January 16, 1993                                                                                                                                         

Annotated Answers to Final Examination

Part A.  Since the position of the correct answer varied in the two versions of the exam, I simply give it in words, not by letter.

1.   At. 0.3 pH unit (log10 2) below the pKa of a buffer, the concentration of the basic form equals one half the total concentration of the buffer, since pH - pKa = log [B]/[BH] = - 0.3, [B]/[BH] = 1/2.

2.   Change of the pH of a buffer as it is diluted occurs mainly when the buffering species include a dianion., as demonstrated in the pH experiment with citrate or phosphate.

3.   Which method for protein determination has a high (and potentially variable) blank absorbance vs. H2O? Coomassie Blue.

4.   Which method for protein determination would be least appropriate for measurement of protein during purification of d-amino acid oxidase in presence of 0.1% Na benzoate?
Ultraviolet absorbance, because the benzoate absorbs at 280 nm.

5.   One tenth of a milliliter of a stock protein solution is diluted to 5 ml with water.  One ml of this first dilution is diluted with 4 ml water to yield the second dilution.  Three tenths of a milliliter of this second dilution is found to contain 0.012 mg protein.  What is the protein concentration of the stock solution?  0.012/0.3 = 0.04 mg/ml in second dilution.  This was a 1:5 dilution, so the concentration of the first dilution was 0.04 x 5 = 0.20 mg/ml.  The first dilution was 1:50 (0.1 ml ->5 ml), so the concentration of the stock solution is 0.2 x 50 = 10 mg/ml.

6.   Which method for protein determination is least sensitive?  Biuret. A memory question.

7.   In which protein determination method would polyglycine give a higher absorbance per mg than polytryptophan?  Biuret, because the biuret reagent reacts with the peptide bond, the molecular weight of a glycine unit in a polypeptide (57) is far less than that of tryptophan (186), so there will be more glycine units per mg protein in polyglycine than there would be tryptopohan units per mg polytryptophan.  This was hard, but I went over it in lecture.  Polytryptophan would react much more in the Lowry method and have a high UV absorbance; I don't know what would happen, if anything, in the Coomassie Blue method (with either polyamino acid).

8.   Treatment of the protein with phenyl isothiocyanate (PITC) after dimethylaminoazoben­zene isothiocyanate (DABITC) is to insure that the a-amino terminus is fully modified, and will be fully cleaved.  Otherwise you would get some of the first amino acid in the second cycle.

9.   What does extraction of the DABITC reaction mixture with heptane-ethyl acetate remove?
a. excess DABITC              b. excess PITC                  c. pyridine                          d. all of these

10. From the fact that DABTH-aspartic acid is first off the hplc column and DABTH-leucine is last off, you should reason that the DABTH-amino acids stick to the column by hydro­phobic interactions, since aspartic acid is the most polar amino acid, least hydro­phobic, whiule leucine is the most hydrophobic.  Simple reasoning, as long as you have a sense of the properties of the amino acids.

11. One of the following amino acids definitely is not cleaved off by carboxypeptidase A:
Proline, since there is another carbon attached to the N.

12. Which of these amino acids should have the lowest mobility in the chromatographic system we used? Asparagine.  Three of the four were in your system; proline, if you remember the structure, would be rather less polar, more mobile, than alanine, certain­ly more than serine or asparagine.

13. Which of the following is not truly an advantage of (NH4)2SO4 in protein precipitation?
Neutral reaction (solution pH = 7)  The others all are advantages, but (NH4)2SO4 solutions have pH = 5.2 unless adjusted.

14. Which purification procedure is likely to achieve greatest purification as a single step?
Hydroxylapatite; an adsorption chromatographic method, in which the elution condi­tions can be carefully chosen, should achieve greater purification than even a gel fil­tration step, which depends on the molecular weights of the proteins present.

15. In gel filtration, the first proteins to elute are the largest, because they are the most excluded from the gel's pores.

16. Specific activity is mmoles/min.mg protein. A definition.

17. Benzoate is added to solutions for purification of d-amino acid oxidase because it stabil­izes FAD on the enzyme.  You just have to know that, but it was emphasized in the lab manual.

