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Multiple Choice

1.   One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis.
     Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be
     referred to as the:
     a. holoenzyme
     b. prosthetic group
     c. apoenzyme
     d. coenzyme
     e. substrate


2.   Compare the two reaction coordinate diagrams below and select the answer that
     correctly describes their relationship. In each case the single intermediate is the
     ES complex.




     a. The ES complex is given by #2 in (a) and #3 in (b).
     b. The activation energy for the catalyzed reaction #5 in (a) and is #7 in (b).
     c. (a) describes a strict “lock and key” model whereas (b) describes a transition-
        state complementary model.
     d. The activation energy for the uncatalyzed reaction is given by #5 + #6 in (a)
        and by #7 + #4 in (b).
     e. The contribution of binding energy is given by #5 in (a) and by #7 in (b).
3.   The concept of “induced fit” refers to the fact that:
     a. when a substrate binds to an enzyme, the enzyme induces a loss of water
         (desolvation) from the substrate.
     b. substrate binding may induce a conformational change in the enzyme, which
         then brings catalytic groups into proper orientation.
     c. enzyme-substrate binding induces an increase in the reaction entropy, thereby
         catalyzing the reaction.
     d. enzyme specificity is induced by enzyme-substrate binding.
     e. enzyme-substrate binding induces movement along the reaction coordinate to
         the transition state.
4.   Which of the following statements about a plot of V0 vs. [S] for an enzyme that
     follows Michaelis-Menten kinetics is false?
     a. Km is the [S] at which V0 = ½ Vmax.
     b. The shape of the curve is a hyperbola.
     c. The y-axis is a rate term with units of µm/min.
     d. As [S] increases, the initial velocity of reaction, V0, also increases.
     e. At very high [S], the velocity curve becomes a horizontal line that intersects
         the y-axis at Km.


5.   Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed
     reaction could be written as
                    k1        k2
            E+S    ↔      ES   →   P
                    k-1

     Using this reaction, the rate of breakdown of the enzyme-substrate complex can
     be described by the expression:
     a. k1 ([Et] - [ES])[S]
     b. k-1 [ES] + k2 [ES]
     c. k2 [ES]
     d. k-1 [ES]
     e. k1 ([Et] - [ES])

6.   An enzyme-catalyzed reaction was carried out with the substrate concentration
     initially 1,000 times greater than the Km for that substrate. After 9 minutes, 1% of
     the substrate had been converted to product, and the amount of product formed in
     the reaction mixture was 12 µmol. If, in a separate experiment, one-third as much
     enzyme and twice as much substrate had been combined, how long would it take
     for the same amount (12 µmol) of product to be formed?
     a. 1.5 min
     b. 3 min
     c. 6 min
     d. 13.5 min
     e. 27 min
7.    In competitive inhibition, an inhibitor:
      a. binds at several different sites on an enzyme.
      b. binds reversibly at the active site.
      c. binds only to the ES complex.
      d. binds covalently to the enzyme.
      e. lowers the characteristic Vmax of the enzyme.

8.    A transition state analog:
      a. resembles the transition-state structure of the normal enzyme-substrate
         complex.
      b. typically yields product more rapidly with an enzyme than the normal
         substrate.
      c. is less stable when binding to an enzyme than the normal substrate.
      d. stabilizes the transition state for the normal enzyme-substrate complex.
      e. resembles the active site of general acid-base enzymes.

9.    Both water and glucose share an -OH that can serve as a substrate for a reaction
      with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however,
      is about a million times more reactive as a substrate than water. The best
      explanation is that:
      a. glucose has more -OH groups per molecule than does water.
      b. the larger glucose binds better to the enzyme; it induces a conformational
          change in hexokinase that brings active-site amino acids into position for
          catalysis.
      c. water normally will not reach the active site because it is hydrophobic.
      d. water and the second substrate, ATP, compete for the active site, resulting in a
          competitive inhibition of the enzyme.
      e. the -OH group of water is attached to an inhibitory H atom while the glucose
          -OH group is attached to C.

10.   Which of these statements about the composition of membranes is generally true?
      a. The lipid composition of all membranes of eukaryotic cells is essentially the
         same.
      b. All biological membranes contain cholesterol.
      c. Free fatty acids are major components of all membranes.
      d. The inner and outer membranes of mitochondria have different protein
         compositions.

11.   The fluidity of a lipid bilayer will be increased by:
      a. decreasing the number of unsaturated positions.
      b. increasing the length of the alkyl chains.
      c. increasing the temperature.
      d. decreasing the temperature.
12.   Which of the following statements about fatty acids is true?
      a. Fatty acids with longer chain lengths have a higher melting point than fatty
         acids with shorter chain lengths.
      b. Saturated fatty acids have a lower melting point than unsaturated fatty acids.
      c. Cis double bonds of unsaturated fatty acids cause tighter packing of
         hydrophobic tails.
      d. Double bonds in polyunsaturated fatty acids are almost always conjugated.
      e. Fatty acids are found mostly as free acids.

