Problem 5. Protein Concentration in Cells
Consider a growing bacterial cell that is producing a protein involved in binary fission. Suppose that, at a particular point in time, a bacterial cell has a volume of 1000 μm3 (microns cubed) and is growing at a rate of 5 μm3/min. Furthermore, suppose that there are 1000000 molecule of the protein in the cell, and the protein is accumulating at a rate of 10000 molecules/min.
Find the concentration of the protein in the cell and the rate at which the con- centration of protein is changing within the cell (include units). Is the concentra- tion increasing or decreasing? Thoroughly justify your answer with the appropriate derivative rule.
I think that we have to find the derivative of the exponential growth equation A=Pe^(rt), but i am not sure please help?
There is no exponential growth here. The population and volume are not growing by a fixed percentage, but just as a linear amount.
the concentration is measured in molecules/μm^3, so at t minutes later,
c = (10^6 + 10^4t)/(1000+5t)
= 2000(100+t)/(200+t)
= 2000(1 - 100/(200+t))
so, find dc/dt; you will see that the concentration is increasing
To solve this problem, we need to find the concentration of the protein in the cell and the rate at which the concentration is changing. We can do this by using the given information about the volume and the number of proteins in the cell.
The concentration of the protein is defined as the amount of protein per unit volume. In this case, it is the number of molecules of protein per cubic micron (μm3). To find the concentration, we divide the number of molecules of protein by the volume of the cell:
Concentration = Number of molecules of protein / Volume of the cell
Given that there are 1000000 molecules of protein and a cell volume of 1000 μm3, the concentration is:
Concentration = 1000000 molecules / 1000 μm3 = 1000 molecules/μm3.
So, the concentration of the protein in the cell is 1000 molecules/μm3.
Now, to find the rate at which the concentration of protein is changing, we need to consider the growth of the cell and the accumulation of the protein over time.
The cell is growing at a rate of 5 μm3/min, which means the volume of the cell is increasing by 5 μm3 every minute. The protein is accumulating at a rate of 10000 molecules/min.
To find the rate at which the concentration of protein is changing, we need to differentiate the concentration function with respect to time (t). The derivative of concentration with respect to time represents the rate of change of concentration over time.
Let's call the concentration of protein within the cell C(t), where t is the time in minutes.
Since the volume of the cell is growing linearly with time, we can express it as:
Volume = 1000 + 5t μm3.
Therefore, the concentration of the protein can be written as:
C(t) = Number of molecules of protein / Volume of the cell
= 1000000 molecules / (1000 + 5t) μm3.
Now, let's differentiate C(t) with respect to t using the quotient rule:
dC(t)/dt = (d/dt)(1000000 / (1000 + 5t))
= [(0 - 1000000)(5)] / (1000 + 5t)2
= -5000000 / (1000 + 5t)2.
So, the rate at which the concentration of protein is changing is given by:
dC(t)/dt = -5000000 / (1000 + 5t)2 molecules/μm3/min.
To determine if the concentration is increasing or decreasing, we need to evaluate the sign of the derivative. Since the derivative is negative (-5000000 / (1000 + 5t)2), the concentration is decreasing over time.
Hence, the concentration of the protein in the cell is 1000 molecules/μm3, and the rate at which the concentration is changing is given by -5000000 / (1000 + 5t)2 molecules/μm3/min.