Skip to main content
Chemistry LibreTexts

2.10: Solutions to Selected Problems

  • Page ID
    200454
  • Exercise 2.4.1:

    a) non-competitive

    b) competitive

    c) competitive

    d) non-competitive

    Exercise 2.5.1:

    a) \(V_{max} = 1.8 \times 10^{-5} \frac{mol}{Ls}\)

    \(\frac{V_{max}}{2} = 9 \times 10^{-6} \frac{mol}{Ls} \: so \: K_{m} = 6 \frac{mol}{L}\)

    b) \(V_{max} = 6.5 \times 10^{-7} \frac{mol}{Ls}\)

    \(\frac{V_{max}}{2} = 3.25 \times 10^{-7} \frac{mol}{Ls} \: so \: K_{m} = 7 \frac{mol}{L}\)

    c) \(V_{max} = 2.6 \times 10^{-5} \frac{mol}{Ls}\)

    \(\frac{V_{max}}{2} = 1.3 \times 10^{-5} \frac{mol}{Ls} \: so \: K_{m} = 6 \frac{mol}{L}\)

    d) \(V_{max} = 1.2 \times 10^{-5} \frac{mol}{Ls}\)

    \(\frac{V_{max}}{2} = 6 \times 10^{-6} \frac{mol}{Ls} \: so \: K_{m} = 6 \frac{mol}{L}\)

    e)

    \(V_{max} = 6.0 \times 10^{-7} \frac{mol}{Ls}\)

    \(\frac{V_{max}}{2} = 3 \times 10^{-7} \frac{mol}{Ls} \: so \: K_{m} = 13 \frac{mol}{L}\)

    Exercise 2.5.2:

    a) \(\frac{1}{V_{max}} = 30 \frac{Ls}{mol} \: so \: V_{max} = 3.3 \times 10^{-2} \frac {mol}{Ls}\)

    \(\frac{-1}{K_{m}} = -40 \frac{L}{mmol} \: so \: K_{m} = 2.5 \times 10^{-2} \frac{M}{L}\)

    b) ​​\(\frac{1}{V_{max}} = 50 \frac{Ls}{mol} \: so \: V_{max} = 2.0 \times 10^{-2} \frac {mol}{Ls}\)

    \(\frac{-1}{K_{m}} = -70 \frac{L}{mmol} \: so \: K_{m} = 1.4 \times 10^{-2} \frac{M}{L}\)

    c) \(\frac{1}{V_{max}} = 60 \frac{Ls}{mol} \: so \: V_{max} = 1.7 \times 10^{-2} \frac {mol}{Ls}\)

    \(\frac{-1}{K_{m}} = -70 \frac{L}{mmol} \: so \: K_{m} = 1.4 \times 10^{-2} \frac{M}{L}\)

    d) \(\frac{1}{V_{max}} = 50 \frac{Ls}{mol} \: so \: V_{max} = 2.0 \times 10^{-2} \frac {mol}{Ls}\)

    \(\frac{-1}{K_{m}} = -100 \frac{L}{mmol} \: so \: K_{m} = 1.0 \times 10^{-2} \frac{M}{L}\)

    e) \(\frac{1}{V_{max}} = 30 \frac{Ls}{mol} \: so \: V_{max} = 3.3 \times 10^{-2} \frac {mol}{Ls}\)

    \(\frac{-1}{K_{m}} = -100 \frac{L}{mmol} \: so \: K_{m} = 1.0 \times 10^{-2} \frac{M}{L}\)

    a) ​​​​​​​\(\frac{1}{V_{max}} = 30 \frac{Ls}{mol} \: so \: V_{max} = 3.3 \times 10^{-2} \frac {mol}{Ls}\)

    \(\frac{-1}{K_{m}} = -60 \frac{L}{mmol} \: so \: K_{m} = 1.7 \times 10^{-2} \frac{M}{L}\)

    Exercise 2.5.3:

    1. uncompetitive
    2. mixed
    3. noncompetitive
    4. competitive
    5. uncompetitive
    6. noncompetitive
    7. competitive