Suppose you specifically want to measure the amount of beta-glucose present in a sample. You have an enzyme that reacts specifically with the beta form of the sugar,and produces a signal that is easy to measure. However, every time you try to measure the amount of beta-glucose, mutarotation occurs and the alpha form of the sugar is converted into beta!! Can you suggest a simple way of overcoming this problem so you could measure only the beta form of the sugar? HINT: Is there a way you can prevent mutarotation?

By using Excel or GoogleSheets, graph the Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available. What are the Km and V values for the inhibited and uninhibited reactions? Is the inhibitor competitive or max noncompetitive? [S] (mM) V, No Inhibitor (mmol min-') V, Inhibitor Present (mmol min-') 1× 10-4 5 × 10-4 1.5 × 10-3 2.5 × 10-3 5 × 10-3 0.026 0.092 0.136 0.150 0.010 0.040 0.086 0.120 0.165 0.142 Activate

From the complete oxidation of glucose (glucose  →  6CO2), how many total nucleotide triphosphates are yielded (be sure to deduct payback) as part of substrate level phosphorylation?

Match the enzymatic activity with the proper description within the ubiquitin-proteasome pathway. Drag and drop options on the right-hand side and submit. For keyboard navigation... SHOW MORE ✓ E1 E2 E3 Proteasome = ||| = ||| = ||| Degradation Ubiquitin-activating enzymes Ubiquitin-conjugating enzymes. Ubiquitin ligases

To answer this question, please reference the Problem Solving Video: pH and pK, of an Amino Acid. Proteases are enzymes that cleave peptide bonds using general acid-base catalysis. General acid-base catalysis relies on a protor donor or acceptor other than water. Proteases rely on proton transfer to create strong nucleophiles from active-site amino acid residues. In the protease chymotrypsin, an active-site serine is a potent nucleophile. A nearby residue, His 57, interacts with serine to increase its reactivity. A schematic of chymotrypsin's active site illustrates the active site Ser 195 and His 57 R groups. The Asp 102 residue helps position the His 57 residue via hydrogen bonding. Asp 102 [His] [His+] His 57 6.0 -H-N Chymotrypsin is a digestive enzyme with a catalytic optimum between pH 7.8 and 8.0. However, due to the presence of gastric juice, chymotrypsin often functions in an environment of pH 5.5-7.0. Incorrect The imidazole group of free histidine has a pK₂ of 6.0. Use the…

Which of the following would result in an increase in the amount of F-2,6-BP? (select all that apply) Group of answer choices high blood glucose activation of fructose bisphosphatase 2 increased phosphofructokinase 2 activity increased glucagon secretion increased phosphorylation of the bifunctional enzyme

A solution of [U 14C] glucose-1-phosphate (specific activity = 16,000 cpm/mmole) was  incubated with glycogen and glycogen phosphorylase, an enzyme which adds glucose units on to  glycogen. Radioactivity was incorporated into the glycogen primer at a rate = 2550 cpm/min. The rate  of the enzymatic reaction in units of mmole glucose incorporated per minute is:  (a) 0.016 mmol/min  (b) 0.57 mmol/min  (c) 0.16 mmol/min  (d) 5.7 mmol/min

Draw two reaction coordinate diagrams (DG vs reaction coordinate) that compare the uncatalyzed process of lipid flipping (lipid out à lipid in ) from outer to inner leaflet to the flippase-catalyzed process.

The initial velocities of two different enzyme-catalyzed reactions were measured over a series of substrate concentrations. The following results were obtained: Enyme A: KM = 1.5 mM, Vmax = 10 μM s-1 Enyme B: KM = 5.0 mM, Vmax = 85 µM s-1 (a) Which enzyme binds to its substrate more tightly (assume k.1 >> k₂ in the Michaelis-Menten model)? (b) Calculate the initial velocities of each reaction when the substrate concentration is 2.5 mM. (c) Calculate the Kcat of each enzyme if the total enzyme concentration is 100 nM. (d) Which enzyme is the more efficient catalyst? Explain your answer. The enzyme carbonic anhydrase is strongly inhibited by the drug acetazolamide. A plot of the initial reaction velocity (as a percentage of Vmax) in the absence and presence of the inhibitor is shown below. What type of inhibition is taking place? Explain your reasoning. V (% of Vmax) 100 50 0.2 0.4 No inhibitor Acetazolamide [S] (MM) 0.6 0.8 1

The Lineweaver-Burke plot was originally developed in order to "linearize" the data obtained from enzyme kinetics experiments, in order to facilitate the determination of kinetic parameters. Why is it not considered to be an accurate method for this purpose? It is very difficult to draw a straight line on a computer. It is very difficult to calculate the variables required for the "x" and "y" axis. It is more accurate to use the standard "V versus [S]" plot to determine Vmax and KM- The plot weights the least accurate data points the most heavily. It is no longer considered to be acceptable to extrapolate from known data.