Annotated Answers — 115:413 Experimental Biochemistry exam, 1994

Part A - I give the correct answer, differently placed on the two versions of the exam.

1.     Of the protein determination methods you used in the lab, the one least likely to destroy the biological activity of the protein is UV absorbance.  The other methods use either strong base (biuret, Lowry) or strong acid (Coomassie Blue).

2.     Beer's law states that -log (I/I0) is proportional to concentration.  Since I/I0 is a fraction < 1, its logarithm is negative; -log (I/I0) is therefore positive, = absorbance.

3.     One ml of a solution is diluted with 4 ml H2O; one ml of the diluted solution is further diluted with 19 ml H2O.  The overall dilution is 1:100, 1:5 x 1:20.

4.     In the Lowry method for measuring protein, reagent B (phosphomolybdate) should be added and mixed quickly, as its half-life in the alkaline solution is short.

5.     Which protein determination method varies most (in E1 mg/ml) from protein to protein?
UV, since it depends entirely on the (variable) tyrosine and tryptophan content.

6.     At 0.3 pH unit (log10 2) above the pKa of a buffer, the concentration of the basic form =
twice the concentration of the acidic form.  Half the conc. of the acidic form if the question read 0.3 pH unit below the pKa.

7.     Addition of one drop of 0.5 m KOH to a beaker of indicator-containing solution causes a color change from yellow to blue.  This suggests that there is no buffer present.
Under any of the other conditions it would have taken more KOH to change the pH enough to change the indicator color completely.

8.     A possible reason for seeing several spots in two-dimensional thin layer chromato­graphy of your isolated DABTH-amino acid is
a. presence of DABITC breakdown products.                               
b. presence of two polypeptide chains in the protein.
c. internal peptide bond cleavage in the protein.
d. breakdown of the DABTH-amino acid to several products.
e. all of the above

9.     Ninhydrin reacts with the a-amino group of amino acids.

10. The most important feature of Edman degradation is that it removes one N-terminal amino acid at a time, so that you can identify successive amino acids in sequence.

11. Which of these amino acids should have the highest Rf in the solvent we used?  Tyro­sine, clearly the most hydrophobic.  Remember that leucine and isoleucine, very hyd­rophobic amino acids, had the highest Rf.  For some reason most people had trouble with this!  Keep a sense of the properties of the side chains of the amino acids in your mind.

12. What does extraction of the DABITC-modified protein with heptane-ethyl acetate do?
It removes excess DABITC and PITC.  Protein stays in the tube, the DAB-thiazolin­one is extracted only later, after cleavage, and there are no free amino acids.

13. Only one of the following is not a reason for precipitation of proteins with (NH4)2SO4 :
Proteins may be made ready for ion exchange chromatography by (NH4)2SO4 precipitation.  Salt in the (NH4)2SO4 precipitate may in fact prevent ad­sorption to an ion exchange column unless removed by dialysis (as you did before hydroxylapatite chromatography.

14. Which of the following purification procedures is not used in purifying d-amino acid oxidase?  Gradient elution chromatography; the hydroxylapatite column was eluted stepwise.  (This was a bit sneaky.)  Irreversible denaturation (at pH 5.3, 48°) was used to remove a lot of other protein.  Gel filtration was the last step.

15. Total units of enzyme activity at step n/total units in original extract = yield.  I got tired of seeing the spelling "yeild", though I suppose asking correct spelling is a bit low.

16. These compounds serve the same function in the purification of d-amino acid oxidase:
Crotonate, benzoate; both are inhibitors, competitive with substrate, which serve to keep the FAD on the enzyme.

17. Hydroxylapatite is a form of calcium phosphate. 

18. Decline of the rate of D-amino acid oxidase action in the pyruvate assay after 10 min may be due to using up much of the dissolved O2 in the assay solution.  This was a return to one of the questions on the experiment.  The substrate, d-alanine, is 50 mm; O2 is only 0.26 mm in water.  If not shaken to equilibrate with air this can be used up.

19. Including catalase in the polarograph assay decreases the rate of O2 utilization by regenerating O2 from the H2O2 product, 2H2O2 Ć 2H2O + O2.

