In a laboratory population of fruit flies (Drosophila), the initial population size ($N$) at time $t=0$ was recorded as 100. Over a specific breeding interval, the intrinsic rate of natural increase ($r$) was calculated to be $0.15$ per individual per day. Assuming the population grows according to the exponential growth model, calculate: The population size after 10 days ($N{10}$). (Given: $e^{1.5} = 4.4816$) The time ($t$) required for the population size to reach 800 individuals. (Given: $\ln(8) = 2.079$, $\ln(1.5) = 0.405$) Explain the graphical difference between exponential growth and logistic growth curves.

Q: In a laboratory population of fruit flies (Drosophila), the initial population size ($N$) at time $t=0$ was recorded as 100. Over a specific breeding interval, the intrinsic rate of natural increase ($r$) was calculated to be $0.15$ per individual per day. Assuming the population grows according to the exponential growth model, calculate: The population size after 10 days ($N{10}$). (Given: $e^{1.5} = 4.4816$) The time ($t$) required for the population size to reach 800 individuals. (Given: $\ln(8) = 2.079$, $\ln(1.5) = 0.405$) Explain the graphical difference between exponential growth and logistic growth curves.

Part 1: Calculation of Population Size after 10 days\nUsing the exponential growth formula: $$Nt = N0 \cdot e^{rt}$|\nGiven data: Initial population ($N0$) = 100 Intrinsic rate of natural increase ($r$) = $0.15$ per day Time ($t$) = 10 days $e^{rt} = e^{0.15 \times 10} = e^{1.5} = 4.4816$ \nSubstituting the values: $$N{10} = 100 \times 4.4816 = 448.16$$\nSince population numbers are discrete organisms, we can round it to 448 individuals. --- Part 2: Calculation of Time to reach 800 individuals\nUsing the exponential formula solved for time $t$: $$Nt = N0 \cdot e^{rt}$| $$\frac{Nt}{N0} = e^{rt}$|\nTaking natural logarithm ($\ln$) on both sides: $$\ln\left(\frac{Nt}{N0}\right) = rt$$ $$t = \frac{\ln(Nt / N0)}{r}$| \nGiven data: $Nt = 800$ $N0 = 100$ $r = 0.15$ \nSubstituting values: $$t = \frac{\ln(800 / 100)}{0.15} = \frac{\ln(8)}{0.15}$|\nGiven that $\ln(8) = 2.079$: $$t = \frac{2.079}{0.15} = 13.86 \text{ days}$|\nSo, it takes approximately 13.86 days. --- Part 3: Difference between Exponential and Logistic Growth Curves Exponential Growth Curve: Occurs when resources (food and space) are unlimited. Results in a J-shaped curve. Equation: $\frac{dN}{dt} = rN$ Logistic Growth Curve: Occurs when resources become limiting at a certain point, leading to competition. Shows an initial lag phase, followed by acceleration, deceleration, and finally an asymptote when the population reaches the carrying capacity ($K$). Results in a Sigmoid (S-shaped) curve. Equation: $\frac{dN}{dt} = rN \left(\frac{K - N}{K}\right)$

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Accepted Answer

Part 1: Calculation of Population Size after 10 days
Using the exponential growth formula: $$Nt = N0 \cdot e^{rt}$|
Given data: Initial population ($N0$) = 100 Intrinsic rate of natural increase ($r$) = $0.15$ per day Time ($t$) = 10 days $e^{rt} = e^{0.15 	imes 10} = e^{1.5} = 4.4816$ 
Substituting the values: $$N{10} = 100 	imes 4.4816 = 448.16$$
Since population numbers are discrete organisms, we can round it to 448 individuals. --- Part 2: Calculation of Time to reach 800 individuals
Using the exponential formula solved for time $t$: $$Nt = N0 \cdot e^{rt}$| $$\frac{Nt}{N0} = e^{rt}$|
Taking natural logarithm ($\ln$) on both sides: $$\ln\left(\frac{Nt}{N0}ight) = rt$$ $$t = \frac{\ln(Nt / N0)}{r}$| 
Given data: $Nt = 800$ $N0 = 100$ $r = 0.15$ 
Substituting values: $$t = \frac{\ln(800 / 100)}{0.15} = \frac{\ln(8)}{0.15}$|
Given that $\ln(8) = 2.079$: $$t = \frac{2.079}{0.15} = 13.86 	ext{ days}$|
So, it takes approximately 13.86 days. --- Part 3: Difference between Exponential and Logistic Growth Curves Exponential Growth Curve: Occurs when resources (food and space) are unlimited. Results in a J-shaped curve. Equation: $\frac{dN}{dt} = rN$ Logistic Growth Curve: Occurs when resources become limiting at a certain point, leading to competition. Shows an initial lag phase, followed by acceleration, deceleration, and finally an asymptote when the population reaches the carrying capacity ($K$). Results in a Sigmoid (S-shaped) curve. Equation: $\frac{dN}{dt} = rN \left(\frac{K - N}{K}ight)$