Accuracy in readout of glutamate concentrations by neuronal cells

Abstract

Glutamate and glycine are important neurotransmitters in the brain. An action potential propagating in the terminal of a presynaptic neuron causes the release of glutamate and glycine in the synapse by vesicles fusing with the cell membrane, which then activate various receptors on the cell membrane of the post-synaptic neuron. Entry of Ca$^{2+}$ through the activated NMDA receptors leads to a host of cellular processes of which long-term potentiation is of crucial importance because it is widely considered to be one of the major mechanisms behind learning and memory. By analysing the readout of glutamate concentration by the post-synaptic neurons during Ca$^{2+}$ signaling, we find that the average receptor density in hippocampal neurons has evolved to allow for accurate measurement of the glutamate concentration in the synaptic cleft.

Graphical abstract

Appendix A

We derive how Eq. 15 is got in this appendix. We have

$$\begin{aligned} \langle C_\textrm{Pr} \rangle _n= & {} \frac{1}{(kc)^n} \sum _{\textrm{Permutate}:\tilde{t}_1,\tilde{t}_3,\tilde{t}_5,\tilde{t}_7} \int _{\tilde{t}_1< \tilde{t}_3<\tilde{t}_5<\tilde{t}_7< kcT} d\tilde{t}_1 d\tilde{t}_3 d\tilde{t}_5 d\tilde{t}_7 \nonumber \\{} & {} \times e^{- \tilde{t}_1} e^{- \tilde{t}_3} e^{- \tilde{t}_5} e^{- \tilde{t}_7} C(kc T-\tilde{t}_7)^n\nonumber \\= & {} \frac{4! C}{6 (kc)^n} \int _0^{kcT} e^{ -\tilde{t}_7 } \left( 1-e^{ -\tilde{t}_7 } \right) ^3(kc T-\tilde{t}_7)^n dt_7\nonumber \\= & {} \frac{4!C (kcT)^{n+1}}{6(kc)^n} \int _0^{1} e^{ -kcT x} \left( 1-e^{ -kcT x } \right) ^3(1-x)^n dx \nonumber \\= & {} \frac{4!C (kcT)^{n+1}}{6(kc)^n} \int _0^{1} e^{ -kcT } e^{ kcT y } \left( 1-e^{ -kcT }e^{ kcT y } \right) ^3y^n dy \nonumber \\= & {} \frac{4!C (kcT)^{n+1}}{6(kc)^n} \int _0^{1} e^{ kcT (y-1) } \left( 1-e^{ kcT (y-1) } \right) ^3y^n dy \nonumber \\= & {} \frac{4!C (kcT)^{n+1}}{6(kc)^n} \int _0^{1} e^{ kcT (y-1) } \left( 1-e^{ 3kcT (y-1) }\right. \nonumber \\{} & {} \left. -3e^{ kcT (y-1)}+3e^{ 2 kcT (y-1)}\right) y^n dy \nonumber \\= & {} \frac{4!C (kcT)^{n+1}}{6(kc)^n} \int _0^{1} \left( e^{ kcT (y-1) } -e^{ 4kcT (y-1) }\right. \nonumber \\{} & {} \left. -3e^{ 2kcT (y-1)}+3e^{ 3 kcT (y-1)} \right) y^n dy \end{aligned}$$

(A1)

Now up to $\mathcal {O}((kcT)^3)$

$$\begin{aligned}{} & {} e^{kcT(y-1)}{=}1{+}kcT(y{-}1){+}\frac{(kcT(y{-}1))^2}{2!}{+}\frac{(kcT(y{-}1))^3}{3!} \end{aligned}$$

(A2)

$$\begin{aligned}{} & {} -e^{4kcT(y-1)}=-1-4kcT(y-1)-\frac{(4kcT(y-1))^2}{2!}\nonumber \\{} & {} \qquad -\frac{(4kcT(y-1))^3}{3!} \end{aligned}$$

(A3)

$$\begin{aligned}{} & {} -3e^{2kcT(y-1)}=-3-6kcT(y-1)-3\frac{(2kcT(y-1))^2}{2!}\nonumber \\{} & {} \qquad -3\frac{(2kcT(y-1))^3}{3!} \end{aligned}$$

(A4)

$$\begin{aligned}{} & {} 3e^{3kcT(y-1)}=3+9kcT(y-1)+3\frac{(3kcT(y-1))^2}{2!}\nonumber \\{} & {} \qquad +3\frac{(3kcT(y-1))^3}{3!} \end{aligned}$$

(A5)

