Recitation 1 Solutions
2. This reaction
creatine phosphate + ADP W creatine + ATP
involves the coupling of the hydrolysis of creatine phosphate, a high-energy intermediate in muscle cells,
with the formation of ATP from ADP to replenish ATP supplies during exercise. As in the first problem
we use the equation
)G = )G0' + RT lnQ
where Q for this problem is
Q = [creatine][ATP]/[creatine phosphate][ADP]
We’re given [creatine] and [creatine phosphate] and are asked to solve for [ATP][/[ADP], so we write
Q as 25/13 x r, where r = [ATP][ADP].
We solve for )G0' by using two entries from Table 14.1 as we did in problem 1. Recall that
one of these entries must be reversed. Using this table we have the )G0' values for creatine phosphate
and ATP hydrolysis as =43.1 and -30.5 kJ/mole, respectively. In the equation we have ATP being
formed, not hydrolyzed, so this is the reaction that must be reversed, giving us
)G0' = -43.1 + 30.5 = -12.6 kJ/mole
To find Keq, we note that this is the value of Q at equilibrium, where )G = 0. Thus we have
)G = )G0' + RT lnQ = -12,600 cal/mole + RT ln (25/13 x r) = 0
Note the switch from kJ/mole to J/mole because R is given as 8.314 J/mole. We must use the fact that
ln ab = ln a + ln b, which gives us
0 = -12,.600 J/mole + 2477.6 J/mole(ln 25/13 + ln r)
solving for r
RT ln r = 12,600 J/mole - 2477.6 J/mole(0.6539) = 10980 J/mole
r = ln-1 (10980/2477.6) = inverse ln (10980/2477.6) = 84
Thus the maximum ration of ATP/ADP that be attained is 84:1.
3.
NH2
N
N
O
HO
P
O
O
O
O
P
O
O
N
N
P O CH2 O
O
H
H
H
H
OH OH
High energy phosphoanhydride linkage
High energy mixed anhydride linkage
O
HO
O
P O C
O
O
C
H C
OH
CH2
O
P O3
High energy thio ester linkage
(resonance destabilization)
O
CoA
S
C CH 3
See text, page 385, Figure 14.18 for structure of CoA
3. Continued
O
Unstable enol form
where double bond
shifts to O when
phosphate is hydrolyzed
C
C
O
O
PO3
CH 2
NADH and FADH2 are reduced forms of these redox cofactors. They readily donate their electrons
which can be used for ATP formation during electron transport, so they should also be considered as
high energy molecules.
4. The phosphoryl transfer potential is defined as the magnitude of the )G0' values of hydrolysis of the
phosphorylated intermediates listed in Table 14.1. Thus, for creatine phosphate the phosphoryl transfer
potential is 43.1 and for ATP it is 30.5 (see problem 2). Since 43.1 is a larger number than 30.5,
creatine phosphate will spontaneously transfer a phosphate group to ADP to form ADP, as we also
saw in problem. ATP, in turn, has a greater phosphoryl transfer potential than glucose-6-phosphate
(13.6), so ATP will spontaneously transfer its phosphate to glucose to form glucose-6-phosphate, etc.