Team:Edinburgh/Efficiency

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(Phage display)
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: N = P / ((S * E) - E)
: N = P / ((S * E) - E)
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Each cellulase is roughly 500 amino acids in size, but to produce one phage requires over 200,000 amino acids, as well as a significant amount of DNA synthesis (which we shall ignore). As for the value of synergy, a best-case scenario would be if the synergistic effects of having the enzymes in close proximity made them 50 times more efficient (this number is suggested and critiqued by Fontes and Gilbert, 2010). But lets take a more plausible figure, e.g. 5 times. In that case, N = 100.
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Each cellulase is roughly 500 amino acids in size, but to produce one phage requires over 200,000 amino acids, as well as a significant amount of DNA synthesis (which we shall ignore). As for the value of synergy, a best-case scenario would be if the synergistic effects of having the enzymes in close proximity made them 50 times more efficient (this number is suggested and critiqued by Fontes and Gilbert, 2010). So lets make a graph for all values of S between 2 and 50:
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[[File:Edinburgh-synergy-display-graph.png|center|thumb|600px|caption|]]
==References==
==References==

Revision as of 12:34, 4 August 2011

Obviously, the highest type of efficiency is that which can utilize existing material to the best advantage.

— Jawaharlal Nehru

Is display of cellulases via phage or INP any good, even in theory?

Consider this: for bacteria to produce phage or INP requires energy. This energy could have been spent producing extra copies of the cellulases. In order for the phage and cell display projects to make sense, the benefits of synergy must outweigh the cost of producing all these extra proteins.

This question can probably be investigated using simple maths and back-of-envelope calculations...

Cell display

For the cell display system, the carrier protein INP (which is fused to a cellulase) is about two thirds the size of any of the cellulases. Each cellulase thus has a final size of about 66% above normal.

So if the synergystic system works at over 166% efficiency compared to the standard system where enzymes are free-floating in the media, cell display via INP makes sense.

Phage display

The situation with phage is more complex. Let us define some variables to help us think about the problem:

S - how efficient the synergistic system is compared to the normal system e.g. 5 means 5 times as efficient.
E - the cost to produce one enzyme
N - the number of enzymes displayed on a phage
P - the cost to produce one phage, not counting attached enzymes

The cost of producing a phage with all its displayed enzymes is thus E * N + P. In order for the system to be worthwhile, this cost must be less than the cost of producing enough free-floating enzymes to do the job equally well without synergy; this cost is E * N * S. Thus the system is worthwhile if:

E * N + P < E * N * S

The critical value for N is therefore:

N = P / ((S * E) - E)

Each cellulase is roughly 500 amino acids in size, but to produce one phage requires over 200,000 amino acids, as well as a significant amount of DNA synthesis (which we shall ignore). As for the value of synergy, a best-case scenario would be if the synergistic effects of having the enzymes in close proximity made them 50 times more efficient (this number is suggested and critiqued by Fontes and Gilbert, 2010). So lets make a graph for all values of S between 2 and 50:

Edinburgh-synergy-display-graph.png

References