Question

In: Chemistry

2) The initial rate for an enzyme-catalyzed reaction has been determined at a number of substrate...

2) The initial rate for an enzyme-catalyzed reaction has been determined at a number of substrate concentrations. Data are given below. [S](mu*M) 5 10 20 50 100 200 500 vo (mu*Mmin-1) 21 62 92 126 185 210 217 a) Estimate Vmax and Km from a direct graph of vo vs [S] (Michaelis-Menton plot). b) Use a Lineweaver-Burk plot to analyze the same data. What are the values of Vmax and Km? c) If the total enzyme concentration is 100 pM, what is kcat? d) Calculate kcat/Km for this enzyme. Is the enzyme an efficient enzyme?

Expert Solution

2) (a) Shwon below is the Michaelis-Menton plot for the given data point.

plot, [S] on x-axid and Vo on y-axis

Vmax = 217 mu.M/min

Km = 35 mu.M

(b) Shown below is the Lineraweaver-Burke plot

Plot 1/[S] on x-axis and 1/Vo on y-axis

y-intercept = 1/Vmax = 0.0364

Vmax = 27.47 mu.M/min

x-intercept = -1/Km = -0.0056

Km = 178.57 mu.M

(c) when [E] = 100 pm = 1 x 10^-4 mu.M

Kcat = Vmax/[E] = 27.47/1 x 10^-4 = 274,700 min-1

(d) For this enzyme Kcat/Km = 274,700/178.57 = 1538.33 1/mu.M.min

The enzyme is an efficient enzyme.

Plots given below

[S](mu.M)	Vo(muM/m)	1/[S]	1/Vo
5	21	0.2	0.04762
10	62	0.1	0.01623
20	92	0.05	0.01087
50	126	0.02	0.00794
100	185	0.01	0.005405
200	210	0.005	0.004762
500	217	0.002	0.004608

queen_honey_blossom answered 2 years ago

1. (3%) You measure the initial rate of an enzyme reaction as a function of substrate...

1. (3%) You measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and absence of an inhibitor. The following data are obtained: [S] V0 –Inhibitor +Inhibitor 0.0001 33 17 0.0002 50 29 0.0005 71 50 0.001 83 67 0.002 91 80 0.005 96 91 0.01 98 95 0.02 99 98 0.05 100 99 0.1 100 100 0.2 100 100 a) What is the Vmax in the absence of inhibitor? b) What is...

25) Am enzyme-catalyzed reaction proceeds at a faster rate than an uncatalyzed reaction because the enzyme...

25) Am enzyme-catalyzed reaction proceeds at a faster rate than an uncatalyzed reaction because the enzyme lowers the difference in free energy between the reactants and the products. a. True b. False

The following reaction rate data are collected upon analysis of an enzyme catalyzed hydrolysis. The reaction...

The following reaction rate data are collected upon analysis of an enzyme catalyzed hydrolysis. The reaction is carried out with two diﬀerent inhibitors. Determine the modes of inhibition for each inhibitor. (The rates provided are relative to the rate measured in the absence of inhibitor. The enzyme concentration for each set of data is equal.) No inhibitor [S]0 (10−2 M) 1.25 3.84 5.81 7.13 v0 0.398 0.669 0.859 1.000 [I1]– –2mmol/L [S]0 (10−2 M) 1.25 2.5 4.00 5.50 v0 0.179...

The simple Michaelis-Menten model applies only to the initial velocity of an enzyme-catalyzed reaction, that is,...

The simple Michaelis-Menten model applies only to the initial velocity of an enzyme-catalyzed reaction, that is, to the velocity when no appreciable amount of product has accumulated. What feature of the model is consistent with this constraint? Explain.

30.0 uM(microMolar) solution of enzyme catalyzed breakdown of 0.250M of substrate in 2 minutes. How many...

30.0 uM(microMolar) solution of enzyme catalyzed breakdown of 0.250M of substrate in 2 minutes. How many molecules of substrate were converted to product each second by one enzyme molecule?

Consider the enzyme-catalyzed reaction A → B. You determine experimentally that the maximal initial velocity, Vmax,...

Consider the enzyme-catalyzed reaction A → B. You determine experimentally that the maximal initial velocity, Vmax, for the amount of enzyme you have added is 100 units. You also determine that the Michaelis constant, Km, for this enzyme is 80 μM. What concentration of A will give an initial velocity of 60 units? (Note that the units for velocity can be ignored here; they will be relevant in another question. HINT: you need do NO calculations.) The answer is 120,...

An enzyme (select all that apply) 1) is not altered by reaction with a substrate 2)...

An enzyme (select all that apply) 1) is not altered by reaction with a substrate 2) is changed by reaction with a substrateis 3) made of carbohydrate 4) can be reused if not damaged 5) is made of protein 6) slows the rate of chemical reactions 7) increases the rate of chemical reaction Select all that apply. An enzyme 1) is a carbohydrate 2) is a protein 3) is a biological catalyst 4) is changed by reaction with a substrate...

If the substrate concentration is limiting, how does increasing enzyme concentration affect reaction rate? A.Increasing enzyme...

If the substrate concentration is limiting, how does increasing enzyme concentration affect reaction rate? A.Increasing enzyme concentration has no effect on reaction rate. B.Increasing enzyme concentration results in an increase in reaction rate. C.Increasing enzyme concentration decreases reaction rate.

Imagine that you are observing an enzyme-catalyzed reaction in the laboratory. The reaction is progressing as...

Imagine that you are observing an enzyme-catalyzed reaction in the laboratory. The reaction is progressing as expected. As you periodically add more enzyme, the reaction increases proportionally until suddenly it stops increasing. At this point, no matter how much more enzyme that you add, the reaction rate does not change. Assuming no other chemicals have been added or changed, explain why the reaction rate has leveled off.

Question 2-17 You want to maintain a pH = 7.0 for an enzyme-catalyzed reaction that will...

Question 2-17 You want to maintain a pH = 7.0 for an enzyme-catalyzed reaction that will produce hydrogen ions along with the desired product. At equal concentrations, which weak acid, if any, will serve as the better buffer for the reaction: Acid A, with pKa = 6.5 or Acid B, with pKa = 7.5? a. Acid A b. Water is as good as either of the acid available c. Acid B