Question

The hydrolysis of pyrophosphate to orthophosphate is important in driving forward biosynthetic reactions, such as the synthesis of DNA. This hydrolytic reaction is catalyzed in E. coli by a pyrophosphate that has a mass of 120 kd and consists of six identical subunits. For this enzyme, a unit of activity is defined as the amount of enzyme that hydrolyzes 10 umol of pyrophosphate in 15 min at 37 degrees C under standard assay conditions. The purified enzyme has a Vmax of 2800 units per milligram of enzyme.
a) How many moles of the substrate is hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than Km? (Draw a velocity versus substrate
concentration graph to show your point)

b) How many moles of active sites is there in 1 mg of enzyme? Assume that each subunit has
one active site.

c) What is the turnover number of the enzyme? (Don't bother comparing it to other enzymes;
just show all your own, which equations you use, what numbers you use, where you get them
from, etc.)

Answer 1

Answers:

a) How many moles of the substrate is hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than Km?

Answer:

31.1 µmol PPi s-1 mg -1

b) How many moles of active site are there in 1 mg of enzyme?

Answer:  

5.0 x 10-8 mol

c) What is the turnover number of the enzyme?

Answer:  

3732 s-1 per mol enzyme but there are 6 active sites per enzyme therefore turnover is 622 s-1 (per active site)

Explanation:

i have attached two pictures that explains the methodology as i can not write formulas here that's why i solved it and created a picture for attachment.