Can somebody answer the unsolved ones ?

There are 15 students in math class.at the begning of class everyone races into class as everyone love maths.the professor takes note of the first 5 students who gets class and ranks them according to the order in which they entred as their top 5 favourite students. in how many different ways could the professor rank their top 5 students using formulla from combinations and permutations

a)  Cell concentration when the dilution rate is one-half of the maximum is  0.598g cell/L

b) the substrate concentration when the dilution rate is 0.8   is 5.2g/l

c) the maximum dilution rate is : 0.41 h⁻¹

d)  the cell productivity at 0.8     is 2.40g cell/L

Explanation:

Given data :

Mass doubling time of Pseudomonas sp. = 2.4 hr

Saturation constant = 1.3 g/L

Cell yield  on acetate = 0.46g cell/g acetate

We are to find;

a. Cell concentration when the dilution rate is one-half of the maximum.

Here, cell yield =amount of cell produced / amount of substrate consumed.

[S] at 0.5D max is determined using the Monod's equation.

Using the formula :

, where D is the dilution rate,

[S] is substrate concentration; &

Ks is the saturation constant.

By replacing the values, we get :

[S]=1.3g/L

The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.

=0.46×1.3

= 0.598g cell/L

b)

Substrate concentration when the dilution rate is 0.8    is calculated as:

0.8=[S]/1.3+[S]

1.04+0.8[S]=[S]

[S]= 5.2g/L

Therefore ,  the substrate concentration when the dilution rate is 0.8   is 5.2g/l

c)

Maximum dilution rate is calculated using the expression

=1/2.4

=0.41 h⁻¹

So, the maximum dilution rate is : 0.41 h⁻¹

d)

The cell productivity at 0.8 can be calculated by multiplying the amount  of the cell yield with the amount of the substrate consumed at 0.8  

Cell yield =

Cell productivity at 0.8    = 0.46 × 5.2

=2.40g cell/L

Therefore, the cell productivity at 0.8     is 2.40g cell/L