2006 7.012 Problem Set 1 Due before 5 PM on FRIDAY, September 15, 2006. Turn answers in to the box outside of 68-120. PLEASE WRITE YOUR ANSWERS ON THIS PRINTOUT. 1. For each of the following parts, pick the best answer out of the choices: covalent bonds, hydrogen bonds, ionic bonds, hydrophobic forces. What kinds of bonds hold together: (a) the O and two Hs in one water molecule? Covalent bonds. Oxygen has 2 unpaired electrons in its outer shell to pair with each Hydrogen s single unpaired electron. (b) molecules of water with each other in a glass of water? Hydrogen bonds. The electronegativity of O causes a partial negative charge on the O s and a partial positive on the H s in water; an O from one water molecule and an H from another water molecule interact to form a hydrogen bond. (c) lipids with each other when placed in water? Hydrophobic forces. Nonpolar parts of molecules interact to avoid unfavorable interactions with the polar solvent. This is driven by the desired increase in entropy (decrease in order) because clumping hydrophobic molecules together with each other reduces the number of water molecules in total that must order themselves around the hydrophobic molecules. (d) the atoms in a single 6 carbon sugar? Covalent bonds. Carbon has 4 unpaired electrons and thus forms 4 covalent bonds to complete its outer shell. (e) sugars in a chain of glycogen or starch? Covalent bonds. Dehydration reactions form covalent bonds (called glycosidic linkages) between monosaccharides to create polysaccharides. (f) the primary structure of a protein? Covalent bonds. Dehydration reactions form covalent peptide bonds between amino acids to form a polypeptide chain. (g) an alpha helix? 1
Hydrogen bonds. Each N-H + - hydrogen bonds with an O=C that is located 3.5 amino acids away in the helix. It is these hydrogen bonds that hold an alpha helix into its helical shape. (h) a beta sheet? Hydrogen bonds. Each N-H + hydrogen bonds with an - O=C from another peptide chain. It is these hydrogen bonds that hold the sheet of chains together into a beta sheet. (i) two cysteines in a disulfide bond? Covalent bonds. Two cysteines can form a covalent bond (called a disulfide bond) with each other by being oxidized. The reaction is: SH + HS S-S + 2H. The original SH groups come from the R group on cysteine. 2. Below is a molecule called sodium dodecyl sulfate, which is a detergent. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 O SO 3 Na + (a) Put a circle around the part of this molecule that is hydrophobic. Circle the Carbon chain up to but not including the Oxygen. Long carbon chains are very hydrophobic, because they cannot form hydrogen bonds with water, and thus do not dissolve in water. (b) Draw a box around the part of this molecule that is hydrophilic. Box in the SO 4 and Na. Charged and polar molecules are very hydrophilic because they can form hydrogen bonds with water, thus allowing them to dissolve in water. (c) What do you think would happen to cells if you added detergent to them? The lipid membranes would dissociate, causing the cell to break open. Detergents are amphipathic just like lipids and would therefore insert into the membrane, disrupting the bonds between individual lipids in the bilayer and thus solvating the lipids. 2
3. **For this problem, you will need to use the internet, so as to view the structure of the protein lysozyme using a program that allows you to view protein structures.** To begin, go to the site: http://web.mit.edu/viz/7.01x/ Then click on Problem Set 1 Problems. To print out the instructions for this problem, click on Problem 3 (word document). To do the problem on the internet while following the instructions, click on Problem 3 (pdf with live links). If you do not have a computer easily available, room 37-212 has an Athena PC cluster that will run the program you need. For the 37-212 cluster schedule, see http://web.mit.edu/acis/labs/37-312.html If you would like to do this problem in the presence of technical staff who are familiar with this computer program, there will be staff present on Thursday 9/14/06 from 4-9pm in 26-152. You are welcome to stop by anytime during that interval to use one of the computers in that room to do the problem, or to ask the staff questions about the program. Please write your answers to the protein viewing question entitled Problem 3 in the space below. NOTE: The protein viewing question entitled Ungraded Exercise (that can be found on the same website) will not be graded and is available for your interest. In this problem, we examine the primary, secondary, and tertiary structure (see Fig 3.6 on p. 42 of your text) of a specific protein (lysozyme, see Fig 3.7 on p. 43). (a) Lysozyme consists of 129 amino acids. i) List in order the 12 amino acids numbered 4 through 15 in lysozyme. Gly Arg Cys-Glu-Leu-Ala-Ala-Ala-Met-Lys-Arg-His ii) What level of protein structure does this represent? Primary. Primary structure is defined as the sequence of amino acids. (b) These 12 amino acids also make up an α-helix in lysozyme. i) What level of protein structure do α-helices and β-sheets represent? Secondary. Secondary structure consists of local regions of shape in a protein s structure. ii) Do the side chains of the amino acids in a helix point into or out of the helix? They point out. The geometry of the hydrogen bonds between amino acids in the chain forces the side chains out. 3
