Glyceraldehyde-3-phosphate Step VI: 1,3-Bisphosphoglycerate Step VII: 3-Phosphoglycerate
AG = -0.4 kcal/mole AG = 0.3 kcal/mole
Figure 6-28
Step VI (Oxidation of an aldehyde into a carboxylic acid derivative): Oxidative phosphorylation of G-3-P into 1,3-bisphosphoglycerate is the only oxidation step in glycolysis. It is the only step where the number of bonds from carbon to oxygen increases. Oxidation releases energy, which is "stored" by adding a phosphate group. Oxidation occurs at carbon 1 as it gains a bond to oxygen when converted from an aldehyde into an acid carboxylate. The term P, indicates that there is an inorganic phosphate gained by the reactant. At pH = 7.2, inorganic phosphate exists mostly in its dianion form. Dephosphorylation of 1,3-bisphosphoglycerate into 3-phosphoglycerate in Step VII, releases a phosphate, which serves to regenerate ATP that was invested in earlier steps.
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Organic Chemistry Carbohydrates Biological Applications
0<v°"
H- OH
In the last stage of glycolysis, the compound loses a phosphate to generate more ATP and then undergoes isomerization and elimination to generate the final product, pyruvate. These three steps are shown in figure 6-29.
Step VIII:
AG = 0.2 kcal/mole Phosphoglyceromutase
°^°"
H- OPO,
Step IX:
AG = -0.8 kcal/mole
O. .O"
OPO,
CBjOPO/"
3-Phosphoglycerate
CH2OH 2-Phosphoglycerate
Enolasew
HoO
ICHj
Phosphoenolpyruvate
L w^^ADP +H
o
CH3 Pyruvate
f Step X:
AfP AG = -4.0 kcal/mole
Figure 6-29
Steps IX (Enolate formation): Conversion of 3-phosphoglycerate into 2- phosphoglycerate positions thephosphate for enolformation in Step IX. Because these two structural isomers have the same bonds, this reaction results in a
minimal change in energy. Conversion of2-phosphoglycerate intophosphoenol
pyruvate involves dehydration. The double bond formed from elimination has a
phosphorylated hydroxyl group attached, making the compound an enol.
Conversion ofphosphoenolpyruvate intopyruvate completes glycolysis.
You should not memorize glycolysis, as the amount of time invested to do so would not have a good return in terms of increasing your MCAT performance.
However, you should have a basic idea of what happens overall. It is more important that you recognize the type of reaction than it is to know details of any particular step. A passage may present one or more steps of glycolysis and ask you questions from both an organic chemistry and biochemistry perspective.
You should know that glucose is converted into pyruvate and that there is a net generation of 2 ATP and a release of energy. It is an oxidative process, so a molecule is brokendown. You can apply the generalprinciplesdiscussed here to other pathways,so learn the concepts and insights.
1) Regulatory steps usually involve ATP.
2) Multipstep pathways produce intermediate products that can feed other pathways.
3) Oxidationbreaksdown molecules and requires an oxidizing agent (a species poor in H, such as NAD+). Oxidation releases energy.
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Organic Chemistry Carbohydrates Oxidation and Reduction of Pyruvate
Once formed, pyruvate has a few potential fates. It can undergo either aerobic or anaerobic breakdown. Pyruvate can undergo fermentation, where carbon 2 is reduced from a ketone into an alcohol, resulting in the formation of lactate, which regenerates NAD+. We cannot reduce lactate any further (only yeast and bacteria can convert pyruvate into ethanol), so it is excreted as waste. Pyruvate can also undergo oxidative decarboxylation. Converting a carboxylic acid group into CO2 results in the gain of a bond to oxygen, so it is oxidation and must therefore be coupled with reduction of NAD+ into NADH. To make these reactions easier to grasp, you should focus on the oxidative or reductive nature of the pathway. Figure 6-30 shows glycolysis, followed by three potential pathways that pyruvate can take.
