Biology 403, spring 2004, Key for First Midterm
Thursday 12 February 2004

1. Define an intensive quantity in thermodynamics and give an example. (4 pts)
   An intensive quantity is one that is not proportional to the quantity of substance present. Examples include temperature and pressure.
   [ 2 points for reasonable definition; 2 points for valid example.]
2. Suppose the reaction A -> B has ΔH = -2.4 kcal/mol, ΔS= -0.03 kJ/(mol-deg), such that dΔH/dT = dΔS/dT ~ 0 over a reasonable temperature range.
    (a) Is this reaction spontaneous at 26.84 deg C?
   We write ΔG = ΔH - TΔS = (-2.4 kcal/mol) - (300 deg) * (-0.03 kJ/(mol-deg)) / (4.184 kJ/kCal)
   Therefore ΔG = -2.4 kcal/mol - 300 * (-0.03) / 4.184 kcal/mol
      =-2.4 + 9/4.184 kcal/mol = -2.4 + 2.151 = -0.249 kcal/mol
   This value is less than zero so the reaction is spontaneous.
   [ 2 points for approach; 1 point for recognizing that the units need to be interconverted; 1 point for the right answer to within 10%.
   An answer of "spontaneous" will be given 0.5 points. A student who works out the numerical answer but does not actually answer the question gets one point. ]
    (b) Determine the temperature at which K_eq = 1. (8 pts)
   Recall that K_eq = e^-ΔG/kT so 1 = e^-ΔG/kT.
   Taking the natural logarithm of both sides, log_e1 = log_e(e^-ΔG/kT)
   The left hand side of this equation is zero, and the right hand side is -ΔG/kT, so
   0 = -ΔG/kT, or ΔG = 0. But by the free energy equation ΔG = ΔH - TΔS = 0, so
   ΔH = TΔS, or T = ΔH / ΔS as long as the two quantities on the right are unchanging with temperature. Consequently
   T = (-2.4 kcal/mol) / (-0.03 kJ/(mol-deg) / 4.184 kJ/kcal)
   T = (2.4 * 4.184) / 0.03 deg = (240/3) * 4.184 deg = 80 * 4.184 deg = 334.7 deg
   This temperature is in Kelvins; the equivalent Celsius temperature is 334.7 - 273.16 = 61.54 degrees C.
   [ 2 points for approach; 2 points for correct answer to within 1 degree. The answer may be left in Kelvin or converted to Celsius as long as it's clear which units are being used.]
3. A solution of lactic acid (pK_a = 3.9) has been stabilized at a pH of 4.6. What is the ratio of [proton acceptor] to [proton donor] at that pH?
   The Henderson-Hasselbach equation tells us that
   pH = pK_a + log₁₀([proton acceptor] / [proton donor]) so
   pH - pK_a = log₁₀([proton acceptor] / [proton donor]) = 4.6 - 3.9 = 0.7
   This means 10^0.7 = 5.01 = [proton acceptor] / [proton donor]
   [1 point for correct approach; 1 point for correct numerical answer to within 10%. Leaving the answer in the form 10^0.7 will earn full credit for that portion.]
   What are the proton acceptor and proton donor in this case? (4 pts)
   The proton acceptor is lactate ion; the proton donor is lactic acid.
   [1 point for recognizing that lactic acid is the proton donor and 1 point for recognizing that lactate is the acceptor.]
4. (a) It is very typical for an enzyme involved in acid-base chemistry at physiological pH to employ a histidine in its reaction mechanism. Why?
   Histidine is the only amino acid with a side-chain pK_a value close to physiological pH; therefore, a histidine residue in a protein can be readily protonated and deprotonated at near-neutral pH during a reaction. This makes it an ideal candidate for acid-base reactions, for which donation and acceptance of protons and hydroxide ions is important.
   [2 points for recognizing that histidine has a side-chain pK_a near 7; 2 points for recognizing that that enables histidines to donate and accept protons readily during reactions. Answers that discuss main-chain pK_a values will not receive points.]
   (b)According to table 3.3 in Horton, the frequency of appearance of histidine residues in proteins is about 2.2%. Is this above or below the average for all amino acids? Why might you expect this to be true? (8 pts)
   There are 20 ribosomally-encoded amino acids, so the average frequency of appearance of the amino acids is 100%/20 = 5%. Thus this frequency of 2.2% is significantly below the average. The relative rareness of histidines makes sense in the context of the result noted in (a): if there were a large number of histidine residues in a protein, they might be changing charge state frequently, possibly with concomitant alterations in the structure and activity of the protein. This would be undesirable, so it is unsurprising that histidines, apart from those found in the active site taking part in the protein's activity, are relatively rare.
   [ 1 points for recognizing that the average frequency is 100/20 = 5%; 1 point for recognizing that histidine is rarer than average as a result. 2 points for getting an answer that makes substantial sense in the second part of this.]
5. An enzyme has a Michaelis constant of 10^-6M and a maximum reaction velocity of 10^-4 Ms^-1. At a substrate concentration of 10^-6M, what would expect the initial reaction velocity v₀ to be? (4 pts)
   We use the Michaelis-Menten equation v₀ = V_max[S] / ([S] + K_m)
    v₀ = (10^-4Ms^-1) * (10^-6M) / (10^-6M + 10^-6M)
    v₀ = 10^-4 Ms^-1 * 0.5
    v₀ = 0.5 * 10^-4 Ms^-1
   We can also remember that K_m is defined as the substrate concentration at which the reaction velocity is half of its maximum value and write this answer down by inspection.
   [ 1 point for recognizing that this is a Michaelis-Menten calculation; 2 points for doing the calculation correctly; 1 point for an answer that is correct to within 5%. If the student chooses to use the half-max argument, then he or she gets 3 points for doing so and 1 for the right answer to within 5%. A numerically correct answer with incorrect or missing units will lose a point.]
6. Define quaternary structure and give an example of a protein that exhibits quaternary structure. (4 pts)
   Quaternary structure is the presence of multiple polypeptide subunits in a single biological protein. Examples include hemoglobin, photosynthetic reaction center, lactate dehydrogenase, and glutamine synthetase.
   [ Two points for a plausible definition; two points for a valid example.]
7. Assume an enzyme is capable of converting aspartic acid to X or glutamic acid to Y. For asp, K_m=10^-6M and k_cat = 10³ s^-1; for glu, K_m=10^-5M and k_cat = 4*10² s^-1. Given starting concentrations of [asp] = [glu] = 3*10^-6M, calculate
   (conversion rate of asp to X) / (conversion rate of glu to Y) (6 pts)
   The ratio k_cat / K_m is the specificity constant for a particular substrate in the context of an enzymatic reaction. The sought-for quantity, namely the ratio of the conversion rates if the initial concentrations of reactants are equal, is the ratio of the specificity constants:
   (k_cat/K_m)_asp / (k_cat/K_m)_glu
   The numerator here is (k_cat/K_m)_asp = 10³ s^-1 / (10^-6M) = 10⁹ M^-1s^-1
   The denominator here is (k_cat/K_m)_glu = 4*10² s^-1 / (10^-5M) = 4 * 10⁷ M^-1s^-1
   The ratio is therefore 10⁹ / (4*10⁷) = 25
   Note that the concentrations [asp] and [glu] are not used explicitly in this problem; they are given only to indicate that the concentrations are in the same range as the K_m values.
   [4 points for concept; 2 points for answer. If the student simply writes "25", give full credit.]
8. A helical protein operates such that the axis of the helix lies along the surface of a phospholipid bilayer, so that some of the amino acid side-chains protrude out from the plane of the surface and some protrude into the interior of the bilayer. Suppose that residue 1 in the helix is an isoleucine. What property would you expect residues 4, 5, and 8 in this helix to have? Why? (4 pts)
   The relevant property is hydrophobicity. One full turn of a helix corresponds to 3.4 residues along the chain; accordingly, residue 1, 4 (= 1 + 3.4 - 0.4), 5 = (1 + 3.4 + 0.6), and 8 = (1 + 2 * 3.4) will point in approximately the same direction. Leucine is a hydrophobic amino acid, so it probably is pointing inward--into the interior of the bilayer. Accordingly, so will residues 4, 5, and 8, so they are probably hydrophobic as well.
   [2 points for answer; 2 points for explanation.]
9. Tropomyosin from shrimp and tropomyosin from chickens have amino acid sequences that are about 50 % identical. Would you expect their tertiary structures to be broadly similar or entirely dissimilar? Why? (7 pts)
   Broadly similar; any sequence identity above about 40% corresponds to a high degree of structural identity, i.e. a very similar three-dimensional fold.
   [ 5 points for concept; 2 points for correct answer.]
10. Would you expect to find most aspartate residues on the surface of an aqueous protein or on its interior? Why? (6 pts)
    On the surface; the aspartate residues are charged, so they will tend to be closely associated with the watery exterior of an aqueous protein, rather than with the hydrophobic, nonpolar core of the protein.
    [2 points for right answer; 4 points of explanation.]
11. Protein P operates entirely within a phospholipid bilayer. Protein Q is anchored to a bilayer but protrudes into the cytoplasm. Protein R is entirely soluble. Which will have the largest percentage of hydrophobic residues on its surface? Which will have the fewest? (4 pts)
    Protein P will have the most hydrophobic residues on the surface; protein R will have the fewest.
    [2 points each.]
12. Name the enzyme that catalyzes each of the following reactions and indicate which EC class (1-6) it belongs to:
    

