How does hexokinase couple ATP use to glucose phosphorylation?
- It releases heat from ATP into the cell before glucose reacts with free phosphate
- It transfers a phosphoryl group directly from ATP to glucose
- It changes glucose's equilibrium constant merely by binding ATP without consuming it
Two reactions are chemically coupled under stated conditions. How do you test whether their net forward conversion is thermodynamically favourable?
- Add their signed delta G values and check for a negative total
- Subtract the downhill value from the uphill value and require a positive total
- Compare their reaction speeds and select the direction of the faster reaction
Under standard biochemical conditions, glucose phosphorylation by free inorganic phosphate is thermodynamically unfavourable.
Circle one: True False
A pathway spends 3 ATP in its opening reactions and produces 8 ATP later. What is its net ATP budget?
- A gain of 5 ATP after subtracting the investment
- A gain of 11 ATP after adding investment to production
- A loss of 5 ATP because later production cannot repay investment
An enzyme chemically couples two accounting reactions with delta G values +12 and -28 kJ/mol under the same stated conditions. What follows for the net reaction?
- +40 kJ/mol; net forward conversion is favourable because release values lose their sign
- -16 kJ/mol; the reaction must be fast regardless of its activation barrier
- -16 kJ/mol; net forward conversion is favourable, but speed is unspecified
Compound X has a more negative hydrolysis delta G than ATP. Is that fact alone enough to make X a better cellular energy currency?
- Yes; a larger release makes every ATP-using enzyme accept X without any chemical changes
- No; coupling chemistry, enzyme compatibility and regeneration also matter
- Yes; release size alone determines compatibility and removes the need to regenerate X
A to I and I to B sum to A to B with positive net delta G. Temperature and endpoint activities stay fixed. Only I increases. What happens to the summed delta G?
- It stays positive because I cancels from the net reaction
- It becomes negative because any increase in an intermediate drives every net reaction forward
- It becomes zero because changing I forces A and B into equilibrium regardless of their activities
A student says hexokinase makes glucose plus free phosphate thermodynamically favourable just by lowering its activation barrier. What has the student missed?
- Lowering a barrier always changes the equilibrium constant of the uncoupled reaction
- ATP only heats the cell and does not appear among the reactants of the net reaction
- ATP is a reactant in a different, chemically coupled net reaction