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Review session for exam-III

Review session for exam-III. Lectures 10-12. The concept of “induced fit” refers to the fact that:. Enzyme specificity is induced by enzyme-substrate binding. Enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction.

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Review session for exam-III

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  1. Review session for exam-III Lectures 10-12

  2. The concept of “induced fit” refers to the fact that: • Enzyme specificity is induced by enzyme-substrate binding. • Enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction. • Enzyme-substrate binding induces movement along the reaction coordinate to the transition state. • Substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation. • When a substrate binds to an enzyme, the enzyme induces a loss of water from the substrate.

  3. The steady state assumption, as applied to enzyme kinetics, implies: • Km = Ks. • The enzyme is regulated. • The ES complex is formed and broken down at equivalent rates. • The Km is equivalent to the cellular substrate concentration. • The maximum velocity occurs when the enzyme is saturated.

  4. The Lineweaver-Burk plot is used to: • determine the equilibrium constant for an enzymatic reaction. • extrapolate for the value of reaction rate at infinite enzyme concentration. • illustrate the effect of temperature on an enzymatic reaction. • solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration. • solve, graphically, for the ratio of products to reactants for any starting substrate concentration.

  5. To determine Km from a double-reciprocal plot, you would: • multiply the reciprocal of the x-axis intercept by -1. • multiply the reciprocal of the y-axis intercept by -1. • take the reciprocal of the x-axis intercept. • take the reciprocal of the y-axis intercept. • take the x-axis intercept where V0 = 1/2 Vmax.

  6. The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the _______________.

  7. In competitive inhibition, an inhibitor: • binds at several different sites on an enzyme. • binds covalently to the enzyme. • binds only to the ES complex. • binds reversibly at the active site. • lowers the characteristic Vmax of the enzyme.

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

  9. A small molecule that decreases the activity of an enzyme by binding to a site other than the catalytic site is termed a(n) _______________.

  10. For the reaction E + S  ES  P, the Michaelis-Menten constant, Km, is actually a summary of three terms. • What are they? • How is Km determined graphically?

  11. Fifteen mg of an enzyme of Mr 30,000 working at Vmax catalyzes the conversion of 60 mmol of substrate into product in 3 min. • What is the enzyme's turnover number?  

  12. Answer: • The amount of enzyme present is • 15 X10-6 g, which is • (15 ´ 10-6 g)/(3 ´ 104 g/mol) = 5 X10-10 mol of enzyme. • The rate of product formation is • 60 ´ 10-6 mol/3 min or • 20 X10-6 mol of product per minute. • The turnover number is, therefore, • (20 X10-6 mol/min)/(5 X10-10 mol of enzyme) or • 4 X10-4 min-1.

  13. Why does pH affect the activity of an enzyme?

  14. Answer: The state of ionization of several amino acid side chains is affected by pH and the activity of many enzymes requires that certain of the amino acid residue side chains be in a specific ionization state.

  15. Chymotrypsin belongs to a group of proteolytic enzymes called the “serine proteases,” many of which have an Asp, His, and Ser residue that are crucial to the catalytic mechanism. The serine hydroxyl functions as a nucleophile. • What do the other two amino acids do to support this nucleophilic reaction?

  16. Answer: • In chymotrypsin,histidine functions as a general base, accepting a proton from the serine hydroxyl, thereby increasing serine's reactivity as a nucleophile. • The negatively charged Asp stabilizes the positive charge that develops on the His.

  17. question What is a zymogen (proenzyme)? Explain briefly with an example.

  18. Answer • A zymogen is an inactive form of an enzyme that is activated by one or more proteolytic cleavages in its sequence. • Chymotrypsinogen, trypsinogen, and proelastase are all zymogens, becoming chymotrypsin, trypsin, and elastase, respectively, after proper cleavage.

  19. What is the less active conformational form of an allosteric enzyme?

  20. The enzymes that catalyze protein phosphorylation are:

  21. Aspartyl transcarbamylase catalyzes the first step in the synthesis of ______________.

  22. ____________ is the most common donor molecule in phosphorylation reactions catalyzed by protein kinases.

  23. Protein kinases add phosphoryl groups to serine, threonine, and _______________ residues in target proteins.

  24. The relaxed form of an allosteric enzyme has _________ affinity for the substrates. • higher • equal • lower • no • None of the above.

  25. The common activator of the pancreatic zymogens is • trypsinogen. • trypsin. • chymotrypsin. • elastase. • None of the above.

  26. How is the clotting cascade initiated?

  27. What is the final step in the clotting pathway?

  28. Individuals in a royal family suffered from hemophilia and often died early from bleeding. • What is the cause of this disease? • How is it treated?

  29. Answer • The disease in the royal family was caused by faulty factor VIII of the intrinsic pathway. • It is treated by the addition of supplemental protein, originally isolated from serum but now a recombinant product.

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