Carbohydrate Metabolism: Glycolysis

Question 1

Which one of the statements regarding carbohydrates is false?

1) Carbohydrates are the most abundant biomolecules on earth
2) The oxidation of carbohydrates is the principle source of energy in photosynthetic cells
3) Human body wants to use carbs first, then fats and then proteins as a source of energy

Answer 1

2!

The oxidation of carbohydrates is the principle source of energy in NON-PHOTOSYNTHETIC CELLS. Human brain gets most of its energy from the metabolism of glucose.

Question 2

What are carbohydrates composed of?

Answer 2

One or many CARBONYL GROUPS (C=O); if it’s at the end of a carbon chain, it’s an aldose. if it’s within a carbon chain, it’s a ketose.

One or many HYDROXYL GROUPS (OH) associated to carbon atoms.

Question 3

Match the different types of carbohydrates to their definition:

1) Monosaccharides
2) Disaccharides
3) Polysaccharides
4) Oligosaccharides
5) Proteoglycans
6) Peptidoglycans

A) a short chain of monosaccharides (3-10 units), non-repetitive and complex and they are often bonded to non-carbohydrate molecules (glycoproteins/glycolipids)

B) formed of 2 units (ex: saccharose = glucose + fructose)

C) long chain of monosaccharide units, bonded to proteins

D) formed of a single unit (ex: glucose)

E) long chain of monosaccharide units, bonded to each other by small peptides

F) long monosaccharide chains, their structure is repetitive and simple, they can be linear (cellulose) or branches (glycogen

Answer 3

Monosaccharides: formed of a single unit (ex: glucose)

Disaccharides: formed of 2 units (ex: saccharose = glucose + fructose)

Polysaccharides: long monosaccharide chains, their structure is repetitive and simple, they can be linear (cellulose) or branches (glycogen)

Oligosaccharides: a short chain of monosaccharides (3-10 units), non-repetitive and complex and they are often bonded to non-carbohydrate molecules (glycoproteins/glycolipids)

Proteoglycans: long chain of monosaccharide units, bonded to proteins

Peptidoglycans: long chain of monosaccharide units, bonded to each other by small peptides

Question 4

What is the general formula of monosaccharides?

How are monosaccharides classified?

Answer 4

C(n)H2(n)O(n), where n is at least 3 and no more than 8!

Monosaccharides are classified according to three different characteristics:
1) The number of carbon atoms it contains (triose, tetrose, pentose, hexose, heptose, octose)
2) The position of the carbonyl groups (aldose/ketose)
3) The chirality of the molecule (D/L configuration)

Question 5

(T/F) Each monosaccharides carries AT LEAST one asymmetric carbon (chiral), but most have more than one.

Answer 5

True. There are can be multiple sometimes.

Question 6

1) A monosaccharide will be given a D configuration if the OH group is to the ________ of the chiral carbon in a __________ projection, whereas L configuration is given when the OH group is _______ of the chiral carbon.

2) What happens if there are multiple chiral carbons?

3) In this kind of projection, _________ bonds are coming towards you and __________ bonds are receding.

Answer 6

1) Right; Fisher; Left (these two stereoisomers are called enantiomers)

2) Same rule applies but for the BOTTOM CHIRAL CARBON (the farthest chiral carbon from the aldehyde/ketone)!

3) Horizontal; Vertical

Question 7

Which one of the statements regarding stereoisomers is true?

1) D and L enantiomer differ in their light refraction properties. When light is reflected on the right = levogyre (L), when light is reflected on the left. = dextrogyre (D).

2) Most of the monosaccharide enantiomers found in nature are D-monosaccharides.

3) A diastereomer is synonymous to enantiomer.

Answer 7

2!

For 1, D and L enantiomer differ in their light refraction properties. When light is reflected on the right = dextrogyre (D), when light is reflected on the left. = levogyre (L).

For 3, a diastereomer is a stereoisomer that is NOT AN enantiomer! Enantiomer is a mirror image of a molecule, while diastereomer is not!

Question 8

Which one of the statements regarding monosaccharide is true?

1) Most monosaccharides have a linear structure at neutral pH (such as aldopentoses, aldohexoses and ketohexoses)

2) The structure of 5-6 carbon monosaccharides is linear.

3) Reaction between the aldehyde group at C-1 (or c-2 in a ketopentose) and the hydroxyl group at c-5 forms a hemiacetal linkages, producing either of the two stereoisomers; alpha or beta.

Answer 8

3!

