Why might the atpS mutant grow so slowly? In order to answer this, you need to consider the following pieces of information:
(1) ADP + Pi > ATP ∆G’¡ = +52 kJ/mol
(2) Under physiological conditions in the bacterium, this reaction actually has ∆G = +67 kJ/mol.
(3) The free energy change for a proton to re-enter the cytosol will be related to the pH gradient (∆pH) and the transmembrane electric field (ψ) by this relationship:
∆G = -2.3 RT ∆pH + F∆ψ (R = 8.315 J mol-1 K-1; T = 298K; F = 96.5 kJ mol-1 V-1)
(4) In the case of M. wanabi, Typically, ∆ψ = -100 mV.
(5) The environment in which M. wanabi grows is close to neutral (usually pH = 6) and somewhat buffered with weak organic acids.
Calculate what the minimal ∆pH would have to be, in order to support ATP synthesis, for the different Fo ring sizes. Does this explain why the mutant grew slowly and the partial revertant grew better? Explain.
atpS mutant means the bacteria does not have ATP synthase enzymes that helps in ATP production so, this bacteria won't be able to synthesize ATP on its own.
If this cell can not meet its energy requirements then it can not process its metabolic pathways efficiently hence slower growth would be observed.
ATP synthase mutant grow so slowly on no fermentable C4-dicarboxylates. Mutation prevents proton translocation and establishment of hydrogen bonds which are critical or existence of protons for ATP synthase
Ph 7 is the minimum PH that support ATP synthase. This explain why the mutant grew slowly and the partial revertant grew better. Ph of 7 is the optimal pH for ATP synthase and weak organic acids slows down the production of ATP .On the other hand partial revertant grow better in weak organic acids.