Name: TF: Section Time

Life Sciences 1A Midterm Exam 1

Monday, October 16, 2006 Please write legibly in the space provided below each question. You may not use calculators on this exam. We prefer that you use non-erasable pen when writing your answers. A significant number of completed exams will be photocopied by the teaching staff. Please write your name on each page of the exam. There are SIX multi-part questions in this exam. We recommend that you first read through all questions and begin with the questions that are easiest for you. Be sure to take a look at all questions before the end of the exam! Please write your TF’s name and section time on the front page only. Question 1 /18 Question 2 /24 Question 3 /28 Question 4 /20 Question 5 /36 Question 6 /24 Total /150

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1. Domoic acid is a potent neurotoxin produced by certain species of red algae. If ingested in significant quantity, the toxin can cause neuronal damage, often resulting in bizarre behavior in animals.

Alfred Hitchcock’s movie The Birds is partially based on series of events involving the bizarre behavior of presumably intoxicated birds off of the coast of Northern California, in 1961.

HO

O N O

OHO

OHDomoic acid

pKa = 2.1

HH

pKa = 3.7

pKa = 5.0

pKa = 9.8"Bond X"

a. (6 points) Identify the chiral centers in domoic acid (circle above)

b. (4 points) Is “Bond X” cis or trans? trans

c. (8 points) Ignoring the double bonds, how many potential stereoisomers does domoic acid have? Briefly explain.

2n = 24 = 16. Each stereocenter has two possible positions. Each one can function independently of the others and therefore there are 2n the number of potential stereoisomers.

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2. Scientists often use PCR to amplify genes from double-stranded circular DNA templates (drawn as a solid strand and a complementary dotted strand below). Three sequences from the solid (outer) strand are shown.

a. (6 points) Provide the sequence of two six-base DNA primers that would be needed to amplify the 2,500 base pair region AB. Explain which strand (dotted or solid) each of your primers will bind to.

5’-CCATGC-3’ binds to dotted, 5’-GGTTAA-3’ binds to solid strand

b. (6 points) How many strands of DNA consisting of ONLY AB (either solid

or dotted) would be present after four cycles of PCR using the primers you described in part a?

22

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c. (6 points) An enzyme was added to your DNA while it was being prepared for the PCR reaction. The enzyme cuts both DNA strands in the circular template at the 5’-GGATCC-3’ sequence. If PCR is performed on the cut template using the primers described in part a, draw the newly synthesized DNA strands that would be generated by DNA polymerase in the reaction. For example, the solid-strand spanning regions A and B would be represented as:

A B5' 3' 5'3'

d. (6 points) How many DNA strands of each of the product(s) you drew in

part c would there be after four rounds of PCR under the conditions described in part c?

4 each (4 solid A, 4 dotted B)

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3. An enzyme involved in vitamin metabolism contains a Lysine (Lys) residue in a region that is required for its enzymatic activity. The Lys is positioned close to a glutamic acid (Glu) residue that is important in properly orienting the Lys side chain within the enzyme. Experiments have shown that the activity of the enzyme is dependent on pH, as shown in the graph below.

a. (6 points) Draw both forms (protonated and deprotonated) of Lys on the peptide backbone structures provided below and list the ratio of protonated to deprotonated at pH = 7. Refer to the pKa table provided at the end of your exam.

HN

O

HN

O

NH3+ NH2

Protonated10,000

Deprotonated1

Polypeptide backbone

Lys Glu

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b. (6 points) Draw both forms (protonated and deprotonated) of Glu on the peptide backbone structures provided below and list the ratio of protonated to deprotonated at pH = 7.

HN

O

HN

O

Protonated1

Deprotonated1,000

OH

O

O-

O

c. (8 points) One type of interaction between the side chains of Lys and Glu can best explain the results shown in the above graph. Name the interaction and describe how the interaction is affected at the different pH values shown in the graph above.

The interaction is most likely to be ionic in nature. At pH = 7, the nitrogen on

the Lys side chain carries a formal positive charge, while the oxygen atom of the carboxyl group on Glu carries a formal negative charge. These two charges will attract each other. As the pH is decreased from pH = 7 to pH = 2, the activity of the enzyme gradually declines. At pH = 2, the majority of the Glu side chain is protonated and no longer carries a negative formal charge. Thus, at lower pH values an ionic interaction would be expected to weaken.

Hydrogen bonding would also be consistent with the above data, since the strength of a H-bond acceptor depends on the electron richness of the group. In other words, COO- is a better H-bond acceptor than COOH. In addition, since we don’t provide a detailed structure a student could propose a H-bond involving the O lone pair that disappears when Glu gets protonated—that answer would also be plausible and correct. I think we don’t need to alter the problem so long as the answer key allows H-bonding to be given full credit if there is a good explanation.

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In addition to the wild-type (wt) enzyme containing the Glu residue, a mutant form of the enzyme exists that contains a Glutamine (Gln) residue in place of a Glu. The graph below compares the activities of the wild-type and mutant enzymes.

d. (8 points) Draw the predominant form of the Gln side chain on the peptide backbone structure provided below that you would expect to see at pH = 7, and explain how the above graph best supports the interaction you described in part c.

HN

O

NH2

O

The results support an ionic interaction. When Glu is replaced by Gln, very little enzymatic activity is seen at pH = 7 in the mutant enzyme. Because Gln lacks a formal negative charge at pH = 7, no ionic interaction could take place. Only H-bonds can form, but these can also form to some extent with Glu. Since the Lys at pH 7 or 2 is an obligate H-bond donor, therefore the Glu/Gln would serve as an H-bond acceptor. Again, if a student proposes that the Glu O that is replaced by a NH2 group in the Gln mutant is serving as a H-bond acceptor, this data does not rule out H-bonding.

