Question

In: Chemistry

Explain what Vmax and Km describe

Explain what Vmax and Km describe

Solutions

Expert Solution

Vmax is the reaction rate when the enzyme is fully saturated by substrate, indicating that all the binding sites are being constantly reoccupied.

Km is the concentration of substrate which permits the enzyme to achieve half Vmax.

Vmax : if concentration of substrate is incresed, more and more enzyme molecules are working. At half maximal velocity 50% enzymes are attached to the substrate. As more substrate is added, all enzyme molecule are saturated(utilized). Further increase in substrate cannot any effect in reaction velocity. This maximum velocity obtained is called Vmax.

Km : km is a value of substrate concentration at half maximal velocity.

It denotes that 50% of enzyme molecules are bound with substrate molecules at particular substrate concentration.

Km is independent of enzyme concentration.

If enzyme concentration if doubled, the Vmax will be double. But Km will remain exactly same.

Km is signature of enzyme. Km value is constant for an enzyme.

Km denotes the affinity of enzyme for substrate.. The lesser numerical value of Km, the affinity of enzyme for substrate is more.


Related Solutions

BIOCHEMISTRY: Vmax and Km values For the following problem
BIOCHEMISTRY: Vmax and Km values For the following problem
How will you describe and contrast on enzyme activity based on Km and Vmax from Michaelis...
How will you describe and contrast on enzyme activity based on Km and Vmax from Michaelis menten equation? (low or high Km/Vmax) what does it mean when Km is High or low what does it mean when Vmax is High or Low
Determine the Vmax and the Km of the enzyme in the absence and presence of each...
Determine the Vmax and the Km of the enzyme in the absence and presence of each inhibitor. Calculate the KI for each inhibitor. Are the inhibitors all the same? Which is the best inhibitor and why? What is the catalytic efficiency and turnover number of the enzyme knowing that during the experiments its concentration was 2 nM? What happens to the catalytic efficiency and turnover number of the enzyme in the presence of each of the inhibitor? The following velocity...
The kinetics of chymotrypsin were studied and yielded a KM of 5 mM and a Vmax...
The kinetics of chymotrypsin were studied and yielded a KM of 5 mM and a Vmax of 0.2 mM/min. The enzyme concentration was 0.20 μg/ml. Calculate the turnover number, assuming one binding site per enzyme.
How does a competitive inhibitor change the Km and Vmax values for an enzyme such as...
How does a competitive inhibitor change the Km and Vmax values for an enzyme such as Penicillin?
An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted...
An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted with a irreversible active site specific inhibitor. After reaction with the inhibitor, the enzyme was assayed using a 2 mM concentration of substrate, and it gave a reaction rate of 20 umol/min. What percentage of the enzyme did the inhibitor inactivate?
1-Describe and explain the relationship between number of carriers and Vmax. 2-Describe and explain the relationship...
1-Describe and explain the relationship between number of carriers and Vmax. 2-Describe and explain the relationship between number of carriers and Km.
Q2. An enzyme has a Km of 2.8 * 10-5 M and a Vmax of 53...
Q2. An enzyme has a Km of 2.8 * 10-5 M and a Vmax of 53 μM/s. (i) Calculate Vo if [S] = 3.7 * 10-4 M and [I] = 4.8 *10-4 M for (a) a competitive inhibitor, (b) a noncompetitive inhibitor and (c) an uncompetitive inhibitor. (Assume Ki = 1.7 * 10-5 M.) (ii) The degree of inhibition is given by i (%) = 100(1 - Vi/Vo). Calculate the percent inhibition in each of the three cases above.
An enzyme catalyzed reaction has a Km of 3.76 mM and a Vmax of 6.72 nM/s....
An enzyme catalyzed reaction has a Km of 3.76 mM and a Vmax of 6.72 nM/s. What is the reaction velocity in nM/s, when the substrate concentrations are: A. 0.500 mM B. 15.6 mM C. 252 mM D. Now assume you have added a competitive inhibitor that has a concentration of 15.6 µM with a KI of 7.30 µM to the enzyme in question 9. Calculate the velocity at the same substrate concentrations as above: E. 0.500 mM F. 15.6...
When Vo = 0.5× Vmax, [S] = 1× Km. Which of the following is true when...
When Vo = 0.5× Vmax, [S] = 1× Km. Which of the following is true when [S] = 4× Km? V0 = 40% of Vmax none of these V0 = 8% of Vmax Vo = 80% of Vmax Vo = 4% of Vmax
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT