115:413 Experimental Biochemistry                                                                                                                                      January 17, 1994

EXAMINATION ANSWERS

The following answers to the final exam are provided for your further education (if you didn't already know the answer).  Since there were two versions of the exam, I give the cor­rect answer in words.  The problems existed in two versions; I work the problems for only one, but essentially only the numbers differed between them.

Part A:

1.  The total concentration of the buffer is not a term in the Henderson-Hasselbalch equation, only the concentrations of the individual buffer species - as I said when lecturing on this.

2.  The pH at which the indicator is green is the pKa of the indicator - when it is half in the blue form, half in the yellow form, therefore green.

3.  e = A/cl.  Just remember that A = ecl, and be able to rearrange to solve for any part.

4.  Coomassie Blue is most sensitive.  You were expected to remember.

5.  0.018 mg/0.3 ml = 0.06 mg/ml diluted solution, x dilution factor of 50 (5/0.1) yields an answer 3 mg/ml.

6.  Setting the blank means all of the above.

7.  Since A = -log(I/I0), an A of 2 corresponds to -log(0.01), or 1% transmission.

8.  Extraction of the tissue from which it is derived - you have to do that first.

9.  Many molecules of product are formed per molecule of enzyme.  The other statements are true, but not really the reason a catalytic assay is used.

10. % Yield means total units at a given step/total units at the first step.

11. All but one statement is true; the false one is "the effectiveness of (NH4)2SO4 precipitation is not affected by the total protein concentration" - it cer­tainly is affected, as I lectured.

12. The pyruvate assay is more sensitive that the peroxidase and polarograph assays primarily because the incubation time is longer - the sensitivity per µmole product (or substrate, O2) is about the same.  The indophenol assay also is incubated for a longer time, and the extinction coefficient fo the product is actually higher, but NH4+ in the sub­strate solution was a problem.

13. Gel filtration separates proteins from smaller molecules by excluding the protein molecules from pores in the gel particles - as lectured on.

14. The A280 of pure d-amino acid oxidase is about 8x the A460 - you should remember this, and it is even more or less stated in the manual (pure fractions have A280/A460 <10).

15. Crotonate, like benzoate, is a competitive inhibitor (vs. the amino acid substrate).

16. All of the above.

17. The ion with the lowest mobility in the stacking gel is glycine - that's how stacking works.

18. Mobility in an SDS polyacrylamide gel is inversely proportional to log mol. wt.

19. Color reactions of carbohydrates in strong acid depend on dehydration to furfurals.

20. The indole test indicates it has one fructose, and the hydrolysis by b-galactosidase indicates the other sugar is galactose.  Since the unhydrolyzed sugar does not react in the Nelson-Somogyi reaction, it is non-reducing and must have the glycosidic OH of all sugars involved in a glycosidic bond.  The only structure satisfying this is b-galactosyl-(1->2)-fructose.

21. Pectin (as your experiment showed) reacts in the carbazole method, since it is partial­ly methylated poly(galacturonic acid).

22. Glyceraldehyde is a reducing sugar (since it has an aldehyde group), but is not big enough to form furfural.

23. Glucose oxidase is specific for glucose, while glucose and lactose are both reducing, both react in the phenol-sulfuric acid method (lactose is hydrolyzed), and their specific rotations are essentially the same.

24. Some phenol dissolves in the aqueous phase (and denatures proteins there).  You don't want it there later, so you extract the phenol from the H2O phase with chloroform.

25. The biggest problem in preparation of mRNA - or almost any RNA - is cleavage by ribonucleases, which are much harder to inhibit than DNAses.

26. All of the above.

27. The nucleotides are eluted from the DEAE-Sephadex by competing Cl-.

28. A logarithmic gradient is shallower - and therefore separates better - late in the run, when UMP and GMP are being eluted.

29. Since at pH 9.3 more of the total d-alanine is present as the anion, to achieve the Km concentration it will require less total d-alanine than at pH 8.3, and the apparent Km, in terms of total d-alanine, will be lower at pH 9.3 than at 8.3.  A tough question.

