(a) Estimate the Km and Vmax for the wild-type and mutant enzyme from the graph. (b) Calculate the Keat and Keat/Km for the wild-type and mutant enzyme based on your estimated values in (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.
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Michaelis-Menten Plot Problem Help
See Picture
![3. Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from
the bacterium Staphylococcus aureus. The enzyme is involved in carbohydrate metabolism and is a
potential biocatalyst for the large-scale production of rare sugars.
-1
v, μmol s¹ mg-¹
1.50-
1.25-
1.00-
0.75-
0.50-
0.25-
0.00+
0
100
→ WT
200 300
[s], mM
400 500
→ V105A
(a) Estimate the Km and Vmax for the wild-type and mutant enzyme from the graph.
(b) Calculate the Keat and Keat/Km for the wild-type and mutant enzyme based on your estimated values
in (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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff800a840-0309-4834-92a3-4244363b7afa%2F29cbd316-83b0-41e4-8907-9fc0b5a41f14%2Frrh3oyom_processed.jpeg&w=3840&q=75)
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- 4. Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from the bacterium Staphylococcus aureus. The enzyme is involved in carbohydrate metabolism and is a potential biocatalyst for the large-scale production of rare sugars. (8) mg V₂ μmol s1 1.50- 1.25- 1.00- 0.75- 0.50- 0.25- 0.00 0 100 200 300 [s], mM 400 500 WT (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 values in (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.3. Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from the bacterium Staphylococcus aureus. The enzyme is involved in carbohydrate metabolism and is a potential biocatalyst for the large-scale production of rare sugars. -1 v, μmol s¹ mg-¹ 1.50- 1.25- 1.00- 0.75- 0.50- 0.25- 0.00+ 0 100 → WT 200 300 [s], mM 400 500 → V105A (a) Estimate the Km and Vmax for the wild-type and mutant enzyme from the graph. (b) Calculate the Keat and Keat/Km for the wild-type and mutant enzyme based on your estimated values in (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.Chitinase 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?
- A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).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-bisphosphatePlease note the following Lineweaver-Burk plot for the enzyme Virbraniumase reacting with a substrate: -0.4 . 1/V 5 4 3 2 -0.2 0 0.2 y = 5.2781x + 1.3338 R² = 0.9967 0.4 0.6 0.8 1/[S] 1 Based on the information provided, what is the Vmax for this reaction?
- Aerobic degradation of an organic compound by mixed cultureof organism in wastewater can be represented by following reaction. C3H6O3 + a O2 + b NH3 → c C5H7NO2 + d H2o + e CO2 A. Determine a, b, c, d and e, if YX/S = 0.4 d X/g S. B. Determine the yield coefficients YX/O2 and YX/NH3. C. Determine the degree of reductions for the substrate, bacteria and RQ for the organismsThe table below shows the purification of liver lactate dehydrogenase. Purification Table of liver LDH Step Crude Extract Ammonium Sulfate DEAE-Sephadex 5'AMP-Sepharose Volume (mL) 200 100 112 40 LDH activity (Units/mL) (mg/mL) [Protein] 47.0 50.00 90.0 63.00 51.0 1.50 137.0/ 0.4 Calculate the LDH yield for the ammonium sulfate step. Provide answer to one decimal only. Show Transcribed Text this is biochemistry and the topic is protein purification could you pleaae show me the full workings even the workings for the comversions(nmol/min) 30 31 32 33 34 35 A student performed a lactate dehydrogenase (LDH) assay multiple times with varying amounts of substrate added to the reaction mixture. All other conditions were kept the same. This graph shows the results observed: 100 75 10 20 30 40 Lactate concentration (nmol/ml) Why is this student not detecting a linear increase in reaction rate at higher substrate concentrations? T TTArial v 3 (12pt) * T.=,三, 日iu 4:03 PM to search 5/6/2021
- The effect of temperature on the hydrolysis of lactose by a ß-galactosidase is shown below in Table 1. The temperature coefficient, Q10 is the factor by which the rate increases by raising the temperature 10°C. The universal gas constant, R is 8.314 J/mol.K. (a) (b) Table 1: Data of Vmax over temperature T (°C) 20 30 35 40 45 Vmax (umoles/min.mg protein) 4.50 8.65 11.80 15.96 21.36 Plot the graph of In Vm vs 1/T using any spreadsheet software (include all appropriate labels and equation). Calculate the activation energy Ea and temperature coefficient Q10.41 The following data describe an enzyme-catalyzed reaction (hydrolysis of carbobenzoxyglycyl-L-tryptophan) Plot these results using the Lineweaver-Burk method, and determine values for KM and Vmax Velocity (mM.sec-) 0 024 0 036 0 053 0 060 0 061 0 062 Substrate Concentration (mM) 25 50 10 0 15 0 200 25 0 42 If the KM of an enzyme for its substrate remains constant as the concentration of the inhibitor increases, what can be said about the mode of inhibition and why? 43 Calculate the turnover number for an enzyme, assumıng Vmax IS 05 M sec1 and the concentration of the enzyme used is 0 002 M Why is it useful to know this? 44 Dıscuss the mechanism of the Bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues Make use of relevant graphs and diagrams to explain your answerThe glucose oxidase used in this experiment has a concentration of 27 U/mg. Calculate the mass (in g) of solid glucose oxidase required to react with all the glucose in 4.00 µL of10.0- mM glucose sample assuming the reaction proceeds for 30 minutes.