20. If the apparent Km of d-amino acid oxidase for d-alanine is 7.5 mm at pH 8.3 and 1.0 mm at pH 9.3, and the pKa of (d- or l-)alanine is 9.87, what form of d-alanine might you con­clude to be the true substrate? CH3CH(NH2)COO-, the basic species.  Since it is a greater proportion of the total d-alanine at higher pH, less total d-alanine will be required to achieve half-saturation of the enzyme at higher pH if the true Km is that for the basic species.  (This was hard, but I have to have some questions that require real thought.)

21. For determination of Km and Vmax, substrate concentration should be calculated as µmoles d-alanine/1 ml (during assay), when the enzyme is acting - not per 3 ml, after 2,4-DNPH and NaOH have been added and the enzyme is no longer active.  Neither time nor enzyme concentration should appear in substrate con­centration.

22. In the indophenol assay, the standard curve worked.  The enzyme assay didn't.  From this you should conclude that the problem is in the enzyme solution.  (NH4)2SO4, phenol-nitroprusside and NaOCl all were involved also in the standard curve, so they could not have been responsible for the failure; either the enzyme solution or alanine-PPi, not present in the standard curve, must be responsible.  Hmm, maybe it was buffering by the PPi and alanine; the K2HPO4 used wouldn't buffer much at pH 9.5, so the pH may not have come up to the level necessary for the indophenol reaction.  We should try using a carbonate-bicarbonate buffer instead.  Unless I can get it to work better, I will drop this assay.  … Perhaps we should run all our assays at pH 9.5.

23. Which part of our gel electrophoresis system has the lowest pH?  The stacking gel, pH = 6.7.  All others are above 8.  This isn't just mindless remembering, it is important for stacking that the stacking gel is at a lower pH.

24. The anion of the upper (reservoir) buffer is glycine.  Again, it is critical for stacking that the anion is a compound actually mostly a zwitterion, especially at the pH of the stacking gel, so that the overall mobility is less than that of proteins.

25. Which of the following is important in making Rm of proteins in SDS gel electrophoresis a dependable function of molecular weight (so that graphs give a straight line)?
a. denaturation of the protein to a random coil.                                      b. constant charge density.
c. reduction of any disulfide bonds.                                                         d. all of the above

26. If the pKa of adenine in AMP is 3.8, the net charge on AMP at pH 3.5 is -0.333.  The phosphate has a charge of -1.0; at a pH 0.3 units (log10 2) below the pKa of the adenine the ratio of Hadenine+ to adenine0 is 2:1, therefore two-thirds of the adenine is proton­ated, it contributes a net charge of +0.667.  -1.0 + 0.667 = -0.333.  A simple pH problem.

27. Most of the RNA in cells is ribosomal.  A simple statement from the Introduction to the experiment, touched on in the question about what you would expect to see following gel electrophoresis of total RNA (only the two ribosomal RNAs will give visible bands).

28. The nucleotide with highest absorbance (for equal concentrations) at 280 nm is CMP; at 260 nm, AMP.  CMP you should certainly have remembered from determining which peak was which; AMP was a little harder to remember.

29. If the average molecular weight of nucleotides is 300, the molar concentration of a 3.6 mg/ml solution of hydrolyzed RNA is 12 µmoles/ml (3.6 mg/ml ∏ 0.3 mg/µmole).  8.0 mm if the question read 2.4 mg/ml.

30. In purification of RNA, the most important compound denaturing RNAses is phenol. Tris at high pH only makes the RNAses relatiuvely inactive, it doesn't denature them.

31. EXTRA CREDIT: The Gilson microfractionator head moves as the cow wanders, boustro­fedwn in Alexandrine Greek.  An utterly unimportant bit of information, but a surpris-ing number of people remembered it.

Part B - Problems. I give the calculations for the version headed Final Examina­tion; the other version, headed Final Examination, had the same problems with different numbers.

