Problem 43 - Calculus for Biology and Medicine

Question

We discuss the Monod growth function, which was introduced in Example 6 of this section. The Monod growth function \(r(N)\) describes growth as a function of nutrient concentration \(N\). Assume that $$ r(N)=5 \frac{N}{1+N}, \quad N \geq 0 $$ Find the percentage increase when the nutrient concentration is doubled from \(N=0.1\) to \(N=0.2\). Compare this result with what you find when you double the nutrient concentration from \(N=10\) to \(N=20\). This is an example of diminishing return.

Step-by-Step Solution

Answer

The percentage increase is 83.33% from 0.1 to 0.2, and 4.76% from 10 to 20, demonstrating diminishing returns.

Step 1: Definition of Monod Growth Function

The Monod growth function is defined as \( r(N) = 5 \frac{N}{1+N} \). This function describes the growth rate \( r \) in relation to the nutrient concentration \( N \). Our goal is to compute the percentage change in the growth rate when the nutrient concentration is doubled.

Step 2: Calculate Growth Rate at Initial Concentration (0.1)

We first calculate the growth rate when \( N = 0.1 \): \[ r(0.1) = 5 \frac{0.1}{1 + 0.1} = 5 \frac{0.1}{1.1} \approx 0.4545 \]

Step 3: Calculate Growth Rate at Doubled Concentration (0.2)

Next, calculate \( r(0.2) \): \[ r(0.2) = 5 \frac{0.2}{1 + 0.2} = 5 \frac{0.2}{1.2} \approx 0.8333 \]

Step 4: Calculate Percentage Increase for N=0.1 to N=0.2

The percentage increase is calculated as: \[ \text{Percentage Increase} = \frac{r(0.2) - r(0.1)}{r(0.1)} \times 100 \approx \frac{0.8333 - 0.4545}{0.4545} \times 100 \approx 83.33\% \]

Step 5: Calculate Growth Rate at Initial Concentration (10)

Calculate the growth rate with \( N = 10 \): \[ r(10) = 5 \frac{10}{1 + 10} = 5 \frac{10}{11} \approx 4.5455 \]

Step 6: Calculate Growth Rate at Doubled Concentration (20)

Calculate \( r(20) \): \[ r(20) = 5 \frac{20}{1 + 20} = 5 \frac{20}{21} \approx 4.7619 \]

Step 7: Calculate Percentage Increase for N=10 to N=20

Determine the percentage increase with higher nutrient concentrations: \[ \text{Percentage Increase} = \frac{r(20) - r(10)}{r(10)} \times 100 \approx \frac{4.7619 - 4.5455}{4.5455} \times 100 \approx 4.76\% \]

Step 8: Compare and Conclude

Comparing both percentage increases, we find 83.33% for \( N = 0.1 \) to 0.2, and 4.76% for \( N = 10 \) to 20. This demonstrates diminishing returns, as increases in nutrient concentration lead to smaller relative increases in growth as \( N \) becomes large.

Key Concepts

Diminishing Returns

The concept of diminishing returns in the context of the Monod Growth Function is crucial for understanding how organisms grow in response to increasing nutrient concentrations. When we say "diminishing returns," we refer to the fact that as nutrient concentration increases, the incremental improvement in growth rate starts to decrease.
Essentially, you get less "bang for your buck" as you supply more nutrients.

For example, if you increase nutrient concentration from a small amount, like doubling from 0.1 to 0.2, the growth rate sees a significant increase of around 83.33%.
However, when starting from a higher concentration like 10 and moving to 20, the growth rate only increases by approximately 4.76%.
This stark contrast highlights diminishing returns, meaning that the large initial increase you see when the dose is small doesn't hold as nutrient levels rise. The diminishing returns concept is not only about growth rates; it applies to various fields where input gains decrease as total input increases.
For organisms, it signifies that there is a saturation point beyond which increasing nutrient availability doesn't substantially improve growth.

Nutrient Concentration

Nutrient concentration plays a pivotal role in the Monod Growth Function. It serves as the key driver that determines the rate at which an organism can grow.
The function itself, given by \( r(N) = 5 \frac{N}{1+N} \), shows how the growth rate \( r \) is reliant on the nutrient concentration \( N \) of the environment.

At lower concentrations, every additional unit of nutrient can significantly impact growth, as shown by the 83.33% increase from doubling nutrients from 0.1 to 0.2.
Conversely, at higher concentrations, the benefit of additional nutrients diminishes.
This is seen when the increase from 10 to 20 yields only a 4.76% rise in growth rate.
As the formula suggests, the graph of \( r(N) \) versus \( N \) is non-linear and levels off as \( N \) becomes larger.Understanding nutrient concentration in this context can aid in optimizing growth conditions in various applications, such as agriculture or bacterial growth in laboratories.
A controlled nutrient supply can ensure maximum efficient growth without unnecessary waste.

Growth Rate Percentage Change

The growth rate percentage change gives an insightful measure of how the growth rate responds quantitatively to changes in nutrient levels.
It's a relative metric that helps us understand performance efficiency as nutrients increase.

The Monod Growth Function exemplifies this through its formula:\[ \text{Percentage Increase} = \frac{r(N_{new}) - r(N_{old})}{r(N_{old})} \times 100 \].
By calculating this for different nutrient levels, you can see firsthand how sensitivity to nutrient changes decreases as concentration increases.In our examples, for small increases in nutrient levels, the percentage change is high, like 83.33% from 0.1 to 0.2.
However, for larger base nutrient levels, like 10 to 20, even doubling results in a mere 4.76% change.
This shift represents diminishing returns and highlights why just focusing on absolute growth (\( r(N_{new}) - r(N_{old}) \)) isn't enough.
The calculated percentage gives a complete picture of growth efficiency under varied nutritional conditions.