115:414 Experimental Biochemistry, spring 2001 name___________________

Final Examination (answers indicated by x)

A. Multiple choice. Circle the letter of the correct answer. Each question counts 2 pts.

1. What is important about the CHCl₃:CH₃OH:H₂O ratio 2:1:0.8?
a. x√It yields a single phase, which is optimal for extraction of lipids from tissue.
b. It extracts the lipids into the lower (CHCl₃-rich) phase.
c. It denatures proteins.
d. It selectively extracts phospholipids.

2. The most nonpolar neutral lipid is
a. cholesterol. b. triglyceride. c. free fatty acid. d. x√cholesterol ester.

3. Olive oil is almost entirely
a. sterol ester. b. glycolipid. c. x√triglyceride. d. phospholipid.

4. Phospholipids as a class are separated by
a. lower solubility in CHCl₃. b. xacetone precipitation.
c. digitonin precipitation. d. thin layer chromatography.

5. Dragendorff's reagent reacts with
a. primary amino groups. b. double bonds.
c. x√quarternary nitrogen groups. d. glycolipids.

6. Total cholesterol is determined using
a. digitonin. b. xFeCl₃. c. ninhydrin. d. _-naphthol.

7. Silver ion chromatography of fatty acid methyl esters separates them by interaction of immobilized Ag⁺ with
a. ester groups. b. methyl groups. c. methylene groups. d. √xdouble bonds.

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

9. What the detector of our gas-liquid chromatograph actually measures is
a. increased thermal conductivity of the gas when it contains sample.
b. x√increased electrical current when it contains ions from burnt sample.
c. capture by the sample of electrons emitted by a radioactive source.
d. absorbance of UV light by the sample.

10. Presence of cholesterol in a sample of 'French fried' potatoes indicates that
a. cholesterol is found in all eukaryotic tissues. b. plant roots contain cholesterol.
c. x√animal fat was used in frying the potatoes. d. potatoes are bad for you.

11. The least polar solvent used in this experiment is
a. x√hexane. b. diethyl ether. c. chloroform. d. methanol.

12. Diethyl ether is treated with FeSO₄ solution before use
a. to saturate it with water.
b. to activate it for thin-layer chromatography.
c. to remove peroxides by extraction of their complexes with Fe⁺⁺ into the water.
d. x√to destroy peroxides by reducing them with Fe⁺⁺.

13. Mitochondria are centrifuged down to the middle of a tube containing Percoll because
a. they haven't been centrifuged long enough to pellet at the bottom of the tube.
b. they associate with the Percoll particles.
c.√xtheir density is the same as that of the Percoll solution at this level of the gradient.
d. they break open at this level.

14. The polarograph assay of mitochondrial activity actually measures
a. reduction of cytochrome c. b. electron transport.
c. phosphorylation of ATP. d. x√reduction of oxygen.

15. The real objective of most purifications of RNA is
a. ribosomal RNA. b. x√messenger RNA. c. transfer RNA. d. small nuclear RNA.

16. The principal function of guanidinium thiocyanate in RNA isolation is
a. x√denaturation of proteins. b. high ionic strength.
c. precipitation of DNA. d. precipitation of RNA.

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

18. Glyoxal is added to the RNA sample before electrophoresis to avoid
a. √xself-association of RNA. b. denaturation of RNA.
c. precipitation of RNA. d. hydrolysis of RNA.

19. RNA extracted from green leaves may show four bands of ribosomal RNA because
a. ribosomes have four sizes of RNA.
b. there is so much mRNA for large and small subunits of RUBISCO.
c. x√chloroplasts represent prokaryotic symbionts.
d. the chlorophyll bands will fluoresce like ethidium bromide.

20. Why does RNA appear in agarose gel electrophoresis of DNA as a smear at the far end of the gel?
a. It is still self-associated under these conditions.
b. x√It has been digested by RNAses during preparation.
c. The different sizes of RNA are not well separated on this gel.
d. The extra hydroxyl on each sugar unit increases its mobility.

21. How are plasmids maintained in bacterial cells?
a. In contrast to phages, they do not lyse cells.
b. x√Their antibiotic resistance genes are essential when the bacterium is grown in presence of an antibiotic.
c. They are incorporated into the cell's chromosome.
d. They contain a multicloning site.

22. Why is the plasmid preparation mixed only gently after addition of NaOH/SDS and NH₄⁺acetate?
a. To avoid denaturing proteins. b. To avoid denaturing the plasmid.
c. x√To avoid breaking up the chromosomal DNA. d. To avoid creating an emulsion.

23. Why do we use DAPI fluorescence to distinguish between DNA and RNA?
a. It is more sensitive than absorbance at 260 nm.
b. Its adducts with DNA and RNA fluoresce at different wave lengths.
c. DAPI binds to AT-rich regions of DNA.
d. x√DNA and RNA have similar absorption spectra.

24. Why do all solutions used with DNA contain EDTA?
a. x√It chelates Mg⁺⁺, which DNAses require. b. As a buffer.
c. It helps precipitate the DNA. d. It inhibits RNAses.

