December 21, 2000                                                                                                               name______________________section___

Final Examination - 115:413 Experimental Biochemistry- Answers

Part A: The letter of the correct answer differed for the two versions, I just give the answer.

1.     Which of the following is not a recommendation in the use of pipetters?
Holding the pipetter vertical when sucking up solution into the tip. Nearly everyone got this.

2.     The sample standard deviation is the square root of (sum of squares of deviations from the mean/no. of observations).  I got confused myself as to whether this was it, or that with no. of observations - 1.

3.     0.1 ml of stock solution is diluted with 2.4 ml H2O.  Fifty microliters of this solution is diluted with 0.45 ml H2O.  The overall dilution is 1:250.  1:25 x 1:10.

4.     Adding one drop of 0.5 m KOH to an indicator-containing solution causes a color change from yellow to red and a pH change from 6.5 to 10.  This suggests that a pKa of the indicator lies between 6.5 and 10.  If one drop of KOH changes the color and raises the pH to 10, there cannot be substantial buffer capacity present.

5.     The pKas of glutamic acid are approximately 2.0, 4.0 and 10.0.  What is the pH of an 0.1 M solution of monosodium glutamate? 7.0.  Monosodium glutamate will have two positive charges, on the Na+ and the NH3+ group, therefore both COOH will be ionized to COO-, therefore the pH will be halfway between the second pKa of COOH ionizing and the pKa of the amino group being deprotonated.  I admiyt I didn't spend much time on this sort of problem this year.

6.     The units of an extinction coefficient for a protein determination method would be
ml·mg-1cm-1.  An extinction coefficient has units the inverse of concentration - vol­ume on top, weight on the bottom - times 1/distance.  This is the only choice that fits.

7.     A plot of absorbance vs. concentration in the Lowry method for protein determination has a negative y intercept.  What is a likely explanation of this?  The blank has higher absorbance than it should (tube contaminated with protein?)

8. Which method for protein determination is least subject to interference by non-protein compounds?  Coomassie Blue

9.     A 1 mg/ml solution of lysozyme has about twice the A280 of a 1 mg/ml solution of oval­bumin.  This suggests that it has a higher content (than ovalbumin) of tryptophan, which has high A280.

10. One of the following will not react in the phenol-H2SO4 method: sorbitol, because it does not have a carbonyl group and thus cannot cyclize.

11. Optical rotation of a carbohydrate solution is proportional to all but one of these: number of chiral centers, whose effect of optical rotation is complex and not easily predictable.

12. Which method for carbohydrate determination is carried out at neutral pH? Glucose oxidation.

13. After graduating you get a job with the Istituto Superiore de Sanità, an Italian govern­ment agency whose responsibilities include checking that spaghetti is made only with durum wheat (starch, glucose polymer).  It is reported that one manufacturer is using instead flour from the Jerusalem artichoke root, which contains inulin, a fructose poly­mer ending with a glucose residue attached in 2Ĉ1 linkage, as in sucrose.  Remembering your biochemistry lab experiments, you check this using the indole reaction, which will indicate the presence of fructose.

14. The independently determined values in a protein purification table are volume, units/ml, mg protein/ml.  The other choices include at least one calculated from others.

15. d-Amino acid oxidase is eluted from the DEAE-Sepharose column by increasing the con­centration of chloride, the counterion which competes with the protein for the DEAE+ group.  I was surprised at how few got this - I must lecture more on ion exchange next year.

16. The ratio A460/A280 of fractions from column chromatography of d-amino acid oxidase is a rough measure of their specific activity, with A460 representing d-amino acid oxidase specifically, A280 representing general protein.

17. Which of these procedures did you not use in purification of d-amino acid oxidase? Gel filtration.  Irreversible denaturation occurred in the heating/acid step, (NH4)2SO4 precipitation was used after that, phenyl-Sepharose was hydrophobic chromatography.

18. Those who did experiment #11 saw that the rate of pyruvate production declined after 10 min even if catalase was present and the tubes were shaken.  This indicates that the decline is not due to using up the oxygen in the solution, because shaking and catalase provided more oxygen.

19. Most of you observed that the pH optimum of d-amino acid oxidase is quite high, ≈9.5.  This suggests that the true substrate may be alanine anion (with NH2), of which more would be present at higher pH.

20. Inhibition of d-amino acid oxidase by benzoate is expected to be competitive vs. d-alanine.  I perhaps didn't give you evidence for this, but with a COO- it is likely to bind where the substrate binds.