HOH2C
OHI Glucose
OH
Glycolysis
^ ^ Lactate
H- OH
CH,
C02 + CH3CH2OH C02 + Acetyl CoA
Figure 6-30
Catabolism is the biological process of breaking down a sugar. Given that the generation of ATP is considered critical to energy storage, let us finish by briefly looking at an ATP molecule. It is made of a nucleoside (comprised of adenine and ribose) and a triphosphate group linked at hydroxyl 5. Adenosine triphosphate is shown in Figure 6-31.
r o
II
Triphosphate x
o o ^
Nucleoside (Sugar + Base) A*.
O — P-rO—P—O—P —O—CB
♦ I
O"
O" O"
Hydrolysis of ATP cleaves here and releases -7.3 kcal/mole
Figure 6-31
OH HO
Ribose
A nucleoside is a monosaccharide (usually ribose) bonded at carbon 1 to one of the bases in either DNA or RNA. Linkages between phosphate groups generate 7.3 kcal/mole when they are hydrolyzed, meaning that when ATP gains H2O, it cleaves to form ADP, Pi (inorganic phosphate), and H+ with a AG = -7.3 kcal/mole. This is a valuable tool for storing energy in small increments.
Biological Applications
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OrganiC ChemiStry Carbohydrates Section Summary
Key Points for Carbohydrates (Section 6)
Monosaccharides
1. Sugars are named for carbon count and carbonyl type
a) Tetrose = 4 C, Pentose = 5 C, and Hexose = 6 C
b) Aldose = aldehyde sugar; Ketose= ketone sugar
c) D-sugar has last chiral center = R; L-sugar has last chiral center = S d) Common monosaccharides from biology should be memorized
i. Ribose: "Riboseis all right!"
ii. Glucose:"F*** glucose!", and flip it off.
iii. Mannose: "Man goes with gun."
iv. Galactose: C-4 epimer of glucose v. Fructose: ketose of glucose
2. Cyclic sugars are either furanoses or pyranoses a) Pyranoses are six-membered ring sugars
i. Beta anomerhas last carbon cis to anomeric hydroxyl group ii. Anomeric carbon is the most oxidized carbon (has two bonds to O) iii. 6-Glucopyranose has allsubstituents in the equatorial position b) Furanoses are five-membered rings
i. Ribose and fructose are the two most common examples Oligosaccharides and Polysaccharides
1. Oligosaccharidescontain 2-8 monosaccharides, polysaccharides contain ? a) Glycosidic linkages bind them (water is released when linkage forms)
i. Disaccharides include sucrose, lactose, and maltose ii. Polysaccharides include cellulose, amylose, and glycogen iii. Linkages are a or C, named for the componenthydroxyl groups iv. Enzymes that cleave glycosidic linkages are specific
b) Branching is typically done using hydroxyl 6 of a sugar in the polymer i. Cellulose (6-linked) and amylose (oc-linked) have no branching ii. Amylopectin (a-linked) has branching about 1 out of 30 monomers iii. Glycogen(a-linked) has branching about 1 out of 10 monomers
Chemical Reactions and Tests
1. Chemical reactions and tests determine chirality,add Cs, or remove Cs.
a) Nitricacid (HNO3) oxidizes terminalCs to acids: optical activity test b) Phenylhydrazine (0HNNH2) forms an osazone: C-2epimer test c) Tollen's test (Ag+)oxidizes C-l of aldoses: aldose test
d) Periodate test (IO4") oxidizessugar by cleaving all C-C bonds
e) Kiliani-Fischer synthesis increases an aldoseby one C;forms epimers f) Ruff and Wohl degradation decrease an aldoseby one C
Biological Examples
1. Blood types have glycoproteins with different stereochemistry
a) Type O contains 3 sugars (Glc-Gal-Fuc) and Types A and Bcontain 4 b) L-Fucose is the third sugar, deviating from the natural sugars are D rule 2. Glycolysis breaks down glucoseinto two pyruvate molecules and energy
a) Forms a net of 2 ATP and 2 pyruvates for further reaction
b) Has one oxidation step; mostly isomerization and phosphate exchange
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Carbohydrates Passages
13 Passages 100 Questions
Suggested schedule:
I: After reading this section and attending lecture: Passages I, III, V, VI, & X Grade passages immediately after completion and log your mistakes.
following Task I: Passages II, IV, VII, & XI (28 questions in 36 minutes) Time yourself accurately, grade your answers, and review mistakes.