18. In assaying enzyme activity during purification, we keep all but one of these constant:
Enzyme concentration, because that is the variable we are measuring!

19. What is not an inherent advantage of a stop-time assay? It shows directly whether the initial rate is being observed.  It doesn't, of course; you have to stop the reac­tion at a number of time points to define the initial rate carefully.  All the other answers are advantages of stop-time reactions.

20. To get activity of the stock solution in the peroxidase assay in mmoles/min.ml enzyme, multiply the ∆A500 by x You must divide by the (millimolar) extinction coefficient; multiply by the volume of the assay; divide by the ml diluted enzyme used; multiply by the dilution factor.  You don't have to multiply by 1000 mmoles/1 mmole, because the extinction coefficient given is the millimolar coefficient, 14.25, not the molar, 14,250.

21. The Km is that [S] at which v = Vmax/2.  The other answers are explicitly wrong, and I harped on this.

22. Determination of the Km for O2 by a single run of the polarograph requires that
a. the instrument measure O2                                      b. the assay be continuous
c. Km be not much less than initial [O2]                         d. all of the above are required.

23. What property of a protein is measured in SDS polyacrylamide gel electrophoresis?
Size of denatured subunits.  A major point of SDS gel electrophoresis is that the proteins are denatured to random coils and therefore shape is not a factor.  Subunits are separated.

24. The key property of the stacking gel is pH several units lower than pKa of glycine, so that the mobility of glycine is less than that of the protein, which stacks.

25. Mobility in an SDS polyacrylamide gel is inversely proportional to log molecular weight.  Small proteins move further, so the proportionality is inverse, and you plot mobility vs log mol. wt.

26. Most of the RNA in leaf tissue is ribosomal RNA.  This was stated in the manual.

27. Basic hydrolysis of RNA depends on presence of a free 2'-OH, at which the 3'-phosphate attacks in base.  This is why DNA doesn't hydrolyze in base.

28. Chromatographic separation of the ribonucleotides depends on differences in pKas of the bases, the phosphates being entirely in the -1 state at pH 3.4.  Although separation of the 2' and 3'-phosphates is sometimes seen, even this is due to slight differences in the pKas of the bases due to the proximity of the phosphate.  The affinity of the phosphates for the positive charges of the DEAE-Sephadex is what sticks them there, but it is the dif­ference in pKas of the bases that effects the separation.

29. Our elution of the nucleotides from DEAE-Sephadex uses a linear gradient, as was explained in the manual.

30. Observation that the ratio of mmoles by orcinol to mmoles by UV absorption is lower for CMP and UMP than for AMP and GMP suggests that the glycosidic bond to purines hydrolyzes more easily than that to pyrimidines.  Orcinol reacts with pentoses which have a free glycosidic hydroxyl; purines hydrolyze off in acid, pyrimi­dines don't.

 

Part B - Problems.  The construction of the problems was the same in both versions of the exam.  I give the working of the problem for the version headed Final Examina-tion, plus the answer for the other version.

1.   The data given, for both standard curve and unknown, generated plots which did not quite pass through the origin.  I wanted people to determine the slopes of both plots, by dividing (highest absorbance - lowest abs.) by (highest mg or ml - lowest mg or ml), then to divide (slope of unknown) by (slope of standard curve) to get a value for mg/ml 1:50 dilution; this value received full credit.  Just reading off the standard curve gave a significantly different value, and received 5 points rather than 6.
Standard curve: (0.78 - 0.13)/(0.03 mg/0.005 mg) = 26 A/mg.
Unknown: (0.498 - 0.108)/(0.3 ml - 0.05 ml) = 1.56/ml.  Then 1.56 ml-1/26 mg-1 = 0.06 mg/ml in the diluted solution, x 50 = 3.0 mg/ml in the stock.
If you read from the standard curve, its equation was A = 26mg-1 - 0.015, and the three samples of unknown contained 0.00473, 0.0107 and 0.0197 mg protein.  Dividing these by ml sample used gave values of mg protein/ml dilute solution 0.0946, 0.0713, 0.0658; aver­aging these gave 0.0713 mg/ml, x 50 = 3.86 mg/ml.  You see the difference!
For the other version (A595 = 0.077, 0.178, 0.337) the values done the first way were the same (26 A/mg) for the standard curve, 2.0 mg/ml by comparing slopes, 2.76 mg/ml by reading from the standard curve.