21. You have determined e463 of your d-amino acid oxidase, using a molar concentration of the enzyme based on a Coomassie Blue protein determination, whiuch was based on a standard curve using bovine serum albumin.  Later you find that bovine albumin gives 50% more color (A595/A466) per mg than an average protein.  What is the likely effect of this on your molar extinction coefficient, as originally determined? It is 50% high.  This is complex and requires thought.  If BSA gave more color than an average protein, it took more of the average protein (such as d-amino acid oxidase) to give the same amount of color, therefore there was really more protein present than you thought, therefore when the protein concentration is corrected the extinction coefficient (A/mg) will be lower, therefore the original extinction coefficient was 50% high.

22. In a SDS polyacrylamide gel system, the SDS
a. associates with the peptide bonds.
b. disrupts the native secondary structure of the proteins.
c. disrupts the native tertiary structure of the proteins.
d. makes the proteins have a constant negative charge density.
e. all of the above.  I think everyone got this.

23. The critical feature of the reservoir buffer in stacking gel electrophoresis is that its anion has low mobility at the pH of the stacking gel, so that the protein stacks in front of it in the stacking gel.

24. Our native gel assay for d-amino acid oxidase depends on what reaction?  Reduction of phenazine methosulfate by reduced FAD.

25. Without which component acrylamide will polymerize, but not gel? N,N-methylene­bisacrylamide, the cross-linking reagent.

26. The enzyme-linked antibody in western blot visualization is an antibody to rabbit immunoglobulin, includinmg the rabbit antibody to d-amino acid oxidase.

27. Isoelectric focusing is particularly good at separating isoforms of the same protein which differ in charge, since it separates by isoelectric point.

Part B - Short Answers.

1.  (3 pts)  Write the balanced chemical reaction for the reaction catalyzed by d-amino acid oxidase.  Show structures, not just names.

CH3CH(NH3+)COO- + O2 + H2O Ĉ CH3COCOO- + NH4+ + H2O2

2.  (3 pts)  What are (supposed) advantages of using A595/A466, rather than just A595, for measuring protein with the Coomassie Blue reagent?

A plot of A595/A466 vs. mg protein is more linear (no curvature close to the y axis); has a higher slope (more sensitive); inaccuracy due to inaccurate addition of the reagent is minimized, since this will affect both A466 and A595.

Part C - Problems. Show all work and indicate your answer clearly.

1.  To a malonic acid solution, 100 ml of 0.05 M, is added 1.0 g Tris (tris[hydroxymethyl]­aminomethane, mol. wt. 121, pKa = 8.3).  The pK1 of malonic acid is 2.7, pK2 is 5.7.

        (1 pt) How many millimoles of malonic acid and Tris are present?

0.1 L malonic acid x 0.05 moles/liter = 0.005 moles = 5 mmoles; 10 mmoles of COOH.  Tris: 1 g/0.121 g/mmole = 8.26 mmoles.

        (1 pt) Which pKa will you use to calculate the pH?  The second pKa of malonate, since there is enough Tris to titrate the first carboxyl group fully but not the second, therefore the second pKa, still below that of Tris, will be controlling.  Essentially all the Tris base will be converted to TrisH+, all of the first COOH of malonate will be titrated, but not all of the second.

        (3 pts) What is the pH?  Five mmoles of Tris base will be consumed in convert­ing malonic acid to the monoanion; the remaining 3.26 mmoles Tris will convert this amount of Hmalonate- to malonate=, leaving 1.74 mmoles Hmalon­ate- remaining.  We now have the amounts to put in the Henderson-Hassel­balch equation: pH = 5.7 + log= 5.7 + 0.273 = 5.97.  The other version of the exam used less Tris and got pH 5.38 (1,61 mmoles malonate=, 3.39 mmoles Hmalonate-).

2. (5 points)  A bovine serum albumin standard solution, 10.0 mg/ml, gives the following results in the biuret method for protein determination:

        ml BSA                         0                              0.2                           0.4                           0.6                           1.0
A54540                                                                                          0                              0.121                       0.238                       0.366                       0.595

        A solution of another protein gives the following results:

        ml unknown                0                              0.2                           0.4                           0.6                           1.0
A540                               0                              0.082                       0.158                       0.241                       0.398

Calculate the protein concentration of the unknown solution.  Use graph below if desired.

The slope of the standard is about 0.06 A/mg (0.5966 to be exact), that of the unknown is about 0.4 A/ml (0.3977); = 6.67 mg/ml.  The other version used a similar calculation to get 8.33 mg/ml.

