Need some clarification Re: rate-limiting steps and irreversibility

[kokopuff]

Hello,

 

Just started my metabolism course and it's been about a year since my last biochemistry course so I've got to refresh myself on some of the basics.

In a section of my textbook that was talking about the regulation of metabolic pathways, it said: "The first committed step [of a metabolic pathway], being irreversible, functions too slowly to permit its substrates and products to equilibrate. (if the reaction were at equilibrium, it would not be irreversible)"

 

I'm a little confused, wouldn't an irreversible reaction be faster than a reversible one since the product is not continually being converted back to the initial substrate? Irreversible reactions are highly exergonic, so I assume that they would occur more easily and frequently than reversible reactions, is that not correct?

 

Also, isn't the part in parentheses redundant, or am I missing something?

 

Thanks for your help!

Edited September 11, 2013 by kokopuff

[daniton]

Hello,

 

Wouldn't an irreversible reaction be faster than a reversible one since the product is not continually being converted back to the initial substrate?

 

I don't think the speed of reaction depends on whether the reaction is reversible or not. It depends on different things like type of the reactants, activation energy, temperature.... the reversiablity of the reaction doesn't have anything with the speed of the reaction.

[CharonY]

I agree that it may be expressed a bit confusingly. As mentioned basically all metabolic reactions are reversible, irreversibility usually takes into account the physiological conditions (more or less). Some reactions that are reversible in vitro, may always be irreversible in vivo.

But first, as already mentioned the important bit is the changes of Gibb's free energy, which is related to the equilibrium constant K. Thus at equilibrium for a simple reaction [latex]K= \frac{Product}{Reactant} [/latex].

How does K relate to the rates of the individual reaction k?

Well: [latex]K= \frac{ k_{forward} }{k_{back}} [/latex]

 

So obviously for reactions that go towards completion the forward reaction rate is higher than the return reaction. This seems to be the issue that is mentioned in OP.

 

However one possibility to resolve this is a view at the whole pathway. Here we do not have a single reaction but many out- and influxes into the reaction under investigation. If subsequent reaction rates (or outflows) are higher than the first reaction, we are not in an equilibrium situation anymore. Even with a relatively low K we may get directionality in the metabolite flow. One may take the hexokinase reaction in, say yeast or other eukaryotes (glucose -> glucose-6P) which (if memory serves) has a K of well below 1000. However the concentration of glucose is generally way higher than that of the phosphorylated product. Hence, a reverse reaction is usually not observed.