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2.10: Soluciones a problemas seleccionados

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    Ejercicio 2.4.1:

    a) no competitivo

    b) competitivo

    c) competitivo

    d) no competitivo

    Ejercicio 2.5.1:

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

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

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

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

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

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

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

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

    e)

    \[V_{max} = 6.0 \times 10^{-7} \frac{mol}{Ls} \nonumber\]

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

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

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

    Ejercicio 2.5.3:

    1. no competitivo
    2. mixto
    3. no competitivo
    4. competitivo
    5. no competitivo
    6. no competitivo
    7. competitivo

    This page titled 2.10: Soluciones a problemas seleccionados is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Chris Schaller via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.