Morphogenesis Most embryos start out as lumps of cells. Cells in these lumps
are initially undifferentiated-that is, they start out all in the same state.
Over time cells then commit to different functions, e.g., to becoming legs,
eyes, and so on. To do this chemicals called morphogens are distributed
unequally through the embryo, allowing each cell to tell where in the embryo
it is located. How are unequal distributions of morphogens achieved? One model
for how morphogens can be distributed through the embryo is that morphogens
are continuously produced at one end (also called pole) of the embryo. From
there they diffuse through the embryo. As the morphogens diffuse, they are
constantly broken down by the cells in the embryo.
First let's ignore the process of morphogen degradation, and focus only on
diffusion. We will assume that the pole at which the morphogen is produced is
located at \(x=0 ;\) and for simplicity's sake the cell occupies the interval \(x
\geq 0\). Then our partial differential equation model for the distribution of
morphogen becomes:
$$
\frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}} \quad
\text { for } \quad x>0
$$
with
$$
-D \frac{\partial c}{\partial x}(0, t)=Q \quad \text { and } \quad c(x, t) \rightarrow 0 \text { as } x \rightarrow \infty
$$
where \(Q\) is the rate of morphogen production.
(a) Let's try to find a steady state distribution of morphogen. That is, we
will assume that over time the morphogen concentration reaches some state that
does not change with time, i.e., the concentration is given by a function
\(C(x) .\) Then \(C(x)\) will satisfy the partial differential equation if and
only if:
$$
\begin{aligned}
0 &=D \frac{d^{2} C}{d x^{2}} \quad \text { for } \quad x>0 \\
-D C^{\prime}(0) &=Q \quad \text { and } \quad C(x) \rightarrow 0 \text { as } x \rightarrow \infty
\end{aligned}
$$
Show that there is no function \(C(x)\) that satisfies this differential
equation. [Hint: Start by integrating once \((10.48)\) to find \(d C / d x\) and
then again to find \(\mathcal{C}(x)\), then try to impose the constraints at
\(x=0\), and as \(x \rightarrow \infty\) on your solution.]
(b) Now let's incorporate morphogen degradation into our model. We will assume
that the breakdown of morphogen has first order kinetics (see Section \(5.9\)
for a discussion of the different kinds of kinetics that chemical reactions
may have). This means that in one unit of time a fraction \(r\) of the morphogen
contained in each region of the embryo is degraded. Then our partial
differential equation must be altered to:
$$
\frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}}-r c
\quad \text { for } \quad x>0
$$
with
$$
-D \frac{\partial c}{\partial x}(0, t)=Q \quad \text { and } \quad c(x, t) \rightarrow 0 \text { as } x \rightarrow \infty
$$
Show that this partial differential equation does have a steady state solution
of the form:
$$
C(x)=Q \sqrt{\frac{1}{D r}} \exp \left(-\sqrt{\frac{r}{D}} x\right)
$$
That is, check that this function \(C(x)\) satisfies both the steady state form
of \((10.49)\) as well as the constraints at \(x=0\) and as \(x \rightarrow
\infty\).
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