BIOCHEMISTRY I HOMEWORK I DUE 9/15/03 59 points total The answers are given in bold font. For problems in which points were assigned for different parts, the breakdown of points are given in red for the associated parts. Average = 46 ± 6 points
BIOCHEMISTRY I HOMEWORK I DUE 9/15/03 59 points total 1). 9 points total Fill in the table: For each of the following modified amino acid side chains, identify the parent amino acid from which it was derived by its one-letter code, the chemical classification of the parent amino acid and the type of chemical modification that has occurred. Under modifications, I was looking for the name of the chemical modification. Modified Residue One letter code Chemical classification of parent amino acid Modification - -CH 2 OPO 3 S polar, uncharged phosphorylation -CH 2 CH-(COO - ) 2 E polar, charged (negative) carboxylation of γ- carbon -(CH 2 ) 4 -NH-C(O)CH 3 K polar charged (positive) acetylation of ε-amino group 2). 20 points total T or F (2 points each; if false, briefly state why it is false). The three-dimensional structures of biomolecules combine the properties of flexibility and stability (T). The term "hydrophobic interactions" describes the tendency of hydrophobic groups to sequester away from H 2 0. T Osmotic pressure is the amount of pressure required to prevent the net movement of solute from one side of a semi-permeable membrane to another. False - prevents the net movement of water,not solute. Weak acids are completely ionized in H 2 0. (false - partially ionized).
A reaction with G = - 10 kcal/mol will occur more rapidly than one with a G = -5 kcal/mol. False - thermodynamics tells us whether a reaction will go, not how fast it will go. Free energy change is a constant for a reaction under any condition. False - Free energy change is very dependent upon conditions! e.g., G = -RTlnK Histidine provides the major buffering action of proteins at physiological ph values T At ph 7.0, the net charge on the amino acid lysine is +2 (False; +1) Covalent bonds between amino acids are primarily found as two types - the peptide bond and the disulfide bond. True Amino acids can never have four different protonation or ionization states. False - see page 77. Problem Solving and Short Answer Questions Short answers - short answers consist of no more than 5 sentences! 3). 2 points Why are living organisms open systems, i.e., why is the cell membrane not an absolute barrier between the cytoplasm and the external environment? The short answer is that cell membranes must be semi-permeable to permit nutrients to enter and wastes to exit. 4). 7 points total Aspirin is a weak acid with an ionizable carboxyl group with a pka = 3.5. The structure is drawn below. A. Circle the ionizable hydrogen in the structure below. for this
Question 4 continued: Aspirin functions by being absorbed through the cells lining in the stomach and small intestine. Absorption requires passage through a plasma membrane, which is hydrophobic. It makes sense then, that the rate of passage is determined by the polarity of the molecule: charged and highly polar molecules pass slowly, whereas neutral hydrophobic ones pass rapidly. The ph of the stomach contents is about 2, and the ph of the contents of the small intestine is about 5. B). Calculate the percentage of the un-ionized form of aspirin at ph 2 and ph 5. For ph 2: i). Use the Henderson - Hasselbach equation: ph = pka + log [RCOO-]/[RCOOH] for using Henderson- Hasellbach equation 2=3.5 + log[rcoo-]/[rcooh] [RCOO-]/[RCOOH] =.032 or a molar ratio of.032 to 1 ii). Calculate percentage of uncharged species is the concentration of that species over the total concentration of all species: %RCOOH = [RCOOH]/([RCOO-]+[RCOOH]} %RCOOH = 1/(1+.032) = 97% For ph 5: RCOO-]/[RCOOH] = 32 or a molar ratio of 32 to 1 %RCOOH = 3% C). Based on your answer above, is more aspirin absorbed into the bloodstream from the stomach or small intestine? Stomach.
5). 2 points total Osmotic lysis is a gentle method of breaking open animal cells to free intracellular proteins. In this technique, cells are suspended in a solution that has a total molar concentration of solutes much less than that found naturally inside cells. Explain why this technique might cause cells to burst. There is a larger osmotic pressure inside the cells than outside because the molar concentration of solutes is much greater inside cells than outside. This results in a diffusion of water into the cells (in an attempt to equalize the concentrations of solutes on both sides of the membrane) causing the cells to swell and burst. 6). 2 points total When a hydrophobic substance like a hydrocarbon is dissolved in water, a clathrate cage of ordered water molecules is formed about it. What do you expect the sign of S to be for this process? Explain your answer. This process would correspond to an entropy decrease ( S < 0) because order is being established in the H 2 O structure.
7). 6 points For the reaction HC 2 H 3 O 2!" C 2 H 3 O 2 - + H+, calculate G o and G o '. Assume T = 25C. The ionization constant for acetic acid is 1.8 x 10-5. Is this reaction spontaneous at 25C? i). Keq = ([C 2 H 3 O 2 - ][H + ])/[HC 2 H 3 O 2 ] G o = -RTln Keq note R = 8.314 J/mol-K and T= 25 o +273.15 In a plug and chug G o = +27.1 kj/mol A positive value of G o indicates this reaction is not spontaneous under standard conditions. ii). G o ' = G o + RTln [H + ] = +27.1 kj/mol + RTln [H+] = 27.1 + (8.314)(298) ln (1 x 10-7 ) = -12.8 kj/mol At ph 7.0, this reaction is spontaneous
8). 4 points total A peptide has the sequence Glu-His-Trp-Ser-Gly-Leu-Arg-Pro-Gly A). Write this peptide using one letter code. E-H-W-S-G-L-R-P-G B). What is the net charge of the molecule at ph 3? At ph 8? Write the pkas for all the ionizable groups - don't forget the amino and carboxyl termini! Figure out the charge on each group at each ph value based on their pkas. Remember, if the ph is below the pka, the group is protonated! + H 3 N terminus E H W S G L R P G COO - terminus pka 9.7 4. 6 12.5 2.0 3 Charge +1 0 +1 - - - - +1 - - -1 at ph 3.0 Charge at ph 8.0 +1-1 0 - - - - +1 - - -1 ph 3: +2 ph 8: 0 C). Estimate the pi for this peptide. You need to figure out the ph range where the molecule would have a net charge of 0. Based on the above table, this is between ph 6 and ph 9.7. The pi is then the midpoint of the range or pi = (6+9.7)/2 = 7.85
9). 7 points total Ramachandran plot: The following two Ramachandran plots represent permitted values of ψφ for different amino acids. A). What do the values of ψφ represent? ψ = permitted rotation about the Cα-CO bond φ = permitted rotation about the Cα-NH bond B). Put a G on the line under the plot that best represents the ψφ space for glycine. Put an L on the line under the plot that best represents the ψφ space for leucine. +2 point C).On the appropriate plot: Mark a B on the region where we would expect to find amino acids in beta sheet structures. Mark an A on the region where we would expect to find amino acids in alpha-helical structures. L G The most appropriate plot that shows any type of significant structure is the plot for leucine. The B would go in the upper left corner where there is a high density of points. The A would go in the central left side where there is a high density of points. +2 point D). Would you expect proline to occupy more ψ φ space than alanine - why or why not? No, due to its restricted backbone conformation, it would not.