Question

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 formed?

Homework Answers

Answer #1

The enzyme converts 1% of substrate to product in 5 minutes

product concentration = 30 micromoles

the enzyme concentration is increased twice and substrate is also increased twice of initial amount.

The increase in concentration or amount of substrate will not affect the conversion rate. however the enzyme will do.

an enzym can convert substrate to 30micromoles in 5 minutes. The double amount of enzyme will require = 5/2 minutes

So the time will be =2.5 minutes

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
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 3500 times greater than the...
An enzyme-catalyzed reaction was carried out with the substrate concentration initially 3500 times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 micro-moles. If, in a separate experiment, the same amount of substrate but six times as much enzyme had been combined in the same volume, how long would it take for the same amount (12 micro-moles) of...
An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than...
An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 mmol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 mmol) of product to be formed?
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. 
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
The following parameter(s) of an enzyme-catalyzed reaction depend(s) on enzyme concentration: A) Km (Michaelis constant) B)...
The following parameter(s) of an enzyme-catalyzed reaction depend(s) on enzyme concentration: A) Km (Michaelis constant) B) Vmax (maximum velocity) C) kcat (turnover number) D) A and B E) B and C Km is the equivalent of: A) Substrate concentration at Vo B) Substrate concentration when Vmax is reached C) Substrate concentration when 1/2 Vmax is reached D) Product concentration when 1/2 Vmax is reached
The zero order rate is seen in an enzyme catalyzed reaction when: A. There is too...
The zero order rate is seen in an enzyme catalyzed reaction when: A. There is too much enzyme and too little substrate B. Enzyme reactions are always first order C. The Michaelis constant is very high D. The Vmax of the reaction is twice that which is observed E. The enzyme substrate comples is at its maximum concentration What happens when you add more enzyme to a catalyzed reaction that already is proceeding at Vmax? A. The reaction is inhibited...
(1 point) Two chemicals A and B are combined to form a chemical C. The rate...
(1 point) Two chemicals A and B are combined to form a chemical C. The rate of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially there are 75 grams of A and 24 grams of B, and for each gram of B, 1.7 grams of A is used. It has been observed that 24.75 grams of C is formed in 15 minutes. How much is formed in...
Procedure Reaction 1: Dissolving the Copper 1. Obtain a clean, dry, glass centrifuge tube. 2. Place...
Procedure Reaction 1: Dissolving the Copper 1. Obtain a clean, dry, glass centrifuge tube. 2. Place a piece of copper wire in a weighing paper, determine the mass of the wire and place it in the centrifuge tube. The copper wire should weigh less than 0.0200 grams. 3. In a fume hood, add seven drops of concentrated nitric acid to the reaction tube so that the copper metal dissolves completely. Describe your observations in the lab report. (Caution, Concentrated nitric...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT