Find solution
Why do serum levels of iron increase during an infection?

Is it true to say that AMP levels increase "exponentially" as ATP levels decrease?

  • Garrett & Grisham (5e, 2013) p. 925 state that "due to the nature of the adenylate kinase equilibrium, AMP levels increase exponentially as ATP levels decrease." However, inspection of the equilibrium suggests that AMP levels increase only linearly as ATP levels decrease. What's the right way to think about this?

  • Answer:

    I just saw the Wikipedia article: ATP+AMP→2ADPATP+AMP→2ADPATP + AMP \to 2ADP is the reaction. Therefore K=[ADP]2[ATP][AMP]⟹[AMP]=[ADP]2K[ATP]K=[ADP]2[ATP][AMP]⟹[AMP]=[ADP]2K[ATP]K=\frac{[ADP]^2}{[ATP][AMP]}\implies [AMP]=\frac{[ADP]^2}{K[ATP]}. If we decrease [ATP][ATP][ATP], [AMP][AMP][AMP] ought to rise as 1x1x\frac{1}{x} does close to the origin (assuming you keep [ADP][ADP][ADP] constant). This is not exponential, and it is mathematically incorrect to state otherwise (this is hyperbolic). It is however a very fast and steep increase (which you can argue is steeper than exponential at very low concs), and is certainly not linear. There is a tendency to refer to any rapid increase as exponential in non-scientific literature, and it seems like the authors had resorted to that hyperbole to make a point (which I feel they ought not to have as this is a technical text). Very non-rigorous, and probably meriting censure but whatever works, I guess? My experience with teaching biochem is non-existent and so perhaps my opinion is worh very little. P.S: I would be surprised if some activity coefficient related effect actually led to exponential increase considering the typical millimolar (or submillimolar) concentrations in cells, where I think the dilute solution approach should work. Then again, these are pretty well ionized, and have fairly high ionic strengths (but I would be still surprised if that was the reason that prompted that word)>

Diptarka Hait at Quora Visit the source

Was this solution helpful to you?

Other answers

Though chemically speaking the equillibrium will suggest that AMP levels increase with a decrease in [ATP][ATP][ATP].As said.But while considering the equillibrium we should also take note of the phenomena of repeated phosphorylation of the previously produced ADP. ADP is generally a product of phosphorylation of intermediates in biochemical reactions the most famous of which would be in the phosphorylation of PhosphoGlyceraldehyde to 1,3 Bis Phosphoglyceraldehyde in Calvin's cycle or addition of  CoA to alpha-Keto Glutarate in the Kreb's Cycle.But once produced it is generally noticed that another ADP is consumed for energy storage as ATP in the same cycle.So for most metabolic and biochemical pathways which are energy intensive [ADP]=[ATP][ADP]=[ATP][ADP]=[ATP] because of rephosphorylation of ADP in the concerned cycle or need based ATP production by the mitochondria.Since the Product and reactant concentration in the equillibrium for the reaction ATP+AMP−−−−>2ADPATP+AMP−−−−>2ADPATP+AMP---->2ADP is constant thus there is no change in [ADP][ADP][ADP]. So for a process in a system,let us say, an idealised cell with no protein kinase or cAMP kinases the equillibrium would be virtually non-existent,such as that observed in some strains of E-Coli.But in higher organisms and plants the equillibrium is established where AMP is phosphorylated by ATP thus leading to an increase which is best described as K[AMP]=[ADP]2/[ATP]K[AMP]=[ADP]2/[ATP]K[AMP]=[ADP]^2/[ATP].But one cannot say that the increase is exponential or at all dependent solely on ATP when it is clear mathematically that [ADP][ADP][ADP] contributes more,which does suggest a slight exaggeration or idealization by the book's authors.But for the sake of serving their purpose if one does take [ADP][ADP][ADP] to be constant then the equation reduces to K[AMP]=C/[ATP]K[AMP]=C/[ATP]K[AMP]=C/[ATP] which means that it is a hyperbolic curve where any decrease in [ATP] leads to a sort of exponential increase in AMP though the real relation is K[AMP]=−C(log[ATP])K[AMP]=−C(log[ATP])K[AMP]=-C(log[ATP]).So for most purposes I would say following the exponential growth approach is better.

Rituparno Chowdhury

Related Q & A:

Just Added Q & A:

Find solution

For every problem there is a solution! Proved by Solucija.

  • Got an issue and looking for advice?

  • Ask Solucija to search every corner of the Web for help.

  • Get workable solutions and helpful tips in a moment.

Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.