Question

The velocity of a reaction is 40 µM/min at a substrate concentration of 2.0 mM. The...

The velocity of a reaction is 40 µM/min at a substrate concentration of 2.0 mM. The enzyme is known to reach maximal velocity at 240 µM/min. What would the velocity be at a substrate concentration of 5 mM?

Science   Biology   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Please look in the image for solution:

Velocity = 80 Micromolar/minutes or 0.080 Millimolar minutes.

please rate high.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The velocity of a reaction is 40 µM/min at a substrate concentration of 1.5 mM. The...
The velocity of a reaction is 40 µM/min at a substrate concentration of 1.5 mM. The enzyme is known to reach maximal velocity at 220 µM/min. What would the velocity be at a substrate concentration of 5 mM?
For an enzyme-catalyzed reaction, the initial velocity was determined at two different concentrations of the substrate....
For an enzyme-catalyzed reaction, the initial velocity was determined at two different concentrations of the substrate. Which of the following would be closest to the value of Vmax? For an enzyme-catalyzed reaction, the initial velocity was determined at two different concentrations of the substrate. Which of the following would be closest to the value of Km? [S] (mM) Vo(mM/min) 1.0 2.0 4.0 2.8
a) The Km of an enzyme of an enzyme-catalyzed reaction is 7.5 µM. What substrate concentration...
a) The Km of an enzyme of an enzyme-catalyzed reaction is 7.5 µM. What substrate concentration will be required to obtain 65% of Vmax for this enzyme? (same enzyme was used in part a and b) b) Calculate the Ki for a competitive inhibitor whose concentration is 7 x10-6 M. The Km in the presence of inhibitor was found to be 1.2x10-5 M. 
An enzyme catalyzes the reaction A --> B. The enzyme is present at a concentration of...
An enzyme catalyzes the reaction A --> B. The enzyme is present at a concentration of 2 nM, and the Vmax is 1.2 mm/s. The Km for substrate A is 10 mM. Calculate the initial velocity of the reaction, Vo, when the substrate concentration is a) 4 mM, b) 15 mM, c) 45 mM.
20. You run a reaction with 1 µg of enzyme QB26, buffer and 25 millimolar (mM)...
20. You run a reaction with 1 µg of enzyme QB26, buffer and 25 millimolar (mM) substrate and obtain an initial reaction velocity (vo) of 100micromolar (µM) product per second. You repeat the reaction the same as before but with 50 mM substrate and observe vo = 101µM/sec, repeat at 100mM substrate, get vo =102µM/sec. a. The substrate concentrations in these reactions are very close the Km b. The velocities in these reactions are very close to Vmax c. The...
An enzyme is discovered that converts substrate A to product B. The molecular weight of the...
An enzyme is discovered that converts substrate A to product B. The molecular weight of the enzyme is 10000 Daltons. The Km and Vmax values for the enzyme are shown below: Km (mM) 2 Vmax (mmole/min/mg) 20 What is the reaction initial velocity for this enzyme when the concentration of A is 4 mM? What is the turnover number of this enzyme?
An enzyme catalyzes a reaction with an initial velocity of 50 micromoles/litre-seconds when the substrate concentration...
An enzyme catalyzes a reaction with an initial velocity of 50 micromoles/litre-seconds when the substrate concentration is 5 micromolar and 80 micromoles/litre-seconds when the substrate concentration is 10 micromolar. The Vmax and Km of this enzyme are: A) 50 micromoles/litre-seconds; 5 micromolar B) 80 micromoles/litre-seconds; 10 micromolar. C) 200 micromoles/litre-seconds; 15 micromolar D) 100 micromoles/litre-seconds; 12.5 micromolar E) 10 micromoles/litre-seconds; 1 micromolar
The following data were obtained for the variation of initial velocity and substrate for a reaction...
The following data were obtained for the variation of initial velocity and substrate for a reaction catalyzed by chymotrypsin (5nM) at pH 8, 37 °C. 1. Using a linear plot, determine the KM and Vmax of the chymotrypsin given the data below: (12) [S] (μM) V0 (μmol/min) 0    0 0.5 200 1.0    400 1.5    580 2.0    750 2.5    840 3.0    860 4.0    875 6.0    890 The kcat/KM parameter is a measure of...
An enzyme-catalyzed reaction was carried out with the substrate concentration initially three hundred times greater than...
An enzyme-catalyzed reaction was carried out with the substrate concentration initially three hundred times greater than the Km for that substrate. After five minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 15 micro moles. If, in a seperate experiment, four times less enzyme and twice as much substrate had been combined, how long would it take for the same amount (15 micro moles) of product to be...
An enzyme-catalyzed reaction was carried out with the substrate concentration initially two hundred times greater than...
An enzyme-catalyzed reaction was carried out with the substrate concentration initially two hundred times greater than the Km for that substrate. After 5 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 30 micro mol. If, in a seperate experiment, TWO TIMES MORE enzyme and twice as much substrate had been combined, how long would it take for the same amount (30 micro mol) of product to be...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT