Diisopropylphosphofluoridate (DIPF) inactivates chymotrypsin by covalently modifying serine 195.Which statement is true of DIPF's inhibitory mechanism?DIPF randomly modifies all serine residues on the protein, and if enough is added, the serine in the active site willeventually be modified.DIPF approaches serine 195 more closely than other substrates.DIPF looks like the substrate for chymotrypsin and binds in the active site as a competitive inhibitor.Serine 195 is in an environment that gives it a higher than normal reactivity with respect to DIPF.

Diisopropylphosphofluoridate (DIPF) is an irreversible inhibitor of chymotrypsin. It forms a…

Answered: Diisopropylphosphofluoridate (DIPF)…

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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Describe the features of the first binding site (for allosteric enzyme unprotkb-p39086), including structural features that recognize and complement the first ligand.
The graph displays the activities of wild-type and several mutated forms of subtilisin on a logarithmic scale. 5 The mutations are identified as: • The first letter is the one-letter abbreviation for the amino acid being altered. • The number identifies the position of the residue in the primary structure. The second letter is the one-letter abbreviation for the amino acid replacing the original one. • Uncat. refers to the estimated rate for the uncatalyzed reaction. Log ₁0k cat (S-¹) -5 -10 Wild type D32A S221A H64A İnti. S221A H64A D32A Uncat. How would the activity of a reaction catalyzed by a version of subtilisin with all three residues in the catalytic triad mutated compare to the activity of the uncatalyzed reaction? O It would have more activity, because the reaction catalyzed by the triple mutant is approximately 1000-fold faster than the uncatalyzed reaction. It would have more activity, because the reaction catalyzed by the triple mutant is approximately three-fold faster…
Trypsin contains a/an [aspartate/serine/histidine] in its specificity pocket to attract lysine and arginine side chains. It contains a potent nucleophilic [histidine/serine/aspartate] in its active site that is capable of attacking the electrophilic [carbon/nitrogen/oxygen] of the closest peptide bond. The [acyl-enzyme/specificity pocket/oxyanion] transition state is stabilized by the backbone amide hydrogens of glycine and serine. In this way, the enzyme is able to catalyze the [hydrolysis/ligation/metalysis/hydrogenation] of the peptide bond.
Substrate A has a Km of 45uM and a kcat of 100/s with trypsin and a Km of 540mM and a kcat of 2/s with chymotrypsin. Substrate A is a better substrate of [trypsin/chymotrypsin] and likely contains a/an/some [aromatic/acidic/basic] residues(s) in its sequence. Chymotrypsin contains a [glycine/serine/aspartate] in its specificity pocket which likely [repel/attract] the residues in substrate A.
In the protein adenylate kinase, the C-terminal region has the sequence Val-Asp-Asp-Val-Phe-Ser-Gln-Val-Cys-Thr-His-Leu-Asp-Thr-Leu-Lys-The hydrophobic residues in this sequence are presented in boldface type.Suggest a possible reason for the periodicity in their spacing.
Draw the catalytic triad of a serine protease at the first tetrahedryl intermediate stage. Your drawing should include the amino acids that make up the triad, any part of the substrate that should be present, and any other molecules that are present in the active site.
In the case of normal RNase A, Anfisen found that oxidizing the Cys residues before slowly removing the urea gave a very different result than oxidizing the Cys after slowly removing the urea. When the urea was removed first, Anfisen recovered 100% of the catalytic activity. When the urea was removed after oxidation, only about 1% of the activity (1/105) was recovered. The conclusion was that 104 of every 105 molecules was misfolded, and therefore catalytically inactive. Let's say someone raises an objection to this interpretation, and says that perhaps, instead, something chemically happens to the active site of RNase A when it is oxidized before refolding, and that the ~1% activity measured represents the residual and greatly reduced enzyme activity of all of the molecules, and the number is just coincidently similar to 1/105. (In other words, instead of 104 completely inactive molecules for every 1 completely active molecule, all molecules are equally damaged, and only function at…
In addition to the reactions mentioned in Section 23.5, PLP can catalyze b-substitution reactions. Propose a mechanism for the following PLP-catalyzedb-substitution reaction:
PFK-1 is actually a tetramer with four identical active sites that produced fructose 1,6-bisphosphate. There are also allosteric sites for citrate and ATP/ADP on each subunit as well which influence how well the active sites are binding substrates or producing product. The arginine at position 48 on each PFK-1 subunit normally forms a salt bridge with a negatively charged oxygen on phosphate in the catalytic interface of PFK-1. Speculate on how the R48C mutation might change this function.
“Caspase” describes the catalytic activity of this enzyme. What amino acyl residue participates in the catalytic mechanism? What substrate motif does caspase-8 recognize?
In a certain bacterial species, the amino acid arginine is synthesized by a particular enzyme so the bacterium does not require arginine in its growth medium. A key enzyme, which we will call arginine synthetase, is necessary for arginine biosynthesis. When these bacteria are given arginine in their growth media, they stop synthesizing arginine intracellularly. Based on this observation alone, propose three different regulatory mechanisms to explain why arginine biosynthesis ceases when arginine is added to the growth medium. To better understand the mechanism of regulation, you measure the amount of intracellular arginine synthetase protein when cells are grown in the presence and absence of arginine. Under both growth conditions, the amount of this protein is identical. Which mechanism of regulation would be consistent with this experimental observation?.
Determine the amino acid sequence of Peptide A, which was obtained by hydrolyzing a larger peptide with trypsin, using the data below: Total hydrolysis of Peptide A produces (Ala2, Arg, Glu, Gly, Met, Pro2, Ser, Tyr). Sequential Edman degradation of Peptide A produces a derivative of Ala in the first cycle; a derivative of Serine in the second cycle; and a derivative of Proline in the third cycle. Cleavage of Peptide A with chymotrypsin yields two peptides: a tripeptide III containing Ala, Arg and Gly; and a heptapeptide IV containing Ala, Glu, Met, Pro2, Ser and Tyr. Edman treatment of tripeptide III generated a derivative of Alanine. Cleavage of Peptide A with cyanogen bromide (which cleaves peptides on the C-terminal side of Methionine) yields two pentapeptides labeled I and II. Total hydrolysis or peptide I indicate that it contains Ala, Arg, Gly, Pro and Tyr; pentapeptide II contains Ala, Glu, Met, Pro, and Ser.

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