2.10: Solutions to Selected Problems
- Page ID
- 200454
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise 2.4.1:
a) non-competitive
b) competitive
c) competitive
d) non-competitive
Exercise 2.5.1:
a) \(V_{max} = 1.8 \times 10^{-5} \frac{mol}{Ls}\)
\[\frac{V_{max}}{2} = 9 \times 10^{-6} \frac{mol}{Ls} \: so \: K_{m} = 6 \frac{mol}{L} \nonumber\]
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)
\[V_{max} = 6.0 \times 10^{-7} \frac{mol}{Ls} \nonumber\]
\[\frac{V_{max}}{2} = 3 \times 10^{-7} \frac{mol}{Ls} \: so \: K_{m} = 13 \frac{mol}{L} \nonumber\]
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:
- uncompetitive
- mixed
- noncompetitive
- competitive
- uncompetitive
- noncompetitive
- competitive