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An oligoribonucleotide is treated with T1 and RP enzymes. the fragments obtained from the action of each enzyme are:
a) with RP AGmGGAAU,AAAGU,AAG,GAT,IGC,C,C,C,C,
U,U
b)with T1
ATUCCAGm,CAAAG,CCAAG,AAUI,UUG,G,G,G
——find the sequence of bases in this m-RNA fragment
Describe, in biochemical detail, the structure of functional [name of enzyme]. Include any coenzymes, cofactors, etc
Do our cells have own catalytic activity to mitigate the harsh effects of ROS? Elaborate
Which process is directly affected by DCCD, electron transfer or ATP synthesis?
What happens to the rate of liver glucose metabolism during moderate intensity exercise
Would the use of pH buffers help prevent muscle fatigue during exercise?
What happens to amylose starch that is not fully hydrolyzed in the small intestine? What might be the clinical consequence?
What happens to the rate of liver glucose metabolism during moderate intensity exercise?
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.
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.
