# 8.7 Mathematical modeling of hippocampal spatial memory with place  (Page 6/6)

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${n}_{1}\ge -\frac{\tau }{I}·ln\left(1,-,\frac{{v}_{th}-{v}_{r}}{{w}_{inp}},·,\left(1,-,{e}^{-I/\tau }\right)\right).$

In the computational method, the Matlab program compT.m calculates ${n}_{1}$ by updating $v\left(t\right)$ with equations [link] and [link] . In the analytic method, AnalysisT.m calculates ${n}_{1}$ with equation [link] .

[link] shows that ${n}_{1}$ is clearly a step-wise function where higher input weights allow the cell to spike after fewer input spikes. Comparison of minimum number of input spikes necessary for activity from computation and analysis as a function of w i n p . v ( t ) is calculated by equations [link] and [link] in compT.m and by equation [link] in AnalysisT.m . (Plotted in AnalysisT.m )

## Application of findings Simple 2-cell network with synapses active in Place Field 1. While the rat is in the place field of Cell 1, Cell 1 receives input with weight w e x t from an external source, and when Cell 1 spikes, it gives input with weight w 12 to Cell 2.

We apply the results from the two previous sections to solve a couple simple questions. Consider the situation depicted in [link] : the rat is in Place Field 1, Cell 1 receives input with weight ${w}_{ext}$ from an external source, and, when Cell 1 spikes, it gives internal input to Cell 2 with weight ${w}_{12}$ . We can find the minimum weight of ${w}_{12}$ necessary for Cell 2 to spike in Place Field 1, or where Place Field 1 and Place Field 2 overlap.

For a fixed interspike interval $I$ of external input of fixed weight ${w}_{ext}$ , we calculate ${n}_{1}$ using equation [link] as a function of $I$ and ${w}_{ext}$ . Thus ${n}_{1}$ denotes the minimum number of external input spikes of weight ${w}_{ext}$ necessary for Cell 1 to fire. Let us denote the time of Cell 1's first spike as

${t}_{1}=\left({n}_{1}-1\right)I.$

While the rat is in Place Field 1, Cell 2 only receives input from Cell 1. To find the interspike interval of Cell 1's spikes, or equivalently the interspike interval that Cell 2 receives input in Place Field 1, consider [link] . We know that Cell 1's first spike is at $t={t}_{1}$ . After it spikes, the voltage decays until the next input spike at ${n}_{1}I$ . Then it takes another $\left({n}_{1}-1\right)I\phantom{\rule{4pt}{0ex}}ms$ for Cell 1 to spike at $t={n}_{1}I+\left({n}_{1}-1\right)I$ . Thus we subtract the first spike time from the second spike time

${n}_{1}I+\left({n}_{1}-1\right)I-\left({n}_{1}-1\right)I={n}_{1}I,$

and we see that after the first spike time at ${t}_{1}=\left({n}_{1}-1\right)I\phantom{\rule{4pt}{0ex}}ms$ , Cell 1 fires every ${n}_{1}I\phantom{\rule{4pt}{0ex}}ms$ . Interspike interval of Cell 1 from external input alone, I 1 . Cell 1's first spike occurs at t = t 1 = ( n 1 - 1 ) I . Then the voltage of Cell 1 decays until the next input spike at t = n 1 I . It then takes another ( n 1 - 1 ) I m s for Cell 1 to spike at t = n 1 I + ( n 1 - 1 ) I . Thus, it can be seen that except for the first spike at t = ( n 1 - 1 ) I , Cell 1 fires every n 1 I m s .

Thus, we know that Cell 1 gives input to Cell 2 with weight ${w}_{12}$ every ${n}_{1}I\phantom{\rule{4pt}{0ex}}ms$ . To find the minimum weight of ${w}_{12}$ that would allow for Cell 2 to fire in Place Field 1, we consider two cases where the first is simpler: Place Field 1 is infinitely long or finitely long.

