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
82.21
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…
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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<
Please dont provide handwrittin solution.../.
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…
14) Calculate K’eq and ΔG’0 for the following: A 0.1 M solution of glucose 1-phosphate at 25oC is incubated with a catalytic amount of phosphoglucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are Glucose 1-phosphate (4.5 X 10-3 M) and Glucose 6-phosphate (9.6 X 10-2 M).
5) In the below two step transformation, the first step AG is positive i.e. 1.7 kJ/mol. Yet this two step coupled reaction takes place to form Fructose-1.6-bisphosphate. Explain why?" C-H CH-OH CH,-0 H-C-OH AG"- 1.7 KJ/mol (+0.4 kcalmol) C=0 AG"=-14.2 k/mol(-3.4 kcalmol) HO-C-H HO-C-H HO-C-H H-C-OH H-C-OH H-C-OH H-C-OH H-C-OH ATP ADP H-C-OH CH-0-P. CH,-0-P-o- CH,-o- Glucose-6-phosphate Fructose-6-phosphate Fructose-1,6-bisphosphate
A bacterial culture is grown using either octadecane (C18H38) or pentachlorophenol (C6HOCl5)) as the sole source of carbon and energy. The cell yield value is determined by dry weight analysis to be 1.49 for octadecane and 0.05 for pentachlorophenol. Using either octadecane or pentachlorophenol, please describe the steps taken (and the final result) showing what percentage of the substrate carbon will be found as cell mass and as CO2? Please show all calculations and give correct answers.
PTP1B Substrate kcat Km kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEPYLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 ± 0.22 0.53 + 0.02 34.7 + 0.25 52.7 ± 0.7 0.066 + 0.001 (d) (. ) The units for kcat/KM in the above are given according to standard scientific notation. On this basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?
Describe reaction between AP and PNPP and how is used to determine the kinetic parameters of AP. i. İndicate the value for the activity of AP(which is 3.08) What does it mean? Is your enzyme active or not? What does it say about the purity of your enzyme? ii. How the purification process affects the activity?

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