Nt-1 = Nt + rNt , where t is time, N is the population size, and r is the growth rate such that rNt is
the number of offspring for that time.
Exponential growth describes populations that are growing continuously. You are studying a
population of 1,000 microbes. If the population is growing exponentially at a per capita growth
rate of r=0.695, answer the following questions.
1) What is N0 ?
2) What is N6 ?
3) What is N10 ?
4) How long does it take for this species to double?
5) Find population sizes every two years from N0 through N20 and plot them on a figure of
population over time. What does the plot resemble?
Geometric growth is simplified exponential growth with discrete reproduction (as a simplified
model of reproduction). The geometric growth rate, lambda, is equal to the population size of
one time interval, divided by the population size of the previous time interval, or λ = Nt /Nt-1 .
6) If your population above instead was growing discretely, how could you determine the
geometric growth rate? In other words, is there a way to relate lambda to r?
7) Once you have solved for lambda, find N6 under geometric growth.
8) What is N10 under geometric growth?
9) Find population sizes, using geometric growth, every two years from N0 through N20 and
plot them on a figure of population over time. How does the plot compare to #5?
Most populations do not grow in ideal conditions and therefore do not exhibit exponential or
geometric growth indefinitely. Density dependent controls, like reduced food availability, will
exert themselves and populations with either crash or slow as they approach a carrying
capacity.
10) Define carrying capacity in your own words, indicating what can and cannot limit
carrying capacity.
11) If dN/dt (or the change in population size over a change in time) = rN under ideal
conditions, what does that plot look like?
12) Under non-ideal conditions, where something limits or controls growth, dN/dt = rN (K-
N/K). What is dN/dt when N=K? Hint: try replacing N with K in the right side of the
equation.
13) What is dN/dt when N=2K?
14) What is dN/dt when N=0.5K?
1) N0 is the initial population size, which in this case is 1,000 microbes.
2) N6 can be calculated by substituting t=6 into the exponential growth equation: N6 = N0 * e^(rt), where e is the base of natural logarithms. Plugging in the values, N6 = 1,000 * e^(0.695*6) ≈ 4,289.6 microbes (rounded to the nearest whole number).
3) Similarly, N10 can be calculated by plugging in t=10 into the exponential growth equation: N10 = 1,000 * e^(0.695*10) ≈ 12,182.5 microbes (rounded to the nearest whole number).
4) To determine the time it takes for the species to double, we can use the formula for exponential growth: Nt = N0 * e^(rt) and solve for t when Nt = 2N0. Plugging in the values, 2N0 = N0 * e^(0.695t), which simplifies to 2 = e^(0.695t). Taking the natural logarithm of both sides, ln(2) = 0.695t. Solving for t, t = ln(2) / 0.695 ≈ 1.00 (rounded to two decimal places). Therefore, it takes approximately 1 time unit for the species to double.
5) To find population sizes every two years from N0 through N20, we can use the exponential growth equation. The values can be calculated as follows:
N0 = 1,000
N2 = 1,000 * e^(0.695*2) ≈ 1,988.7 microbes
N4 = 1,000 * e^(0.695*4) ≈ 3,947.3 microbes
N6 = 4,289.6 microbes (calculated in question 2)
N8 = 1,000 * e^(0.695*8) ≈ 8,269.0 microbes
N10 = 12,182.5 microbes (calculated in question 3)
N12 = 1,000 * e^(0.695*12) ≈ 16,053.6 microbes
N14 = 1,000 * e^(0.695*14) ≈ 26,490.4 microbes
N16 = 1,000 * e^(0.695*16) ≈ 43,680.5 microbes
N18 = 1,000 * e^(0.695*18) ≈ 71,957.1 microbes
N20 = 1,000 * e^(0.695*20) ≈ 118,297.6 microbes
Plotting these population sizes on a figure of population over time would resemble an exponential growth curve, where the population size increases rapidly at first and then gradually levels off.
6) If the population is growing discretely, the geometric growth rate, lambda (λ), can be determined by calculating λ = Nt / Nt-1 , where Nt is the population size at time t and Nt-1 is the population size at the previous time interval.
7) To find N6 under geometric growth, we need to know the population size at the previous time interval, N5. Without that information, we cannot determine N6 under geometric growth.
8) Similarly, to find N10 under geometric growth, we need to know the population size at the previous time interval, N9. Without that information, we cannot determine N10 under geometric growth.
9) Without the information about N5 and N9, we cannot calculate the population sizes every two years from N0 through N20 using geometric growth, nor can we plot them on a figure of population over time. Therefore, we cannot compare this plot to the one in question 5.
10) Carrying capacity refers to the maximum population size that an environment can sustainably support given the available resources and conditions. It represents the point at which the population growth rate equals zero, as the resources become limited and cannot support further exponential growth. Limiting factors such as food availability, space, competition, and predation can determine the carrying capacity. The carrying capacity can be limited by both biotic (living) and abiotic (non-living) factors.
11) If dN/dt = rN under ideal conditions, the plot of population size over time would be a straight line with a positive slope. This represents exponential growth, where the population size increases at a constant rate.
12) When N = K (carrying capacity), the equation becomes dN/dt = rN(K-N)/K. Plugging in N = K, the equation becomes dN/dt = rN(0)/K, which simplifies to dN/dt = 0. Therefore, when N = K, the population size does not change, and the plot of population size over time becomes a horizontal line.
13) When N = 2K, the equation becomes dN/dt = rN(1-2)/K, which simplifies to dN/dt = -rN/K. This means that the population size is decreasing at a constant rate, as the population is above the carrying capacity and resources are limited. The plot of population size over time would be a decreasing line with a negative slope.
14) When N = 0.5K, the equation becomes dN/dt = rN(0.5-1)/K, which simplifies to dN/dt = -0.5rN/K. Similar to when N = 2K, the population size is decreasing at a constant rate, as the population is still above the carrying capacity. The plot of population size over time would also be a decreasing line with a negative slope.