    reactants	products	name of enzyme	class#
    glutamate + ATP + NH4+	glutamine	glutamine synthetase	6
    lactic acid + NAD+	pyruvic acid + NADH + H+	lactate dehydrogenase	1
    elastin + nH2O	(n+1)(peptide fragments)	elastase	3
    arginine + MgATP	arginine phosphate + MgADP + H+	arginine kinase	2
    pyrophosphate + H2O	2 (phosphate)	pyrophosphatase	3

     (10 pts)
    [0.5 point for each enzyme name; 1.5 points for each enzyme class. We discussed all but one of these enzymes (elastase) in class, and they're named in the textbook as well. For elastase, I was hoping that you would recognize that the trivial name for many enzymes is taken by replacing the suffix on the substrate with -ase.]
13. Xylose isomerase (XI) catalyzes the interconversion of D-xylose and D-xylulose. Suppose we chemically synthesize a polypeptide that is identical in sequence with XI except that the building blocks are all D-amino acids. What reaction would this polypeptide catalyze? Why? (6 pts)
    This artificial enzyme should interconvert L-xylose and L-xylulose. This protein is an exact mirror image of the natural protein, so it will catalyze a reaction that is an exact mirror image of the reaction catalyzed by its natural counterpart.
    [3 points for the correct answer; 3 points for the explanation.]