iii) What type of bond is primarily responsible for maintaining secondary structure? Hydrogen bonds. See question #1. iv) What part of the amino acid participates in this bond (side chain or backbone)? The backbone. The R groups are facing out, but what is holding the helix into place are the hydrogen bonds between each N-H + and the - O=C that is located 3.5 amino acids away in the helix. (c) The tertiary structure of a protein is formed by bending and folding, with the interactions between the amino acid side chains determining this structure. Below we list four kinds of tertiary interactions between side chains that are possible and four sets of residues. i) Match up which set of residues belongs to which type of interaction (you will need information from Fig 3.2 on p. 39 to answer this). Draw a line from each choice on the left to one choice on the right. A. Hydrogen bond I. Residues 76, 94 B. Ionic bond II. Residues 51, 68 C. Disulfide bridge III. Residues 58, 63, 98, 108 D. Hydrophobic cluster IV. Residues 1, 7 RESIDUES 76 AND 94 ARE BOTH CYS, SO THIS IS A DISULFIDE BRIDGE (choice C ABOVE) RESIDUES 51 AND 68 ARE THR AND ARG. FROM FIG 3.2 on p. 39, ARG HAS A POSITIVELY CHARGED SIDECHAIN AND THR HAS A POLAR BUT UNCHARGED SIDE CHAIN, SO THIS IS A HYDROGEN BOND (choice A ABOVE). 4
RESIDUES 58, 63, 98, 108 ARE ILE, TRP, ILE, TRP, SO THIS IS A HYDROPHOBIC CLUSTER (choice D ABOVE). RESIDUES 1 AND 7 ARE LYS AND GLU. LYS HAS A POSTIVELY CHARGED SIDE CHAIN AND GLU HAS A NEGATIVELY CHARGED SIDE CHAIN, SO THIS IS AN IONIC BOND (B ABOVE). ii) Which set of residues represent the strongest interaction? Disulfide. Covalent bonds are the strongest of the bonds, at about 80 kcal/mol. iii) Which set of residues represent the weakest interaction? Hydrogen bonds or Hydrophobic interactions. Both are rather weak and depend on such variables as temperature, distance between atoms, and surface area (for hydrophobic surfaces). Either answer was accepted for this part. (Ionic or covalent were not accepted, as these are the two strongest of the four types of bonds.) 4. (a) Draw the peptide leu ala arg in the form it would be at ph = 7. Peptide bonds O O O H ll H ll H ll N-terminus H 3 N + C C N C C N C C O H H CH 2 CH 3 (CH 2 ) 3 CH + HN-C=NH 2 CH 3 CH 3 NH 2 C-Terminus 5
(b) Label the N and C termini. (c) Circle each peptide bond in your drawing above. (d) Which of the amino acids listed below do you think could have a phosphate group covalently added onto their R groups? Circle all that apply. All those with free hydroxyl (OH) groups can have a PO 4 group added onto them. Serine Alanine Threonine Valine Glycine Phenylalanine Methionine Leucine Isoleucine Tyrosine 5. Below is drawn the energy profile of the reaction: A + B C + D where A and B are the reactants, and C and D are the products. 22 20 Free energy (kcals per mole) 18 16 14 12 10 8 6 4 A + B C + D 2 Reaction Progress (a) What is the value of ΔG for the forward reaction (reactants to products)? 3 kcal/mol. (G products G reactants = 9 6 = 3). G = free energy (b) What is the value ΔG for the backward reaction (products to reactants)? 6
3 kcal/mol. (G reactants G products = 6 9 = 3) (c) What is the value of the activation energy for the forward reaction? 14 kcal/mol. G transition state G reactants = 20 kcal/mol 6 kcal/mol = 14 kcal/mol. (d) What is the value of the activation energy for the backward reaction? 11 kcal/mol. G transition state G products = 20 kcal/mol 9kcal/mol = 11 kcal/mol. (e) Which of the reactions would occur spontaneously: forwards only, backwards only, both, or neither? Only the backwards reaction could occur spontaneously under the correct conditions since ΔG is negative for this reaction. It would be favorable for C and D to convert to A and B since A and B are at a lower free energy. This reaction is catalyzed by the presence of a specific enzyme. (f) Would the enzyme affect ΔG for the forward reaction (reactants to products)? No. ΔG remains constant when enzymes are involved since you are only affecting the transition state, not the free energy states of the products and reactants. (g) Would the enzyme affect ΔG for the backward reaction (products to reactants)? No. Same reasoning as above. (h) Would the enzyme affect the activation energy for the forward reaction? Yes. Enzymes work by lowering the activation energy to make the reaction more favorable. (i) Would the enzyme affect the activation energy for the backward reaction? Yes. The law of microscopic reversibility says that the reaction must go back the same way it goes forward. This applies to enzyme-catalyzed reactions. (j) Which of the reactions are catalyzed by this specific enzyme: forwards only, backwards only, both, or neither? Both. The enzyme must lower the activation energy for the reaction in both directions. 7
6. (a) Draw the di-nucleotide sequence 5 -GC-3. phospho diester bond (b) Label the 5 and 3 ends in your drawing above. (c) Circle each phosphodiester bond in your drawing above. (d) Which of these secondary structures are legitimate folding structures that could be formed by RNA? (The sets of four short dashes indicate hydrogen bonds.) Circle all that are possible. All structures are legitimate in which hydrogen bonding is between anti-parallel strands. One strand must be oriented 5 3 and the other strand must be oriented 3 5 for the two strands to hydrogen bond. 8