Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

The polymerase chain reaction is a means of making multiple copies of a DNA segment from only a minute amounts of original DNA. The procedure consists of a sequence of, say, 30 cycles in which each segment present at the beginning of a cycle is duplicated once; at the end of the cycle that segment and one copy is present. Introduce notation and write a difference equation with initial condition from which the amount of DNA present at the end of each cycle can be computed. Suppose you begin with 1 picogram \(=0.000000000001 \mathrm{~g}\) of DNA. How many grams of DNA would be present after 30 cycles.

Step-by-Step Solution

Answer

1.073741824 grams of DNA would be present after 30 cycles.

Step 1: Define Notation

Let's define notation for the problem. Let \( N(n) \) represent the amount of DNA (in grams) present after cycle \( n \). We start with 1 picogram of DNA, so initially, we have one copy of the DNA segment. At the beginning, \( N(0) = 1 \, \text{picogram} \).

Step 2: Set Up the Recurrence Relation

In each cycle, each existing segment duplicates, thereby doubling the amount of DNA. Hence, the rule for changing DNA after each cycle can be expressed as a recurrence relation: \[ N(n+1) = 2 \cdot N(n) \] This reflects the duplication of each DNA segment every cycle.

Step 3: Apply Initial Conditions

We know from the problem statement that at the beginning (cycle 0), we start with 1 picogram of DNA. Therefore, \( N(0) = 1 \, \text{picogram}\).

Step 4: Calculate DNA Amount After 30 Cycles

Utilizing the recurrence relation derived, calculate \( N(30) \), the amount of DNA after 30 cycles. Using the formula, we find:\[ N(n) = 2^n \times N(0) \]Substituting the known values, we find:\[ N(30) = 2^{30} \times 1 \, \text{picogram} \approx 1,073,741,824 \, \text{picograms} \]

Step 5: Convert Picograms to Grams

Convert 1,073,741,824 picograms to grams using the conversion rate (1 picogram = \( 10^{-12} \, \text{grams} \)):\[ 1,073,741,824 \, \text{picograms} \times 10^{-12} \, \text{grams/picogram} = 1.073741824 \, \text{grams} \]

Key Concepts

Difference Equation

To understand how DNA amplification through polymerase chain reaction (PCR) can be modeled mathematically, we need to introduce the concept of a difference equation. A difference equation is a mathematical expression that relates a function or sequence to its values at various points. In the case of PCR, this sequence represents the amount of DNA present at each cycle.

Difference equations are pivotal in expressing changes over discrete intervals—cycles, in this scenario. Instead of dealing with continuous processes, they allow us to break down changes cycle by cycle, making them ideal for processes like DNA amplification where incremental duplication occurs distinctly one cycle at a time.

In PCR, after each cycle, the existing DNA segments are doubled. This simple rule of doubling can be captured by the same difference equation used in many biological growth models. Understanding this foundation is crucial when translating the physical duplication of DNA into mathematical terms.

Recurrence Relation

Connected to the difference equation is the concept of a recurrence relation. A recurrence relation defines a sequence of numbers in terms of preceding numbers in the sequence. For PCR, where DNA segments double every cycle, the recurrence relation expresses exactly how the amount of DNA in the next cycle depends on the current amount.

The recurrence relation for our problem is given by:

• \[ N(n+1) = 2 \cdot N(n) \]

Here, \( N(n) \) stands for the amount of DNA after cycle \( n \). Each term is derived directly from its predecessor, reflecting the doubling process. This simple multiplicative relation is common in modeling exponential growth scenarios, like the geometric population growth of cells or the doubling of DNA strands.

Effectively understanding the recurrence relation allows us to predict the sequence of DNA increase across multiple cycles. It's a succinct encapsulation of the PCR process, creating a powerful tool for both academic and practical applications.

DNA Amplification

DNA amplification is the core objective of the polymerase chain reaction (PCR) technique. It enables scientists to generate multiple copies of a specific DNA segment, even when they start with minute quantities of original DNA. This exponential copying is what makes PCR a revolutionary tool in molecular biology for various applications, from research to medicine.

The process involves a cyclical procedure, typically consisting of about 30 cycles. During each cycle, every existing DNA segment is copied once, meaning that by the end of each cycle, the DNA amount is doubled. This exponential growth is calculated effectively using the recurrence relation \( N(n+1) = 2 \cdot N(n) \), demonstrating a clear pattern of how DNA quantity increases cycle by cycle.

The art of DNA amplification lies in harnessing this power of doubling through cycles managed by specific conditions and reagents, creating countless copies swiftly. This makes PCR an indispensable method in DNA-based experiments, diagnostics, and even forensic investigations, where having larger amounts of DNA can significantly influence outcomes.

Initial Conditions

In mathematical modeling such as with DNA amplification, initial conditions play a crucial role in determining the sequence's progression. Initial conditions refer to the starting values of a system or sequence, which directly impact all subsequent calculations or predictions.

For the PCR scenario, the initial condition is that the DNA amount at the beginning of the process (cycle 0) is 1 picogram. This initial measurement sets the stage for all subsequent cycles. Understanding and accurately setting these conditions is vital for achieving a valid model and ensuring that predictions remain accurate.

Applying the initial condition to our recurrence relation, we begin with \( N(0) = 1 \, \text{picogram} \). This starting point translates into the equation \( N(n) = 2^n \times N(0) \), which can be used to compute the DNA present after any cycle \( n \). Here, knowing the initial condition makes it possible to outline the entire amplification process, highlighting how the sequence develops at each stage.