Concept explainers
At 37 oC, the serine protease subtilisin has kcat = 50 S-1 and KM = 1.4 x 10-4 M. It is proposed that the N155 side chain contributes a hydrogen bond to the oxyanion hole of subtilisin. J. A. Wells and colleagues repotted (1986, Phil Trans. R Soc. Lond. A 317: 415-423) the following kinetic parameters for the N155T mutant of subtilisin:
kcat = 0.02 S-1 and
KM = 2 x 10-4 M.
a. Subtilisin is used in some laundry detergents to help remove protein-type stains. What unusual kind of stability does this suggest for subtilisin?
b. Subtilisin does have a problem in that it becomes inactivated by oxidation of a methionine close to the active site. Suggest a way to make a better subtilisin.
c. Is the effect of the N155T mutation what you would expect for a residue that makes up part of the oxyanion hole? How do the reported values of kcat and KM support your answer?
d. Assuming that the T 155 side chain cannot H-bond to the oxyanion intermediate, by how much (in kJ / mol) does N155 appear to stabilize the transition state at 37 oC?
e. The value you calculated in part (d) represents the strength of the H-bond between N155 and the oxyanion in the transition state. This value is higher than typical H-bonds in water. How might this observation be rationalized? Hint: Consider Equation 2.2 (Coulomb's Law).
Want to see the full answer?
Check out a sample textbook solutionChapter 8 Solutions
Biochemistry: Concepts and Connections
- Shown below is a proposed mechanism for the cleavage of sialic acid by the viral enzyme neuraminidase. The kcat for the wild-type enzyme at pH =6.15, 37 °C is 26.8 s-1.(a) Describe the roles of the following amino acids in the catalytic mechanism: Glu117, Tyr409, and Asp149. List all of the following that apply:general acid/base catalysis (GABC), covalent catalysis, electrostaticstabilization of transition state.(b) Based on the information shown in the scheme, would you expect mutation of Glu 117 to Ala to have a greater effect on KM or kcat?(c) For the R374N mutant at pH = 6.15, 37 °C, kcat is 0.020 s-1, and KMis relatively unaffected. Based on this result, it seems that R374 is morecritical for catalysis than for substrate binding. Explain how R374 stabilizesthe reaction transition state more than the substrate (i.e., what feature of this reaction would explain tighter binding to the transition state vs. substrate?).arrow_forwardAntifungal Activity of Red Dragon Peel (Hylocereus polyrhizus)ⓒ Rudi Hendra et al 2020 IOP Conf. Ser.: Mater. Sci. Eng. 833 012014 STATE THE HYPOTHESIS (In if-then format): Plants, when extracted, can produce dyes and the extracts have been used for different purposes such as the textile or food industry [1]. One potential plant that can be used as a natural dye is dragon fruit (Hylocereus cacti) or by another name Pitaya. Wybraniec and coworkers reported that dragon fruit has a potential as a source to produce red pigment and it's called betalains [2]. Betalain is a water-soluble pigment that gives color to flowers and fruits. Betalain pigments are divided into two groups, namely betacyanin which produces purplish red and betaxanthin which produce yellow-orange colors [2, 3]. In the dragon fruit peel, it is also containing pigments which are usually only discarded as food waste and have not been used optimally. This is very unfortunate because the fruit of the dragon fruit itself has…arrow_forwardCalculate the mass of invertase (in mg) and concentration of invertase (in mM) contained in a 25.0mL sample of yeast extract that has 3,000 total units of activity, assuming that pure invertase has a specific activity of 1,000 units/mg with a mass of 270kD.arrow_forward
- The objective is to study a novel protease P isolated from the digestive tract of an Amazonian insect. This protease can exist into two forms Pi and Pa which have identical amino acid sequences (both of 80 kDa). However, only Pa shows proteolytic activity. To better understand the activation mode of Pi (inactive form) in Pa (active form), the following experiment was done using DIPF. DIPF (diisopropylphosphofluoridate) is a well-known irreversible inhibitor of serine proteases. It reacts with the catalytic serine residue of the active site of proteases as shown below: Enzyme -CH₂OH + CH(CH3)2 O F-P=0 O CH(CH3)2 Diisopropylphospho- fluoridate (DIPF) Enzyme -CH,—O CH(CH3)2 O <=0 O CH(CH3)2 DIP-Enzyme Both proteases Pa and P₁ were incubated with 32P-DIPF for 30 min at 37°C, and then dialysed to remove excess of unreacted radiolabelled reagent. The two proteases were then analyzed in Sodium Dodecyl Sulphate-Polyacrylamide Gel Electrophoresis (SDS-PAGE), with and without 2-mercaptoethanol.