13.   Which of the following statements is not indicative of passive transport? Check
      all that apply.
      a. No energy is required.
      b. ATP is required for transport.
      c. Small organic molecules will diffuse based on molecular weight and solubility
          in lipids.
      d. Molecules are transported from low concentration to high concentration.
      e. Molecules are transported from high concentration to low concentration.

14.   Check all of the following that apply.
      Enzyme catalysts:
      a. shift the equilibrium of a reaction to favor products.
      b. form asymmetric active sites due to their inherent chirality.
      c. are sometimes permanently altered after releasing product.
      d. often require prosthetic groups, which are transiently associated with a given
         enzyme.
      e. are often regulated by covalent modifications.

15.   Lysozyme has a pH centered around pH 5.0. The active site of lysozyme contains
      a glutamic acid residue (pKa = 5.5) and an aspartic acid residue (pKa = 4.0).
      Which of the following statements is correct about the mechanism of lysozyme?
      a. The glutamic acid residue is in a more polar environment than the aspartic
         acid.
      b. During the entire catalytic mechanism, the aspartic acid residue remains
         unprotonated.
      c. During the mechanism, an oxyanion transition state forms.
      d. The glutamic acid residue acts as a general base catalyst.

16.   The conversion of NAD+ to NADH is an example of reduction because
      a. the pyridine ring loses electrons (and a hydrogen).
      b. the pyridine ring gains electrons (and a hydrogen).
      c. the adenine ring loses electrons.
      d. the adenine ring gains electrons.
17.   The process of glycolysis
      a. requires a pathway of chemically coupled phosphoryl-transfer reactions.
      b. uses 2 ATP molecules and forms 2 ATP molecules and one NADH molecule.
      c. occurs in the mitochondria.
      d. converts glucose into two glycerate molecules.


18.   Which of the following is not true of the reactions and enzymes in the glycolytic
      pathway?
      a. Triose phosphate isomerase is referred to as a “perfect enzyme” because
         product is formed as soon as enzyme and substrate collide.
      b. Phosphoglucose isomerase converts the glucose ring to a fructose ring.
      c. Domains of phosphoglycerate kinase clamp down on the substrate so as to
         exclude water from the reaction.
      d. The reaction converting 2-phosphoglycerate to phosphoenolpyruvate is a
         dehydration reaction.
      e. Phosphofructokinase is stimulated by ATP and citrate.

19.   Which statement about the gluconeogenesis pathway is false?
      a. Most of the same glycolytic enzymes are utilized in gluconeogenesis.
      b. The decarboxylation of oxaloacetate provides the free energy required for
          synthesis of phosphoenolpyruvate.
      c. Pyruvate carboxylase contains a biotin prosthetic group, which functions as a
         CO2 carrier.
      d. Oxaloacetate is generated in the mitochondria.
      e. The hexokinase and phosphoglycerate kinase reactions of glycolysis are
         reversed in gluconeogenesis via hydrolytic reactions that release Pi in
         exorgonic processes.

20.   Which of the following statements about glycogen metabolism is/are false?
              I. Release of epinephrine leads to increased glycogen breakdown.
              II. cAMP is a second messenger that triggers an increase in glycogen
                   synthesis.
              III. Insulin results in dephosphorylation and subsequent activation of
                    glycogen synthase.
              IV. Glucagon promotes glycogen breakdown.
              V. Glucose-6-phosphate is an allosteric inhibitor of phosphorylase.
      a. II only
      b. V only
      c. III and IV
      d. II and V
      e. III and V
Written Answers
1.)   a. Draw the basic chemical structure of all glycophospholipids, (2 points)




      b. (4 points) (i) Explain why phosphoglycerides are capable of spontaneously
                assembling into the bilayer structure found in biological membranes
                but triacylglycerols are not.

             (i) Triacylglycerols have three fatty acyl groups in ester linkage with
             glycerol; they are very hydrophobic because the carboxyl groups, which
             are involved in the ester linkages, cannot ionize. Phosphoglycerides have a
             polar region at their head group, where a phosphate in a phosphodiester
             linkage bears a full negative charge. The head group itself (serine,
             ethanolamine, choline, etc.) may also be charged and is in any case polar.
             Thus the phospholipids is amphipathic, having both polar and nonpolar
             regions, and it forms lipid bilayers spontaneously in water.

             (ii) What are the forces that drive bilayer formation?

             (ii) These lipid bilayers are stabilized by the energy gained from burying
             hydrophobic groups out of contact with water. A hydrophobic chain in
             water forces the formation of a cage of immobilized water molecules
             around it. When several hydrophobic regions cluster in a bilayer, the
             surface area exposed to water decreases, and the water molecules in the
             cage are released, accompanied by a gain in entropy that drives the
             formation of the bilayer.
c. (4 points) A protein is found to extend all the way through the membrane of a
       cell.
          Describe this protein in terms of the location of particular types of amino acid
          side chains in its structure and its ability to move within the membrane.

