BICHEM: AAMC 1D: CARBOHYDRATES

Question 1

Sugars, monosaccarides, disaccharides, polysaccharides are called?

deoxyribose has an H or OH?

What is D/L configuration used for?

α/β = are used for what configuration?

What are the suffixes for sugars?

Answer 1

• carbohydrates
• H
• absolute configuration, assigned based on chirality of the carbon atom furthest from the carbonyl group
• α/β = anomeric configuration
• all sugars end with -ose

Question 2

Classify these sugars:

________= sugars with an aldehyde group

___________ = sugars with a ketone group

____________ = sugars in a 6-membered ring structure = hexagonal shaped

_____________ = sugars in a 5-membered ring structure = pentagonal shaped

hexose sugar has _______carbons

Answer 2

• Aldose
• Ketose
• Pyranose
• Furanose
• 6 carbons, # ose = sugar with # carbon atoms

Question 3

In order to e classified as a carbohydrate, a molecule must have:

Answer 3

• at least a 3-carbon backbone
• an aldehyde or ketone group
• at least 2 hydroxyl groups

Question 4

what is the simplest smallest carb called?

Answer 4

• glyceraldehyde and dihydroxyacetone

Question 5

name 3 common monosaccharides:

Answer 5

• glucose, fructose, and galactose.
• Glucose isour blood sugar and the product of photosynthesis.
• Fructose is the sugar in fruits, and it is sweeter than glucose.
• Galactose is one of the monomers that make up lactose, which is the sugar in milk; it is less sweet than glucose.

Question 6

name the sugars that make up the RNA & DNA

Answer 6

• ribose
• deoxyribose ~ “​2’-deoxyribose”

Question 7

Name diasaccharides:

Answer 7

• Sucrose is a disaccharide made from α-glucose and β-fructose joined at the hydroxyl groups on the anomeric carbons (making acetals). Sucrose is table sugar, the sugar we buy in stores.
• Lactose is a disaccharide made from β-galactose and α/β-glucose joined by a 1-4 linkage
• Starch = glucose molecules joined by α 1-4 linkage (amylose = unbranched; amylopectin = branched), main energy form for plants
• Cellulose: main structurla component for plant cell walls. Main source of fiber in human diet
• Glycogen = same as starch, but with additional α 1-6 linkages for branching

Question 8

Absolute Configuration of Carbohydrates: how to calssify them

Answer 8

• The chiral carbon furthest from carbonyl group determines the absolute configuration L or D of the sugar
• If in the Fischer projection, the OH group on the chiral carbon furthest from the carbonyl is pointing left, then it’s L. If it’s pointing right, then it’s D.
• Note: L and D are enantiomers, not epimers. So, every chiral carbon center inverts. It’s just that you assign L and D based on the chiral carbon furthest from the carbonyl.

Question 9

How to convert a Fischer projection to Haworth (cyclic) form?

Answer 9

• -OH groups that point Left on Fischer become Up on Haworth
• -OH groups that point Right on Fischer become Down on Haworth
• -OH group on anomeric carbon (Fischer carbonyl) can be either Up (beta) or Down (alpha)
• The CH2OH group on absolute configuration carbon (carbon 5) points up for D, and down for L

Question 10

Hydrolysis of the glycoside linkeage

write down the reaction:

what catalysts are included for starch & for nucleotides?

Answer 10

• Glycoside linkage = acetal linkage = linkage involving the hydroxyl group of the anomeric carbon
• Glycoside linkage can also mean linkage between sugar and base in nucleotides
• Examples of glycosidic linkages = starch, glycogen, nucleotide
• Hydrolysis of the glycosidic bond has the same mechanism as hydrolysis of the acetal bond
• Glycoside + H2O + catalyst—>hydrolysis
• Catalysts include: Amylase for starch and glycosylase for nucleotide

Question 11

Monosacharides

Answer 11

• Simple sugars, the most common of which is glucose. Have a formula (CH2O)n, and typically contain three to seven carbon atoms.
• Most oxygens in monosaccharides are found in hydroxyl (OH) groups, but one of them is part of a carbonyl (C=O) group. The position of the carbonyl (C=O) group can be used to categorize the sugars:
  
  • If sugar has an aldehyde group, meaning that carbonyl C is last one in chain, it is known as an aldose.
  • If carbonyl C is internal to the chain, so there are other carbons on both sides of it, it forms a ketone group and the sugar is called a ketose.
  • Sugars are named according to their number of carbons: some of the most common types are trioses (three carbons), pentoses (five carbons), and hexoses (six carbons)

Question 12

Disacchrides

Answer 12

• Form when two monosaccharides join together via a dehydration reaction, also known as a condensation reaction or dehydration synthesis. In this process, the hydroxyl group of one monosaccharide combines with the hydrogen of another, releasing a molecule of water and forming a covalent bond known as glycosidic linkage.
• Common disaccharides include lactose, maltose, and sucrose.
• Lactose consists of glucose and galactose and found naturally in milk. Many people can’t digest lactose as adults, resulting in lactose intolerance.
• Maltose, or malt sugar, is made of two glucose molecules.
• Most common disaccharide is sucrose (table sugar), made of glucose and fructose.