Review: Passages VIII, IX, XII, XIII, & Questions 92 - 100 focus on reviewing the concepts. Do not worry about timing.
II
m
R.E-V-I.E-W .
Specializing in MCAT Preparation
I. Glucose and Glucopyranose II. Fischer and Haworth Projections III. Sugar Conventions
IV. Monosaccharides versus Disaccharides
V. Amylose, Amylopectin, and Cellulose VI. Unknown L-Aldopentose Elucidation
VII. Unknown D-Aldohexose Elucidation
VIII. Kiliani-Fischer Synthesis
IX. Osazone Derivative Test
X. Glycolysis Reactions XI. Blood Types
XII. Biochemistry of Sugars XIII. Combustion of Sugars
Questions not Based on a Descriptive Passage
Carbohydrates Scoring Scale
Raw Score MCAT Score
85 - 100 13- 15
6 5 - 8 4 1 0 - 1 2
47 -64 7 - 9
33-46 4 - 6
1 - 3 2 1 - 3
(1 -7) (8 - 14) (15 - 22) (23 - 29) (30 - 36) (37 - 43) (44 - 50) (51 -57) (58 - 64) (65-71) (72 - 78) (79 - 85) (86-91) (92 - 100)
Passage I (Questions 1 - 7)
The most stable form of glucose is a hemiacetal ring structure. The ring is formed when the hydroxyl group on carbon 5 attacks the aldehyde group to form a hemiacetal.
Thehydroxyl can attack the carbonyl carbon (sp2-center) from either the top or bottom of the rc-bond. As a result of this attack, there are two possible diastereomers that can form.
The aldehyde carbon forms the new stereocenter. These two possible diastereomers are referred to as the alpha and beta anomers of the glucopyranose. The beta isomer (anomer) is defined as having the anomeric hydroxyl (on carbon one) cis with respect to carbon number six in the ring form. Because most natural sugars have D-configuration, it is perhaps easier to think of the beta anomer as the one with the hydroxyl group up (equatorial) in the ring form. Both anomers of D- glucopyranose are drawn in Figure 1 below.
OH
6-D-glucopyranose a-D-glucopyranose Figure 1 Anomeric forms of D-glucopyranose The 6-anomer is favored by a ratio of 64% to 36% over the oc-anomer. Hydroxyl groups on adjacent carbons in the ring can form hydrogen bonds from all orientations except diaxial. When the hydroxyl groups are equatorial they experience less steric repulsion than when they have axial orientation, but hydrogen bonding is slightly reduced.
Because the 6-anomer is more abundant, the reduced steric hindrance of the equatorial orientation must be of greater importance than the slight loss in hydrogen bonding.
Hydrogen bonding is possible from the gauche orientation associated with diequatorial hydroxyl groups.
1. Which is the more stable anomer of D-glucose?
A. The alpha-anomer, becausethe hydroxyl group has equatorial orientation.
B. The beta-anomer, because the hydroxyl group has equatorial orientation.
C. The alpha-anomer, becausethe hydroxyl group has
axial orientation.
D. The beta-anomer, because the hydroxyl group has
axial orientation.
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2. D- Glucose and D-galactose are best described as which of the following?
A. Anomers
B. C-2 epimers C. C-4 epimers
D. Enantiomers
3. D-Glucose and L-glucose are:
A. anomers.
B. C-5 epimers.
C. diastereomers.
D. enantiomers.
4. D-Glucose has which of the following stereochemical arrangements?
A. 2R, 3R, 4R, 5R B. 2R, 3S, 4R, 5R C. 2S, 3R, 4R, 5R D. 2S, 3S, 4R, 5R
5. How many stereoisomers are possible for a linear
aldohexose?
A. 4 B. 8 C. 16 D. 32
6. In the beta-anomer of D-glucopyranose, the first carbon has the hydroxyl with what orientation?
A. Cis B. Planar C. Axial
D. Equatorial
7. The oxygen present in the 1,4-linkage in the disaccharide below is part of what functionality?
HOH2C