              This integral membrane protein associates with the lipid bilayer through
              hydrophobic interactions between domains containing many hydrophobic
              amino acids and the fatty acyl chains of membrane lipids. Polar and
              charged residues are located on portions of the protein that protrude out of
              either face of the membrane. The protein is free to diffuse laterally in the
              plane of the membrane, but cannot move across the lipid bilayer.




2.)     a. For the reaction E + S ↔ ES → E + P, the Michaelis-Menten constant, Km, is
actually
           a summary of three terms. What are they? (2 points) How is Km determined
        graphically? (2 points)

       Km = (k2 + k-1)/k1 where k-1 and k1 are the rate constants for the breakdown and
       association, respectively, of the ES complex and k2 is the rate constant for the
       breakdown of ES to form E + P. Km can be determined graphically on a plot of
       V0 vs. [S] by finding the [S] at which V0 = ½ Vmax. More conveniently, on a
       double plot, the x-axis intercept = -1/Km.



       b. A biochemist obtains the following set of data for an enzyme that is known to
          follow Michaelis-Menten kinetics. (4 points)
                     Substrate                                      Initial
                     Concentration                                  velocity
                     (µM)                                           (µmol/min)
                              1                                          49
                              2                                          96
                              8                                         349
                             50                                         621
                            100                                          676
                           1000                                          698
                           5000                                          699
(i) Vmax for the enzyme is _____________________. Explain in one
           sentence how you determine Vmax.

       Vmax is about 700. In a plot of V vs. [S], the asymptote is Vmax. Simple
       inspection of the data shows the approach to Vmax ---the rate increases by
       only 1 unit when [S] increases fivefold.




       (ii) Km for the enzyme is __________________________-. Explain in one
            sentence how you determined Km.

       Km is about 8 µM, the [S] at which the velocity is half-maximal. Because
       Vmax is about 700, 12 Vmax is about 350. The [S] at that rate is about
       8 µM.


c. Why is the Lineweaver-Burk (double-reciprocal) plot more useful than the
   standard V vs. [S] plot in determining kinetic constants for an enzyme? (Your
   answer should show typical plots with labeled axes). (3 points)

       The plot of V vs. [S] is hyperbolic; maximum velocity is never achieved
       experimentally, because it is impossible to do experiments at infinitely
       high [S]. The Lineweaver-Burk transformation of the Michaelis-Menten
       equation produces a linear plot that can be extrapolated to infinite [S]
       (where 1/[S] becomes zero), allowing a determination of Vmax.
3.)    You are faced with the task of synthesizing an effective catalytic antibody or
       abzyme for the breakdown of the drug oxycontin which has the following
       structure:




       a. Briefly describe the steps that you would take to make such an antibody.

        (i) Decide first in what manner the oxycontin might be broken down. For
        example, one possibility you might think of is hydrolysis, i.e. the breaking of
        bonds with the participation of water. Are there any regions of oxycontin that
        might be susceptible to such a reaction? There are many other sorts of reactions
`       you might imagine. The main thing is that you first envision the type of reaction
        to be accomplished.
        (ii) Given a particular type of reaction to be accomplished one then must design
        and synthesize a stable transition state analog. Part of the design of such an analog
        will be the incorporation of a group or region on the molecule that will enable it to
        be attached to a carrier molecule (see part (b) below). This attachment site needs
        to be designed in such a manner as to not interfere drastically with the availability
        of the major transition state part of the molecule that will elicit antibodies.
        (iii) collect blood & purify antibodies
        (iv) test enzymatic activity of antibody preparation
(an alternative to the preparation of polyclonal antibodies described above is to prepare
monoclonal antibodies)                                         (5 points)
b. Very often the antigen used to elicit a catalytic antibody is conjugated to a
   carrier protein prior to injection into the animal that will be used to generate the
   antibodies. Why is this?

Small antigens are typically combined with large carrier molecules such as
proteins because (i) the small molecules are very inefficient in generating
antibodies and the small molecules are often very rapidly removed from tissues in
the bloodstream and (ii) the carrier protein can often act as an adjuvant or
stimulant for antibody production                              (2 points)
(for grading part (i) is sufficient for 2 points)

c. You finally succeed in making a catabolic antibody of the desired
   characteristics. How could you detect whether the antigen you used (minus the
   carrier) was a competitive inhibitor of this enzyme?

