Download Set 5 - UC Davis Plant Sciences

PRACTICE SET 5
Gluconeogenesis, Polysaccharide Synthesis, Pentose Phosphate Pathway, Photosynthetic
Carbon Fixation (Calvin Cycle)
RECOMMENDED PROBLEMS FROM THE TEXTS:
LNC:
G&G:
VV&P:
1.
Ch 20 (1-10, 14, 15)
Ch 22 (4, 5, 8); Ch 23 (1-7, 11, 12)
Ch 14 (8-10); Ch 15 (1-10); Ch 18 (7-9)
List the biologically important functions of the pentose phosphate pathway.
2.
Draw the structure of the products formed when fructose-6-phosphate reacts with
erythrose-4-phosphate in the presence of transketolase.
3.
Draw the structures of the products formed when ribose-5-phosphate reacts with
sedoheptulose-7-phosphate in the presence of transaldolase.
4.
Explain the metabolic fate of glucose 6-P under each of the following metabolic
conditions: (a) much more NADPH than ribose-5-P is required, (b), much more ribose 5P than NADPH is required, and (c) the needs for both ribose 5-P and NADPH are such
that 1 mole of ribose-5-P is needed for every 2 moles of NADPH produced.
5.
How many ATPs are used for each CO2 fixed during each of the following phases of the
dark reaction in photosynthesis?
a.
carboxylation phase
b.
reduction phase
c.
regeneration phase
6.
A wheat plant was placed in an atmosphere containing radioactive carbon dioxide
(14CO2). After 10 seconds of photosynthesis, the plant was killed and the
monosaccharide glucose was isolated. Radioactive carbon (14C) was found
predominantly in two of its six carbons. Indicate which carbon atoms of glucose were
labeled and indicate the most probable sequence of reactions that can account for the
labeling.
-1-
Answers
1.
2.
a). generate NADPH that is used for a variety of purposesnet biosynthetic, reductive pathways & glutathione reduction
b). generate 4, 5, and 7 carbon sugars for biosynthetic purposes
c). allow 4, 5, and 7 carbon sugars to be utilized as either an energy source, source of
glucose for storage, or as biosynthetic precursors for whatever the cell needs.
CH2OH
|
C=O
|
HO—C—H
|
H—C—OH
|
H—C—OH
|
CHO
|
H—C—OH
|
H—C—OH
|
CH2OPO3=
+
same as reactants
CH2OPO3=
3.
CH2OH
|
C=O
|
HO—C—H
|
H—C—OH
|
H—C—OH
|
H—C—OH
|
+
CHO
|
H—C—OH
|
H—C—OH
|
H—C—OH
|
CH2OPO3=
CH2OPO3=
CHO
|
H—C—OH
|
H—C—OH
|
CH2OPO3=
CH2OH
|
C=O
|
HO—C—H
|
H—C—OH
|
H—C—OH
|
H—C—OH
|
H—C—OH
|
CH2OPO3=
+
erythrose-4-P
-2-
octolose-8-P
4.
5.
a) Glc-6-P will be converted in the oxidative phase of the PPP to generate NADPH, and
then the products of the oxidative phase not needed will be converted to F-6-P and G-3-P
via the two non-oxidative phases of the PPP. 2 G-3-Ps can be converted into F-6-P by
gluconeogenic reactions. All F-6-P produced can be isomerized to GLC-6-P which can
then go through PPP.
b). The two non-oxidative phases (II and III) of the PPP will operate in “reverse” starting
from F-6-P and G-3-P.
c). The oxidative phase and the isomerization reaction converting ribulose-5-P to ribose-5P of the non-oxidative phase II of the PPP will operate as necessary to produce both
NADPH and ribose-5-P.
a. none
b. 2 ATP/CO2
c. 1 ATP/CO2
6.
CHO
H
HO
OH
H
H
OH
H
OH
14
C would be found in both positions
CH2 OH
-3-