LS1a Fall 2014 Problem Set #2 Due Monday 10/6 at 6 pm in the drop boxes on the Science Center 2 nd Floor

Transcription

1 LS1a Fall 2014 Problem Set #2 Due Monday 10/6 at 6 pm in the drop boxes on the Science Center 2 nd Floor Note: Adequate space is given for each answer. Questions that require a brief explanation should be answered only in the provided space. I. Basic Concept Questions 1. (12 points) Fill out the table below using the filled-in boxes as guides. Draw the structures of the indicated amino acids (using standard drawings) at ph = 7 and include the one-letter and three-letter codes where appropriate, as well as a description of their side chain properties (i.e., polar, nonpolar, positive at ph 7.0, or negative at ph 7.0). 1

2 2. (7 points) Shown below is the structure of the amino acid arginine (drawn at ph 7). a. (3 points) On the structure above, identify this amino acid s side chain, amine, and carboxylate. b. (4 points) Shown below is a tripeptide. The resonance structure of a peptide bond causes the nitrogen atom of the peptide bond to adopt a trigonal planar molecular geometry (just like in the week #4 section handout and in the example of ethylene on pages of the notes for lecture 7). Consider the peptide bond between amino acids 1 and 2 (amino acid 1 has R1 as its R-group, and amino acid 2 has R2 as its R-group) and put a box around each of the six atoms that are coplanar (i.e., those that must reside in the same plane) due to the resonance form of the peptide bond between amino acids 1 and 2. Similarly, draw circles around each of the six atoms that are coplanar due to the peptide bond linking amino acids 2 and 3. 2

3 3. (22 points) The complex series of reactions through which organisms build amino acids often involve other amino acids as intermediates. A number of these reactions are outlined below. Each species is identified by a number. Multi-step processes are shown with multiple arrows ( ). a. (14 points) Indicate whether each numbered species is one of the 20 naturally occurring amino acids that is found in proteins (yes or no). If the answer is yes, write its name. Amino Acid? Name (if Yes) Amino Acid? Name (if Yes) 1. Yes No 8. Yes No Aspartic Acid 2. Yes No _Threonine_ 9. Yes No _Glutamic Acid_ 3. Yes No 10. Yes No _Glutamine 4. Yes No 11. Yes No _Asparagine 5. Yes No Lysine 12. Yes No 6. Yes No 13. Yes No 7. Yes No 14. Yes No Arginine 3

4 b. (8 points) The molecule shown below can undergo one of two processes, each leading to a different amino acid (after additional steps). The transition state from each reaction is shown below. In the space provided, draw the product of each transition state and give the name of the amino acid that is the most likely end product of each transition state by taking into account amino acid side chain structures. 4

5 4. (8 points) As we learned in lecture, keratin, the protein in hair, contains disulfide bonds between cysteine side chains that must be broken when hair stylists give perms. Shown below is the arrow pushing mechanism describing how thioacetic acid (CH 3COSH; shown in bold) is used to accomplish this. a. (4 points) In the space provide above, draw the products of the arrow pushing mechanism. b. (4 points) In the space below, draw the transition state for this reaction. 5

6 5. (11 points) The reaction of the cysteine side chain with a chemical reagent called N- ethylmaleimide ( NEM ) is shown below. a. (5 points) Onto the structures of the reactants, draw the arrow pushing mechanism that leads from the reactants to the products via the transition state that is shown. b. (6 points) Consider two unfolded samples of a protein whose folded structure is stabilized by disulfide bonds (for example: keratin). One of the two unfolded protein samples is treated with NEM and the other unfolded protein sample is left untreated. If both samples are then allowed to fold, which protein sample (the NEM-treated or untreated ) would feature a larger ratio of folded to unfolded protein at equilibrium? Briefly explain your answer. NEM treatment prevents cysteine residues from being able to form disulfide bonds. You can consider the reaction between cysteines to form a disulfide bond and the reaction of a cysteine with NEM to be two coupled reactions: a cysteine can EITHER react with NEM to form a covalent bond OR two cysteines can react with each other to form a covalent disulfide bond. Disulfide bonds can only form between cysteines with available SH ( sulfhydryl ) groups at the ends. Given a protein folding reaction, the reactants are the unfolded protein and the products are the folded protein. Since disulfide bonds stabilize the folded structure, the folded protein is lower in energy when the disulfide bonds are made than when they are not made. Without the disulfide bonds, ΔG folding is therefore less favorable (since the folded protein product is now higher in energy without the disulfide bonds). Since ΔG folding is less negative, K eq for folding will be smaller for the NEM treated sample. Since K eq is larger for the sample without NEM treatment, the ratio of folded to unfolded protein will be higher for the untreated sample. 6