1. (5 points)  A bovine serum albumin standard solution, 0.1 mg/ml, gives the following results in the Coomassie Blue method for protein determination:

        ml BSA                         0                              0.05                         0.1                           0.2                           0.3
A54540                                                                                          0                              0.102                       0.199                       0.401                       0.595

        Another protein solution is diluted 1:50.  Samples of the dilute solution give the following results:

        ml unknown                0                              0.05                         0.1                           0.15                         0.2
A540                               0                              0.062                       0.118                       0.177                       0.242

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

        Observe that the slope of the standard curve is approximately 2 A/ml standard solution, or 20A/mg.  The slope of a plot of results with the unknown protein is approximately 1.2A/ml.  Dividing this by the slope of the standard gives a value of 0.06 mg protein/ml dilute solution.  Multiplying by the dilution factor, 50, gives a final value of 3.0 mg/ml.  In the other exam the standard slope was 18 A/mg, the unknown slope 1.5 A/ml, the con­centration of the dilute solution 0.0833 mg/ml, and the stock solution 5.0 mg/ml.

2. (4 points) Fifty microliters of a 1:5 dilution of stock d-amino acid oxidase is added to 3 ml assay mixture in the polarograph and yields a rate of decrease of [O2] = 0.256 chart widths (2.56 cm) per minute.  What is the d-amino acid oxidase activity of the stock solution, in µmoles O2 used per minute?

        This was unfair, I admit; I forgot to include the conversion factor, 0.78 µmoles O2/full scale on the chart (since the assay volume is 3 ml and each ml contains 0.26 µmoles O2).  I gave full credit to anyone who picked a value and calculated through correctly.

        Using this factor, 0.256 x 0.78 x = 20 .  For the other exam, the calculation is similar; the final value is 24 .

3. A b-alanine acetate buffer for gel electrophoresis of cationic proteins, pH 5.0, is 0.025 M in total acetate/acetic acid (pKa at 20° 4.76).  The pK1 of b-alanine is 3.58.
a. (3 points) What is the concentration of acetate- in the buffer?  (Remember that the Henderson-Hasselbalch equation holds for each ionizing compound in the solution.)

        A heartening number of people got this.  Set up the Henderson-Hasselbalch equation:
5.0 = 4.76 + log 0.24 = log 1.74 = [HAc] = = 0.575 [Ac-].   [HAc] + [Ac-] = 0.025 m = [Ac-] + 0.575 [Ac-]   [Ac-] = = 0.0159 m.  For the other exam the calculation is similar; [Ac-] = = 0.01064 m.

        b. (7 points)  What is the total concentration of b-alanine (all forms) in the buffer?  (The concentration of the basic form ala- can be assumed to be negligible.)

        Few people got this; it was not easy, but a similar problem was on reserve in last year's exam.  The key insight is that the concentration of b-alanine cation is equal to the con­centration of acetate anion (it might have helped if I said that the buffer was pre­pared by adding b-alanine [zwitterion] to 0.025 m acetic acid until the pH was 5.0).  From [b-ala+] = [Ac-] = 0.0159 m and the Henderson-Hasselbalch equation, 5.0 = 3.58 + log , 1.42 = log, 26.3 = , [b-ala± ]= 26.3 [b-ala+] = 26.3 x 0.0159 = 0.418 m.  Since the question asks for total b-alanine, the concentra­tion of b-ala+ = 0.0159 m must be added, 0.418 + 0.016 = 0.434 m total.  For the other exam the form of the calculation is the same, but because the pH is lower the concentration is much smaller: b-ala± = 8.31[b-ala+] = 0.0885 m, total = 0.0885 + 0.01064 = 0.099 m.

4.  (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 0.46 on the same gel.  Calculate its molecular weight (use graph below, or fit the standard molecular weights to an appropriate equation).

I set this up from an equation Rm = 3 - 0.555 log mol. wt.  Substituting Rm = 0.46 into this gives mol. wt. = 37,325 (in the other exam Rm = 0.39 Ć mol. wt. 50,432)  The graph should give these values ± 2,000 or so, and most people got it.

