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

The answer is "
`\bold {3.3 \ (g)/(L)}`"

Step-by-step 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) = (0.693)/(6.5) = (0.1)/(hours).`

Know we calculate the two-equation,

Initially, the maximum dilution frequency for
`D_(max)` that is:

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

In secondary the steady concentration of state cells X,

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

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

For equation (a):

`D_(max) = 0.1 * (10)/( (12 * 10^(-3) + 10))`

`= 0.1 * (10)/( ((12)/(10^(3)) + 10))\\\\= 0.1 * (10)/( 0.012 + 10))\\\\= 0.1 * (10)/( 10.012)\\\\=0.1 * 0.9988\\\\= 0.09988\\\\= 0.1 \ \ (1)/(h)`

In the Operating value is equal to
`D =(1)/(2)` of
`D_(max)`, so D is =
`(0.05)/(h)` in our case.

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

`X = 0.33 * (10 - 12 * 10^(-3) * (0.05)/((0.1 - 0.05)))\\\\X = 0.33 * (10 - 12 * 10^(-3) * (0.05)/(-0.05))\\\\X = 0.33 * (10 - 12 * 10^(-3) * -1 )\\\\X = 0.33 * (10 + (12)/(10^(-3)) )\\\\X = 0.33 * ( 10.012 )\\\\X= 3.303\\`