For 1; Most monosaccharides have a CYCLIC structure at neutral pH (such as aldopentoses, aldohexoses and ketohexoses)

for 2; The structure of 5-6 carbons monosaccharides is CYCLIC.

Question 9

What is the difference between the two stereoisomers (anomer α and anomer β) formed when a monosaccharide forms a cyclic structure?

Answer 9

Anomer α: the OH from the first carbon and distal CH2OH (carbon #6) are on opposite sides

Anomer β: the OH from the first carbon and distal CH2POH (carbon #6) are on the same side

Question 10

Disaccharides are made of ____ monosaccharides held together by a __ _______ bond.

The bond forms between the ________ group of one monosaccharide and the _______ group of the other monosaccharide.

Answer 10

two, O-glycosidic

hydroxyl; hydroxyl

Question 11

(T/F) Anomer α and anomer β have the same energy but form different bonds. Anomer β’s bonds are harder to break because there are no enzymes to break them in the human body.

Answer 11

True!

*It is an α-glycosidic bond, if the first glucose is in an α configuration

*it is a β-glycosidic bond, if the first glucose is in a β configuration

Question 12

(T/F) Maltose, trehalose, and sucrose are disaccharides with α connections, while cellobiose, lactose, and gentiobiose are disaccharides with β connections.

Answer 12

True!

Question 13

What are the two classifications of polysaccharides according to their structure? Briefly describe them.

Answer 13

1) Homopolysaccharides: formed of one type of unit only. Ex. cellulose, glycogen and starch are polymers of glucose.

2) Heteropolysaccharides: formed of at least 2 different types of monosachharide.

Question 14

What are the three physiological roles of carbohydrates?

Answer 14

1) STRUCTURAL; they support and protect biological structures. Ex. fiber shaped carbs (cellulose), carbs in the forms of gels (glycosaminoglycan in cartilage + tendons)

2) ENERGY SOURCE; principle source of fuel and this energy can be stored (glycogen).

3) METABOLIC: carbohydrates can be changed into other types of molecules (amino acids, fatty acids and nucleotides).

Question 15

What is the cause of the obesity crisis?

Answer 15

SUGAR everywhere!

It can be stored as fat! Our body still behaves as cavemen; every metabolic pathway is to STORE, STORE, ETC.

Question 16

Briefly describe glycolysis and the conditions it occurs in.

Answer 16

Glycolysis is the pathway by which six-carbon sugars are split to yield two three-carbon compoounds (pyruvate).

During this process, the potential energy store in these six-carbon sugars is used in the synthesis of ATP from ADP.

Glycolysis can occur under aerobic and anaerobic conditions.

Question 17

While reactions 1-5 of glycolysis are part of the energy ___________ phase, reactions 6-10 form the energy ____________ phase.

Answer 17

investment; generation

Question 18

Match the energy investment phase reactions to their descriptions:

1) Reaction 1
2) Reaction 2
3) Reaction 3
4) Reaction 4
5) Reaction 5

A) D-fructose-6-phosphate (F6P) is PHOSPHORYLATED at C-1 by the PHOSPHPOFRUCTOKINASE 1 (PKF) to generate D-fructose-1,6-bisphophate (FBP).

B) D-fructose-1,6-bisphophate (FBP) is CLEAVED to generate two 3C molecules: glyceraldehyde-3-phosphate (GAP) and dihydroxyacetone phospahte (DHAP) by FRUCTOSE-1,6-BISPHOSPHATE ALDOLASE.

D) α-D-glucose is PHOSPHORYLATED to form α-D-glucose-6-phosphate (G6P) by HEXOKINASE.

E) ISOMERIZATION of the dihydroxyacetone phospahte (DHAP) to glyceraldehyde-3-phosphate (GAP) by the TRIOSE PHOSPHATE ISOMERASE (TPI).

F) α-D-glucose-6-phosphate (G6P) is ISOMERIZED into D-fructose-6-phosphate (F6P) by the PHOSPHOHEXOSE ISOMERASE.

Answer 18

Reaction 1: α-D-glucose is PHOSPHORYLATED to form α-D-glucose-6-phosphate (G6P) by HEXOKINASE.

Reaction 2: α-D-glucose-6-phosphate (G6P) is ISOMERIZED into D-fructose-6-phosphate (F6P) by the PHOSPHOHEXOSE ISOMERASE.

Reaction 3: D-fructose-6-phosphate (F6P) is PHOSPHORYLATED at C-1 by the PHOSPHPOFRUCTOKINASE 1 (PKF) to generate D-fructose-1,6-bisphophate (FBP).