30. As noted above (Q#12), NH4+ in the sub­strate solution was a problem.  You use 0.5 ml of it, but only a very small amount (after dilution) of enzyme solution.

31. Higher apparent activity of crude extract in the pyruvate assay than in the peroxidase assay could be due to catalase, which would hydrolyze some of the H2O2 formed by d-amino acid oxidase before it could be used by peroxidase to produce colored product; the production of pyruvate would not be affected.  l-amino acid oxidase would be active in both assays, hence is not the answer.  Substrates for pyruvate kinase and lactate oxidase (which would also produce pyruvate) would not be present in appreciable amount, at least if the enzyme is well diluted.

Part B - Problems (the first answers are for the exam with a non-bold FINAL EXAMINATION on the first page, the second for the exam with a FINAL EXAMINATION):

1.     Upper reservoir buffer for gel electrophoresis is prepared by adding solid Tris base (pKa of TrisH+ = 8.3; mol. wt. = 121) to one liter of 0.192 m glycine (pK2 = 9.87) until the pH reaches 8.8.  (The Henderson-Hasselbalch equation must be satisfied for both Tris and glycine at this pH.)

        a) Calculate the concentration of glycine anion at this pH. (4 pts)

        Use the Henderson-Hasselbalch equation with the pK2 of glycine, 9.87.  Set the concen­tration of glycine anion = x.  Then 8.8 = 9.87 + log ,   -1.07 = log ,
0.085 = , (0.085)(0.192-x) = x, 0.01634 - 0.085x = x, 0.01634 = 1.085x, x = 0.01506 m (for the other exam, pH 9.0, x = 0.0228 m).

        b) Calculate how much total Tris base was added to reach this pH.  (6 pts)  (Hint: What is the concentration of TrisH+ equal to?)

        [TrisH+] = [gly-] as calculated above = 0.01506 m, since both are formed by the reaction Tris base + gly± -> TrisH+ + gly-.  Then use the H-H equation with the pKa of Tris:
8.8 = 8.3 + log , 0.5 = log , 3.16 = , [TrisH+]= 0.01506 = , [Tris] = 3.16 x 0.01506 = 0.0476 m, + the [TrisH+] = 0.01506 m, total [Tris] = 0.0476 + 0.01506 = 0.0627 m.

        0.0627 moles/L x 1 L x 121 g/mole = 7.58 g (for the other exam 16.58 g).

2. The [a]d of d-sucrose is +66.5°, of d-glucose is +52.8°, of d-fructose is -92.0°.  The molecu­lar weights are 342 g/mole for sucrose, 180 for glucose and fructose.
a)  What will be the optical rotation of an 0.5 m solution of sucrose in a 20 cm cell? (3 pts)

        0.5 m sucrose = 171 g/L = 17.1 g/100 ml (the concentration units for optical rotation calculations), l = 20 cm = 2 dm    Optical rotation = = = 22.74°. 
For the other exam, 0.5 m lactose, OR = 17.89°.

       

        b) What will be the optical rotation of the solution if the sucrose is completely hydro­lyzed to glucose + fructose?

        The solution is now 0.5 m in glucose and in fructose, 0.5 m = 90 g/L = 9.0 g/100 ml.  Since the concentrations are the same one can simply add together their [a]ds,
OR = =  = -7.056°.  For the other exam +24.21°.

3. The millimolar extinction coefficients (units: ml/mmole.cm) for CMP and AMP are:

                                            e260                                              e280                                    e260/e280

        CMP                             6.2                             13.0                          0.48

        AMP                          14.7                               3.3                           4.5

        A solution containing CMP and AMP has A260 = 0.960, A280 = 0.912.  Calculate the molar concentration of the two nucleotides present. (6 pts)

         A260 = 6.2[CMP] + 14.7 [AMP]
 A280 = 13.0[CMP] + 3.3[AMP]   6.2/13 = 0.477 ≈ 0.48.  Multiply the second equation by 0.48:

0.48A280 = 6.2[CMP] + 1.57[AMP]   Subtract this from the first equation:

A260 - 0.48 A280 = 13.13 [AMP], or in actual numbers 0.960 - 0.435 = 0.525 = 13.13[AMP], then [AMP] = 0.040 mm.  A280 of AMP = 3.3 x 0.04 = 0.132. Thus A280 of CMP = 0.912 - 0.132 = 0.780, [CMP] = 0.78/13 = 0.06 m. In the other exam the numbers were halved, [AMP] = 0.020 m, [CMP] = 0.030 m.