5.(6 pts) The millimolar extinction coefficients (units: ml/mmole.cm) for AMP and CMP are:

                                            e260                                            e280                                 e260/e280

        AMP                          14.7                               3.3                           4.5

        CMP                             6.2                             13.0                          0.48

        A solution containing AMP and CMP has A260 = 0.666, A280 = 0.846.  Calculate the molar concentration of the two nucleotides present.

        Use the equations A260 = 14.7[AMP] + 6.2 [CMP], A280 = 3.3[AMP] + 13.0[CMP]; multiply the latter equation by 6.2/13.0 = 0.477, it becomes 0.477A280 = 1.57[AMP] + 6.2[CMP].  Subtract it from the first equation; the difference is A260 - 0.477 A280 = 13.13[AMP].  Putting in the numbers given, 0.666 - 0.4035 = 0.2635 = 13.13[AMP], [AMP] = 0.02 mm.  Substitute this value in either equation above and solve for [CMP] = 0.06 mm.  In the other exam [AMP] = 0.06 mm, [CMP] = 0.02 mm.  Many people got the right answers here.

6.  a) (8 pts) The following rates of enzymatic oxidation of samples of 0.02 m d-leucine are observed in the peroxidase assay (assay volume 3.0 ml):

        ml d-leucine                              0.025                0.05                 0.10               0.20                  0.375                0.75                  1.5

        [d-leucine], mm:                    0.1667             0.333              0.667               1.33                 2.5                  5.0                 10.0           

        ∆A500/min                                  0.065               0.107               0.158               0.207               0.242               0.268               0.283

        Calculate the leucine concentrations, the Km for this substrate, and Vmax in mmoles/ min for this amount of enzyme (hint for speed: calculate in ∆A500, then convert Vmax to mmoles/min [e500 of peroxidase pro­duct = 14,250].  Use graph below if desired.)

        The d-leucine concentrations are found by multiplying the stock concentration (0.02 m d-leucine) by the ml used/3 ml, thus: 0.025/3 x 0.02 m (= 20 mm) = 0.1667 mm. 

        An important point here is to estimate first, to be sure that your answers make sense.  The highest v given is ∆A500/min = 0.283, hence Vmax must be slightly larger than this, say 0.300 A/min (this was in fact the value I used to construct the problem.)  Then Km = [d-leu] at v = Vmax/2 = 0.150 A500/min.  This is a little less than the rate at 0.667 mm d-leu, so Km ≈ 0.6 mm (the value I used in setting up the problem).

        For the real calculation assume I used exact values for at least the highest and lowest rates and set up either the Lineweaver-Burk or Woolf plot: [S] = 0.1667, 10.0, 1/[S] = 6, 0.1, ∆1/[S] = 5.9; v = 0.065, 0.283, 1/v = 15.38, 3.53, ∆1/v = 11.85; slope = Km/Vmax = 11.85/5.9 = 2.0; intercept = 1/Vmax = 3.53 - (0.1 x 2.0 = 0.20) = 3.33, Vmax = 0.300 A500/min, Km = 2.0 x 0.30 = 0.6 mm.  Now convert 0.300 A500/min to µmoles/min: divide by the millimolar extinction coefficient, 14.25 ml/µmole.cm, and multiply by 3 ml, Vmax = 0.0632 µmole/min.  For the other exam, where the substrate is stated to be isoleucine, Km = 0.55 mm, Vmax = 0.320 A500/min = 0.0674 µmole/min.

(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?

        Divide by the volume used and multiply by the dilution factor: 0.0632 x = 12.64 µmol/min.ml stock (13.74 µmole/min,ml for the other exam).

(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?

        12.64 x = 15.8 µmoles/min.mg, x 50 mg/µmole = 790 min-1 = 13.17 s-1.  For the other exam, 13.47 x = 16.84 µmoles/min.mg, x 50 mg/µmole = 842 min-1 = 14.0 s-1.

Not too many people really got this question (of course I gave credit for correct calculations of b and c even if the Vmax calculated in a was incorrect), even though I have had essentially this question on the exam for several years.  But overall you did well, with average and median = 58, 3 points above the usual 55, even though therew were some decided curve balls thrown (A.14, 20, B.2).