Question

Can i Get some help on these questions?

For each of the following questions, show all work 7. Phosphofructokinase reaction The main control step of glycolysis synthesizes fructose-1,6-bisphosphate (F1,6BP) using the enzyme a) phosphofructokinase. The AG°* for this reaction is -17.2 kJ/mol. Based on this standard state free energy, which of the following is true? OKeg 1 OKeq1 O Keg 1

Answer 1

7

(a) Phosophofructokinase reaction :

  $\Delta$G^o = -17.2 kJ/mol ( T = 298 K)

since, $\Delta$G^o = -RT ln Keq and ln Keq = - $\Delta$G^o /RT

so, ln Keq = +6.93 ; Keq =1022.5 and Keq >1

(b) Hydrolysis of phosphate from C1 of F1,6 BP :

(you have not provided data ) :

F,6 P ------>  F1,6 BP $\Delta$G^o = -17.2 kJ/mol

F1,6 BP -----> F, 6 P +Pi

we take phosphate hydrolysis : $\Delta$G^o = x (?)

so, F1,6 BP -----> F, 6 P +Pi $\Delta$G^o = x +17.2 kJ/mol  

(c) compare data

(d)  

Phosophofructokinase reaction in mucles  :

  $\Delta$G = -25.9 kJ/mol   

Under standard state :

Phosophofructokinase reaction :

  $\Delta$G^o = -17.2 kJ/mol

yes , reaction under cellular condition is further from equilibrium than standard state , since

it has much larger negative free energy than standard state.

(e)

$\Delta$G = -25.9 kJ/mol ( T = 310 K or 37 C)

since, $\Delta$G = -RT ln Q and ln Keq = - $\Delta$G /RT

so, lnQ = +10.05

Q = 2.3 *10⁴

(f)   F,6 P + ATP ------>  F1,6 BP + ADP

$\Delta$G = -25.9 kJ/mol    ( T = 310 K or 37 C)

Q = ( [ F1,6 BP ][0.2*10^-3]./ [ F,6 P][0.90*10^-3] ) = 2.3 *10⁴

[ F1,6 BP ]/ [ F,6 P] = 1.0 *10⁵.