4. (a) The following A560 were observed for aliquots of a standard pyruvate solution in the dinitrophenylhydrazone assay:

        mmoles                                          0                    0.05                 0.10                 0.15                 0.20                 0.25
A560                                                0                   0.176               0.347               0.525               0.701               0.875

        Calculate the slope (factor to convert observed A560 to mmoles pyruvate.) (2 pts)  Use the graph if you wish.

        Use =  3.5 A/mmole; in the other exam, 0.813/0.25 = 3.25 A/mmole.

(b) Calculate the molar extinction coefficient of pyruvate dinitrophenylhydrazone at 560 nm (assuming the same procedure used in the lab) (2 pts).

        3.5 A/µmole x 3 ml x 106 µmole/mole x 1 L/103 ml = 10,500 L/mole.cm.  In the other exam, 9750 L/mole.cm.

(c) The following A560 values were observed in assay of samples of a 1:400 dilution of purified d-amino acid oxidase:

        ml enzyme                                      0.05                  0.1                   0.2
A560                                                0.085               0.169               0.342

        Using the standard curve in (a), calculate the mmoles pyruvate formed in each sample. (2 pts).  Divide by the slope in a:
                                                       0.0243             0.0483             0.0977                µmoles    (same in other exam)

(d) From these results, calculate an average value for activity (mmoles pyruvate/min.ml) of the stock enzyme solution. (2 pts)

        µmoles/min                                  .00243             .00483             .00977
µmoles/min.ml dil E                      .0486               .0483              .04886             Average =            0.04857
Multiply by the dilution factor, 0.04857 x 400 = 19.43 µmole/min.ml stock enzyme

5.  a) The following rates of enzymatic oxidation of samples of 0.04 m d-norvaline, diluted to an assay volume of 3.0 ml, are observed in the peroxidase assay:

        ml d-norvaline                           0.03                 0.06                 0.12               0.24                   0.50                  1.0                   1.5

        [d-norvaline], mm:                    0.4                  0.8                  1.6                  3.2                 6.67               13.33                20             

        ∆A500/min                                  0.091               0.158               0.250               0.353               0.444               0.486               0.511

        Calculate the norvaline concentrations, the Km for this substrate, and Vmax in mmoles/ min for this amount of enzyme (hint for speed: calculate in A560, then convert Vmax to mmoles/min [e500 of peroxidase pro­duct = 14,250].  Use graph below if desired.) (8 pts)

        I use the Woolf plot, [S]/v vs. [S], using ∆A500/min values directly with [d-norvaline]:

        [S]/v                                           4.39                5.06                 6.4                 9.06               15.02              27.43              39.14

        From this slope = 1/Vm = 1.78, Vm = 0.562 A500/min, x 3 ml = 0.118 µmole/min.
int = = 3.517, divided by the slope = Km = 1.98 mm.  These were not the numbers I intended, and my grading of some exams will show my original expectations, 0.10 µmole/min and 1.2 mm, but I had forgotten to change the actual ∆A500/min values from last year's exam!

        In the other exam slope = 1/Vm = 1.684, Vm = 0.5937 A500/min = 0.125 µmole/min;
int = = 2.0227, divided by slope = Km = 1.20 mm.

(b)(1 pt)  If the enzyme used was 0.1 ml of a 1:20 dilution, what is the Vmax in mmoles/min. stock enzyme?

        0.118 µmole/min x x 20 = 23.6 u/ml; in the other exam, with a 1:50 dilution,
0.125 µmole/min x = 62.5 u/ml.

(c)(1 pt) If the stock enzyme had a protein concentration of 0.8 mg/ml, and the molecular weight (per subunit) is 50,000, what is the turnover number?

= 25 u/mg; x = 1250 min-1. The answer is the same in the other exam.