115:414 Experimental Biochemistry, spring 1999                                                name___________________

Final Examination - Answers

A. Multiple choice.  Circle the letter of the correct answer.  Do questions C7-15 or L7-15, depending on whether you worked with carotenoids or lipids (not both).  Question 6 can be answered by everyone.  Each question counts 2 points, toward a total of 60 for this part of the examination.

1. An aldohexose contains how many asymmetric C atoms?

a. one                                 b. two                                c. four                                d. five

2. An aldohexose could have how many conformations in solution (as might be shown by silylation and gas chromatography):

a. one                                 b. two                                c. four                                d. five

3. Kinetic control is shown in the specificity of the
a. orcinol method.               b. ferricyanide method  c. indole method d. glucose oxidase method

4. The compound actually measured in the ferricyanide method for reducing sugars is
a. Fe(CN)63-                       b. Fe(CN)64-                       c. acid lactone                    d. H2O2

5. What method would be best for determination of hydrolysis of maltose?
a. phenol- H2SO4               b. Nelson-Somogyi             c. ferricyanide                     d. glucose oxidase

6. A saponifiable lipid has what linkage?
a. RCOOR'                       b. ROPO2OCH2CH2N∫    c. RCHCHR'                 d.RCH2OR’

L7. Plant lipids never contain
a. saturated fatty acids        b. cholesterol                    c. glycolipids                       d. phospholipids

L8. The method for lipid extraction initially extracts all soluble cell components into one phase.  To get two phases, with the lipids in one, you then add
a. CH3OH                          b. CHCl3                            c. H2O                               d. CHCl3 and H2O

L9. Which component will have lowest Rf in thin layer chromatography of neutral lipids?
a. cholesterol                      b. free fatty acid            c. monoacylglycerol         d. phosphatidylcholine

L10. Phospholipids are isolated from total lipids by
a. thin layer chromatography.                                        b. acetone precipitation.
c. extraction into chloroform.                                         d. extraction into methanol.

L11.  Dragendorff's reagent reacts with
a. quarternary ammonium groups.                            b. primary amino groups.
c. carbohydrates.                                                          d. dragons.

L12. Silver ion chromatography separates fatty acid methyl esters by
a. the number of silver ions associated with ester groups.
b. the number of silver ions associated with double bonds.
c. the conformation of the fatty acids in presence of silver ions.
d. the adsorption of double-bond-containing fatty acid esters on precipitated silver.

L13. In analysis of fatty acid methyl esters by gas chromatography, a straight line should be obtained by plotting
a. retention time vs. number of C atoms.                        b. retention time vs. log number of C atoms.
c. log time vs. number of C atoms.                            d. log retention time vs. log. no. of C atoms.

C7. The most important structural feature for the absorption spectrum of a carotenoid is
a. the number of hydroxyl or keto groups.
b. the number of double bonds.
c. the number of in-plane conjugated double bonds.
d. the presence or absence of epoxide groups.

C8. We treated diethyl ether with FeSO4 before use as a solvent for carotenoids because of
a. its peroxide content.  b. its flammability.  c. its volatility                               d. its anaesthetic effects

C9. The "epoxide shift" in the spectrum of a carotenoid on exposure to acid represents
a. the hydrolysis of a 5,6-epoxide group.
b. the isomerization of a 5,6-epoxide to a 5,8-epoxide.
c. the isomerization of a 5,8-epoxide to a 5,6-epoxide.
d. the oxidation of the carotenoid, introducing a 5,6-epoxide group.

C10. How do you recover carotenoids from 90% methanol?
a. Evaporate off the solvent.                                          b. Add water to get two phases.
c. Extract with pet. ether.    d. Extract with diethyl ether.

C11. Mono- and dihydroxycarotenoids may best be distinguished by
a. phase separation                                              b. mobility in thin layer chromatography. 
c. HCl addition to ethanol solution.                       d. absorption spectroscopy.

C12. The adsorbent we used for column chromatography of carotenoids was
a. alumina.                          b. cellulose.                  c. magnesium oxide/Celite     d. sucrose

C13. The last carotenoid to elute from our hplc column would be
a. b-carotene.                    b. b-cryptoxanthin.              c. violaxanthin.                    d. auroxanthin.

14. What is the function of guanidinium thiocyanate in RNA extraction?
a. buffering the extraction solution.                                 b. dissolving the RNA.
c. denaturing the RNA.       d. denaturing the RNAses.

15. What is the function of glyoxal in 'northern' gels?
a. denaturing the RNA.    b. denaturing the RNAses.
c. modifying the agarose.     d. visualizing the RNA.

16. Why are 'northern' gels called by that name?
a. After a man named Northern, who first used them.
b. Because they were developed in a lab in Edinburgh, Scotland.
c. Because they are run on ice.
d. By analogy with Southern gels.

17. What kind of RNA are the most scientists interested in?
a. ribosomal RNA.             b. transfer RNA.                 c. messenger RNA.          d. small nuclear RNA.

