i) Re-arrange the Michaelis Menten equation so it involves the ratio. Show all steps beginning Km with noting any assumptions or required conditions. d[P] = dt = K₂ [E],[S] Km+[S]' [S]O ii) Calculate the ratio for the case when the rate of product formation is 68% of Vmax and the Km substrate is in great excess. V Vmax Vmax Begin by explaining the meaning of stating simply "it's the ratio...." is not sufficient. Include in your explanation the factors that effect v and Vmax- Consider what factors make v different from or equal to Vmax. Consider what Km represents concerning processes involving ES. [S]O Km iii) Explain, in a few sentences, why the ratio determines the ratio V

Biochemistry
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Chapter1: Biochemistry: An Evolving Science
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i) Re-arrange the Michaelis-Menten equation so it involves the ratio \(\frac{[S]_0}{K_m}\). Show all steps beginning with \[\frac{d[P]}{dt} = \frac{k_2 [E]_0 [S]}{K_m + [S]}\], noting any assumptions or required conditions.

ii) Calculate the ratio \(\frac{[S]_0}{K_m}\) for the case when the rate of product formation is 68% of \(v_{\text{max}}\) and the substrate is in great excess.

iii) Explain, in a few sentences, why the ratio \(\frac{[S]_0}{K_m}\) determines the ratio \(\frac{v}{v_{\text{max}}}\).
- Begin by explaining the meaning of \(\frac{v}{v_{\text{max}}}\), stating simply “it’s the ratio...” is not sufficient.
- Include in your explanation the factors that affect \(v\) and \(v_{\text{max}}\).
  - Consider what factors make \(v\) different from or equal to \(v_{\text{max}}\).
  - Consider what \(K_m\) represents concerning processes involving ES.

iv) Calculate \(K_M\) at 310K given the following rate constant information:
- \(k_1 = 17 \, \text{s}^{-1} \text{M}^{-1}\) at 300K with \(A = 7300 \, \text{s}^{-1} \text{M}^{-1}\)
- \(k_{-1} = 6 \, \text{s}^{-1}\) at 300K with \(A = 14500 \, \text{s}^{-1}\)
- \(k_2 = 31 \, \text{s}^{-1}\) at 300K with \(A = 600 \, \text{s}^{-1}\)
Transcribed Image Text:i) Re-arrange the Michaelis-Menten equation so it involves the ratio \(\frac{[S]_0}{K_m}\). Show all steps beginning with \[\frac{d[P]}{dt} = \frac{k_2 [E]_0 [S]}{K_m + [S]}\], noting any assumptions or required conditions. ii) Calculate the ratio \(\frac{[S]_0}{K_m}\) for the case when the rate of product formation is 68% of \(v_{\text{max}}\) and the substrate is in great excess. iii) Explain, in a few sentences, why the ratio \(\frac{[S]_0}{K_m}\) determines the ratio \(\frac{v}{v_{\text{max}}}\). - Begin by explaining the meaning of \(\frac{v}{v_{\text{max}}}\), stating simply “it’s the ratio...” is not sufficient. - Include in your explanation the factors that affect \(v\) and \(v_{\text{max}}\). - Consider what factors make \(v\) different from or equal to \(v_{\text{max}}\). - Consider what \(K_m\) represents concerning processes involving ES. iv) Calculate \(K_M\) at 310K given the following rate constant information: - \(k_1 = 17 \, \text{s}^{-1} \text{M}^{-1}\) at 300K with \(A = 7300 \, \text{s}^{-1} \text{M}^{-1}\) - \(k_{-1} = 6 \, \text{s}^{-1}\) at 300K with \(A = 14500 \, \text{s}^{-1}\) - \(k_2 = 31 \, \text{s}^{-1}\) at 300K with \(A = 600 \, \text{s}^{-1}\)
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