ular features that stabilize structure. 4. You perform Michaelis-Menten kinetics on (i) trypsin and (ii) a mutant form of the same enzyme (a single amino acid has been changed). The specificity constant for the mutant was 10 times larger than for trypsin. 4. (a) Define the specificity constant and explain using this definition how a larger value might occur. 4. (b) Explain how, if at all, a non-competitive inhibitor of trypsin would affect the specificity constant.

ular features that stabilize structure. 4. You perform Michaelis-Menten kinetics on (i) trypsin and (ii) a mutant form of the same enzyme (a single amino acid has been changed). The specificity constant for the mutant was 10 times larger than for trypsin. 4. (a) Define the specificity constant and explain using this definition how a larger value might occur. 4. (b) Explain how, if at all, a non-competitive inhibitor of trypsin would affect the specificity constant.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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For an enzyme-catalyzed reaction, the presence of 10 nM of a reversible inhibitor yields a Vmax value that is 80% of the value in the absence of the inhibitor. The KM is unchanged by the inhibitor. (a) What type of inhibition is taking place? (b) Calculate the dissociation constant for the inhibitor (Ki). (c) What percentage of the enzyme molecules have bound inhibitor?
5. An enzyme catalyzes the reaction M concentration of 1 nM, and the Vmax is 2 µM s-1. The Km for substrate M is 4 µM. N. The enzyme is present at a (a) Calculate kcat. (b) What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 2.0? Increased or decreased?
Given the following data in enzyme-catalyzed reaction, what is the Vm, Km and type of inhibition of Experiment B?
How can distance validate non-covalent interactions? Also, please explain if the distance between c-alpha of a particular residue in an enzyme and c-alpha of its ligand are 12.4 Å, 11.7 Å, 6.5 Å, 7.6 Å, and 7.7 Å, what does that mean? What is the meaningful distance, what is the legth of hydrophobic,hydrophilic, h-bond...?
4-1Based on the experimental results shown in the table, Identify whether the inhibitor acting on this enzyme is acompetitive inhibitor or a non-competitive inhibitor, and Explain the reason logically. 4-2In 4-1, the concentration of the enzyme competition inhibitor was 1×10-3M. If 1 µmol of the inhibitor is present inthe 1 mL reaction mixture, Indicate how much the initial degree of hydrolysis decreases as a proportion(in thepresence of inhibitors/in the absence of inhibitors) with respect to the absence of inhibitors.
The catalytic activity of an enzyme is being measured at pH 7.5 and requires that the side chain of a specific Cys (C) residue be in its unprotonated (conjugate base) form for the enzyme to be active. That specific Cys side chain in the protein has a pKa of 7.8. a) For this Cys side chain with a pKa of 7.8, what is the base/acid ratio at pH 7.5? b) What fraction of the total enzyme molecules in a solution of the enzyme at pH 7.5 have that Cys side chain in the active form? (Express the answer as a %) c) The name of the Cys side chain functional group is? d) The active form of the Cys side chain in this enzyme would be? i. positively (+) charged ii. uncharged (neutral) iii. negatively (–) charged
How can the Michaelis constant (Km) be identified and used to represent the affinity between the enzyme and the substrate?
Enz-SH + Ag* - Enz-S-Ag + H* The affinity of Ag* for sulfhydryl groups is so great that Ag* can be used to titrate -SH groups quantitatively. To 10.0 mL of a solution containing 3.5 mg/mL of a pure enzyme, an investigator added just enough AGNO, to completely inactivate the enzyme. A total of 0.536 umol AGNO, was required. Calculate the minimum molecular weight (M,) of the enzyme. minimum M, = The determined M, is not exact, but it is a minimum value for the M, of the enzyme. Why does the M, value obtained this way give only the minimum molecular weight? There is only one polypeptide chain. There are no disulfide bonds in the protein. There is only one titratable -SH group per protein molecule. The Ag* would also react with the terminal carboxyl group.
Could someone explain what this is saying? I would like a better understanding. FIGURE D CPP32 (Caspase-3) protease activity. Aliquots of cell lysates (50μL) were incubated with an equal volume (50 μL) of the reactionbuffer and 5 μL of 7-amino-4-trifluoromethyl coumarin (AFCDEVD) for 1 hour at 37˚C (Clontech). The shift in fluorescenceemission of AFC-DEVD, on its proteolysis to free AFC by theprotease, was detected in a fluorometer, using a 400 nm excitationfilter and 505 nm emission filter.
Bilayer leaflet CO COON 1. Unfolded peptide aqueous phase 2. Unfolded peptide in membrane 3. α-helix in membrane Figure 1: Membrane binding peptide of a glycosyltransferase is shown interacting with a membrane. Positive charges are shown as blue +, negative charges are shown as red -. 7. (3%) While the membrane binding peptide is in solution it is shown as being "un- folded" in Figure 1. (a) What is the name we give to this type of secondary structure? (Answer with 1 sentence) (b) What is a general role of this type of secondary structure in a protein? Answer with proy
Following data were obtained for an enzyme in the absence and presence of an inhibitor. What are the Km and Vmax values for the inhibited and uninhibited enzymes? What is the type of inhibitor?
You performed a kinetic study of the enzyme lipase and you obtained the following double reciprocal (Lineweaver-Burk) plot: 4 y = 4.3511x + 1.8889 %3D 1 -0.5 -0.3 -0.1 0.1 0.3 0.5 0.7 0.9 1/[S] When you performed the same kinetic study in the presence of an inhibitor, you obtained a new line that had y-intercept 3 and x-intercept -0.3. Based on this information select all correct statements about this inhibitor. This is mixed inhibitor. When this inhibitor is present apparent Km of enzyme decreases. This inhibitor prevents substrate from binding to the enzyme. This is uncompetitive inhibitor. When this inhibitor is present apparent Vmax of enzyme decreases. When this inhibitor is present apparent Km of enzyme increases. This inhibitor can bind to either enzyme or enzyme-substrate complex.