7 II. Applied Concept Questions 6. (17 points) Later this semester we will be using a technique called chromatography to separate proteins in order to purify a protein called the Green Fluorescent Protein (or GFP ). One type of chromatography involves separating peptides based on their net charges. One way of doing this is to pass a mixture of peptides through a column that is lined with positive charges. As the mixture flows through the column, peptides with negative charges interact with the positively-charged column and pass through the column more slowly. The greater the negative charge of the peptide, the slower it will move through a positively-charged column. Because a positively-charged column separates different peptides based on the differences between their net charges, we first need to choose an appropriate ph to maximize the difference between the peptides' net charges. A mixture of two peptides, NH 2-C-D-R-Y-COOH (peptide 1) and NH 2-K-H-E-W-COOH (peptide 2), are both dissolved in three different buffered solutions, A, B, and C, shown below. Solution A: ph 1.0 Solution B: ph 5.0 Solution C: ph 13.0 a. (10 points) In the space below, draw both peptides as they would predominately exist in solution A. Refer to the lecture notes for pk a values of the amino acid side chains. The pka of the carboxyl terminus is ~ 2.3, and the pka of the free amino terminus is ~ 9.8. In your diagram, draw all amino acids in the trans conformation. You do not need to show stereochemistry. 7

8 b. (4 points) Calculate the net charges for each peptide in each of the solutions (A, B, and C). In which solution (A, B, or C) would there be the greatest difference in net charge between the two peptides? Solution A: Peptide 1, +2; Peptide 2, +3 Solution B: Peptide 1, 0; Peptide 2, +1 Solution C: Peptide 1, -4; Peptide 2, -2 Solution C gives the greatest difference in net charge. c. (3 points) Which solution (A, B, or C) would allow for the best separation of the two peptides by a positively charged column? Which peptide would spend more time bound to the column? Briefly explain your reasoning. Solution C because it generates the greatest possible difference in net charge that is possible between the two peptides. Peptide 1 is more negative and would spend more time bound to the column than the less negative peptide. 8

9 7. (8 points) Three-dimensional representations of two tripeptides are shown below. These two images were generated using PyMOL, the software used in lab to visualize 3- D molecules. [Note: green is carbon; blue is nitrogen; red is oxygen; yellow is sulfur; hydrogen atoms are not shown. If you print this question out in black and white, it will probably help to at least look at the color picture on the computer monitor while considering this question.] a. (4 points) Write the amino acid sequence of each peptide starting from the N- terminus. Label the N- and C-termini in your answer. Peptide 1: N-terminus Glu Trp Pro C-terminus Peptide 2: N-terminus Met Pro Thr C-terminus b. (4 points) Circle all α-carbons present in the structures shown above. Draw an arrow pointing to any cis peptide bond(s) that are present. 9

10 8. (5 points) A schematic diagram of a 372 amino acid protein is shown below. Two stretches of amino acids that interact with each other in the folded protein to form a β- sheet are highlighted: a. (3 points) The diagram below indicates two amino acids, glycine 181 and tyrosine 164. The numbers indicate the amino acid s position in the protein s primary sequence (where 1 is the amino acid at the N-terminus and 372 is the amino acid at the C-terminus of this particular protein). Based on these numbers, fill in the blank next to each of the two indicated amino acids to show each amino acid s number in the primary sequence. Briefly explain how you arrived at your answer. First one must identify the N-/C-termini; on the top strand, the N terminus is to the left and on the bottom strand, the N-terminus is to the right. The Ile is two positions closer to the N-terminus than Tyr 164, so it must be in position number 162. The Ala is three positions further from the N-terminus than Gly 181, so it must be in position number 184. b. (2 points) In the figure β-sheet shown above, draw dashed lines ( ) to indicate the hydrogen bonds that stabilize this segment of secondary structure. 10