Adding the above four equations gives

$$\begin{aligned}{} & {} \left( e^{ kcT (y-1) } -e^{ 4kcT (y-1) }-3e^{ 2kcT (y-1)}+3e^{ 3 kcT (y-1)} \right) =\\{} & {} \qquad (1-1-3+3)+kcT(y-1)(1-4-6+9)\\{} & {} \qquad +\frac{(kcT(y-1))^2}{2!}(1-16-12+27)\\{} & {} \qquad +\frac{(kcT(y-1))^3}{3!}(1-64-24+81)\\{} & {} \quad = -(kcT(y-1))^3\\{} & {} \quad = (kcT(1-y))^3 \end{aligned}$$

Hence,

$$\begin{aligned} \langle C_\textrm{Pr} \rangle _n= & {} \frac{4!C (kcT)^{n+1}}{6(kc)^n} \int _0^{1} \left( e^{ kcT (y-1) } -e^{ 4kcT (y-1) }\right. \\{} & {} \left. -3e^{ 2kcT(y-1)}+3e^{ 3 kcT (y-1)} \right) y^n dy \\= & {} \frac{4!C (kcT)^{n+1}}{6(kc)^n} \int _0^{1} (kcT(1-y))^3 * y^n dy \\= & {} \frac{4!C (kcT)^{n+4}}{6(kc)^n} \int _0^{1} (1-y^3-3y+3y^2) * y^{n} dy, \\ kcT&<<&1\\= & {} \frac{4!C (kcT)^{n+4}}{6(kc)^n}\left( \frac{1}{n+1}-\frac{1}{n+4}-\frac{3}{n+2}+\frac{3}{n+3}\right) \\= & {} \frac{4!C (kcT)^{n+4}}{6(kc)^n}\left( \frac{n+4-n-1}{(n+1)(n+4)}+\frac{-3n-9+3n+6}{(n+2)(n+3)}\right) \\= & {} \frac{4!C (kcT)^{n+4}}{6(kc)^n}\left( \frac{3}{(n+1)(n+4)}-\frac{3}{(n+2)(n+3)}\right) \\= & {} \frac{4!C (kcT)^{n+4}}{2(kc)^n}\left( \frac{1}{(n+1)(n+4)}-\frac{1}{(n+2)(n+3)}\right) \\= & {} \frac{4!C (kcT)^{n+4}}{2(kc)^n}\left( \frac{(n+2)(n+3)-(n+1)(n+4)}{(n+1)(n+4)(n+2)(n+3)}\right) \\= & {} \frac{4!C (kcT)^{n+4}}{2(kc)^n}\left( \frac{n^2+5n+6-n^2-5n-4}{(n+1)(n+4)(n+2)(n+3)}\right) \\= & {} \frac{4!C (kcT)^{n+4}}{(kc)^n(n+1)(n+2)(n+3)(n+4)}\\ \end{aligned}$$

Similarly,

$$\begin{aligned} \langle C_\textrm{Pr}^2 \rangle _n{} & {} = \frac{4!C^2 (kcT)^{2n+4}}{(kc)^{2n}(2n+1)(2n+2)(2n+3)(2n+4)} \end{aligned}$$

Hence

$$\begin{aligned}{} & {} \frac{\sqrt{\langle C_\textrm{Pr}^2 \rangle _n -\langle C_\textrm{Pr} \rangle ^2_n }}{c\frac{d \langle C_\textrm{Pr} \rangle _n}{dc}}=\frac{C*(kcT)^{n+2}}{kc^n} \\{} & {} \qquad \sqrt{ \frac{4!}{(2n+1)(2n+2)(2n+3)(2n+4)}-\frac{4!*4!(kcT)^{4}}{(n+1)^2(n+2)^2(n+3)^2(n+4)^2}}*\\{} & {} \qquad \frac{(kc)^n(n+1)(n+2)(n+3)(n+4)}{4*4!C(kcT)^{n+4}}\\{} & {} \quad =\sqrt{\frac{4!}{(2n+1)(2n+2)(2n+3)(2n+4)}-\frac{4!*4!(kcT)^{4}}{(n+1)^2(n+2)^2(n+3)^2(n+4)^2}}*\\{} & {} \qquad \frac{(n+1)(n+2)(n+3)(n+4)}{4*4!(kcT)^{2}}\\{} & {} \quad =\sqrt{ \frac{(n+1)^2(n+2)^2(n+3)^2(n+4)^2}{4!(2n+1)(2n+2)(2n+3)(2n+4)}-(kcT)^{4}}*\frac{1}{4(kcT)^{2}}\\{} & {} \quad = \frac{1}{4} \sqrt{ \frac{(n+1)^2(n+2)^2(n+3)^2(n+4)^2}{4!(kcT)^{4}(2n+1)(2n+2)(2n+3)(2n+4)}-1}\\{} & {} \quad = \frac{1}{4} \sqrt{ \frac{(n+1)(n+2)(n+3)^2(n+4)^2}{4!4(kcT)^{4}(2n+1)(2n+3)}-1} \end{aligned}$$