Run a series of enzyme assays, varying the substrate concentration, in the
presence and absence of the antigen (the possible competitive inhibitor). Plot the
data using a Lineweaver-Burk plot (or equivalent). If the transition state analog is
a competitive inhibitor then the Lineweaver-Burk plot should look like Fig. 14-12
(below) with the lines with and without inhibitor intersecting on the Y axis.
                                                      (3 points)
4.) Here’s the BLOSUM50 amino acid substitution matrix used to align protein
    sequences:

     A    R    N    D    C    Q    E    G    H    I    L    K    M    F    P    S    T    W    Y    V
A    5   -2   -1   -2   -1   -1   -1    0   -2   -1   -2   -1   -1   -3   -1    1    0   -3   -2    0
R   -2    7   -1   -2   -4    1    0   -3    0   -4   -3    3   -2   -3   -3   -1   -1   -3   -1   –3
N   -1   -1    7    2   -2    0    0    0    1   -3   -4    0   -2   -4   -2    1    0   -4   -2   -3
D   -2   -2    2    8   -4    0    2   -1   -1   -4   -4   -1   -4   -5   -1    0   -1   -5   -3   -4
C   -1   -4   -2   -4   13   -3   -3   -3   -3   -2   -2   -3   -2   -2   -4   -1   -1   -5   -3   -1
Q   -1    1    0    0   -3    7    2   -2    1   -3   -2    2    0   -4   -1    0   -1   -1   -1   -3
E   -1    0    0    2   -3    2    6   -3    0   -4   -3    1   -2   -3   -1   -1   -1   -3   -2   -3
G    0   -3    0   -1   -3   -2   -3    8   -2   -4   -4   -2   -3   -4   -2    0   -2   -3   -3   -4
H   -2    0    1   -1   -3    1    0   -2   10   -4   -3    0   -1   -1   -2   -1   -2   -3    2   –4
I   -1   -4   -3   -4   -2   -3   -4   -4   -4    5    2   -3    2    0   -3   -3   -1   -3   -1    4
L   -2   -3   -4   -4   -2   -2   -3   -4   -3    2    5   -3    3    1   -4   -3   -1   -2   -1    1
K   -1    3    0   -1   -3    2    1   -2    0   -3   -3    6   -2   -4   -1    0   -1   -3   -2   -3
M   -1   -2   -2   -4   -2    0   -2   -3   -1    2    3   -2    7    0   -3   -2   -1   -1    0    1
F   -3   -3   -4   -5   -2   -4   -3   -4   -1    0    1   -4    0    8   -4   -3   -2    1    4   -1
P   -1   -3   -2   -1   -4   -1   -1   -2   -2   -3   -4   -1   -3   -4   10   -1   -1   -4   -3   -3
S    1   -1    1    0   -1    0   -1    0   -1   -3   -3    0   -2   -3   -1    5    2   -4   -2   -2
T    0   -1    0   -1   -1   -1   -1   -2   -2   -1   -1   -1   -1   -2   -1    2    5   -3   -2    0
W   -3   -3   -4   -5   -5   -1   -3   -3   -3   -3   -2   -3   -1    1   -4   -4   -3   15    2   -3
Y   -2   -1   -2   -3   -3   -1   -2   -3    2   -1   -1   -2    0    4   -3   -2   -2    2    8   -1
V    0   -3   -3   -4   -1   -3   -3   -4   -4    4    1   -3    1   -1   -3   -2    0   -3   -1    5

a. Some amino acids are “specialists” and play very specific roles in proteins that can be
difficult to fill using other amino acids. From the table, which single amino acid is most
likely not to be substituted for by another? How can you tell?

W (has the highest score on the diagonal, meaning that other residues substitute poorly
for it.) (2 points)

b. Some amino acids are more “generic” in their properties than others. From the table,
which 6 amino acids are most easily substituted for by others? How can you tell?

A, I, L, S, T, V (have lowest scores on the diagonal) (3 points)

c. What are the 3 most disfavored amino acid substitutions?
W for D, W for C, F for D      (3 points)


d. Sequence alignment programs like BLAST work well for proteins and DNA, but very
poorly for aligning RNA molecules. Why is that the case?

BLAST assumes that each position in the alignment is independent of positions far away,
and ignores correlations between the sites. RNA molecules have many interactions in the
2D and 3D structure that require base-pairing between distant bases, so the positions are
strongly correlated, and accounted for poorly by this approximation. (2 points)
5.)   Consider the process of fermentation (13 metabolites / 12 enzymes) as
      summarized below. Then answer the questions at right. Identify an enzyme by
      its two metabolites.