18. What is removed from precipitated RNA by 'washing' with 75% ethanol?
a. Na acetate.                     b. phenol.               c. guanidinium thiocyanate               d. all of the above.

19. Plasmid DNA is normally in what form?
a. superhelical                  b. relaxed                           c. linear                              d. single-stranded

20. DNA is determined specifically (distinguished from RNA) by
a. UV spectroscopy.                                                     b. fluorescence of bound ethidium bromide.
c. fluorescence of bound DAPI.                                 d. the orcinol reaction.

21. Only one of the following would be a site for restriction digestion of DNA:
a. CGGTTA                       b. AGGCCA                      c. ACTTCA                       d. CAATTG

22. At which step is chromosomal DNA separated from plasmid DNA?
a. hydrolysis in SDS-NaOH                                          b. precipitation with NH4+ acetate
c. precipitation with isopropanol                                    d. 'washing' the precipitate with 75% EtOH

23. What is the function of incubation of the agarose gel in 0.5 M NaOH - 1.5 M NaCl after electrophoresis?
a. loosening up the gel for easier transfer of DNA                   b. making the DNA single-stranded
c. partially hydrolyzing the DNA for easier transfer  d. relaxing supercoiled DNA

24. Why is the blot washed in 0.1xSSC - 1% SDS after hybridization?
a. To remove excess probe adsorbed to the membrane.
b. To remove excess probe weakly hybridized to slightly homologous DNA.
c. To remove prehybridization solution components from the membrane.
d. To remove excess salts (20X SSC) from the membrane.

25. For what purpose is the Maxam-Gilbert sequencing method still useful?
a. sequencing RNA with reverse transcriptase b. sequencing very GC-rich DNA
c. non-radioactive sequencing                           d. 'footprinting' where proteins bind to DNA

26. In the Qiaprep procedure, the DNA is adsorbed onto the Qiaprep spin column after adding 350 µl 3M Na acetate (total volume 850 µl; 1.23 M Na acetate).  It is eluted in TE (20 mM Tris Cl pH 8.0 - 1 mM EDTA).  What sort of separation process do you thing is being used?
a. hydrophobic adsorption.                                         b. ion exchange
c. gel filtration                                                               d. DNA phosphates binding to Ca++

27. If sequence you have determined contains E. coli plasmid sequence beyond the insert sequence (samples G, I), the most likely explanation is that
a. The organism from which the insert came had incorporated plasposon DNA into its chromosome.
b. Sequence has been determined right through the insert to vector in the other side.
c. The wrong primer was used, leading to DNA synthesis complementary to the vector rather than the insert.
d. The sample was contaminated with plasmid DNA.

28.  Homology between a newly determined sequence and one in GenBank is considered sig­nificant if the e value is less than
a. 32.                                 b. 1.                                   c. 0.01.                              d. 10-6.

29. Which component of those listed below is the most essential to the polymerase chain reaction method of DNA amplification?
a. two primers, binding to the two strands on either side of the region to be amplified
b. a thermal cycler which rapidly heats and cools the sample
c. a thermophilic DNA polymerase, which withstands 95° temperature for a short time
d. all of the above

30. Polymerase chain reactions usually do not produce as much DNA as predicted from the appropriate power of 2 (for 25 cycles, amplification by 225 = 33,554,432).   Which of the following explanations is most likely?  (Hint: would we use an insufficient condition if it was insufficient?)
a. Many molecules synthesized are not full length, and thus do not bind the other primer and are not replicated and amplified.
b. The DNA polymerase does not last 25 cycles.
c. Heat transfer to the tubes is inefficient.
d. The incubation time at 52° is not long enough for the primers to bind.

B. Problems.  Do C1 or L1.

C1.  A carotenoid was isolated by chromatography on MgO/Celite (eluted in 5% acetone in pet. ether) from the 95% CH3OH fraction from phase separation.  In thin layer chromato­graphy on silica, CH2Cl2:ethyl acetate 4:1 as solvent, its Rf was 0.75, vs. 0.35 for lutein stand­ard.  Its absorption spectrum in column solvent shows peaks at 421, 445, 474 nm; in carbon disulfide at 479, 512 nm; in ethan­ol at 424 (shoulder), 445, 477 nm.  Addition of HCl to the ethanol solution caused a shift of the spectrum to peaks at 400, 424, 450 nm.  Suggest a likely structure for the carotenoid.  (To give you a reference point, the peaks for b-carotene in pet. ether are at (425), 449, 475 nm; in CS2, 450, 485, 520 nm; in ethanol, (427), 449, 475 nm.)  (10 pts)

b-cryptoxanthin monoepoxide (monohydroxy, by phase separation and tlc Rf; mono-epoxide, by epoxide shift; b-carotene as parent compound, because lmax is only 4 nm below b-carotene)

L1.  The retention times of fatty acid methyl ester standards in gas-liquid chromatography are as follows (12:0 means 12 C atoms, no double bonds):

No. of C atoms         12:0           14:0           16:0           18:0           20:0           18:1           18:2           18:3
time, min                   0.74           1.38           2.57           4.79           8.91           5.62           6.60           7.76

A fatty acid methyl ester from a transgenic tobacco plant has retention time 3.55 min.  Assuming it has an even number of C atoms and nothing more complicated than double bonds, calculate the number of carbon atoms and double bonds.  Use graphs below if desired. (10 pts)

Version A: 16:2.  They should be able to tell by inspection that it is a 16 C fatty acid!
Version B: 16:1  The equation for standards was log t = 0.135(#C) - 1.75; -1.68 for 1 double bond, -1.61 for two, -1.54 for three.

