Acetazolamide is a drug which inhibits carbonic anhydrase. Carbonic anhydrase participatesin regulation of the pH and bicarbonate content of a number of body fluids. Figure 2 showsthe experimental curve of initial reaction velocity (as percentage of Vmax) versus [S](concentration) for the carbonic anhydrase reaction. The graph also shows the curve in thepresence of acetazolamide.100No inhibitor50Acetazolamide0.20.40.60.8(S] (mM)Figure 2(i)Compare the maximal velocities and Michaelis Menten constants of the enzyme inthe absence and the presence of the inhibitor acetazolamide. Determine the nature ofinhibition by acetazolamide. Explain your answer.(*"A JO %) A

Carbonic anhydrase is inhibited by acetazolamide. A graph has been given in the question where…

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Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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A junior student performed an experiment to found Ki values for 3 different competitive inhibitors. The results are 5, 1 and 0,2 . Please answer the questions:a. Which inhibitor has higher affinity for the enzyme?b. If the student uses equal concentrations of inhibitor in each trial, which inhibitor gives the smallest Km value?
Do not give handwriting solution.
Explain the molecular basis behind the reported observation in the parameter measured below:
This is a visible spectra between 390-590 nm obtained during the protein separation process of haemoglobin and cytochrome c using CM Sephadex chromatography. I'd like the results shown on the image interpreted. Look for characteristic peaks or patterns that correspond to the absorption properties of these proteins in the visible range.
Consider the following experimental data from another experiment: [S] 1.5 2.00 2.50 5.00 10.00 V (No inhibitor) mmol ml¹ min¹ 0.167 0.204 0.232 0.313 0.385 V (inhibitor) mmol ml¹¹ min¹¹ 0.115 0.143 0.167 0.250 0.333 Calculate Km and V max and determine whether this inhibitor is competitive, non-competitive or uncompetitive.
A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).
Which of the following is incorrect about blood glucose test interpretation A) fasting plasma glucose 6.1-6.8 defined as having IGT and should probably be screened again after a year or so. B) 2 results>7mmmol/l defined as having diabetes C) fasting plasma glucose >6mmol, recall for further test. D) fasting or random plasma glucose, on one reading of <11.1mmol/l is diagnostic of diabetes
The Ki for a competitive inhibitor pyrrole-2-carboxylate is 0.05M.Calculate the initial velocity if the substrate concentration is 0.25M and 1.0 mole/liter of the inhibitor is added to thereaction.
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…
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Following is a Dixon plot for PNPP inhibition by inorganic phosphates. What answer choice shows the correct combination of the mode of inhibition and Ki of the EI complex? - Mixed inhibition, Ki= -21.2 uM - Mixed inhibition, Ki= 1/16.5 uM - Competitive inhibition, Ki= 16.5 uM - Competitive inhibition , Ki= -1/16.5 uM - None of the above
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