enzymatic reaction, the equations that correspond with and without inhibition according to linewear-burk
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In an enzymatic reaction, the equations that correspond with and without inhibition according to linewear-burk are calculated, these are with inh: 3.5x + 6 = y without inh: 3.5 + 10 = y. calculate alpha factor
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- Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. (inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM) 0.20 per minute 0.50 per minute 1.0 per minute 2.0 per minute 5.0 per minuteThe enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.An enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.
- Staphylococcal nuclease has a ΔΔG‡ of -84.1 kJ mol-1 at 25.0 °C. If the uncatalyzed rate is 0.630x10-13 µmol s-1, calculate the enzyme-catalyzed rate in µmol s-1. (Use R = 8.3145 J mol-1 K-1)A marine microorganism contains an enzyme that hydrolyzes glucose-6-sulfate. The activity test is based on the measurement of the rate of glucose formation. The enzyme in a cell-free extract has a KM of 6.7 x 10-9 M and a Vmax of 300 nM min-1 . Galactose-6-sulfate is a competitive inhibitor of the enzyme. To a concentration of glucose-6-sulfate 2 x 10-5 M and galactose-6-sulfate 10-5 M, the speed initial is 1.5 nM min-1 . Calculate the Ki for galactose-6-sulfate.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
- The KM for the reaction of chymotrypsin with Substrate A is 8.8 x 10-4 M, while the KM for the reaction of chymotrypsin with Substrate B is 8.7 x 10-3 M. Which of the following statements are likely true? Chymotrypsin has a higher apparent affinity for Substrate A. The Vmax would be higher in the presence of Substrate A. The kcat would be higher with Substrate A. The V max would be higher in the presence of Substrate B. Two of the above are true.Shown below are Km, and Vmax values obtained for an enzyme A which catalyze the transformation of the following substrates. Enzyme concentration used was 0.01 M. Km, mM 0.02 Vmax, mM/min 5.3 Substrate 1 2 1.5 13.7 3 2.6 100 4 0.1 25 0.05 62 1. Which substrate have the highest affinity for the enzyme? Explain. 2. Which will show higher efficiency of converting the substrate to the product? Show solutions and еxplain.Please note the following Lineweaver-Burk plot for the enzyme Virbraniumase reacting with a substrate: -0.4 . 1/V 5 4 3 2 -0.2 0 0.2 y = 5.2781x + 1.3338 R² = 0.9967 0.4 0.6 0.8 1/[S] 1 Based on the information provided, what is the Vmax for this reaction?
- The KM values for the reaction of chymotrypsin with two different substrates are given in the table below. Considering this information, which substrate has the lower apparent affinity for the enzyme? Which substrate is likely to give a lower value for Vmax? Substrate N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester KM (M) 8.8 X 10-² 6.6 X 10-4 N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give a lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give the lower V₁ max. N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; N- acetyltyrosine ethyl ester is likely to give the lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; N- acetylvaline will likely to give the lower Vmax. None of the above statements are correct.The equil ibrium constant for the attachment of a substrate to the active site of an enzyme was measured as 200.In a separate experiment, the rate constant for the secondorder attachment was found to be 1.5 x 108 dm3 mol-1 s- 1.What is the rate constant for the loss of the unreacted substrate from the active site?Give a possible set of bacterial factors (or lack thereof) and what properties they have that would explain an increase in the 50% inhibitory concentration for tetracycline by 1,000-fold in an organism.