[ranma]

Hi, I have a new question:

The hydrolysis of pyrophosphate to orthophosphate is important in driving forward biosynthetic reactions, such as the synthesis of DNA. The hydrolytic reaction is catalyzed in E.Coli by a pyrophosphatase that has a mass of 120 kD and consists of 6 identical subunits.

For this enzyme, a unit of activity is defined as the amount of enzyme that hydrolyzes 10 umoles of pyrophosphate in 15 minutes at 37°C under standard assay conditions. The purified enzyme has a V_max of 2800 Units per milligram of enzyme.

A. How many moles of substrate are hydrolyzed per second per milligram of enzyme  when the substrate concentration is much greater than the K_m?

B. How many moles of active site are there inn 1.0 mg of enzyme? Assume that each subunit has 1 active site.

C. What is the turnover number of this enzyme (reactions per second per enzyme molecule)?

So far, I did the math for A and got this answer:
         (10umol substrate/15min.)(1min./60sec.)= 0.01umol substrate hydolyzed/sec.
         
         Becausethe substrate concentration is much greater than the K_m, my assumption is that this must be occurring at Vmax. Therefore:   
         2800 Units(0.01umol)=28umol= 28x10^-6 mol substrate

Is this correct, and if so, how do I complete parts B. and C.? 

[Yggdrasil]

So far, I did the math for A and got this answer:
         (10umol substrate/15min.)(1min./60sec.)= 0.01umol substrate hydolyzed/sec.
         
         Becausethe substrate concentration is much greater than the K_m, my assumption is that this must be occurring at Vmax. Therefore:   
         2800 Units(0.01umol)=28umol= 28x10^-6 mol substrate

Is this correct, and if so, how do I complete parts B. and C.? 

Part A looks correct to me.  For part B, you just need to use the definition of molecular weight to convert mg of enzyme into moles.  For part C, you just need to know what turnover number means in enzymology.

Let me know if this information helps, or if you need more help.

[ranma]

For Part B:

Wouldn't the number be the same as in part A (28x10-6 moles) because there's only 1 substrate for each active site, and part A was asking about the the moles of substrate hydrolyzed per second per milligram of enzyme at K_max?

[Yggdrasil]

No.  Although only one active site can bind only one substrate at a time, once the product dissociates from the active site, the active site is free to bind another substrate and repeat the reaction.  Therefore, each active site can catalyze hydrolysis of numerous substrates.

[ranma]

Okay,
For part B this is what I got:

Since pyrophosphatase weighs 120kD,

(120,000g enzyme/1 molecule enzyme)x(6.022x10²³ molecules/1 mole)= 7.23x10²⁸g/mol

(1 mole enzyme/7.23x10²⁸g)(1g/1000mg)-1.38x10^-32 moles enzyme/mg

(1.38x10^-32 moles enzyme/mg)x6= 8.3x10^-32 moles active site

For Part C:

V_max=K_cat x E_T

V= 2800 Units/mg
E_T = 1.38 x 10^-32 moles

2800/ 1.38 x 10^-32 = 2.02 x 10³⁵

BTW, I've been going back and forth on this problem, which is why its taking a while to get to an answer. 

[Yggdrasil]

Okay,
For part B this is what I got:

Since pyrophosphatase weighs 120kD,

(120,000g enzyme/1 molecule enzyme)x(6.022x10²³ molecules/1 mole)= 7.23x10²⁸g/mol

This is incorrect.  One molecule weighs 120 kDa (1 Da = 1.66x10^-22 g).  This also means that the molar mass is 120,000 g/mol.  (The molecular weight of a molecule in daltons is always equal to the molar mass of the molecule in grams per mole).

The other calculations are set up correctly, but they use the wrong numbers because of the above error.

[ranma]

This is what I now have for Part B:

(1 mole enzyme/120,000g enzyme)x(1g enzyme/1000mg enzyme)x(6 moles active site/1 mole enzyme)= 720 moles active site 

[Yggdrasil]

The calculations are setup correctly, but your math is off.  Try punching the numbers into the calculator a little more carefully.

[ranma]

Okay, now I got 5x10^-8 moles. 

Thanks for all your help on this problem and the last one! I will move on to figure out part C  .

[ranma]

For Part C:
Vmax = k_cat x E_t

V_max = 2800 units/mg
Since there is 1 active site for each unit, and the enzyme has 6 subunits:

E_t = (5x10^-8 moles subunits)/6 = 8.3x10^-9 moles enzyme

k_cat = (2800 Units/mg)/ (8.3x10^-9 moles enzyme) = 3.4 x 10¹¹

[Yggdrasil]

For part C, you should express V_max in units of mol s^-1 mg^-1.  Right now, your turnover number is given in units of Units/mol, which is not incorrect, but turnover numbers are generally expressed with units such as s^-1.  That way, the turnover number will directly tell you the average number of reactions performed by a molecule of enzyme per second.

[ranma]

Okay, that makes sense. Thanks for pointing that out.