Equation [link] tells us that if

${w}_{12}\ge \left({v}_{th}-{v}_{r}\right)\left(1-{e}^{-{n}_{1}I/\tau }\right),$

then Cell 2 would fire given a sufficiently long Place Field 1. Thus, if Place Field 1 is infinitely long, we simply require that equation [link] be true, then Cell 2 fires in Place Field 1. As we do in "Number of input spikes versus input weight" , we choose an interspike interval: let ${n}_{1}I=40\phantom{\rule{4pt}{0ex}}ms$ . For this value of ${n}_{1}I$ , App.m calculates the minimum of equation [link] to be $13.83\phantom{\rule{4pt}{0ex}}mV$ , as marked in [link] . Minimum w 12 necessary for Cell 2 to fire as a function of n 1 I . It is clear that the larger the interspike interval, the more weight is required for activity. We note the value of w 12 at n 1 I = 40 because these values are used in the case where Place Field 1 is finite. ( App.m )

Now assume Place Field 1 is finitely long. We apply equation [link] to find ${n}_{2}$ , where ${n}_{2}$ is the minimum number of input spikes of weight ${w}_{12}$ necessary for Cell 2 to fire. We let ${n}_{1}I=40\phantom{\rule{4pt}{0ex}}ms$ and consider only the values of ${w}_{12}\ge 13.83\phantom{\rule{4pt}{0ex}}mV$ as calculated for ${n}_{1}I=40\phantom{\rule{4pt}{0ex}}mV$ in equation [link] . [link] shows ${n}_{2}$ as a function of these values of ${w}_{12}$ . As in "Number of input spikes versus input weight" , ${n}_{2}$ is step-wise like ${n}_{1}$ . Minimum number of input spikes necessary for Cell 2 to fire, denoted as n 2 , as a function of input weight w 12 . We choose n 1 I = 40 m s and consider only the values of w 12 ≥ 13 . 83 m V (the minimum w 12 necessary for Cell 1 to fire from equation [link] ). Like n 1 from "Number of input spikes versus input weight" , n 2 is step-wise. We note the minimum weight necessary for Cell 2 to fire in Place Field 1, w 12 = 16 . 01 m V . ( App.m )

The first spike of Cell 2 occurs at time ${t}_{2}=\left({n}_{2}\right)\left({n}_{1}I\right)-I$ . We find the value of ${w}_{12}$ necessary for ${t}_{2}\le$ the time spent in Place Field 1. Suppose the time spent in Place Field 1 is $50\phantom{\rule{4pt}{0ex}}ms$ . We see from [link] that

${n}_{2}=\left\{\begin{array}{cc}2\hfill & \phantom{\rule{4.pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}13.83\le {w}_{12}\le 15.95\hfill \\ 1\hfill & \phantom{\rule{4.pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}15.95<{w}_{12}\le 20.00\hfill \end{array}\right)$

and calculate that

${t}_{2}=\left\{\begin{array}{cc}60\hfill & \phantom{\rule{4.pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}13.83\le {w}_{12}\le 15.95\hfill \\ 20\hfill & \phantom{\rule{4.pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}15.95<{w}_{12}\le 20.00\hfill \end{array}\right)$

Thus, for $13.83\le {w}_{12}\le 15.95\phantom{\rule{4pt}{0ex}}mV$ , ${t}_{2}=60\phantom{\rule{4pt}{0ex}}ms$ and Cell 2 will not fire in Place Field 1, but for $16.01\le {w}_{12}\le 20.00\phantom{\rule{4pt}{0ex}}mV$ , ${t}_{2}=20\phantom{\rule{4pt}{0ex}}ms$ and Cell 2 will fire in Place Field 1. The minimum value of ${w}_{12}$ for ${t}_{2}\le 50\phantom{\rule{4pt}{0ex}}ms$ is marked on [link] ( App.m ).

## Future work

Our goal is to better understand the relation between input weights and backward shift of place fields. We modeled the Double Rotation Experiment using a simple 120-cell ring and calculated the backward shift of the place fields as a function of input weights of that model [link] , [link] . We have computationally and analytically found the minimum input weight necessary for activity as a function of the interspike interval as well as the minimum number of input spikes necessary for activity as a function of input weight.We applied our findings analytically to find the minimum internal input weight necessary for place fields to overlap in a simple 2-cell model.

Future work may include constructing a code to compare our analytical results from "Application of findings" for the infinite and finite place field cases that could give us the minimum internal input weight necessary for place fields to overlap. We could also couple the equations regarding input weight changes by spike timing-dependent plasticity with the equation that gives us the minimum number of input spikes necessary for activity that is dependent upon input weights. Since the time of the first spike of a cell can be given in terms of the minimum number of input spikes necessary for activity as a function of input weight, we may be able to find a value that the time approaches. Ultimately, given a set maximum for the input weight, we would like to be able to predict the amount of backward shift of a place field using spike timing-dependent plasticity.

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