 

 

2. A carbohydrate, 0.6 g, is dissolved in 1 mM HCl (to hasten mutarotation) and diluted with this to a volume of 15.0 ml.  The observed optical rotation (in a 2 dm cell) is +4.2°.  Calculate [a]d and make a possible identification of the carbohydrate from the table below.  (5 pts for [a]d, 1 for identification.)

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.

Both versions: [a]d = +52.5°, so it might be either lactose or glucose.

What further information might you seek to aid your identification? (2 points)

Therefore, use the glucose oxidase method to tell which!

3.  A DNA insert has been cloned in a vector, original size 2.3 kb.  Inserting this piece between the Sac I and Kpn I sites, which are 102 bp apart, has removed this amount from the vector.  The complete plasmid is digested with Sac I, with Kpn I, with Sac I and Kpn I together, with Sac I and Hind III, and with Kpn I and Hind III.  These digests are electro­phoresed on a gel together with a "1 kb ladder" marker.  Sizes and mobilities of bands in the standard lane are:

Size, kbp        10          8            6           4            3           2          1.6        1.0        0.51

cm moved     2.33      2.66       3.07      3.66       4.08      4.66       4.99      5.66       6.64

Mobilities of bands in the digest lanes are as follows:

Enzymes             Sac I                   Kpn I            Sac I + Kpn I           Sac I +               Kpn I +
                                                                                                         Hind III               Hind III

cm moved           2.85                    2.85                 3.4, 4.53             3.2, 5.08            3.77, 3.94

a.  Calculate the size, in bp or kbp, of the whole plasmid, of the complete insert, and of the smaller pieces from the double digests with Hind III (2.5 pts each - 10 total).  Use one of the graphs below if desired.

The whole plasmid is 7.0 kb; the vector is 2.2 kb, the insert 4.8 kb (Sac I + KpnI digestion); in version A SacI + Hind III gives 5.5 + 1.5 kb pieces, KpnI + Hind III gives 3.7 + 3.3 kb pieces, \ the Hind III site is 1.5 kb from the SacI site, into the insert.  In version B the site is 1.5 kb from the KpnI site, the pieces are the same size but exchanged between the two double digests.

 

 

b. Draw a map of the insert from the Sac I site to the Kpn I site, showing the location of the Hind III site (indicate the distances from it to the Sac I and Kpn I sites) (2 pts.)

4. After a 30 cycle PCR amplification reaction, the DNA synthesized is precipitated with isopropanol and washed with 75% ethanol to remove unincorporated nucleotides, primer, enzyme, etc.  The DNA pellet is dissolved in 30 µl TE.  Five microliters of this solution is dil­uted to 1.0 ml and an absorbance spectrum taken.  The A260 is 0.050.  E260 = 20 L/mole·cm.

a) What is the DNA concentration in the cuvette? (2 pts)

version A : 0.05/20 = 0.0025 mg/ml                          version B: 0.03/20 = 0.0015 mg/ml

b). What is the DNA concentration in the stock solution?  What is the total amount of DNA present? (2 pts)

version A : 0.0025 x 200 = 0.5 mg/ml = 0.5 µg/µl, x 30 µl = 15 µg total
version B: 0.0015 x 200 = 0.3 mg/ml = 0.3 µg/µl, x 30 µl = 9 µg total

c) What was the minimum amount of DNA template present at the beginning of the reac­tion?  (If you don't remember how to calculate 230, multiply the log of 2 by 30, take the antilog.) (2 pts)

230 = 1,073,741,824.         Version A: divide 15 x 10-6 g by this = 1.4 x 10-14 g
                                                        Version B: divide 9 x 10-6 g by this = 8.4 x 10-15 g

d) How many moles was this? (2 pts)

If the DNA was 1 kb = 1000 base pairs long, x x 2 = 6 x 105 g/mole

e) How many molecules was this? (Avogadro's number: 6.023 x 1023 molecules/mole)

Version A: 1.4 x 10-14g ∏  6 x 105 g/mole = 2.33 x 10-20 moles; x 6.023 x 1023 molecules/mole =
                                          14,053 molecules

Version B: 8.4 x 10-15g ∏ 6 x 105 g/mole = 1.4 x 20-20 moles; x x 6.023 x 1023 molecules/mole =
                                          8,432 molecules