Hydrolytic driving force. The hydrolysis of pyrophosphate to orthophosphate is i

Question

Hydrolytic driving force. 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 Escherichia coli by a pyrophosphatase 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 mu mol of pyrophosphate in 15 minutes at 37 degree C under standard assay conditions. The purified enzyme has a V_max of 2800 units per milligram of enzyme. How many moles of substrate is hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than km? How many moles of active sites is there in 1 mg of enzyme? Assume that each subunit has one active site. What is the turnover number of the enzyme? Compare this value with others mentioned in this chapter.

Explanation / Answer

Ans. Part 1:

Given, 1 unit activity = 10 umol substrate hydrolyzed in 15 minute

At [S] >> Km , the enzyme exhibits maximal activity equivalent to that under standard conditions.

Now,

Convert 10 umol substrate / 15 minute in moles per second as follow-

            10 umol = 10 x 10-6 moles                      ; [1 mol = 106 umol]

            15 minute = 15 x 60 seconds = 900 s      ; [1 min = 60 s]

So, 10 umol substrate / 15 minute = 10 x 10-6 moles / 900s

                                                = 1.11 x 10-9 moles / second

Thus, the enzyme hydrolyzes .11 x 10-9 moles substrate per second

Ans. b. 1 kDa = 1000 Da = 1000 x (1 g /mol) = 1000 g / mol       ; [1 Da = 1 g/mol]

Given, molar mass of enzyme = 120 kDa

                                                = 120000 g/mol

Moles of enzyme in 1.0 mg enzyme sample = mass of enzyme in g / molar mass of enzyme

                                    = 1x 10-3 g / (120000 g/mol)                   ; [1 g = 10-3 mg]

                                    = 8.33 x 10-9 moles

So, 1 mg enzyme sample has 9.33 x 10-9 moles enzyme.

Total moles of active site = Moles of enzyme x number of active site in one enzyme molecule

                                    = 8.33 x 10-9 moles x 6

                                    = 5.0 x 10-8 moles

Ans. c. Using Turnover number, Kcat = Vmax / [Et]        - equation 1

                        Where, Vmax = maximum velocity          ; [Et] = concentration of enzyme

Assuming reaction volume be 1.000 mL   (standard value).

Now, because each active site hydrolyzes the substrate, the concentration of enzyme [Et] is equal to the concentration of total active sites in 1 mg enzyme sample dissolved in 1.0 mL volume solution.

Now, [Et] = Moles of enzyme / volume of solution in Liters

= 5.0 x 10-8 moles / 0.001 L                  ; [1 mL = 10-3 L]

= 5.0 x 10-5 M

Given, Vmax = 2800 units per mg of enzyme / (reaction volume in Liters)

                        = [ 2800 x (1 unit activity) ] / (reaction volume in Liters)

                        = (2800 x 1.11 x 10-9 moles / second ) / 10-3 L

= 3.11 x 10-9 M /s        

So, Vmax = 3.11 x 10-9 M /s

Note: 1.11 x 10-9 moles substrates are hydrolyzed by 1.0 mg enzyme- but this values does not indicates the volume of reaction mixture. Thus, it’s assumed that 1.0 mg enzyme is present in 1.0 mL reaction mixture. The calculation thus gives Vmax in terms of substrate concentration per second i.e [S] catalyzed / s.

Putting the values in equation 1

            Kcat = Vmax / [Et] = (3.11 x 10-9 M /s) / 5.0 x 10-5 M

                        = 6.2 x 10-5 s-1

Thus, turnover number = 6.2 x 10-5 s-1

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.