2.1 Cell Composition

Cells primarily contain water! Typically 70% of cell mass is water and the remaining is dry matter. Therefore it is conventional to express cell composition on a dry basis. The microorganism Eschericia coli is widely used in genetic engineering. Typical elements found in Eschericia coli are given below:

Table 1 Elemental Composition of E. coli
(after Stanier et al)

Element

% Dry Weight

C

50

O

20

N

14

H

8

P

3

S

1

K

1

Na

1

Ca

0.5

Mg

0.5

Cl

0.5

Fe

0.2

others

0.3

Nearly half of the dry matter in cells is carbon and the elements carbon, oxygen, nitrogen and hydrogen total up to about 92% of the total. This observation for E. coli is also found to be generally true for other cellular organisms.

Table 2 Elemental Composition of Microorganisms

Microorganism Carbon Source Growth Rate Composition Empirical Formula Molecular Weight
 C  H  N  O
Klibsiella aerogenes Glycerol 0.1 50.6 7.3 13.0 29.0 CH1.74 O0.43 N0.22 23.7
Aerobacter aerogenes Complex 48.7 7.3 13.9 21.1 CH1.78 O0.33 N0.24 22.5
Aerobacter aerogenes Complex 0.9 50.1 7.3 14.0 28.7 CH1.73 O0.24 N0.43 24.0
Saccharomyces cerevisiae 47.0 6.5 7.5 31.0 CH1.66 O0.49 N0.13 23.5
Sachromyces cervisiae 50.3 7.4 8.8 33.5 CH1.75 O0.15 N0.5 23.9
Candida utilis Glucose 0.45 46.9 7.2 10.9 35.0 CH1.84 O0.56 N0.2 25.6
Candida utilis Ethanol 0.43 47.2 7.3 11.0 34.6 CH1.84 O0.55 N0.2 25.5

Table 2 above shows that in different microbes, the carbon content varies from 46 to 50%, hydrogen from 6 to 7%, nitrogen 8 to 14% and oxygen from 29 to 35%. These are small variations and the variations appear to depend on substrate and growth conditions. For many engineering calculations, it is reasonable to consider cell as a chemical species having the formula

This engineering approximation is a good starting point for many quantitative analysis while a more carefully formulated empirical formula based on proximate analysis may be necessary for complete material flow analysis. The cell molecular weight for the above cell formula is 12+1.8 + 0.5(16) +0.2 (14) = 24.6.

Example 2-1


Suppose we want to produce 10 g of cells using glucose as a carbon source. What is the minimum amount of glucose that would be needed?

Solution

Assume cell composition as CH1.8 O0.5 N0.2

Glucose is C6 H12 O6

MW of glucose is 180

Since glucose has 6 moles of carbon per mole of glucose,