Question 13

Polysaccharides

Answer 13

• Long chain of monosaccharides linked by glyosidic bonds. The chain may be branched or unbranched and may contain different types of monosaccharides.
• Common examples in living organisms include starch, glycogen, cellulose, and chitin.
• Starch: main energy storage for plants, is the stored form of sugars in plants and made up of a mixture of two polysaccharides, amylose (unbranched) and amylopectin (branched) (both polymers of glucose).
  
  • Plants able to synthesize glucose using light energy gathered in photosynthesis, and excess glucose, beyond the plant’s immediate energy needs, is stored as starch in different plant parts.
• Cellulose is major component of plant cell walls, rigid structures that enclose the cells.

Question 14

Glycolysis is an (aerobic process)

Define glycolysis

what is glycolysis inhibited by?

Key enzymes?

Total ATP?

Locations?

Answer 14

• covverts glucose (6 carbons) to 2 molecules of pyruvate (3 carbons)
• Key enzymes: Hexokinase, Phosphofructokinase, Pyruvate Kinase
• Location: cytosol
• Occurs under both aerobic and anaerobic conditions
• Glycolysis inhibited by ATP
• Products:
  
  • 2 net ATP made for every glucose (2 input ATP, 4 output ATP)
  • 2 NADH made for every glucose

Question 15

Aerobic decarboxylation

location?

key enzyme

products?

Answer 15

• Aerobic decarboxylation (mitochondrial matrix) = convert pyruvate (3 carbons) to an acetyl group (2 carbons)
• Key enzyme: Pyruvate dehydrogenase
• Products: 1 NADH made for every pyruvate
• Only occurs in the presence of oxygen (asK)
• Acetyl group attaches to Coenzyme A to make acetyl CoA

Question 16

Fermentation (anaerobic glycolysis)

key enzymes

products

purpose?

name the 2 types?

Answer 16

• Key enzyme: Lactate dehydrogenase
• Products: 1 NAD+ made for every pyruvate
• Alcohol fermentation = pyruvate reduced to ethanol
• Lactic acid fermentation = pyruvate reduced to lactate
• Purpose: to regenerate NAD+, which is needed for glycolysis

Question 17

Gluconeogenesis:

• pyruvate combines with what to produce OAA, and through which enzyme? what is this enzyme activated by?
• GTP is then used to transform OAA into __________ via phosphoenolpyruvate carboxykinase (PEPCK)

Answer 17

• reversal of glycolysis
  
  • Instead of using carbohydrates to produce glucose, our body converts non-carbohydrate sources (like amino acids) in our liver into glucose.
• pyruvate combines with HCO3 to form OAA via pyruvate carboxylase (rxn requires ATP)
  
  • pyruvate carboxylase ​activated by: Acetyl Co-A, begins in mitochondira and uses malate as intermediate to exit the mitochondria
• transforms into PEP (phosphoenolypyruvate (PEP), a 3- carbon molecules with phosphate attached via phosphoenolpyruvate carboxykinase
  
  • ​ PEPCK activated by glucagon & cortisol

Question 18

Pentose Phosphate Pathway:

Key enzyme?

Location:

Products:

Answer 18

• aka hexose monophosphate (HMP) shunt
• Key enzyme: Glucose-6-phosphate dehydrogenase (activated by NADP+ & insulin, and inhibited by NADPH)
• Can use any available molecules of glucose-6-phosphate, whether they are produced by glycolysis or other methods.
• Takes place in the cytosol of the cell, same location as glycolysis.
• Two most important products are ribose-5-P sugar used to make DNA and RNA, and the NADPH molecules which help with building other molecules.

Question 19

Pentose Phosphate Pathway: OXIDATIVE PHASE

How many irreversible steps?

