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3-2-1 Code It

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ISBN:9781337660549

Author:GREEN

Publisher:GREEN

Chapter17: Cpt Pathology And Laboratory

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The conversion of glucose into glucose 6-phosphate, which must occur in the breakdown of glucose, is thermodynamically unfavorable (endergonic). How do cells overcome this problem?
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m BIO101 Short EXAM First session X G Google om/moodle/mod/quiz/attempt.php?attempt%3D1313727&cmid%3D298536 e on the bookmarks bar. Import bookmarks now... AL BIOLOGY I/ ill courses GENERAL BIOLOGY I BIO101 Short EXAM First sess Time left 0:21:42 If an enzyme in solution is saturated with its substrate, the most effective way to obtain a higher reaction rate is to: O a. add more substrate O b. heat the solution to 95°C O c. add more of the enzyme O d. All of the options are correct O e. add a noncompetitive inhibitor Next page vity Jump to.. ents 51°F Light rair
1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…
Briefly comment on the differences of using a fixed-time assay versus a kinetic assay to measure enzyme activity. Is it reasonable to assume that the reaction velocity obtained by measuring the amount of product after 30 minutes in a fixed-time assay is directly proportional to absorbance? How could you determine whether this was the case? Word limit 180 words including citation and reference
Part 1: Assess the following partial results section below by editing it for brevity by omitting any unnecessary parts (1 point), explain why you decided to remove certain sections (1 point): To evaluate inhibitory effects of the selected molecules, 10mM stock solutions of each molecule were prepared in DMSO. A reaction mixture (200μl) was prepared with the same formula optimized for the enzyme activity assay (0.1 M Tris-HCl ph 8, 0.1 M KCI, 25 mM NaCl, 0.25 mM ATP, and two units of inorganic yeast pyrophosphatase) with 10 µM of the sample molecule. The reaction mixture was incubated for 20 minutes at ambient temperature. Enzymatic reaction was triggered by addition of the substrate B (0.2 mM) and the absorbance of the product was monitored at 290 nm for 10 minutes. Six out of 15 sample molecules showed appreciable inhibition at 10 μM (Figure 5). Three of the molecules, A3, A6, and A7 exhibited more than 50% inhibition of the enzyme activity and were further diluted to find the minimal…
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-bisphosphate
Below is kinetic data obtained for an enzyme-catalyzed reaction. The enzyme concentration is fixed at 100 nM. Using a Lineweaver-Burke plot, calculate the Vmax value for this reaction. Report your answer to four significant figures in units of uM/min.
Asap I need both answer. Please explain both why answer is correct. Advance thank you
(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
Using the attachment, Answer the following questions: Prepare a double reciprocal plot with all three experiments (lines) on the same graph. Use your graph, and then answer items 2, 3, and 4 below. 1. Calculate the Vmax or apparent Vmax for all three sets of data. Likewise, calculate the Km or apparent Km for each set. 2. For Inhibitor X, what is the mode/type of inhibition? 3. For Inhibitor Z, what is the mode/type of inhibition? By comparison of the apparent Vmax to the control Vmax, what is the value of α’, as defined in class? If Ki’ = 10 mM for this inhibitor, then what must the inhibitor concentration [Z] be?
You are required to isolate, purify and verify the purity of an enzyme from cytosol of a plant species for industrial use. Briefly describe how the desired enzyme can be obtained for the intended purpose stating the techniques involved. (MW of enzyme = 60 kbp)
J. C. Servaites, in Plant Physiol. (1985) 78:839–843, observed that Rubisco from tobacco leaves collected before dawn had a much lower specific activity than the enzyme collected at noon. This difference persisted despite extensive dialysis, gel filtration, or heat treatment. However, precipitation of the predawn enzyme by 50% (NH4)2SO4 restored the specific activity to the level of the noon-collected enzyme. Suggest an explanation.
An enzyme has a V of 1.2 uM s The Km for its substrate is 10 µM. max Calculate the initial reaction velocity, Vo, for each substrate concentration, [S]. Calculate Vo when [S] is 2 µM. Vo = µM s Calculate Vo when [S] is 10 µM. Vo = µM s-1 Calculate Vo when [S] is 30 µM. Vo = µM s-1

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