Between the following 4 Km values, select the one that indicates binding of the enzyme to its substrate with the highest affinity: Group of answer choices 10 nM 1000 uM 1 mM 10 pM

1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…

200 ml of a 2% protein solution containing an enzyme that you want to purify. Half of the sample is subjected to method A, consisting of fractionated precipitations and 5 ml of final solution are obtained, with a concentration protein equal to 5 mg / ml and enzymatic activity equal to 2000 U / ml. The other half is subjected to method B, consisting of ion exchange chromatography, and a final solution of 10 ml, with protein richness equal to 10 mg / ml and with an activity enzymatic also equal to 2000 U / ml. You want to know: a) Which of the two methods has provided the purest enzyme. b) By which of the methods the greatest amount of enzyme has been obtained.

The enzyme triosephosphate isomerase (TIM) catalyzes the following reaction in glycolysis, where it converts dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (GAP). CH₂OH C=O CH₂OPO²- DHAP triose phosphate isomerase [DHAP] (MM 1.00 2.00 3.00 6.00 O V. (mM/s) 3.700 6.727 9.250 14.800 = H HCOH CH₂OPO²- Kinetics experiments were performed on TIM. Enzyme activity (initial velocity, V.) was measured at varying concentrations of DHAP. The enzyme kinetics were also measured in the presence of two inhibitors, A and B. The enzyme concentration used in all experiments was 25.00 μM. The data are shown below. GAP V. (mM/s) + A 1.452 2.794 4.038 7.281 V₁ (mM/s) + B 0.755 1.379 1.905 3.077

An enzyme catalyzes a reaction with a Km of 7.50 mM and a Vmax of 2.90 mM - s-1. Calculate the reaction velocity, un, for each substrate concentration. [S] = 2.75 mM mM · s- * TOOLS x10 [S] = 7.50 mM mM · s-! [S] = 11.0 mM mM - s-

Answer bi) and bii) only

A purified protease enzyme from the fungus Aspergillus sp. Is tested in a laboratory. This enzyme was lyophilized as a white powder. When reconstituting with phosphate buffer pH 7.2 the active enzyme is obtained. To check its purity, an electrophoresis is performed where a single band of approximately 70,000 molecular weight is observed. The solution with enzymatic activity was stored at 4ºC for later use. A few days later it was found that the enzymatic activity had been lost and in the electrophoretic analysis, instead of a single band there were three bands of weights 40,000, 20,000 and 10,000. Come up with a reasoned explanation of what might have happened to the enzyme

Km value should be more than 10% apart, use non-competitive inhibitor

Please dont provide handwrittin solution.../.

Data from enzyme inhibition are used to determine a Kmapp and Vmax PP. Comparison of these values with assays run without inhibitor are used to understand how the inhibition is occurring. This is useful for better understanding the active site as well as the practical aspect of pharmaceutical drugs. Below are idealized Line-Weaver Burke plots of different types of inhibitors. Comnetitive Uncomnetitive Mixed +Inh +Inh 4Inh Anh Inh Anh [S] [S] [S] a. How does the value of Vmax for the enzyme compare to the Vmax PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed b. How does the value of Km for the enzyme compare to the Km PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed c. For each situation in Model 1, consider an inhibitor that is better than the one shown on the graph. Answer the following questions for each type of inhibition: i. How would the KmPP change? ii. How would the Vmax PP change?