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Theoretical and Experimental DNA Computation (Natural ...

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PROMOTER STRUCTURAL GENE(S)<br />

TERMINATOR<br />

Fig. 6.1. Major regions found within a bacterial operon<br />

lacI Operator lacZ lacY lacA<br />

Repressor<br />

Promoter<br />

6.1 Introduction 149<br />

Fig. 6.2. Repressed state of the lac operon by the lacI gene product<br />

Promoter<br />

lacI lacZ lacY lacA<br />

Repressor<br />

¡ ¡ ¡ ¡ ¡ ¡ ¡<br />

¡ ¡ ¡ ¡ ¡ ¡ ¡<br />

¡ ¡ ¡ ¡ ¡ ¡ ¡<br />

¡ ¡ ¡ ¡ ¡ ¡ ¡<br />

RNA polymerase<br />

CAP-cAMP<br />

complex<br />

¦<br />

¦ ¦<br />

¦ ¦<br />

§ §<br />

§ §<br />

§ §<br />

¦<br />

¨<br />

©<br />

©<br />

¨<br />

�<br />

� �<br />

� �<br />

� �<br />

�<br />

Inducer<br />

Cap binding site<br />

¤<br />

¤ ¤<br />

¤ ¤<br />

¥ ¥<br />

¥ ¥<br />

¥ ¥<br />

¤<br />

¢<br />

¢<br />

£<br />

£<br />

£<br />

¢<br />

Fig. 6.3. Induced state of the lac operon<br />

complex, whose level increases as the amount of available glucose decreases.<br />

Therefore, if lactose were present as the sole carbon source, the lacI repression<br />

would be relaxed <strong>and</strong> the high CAP-cAMP levels would activate transcription,<br />

leading to the synthesis of the lacZYA gene products (Fig. 6.3). Thus,<br />

the promoter is under the control of two sugars, <strong>and</strong> the lacZYA operon is<br />

only transcribed when lactose is present <strong>and</strong> glucose is absent.<br />

It is clear that we may view the lac operon in terms of the Boolean AND<br />

function, in that it has output value 1 if transcribed, <strong>and</strong> 0 otherwise. The<br />

presence/absence of glucose corresponds to the gate’s first input being equal<br />

to 0/1, <strong>and</strong> the presence/absence of lactose corresponds to the gate’s second<br />

input being 1/0 (note the difference in representation between the two sugars).

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