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.?