203f99_e2

203f99_e2 - Last Name: Z “(AW First Name: 9 ’ Chemistry...

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Unformatted text preview: Last Name: Z “(AW First Name: 9 ’ Chemistry 203 Lg (Fall 99) Exam #2 October 26, 1999; 11:00-11:50 am. Social Security # (M «ME T.A: 2L« 965‘"! If Prof. Charles E. McKenna INITIAL OF LAST NAME Please do not open this exam until you are told to do so. Write legibly to avoid confusion. GOOD LUCK ! ! V Questlon # Points “'4” M I ' . . I will observe the rules of Academic integrity while taking, this exam. Student Signature McK Sys/AppszDesktop Folder:exam2z203, F99, Ex #2 Circle the most nearly correct answer. (4 Pts.) (1) The number of electrons on a carbon atom (C) available for bonding is: (D Q-hUJN (2) In H—O-H, to complete its ‘octet’ the oxygen atom has: two bonding, two non bonding electrons r) g four bonding, four non-bonding electrons ‘ eight bonding electrons eight non bonding electrons (3) The compound Li+F' has: - hydrogen bond only - covalent bond only 0 ionic bond - both covalent and ionic bonds (4) Which among the following 13% a polar covalent bond: 0 F—CH3 — H—CH3 - o: o (5) The compound shown HaCO‘Q-CHZCHzNHz contains: a both an ether and an amine group - both an amine and an ester group - both an amide and an acid group - both a ketone and an aldehyde group G. - n (6) In a bond-making process like 2H’ --> H—-—H r I :r G ‘ u - the product (H2) is less stable than the starting material (2H atoms) L " “ 1"” - the reaction happens more slowly at higher temperatures C? energy is released - energy must be supplied (7) In the reaction H2+ HZC = CH2 —(Pd/C)—-> H3C-CH3 the total number of (old) bonds broken + (new) bonds made is: (8) Which one of the following molecules is chiral? CH3 " H"J\CI H CH3 “ “4‘0 CI CH3 #0. H \ 6;}\CH20H3 (9) How many different dipeptides HZN-aal-aa2-C02H are possible, assuming aan could be any of " H3O the common amino acids? — 2 _ fl,» 4 400 - 50,000 (10) An e ample of a drug that works by m-competitive inhibition of a bacterial enzyme is: @ penicillin - sulfa - aspirin - AZT (11) Myoglobin, the O2 — storage protein of muscle, (shown to the right): - has several extensive 0t- helical regions - has an embedded Fe atom which actually binds the O2 - has no extensive 4/“ "beta sheet” regions f the above are tru (12) Primary vs. tertiary structure of a protein: [—fl primary = coils or sheets, tertiary = random structure 0 primary = amino acid sequence, tertiary = overall conformation in space i - primary = overall conformational in space, tertiary amino acid sequence - primary = overall conformation in space, tertiary = grouping of > 2 protein molecules (13) Energy Transition State The reaction diagram above shows the change in energy when the substrate S is converted to the product P with (B) and without (A) enzyme E. According to the diagram: Q the enzyme lowers the reaction energy barrier, speeding up the reaction - higher temperature will slow down the reaction - adding the enzyme has no effect on the reaction rate constant (transition state) — adding the enzyme changes the net amount of energy released. (14) Protein folding in H20 solutions: - creates the protein primary structure 0 places non—polar R groups like —CH3, ——CH(CH3)2 in the interior of the protein — places non—polar R groups like ——CH3, ——CH(CH3)2 on the surface, contacting H20 - places polar —-C(O)NH2, ——OH, and charged —NH3+, —C02‘ groups in the interior of the protein (15) Catalysis by enzymes: 0 can distinguish two substrates which differ only as image and mirror-image (chiral) ~‘ - does not require special ionic, polar and/or hydrophobic interactions between the enzyme and the substrate - is not involved in the multiplication of HIV viruses - all of the above are true (16) A “mechanism” of an enzyme-catalyzed reaction is a sequential description of the bond-making —breaking steps that convert substrate to product. In the mechanism for chymotrypsin shown above, the label for the histidine base group that removes a proton from water, activating it to attack the bound peptide fragment, is : A if? D (17) For an enzyme protein, a “mutation” means: one nucleoside base changes into an amino acid gone amino acid changes into a different one one amino acid changes into a nucleoside one nucleoside base changes into another (18) K1- is the inhibition constant for a drug that inhibits an enzyme E. K = I ~ E/EI, where E1 is : ~ ' active enzyme—drug complex. the larger the K, the weaker the inhibition for a given drug concentration (1) K1. does not depend on the type of enzyme involved Ki does not depend on the type of inhibitor involved the larger the K, the more potent the inhibition 7 (19) The figure below shows a pyruvate molecule binding to the active site of enzyme lactate dehydrogenase by: - like—charge repulsion forces only 0 weak (van der Waals) forces, H—bonding and ionic bonding - H-bonding and ionic bonding only — weak forces, H—bonding and covalent (disulfide) bonds (20) Amino acids are linked together to form a protein using: 0 -C(O)-NH— groups - phosphate diester groups - —C(O)—O— groups - benzene groups (21) The Michaelis~Menten equation is: V = Vm - S /(Km + S), where Vm = k2 E0 (E0 = total enzyme concentration). It says that: - adding more enzyme has no effect on the reaction rate (V) a as more and more S is added, the rate (V) reaches a limit (Vm) - the rate depends on the concentration of S, even if S>> Km - the rate never depends on the concentration of S (22) The electronegativity of atoms tends to (see excerpt from Periodic Table) 0 increase from left to right - increase from right to left - peak in mid-row (C) - is the largest at the left/right extremes (Li, Ne) (23) 1N + inhibitor I no inhibitor 1/S The above plot for a non-competitive inhibitor of an enzyme (e.g., new AIDS drug) says that: - as the I concentration is increased, the inhibition is smaller - the apparent Km of the enzyme is increased by the presence of the inhibitor 0 the enzyme is “killed” by the inhibitor, lowering Vm, but not changing Km - all of the above are true (24) Osteo orosis: causes a shrunken skeleton and bones prone to fracture — is a more serious problem for women than for men a can be prevented by a class of drug called bisphosphonates @ all of the above are true (25) According to the induced fit theory: — a substrate can bind to any part of an enzyme equally well a an enzyme active site adjusts itself to enhanced binding to a substrate - a substrate fits into a rigid enzyme active site like a lock into a key - none of the above are true TWO BONUS QUESTIONS! (plus 4 x 2 points) (a) Identify the following amino acids by name or symbol: .CO H (i), H2N-C'H 2 (ii). HgN-C'H'COZH M A 4A W‘M CH3 (b) Using your own L/R feet and L/R shoes as the examples, define: - racemate L/KW WM g: . L, [/5 {Lurk WM — enantiomers L W M Ql//" A / Lm flfzvrf 10 ...
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This note was uploaded on 07/22/2008 for the course CHEM 203Lxg taught by Professor Mc kenna during the Fall '07 term at USC.

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203f99_e2 - Last Name: Z “(AW First Name: 9 ’ Chemistry...

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