Question

1. Carefully explain how to complete the table below to determine the SDH specific activity in the post nuclear supernatant using the data supplied. Assume the same volumes were used as in the practical and tubes 7 and 9 had a negligible change in DCPIP concentration over 30 minutes. tube A600nm 5 mins A600am 35 mins AA600nm 5-35 mins A[DCPIP) UM 5-35 mins 1 tube umoles DCPIP.min volume of extract assayed (ml) activity umoles.min l.ml1 specific activity umoles.min-1.mg 1 1 Note: umoles is an amount while uM is a concentration. Protein Assay dilution factor Absorbance protein conc (minus blank) from curve actual protein conc. of extract PNS 0.22 0.7 diluted 9/10 and then 1/5 Time (mins) DCPIP standard curve 0.45 0.4 0 0.35 0.3 5 0.25 ..... Absorbance 600nm y = 0.0081% R = 0.9857 10 Tube 1 (PNS) Blanked with tube 2 0.778 0.741 0.724 0.712 0.698 0.676 0.671 0.62 0.2 0.15 0.1 15 20 0.05 0 0 10 20 30 40 50 60 25 30 35 DCPIP (UM) unknowns Absobance (minus blank) protein conc from curve actual protein conc of extract mitos 1/5 0.35 1.04 1.3 diluted 1/5 and concentrated 4/1

Answer 1

Given the DCPIP standard curve and the equation of a straight line obtained for the curve is:

y = 0.0081x

we would be able to determine the uM concentration corresponding to the absorbance obtained by solving for x in the above formula:

DCPIP (uM) : x = y / 0.0081, where y would be the absorbance value.

Now, let us look at table 1:

tube	A600nm 5 mins	A600nm 35 mins	$\Delta$ A600nm 5 - 35 mins	$\Delta$ [DCPIP] uM 5-35 mins
1	0.741	0.62	0.121	14.9

From the absorbance values, we can subtract the value obtained for 35 min from the 5 min value to get a final absorbance of 0.121

The corresponding DCPIP concentration can be determined by solving for x:

x = 0.121 / 0.0081 = 14.9 uM DCPIP

Now, let us look at the second table:

tube	umoles DCPIP / min.	Volume of extract assayed mL	Activity umoles/mL/min.	Specific activity umoles/min./mg
1	0.0005	0.01	0.0005	0.0014

Firstly, let us calculate the protein concentration of PNS:

The concentration is given as 0.7 mg / mL of the extract diluted 1 / 10 and then 1 / 5.

The total dilution of the extract would be: 1 / 10 * 1 / 5 = 1 / 50, which means the extract is diluted 50 times.

The dilution factor would thus be 50.

The actual protein concentration of the extract is: 50 * 0.7 = 35 mg / mL.

Now, let us go back to table 2:

The volume of extract is not mentioned here, but let us assume 0.01 mL or 10 uL is used for the assay. The amount of protein used would then be 35 * 0.01 = 0.35 mg

We are asked to calculate umoles DCPIP per minute.

Molarity = moles / liter. A molar solution would contain 1 mole of the substance per liter.

1 uM = 1 umole / L; 1 nmole / mL

We obtained 14.9 uM DCPIP per 30 min. reaction (35 - 5 min.)

concentration of DCPIP per min. would be : 14.9 / 30 = 0.5 uM DCPIP = 0.5 nmoles DCPIP / min.

Assuming the reaction volume as 1 mL, the umoles DCPIP would be :0.5 / 1000 = 0.0005 umoles /mL/min.

Given the protein used is 0.35 mg (10 uL extract), the specific activity is determined by finding the umoles DCPIP per mg protein.

Specific activity (umoles/mg protein) = 0.0005 / 0.35 = 0.0014 umoles/min./mg

The values may vary depending upon the actual volume of enzyme used, and the reaction volume..