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4. Proteases are enzymes responsible for breaking peptide bonds. The mechanism for the first step of amide bond hydrolysis is shown below for one type of protease using arrow pushing notation.

a. (8 points) Draw the product(s) of the reaction in the box provided, indicating any formal charges that might arise.

HN NO

HN

H

O

HN NH

O HN

O

b. (2 points) What is the geometry of the circled carbon atom?

Trigonal planar c. (2 points) Does the geometry of this carbon change in the product? If so,

what is the new geometry of the carbon atom in the product? tetrahedral d. (8 points) The pKa of the histidine side chain is 6.5. What is the value of

the following ratio (see picture below)? You may leave your answer as a mathematical expression.

HN N

HN NH+

H+

The answer is 10^(6.5) (a little over a million). Alternatively, you could invert my expression to make the question simpler—then the answer would be 10^(-6.5)

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5. Histones are proteins that help facilitate the efficient and highly compact packaging of DNA within cells. The peptide shown below comes from the N-terminus of one of the histone proteins, H4. The first and 55th amino acids are labeled below.

N-Ser-Gly-Arg-Gly-Cys-Gly-Gly-Cys-Gly-Lys-Gly-Lys-Gly-Gly-Ala-Lys-Arg-C

a. (6 points) Draw the peptide structure for the first 3 amino acids (shown in bold) of the above sequence at physiological pH (~7.5) using the pKa table at the end of the exam. You do not need to designate stereochemistry.

+H3NNH

HN

O

OOH

NH

O

O-

NH2

NH2+

+H3NNH

HN

O

OOH

NH

O

NH2

NH2+

or

The c-term can be labeled either as a carboxylate or end just before it.

b. (6 points) This peptide has a very flexible secondary structure that can adopt many different conformations. Based on the primary sequence, explain this observation.

Nearly half of the residues are glycine residues. The small size of this amino acid enables the formation of many different structures as there would be less steric hindrance to restrict potential bond orientations and angles.

c. (4 points) Histone H4 adopts a well-defined secondary structure starting around amino acid 55. What is the name of the secondary structure indicated by the arrows?

Alpha-helix

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d. (8 points) State the intramolecular force(s) most important for stabilization of this secondary structure and which part(s) of the peptide are involved.

Hydrogen bonds formed between the peptide backbone carbonyl oxygens and peptide backbone amine hydrogens. e. (4 points) Histones are capable of both binding DNA and compacting it. At

a pH of 7, what is the net charge of the histone peptide redrawn below? N-Ser-Gly-Arg-Gly-Cys-Gly-Gly-Cys-Gly-Lys-Gly-Lys-Gly-Gly-Ala-Lys-Arg-C The net charges is +5. The positive charges have been labeled red, and the negatives in blue.

f. (8 points) What is the likely role that this charge plays in the function of the histone protein?

This positive charge promotes interaction with the negatively charged DNA. This neutralizes the negative charges on the phosphate backbone of DNA, allowing tighter compression.

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6. Some proteins adopt their native conformation in non-polar environments. A major class of these includes proteins found in biological membranes. In these proteins the polar and charged residues are found in the interior of the protein whereas the non-polar residues face the non-polar surroundings of the non-polar membrane. Consider the following scenario where a protein is placed in a non-polar hydrocarbon solvent such as hexane

Legend: hexane =

Unfolded

Protein

Folded

Protein

a. (8 points) The folded protein forms more ionic interactions and hydrogen bonds between amino acid groups than the unfolded protein. What is the sign of ΔH for the folding of this protein? Briefly explain your reasoning.

ΔH < 0 only. Forming bonds is favorable. The polar and charged residues form hydrogen bonds and other ionic interactions only in the folded form of the protein.

b. (8 points) What is the sign of ΔSprotein for the unfolding of this protein in hexane? Briefly explain your reasoning. ΔS > 0 as more conformation is available in the unfolded versus unfolded state.

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Scientists have recently discovered organisms living near deep sea hydrothermal vents at temperatures above 100º C. The enzymes found in these organisms are called extremozymes because they can function at such high temperatures. The following observations have been made for the folding of one extremozyme:

@ 25º C ΔG > 0 @ 90º C ΔG < 0

c. (8 points) What is the sign of ΔS for folding at 90ºC? Rationalize your answer using ΔG = ΔH - TΔS.

ΔS is greater than 0 (positive). Consider the four cases: 1. ΔH > 0, ΔS > 0 2. ΔH > 0, ΔS < 0, ΔG always greater than 0 3. ΔH < 0, ΔS > 0, ΔG always less than 0 4. ΔH < 0, ΔS < 0 Case 2 and 3 are temperature independent and thus cannot explain the folding of the extremozyme. Thus the remaining cases to consider are 1 and 4. From the question, the process becomes favorable as temperature increases. Based on the equation for free energy, only equation 1 makes sense. As temperature increases, the value of –TΔS becomes more negative, and thus helps contribute to ΔG become more negative as well. Case 4 would not work as when temperature increases, the term –TΔS would become more positive and thus ΔG would become more positive and the reaction should become less favorable.

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Ionizable group pKa Asp side chain 4.0 Glu side chain 4.0 His side chain 6.5 Cys side chain 9.3 Tyr side chain 10.2 Lys side chain 10.9 Arg side chain 12.0 Ser side chain 13.0 Gln side chain 16.0 Amine terminus 9.8 Carboxy terminus 2.3