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Chemistry LibreTexts

Solutions 8

  • Page ID
    47382
  • S8.1

    This is incorrect. 

    Because the activation energy is the energy hill between reactants and products, enzymes decreasing the size of the hill also decreases the amount of energy needed for reactions to go in either direction. A smaller energy hill allows reactants and products to overcome the barrier quicker, resulting a faster reaction rate.

    S8.2

    We already know the turnover number (\(k_{cat}\)). The amount of time necessary to cleave the substrate is the reciprocal of the turnover rate.

    \[t= \dfrac{1}{k} = \dfrac{1}{30,000 \;s^{-1}} = 3.33 \times 10^{-5} s\] 

    S8.3

    Assuming rapid equilibrium is appropriate when k-1 >> k2, so that KM = KS. In this case, the values of k-1 and k2 are not dramatically different, so we can calculate KS and KM and compare.

    1. Calculate KM

    $$ K_M = \dfrac{k_{-1} + k_2}{k_1} $$

    $$ K_M = \dfrac{8 \times 10^5\ s^{-1} + 5 \times 10^4\ s^{-1}}{7 \times 10^7\ M^{-1}\ s^{-1}} $$

    $$ K_M = 0.01\ M $$

    2. Calculate KS

    $$ K_S = \dfrac{k_{-1}}{k_1} $$

    $$ K_M = \dfrac{8 \times 10^5\ s^{-1}}{7 \times 10^7\ M^{-1}\ s^{-1}} $$

    $$ K_M = 0.01\ M $$

    Since \(K_M = K_S\), it is appropriate to assume rapid equilibrium.

    S8.4

    1. Ans: Kcat=Vmax/[E]0, so we can change the initial enzyme concentration and measure the maximum rate of reaction, and then calculate the Kcat.

    2. Ans: Although the rate of change of the ES complex is 0, there can still be a consistent production of products. As the ES complex transforms to product, it also regenerated the enzyme. Then the enzyme can enter into the catalytic cycle again to form new ES complex. 

    3. How would the rate of product formation change if:

      a) The substrate concentration was doubled? 

        Ans: No change, because the substrate is already in excess and the rate of product formation will be limited by the concentration of the enzyme.

      b) The enzyme concentration was doubled?

        Ans: The maximum rate of product formation will also double: Vmax=K2[E]0

      c) The reaction was carried out in deuterated water (D2O) instead of H2O (comment qualitatively)?

        Ans: The reaction involves proton tunneling. If the hydrogen is replaced with deuterium, its tunneling will be slowed down because of deuterium's larger molecular weight. So the reaction will also be slower.

    4. Hydrogen bonding is essential to the hydrolysis of protein catalyzed by chymotrypsin. It is involved in the process of breaking bonds and forming new bonds. (refer to the chemwiki texts for details)

    5. They will look essentially the same.