In muscle cells, the AG for glucose+ATP2 glucose-6-P+ADP is –33.5 kJ•mol-'. In contrast,the AG for glucose – 6 – P fructose – 6 - P is –2.5 kJ · mol-1.(a) In a physiological context, which reaction is faster? How do you know?(b) The enzyme that catalyzes glucose + ATP 2 glucoseimportant point of regulation in glycolysis, while the enzyme that catalyzes glucose – 6 – PZfructose – 6 – P (phosphoglucose isomerase) is not. Why is hexokinase a good step at whichto regulate glycolysis relative to phosphoglucose isomerase?6 – P + ADP (hexokinase) is an|-|

During glycolysis, Glucose is converted into Pyruvate. The first step of glycolysis is the…

Answered: In muscle cells, the AG for glucose+ATP…

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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The hydrolysis of sucrose has the following data. Time (min) 0 30 60 90 130 180 [Sucrose] (M) 0.5011 0.4511 0.4038 0.3626 0.3148 0.2674 Sucrose + H₂O → glucose + fructose Unistanit. Determine the half-life. (Use the order from a.) You will need to take into account the order!
Consider the graphic representation shown below for the free energy change for the overall reaction of Fructose-6-phosphate and ATP to form the products Fructose-1,6-biphosphate and ADP in the presence (blue line) or absence (red line) of enzyme. Match each of the arrows (labeled A, B, C, or D) indicated in the graph with an apropriate description by dragging the letters into their corresponding box. C D reactants Fructose-6-phosphate + ATP A products Fructose-1,6-biphosphate + ADP Progress of the Reaction Reset Help D BAc В C Energy of activation for the forward reaction in the absence of enzyme Energy of activation for the forward reaction Energy of activation for the reverse reaction in absence AG for the reaction the presence of enzyme of enzyme Energy
3-2.22.1
What is the molecular weight of a linear polysaccharide consisting of 7 galactose monomers and 1 terminal glucose monomer? The molecular weight of glucose (C6H12O6) is 180.156 g/mol. Remember that the types glycosidic of linkages are irrelevant for calculating the molecular weight. Write your answer in terms g/mol.
An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-s
1. a. Calculate the physiological DG of the reaction shown below at 37°C, as it occurs in the cytosol ofneurons, with phosphocreatine at 4.7 mM, creatine at 1.0 mM, ADP at 0.73 mM, and ATP at 2.6mM. The standard free energy change for the overall reaction is –12.5 kJ/mol. Phosphocreatine + ADP ® creatine + ATP b. The enzyme phosphoglucomutase catalyzes the conversion of glucose 1-phosphate to glucose6-phosphate. Calculate the standard free energy change of this reaction if incubation of 20 mMglucose 1-phosphate (no glucose-6 phosphate initially present) yields a final equilibrium mixtureof 1.0 mM glucose 1-phosphate and 19 mM glucose 6-phosphate at 25°C and pH 7.0. c. If the rate of a nonenzymatic reaction is 1.2 x 10–2 μM s–1, what is the rate of the reaction at 37℃ inthe presence of an enzyme that reduces the activation energy by 30.5 kJ/mol?
If the enzyme lactase has a Vo of 0.40 mM per minute when [S] = 1.25 mM, and a Vo of 1.0 mM per minute when [S] = 5.0 mM, what is its Vmax? Calculate the Km of the above enzyme. Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the above enzyme.
(b) You are investigating the effects of several agents on the activity of alcohol dehydrogenase. The enzyme activity data are shown in the table below. Construct a [substrate] vs. activity plot and a double-reciprocal plot for this enzyme. Be sure to label all axes. Determine the Vmax and KM for AD from the graphs in each type of plot. AD activity (nM/min) AD activity + agent A (nM/min) AD activity + agent B (nM/min) [Alcohol] (nM) 0.1 14 2 0.5 50 7 8. 1.0 65 10 30 2.0 72 12 45 4.0 80 14 62 8.0 85 15 75 32.0 90 16 90
Chymotrypsin is an enzyme that is optimized to function in the small intestine (at a pH of ~7.4). Explain, given what you know about the chymotrypsin mechanism and what we learned about acid/base chemistry, why chymotrypsin would not be able to initiate catalysis if it were secreted into the stomach (pH ~2.5).
You run an experiment using 0.5 µM of an enzyme. The kinetic parameters for your enzyme are kcat = 1800 s-1 and KM = 25 µM. (a) If k-1 is 240 s -1, what is the value of k1? (Report your answer in units of M-1 s -1) (b) What is Vmax for the enzyme (Don't forget to include units in your answer)? (c) You set up an experiment with the following initial concentrations of substrate. Calculate the rate of the reaction at each substrate concentration (report your answer in units of µM sec-1 [S](μM) 1 100 1000 [S][μM) 1 100 1000 v (μMsec ¹) (d) You now add 10 nM of a competitive inhibitor with a KI = 2 nM. What is the rate of the reaction (v) for the following substrate concentrations and how does it compare to the rate without inhibitor (VO)? Vinhib (uMsec-¹) Vinhih/V
Assume that in a certain cell, the ratio of products/reactants or Keg = 809.5 (Keq is dimensionless) for the reaction Glucose + 2ATP > Glucose-1,6-diP + 2ADP, at a particular instant, the concentrations of each compound were Glucose =2.4M, ATP =11.1M, ADP -12.8M and G-6-P -28.4M. Calculate the difference (dimensionless) between Keq and the ratio of products/ractants at this instance, in this cell, to five decimal places
When 10 micrograms of an enzyme with a molecular mass of 80,000 Daultons (grams/mol) is added to a solution containing its substrate at a concentration 100 times the Km, it catalyzes the conversion of 65 micromoles of substrate into product in 3 minutes. What is the enzyme's turnover number (in units of min-1)?

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