Quiz 2, BICH 410 (TuTh 12:45-2), Tuesday, Oct. 8, 1996
Clearly circle the letter corresponding to your answer for each problem. There are seven problems, six of which count 3 points, and one is worth 2 points. No partial credit given for multiple-choice questions.
1. A hypothetical purification table for a protein is shown below.
|Step in protein purification|| total activity (Units)|| total mass of protein (mg)|
|crude extract|| 2000 U|| 1000 mg|
|after step A|| 2000 U|| 500 mg|
|after step B|| 1600 U|| 400 mg|
|after step C|| 1000 U|| 100 mg|
|after step D|| 500 U|| 25 mg|
|after step E|| 400 U|| 20mg|
What step gives the greatest purification?
- A. step A
- B. step B
- C. step C
- D. step D
- E. step E
Answer: C; calculate specific activity (U/mg) for each row; step with greatest purification is one that shows greatest fold-increase in specific activity
2. Two alanine residues are separated by 12 other amino acids within an alpha helix of a protein. Approximately how far apart are these two alanine residues?
- A. 0.5 nm
- B. 1 nm
- C. 2 nm
- D. 5 nm
- E. 20 nm
Answer: C; residues are separated by 14 amino acids; divide 14 by 3.6 amino acids per turn = 3.9 turns of helix; then (3.9 turns)(0.56 nm/turn) = 2.2 nm
3. If the same two alanine residues were separated by 12 other amino acids within a single beta strand instead of an alpha helix, how would their spacing be changed relative to that in your answer to question 2?
- A. longer distance than in alpha helix
- B. same distance as in alpha helix
- C. shorter distance than in alpha helix
- D. cannot predict this answer unless you know the composition of all other amino acids in this beta strand
- E. alanine residues cannot exist in a beta strand
Answer: A; the beta strand is a more extended structure than the alpha helix
4. (2 pts) Performic acid:
- A. cleaves a polypeptide backbone specifically at methionine residues.
- B. oxidizes disulfide bonds.
- C. denatures proteins by disrupting hydrophobic interactions.
- D. reacts with proline residues in proteins to change their configuration from trans to cis.
- E. none of the above
5. The most numerous amino acid in collagen is:
- A. proline
- B. hydroxyproline
- C. lysine
- D. glycine
- E. alanine
Answer: D; collagen repeat is gly-X-Y, so gly is at least 33 percent of total
6. Which of the following statements concerning X-ray crystallography is not true?
- A. Only crystallized proteins can be analyzed.
- B. The electron density maps are obtained by applying the Fourier transform to the scattered electron intensities.
- C. This technique can be used only to determine structures of proteins of molecular weight less than approximately 30,000.
- D. The first protein whose structure was determined by this technique was myoglobin.
- E. A structure with a resolution of 2 Å gives atomic details about the protein.
Answer: C; this statement refers to the size limitation for determination of structures by NMR
7. Determine the sequence of an octapeptide using the following information.
i) The amino acid composition of the peptide is: alanine, arginine (2 moles per mole peptide), leucine, lysine, methionine (2 moles per mole peptide), valine.
ii) Upon treatment of the octapeptide with fluorodinitrobenzene, dinitrophenylmethionine was detected after hydrolysis and analysis by HPLC.
iii) Treatment of the octapeptide with trypsin resulted in two dipeptides and a tetrapeptide. The compositions of the dipeptides were as follows: dipeptide 1: lysine, methionine; dipeptide 2: arginine, leucine.
iv) Treatment of the octapeptide with cyanogen bromide resulted in free methionine, a dipeptide and a pentapeptide. The composition of the pentapeptide was: alanine, arginine, leucine, lysine, methionine.
The sequence of the octapeptide is:
- A. methionine-lysine-leucine-arginine-alanine-methionine-valine-arginine
- B. methionine-arginine-lysine-methionine-arginine-leucine-alanine-valine
- C. methionine-lysine-methionine-arginine-leucine-alanine-lysine-arginine
- D. arginine-valine-methionine-alanine-arginine-leucine-lysine-methionine
- E. methionine-leucine-arginine-lysine-alanine-methionine-valine-arginine
From statement ii), met must be on N-terminus
From statement iii), three peptides are obtained after trypsin digestion when there are three basic amino acids (2 arg, one lys); hence, arg must be at the carboxyl end; sequence of peptides are: met-lys, leu-arg, and X-X-X-arg
From statement iv), since there are two methionines, and the fragments are a single met, a pentapeptide that must end with met, and a dipeptide, the dipeptide must be on the C-terminus. Hence, it is val-arg, given that val is not in the pentapeptide. Using the information gleaned from statements iii) and iv), the sequence of the pentapeptide must be: lys-leu-arg-ala-met.
Therefore, putting it all together, the answer is: