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Substrate ($m$) concentration in a reaction scheme is governed by the equation $$\frac{dm}{dt} = -\dfrac{r_1c_1m}{1+\dfrac{r_1}{r_2}m + \dfrac{r_1r_3}{r_2r_4}m^2}$$ where $c_1$ is the initial enzyme concentration and each $r_i$ is the reaction rate of the reactions present within this reaction scheme (which are positive). Now additionally suppose the substrate is being continuously replenished by a reaction $$\ce{A->[r_5]S,}$$ where $A$ has constant concentration $a_0$. Modify $dm/dt$ and show there are two distinct positive equilibria $0<m_1<m_2$ if $r_5<r_5^C$ where $r_5^C$ is a critical value to be determined.


With this information we have

$$\frac{dm}{dt} = -\dfrac{r_1c_1m}{1+\dfrac{r_1}{r_2}m + \dfrac{r_1r_3}{r_2r_4}m^2}+r_5a_0.$$

Equilibria when $dm/dt=0$. We rearrange this to produce the quadratic

$$r_2r_4r_5a_0 + r_1r_4(r_5a_0 - r_2c_1)m + r_1r_3r_5a_0m^2=0$$

So the equilibria occur at

$$m_{1,2} = \dfrac{r_1r_4(r_2c_1-r_5a_0)\pm\sqrt{r_1^2r_4^2(r_5a_0-r_2c_1)^2-4r_1r_2r_3r_4r_5^2a_0^2}}{2r_1r_3r_5a_0}$$

Now I need to justify these are positive and find the critical value for $r_5$ but I have no idea how to continue here.

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  • $\begingroup$ If i am following you, $m$ is the substrate concentration. So $m = r_5a_0$ which is substituted into your first equation. $\endgroup$ – MaxW Nov 7 '15 at 21:24
  • $\begingroup$ $m$ is the substrate concentration but I have not made that substitution $\endgroup$ – Michael Howlard Nov 7 '15 at 21:46
  • $\begingroup$ How did you get from your equation 1 to your equation 3 ?!? $\endgroup$ – MaxW Nov 7 '15 at 21:48
  • $\begingroup$ I have added $r_5a_0$ to the rate of change for concentration of the substrate as it is being increased by $a_0$ at reaction rate $r_5$. Is this incorrect? $\endgroup$ – Michael Howlard Nov 7 '15 at 22:00
  • $\begingroup$ You have $\ce{A->[r_5]S}$ which means the concentration of S is given by $r_5$ times the concentration of A. In the first sentence you wrote "Substrate ($m$) concentration" so $m$ is the concentration of S. $\endgroup$ – MaxW Nov 7 '15 at 22:18
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I can't believe the way we solved the equation for the discriminant equal to zero. You and I both did the same thing. All we really needed to do is solve $R^2-\lambda = 0$ by writing $R^2=\lambda$, and taking the square root of both sides. The resulting equation is then linear in r5.

Anyway, if I did the algebra correctly, I get the double root as: $$m_1=m_2=\sqrt{\frac{r_2r_4}{r_1r_3}}$$ So, I guess one has to conclude that, in order to guarantee that the roots are different, we have to take as the critical value of $r_5$ the value that we determined by setting the discriminant equal to zero.

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  • $\begingroup$ When you square root both sides why do we take $R>0$, i.e. $R = +\sqrt{\lambda}$ over $R = -\sqrt{\lambda}$? $\endgroup$ – Michael Howlard Nov 8 '15 at 21:15
  • $\begingroup$ For the same reason we said before. That's the only way to guarantee a positive r5. $\endgroup$ – Chet Miller Nov 8 '15 at 21:31
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The first step is to set the discriminant equal to zero, and thereby solve for the critical value of $r_5$. When I did this, I got $$r_5a_0=\frac{r_2c_1}{1+2\sqrt{\frac{r_2r_3}{r_1r_4}}}$$. The next step is to substitute this into the equation for m, and find the two corresponding equal values of m.

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  • $\begingroup$ Actually it's not. All that's required is a little more mathematical manipulation. Factor out the $r_2c_1$ and, in the denominator, use the identity $a^2-b^2=(a+b)(a-b)$ $\endgroup$ – Chet Miller Nov 8 '15 at 15:02
  • $\begingroup$ Yes. I left out the 2. Thanks. I'm going to go back and correct it. We choose the negative numerator because that choice will guarantee that the answer is always positive. $\endgroup$ – Chet Miller Nov 8 '15 at 15:57
  • $\begingroup$ Also replace all the $\ge$ signs with $>$ signs. $\endgroup$ – Michael Howlard Nov 8 '15 at 16:19
  • $\begingroup$ I like what you did much better than what I was suggesting. Very nice reasoning. $\endgroup$ – Chet Miller Nov 8 '15 at 16:51
  • $\begingroup$ Which would be the correct answer in the end then? $\endgroup$ – Michael Howlard Nov 8 '15 at 17:37

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