Question

Tetrahymena thermophila protozoa have a minimum doubling time of 6.5 hours when grown using bacteria as the limiting substrate. The yield of protozoal biomass is 0.33 g per g of bacteria and the substrate constant is 12 mg/ l. The protozoa are cultured at steady state in a chemostat using a feed stream containing 10 g/ l of nonviable bacteria. What is the maximum dilution rate for operation of the chemostat?

Answer 1

Answer:

The answer is "$\bold {3.3 \ \frac{g}{L}}$"

Explanation:

If the endogenous metabolic rate wasn’t substantial after, which $\mu_{net} = \mu_{max} = \frac{0.693}{ \text{ doubling period}}$ .

Throughout the calculating of doubling the time is set at 6.5 hours, consequently $\mu_{max} = \frac{0.693}{6.5} = \frac{0.1}{hours}.$

Know we calculate the two-equation,  

Initially, the maximum dilution frequency for $D_{max}$ that is:  

$D_{max}= \mu _{max}\times \frac{[S_0]}{{K_s + [S_0]}} .....(a)$

In secondary the steady concentration of state cells X,  

$X = \frac{Y_{\frac{x}{s}} (s[S_0]-K_sD)}{(\mu _{max}-D)}...... (b)$

In this section, we will display the $Y_{\frac{x}{s}}$ , that is equal to $0.33\ \frac{g}{g-substrate}$, and the value of the $K_s = 12 \times 10^{-3} \ \frac{g}{L}$.  

For equation (a):

$D_{max} = 0.1 \times \frac{10}{ (12 \times 10^{-3} + 10)}$

         $= 0.1 \times \frac{10}{ (\frac{12}{10^{3}} + 10)}\\\\= 0.1 \times \frac{10}{ 0.012 + 10)}\\\\= 0.1 \times \frac{10}{ 10.012}\\\\=0.1 \times 0.9988\\\\= 0.09988\\\\= 0.1 \ \ \frac{1}{h}$

In the Operating value is equal to $D =\frac{1}{2}$ of $D_{max}$, so D is =$\frac{0.05}{h}$ in our case.  

Finally, the amount of protozoa cells in equation (b):

$X = 0.33 \times (10 - 12 \times 10^{-3} \times \frac{0.05}{(0.1 - 0.05)})\\\\X = 0.33 \times (10 - 12 \times 10^{-3} \times \frac{0.05}{-0.05})\\\\X = 0.33 \times (10 - 12 \times 10^{-3} \times -1 )\\\\X = 0.33 \times (10 + \frac{12}{10^{-3}} )\\\\X = 0.33 \times ( 10.012 )\\\\X= 3.303$