      1. Glucose (1)a) Identify compound Z=___acetaldehyde_________________
      2. Compound “U”      (2)     b) F-2,6-BP (activates / inactivates) the
      3. Fructose-6-P                 enzyme that converts metabolites _3__ to _4__.
      4. Compound “V”      (1)     c) ATP is produced in glycolysis at step(s).
      5. 3-P-Glycerate + DHAP          _6_ to _7_, (and _10_ to _11_, __ to __, __ to
__)
                                            +
      6. Compound “W” (1)           d) NAD is produced as _12___ goes to _13___.
      7. 1,3 BisP-glycerate (1)     e) The conversion of _6__ to _7__ involves the
      8. Compound “X”                  a thiohemiacetal intermediate.
      9. 2-Phosphoglycerate (1)     f) The conversion of _8__ to _9__ involves the
      10. Compound “Y”                 a phosphoryl-His group intermediate
      11. Pyruvate          (1)     g) Oxidation during glycolysis takes place
      12. Compound “Z”                  in converting _6__to _7__.).
      13. Ethanol




      Where (which carbon position) would 14C have to be located in glucose to ensure
      that all the 14C activity was liberated as 14CO2 during alcoholic fermentation into
      ethanol? __3 or 4___ (2 points)
+
6.)   The standard reduction potential for the reduction of NAD to NADH is –0.315V
      and that for the reduction of pyruvate to lactate is –0.185V. (1 F = 96,500 J / V)

      a. Calculate the standard free energy (∆Go’) for the oxidation of lactate to
                         +
      pyruvate by NAD .         (4 points)




      b. What is the equilibrium constant for this reaction?     (2 points)




