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I came across the following sentence while reading the paper:

Nanchen, Annik, et al. "Cyclic AMP-dependent catabolite repression is the dominant control mechanism of metabolic fluxes under glucose limitation in Escherichia coli." Journal of bacteriology 190.7 (2008): 2323-2330.

These in vivo flux data are qualitatively corroborated by in vitro enzyme activities in the Crp and Cya mutants, which exhibited significantly reduced activity of the key glyoxylate shunt key enzyme isocitrate lyase ...

My understanding of in vitro experiments is: those performed in artificial set up , possibly only with part of the cell, where the complete cell is not potentially alive. For, example we can test the inhibition activity of a particular enzyme by a metabolite.

In the paper cited, experiments from which flux are derived (chemostat) are considered in vivo. What would it mean to say the enzyme activity was tested in vitro on the mutant? When we say the response of the a single gene knock out mutant, the cell should be intact except for lacing one gene, which would mean that the response is in vivo.

Please help me if I am misunderstanding some thing.

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Answer clarified based on comment by @DurgaDatta

What would it mean to say the enzyme activity was tested in vitro on the mutant? When we say the response of the a single gene knock out mutant, the cell should be intact except for lacing one gene, which would mean that the response is in vivo.

The knockouts were performed inside the E. coli cells and, as you noted, the flux measurements were in vivo (in the cell) inside of the chemostat culture. They identified two mutants, Crp and Cya, that yielded much higher biomass. On page 2326, the authors write,

In the Cya (encoding adenylate cyclase) and Crp mutants that lack the cAMP-CRP complex, in vivo glyoxylate shunt activity was essentially abolished....

I am not familiar with these genes but the implication is that the wild type regulators have the cAMP-CRP complex. Because this complex is missing from the mutants, the enzymes no longer function the same. However, it would be possible that some type of interaction of the mutants with other enzymes caused the observed result in vivo. They therefore verified the result by extracting the Crp and Cya mutant enzymes from the cells and assaying their enzyme activity in isolation from the other enzymes (in vitro). The technique is explained in the answer by @dgg32.

In your comment to that answer, you ask.

what difference does it make whether the enzyme was taken from wild-type or the mutant one, since it is the same enzyme?

It is the same enzyme in name but not in function. The mutant enzyme lacks the cAMP-CRP complex so it cannot perform the same function as the wild type. In vitro, they first showed that mutant enzyme activity was altered relative to the wild type. They then added the wild type enzyme to the in vitro test and restored normal wild type activities:

To verify that the observed changes were not due to polar effects of the still present marker gene or secondary-site mutations, we complemented the Crp mutant with the plasmid-based Crp gene and cultivated the complemented mutant under the same chemostat conditions. In vitro enzyme activities demonstrated that both the isocitrate lyase and isocitrate dehydrogenase activities were restored to the wild-type level

Yes, this was performed inside the chemostat cultures but only with the enzymes, not the living E. coli cells.

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  • $\begingroup$ But the flux measurement were also performed in chemostat, which are considered in vivo here. $\endgroup$ – DurgaDatta Sep 7 '14 at 15:01
  • $\begingroup$ @DurgaDatta I fixed my answer in response to your comment. $\endgroup$ – Michael S Taylor Sep 8 '14 at 0:32
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It is possible to conduct an in vitro experiment with cell extracts. Meaning that you break up the cells and put their pieces into a reaction glass and do your experiment. It is assumed that the cell molecules (enzyme, sugar or cofactors) will still carry out their functions even though the cell is no more.

In other words, you used the mutant to produce the enzymes you need (minus the knock-outed one) to set up the desired reaction requirements. Once the products are there, you don't need the producers alive or intact any more.

Check the method part in your paper and see whether it is so.

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  • $\begingroup$ If we would like to do test the activity of an enzyme (not encoded by the knocked out gene), what difference does it make whether the enzyme was taken from wild-type or the mutant one, since it is the same enzyme? $\endgroup$ – DurgaDatta Sep 6 '14 at 15:47
  • $\begingroup$ But based on your article excerpt, the authors seemed to knock-out a gene and see how the metabolism flux is affected, just like a police shuts down a street block to see how the traffic is diverted. Their approach gives them a more systematic view of the pathway rather than a single enzyme's activity. In that case, you need the cell extract. $\endgroup$ – dgg32 Sep 7 '14 at 11:14

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