25. Which conformation of plasmid DNA migrates furthest in agarose electrophoresis?
a. x√supercoiled b. relaxed c. linear d. all the same

26. Which of these is least likely to be a safety hazard of nucleic acid preparation and analysis?
a. phenol burns b. ethidium bromide c. UV radiation when looking at the gel
d. x√transfer of antibiotic resistance to microorganisms of your intestine
e. none, all are hazards

27. Which procedure used in your plasmid preparation procedure is obviated (made unnecessary) in the Qiagen procedure?
a. cell lysis by NaOH-SDS b. neutralization with high-conc. acetate
c. presence of EDTA d. x√precipitation with isopropanol

28. What is actually measured in automated sequencing of DNA?
a. Silver metal, from Ag⁺ reduced on exposure to electrons emitted by ³²P.
b. Silver metal, from Ag⁺ reduced on exposure to light in presence of nucleic acid.
c. x√Fluorescence, of dye molecules attached to terminal dideoxynucleotides.
d. Fluorescence, of dye molecules attached to nucleotides incorporated along the chain.

29. The e value of a comparison of a new sequence with a sequence from GenBank represents
a. the chance that homology between the sequences is significant.
b. √xthe chance that homology between the sequences is random, not significant.
c. the per cent identity between the sequences.
d. the number of identical residues in the sequences.

30. What aspect of polymerase chain reaction amplification of a DNA sequence is not shared with DNA synthesis for automated sequencing?
a. a DNA template b. thermal cycling c. x√use of two primers d. Taq DNA polymerase

B. Short answers (one sentence) - 5 points each

1. We collect the mitochondria from the Percoll gradient, then dilute them with sucrose medium. What is made possible by the dilution?

Diluting the Percoll so that the density is never as high as that of mitochondria makes it possible to pellet them (centrifuge them to the bottom).

2. How would the presence of a specific mRNA in an RNA preparation be detected?

By transfer to a membrane and hybridization with a specific, labeled probe.

C. Problems

1. 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 0.80 1.52 2.90 5.53 10.55 6.53 7.64 8.89

A fatty acid methyl ester from a plant has retention time 1.80 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. (9 pts)

The data were constructed from log t = 0.14 #C -1.777 (saturated), -1.705 (:1), -1.637 (:2), -1.571 (:3). Since the elution time is between 1.52 (14:0) and 2.90 (16:0), the methyl ester probably has 14 C atoms. Inserting this in the above equations (or locating it on parallel lines on the graph), the time comes out right for 14:1.

2. A DNA insert has been cloned in a vector, original size 2.3 kb. Inserting this piece between the Xba I and Kpn I sites, which are 84 bp apart, has removed this amount from the vector. The complete plasmid is digested with XbaI, with Kpn I, with Xba I and Kpn I together, with Xba 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 XbaI Kpn I XbaI + Kpn I XbaI + Kpn I +
Hind III Hind III

cm moved 2.89 2.89 3.49, 4.51 3.3, 4.9 3.66, 4.18

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 (1.5 pts each - 9 total). Use one of the graphs below if desired.

The data for the standard define a line cm = 5.665 - 3.33 log kb. Using this, the complete linear plasmid is 6.8 kb. The double digest with Kpn1and Xba1 yields pieces 4.5 kb (insert) and 2.3 kb (vector). The double digest pieces with Xba1 and HindIII are 5.1 and 1.7 kb. The double digest pieces with Kpn1 and HindIII are 4.0 and 2.8 kb.

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

The small pieces from the double digests with HindIII add up to 4.5 kb, therefore the HindIII site lies in the insert, 1.7 kb from the Xba1 site, 2.8 kb from the Kpn1 site.

3. Five microliters of a 50 µL PCR reaction mixture, after 25 cycles of reaction, is diluted to 0.14 ml in 0.3 M Na acetate, and 0.36 ml (2.5 vol) ethanol added to precipitate the synthesized DNA (to separate it from primers and unincorporated deoxynucleotide triphosphates). The precipitated DNA is redissolved in 0.25 ml TE. The absorbance at 260 nm is 0.054.

a) What DNA concentration is this? (2 points) 0.0027 mg/ml = 2.7 µg/ml.

b) What total amount of DNA is in the cuvette? (1 point) 2.7 µg/ml x 0.25 ml = 0.675 µg.

c) What total amount of DNA was synthesized? (1 point) x0.675 = 6.75 µg.

d) If the DNA synthesized is 1.6 kbp long, how many moles is this? (Assume mol. wt. of a deoxynucleotide = 300) (1 point) 1600 nucleotide pairs x 300 g/mole nucleotide x 2/pair = 960,000 g/mole. 6.75 x 10^-6 g ∏ 9.6 x 10⁵ g/mole = 7.03 x 10^-12 moles.

e) If the efficiency of the reaction was 90%, i.e. the amount of DNA after each cycle is 1.9x the amount present at the start of the cycle, what was the amount of template present in the original sample which has been amplified? (5 pts) 

Calculate the amplification factor of 25 cycles at 90% efficiency as antilog (25 log 1.9) = antilog 25·0.2788 = antilog 6.969 = 9,307,650. 6.75 µg ∏ 9,307,650 = 7.25 x 10^-7 µg, or 7.55 x 10^-19 moles. (This is up from 2.03 x 10^-19 moles if amplification is perfectly efficient, 2.0/round.)