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778 Chapter 24
Copyright © 2017 Pearson Education, Inc.
Solutions to Problems
1.
Solved in the text.
2.
a.
The
p
K
a
values of the three OH groups of ADP are 0.9, 2.8, and 6.8. At pH 7.4, two of the groups will
be in their basic forms, giving ADP two negative charges.
We can determine the fraction of the group with a
p
K
a
of 6.8 that will be in its basic form at pH 7.4
using the method shown in Problem 1.
fraction of group present in basic form
=
K
a
K
a
+
[H
+
]
=
1.6
*
10
-
7
1.6
*
10
-
7
+
4.0
*
10
-
8
=
1.6
*
10
-
7
1.6
*
10
-
7
+
0.4
*
10
-
7
=
0.8
total negative charge on ADP
=
2.0
+
0.8
=
2.8
b.
The p
K
a
values of the alkyl phosphate are 1.9 and 6.7. At pH 7.4, the OH group with a p
K
a
of 1.9 will
account for one negative charge. We need to calculate the fraction of the group with a p
K
a
value of 6.7
that will be negatively charged at pH 7.4.
fraction of group present in basic form
=
K
a
K
a
+
[H
+
]
=
2.0
*
10
-
7
2.0
*
10
-
7
+
0.4
*
10
-
7
=
0.8
total negative charge on the alkyl phosphate
=
1.0
+
0.8
=
1.8
3.
CH
2
OH
CHOH
CH
2
OH
O
−
O
O O
Ad
O
−
O
−
O
−
+
glycerol
CH
2
OH
CHOH
CH
2
OPO
3
2
−
glycerol-3-phosphate ADP
+
−
O
+
H
+
P
O
P
O
P
O
O O
Ad
O
−
O
−
P
O
P
O
4.
CH
2
OH
H
HO
CH
2
OPO
3
2
−
(
R
)-glycerol-3-phosphat
e
C
OH
H
CH
2
1
3
2
OPO
3
2
−
HOCH
2
or
5.
The resonance contributor on the right shows that the
b
-carbon of the
a
,
b
-unsaturated carbonyl compound
has a partial positive charge. The nucleophilic OH group, therefore, is attracted to the
b
-carbon.
_
RCH CH SCoA
+
RCH CH SCoA
C
O
C
O
6.
Because palmitic acid has 16 carbons and the acyl group of acetyl-CoA has 2 carbons, 8 molecules of
acetyl-CoA are formed from 1 molecule of palmitic acid.