1. Calculate Vmax and Km???
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![Substrate
Initial velocity Vo
concentration [S] µM s-1
10
0.13
25
0.27
50
0.45
100
0.65
150
0.77
200
0.85
300
0.94
500
1.03
1. Calculate Vmax and Km???
The enzyme is dissolved at a concentration of 1
mg/ml and has a molecular weight of 50 kDa.
a) What is the turnover number of the enzyme?
b) What is the specificity constant of the
enzyme?
c) Is the enzyme operating near the maximum
rate possible, i.e. is the enzyme diffusion
controlled?
please show all working and formulas as I am
trying to learn. thank you](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe97b099-d6ed-461e-99fb-0b0ad0d97cf4%2F1a7f26a8-57dc-425b-abe9-10fe9e84cf01%2Fgpm68un_processed.jpeg&w=3840&q=75)
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- The turnover number for an enzyme is known to be 5000min-1. Given the following set of data, Substrate concentration(mM) 1, 2, 4, 6, 100, 1,000 Initial Rate(micromol/min) 167, 250, 334, 376, 498, 499 a) What is the Km of the enzyme for the substrate? (do this without using a calculator) b) What is the total amount of enzyme present in the assay?Calculate the Km of the enzyme with these parameters. kcat = 130s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µMA 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).
- For some Enzyme, the Vmax is 18 micromols/min, Km is 400 microM. If the substrate concentration is 100 microM, what is the velocity of the reaction?What is the concentration of enzyme (in mM) needed to achieve a Vmax of 8.00 mM/s if the enzyme has a turnover number of 1.50s-1?If Vmax for a reaction is 10 μM ⋅ s-1 and the KM is 0.5 μM, what is the reaction velocity when the substrate concentration is 2 μM?
- An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min-1 when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme in µmol⋅min−1?The turnover number for an enzyme that approximates Michaelis-Menten kinetics is known to be 500 min^-1. From the results shown in the table, enumerate Km and total amount of enzyme present. What is the Km for this enzyme? What is the Vmax for this enzyme? And what is the [E]T for this enzyme?An enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :
- 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 %24Answer the following questions, showing work in the form of calculations. a) At what substrate concentration would an enzyme with a Vmax of 102.5 M/min and a Km of 0.0025 M operate at one-quarter of its maximum rate? b) Determine the fraction of Vmax that would be obtained at the following substrate concentrations [S]: 0.5Km, 2Km, and 10Km.Calculate the turnover number for an enzyme, assuming Vmax is 0.5 M.sec-1 and the concentration of the enzyme used is 0.002 M. Why is it useful to know this?