The following chemical formula is often used for representation of a bacterial c

Question

The following chemical formula is often used for representation of a bacterial cell, C60H87O23N12P1

(a) Determine the mass (mg) of each element in 3000 mg of bacterial cells.

(b) Suppose 25 mg/L of ammonia nitrogen and 5 mg/L of orthophosphate as phosphorus are available for growing bacteria. If other nutrients are in abundance, which is the limiting nutrient, nitrogen or phosphorus?

(c) What mass of bacterial cells could be produced in terms of milligrams of bacterial cells per liter of water based on the limiting nutrient in Part b?
mg/L

(d) Suppose the nitrogen source was cut to 15 mg/L of ammonia nitrogen. How much bacterial cell mass (mg/L) could be produced?
mg/L

(e) Suppose the phosphorus source was cut to 4 mg/L of orthophosphate as phosphorus. How much bacterial cell mass (mg/L) could be produced?
mg/L

Explanation / Answer

a)

Ans: The molecular weight of C60H87O23N12P1 is calculated as follows:

Molecular weight = 60 C×12 + 87 H ×1+ 23 O×16 +12 N ×14 +1 P ×31 = 1374

The mass of each element is calculated as below:

C = 3000 mg (60×12/1374) = 1572 mg H = 3000 mg (87×1/1374) = 190 mg O = 3000 mg (23×16/1374) = 803 mg N = 3000 mg (12×14/1374) = 367 mg P = 3000 mg (1×31/1374) = 68 mg

Now addtion of of all mass of elements = 1572 + 190 + 803 + 367 + 68

=3000 mg.

b) 25 mg N/ L X (3000 mg bacterial cells/67 mg N) = 204 mg cells/L

5mg P/L X (3000 mg bacterial cells/68 mg P) = 221 mg cells/L

But it is given that Nitrogen is limiting cell growth

So 204 mg/L bacterial cells are produced from Nitrogen versus 221 mg/L bacterial cells from Phosphurus.

c)

Ans: 204 mg/L of bacterial cells could be produced.

d)

Ans: 15 mg N/L X (1374 mg bacterial cells/12 X14 Mg N) = 123 mg cells/L

Hence 123 mg cells/ L bacterial cell mass (mg/L) could be produced.

e)

Ans: 4 mg N/L X (1374 mg bacterial cells/1 X31 Mg N) = 177 mg cells/L

Hence 177 mg cells/ L bacterial cell mass (mg/L) could be produced.

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