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### Topic: Units of activity, Km, and Vmax!  (Read 33277 times)

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#### ranma

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##### Units of activity, Km, and Vmax!
« on: July 31, 2009, 10:53:04 PM »
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 Vmax 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 Km?

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 Km, 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

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##### Re: Units of activity, Km, and Vmax!
« Reply #1 on: July 31, 2009, 11:00:05 PM »
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 Km, 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

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##### Re: Units of activity, Km, and Vmax!
« Reply #2 on: August 04, 2009, 06:10:09 PM »
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 Kmax?

#### Yggdrasil

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##### Re: Units of activity, Km, and Vmax!
« Reply #3 on: August 04, 2009, 08:08:00 PM »
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

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##### Re: Units of activity, Km, and Vmax!
« Reply #4 on: August 04, 2009, 10:11:28 PM »
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:

Vmax=Kcat x ET

V= 2800 Units/mg
ET = 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

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##### Re: Units of activity, Km, and Vmax!
« Reply #5 on: August 04, 2009, 10:38:47 PM »
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

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##### Re: Units of activity, Km, and Vmax!
« Reply #6 on: August 09, 2009, 04:51:07 PM »

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

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##### Re: Units of activity, Km, and Vmax!
« Reply #7 on: August 09, 2009, 07:38:37 PM »
The calculations are setup correctly, but your math is off.  Try punching the numbers into the calculator a little more carefully.

#### ranma

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##### Re: Units of activity, Km, and Vmax!
« Reply #8 on: August 09, 2009, 08:09:48 PM »

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

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##### Re: Units of activity, Km, and Vmax!
« Reply #9 on: September 02, 2009, 05:22:37 PM »
For Part C:
Vmax = kcat x Et

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

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

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

#### Yggdrasil

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##### Re: Units of activity, Km, and Vmax!
« Reply #10 on: September 07, 2009, 10:50:18 PM »
For part C, you should express Vmax 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

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##### Re: Units of activity, Km, and Vmax!
« Reply #11 on: September 08, 2009, 08:30:33 PM »

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