Two yeast cells were placed into a special container to which food was continually added, to keep it at a constant concentration. All other factors were set for optimal yeast growth (for example, temperature, oxygen, and pH). The population was sampled every hour for 21 hours and the results of the estimated population size were recorded in the table below.
Time (hour), Number of yeast cells
(0, 2)
(1, 10)
(2, 15)
(3, 20)
(4, 40)
(5, 60)
(6, 100)
(7, 190)
(8, 260)
(9, 350)
(10, 450)
(11, 530)
(12, 580)
(13, 600)
(14, 600)
(15, 600)
(16, 600)
(17, 600)
(18, 600)
(19, 600)
(20, 600)
What population growth model does the population appear to follow? Write out the mathematical equation it follows.
The attempt at solving:
Looking at the values, I recognize the carrying capacity is reached at hour 13 with 600 cells, signifying a logistic growth model. The formula given to me for logistic growth is:
dN/dt = rN[(K - N)/K]
From what I have researched:
dN = population size change
dt = time interval
r = growth rate
N = starting population size
K = carrying capacity
So:
dN = 600 - 2 = 598?
dt = 21 hours
r = ?
N = 2
K = 600
Plugging the values in gives me:
598/21 = r2[(600 - 2)/600]
And I have no idea what I'm supposed to do from here.
Do I solve for r? What is the completed formula supposed to look like? There is not sufficient explanation from the teacher's lesson and I don't understand other examples online.
dN/dt = rN[(K - N)/K]
where (Note the MAX I have inserted)
dN = population size change
dt = time interval
r = MAX growth rate
N = starting population size
K = carrying capacity
you solve for rmax by looking at the data.
rmax occurs between time 8 and 10 (around time 9)
rmax= (90+100)/2= 95/unittime Notice this rmax is the average of each time before and after.
Now knowing rmax
dN/dt=95(K-350)K=(450-260)/2
solve for K
@bobpursley
I have understood how to figure out rmax value, however shouldn't the formula with rmax value be:
dN/dt=95N[(K-N)/K]
With N= 2
dN/dt=95(2)[(K-2)/K]
and if K value is carrying capacity, can't we just substitue K for 600?
dN/dt=95(2)[(600-2)/600]
dN/dt=190(598/600)
dN/dt=190(0.997)
dN/dt=189?
Or because the question asks me to "Write out the mathematical equation it follows." am I supposed to leave a value unknown?
To find the growth rate (r) in the logistic growth model, you need to solve for it using the provided data.
The logistic growth model equation is dN/dt = rN[(K - N)/K]
Let's use the data from the table to find the growth rate (r):
(1, 10) - (0, 2) = 8
(2, 15) - (1, 10) = 5
(3, 20) - (2, 15) = 5
(4, 40) - (3, 20) = 20
(5, 60) - (4, 40) = 20
(6, 100) - (5, 60) = 40
(7, 190) - (6, 100) = 90
(8, 260) - (7, 190) = 70
(9, 350) - (8, 260) = 90
(10, 450) - (9, 350) = 100
(11, 530) - (10, 450) = 80
(12, 580) - (11, 530) = 50
(13, 600) - (12, 580) = 20
(14, 600) - (13, 600) = 0
(15, 600) - (14, 600) = 0
(16, 600) - (15, 600) = 0
(17, 600) - (16, 600) = 0
(18, 600) - (17, 600) = 0
(19, 600) - (18, 600) = 0
(20, 600) - (19, 600) = 0
Now, let's calculate r for each set of data:
r = (8/1) / 2 = 4
r = (5/1) / 10 = 0.5
r = (5/1) / 15 = 0.333
r = (20/1) / 20 = 1
r = (20/1) / 40 = 0.5
r = (40/1) / 60 = 0.666
r = (90/1) / 100 = 0.9
r = (70/1) / 90 ≈ 0.778
r = (90/1) / 100 ≈ 0.9
r = (100/1) / 100 = 1
r = (80/1) / 80 = 1
r = (50/1) / 50 = 1
r = (20/1) / 20 = 1
r = (0/1) / 20 = 0
Now, calculate the average r from these values:
r_average = (4 + 0.5 + 0.333 + 1 + 0.5 + 0.666 + 0.9 + 0.778 + 0.9 + 1 + 1 + 1 + 1 + 0) / 14 ≈ 0.748
Therefore, the growth rate (r) in the logistic growth model is approximately 0.748.
Using this value, the completed formula for the logistic growth model with the given data is:
dN/dt = 0.748N[(600 - N)/600]