Box 2.

Mathematical proof.

If we start, at baseline, with a given tsh level, we can let p represent the proportion of this level remaining after a reduction in total thyrotroph function/number. In this situation, $0\le \mathrm{p}\le 1$, with 1 representing the full, original response. As we want to change the value of p through the argument, we let all quantities be a function of p, so the TSH level is $\mathrm{T}\mathrm{S}\mathrm{H}\left(\mathrm{p}\right),$ the tsh level is $\mathrm{t}\mathrm{s}\mathrm{h}(\mathrm{p})$ and inhibition level is $\mathrm{r}(\mathrm{p})$. We know, for all $0\le \mathrm{p}\le 1,$ that $\mathrm{t}\mathrm{s}\mathrm{h}\left(\mathrm{p}\right)=\mathrm{p}\times \mathrm{t}\mathrm{s}\mathrm{h}\left(1\right)$, and further that $\mathrm{T}\mathrm{S}\mathrm{H}\left(\mathrm{p}\right)=\mathrm{r}\left(\mathrm{p}\right)\times \mathrm{t}\mathrm{s}\mathrm{h}(\mathrm{p})$, (as per “Results”, 1st paragraph)).

In the absence of autoregulation

$\mathrm{T}\mathrm{S}\mathrm{H}(1)=\mathrm{t}\mathrm{s}\mathrm{h}(1)$

$\mathrm{T}\mathrm{S}\mathrm{H}(\mathrm{p})=\mathrm{t}\mathrm{s}\mathrm{h}(\mathrm{p})=\mathrm{p}\times \mathrm{t}\mathrm{s}\mathrm{h}(1)=\mathrm{p}\times \mathrm{T}\mathrm{S}\mathrm{H}(1)$

In the presence of autoregulation

$\mathrm{T}\mathrm{S}\mathrm{H}(1)=\mathrm{t}\mathrm{s}\mathrm{h}(1)\times \mathrm{r}(1)$

$\begin{array}{r}\mathrm{T}\mathrm{S}\mathrm{H}(\mathrm{p})=\mathrm{t}\mathrm{s}\mathrm{h}(\mathrm{p})\times \mathrm{r}(\mathrm{p})=\mathrm{p}\times \mathrm{t}\mathrm{s}\mathrm{h}(1)\times \mathrm{r}(\mathrm{p})=\mathrm{p}\times \mathrm{t}\mathrm{s}\mathrm{h}(1)\times \mathrm{r}(\mathrm{p})\times (\mathrm{r}(1)/\mathrm{r}(1))\end{array}$

$=\mathrm{p}\times (\mathrm{t}\mathrm{s}\mathrm{h}(1)\times \mathrm{r}(1))\times \mathrm{r}(\mathrm{p})/\mathrm{r}(1)=\mathrm{p}\times \mathrm{T}\mathrm{S}\mathrm{H}(1)\times \mathrm{r}(\mathrm{p})/\mathrm{r}(1)$

Therefore, $\frac{\mathrm{T}\mathrm{S}\mathrm{H}(\mathrm{p})\mathrm{i}\mathrm{n}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{u}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{g}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{T}\mathrm{S}\mathrm{H}(\mathrm{p})\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{u}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{g}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}=\frac{\mathrm{p}\times \mathrm{T}\mathrm{S}\mathrm{H}(1)\times \mathrm{r}(\mathrm{p})/\mathrm{r}(1)}{\mathrm{p}\times \mathrm{T}\mathrm{S}\mathrm{H}(1)}=\frac{\mathrm{r}(\mathrm{p})}{\mathrm{r}(1)}$

A smaller value of p results in a smaller value of tsh, while a smaller value of tsh results in a decrease in autoregulation, and a decrease in autoregulation results in an increase in r (as per “Results” 1stand 3rd paragraphs).Therefore, if p<1, then r(p)/r(1) >1.

And so we can conclude that, TSH(p) in the presence of autoregulation > TSH(p) in the absence of autoregulation.

Thus in the presence of autoregulation the TSH level is higher i.e. the drop in TSH consequent upon the reduction in tsh from tsh(1) to tsh(p)is less, as compared to the changes in the absence of autoregulation.

The proof works similarly if we consider an increase rather than a decrease in thyrotroph number and/or function (i.e. an increase in the tsh curve). In this situation the effect of autoregulation would be to result in a reduced increase in the TSH level, i.e. there would be a lower TSH level. The proof also works, with some modification, if we consider the 2 initial curves to be not identical (as per “Results” 4th paragraph,).