Reaction 4: D-fructose-1,6-bisphophate (FBP) is CLEAVED to generate two 3C molecules: glyceraldehyde-3-phosphate (GAP) and dihydroxyacetone phospahte (DHAP) by FRUCTOSE-1,6-BISPHOSPHATE ALDOLASE.

Reaction 5: ISOMERIZATION of the dihydroxyacetone phospahte (DHAP) to glyceraldehyde-3-phosphate (GAP) by the TRIOSE PHOSPHATE ISOMERASE (TPI).

Question 19

Match the energy investment phase reactions to their descriptions:

1) Reaction 1
2) Reaction 2
3) Reaction 3
4) Reaction 4
5) Reaction 5

A) REVERSIBLE; ΔG°’ is close to 0 (anything -/+5 is reversible). Because previous reaction yields lots of reactant for this reaction as it is irreversible, this reaction is pushed in a forward direction.

B) Strongly ENDERGONIC (requires energy to proceed). ΔG is close to 0 in cell conditions, thus REVERSIBLE.

C) Investment of an ATP; so highly favourable. It is IRREVERSIBLE!

D) Weakly ENDERGONIC (requires energy to proceed). ΔG is close to 0 in cell conditions, thus REVERSIBLE.

E) Investment of an ATP; highly favourable. It is IRREVERSIBLE. Most important and regulated reaction in glycolysis.

Answer 19

Reaction 1: Investment of an ATP; so highly favourable. It is IRREVERSIBLE!

Reaction 2: REVERSIBLE; ΔG°’ is close to 0 (anything -/+5 is reversible). Because previous reaction yields lots of reactant for this reaction as it is irreversible, this reaction is pushed in a forward direction.

Reaction 3: Investment of an ATP; highly favourable. It is IRREVERSIBLE. Most important and regulated reaction in glycolysis.

Reaction 4: Strongly ENDERGONIC (requires energy to proceed). ΔG is close to 0 in cell conditions, thus REVERSIBLE.

Reaction 5: Weakly ENDERGONIC (requires energy to proceed). ΔG is close to 0 in cell conditions, thus REVERSIBLE.

*in the first five reactions, two ATP molecules are used to convert one molecule of glucose to two molecules of glyceraldehyde-3-phosphate (GAP).

Question 20

Match the FOUR different hexokinases found in mammals to their definitions:

1) Hexokinase I, II, and III
2) Hexokinase IV

A) also called GLUCOKINASE. Found in the LIVER and PANCREAS. Glucose specific! High Km enzyme; low affinity (~7.5mM). Allow the liver to adjust its rate of glucose usage to the variations in blood glucose levels. Not inhibited by glucose-6-phosphate.

B) Found in multiple tissues but mainly located in SKELETAL MUSCLE. Not specific to glucose! Low Km enzymes; strong affinity (~0.04nM). Strongly inhibited by the product of the reaction, glucose-6-phosphate.

Answer 20

Hexokinase I, II, and III: Found in multiple tissues but mainly located in SKELETAL MUSCLE. Not specific to glucose! Low Km enzymes; strong affinity (~0.04nM). Strongly inhibited by the product of the reaction, glucose-6-phosphate.

Hexokinase IV: also called GLUCOKINASE. Found in the LIVER and PANCREAS. Glucose specific! High Km enzyme; low affinity (~7.5mM). Allow the liver to adjust its rate of glucose usage to the variations in blood glucose levels. Not inhibited by glucose-6-phosphate.

Question 21

Why are hexokinases I, II, and III strongly inhibited by the product of the reaction, glucose-6-phosphate?

Answer 21

In skeletal muscles, where these hexokinases are found, glycolysis occurs to produce Energy.

Muscles need energy when they are working. When they stop working, everything accumulates. Now, there is less need for glucose-6-phosphate. Therefore, as its concentration increases, it inhibits the enzymes that make it.

Question 22

The Km of the hexokinases in skeletal muscles is very low. However, the glucose concentration is around 2-15 mM. Therefore, these hexokinases work at Vmax all the time. Why is there a need for this?

Answer 22

This is because the muscle wants energy right away if the muscle is about to be used. Glucose is the source of energy, allowing the enzymes to be at full speed right away!

Question 23

Describe the phenomenon behind intermittent fasting.

Answer 23

Intermittent fasting is when you do not eat for a long periods of time to lose weight.

