4. Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated fromthe bacterium Staphylococcus aureus. The enzyme is involved in carbohydrate metabolism and is apotential biocatalyst for the large-scale production of rare sugars. (8)mgV₂ μmol s11.50-1.25-1.00-0.75-0.50-0.25-0.000100200 300[s], mM400500WT(a) Estimate the KM and Vmax for the wild-type and mutant enzyme from the graph.V105A(b) Calculate the keat and keat/KM for the wild-type and mutant enzyme based on your estimated valuesin (a) if the total enzyme concentration is 0.5 µmol/mg.(c) Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.

For a one-substrate enzyme catalyzed reaction, the Michaelis-Menten equation shows the quantitative…

Answered: 4. Below is a Michaelis-Menten plot for…

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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1 %09 l11. 9:1A Nucleotide Metabolism (Q) |.. biochemistryquestions.wordpress.com transferase e) Nucleotides Answer to NcM-01 Question NcM-02 Both the salvage and de novo synthesis pathways of purine and pyrimidine biosynthesis lead to the production of nucleoside-5'-phosphates through the utilization of an activated sugar. The activated sugar intermediate used for this purpose is: a) Glucose 6 (P) b) Ribose 5 (P) 5 Phosphoribosyl 1 pyrophosphate d) Ribosyl pyrophosphate e) Deoxyribosyl5 pirophosphate c) Question NcM-03 The first purine nucleotide that is fully formed in 1 Privacy & Cookies: This site uses cookies. By continuing to use this website, you agree to their use. To find out more, including how to control cookies, see here: Cookie Policy Close and accept II
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TABLE 3-LACTATE PRODUCTION IN FORTIFIED HEMOLYSATES OF HUMAN ERYTHROCYTES* Substrate Glucose Glucose Lactate production† No. of experiments pH 6 7.1 2.03 ± 0.91 6 7.8 4.76 ± 1.09 7-1 10-73 +1-88 5 7.8 12.34 ±2.92 5 7.0 7-15±0.73 5 7-7 (b)( ) In mature erythrocytes (red blood cells) the end product of glycolysis is lactate because of the absence of mitochondria. On the right is a table comparing the rate of lac- tate production in hemolysates (lysed cells) of human RBCs as a function of pH with dif- ferent substrates introduced into the glyco- lytic pathway. The hemolysate was fortified with 30 μmoles substrate, 7.5 μmoles MgCl2, 10 μmoles disodium phosphate, 1.5 μmoles NAD and 5 μmoles ATP in a volume of 5 mL. The rate of lactate production is given as μmoles of lactate/g Hb/hr at 37° C, buffered to either pH 7.1 or 7.8, as indicated. According to the results in the table which glycolytic enzyme is rate-limiting? Explain. Glucose-6-phosphate Glucose-6-phosphate Fructose-1,6-diphosphate…
Please determine the template sequence used in a pyrosequencing reaction in the following graph. A 4G T 2G C 3T 4G 2T G CA G 2T G 3 2 min AGT CA G T C A G T C A G T C A GT C AG T c A G 0.5 pmol
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While furthering your studies of iocane powder you determine that a ceular enzyme, iocase will degrade it. Under normal conditions, iocase shows km 0f 1.85 uM and a kcat of 76. The drug sildenafil was found to have no effect on the km of iocanse yet the kcat was found to drop 58 minutes^-1, when the drug was added and no other changes were made in the reaction conditions. What can be inferred from this data? (remember kcat = vmax/et) Sildenafil acts as an uncompetitive inhibitor of iocanse Sildenafil acts as a noncompetitive inhibitor of iocanse The effect of sildenafil could be countered by adding more iocane to the reaction The y-intercept of the sildenafil inhibited iocanse would appear to be lower on a Lineweaver burke plot as compared to uninhibited iocanase The slope of a Lineweaver burke plot of the inhibited versus uninhibited iocanase would remain unchanged
Table 1. Kinetic parameters of a chosen model wild-type (WT) enzyme and three catalytically-improved mutants (M1 to M3). Enzyme variants k (8¹) WT MI M2 M3 50 500 1000 50 K₁, (μM) 10 100 100 1 k/K (HM¹¹ s¹) 5 The study above was done on a wildtype enzyme and three genetically-engineered mutants. The purpose was to find a mutation to improve the catalytic efficiency of the enzyme for industrial use. Which is the most accurate statement concerning the data? M2 is the optimal choice when substrate concentrations are saturating. The wildtype enzyme is still preferable to M3 when S<
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You want to prepare 500 mL of the macronutrient and micronutrient stock solutions which have 100X and 1000X higher concentration, respectively, than the required concentration in the culture mediuma)If 0.17 g/L KH2PO4 is required in MS medium, determine the mass (g) of KH2PO4 that is required to prepare the stock solution and the volume required (mL) to prepare 200 mL MS medium. Show your work.
this drug contains one or more building blocks derived from either ethylene oxide or epichlorohydrin.Identify the part of each molecule that can be derived from one or the other of the building block and propose structural formulas for the nucleophile(s) that can be used along with either ethylene oxide or epichlorohydrin to synthesize each molecule.
1. You need 0.5 liter of a 0.1 M buffer for a biochemical reaction that you plan to perform at pH 2.0. On the shelf in your laboratory you have available phosphoric acid, and the amino acids aspartate, glycine, glutamate, tyrosine, and histidine (all you have available are the fully deprotonated forms of the molecules); these are the only buffer materials you have available. (The pKa values for these compounds are given below.) For phosphoric acid, pKal1 = 2.12, pKa2 = 7.21, and pKa3 = 12.32 For aspartate, pKaCOOH = 1.88, pKaNH3+ = 9.60, pKa sidechain = 3.65 For glycine, pKaCOOH = 2.34, pKaNH3+ = 9.60 For glutamate, pKaCOOH = 2.19, pKaNH3+ = 9.67, pKa sidechain = 4.25 For histidine, pKɑCOOH = 1.82, pKɑNH3+ = 9.17, pKa sidechain = 6.00 For tyrosine, pKaCOOH = 2.20, pKaNH3+ = 9.11, pKa sidechain = 10.07 a. Of these possible buffers, which is likely to yield the greatest resistance to pH change from the starting pH of 2.0, assuming that you were going to run a biochemical reaction that…

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