NLI MeCO₂Me Me c THF 0 °C to RT, 90 min, 87% MeO. Me Me Me 'MeDetermine the kcat of the reaction given [E] = 2.5E-7 and the enzyme has a mass of kDa. (Km = 9.86E-6 micromol; Vmax 140.85 micromol/minute) HO.HO [S] (M) 1/[S] vo (μmol/min) 1/vo OHO (M) (μmol/min) 2.50E-06 28 4.00E+05 3.57E-02 4.00E-06 40 2.50E+05 2.50E-02 1.00E-05 70 1.00E+05 1.43E-02 2.00E-05 95 5.00E+04 1.05E-02 4.00E-05 112 2.50E+04 8.93E-03 1.00E-04 128 1.00E+04 7.81E-03 2.00E-03 139 5.00E+02 7.19E-03 1.00E-02 140 1.00E+02 7.14E-03 y = 7.0 × 10-8 x + 0.0071The image below contains a Michaelis Menten plot for an enzyme catalysed reaction. Based on this plot, estimate the Vmax and Km. Velocity, uM/sec 40 30 20 10 0 0 ENZYME CATALYSED REACTION 20 60 Substrate concentration, mM 40 Vmax= 39 umol/sec; Km = 14 mM O Vmax = 39 umol/sec; Km = 20 mM O Vmax = 14 umol/sec; Km = 39 mM O Vmax = 20 umol/sec; Km = 14 mM 80
- An orange juice is thermally processed for determining ascorbic acid degradation kinetics. After different temperature experiments, the following ascorbic acid concentration values were obtained: Ascorbic Acid (mg/L) Processing Time (second) 0 15 30 60 90 120 Temperature (oC) 75 558.3 546.5 535 526.5 519.6 508.6 85 608.6 595.2 587.4 572.7 557.5 542.3 95 612.6 601.5 588.2 546.1 521.8 509.5 According to the results; Determine ascorbic acid degradation order Derive the equations related to the degradation order (general equation and t1/2) Calculate k-values at the constant processing temperatures Calculate activation energy (Ea) of the degradation reaction Calculate half-life time of ascorbic acid for all processing (For better evaluation, add figures for the addressed questions, if necessary)What mutant of Dihydroorotase affected the most it's turnover rate? Estimate its impact on the enzyme's turnover rate. Table 3: Kinetic Parameters for Mutants of Dihydroorotase kcat (s-1) Km (mM) kca/ K„ (M¯' s-') wild type D250A 160 +8 <0.01% 12 +1 <0.01 1.7 + 0.2 1.5 (0.2) × 10$ nd 6.3 (0.3) x 103 nd nd 4.3 (0.1) x 10! nd 1.7 (0.1) x 104 ndº D250E D250H 1.9 + 0.3 nd ndº 0.51 +0.1 nd 0.9 + 0.1 D250N <0.01 0.022 + 0.002 <0.01 15 +1 D250S R20Q R20K1. For enzymatic reaction, a mechanism was proposed by Michaelis and Menten as follows: E + S == ES k, and k,' ES → E and P k2. a. Use steady state assumption, derive expression for the reaction rate. Where E is concentration of enzyme, S substrate, ES complex of E and S, E = E, – ES. (If you have difficulty in doing it, please consult lecture note) b. Assume Km = 0.038 mol.L1 at 25 °C, when the substrate concentration is 0.156 Mol.L1, the rate of the reaction is 1.21 m mol/L.s. The maximum rate of conversion reaction is reached at high substrate concentrations. Calculate the maximum rate of this enzyme catalyzed reaction.