Fall 2006 - Prelim 1 Solutions

Fall 2006 - Prelim 1 Solutions - Prelim regrade requests...

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Unformatted text preview: Prelim regrade requests must be submitted in writing to Shirley Soule, 217 Stimson, no later than 2:00 PM on Friday, 10/27. (Shirley’s hours are M,W 7:30 AM – 4:00 PM, F 10:00 AM – 2:00 PM). 1. a. (6 points) Draw the structures of leucine and isoleucine at pH 7.0. See structures on Fig. 3-5 (Nelson & Cox, p. 79). b. (2 points) Poly-L-leucine in an organic solvent such as dioxane is a helical, whereas poly-L-isoleucine is not. Why do these amino acids have different helix-forming tendencies? The methyl group in isoleucine is closer to the main chain which causes it to interfere sterically with alpha helix formation. 2. (2 points) A mutation that changes an alanine residue in the interior of a protein to a valine is found to lead to a loss of activity. However, activity is regained when a second mutation at a different position changes an isoleucine residue to glycine. How might this second mutation lead to a restoration of activity? The first mutation destroys activity because valine occupies more space than alanine does, and so the protein must take a different shape, assuming that this residue lies in the closely packed interior. The second mutation restores activity because of a compensatory reduction of volume; glycine is smaller than isoleucine. 3. (2 points) Histones are proteins of eukaryotic cell nuclei. They are tightly bound to DNA, which has many phosphate groups. The pI of histones is very high, about 10.8. Considering these two facts, what amino acids must be present in relatively large numbers in histones? Lysine and Arginine have pKa’s of 10.9 and 12.5 respectively. Since they are abundant, they contribute greatly toward the pI of 10.8. Because they are positively charged at neutral pH, they bind the negatively charged phosphate groups on DNA. 4. (2 points) In the course of purifying an enzyme, a researcher performs a purification step that results in an increase in the total activity to a value greater than that present in the original crude extract. Explain how the amount of total activity might increase? (Nelson & Cox, pp. 89-92) An inhibitor of the enzyme being purified might have been present and subsequently removed by a purification step. This removal would lead to an apparent increase in the total amount of enzyme present. BioBM 330 prelim 1, Fall '06 Answer Key page 2 5. You have been provided with an octapeptide with the following sequence: A V G F R V K S a. (2 points) On the above sequence, draw an arrow(s) labeled with a “ T ” to show the bond(s) that would be broken when this octapeptide is treated with trypsin . (specificity from protein evolution activity) Because its specificity is for cleavage at positively charged residues, trypsin cleaves on the carboxyl (right hand side) of R and K b. (2 points) On the above sequence, draw an arrow(s) labeled with a “ C ” to show the bond(s) that would be broken if the original octapeptide is treated instead with chymotrypsin ....
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Fall 2006 - Prelim 1 Solutions - Prelim regrade requests...

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