Question 1. The following substrates are metabolized by the same enzyme, Acetylcholinesterase. Use the table below to answer the following questions: Substrate Km (M) kcat (S-1) kz/Km A 4.0 x 10-4 1.6 x 10+ 4.0 x 107 B 6.7 x 10-2 1.75 x 104 2.6 x 105 C 1.1 x 10-3 9.5 x 103 8.6 x 106 a. Organize the substrates from lowest to highest binding affinity. Briefly explain your answer. b. Organize the substrates from slowest to fastest product formation. Briefly explain your answer. c. For which substrate does the enzyme have the highest catalytic efficiency and why?
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- Consider the following set of data and answer the foStaphylococcal nuclease has a ΔΔG‡ of -84.1 kJ mol-1 at 25.0 °C. If the uncatalyzed rate is 0.630x10-13 µmol s-1, calculate the enzyme-catalyzed rate in µmol s-1. (Use R = 8.3145 J mol-1 K-1)Question 2. Answer the following questions: A. The following experimental data was collected during a study of the catalytic activity of anintestinal peptidase with the substrate glycylglycine. Plot the data as a graph, and use it toestimate the Km and the Vmax for this enzyme. B. Now transform this data to plot it as a straight line (Lineweaver-Burk plot). Determine Km andthe Vmax for this enzyme using this new plot. Do your results agree with the estimates made fromthe first graph of the raw data (from 2A)? C. Now assume that the activity of this intestinal peptidase is regulated by covalent modificationof its catalytically active amino acid. Upon phosphorylation, the Km of the catalyzed reaction has been observed to increase by a factor of 3 without any effect on its Vmax. Is the enzyme getting activated or inhibited upon phosphorylation? Justify your answer. D. How will the Lineweaver-Burk plot of the phosphorylated enzyme differ from the plot of the unmodified enzyme (from 2B)?…
- Please help me with this question immediately within an hourcan you explain what is happening in the attached graph? Figure 3: Lineweaver-Burke plot of alkaline phosphatase activity without inhibitorOne of your colleagues has obtained a sample of muscle phosphorylase b that is known to be relatively inactive. She has approached you for advice on how to set up an appropriate assay. She has the following items available, not all of which are appropriate for this study. Help her out by selecting the items that she should use and what their purpose is in the assay, and then explain why each of the other items would not be useful. 1. 100 UM AMP 2. 100 UM GTP 3.100 uM glucose 4. 100 uM glucose 6-phosphate 5. Branched glycogen 6. Amylose (i.e. unbranched glycogen) 7.50 mM HEPES buffer, pH 7.5 8. 50 mM potassium phosphate buffer, pH 7.5 Write out a short explanation for each of the above items, and upload your answers by the due date.
- National Board of Medical Examiners Biochemistry Mark 36. In the presence of a metabolite (X), 6-phosphofructokinase is assayed at a fixed concentration of ATP and varying concentrations of fructose 6-phosphate. The resulting data are shown in the table. Fructose 6-phosphate (pM) 5 10 20 40 75 100 200 Velocity umoles/min 0.05 0.15 0.25 0.70 1.7 2.2 3.1 3.1 Velocity (+X) umoles/min 0.006 0.025 0. 10 0.35 1.03 16 2.9 3.1 400 Metabolite (X) is most likely which of the following substances? O A) ADP O B) AMP OC) CAMP D) Citric acid O E) Fructose-2,6-bisphosphateAnswer for D, and E. The answers for A, B, and C are down below.Michaelis Menton Plot and a Lineweaver-Burke plot using this data
- Question Is number 2 in photosCalculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. (inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM) 0.20 per minute 0.50 per minute 1.0 per minute 2.0 per minute 5.0 per minuteTABLE 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…