After you eat, the body uses the glucose from the food as a source of energy. Then, it uses the stored glycogen.
There is a limited amount of sugar the body can store as glycogen, then the rest will be stored as fats, proteins, etc.

When there is no glycogen left, it uses fat and protein as a source of energy.

So not only are you losing fat, you are also losing muscles!

Question 24

(T/F) Muscle enzymes (hexokinases I, II and III) are more responsive to changes in glucose concentration than liver enzymes (hexokinase IV).

Answer 24

False!

In the muscles, hexokinases I, II, or III run at Vmax on little glucose so changes in glucose concentration do not affect the enzyme.

However, in the liver, hexokinases require much higher level of glucose concentration to run at Vmax; therefore they are responsive to changes in glucose concentration.

Question 25

Briefly answer the following questions about GLUT?

1) What is GLUT? What does it do?

2) How many GULTs are there?

3) Describe the differences between GLUT2 and GLUT4.

Answer 25

1) GLUT is an acronym for GLUcose Transporter. Glucose transporters move glucose molecules across the plasma membrane.

2) There is more than 12 glucose transporters expressed in various tissues and having slightly different roles.

3)
GLUT2: found in the LIVER, pancreas and kidney. It is insulin INDEPENDENT. It quickly equilibrates concentration of glucose across plasma membrane. It allows HEXOKINASE IV to adjust its rate to the concentration of glucose in the blood (like a thermostat).

GLUT4: found in SKELETAL MUSCLE, adipose tissues, and heart. REGULATED BY INSULIN/epinephrine. In absence of sugar, this transporter is sequestered from the membrane to the cell. It is released upon insulin presence to allow glucose to come inside.

Question 26

Can you explain the discrepancy between the standard and actual ΔG of reaction 4?

ΔG°’ = +23.9kJ/mol
ΔG = -1.3kj/mol in cell conditions

Answer 26

ΔG = ΔG°’ + RT ln[DHAP][GAP]/[FBP]

Reaction 3 had an investment of ATP, making it a highly favourable, IRREVERSIBLE reaction. This produced a lot of reactant for reaction 4 (FBP).

Reaction 4 moves in the forward direction since there is an increase in the reactant (even a decrease in the product would explain the discrepancy).

Question 27

Match the energy generation phase reactions to their descriptions:

1) Reaction 6
2) Reaction 7
3) Reaction 8
4) Reaction 9
5) Reaction 10

A) ISOMERIZATION of 3-phosphoglycerate (3PG) in 2-phosphoglycerate (2PG) by the PHOSPHOGLYCERATE MUTASE.

B) DEHYDRATION of 2-phosphoglycerate (2PG) in phosphoenolpyruvate (PEP) by the ENOLASE.

C) OXIDATION and PHOSPHORYLATION of Glyceraldehyde-3-phosphate (GAP) to generate 1,3- bisphosphoglycerate (BPG) by GLYCERALDEHYDE-3-PHOSPHATE DEHYDROGENASE (GAPDH) and NAD+

D) Synthesis of an ATP by the TRANSFER OF A PHOSPHORYL GROUP from phosphoenolpyruvate (PEP) to pyruvate by the PYRUVATE KINASE.

E) Synthesis of an ATP by the TRANSFER OF A PHOSPHORYL GROUP from 1,3-bisphosphoglycerate (BPG) to 3-phosphogylcerate (3PG). Reaction catalyzed by PHOSPHOGLYCERATE KINASE.

Answer 27

Reaction 6: OXIDATION and PHOSPHORYLATION of Glyceraldehyde-3-phosphate (GAP) to generate 1,3- bisphosphoglycerate (BPG) by GLYCERALDEHYDE-3-PHOSPHATE DEHYDROGENASE (GAPDH) and NAD+

Reaction 7: Synthesis of an ATP by the TRANSFER OF A PHOSPHORYL GROUP from 1,3-bisphosphoglycerate (BPG) to 3-phosphogylcerate (3PG). Reaction catalyzed by PHOSPHOGLYCERATE KINASE.

Reaction 8: ISOMERIZATION of 3-phosphoglycerate (3PG) in 2-phosphoglycerate (2PG) by the PHOSPHOGLYCERATE MUTASE.

Reaction 9: DEHYDRATION of 2-phosphoglycerate (2PG) in phosphoenolpyruvate (PEP) by the ENOLASE.

Reaction 10: Synthesis of an ATP by the TRANSFER OF A PHOSPHORYL GROUP from phosphoenolpyruvate (PEP) to pyruvate by the PYRUVATE KINASE.