      c. Draw the structure of NADH. (4 points)
Exam 2 fall03

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Exam 2 fall03

  • 1. Multiple Choice 1. One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the: a. holoenzyme b. prosthetic group c. apoenzyme d. coenzyme e. substrate 2. Compare the two reaction coordinate diagrams below and select the answer that correctly describes their relationship. In each case the single intermediate is the ES complex. a. The ES complex is given by #2 in (a) and #3 in (b). b. The activation energy for the catalyzed reaction #5 in (a) and is #7 in (b). c. (a) describes a strict “lock and key” model whereas (b) describes a transition- state complementary model. d. The activation energy for the uncatalyzed reaction is given by #5 + #6 in (a) and by #7 + #4 in (b). e. The contribution of binding energy is given by #5 in (a) and by #7 in (b).
  • 2. 3. The concept of “induced fit” refers to the fact that: a. when a substrate binds to an enzyme, the enzyme induces a loss of water (desolvation) from the substrate. b. substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation. c. enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction. d. enzyme specificity is induced by enzyme-substrate binding. e. enzyme-substrate binding induces movement along the reaction coordinate to the transition state. 4. Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false? a. Km is the [S] at which V0 = ½ Vmax. b. The shape of the curve is a hyperbola. c. The y-axis is a rate term with units of µm/min. d. As [S] increases, the initial velocity of reaction, V0, also increases. e. At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km. 5. Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed reaction could be written as k1 k2 E+S ↔ ES → P k-1 Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression: a. k1 ([Et] - [ES])[S] b. k-1 [ES] + k2 [ES] c. k2 [ES] d. k-1 [ES] e. k1 ([Et] - [ES]) 6. An enzyme-catalyzed reaction was carried out with the substrate concentration initially 1,000 times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 µmol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 µmol) of product to be formed? a. 1.5 min b. 3 min c. 6 min d. 13.5 min e. 27 min
  • 3. 7. In competitive inhibition, an inhibitor: a. binds at several different sites on an enzyme. b. binds reversibly at the active site. c. binds only to the ES complex. d. binds covalently to the enzyme. e. lowers the characteristic Vmax of the enzyme. 8. A transition state analog: a. resembles the transition-state structure of the normal enzyme-substrate complex. b. typically yields product more rapidly with an enzyme than the normal substrate. c. is less stable when binding to an enzyme than the normal substrate. d. stabilizes the transition state for the normal enzyme-substrate complex. e. resembles the active site of general acid-base enzymes. 9. Both water and glucose share an -OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The best explanation is that: a. glucose has more -OH groups per molecule than does water. b. the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis. c. water normally will not reach the active site because it is hydrophobic. d. water and the second substrate, ATP, compete for the active site, resulting in a competitive inhibition of the enzyme. e. the -OH group of water is attached to an inhibitory H atom while the glucose -OH group is attached to C. 10. Which of these statements about the composition of membranes is generally true? a. The lipid composition of all membranes of eukaryotic cells is essentially the same. b. All biological membranes contain cholesterol. c. Free fatty acids are major components of all membranes. d. The inner and outer membranes of mitochondria have different protein compositions. 11. The fluidity of a lipid bilayer will be increased by: a. decreasing the number of unsaturated positions. b. increasing the length of the alkyl chains. c. increasing the temperature. d. decreasing the temperature.
  • 4. 12. Which of the following statements about fatty acids is true? a. Fatty acids with longer chain lengths have a higher melting point than fatty acids with shorter chain lengths. b. Saturated fatty acids have a lower melting point than unsaturated fatty acids. c. Cis double bonds of unsaturated fatty acids cause tighter packing of hydrophobic tails. d. Double bonds in polyunsaturated fatty acids are almost always conjugated. e. Fatty acids are found mostly as free acids. 13. Which of the following statements is not indicative of passive transport? Check all that apply. a. No energy is required. b. ATP is required for transport. c. Small organic molecules will diffuse based on molecular weight and solubility in lipids. d. Molecules are transported from low concentration to high concentration. e. Molecules are transported from high concentration to low concentration. 14. Check all of the following that apply. Enzyme catalysts: a. shift the equilibrium of a reaction to favor products. b. form asymmetric active sites due to their inherent chirality. c. are sometimes permanently altered after releasing product. d. often require prosthetic groups, which are transiently associated with a given enzyme. e. are often regulated by covalent modifications. 15. Lysozyme has a pH centered around pH 5.0. The active site of lysozyme contains a glutamic acid residue (pKa = 5.5) and an aspartic acid residue (pKa = 4.0). Which of the following statements is correct about the mechanism of lysozyme? a. The glutamic acid residue is in a more polar environment than the aspartic acid. b. During the entire catalytic mechanism, the aspartic acid residue remains unprotonated. c. During the mechanism, an oxyanion transition state forms. d. The glutamic acid residue acts as a general base catalyst. 16. The conversion of NAD+ to NADH is an example of reduction because a. the pyridine ring loses electrons (and a hydrogen). b. the pyridine ring gains electrons (and a hydrogen). c. the adenine ring loses electrons. d. the adenine ring gains electrons.