11 9. (10 points) Below is an image of two proteins (one green and the other cyan) binding to one another via their β-sheets. The colors are important for this question, so you may want to do this problem looking at a color print out or online through the course website. Protein A Protein B a. (4 points) The image to the left shows the same two proteins in the same orientation, but hides much of the two proteins and focuses the regions of through which these two proteins interact. The image shows only the polypeptide backbone (i.e., the side chains are hidden), and intermolecular interactions are shown in magenta. Use your understanding of the interactions that stabilize secondary structure to briefly explain the types of interactions proteins exploit to bind one another by aligning the edges of their β- sheets. In particular, what makes the exposed edges of β-sheets productive places for proteins to interact? Protein A Protein B β-strands adopt an extended shape in which the polypeptide backbone carbonyls and amines of successive residues point in opposite directions. These amide carbonyls and amides serve as hydrogen bond acceptors and donors with adjacent β -strands. β-sheets are held together by the many hydrogen bonds that occur between adjacent β-strands. 11

12 While the strands within the sheets make hydrogen bonds to adjacent strands on both of sides, the β-strands at the edge of a sheet are only adjacent to one β- strand, leaving the potential hydrogen bonds that could be made to another β- strand unsatisfied. β-strands at an exposed edge of a sheet are therefore capable of forming hydrogen bonds with another β-strand from another protein that is stretched out in the same conformation (with its amines and carbonyls in the appropriate spatial arrangement) because it is also a β-strand. The structure of β-strands optimally orients each carbonyl and amine group to hydrogen bond with an adjacent strand, such that a β-strand at the edge of a β-sheet in one protein is optimally oriented to form as many hydrogen bonds as possible with a β-strand at the edge of a β-sheet from a different protein. b. (6 points) In addition to interacting using the edges of their β-sheets, the proteins also interact through several side chains. One example, in which lysine, the 84 th amino acid on the cyan protein, interacts with glutamate, the 31 st amino acid on the green protein, as shown below: K84 E31 To determine the importance of this interaction between lysine and glutamate, scientists conducted three experiments in which they changed which amino acids are present at these same positions. In experiment 1, scientists replaced lysine with glutamate at amino acid position 84 of the cyan protein. In experiment 2 the lysine at position 84 of the cyan protein and the glutamate at position 31 of the green protein were both swapped for alanines. In experiment 3, the lysine of the cyan protein is replaced by a glutamate and the glutamate on the green protein is replaced with a lysine. These experiments, along 12

13 with the ΔG binding describing the favorability of the two proteins binding one another, are summarized in the table below: Experiment Protein ΔG binding Control Unaltered amino acid sequence -7 kj/mol 1 K84E +3 kj/mol 2 K84A, E31A -1 kj/mol 3 K84E, E31K -7 kj/mol This chart uses a formalism when describing changes to a protein s amino acid sequence, in which the first letter indicates what the amino acid started out as, the adjacent number details where in the primary sequence the amino acid occurs, and the final letter indicates which amino acid the original one has been changed to: Original amino acid K84E Final amino acid Position in primary structure This example of K84E translates to amino acid 84, which is normally lysine, has been changed to glutamate. Using the information in the chart above, briefly explain why each experiment makes the protein binding interaction more or less favorable. In the control experiment with wild-type proteins, the positively-charged lysine makes an ionic bond with the negatively-charged glutamate. In experiment 1, lysine is replaced by a negatively-charged glutamate, which will be repelled by the glutamate on the green protein, making the binding reaction less favorable. In experiment 2, by exchanging both lysine and glutamate for alanines, the favorable ionic bond is lost, so binding is weaker than between the two unaltered proteins. However, since no repelling charges are introduced at the protein interface as in experiment 1, the binding interaction between the two proteins in experiment 2 is still more favorable than in experiment 1. In experiment 3, the two amino acids are swapped between the two proteins: the cyan protein now has glutamate instead of lysine, and the green protein has lysine instead of glutamate, such that the interaction between these two side chains is equivalent to the original, unaltered scenario. 13