3. (5 points) Observed Rms and molecular weights of the standard proteins for molecular weight determination by SDS gel electrophoresis are as follows:

Protein                                       Rm                       mol. wt.              Protein                                                   Rm                   mol. wt.

aprotinin                                    0.88                            6,500            ovalbumin                                             0.415                        45,000
a-lactalbumin                            0.69                          14,200            albumin, bovine serum                        0.32                          66,000
trypsin inhibitor                       0.61                          20,000            b-galactosidase                                    0.186                      116,000
carbonic anhydrase                 0.52                          29,000            myosin                                                   0.05                        205,000

An unknown protein has an Rm of 0.55 on the same gel.  Calculate its molecular weight (use a graph below, or fit the standard molecular weights to an appropriate equation) .

Plot Rm vs log mol. wt. on left graph, or vs mol.wt. on right (semilog) graph.  By construction of a graph, I got mol. wt. = 26,000.  The actual equation used to determine the Rms of the standards was Rm = 2.991 - 0.554(log mol. wt.).  From this a mol. wt. of the unknown = 25,476 was calculated (log mol. wt. = ).  For the other version, Rm = 0.38, log mol. wt. = 4.71, mol. wt. = 51,641.

4.  The specific rotations of carbohydrates used in our experiment are:

Sugar                           [a]d      Sugar                                          [a]d            Sugar                                  [a]d

d-glucose                +52.8°      l-fucose@                                -75.6°      d-trehalose∂                       +178°

d-fructose#              -92.0°      l-rhamnose@                            +8.9°      d-sucrose∂#                       +66.5°

d-galactose             +81.7°      d-sorbitol (glucitol)◊                   -2.0°      d-maltose∂                      +129.0°

l-sorbose#              -42.7°      l-arabinose*                        +104.0°      d-lactose∂                         +52.3°

d-mannose              +14.1°      d-xylose*                               +18.6°      d-cellobiose∂                                        +34.5°

*Pentose   #Ketose   ∂Disaccharide   @6-deoxyhexose   ◊Sugar alcohol   The others are aldohexoses.

The unknown (90 mg/ml, 2 dm path length) shows an optical rotation of +9.0°; also observed were +9.1° for d-glucose, -17.4° for d-fructose, +23.2° for d-maltose, +33.2° for d-trehalose, +13.3° for l-arabinose, +9.0° for lactose, +14.6° for galactose (all at 90 mg/ml, 2 dm).

The following data are determined in the colorimetric measurements of carbohydrates (data are shown for only one level of standard sugars):

Phenol-H2SO4: 0.4 ml glucose A480 = 1.704, maltose 1.718, fructose 1.931, unknown 1.878.

Orcinol: 0.3 ml arabinose A660 .= 0.557, glucose 0.047, fructose 0.195, unknown 0.098.

Indole: 0.3 ml fructose A480 = 0.499, sucrose 0.262, glucose 0.058, lactose 0.054, unknown 0.061

Nelson-Somogyi: 0.4 ml glucose A660 = 1.919, maltose 0.888, lactose 1.087, unknown 1.01.

Alk. ferricyanide: no sugar A420 = 1.64, 1.0 ml glucose 0.721, lactose 1.09, maltose 1.37, sucrose 1.63, unknown 1.18.

Glucose oxidase: 1.0 ml glucose A500 = 1.038, sucrose 0.034, lactose 0.003, unknown 0.002.

(5 pts) Identify the unknown sugar. 

(5 pts) Explain its behavior, compared to standard sugars, in each test; does it or does it not react?  Why?  Which test tells you the most about its identity?

5.  Below is a chart trace from a polarograph assay (volume 3 ml) of d-amino acid oxidase.

 

  (2 pts) Draw a straight line and determine initial velocity (chart width/min or cm/min) of the reaction observed.

  (3 pts) Calculate the activity of the stock enzyme in µmoles/min·ml stock.  Amount and dilution of enzyme used are on the chart.  Solubility of oxygen in water at 37° = 0.26 mM.

6.  a) The following rates of enzymatic oxidation of samples of 0.02 m dl-norleucine, diluted to an assay volume of 3.0 ml, are observed in the peroxidase assay:

        ml dl-norleu/PPi                       0.03                 0.06                0.125              0.25                   0.50                  1.0                   2.0

        [d-norleu], mm:

        ∆A500/min                                  0.083               0.143               0.227               0.312               0.385               0.435               0.465

        a. (3 pts) Calculate D-norleucine concentration, and either [S]/v (Woolf plot) or 1/[S] and 1/v (Lineweaver-Burk plot) and plot on the graph below.  The rate (v) is most easily plotted as ∆A500/min.  Label the axes clearly.  e500 of peroxidase product = 14,250 L/mole·cm.

        1/[S]

        1/v

        [S]/v

 

b. (5 pts) Determine Vmax (in whatever units you used above) and Km (mM) from the data above.

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

d. (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?