…arrow_forwardChymotrypsin has the highest affinity for which of the following substrates: Table. The values of KM and kcat for some Enzymes and Substrates Enzyme Chymotrypsin Ки (М) 4.4 x 10-1 8.8 x 10-2 6.6 x 104 Kcat (S-1) 5.1 x 10-2 1.7 x 10-1 1.9 x 102 Substrate N-acetylglycine ethyl ester N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester Catalase H2O2 2.5 x 10-2 1.0 x 107 Urease Urea 2.5 x 10-2 4.0 x 105 OA. N-acetylglycine ethyl ester OB. N-acetylvaline ethyl ester OC. N-acetyltyrosine ethyl ester D. Ureaarrow_forwardb. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein E. To determine where the block in protein-E production occurred in each individual, the various intermediates were given to each individuals cel Is in culture. After a few weeks of growth with the intermediate, the cells were assayed for the production of protein E. The results for each individuals cells are given in the following table. A plus sign means that protein E was produced after the cells were given the intermediate listed at the top of the column. A minus sign means that the cells still could not produce protein E even after being exposed to the intermediate at the top of the column. Denote the point in the pathway in which each individual is blocked.arrow_forward
- 3. Subtilisin (Mol. weight 27,600) is a protease that can catalyze hydrolysis of certain amino acid esters and amides. For the synthetic substrate N-acetyl-L-tyrosine ethyl ester (Ac-Try-OEt), subtilisn exhibits Km and kcat values of 0.15 M and 550 s-¹, respectively. a) What is the Vmax when the subtilisin concentration is 0.4 mg/ml? b) Indole is a competitive inhibitor of subtilisin with a Ki of 0.05 M. What is the Vmax for Ac-Try-OEt hydrolysis by 0.4 mg/ml subtilisin in the presence of 6.25 mM indole? c) What is the Vo when 0.4 mg/ml subtilisin is incubated with 0.25 M Ac-Try-OEt and 1.0 M indole?arrow_forwardThe ESI-MS spectrum in positive ionization mode for lysozyme is obtained. a. What is the molecular weight of the protein to 5 significant figures based on the two highlighted ion species? b. What is the charge of the peaks at 1101.5 and 1789.2.arrow_forwardConsider the role of Histidine in the Serine protease mechanism and sketch a plot showing the predicted pH profile of chymotrypsin which has a pH optimum of approximately ~8. The pk, for the His in the catalytic triad is 7.3 in free chymotrypsin which increases to greater than 8 with a bound peptide. Be sure to label the plot axes and indicate the pka of His on the plot,arrow_forward
- Given the following data, calculate Keq for the denaturation reaction of the protein β-lactoglobin at 25oC: ΔH° = –88 kJ/mol ΔS° = 0.3 kJ/mol. The free energy of hydrolysis of ATP in systems free of Mg2+ is −35.7 kJ/mol. When the concentration of this ion is 5 mM, ΔG°observed is approximately −31 kJ/mol at pH 7 and 38°C. Suggest a possible reason for this effect.arrow_forwardChitinase is a protein that breaks down chitin, a primary component of the cell wall in fungi, scales in fish and exoskeletons of arthropods. The activity of chitinase extracted from a plant was shown to be optimum at pH 5. You were tasked to prepare 300 mL of 150 mM buffer solution for further analysis of the extracted chitinase. REAGENTS Ka 2.5M Acetic acid Solid NaOAc•3H2O [136.08g/mol] 1.76 x 10-5 2.5M NH3 Solid NH4Cl [53.49g/mol] 5.6 x 10-10 2.5M Lactic acid Solid sodium lactate [112.06g/mol] 4.0 x 10-5 5 M HCl 5M NaOH Pls show sol'ns 1. Given the following reagents, give the moles of each component (acid & base).2. What are the mass/volume of the components needed to prepare the buffer? 3. What will the pH of the buffer be if 1mL of 5 M NaOH was added?arrow_forwardThe kcat and KM for chymotrypsin-catalysed cleavage of a synthetic substrate, S, were determined to be 60 s-1 and 0.5 mM, respectively, using steady-state kinetics. The concentration of enzyme in the assay was 5 x 10-8 M. ii) Draw a graph on the axes shown in Figure 1 below showing how the initial rate of the reaction varies with [S], making use of the values for KM and Vmax. Label the axes with appropriate titles and units.arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage LearningBiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning