Short answer 53. Carbohydrates, proteins, nucleotides, and nucleic acids are important groups of molecules found in living organisms. Which 6 elements do living organisms need to build these molecules? Ans: C, H, O, N, S, P Level of Difficulty: Moderate Section 1.1.A Learning objective: The Origin of Life 54. Explain in one or two sentences why scientists believe that all living organisms are related (that they have all evolved from a common ancestor)? Ans: Biochemically, all living organisms are very similar. Level of Difficulty: Moderate Section 1.2.C Learning objective: Cellular Architecture 55. Consider a reaction in which H = -20. kJ/mol and S = 10. J/mol · K. a. Calculate the G for this reaction at 25°C. b. Is the reaction spontaneous (explain your answer)? Ans: a. G = H − TS  G = -20,000 J/mol -(298 × 10 J/mol) = −23,000 J/mol b. The reaction is spontaneous because the G is negative. Level of Difficulty: Moderate Section 1.3.C Learning objective: Thermodynamics G6P). A student SDSU 56. Consider the isomerization reaction R  P, in which R is converted to P. atThe G°’incubates for this a reaction is −10. kJ/mol. Calculate the [P]/[R] at equilibrium at 25°C. At equilibrium [P]/[R] = Keq = e−G°/RT  [P]/[R] = e− -(−-10000 J/mol)/8.3145 ·x 298 J/mol = e10000/8.3145 ×x 298 = 57 Level of Difficulty: Moderate Section 1.3.D Learning objective: Thermodynamics 57. Phosphoglucomutase catalyses the reaction in which a phosphate group is transferred from the 1 carbon of glucose to the 6 carbon of glucose (G1P  0.2 M solution of glucose-1-phosphate overnight with a small amount of the enzyme. At equilibrium the concentration of glucose 1-phosphate is 9.0 × 10−3 M and the concentration of glucose 6-phosphate is 19.1 × 10−2 M. 19 Calculate the equilibrium constant (Keq) and the standard state free energy (∆G°’) for this reaction at 25°C. Ans: The equilibrium constant Keq = [G6P]eq/[G1P]eq = 19.1 × 10−2 M/ 9.0 × 10−3 M = 21. G°ꞌ = −RTlnKeq = −8.3145 × 298 J/mol × ln 21.2 = −7,600 J/mol. Level of Difficulty: Difficult Section 1.3.D Learning objective: Thermodynamics 58. The reaction pyruvate + ATP  PEP + ADP is catalyzed by the enzyme pyruvate kinase. The G°ꞌ for this reaction is 31.4 kJ/mol. Use this information to answer the following questions. a. Does this reaction move forward (from left to right) or backward under standard state conditions? Explain you answer in one sentence. b. Calculate the Keq for this reaction at 25°C. c. After the reaction has reached equilibrium at 25°C [ATP] = 5 mM, [ADP] = 0.5 mM, [pyruvate] = 0.2 mM. Calculate the concentration PEP. Ans: a. The reaction goes backward (from right to left) because G°ꞌ for the forward reaction is positive. b. Keq = e−G°ꞌ/RT = e−31400/8.3145 × 298 = 3.13 × 10−6 c. Keq = [ADP]eq × [PEP]eq/[ATP]eq × [Pyruvate]eq  3.1 × 10−6 = 0.5 mM × [PEP]eq/0.2 mM × 5 mM  [PEP]eq = (3.1 × 10−6 × 0.2 mM × 5 mM)/0.5 mM = 6.2 × 10−6 mM (6.2 × 10−9 M) Level of Difficulty: Difficult Section 1.3.D Learning objective: Thermodynamics 20

Chapter 1 Introduction to the Chemistry of Life Short answer

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