2a. First, remember that dl-alanine is half d-alanine, so the total d-alanine concentration is 0.071/2 = 0.0355 m.  The leftover pK3 in the question may have confused some.

pH = pK2 + log 9.40 = 9.87 + log -0.47 = + log 0.3388 =
[ala±] = [ala-]/0.3388    0.3388[ala±] = [ala-]    [ala±] + 0.3388[ala±] = 1.3388 [ala±] = 0.0355 m
[ala±] = 0.0355/1.388 = 0.0265 m    [ala-] = 0.0355 - 0.0265 = 0.009 m.  The other version has the same answer.

b.   8.5 = 9.87 + log -1.37 = log 0.04266 = [ala±] = = = 0.211  Now remember to add the [ala-], 0.211 + 0.009 = 0.22 m total d-alanine = 0.44 m dl-alanine; 0.686 m in the other version.

3.  I did this type of problem in lecture, and it's in the manual.  The absorbance at each wave length is the sum of the absorbances due to the GMP ([G]) and UMP ([U]) present:

                  A260 =     11.7[G] + 9.8[U]

                 A280 =       8.0[G] + 3.5[U]   Multiply by e260/e280 of [G] so that the [G] terms are equal.

      1.46 A280 = 11.7[G] + 5.11[U] Subtract this from the first line:

      A260 - 1.46 A280 = 4.69[U].  Now put in the absorbance values given:  0.86 - (1.46)(0.46) = 0.86 - 0.6716 = 0.1884 = 4.69[U], 0.1884/4.69 = 0.0402 ≈ 0.04 mm = [U]

      Then A260 due to UMP = 0.0402 x 9.8 = 0.394, A260 due to GMP = 0.86 - 0.394 = 0.466, 0.466/11.7 = 0.0398 ≈ 0.04 mm = [G].  In the other version  the concentrations are 0.03 mm U, 0.05 mm G.

4. (a) Use the end points (these are perfect data, and the line does pass through the origin) to approximate: 0.833 A560/0.25 mmoles = 3.33 A/mmole (or, 0.307 x A560 = mmoles).  This is the same in both versions.

(b) Relatively few people really faced the fact that this requires a Lineweaver-Burk plot or something similar.  I do it, as I usually do, as a Woolf plot, [S]/v vs [S], using A560 as v (and calculating only the Vmax, not the individual A560 values, through to mmoles/min), because it is statis­tically better and because it requires only one calculation from the data, not two; but the result should be the same no matter how you do it.  Simply plotting v vs [S] and using the highest v value as Vmax is insufficient, the highest v (A560) is quite a bit short of the real Vmax; but it's on the right track.

      ml d-norvaline                  0.025           0.05            0.10            0.20            0.40            0.60            0.80

      [d-norvaline], mm:            0.625           1.25             2.5              5.0             10.0            15.0            20.0

      A560                                 0.067          0.120          0.200          0.300          0.400          0.450          0.480

      [S]/A560                             9.37           10.41           12.5           16.67            25            33.33          41.67

      The slope = 1/Vmax = 1.667, Vmax = 0.60 A560 = 0.18 mmole/10 min = 0.018 mmole/min.  Remember that the assay lasts for 10 min!

      The intercept = Km/Vmax = 8.33, Km = 5.0 mm.  For a Lineweaver-Burk plot the slope and intercept are interchanged, but the numbers are the same.
For the other version (with d-valine), the Vmax is the same, but Km is 3.5 mm.

(c) 0.018 mmole/min x 200/0.1 ml = 36 mmole/min.ml, in both versions.

(d) x x = 500 min-1 in both versions.