Question

THANK YOU 30- A species of Shewanella bacteria contains an enzyme that catalyzes the dehalogenation of...

THANK YOU

30- A species of Shewanella bacteria contains an enzyme that catalyzes the dehalogenation of tetrachloroethene. The KM is 120 μM and the Vmax is 1.0 nmol · min−1 · mg−1. What is the substrate concentration when the velocity is 0.75 nmol · min−1 · mL−1?

Homework Answers

Answer #1

Michaelis–Menten equation is used typically in enzymatic reactions.

V = Vmax*[S] / (Km + [S])

Where

Vmax = max rate velocity

[S] = substrate concentration

Km = Michaelis–Menten constant

V = reaction rate


Michaelis–Menten equation is used typically in enzymatic reactions.

V = Vmax*[S] / (Km + [S])

Where

Vmax = max rate velocity

[S] = substrate concentration

Km = Michaelis–Menten constant

V = reaction rate

then...

if we need

[S] = ? for V = 0.75; get

V = Vmax*[S] / (Km + [S])

change KM units from micromol to nanomol --> (120 microm = 120*10^3 nanom)

0.75 = 1*[S] / (120000+ [S])

0.75 *120000 + 0.75 *[S] = [S]

0.25*[S] = 90000

[S] = 90000 / 0.25

[S] = 360000 nanoM = 360.000 micro M = 360 microM

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 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.
An enzyme catalyzes the reaction A -> B +H2O. If km for this reaction is 100uM...
An enzyme catalyzes the reaction A -> B +H2O. If km for this reaction is 100uM and vmax is 4umol/min/mg enzyme, calculate the amount of B formed in 2.0 min by 1.0 mg enzyme with 50uM A in solution.
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
An enzyme has a KM of 47 μM. If the Vmax of the preparation is 22...
An enzyme has a KM of 47 μM. If the Vmax of the preparation is 22 μmol∙L-1∙min-1, what velocity would be observed in the presence of 7 μM substrate and 4 μM of an uncompetitive inhibitor that has a KI of 3 μM.
An enzyme is found that catalyzes the reaction A ↔B. In an experiment with total enzyme...
An enzyme is found that catalyzes the reaction A ↔B. In an experiment with total enzyme concentration of 5 nM, Vmax is 1.5 μM s^-1. What is kcat for the enzyme. Give your answer as a number only in units of s^-1.
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
(A) Consider the process: Bruce Wayne ⇌ Batman This reaction is catalyzed by the enzyme GOTHAM,...
(A) Consider the process: Bruce Wayne ⇌ Batman This reaction is catalyzed by the enzyme GOTHAM, which, for this reaction has a kcat of 500 s-1. If the concentration of GOTHAM is 10 nM and the concentration of Bruce Wayne is 20 μM, what is the KM is the initial reaction rate is 3.1 μM s-1? (B) The enzyme GOTHAM also catalyzes the following reaction: Selena Kyle ⇌ Catwoman The KM for this reaction is 120 μM and kcat is...
A biochemist obtains the following set of data for a mouse enzyme that is known to...
A biochemist obtains the following set of data for a mouse enzyme that is known to follow Michaelis- Menten kinetics. The experimental conditions were with 0.1 μmol of enzyme and pH 6.0. Substrate Initial concentration velocity (μM) (μmol/min) and Initial velocity 1 49 2 96 8 349 50 621 100 676 1,000 698 5,000 699 Question 1. Estimate what would be the initial velocity in μmol/min at a substrate concentration of 10,000 μM. Question 2. What is the approximate Vmax...
The following rate constants were determined for a Michaelis-type enzyme: k1 = 2.5 x 10^9 M-1...
The following rate constants were determined for a Michaelis-type enzyme: k1 = 2.5 x 10^9 M-1 s -1 , k-1 = 5.0 x 10^3 s -1 , and k2 = 5.0 x 10^4 s -1 . a. What are the values of Km, Ks, and kcat for this enzyme? Is this a rapid equilibrium enzyme or does it just follow steady state kinetics? Explain the basis for your answer. b. What substrate concentration is needed to achieve half maximal velocity?...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT