Many biochemists go bananas, and justifiably, when they see a Michaelis-Menten plot like the one shown here.Vmaxол[S]To better understand their outrage, calculate V as a fraction of Vmax when [S] = 10 Kè and [S] = 20 KM.MWhen [S] = 10 KM, Vo=0.91VmaxWhen [S] = 20 KM, Vo =95IncorrectVmax.

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

Answered: Many biochemists go bananas, and…

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 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…
Does this calculation look correct? My goal isto have a target mass of 10 µg of my protein with a total volume of 30 µl. Protein was measured and found to have a concentration of at 308.35 µg/ml Recall, mass = concentration * volume 10 µg = 308.35 µg/ml * V V = 0.01 mg/ 0.30835 mg/mL = 0.032430679 mL = 32.43067942 µL Note: cannot measure this amount; two dilutions required. Dilute by 1/40: measure 2.5 µl of protein at 0.30835 µg/µl and add it to 97.5 diluent to prepare a solution of protein at 7.71 mg/ml. This diluted sample will be used further. Mass = concentration * volume 10 µg = 7.71 mg/ml * V V = 0.01 mg/ 7.71 mg/ml = 1.3 µL Therefore, will measure 1.3 µl of protein at 7.71 mg/ml and add it to 28.7 µl of diluent.
you are trying to come up with a drug to inhibit the activity of an enzyme thuogth to have a role in a liver disease. in the lab the enzyme was shown to have a Km of 1x10-6 M and Vmax of 0.1 micromoles/min.mg measruing at room temperature. you developed a mixed non-competitve inhibitor with a ki=0.4x10-6M and a Ki` pf 0/2 x 10-5 What will be the apparent Km in the presence of 1.0x10-6 M?
You are super jazzed because your protein is purified and at a concentration of 600 μg in 1.75 ml of buffer. Convert this concentration to mM. Note that biochemisfunase has a molecular weight of 54,321 Daltons (which is just another unit for g/mol). Show your calculations.
Why is ΔG a useful quantity for predicting the favorability of biochemical reactions?
If 287.9 umol of enzyme X has a Vmax = 47.8 mmol/sec, what is the value of kcat %3D sec-1? Please report answer with 1 decimal place. Please do not report units. Your Answer: Answer units MacBook Air 888 F5 F4 F3 F2 %23 %24 %24
You are interested in purifying an enzyme X and decide to use an affinity chromatography followed by an enzymatic assay. You obtain the following data: Total volume of [protein] the sample (ml) (mg/mL) (U) |Total activity Sample name Specific activity |(U/mg) |Affinity | chromatography 2 0.12 5000 20833 You have obtained a 10X fold purification and a yield of 10% at the Affinity step. What was the protein concentration of your crude lysate sample, if its volume was 15 mL? Detail your answer by showing the calculations, and do not forget the UNITS!
Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km?
Quantitative Estimation of Amino Acids by Ninhydrin http://vlab.amrita.edu/?sub=3&brch=63&sim=156&cnt=2 can u help me with question 2 of the assignment questions Based on the experimental data provided, estimate the amount of amino acid in the given unknown solution by Ninhydrin method. SI No. Volume of standard amino acid solution (ml) Amount of amino acid (µg) OD at 570nm 1 Blank 0 2 0.2 0.12 3 0.4 0.25 4 0.6 0.45 5 0.8 0.55 6 1.0 100 0.68 7 Unknown (0.5ml) 0.41
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.
Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km? Vmax = 5 umol min^-1, Km = 2.5 mM? Vmax = 5 mmol min^-1, Km = 25 M? Vmax = 5 umol min^-1, Km = 25 mM? Vmax = 5 mol min^-1, Km = 2.5 mM? Vmax = 5 mol min^-1, Km = 25 mM?
Prepare a schematic diagram and present it as though it were a Figure in a publication (scientificjournal or textbook) (20%). The Figure should illustrate the interactions made between thekey components of (a) total and (b) non-specific binding reactions. In preparing your figure,you should reflect on the role of each of the components of the reaction mixtures, and whythe subtraction of non-specific from total binding allows us to calculate specific binding.

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