  • 5. 17. The process of glycolysis a. requires a pathway of chemically coupled phosphoryl-transfer reactions. b. uses 2 ATP molecules and forms 2 ATP molecules and one NADH molecule. c. occurs in the mitochondria. d. converts glucose into two glycerate molecules. 18. Which of the following is not true of the reactions and enzymes in the glycolytic pathway? a. Triose phosphate isomerase is referred to as a “perfect enzyme” because product is formed as soon as enzyme and substrate collide. b. Phosphoglucose isomerase converts the glucose ring to a fructose ring. c. Domains of phosphoglycerate kinase clamp down on the substrate so as to exclude water from the reaction. d. The reaction converting 2-phosphoglycerate to phosphoenolpyruvate is a dehydration reaction. e. Phosphofructokinase is stimulated by ATP and citrate. 19. Which statement about the gluconeogenesis pathway is false? a. Most of the same glycolytic enzymes are utilized in gluconeogenesis. b. The decarboxylation of oxaloacetate provides the free energy required for synthesis of phosphoenolpyruvate. c. Pyruvate carboxylase contains a biotin prosthetic group, which functions as a CO2 carrier. d. Oxaloacetate is generated in the mitochondria. e. The hexokinase and phosphoglycerate kinase reactions of glycolysis are reversed in gluconeogenesis via hydrolytic reactions that release Pi in exorgonic processes. 20. Which of the following statements about glycogen metabolism is/are false? I. Release of epinephrine leads to increased glycogen breakdown. II. cAMP is a second messenger that triggers an increase in glycogen synthesis. III. Insulin results in dephosphorylation and subsequent activation of glycogen synthase. IV. Glucagon promotes glycogen breakdown. V. Glucose-6-phosphate is an allosteric inhibitor of phosphorylase. a. II only b. V only c. III and IV d. II and V e. III and V
  • 6. Written Answers 1.) a. Draw the basic chemical structure of all glycophospholipids, (2 points) b. (4 points) (i) Explain why phosphoglycerides are capable of spontaneously assembling into the bilayer structure found in biological membranes but triacylglycerols are not. (i) Triacylglycerols have three fatty acyl groups in ester linkage with glycerol; they are very hydrophobic because the carboxyl groups, which are involved in the ester linkages, cannot ionize. Phosphoglycerides have a polar region at their head group, where a phosphate in a phosphodiester linkage bears a full negative charge. The head group itself (serine, ethanolamine, choline, etc.) may also be charged and is in any case polar. Thus the phospholipids is amphipathic, having both polar and nonpolar regions, and it forms lipid bilayers spontaneously in water. (ii) What are the forces that drive bilayer formation? (ii) These lipid bilayers are stabilized by the energy gained from burying hydrophobic groups out of contact with water. A hydrophobic chain in water forces the formation of a cage of immobilized water molecules around it. When several hydrophobic regions cluster in a bilayer, the surface area exposed to water decreases, and the water molecules in the cage are released, accompanied by a gain in entropy that drives the formation of the bilayer.
  • 7. c. (4 points) A protein is found to extend all the way through the membrane of a cell. Describe this protein in terms of the location of particular types of amino acid side chains in its structure and its ability to move within the membrane. This integral membrane protein associates with the lipid bilayer through hydrophobic interactions between domains containing many hydrophobic amino acids and the fatty acyl chains of membrane lipids. Polar and charged residues are located on portions of the protein that protrude out of either face of the membrane. The protein is free to diffuse laterally in the plane of the membrane, but cannot move across the lipid bilayer. 2.) a. For the reaction E + S ↔ ES → E + P, the Michaelis-Menten constant, Km, is actually a summary of three terms. What are they? (2 points) How is Km determined graphically? (2 points) Km = (k2 + k-1)/k1 where k-1 and k1 are the rate constants for the breakdown and association, respectively, of the ES complex and k2 is the rate constant for the breakdown of ES to form E + P. Km can be determined graphically on a plot of V0 vs. [S] by finding the [S] at which V0 = ½ Vmax. More conveniently, on a double plot, the x-axis intercept = -1/Km. b. A biochemist obtains the following set of data for an enzyme that is known to follow Michaelis-Menten kinetics. (4 points) Substrate Initial Concentration velocity (µM) (µmol/min) 1 49 2 96 8 349 50 621 100 676 1000 698 5000 699
  • 8. (i) Vmax for the enzyme is _____________________. Explain in one sentence how you determine Vmax. Vmax is about 700. In a plot of V vs. [S], the asymptote is Vmax. Simple inspection of the data shows the approach to Vmax ---the rate increases by only 1 unit when [S] increases fivefold. (ii) Km for the enzyme is __________________________-. Explain in one sentence how you determined Km. Km is about 8 µM, the [S] at which the velocity is half-maximal. Because Vmax is about 700, 12 Vmax is about 350. The [S] at that rate is about 8 µM. c. Why is the Lineweaver-Burk (double-reciprocal) plot more useful than the standard V vs. [S] plot in determining kinetic constants for an enzyme? (Your answer should show typical plots with labeled axes). (3 points) The plot of V vs. [S] is hyperbolic; maximum velocity is never achieved experimentally, because it is impossible to do experiments at infinitely high [S]. The Lineweaver-Burk transformation of the Michaelis-Menten equation produces a linear plot that can be extrapolated to infinite [S] (where 1/[S] becomes zero), allowing a determination of Vmax.
  • 9.
