Problem 6 - Calculus for Biology and Medicine

Question

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 \(x \rightarrow \infty\).

Step-by-Step Solution

Answer

Part (a): No solution for \(C(x)\) exists. Part (b): The proposed function satisfies the steady-state equation with degradation.

Step 1: Define steady state assumptions

For a steady-state solution, the concentration does not change with time. Therefore, setting \(\frac{\partial c}{\partial t} = 0\), the equation simplifies for the initial condition (a) to:\[0 = D \frac{d^2 C}{dx^2} \quad \text{for} \quad x>0\]

Step 2: Integrate the differential equation

Integrate \(0 = D \frac{d^2 C}{dx^2}\) once:\[D \frac{dC}{dx} = A\]where \(A\) is a constant of integration. Integrate again to obtain \(C(x)\):\[C(x) = Ax + B\]where \(B\) is another constant of integration.

Step 3: Apply boundary conditions for part (a)

Using the boundary conditions:1. \( -D C'(0) = Q \Rightarrow -DA = Q \Rightarrow A = -\frac{Q}{D} \).2. \( C(x) \rightarrow 0 \) as \( x \rightarrow \infty \Rightarrow C(x) = 0 \) but this does not satisfy the condition derived, resulting in a contradiction.

Step 4: Modify equation with degradation (part b)

Include degradation in the model:\[\frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2} - rc \quad \text{for} \quad x>0\]For steady-state, this equation simplifies to:\[0 = D \frac{d^2 C}{dx^2} - rC\]

Step 5: Solve steady-state equation

Solve \(D \frac{d^2 C}{dx^2} - rC = 0\) using the proposed solution form:Assume \(C(x) = Q \sqrt{\frac{1}{Dr}} e^{-\sqrt{\frac{r}{D}} x}\).For \(C(x)\), check derivatives:\[C'(x) = -\sqrt{\frac{r}{D}} Q \sqrt{\frac{1}{Dr}} e^{-\sqrt{\frac{r}{D}} x}, \quad C''(x) = \frac{r}{D}Q \sqrt{\frac{1}{Dr}} e^{-\sqrt{\frac{r}{D}} x}\]Check that \(D C''(x) - rC(x) = 0\).

Step 6: Verify boundary conditions with proposed solution

Verify:1. Boundary at \(x = 0\):\[-D C'(0) = Q \quad \rightarrow\quad -D \left(-\sqrt{\frac{r}{D}} Q \sqrt{\frac{1}{Dr}} \right) = Q\] confirms \(Q=Q\).2. As \(x \to \infty\), \(C(x) \to 0 \) due to the exponential decay, confirming the boundary condition \(C(x) \to 0\).

Key Concepts

Partial Differential Equation

Partial differential equations (PDEs) are essential tools for modeling dynamic processes where quantities like temperature, pressure, or chemical concentrations change not only over time but also across different spatial dimensions. In the context of morphogenesis, a PDE can help describe how morphogens, the chemicals that guide the formation of an organism's structure, distribute throughout the embryo over time.

For our purpose, we begin with the equation:

• \(\frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2} \)

where:

• \(c\) is the concentration of the morphogen,
• \(t\) is time,
• \(x\) is the position along the embryo, and
• \(D\) is the diffusion coefficient, reflecting how quickly the morphogens spread through the embryo.

Initially, the solution of this equation gives us how morphogens diffuse without considering their breakdown or degradation. This forms the starting point for understanding morphogen distribution before moving on to more complex scenarios involving degradation or other biological factors.

Steady State Solutions

Steady state solutions occur when a system reaches a point where its properties no longer change over time. In the context of morphogen distribution, finding a steady state means identifying a morphogen concentration distribution that remains constant despite ongoing production and diffusion processes.

To find a steady state solution, we set the time derivative to zero, indicating no change with time:

• \(0 = D \frac{d^2 C}{dx^2} \)

Here, \(C(x)\) is the steady-state concentration. Initial explorations without considering morphogen degradation show that such a distribution becomes constant, contradicting boundary conditions. When degradation is accounted for, the solution takes the form \(C(x) = Q \sqrt{\frac{1}{Dr}} e^{-\sqrt{\frac{r}{D}} x}\).

This demonstrates how incorporating additional biological processes like degradation allows a more complex and realistic steady state to be achieved, accurately mirroring the biological phenomenon.

Morphogen Degradation

Morphogen degradation is a critical component in modeling morphogen dynamics as it reflects the natural process by which cells break down these guiding chemicals. In our PDE, morphogen degradation is represented as a first-order kinetic reaction, meaning that the rate of degradation is proportional to the morphogen concentration.

When degradation is included, the equation becomes:

• \(\frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2} - rc\)

where:

• \(r\) is the degradation rate, determining how quickly the morphogens are broken down.

The introduction of the degradation term balances against diffusion and production to allow a steady state to be reached.

This modification leads to a decaying exponential solution, \(C(x) = Q \sqrt{\frac{1}{Dr}} e^{-\sqrt{\frac{r}{D}} x}\), fulfilling the conditions set by production at one pole and ensuring that morphogen concentrations taper off to zero at a distance. This balance is crucial for the complex patterns of structure development found in nature.