Question 27

Match the energy phase reactions to their definitions:

1) Reaction 6
2) Reaction 7
3) Reaction 8
4) Reaction 9
5) Reaction 10

A) Standard state conditions ~-17.2 kJ/mol, while cell conditions is around 0. Thus, REVERSIBLE.

B) REVERSIBLE; product concentration is kept low by next reaction’s enzyme to keep reaction forward.

C) Weakly endergonic; REVERSIBLE (Nad+ –> NADH; loss of electrons)

D) IRREVERSIBLE reaction, with a -29.7kJ/mol as standard state conditions.

E) Weakly endergonic; REVERSIBLE. Reactant concentration is kept high by reaction 6 and 7, driving this reaction forward

Answer 27

Reaction 6: Weakly endergonic; REVERSIBLE (Nad+ –> NADH; loss of electrons)

Reaction 7: Standard state conditions ~-17.2 kJ/mol, while cell conditions is around 0. Thus, REVERSIBLE.

Reaction 8: Weakly endergonic; REVERSIBLE. Reactant concentration is kept high by reaction 6 and 7, driving this reaction forward

Reaction 9: REVERSIBLE; product concentration is kept low by next reaction’s enzyme to keep reaction forward.

Reaction 10: IRREVERSIBLE reaction, with a -29.7kJ/mol as standard state conditions.

Question 28

What is the metabolic yield of glycolysis?

Answer 28

Glucose –> 2 Pyruvate
2 ADP –> 2 ATP
2 NAD+ –> 2 NADH

In the first five reactions, 2 ATP are used to produce 2 ADP.

In the last five reactions, 4 ATP are produced using 4 ADP and 2 NAD+ are used to produce 2 NADH.

Question 29

Glycolysis was probably the first energy yielding pathway utilized by the earliest known organisms, when the atmosphere was still anaerobic.

Anaerobic glycolysis of one glucose molecule yields ____ ATPs (low energy efficiency), while aerobic glycolysis of one glucose molecule generates ______ ATPs.

Answer 29

2; 30-32

Question 30

What is fermentation?

What are the two types? Describe each.

Answer 30

Fermentation: anaerobic degradation of a nutrient (glucose), with no net change in the oxidation state, to produce energy.

1) Lactic fermentation: reduction of pyruvate to lactate
2) Alcoholic fermentation: reduction of pyruvate to ethanol

Question 31

In lactic fermentation, there is a generation of ATP (energy) without consuming oxygen. It also regenerates _______ from ______ for further glycolysis under anaerobic conditions.

Answer 31

NAD+; NADH

(1 NADH is about 2.5ATP!)

The 2 NADHs produced in glycolysis is used when the two pyruvate turn into two lactase, yielding 2 NAD+s. This yields NO NET CHARGE!

Question 32

In the early moment of a strenuous exercise, lactate builds up in the ________. This is because the amounts of oxygen required doesn’t match the amount needed for ________ glycolysis.

The lactate produced will diffuse through the tissue and _______ to the ______ via the _______ where it will be catabolized through _______ glycolysis or be converted to __________ through the _____ cycle (precursor for gluconeogenesis).

Answer 32

muscle; aerobic

travel; liver; bloodstream; aerobic; glucose; cori

*the first minute of exercise is always anaerobic; debt
*converting to glucose through the cori cycle is very costly! therefore, breathing heavily after exercising; producing extra ATP (pay back time).

Question 33

(T/F) White blood cells rely on anaerobic glycolysis since they lack mitochondria.

Answer 33

False! It’s RED BLOOD CELLS that rely on anaerobic glycolysis since they lack mitochondria.

They produce lactate, which goes to the liver to be recycled.

Question 34

(T/F) In lactic fermentation, two pyruvates are turned into two lactates directly, resulting in 2 NAD+.

In alcoholic fermentation, two pyruvates are turned into two ACETALDEHYDES first and then into two ETHANOLS after. One pyruvate turning into acetaldehyde results in one CO2, and one acetaldehyde turning into ethanol results in one NAD+. Overall, two CO2, two NAD+ and two ethanol is produced!

Answer 34

True!

Question 35

Which reactions of glycolysis are irreversible? What do these do?

Answer 35

Reaction 1: glucose –> G6P
Reaction 3: F6P –> FBP
Reaction 10: PEP –> Pyruvate
are the three reactions that are irreversible in glycolysis

The irreversible reactions allow other reversible reactions to go forward either by increasing their reactants or decreasing their products!