  • 10. 3.) You are faced with the task of synthesizing an effective catalytic antibody or abzyme for the breakdown of the drug oxycontin which has the following structure: a. Briefly describe the steps that you would take to make such an antibody. (i) Decide first in what manner the oxycontin might be broken down. For example, one possibility you might think of is hydrolysis, i.e. the breaking of bonds with the participation of water. Are there any regions of oxycontin that might be susceptible to such a reaction? There are many other sorts of reactions ` you might imagine. The main thing is that you first envision the type of reaction to be accomplished. (ii) Given a particular type of reaction to be accomplished one then must design and synthesize a stable transition state analog. Part of the design of such an analog will be the incorporation of a group or region on the molecule that will enable it to be attached to a carrier molecule (see part (b) below). This attachment site needs to be designed in such a manner as to not interfere drastically with the availability of the major transition state part of the molecule that will elicit antibodies. (iii) collect blood & purify antibodies (iv) test enzymatic activity of antibody preparation (an alternative to the preparation of polyclonal antibodies described above is to prepare monoclonal antibodies) (5 points)
  • 11. b. Very often the antigen used to elicit a catalytic antibody is conjugated to a carrier protein prior to injection into the animal that will be used to generate the antibodies. Why is this? Small antigens are typically combined with large carrier molecules such as proteins because (i) the small molecules are very inefficient in generating antibodies and the small molecules are often very rapidly removed from tissues in the bloodstream and (ii) the carrier protein can often act as an adjuvant or stimulant for antibody production (2 points) (for grading part (i) is sufficient for 2 points) c. You finally succeed in making a catabolic antibody of the desired characteristics. How could you detect whether the antigen you used (minus the carrier) was a competitive inhibitor of this enzyme? Run a series of enzyme assays, varying the substrate concentration, in the presence and absence of the antigen (the possible competitive inhibitor). Plot the data using a Lineweaver-Burk plot (or equivalent). If the transition state analog is a competitive inhibitor then the Lineweaver-Burk plot should look like Fig. 14-12 (below) with the lines with and without inhibitor intersecting on the Y axis. (3 points)
  • 12. 4.) Here’s the BLOSUM50 amino acid substitution matrix used to align protein sequences: A R N D C Q E G H I L K M F P S T W Y V A 5 -2 -1 -2 -1 -1 -1 0 -2 -1 -2 -1 -1 -3 -1 1 0 -3 -2 0 R -2 7 -1 -2 -4 1 0 -3 0 -4 -3 3 -2 -3 -3 -1 -1 -3 -1 –3 N -1 -1 7 2 -2 0 0 0 1 -3 -4 0 -2 -4 -2 1 0 -4 -2 -3 D -2 -2 2 8 -4 0 2 -1 -1 -4 -4 -1 -4 -5 -1 0 -1 -5 -3 -4 C -1 -4 -2 -4 13 -3 -3 -3 -3 -2 -2 -3 -2 -2 -4 -1 -1 -5 -3 -1 Q -1 1 0 0 -3 7 2 -2 1 -3 -2 2 0 -4 -1 0 -1 -1 -1 -3 E -1 0 0 2 -3 2 6 -3 0 -4 -3 1 -2 -3 -1 -1 -1 -3 -2 -3 G 0 -3 0 -1 -3 -2 -3 8 -2 -4 -4 -2 -3 -4 -2 0 -2 -3 -3 -4 H -2 0 1 -1 -3 1 0 -2 10 -4 -3 0 -1 -1 -2 -1 -2 -3 2 –4 I -1 -4 -3 -4 -2 -3 -4 -4 -4 5 2 -3 2 0 -3 -3 -1 -3 -1 4 L -2 -3 -4 -4 -2 -2 -3 -4 -3 2 5 -3 3 1 -4 -3 -1 -2 -1 1 K -1 3 0 -1 -3 2 1 -2 0 -3 -3 6 -2 -4 -1 0 -1 -3 -2 -3 M -1 -2 -2 -4 -2 0 -2 -3 -1 2 3 -2 7 0 -3 -2 -1 -1 0 1 F -3 -3 -4 -5 -2 -4 -3 -4 -1 0 1 -4 0 8 -4 -3 -2 1 4 -1 P -1 -3 -2 -1 -4 -1 -1 -2 -2 -3 -4 -1 -3 -4 10 -1 -1 -4 -3 -3 S 1 -1 1 0 -1 0 -1 0 -1 -3 -3 0 -2 -3 -1 5 2 -4 -2 -2 T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 2 5 -3 -2 0 W -3 -3 -4 -5 -5 -1 -3 -3 -3 -3 -2 -3 -1 1 -4 -4 -3 15 2 -3 Y -2 -1 -2 -3 -3 -1 -2 -3 2 -1 -1 -2 0 4 -3 -2 -2 2 8 -1 V 0 -3 -3 -4 -1 -3 -3 -4 -4 4 1 -3 1 -1 -3 -2 0 -3 -1 5 a. Some amino acids are “specialists” and play very specific roles in proteins that can be difficult to fill using other amino acids. From the table, which single amino acid is most likely not to be substituted for by another? How can you tell? W (has the highest score on the diagonal, meaning that other residues substitute poorly for it.) (2 points) b. Some amino acids are more “generic” in their properties than others. From the table, which 6 amino acids are most easily substituted for by others? How can you tell? A, I, L, S, T, V (have lowest scores on the diagonal) (3 points) c. What are the 3 most disfavored amino acid substitutions? W for D, W for C, F for D (3 points) d. Sequence alignment programs like BLAST work well for proteins and DNA, but very poorly for aligning RNA molecules. Why is that the case? BLAST assumes that each position in the alignment is independent of positions far away, and ignores correlations between the sites. RNA molecules have many interactions in the 2D and 3D structure that require base-pairing between distant bases, so the positions are strongly correlated, and accounted for poorly by this approximation. (2 points)
  • 13. 5.) Consider the process of fermentation (13 metabolites / 12 enzymes) as summarized below. Then answer the questions at right. Identify an enzyme by its two metabolites. 1. Glucose (1)a) Identify compound Z=___acetaldehyde_________________ 2. Compound “U” (2) b) F-2,6-BP (activates / inactivates) the 3. Fructose-6-P enzyme that converts metabolites _3__ to _4__. 4. Compound “V” (1) c) ATP is produced in glycolysis at step(s). 5. 3-P-Glycerate + DHAP _6_ to _7_, (and _10_ to _11_, __ to __, __ to __) + 6. Compound “W” (1) d) NAD is produced as _12___ goes to _13___. 7. 1,3 BisP-glycerate (1) e) The conversion of _6__ to _7__ involves the 8. Compound “X” a thiohemiacetal intermediate. 9. 2-Phosphoglycerate (1) f) The conversion of _8__ to _9__ involves the 10. Compound “Y” a phosphoryl-His group intermediate 11. Pyruvate (1) g) Oxidation during glycolysis takes place 12. Compound “Z” in converting _6__to _7__.). 13. Ethanol Where (which carbon position) would 14C have to be located in glucose to ensure that all the 14C activity was liberated as 14CO2 during alcoholic fermentation into ethanol? __3 or 4___ (2 points)
  • 14. + 6.) The standard reduction potential for the reduction of NAD to NADH is –0.315V and that for the reduction of pyruvate to lactate is –0.185V. (1 F = 96,500 J / V) a. Calculate the standard free energy (∆Go’) for the oxidation of lactate to + pyruvate by NAD . (4 points) b. What is the equilibrium constant for this reaction? (2 points) c. Draw the structure of NADH. (4 points)