Answer 19

• made up of 2 irreversible steps:

1. G6P oxidized to form lactone. NADPH produces as a byproduct of this reaction as NADP+ is reduced as G6P is oxidized. Another reaction, catalyzed by different enzyme, uses water to form 6-phosphogluconate, the linear product. NADPH often used in reactions that build molecules and occurs in a high concentration in the cell.
2. A carbon is cleaved, and CO2 is released. Once again, the electrons released from this cleavage is used to reduce NADP+ to NADPH. This new 5-carbon molecule is called ribulose-5-phosphate.

Question 20

Pentose Phosphate Pathway: NON-OXIDATIVE PHASE

Answer 20

• reactions are reversible.
• Allows different molecules to enter the PPP in different areas of the non-oxidative phase and be transformed up until the first molecule of the non-oxidative phase (ribulose-5-P).
  
  • Ribulose-5-P is the precursor to the sugar that makes up DNA and RNA, and is also a product of oxidative stage
    
    3. Ribulose-5-P can be converted into two different 5-carbon molecules. One is the sugar used to make up DNA and RNA called, ribose-5-phosphate. Ribulose-5-P isn’t being divided because the carbon count is the same in the next step.
    4. Rest of the cycle depends on the cell’s needs. Ribose-5-P is combined with another molecule of ribose-5-P to make one, 10-carbon atom. Excess ribose-5-P, which may not be needed for nucleotide biosynthesis, is converted into other sugars that can be used by cell for metabolism.
    5. The 3-carbon molecule and 7-carbon molecule, from interconversion in step 4, interconvert again to make a new 4-carbon and 6-carbon molecule. The 4-carbon is a precursor for amino acids, while the 6-carbon molecule can be used in glycolysis. The same reversal of steps in option 4 can happen here as well.

Question 21

SUMMARY OF oxidative/nonoxidative phases, glycolysis

Answer 21

• Oxidative phase:
  
  • -1 H2O
  • +2 NADPH
  • +1 CO2
• Non-oxidative phase:
  
  • Ribose-5-P for DNA/RNA building
  • Net molecular and energetic results of respiration processes
• Glycolysis:
• Net yield of 2 ATP per glucose molecule (4 total made but 2 used)
• Net yield of 2 NADH per glucose

Question 23

Maintenance of a dynamic steady state:

Answer 23

• The body continually degrades and synthesizes proteins in order to keep the organism functioning properly at a steady level.
  
  • e.g. Hemoglobin: body continually and simultaneously degrades older hemoglobin and synthesizes new hemoglobin.
  • This keeps metabolic enzymes fresh and new so there will be very few issues, if any, with aging and degrading enzymes in the pathway

Question 24

Regulation of glycolysis & gluconeogenesis

• what happens if we have a large influx of glucose? large influx of OAA? if we have lots of ATP?
• these are fast acting forms of regulations ^^^*

Answer 24

• Think about Le Châtlier’s Principle: if there’s any change to something in equilibrium, the equilibrium will adjust to counter that change and return the system back to equilibrium
  
  • e.g. if we have large influx of glucose, this promotes glycolysis
  • e.g. if we have large influx of oxaloacetate, the equilibria will be pushed towards the opposite direction towards the production of glucose
  • Allosteric regulation: if we have lots of ATP in the cell, gluconeogenesis would be favored, and glycolysis would be inhibited.

o These are fast-acting forms of regulation

Question 25

Regulation of glycolysis & gluconeogenesis

Slow acting forms of regulations?

Answer 25

• often take advantage of transcriptional changes within the cell.
• Imagine an organism is in a long-term fasting state. It will want to up-regulate the transcription of enzymes that promote things like gluconeogenesis to dump glucose into the blood. It’s implied that the process of going from DNA to mRNA to enzymes will take much longer than simple Le Châtlier or allosteric regulation. This is more of an adaptive process that allows the organism to adapt to more long-term changes that it experiences in its environment.

Question 26

Regulation of glycolysis & gluconeogenesis

• Hormonal Regulations?*
• which promotes glycolysis & which promotes gluconeogenesis?*

Answer 26

• If blood glucose level rises, it stimulates the body to release the hormone insulin, and if the blood glucose levels decrease, it stimulates the body to release the hormone glucagon.
• Thus, insulin promotes glycolysis and glucagon promotes gluconeogenesis.

Question 27

Glycogenolysis:

rate limiting enzyme is_____________, which converts glucose-1-phosphate to glucose-6-phosphate

_____________ is a two-enzyme complex that deconstructs the branches in glycogen that have been exposed by glycogen phosphorylase

Answer 27

• degradation of stored glycogen occurs through action of glycogen phosphorylase and glycogen debranching enzyme.
• Rate limiting enzyme is glycogen phosphorylase, which converts the glucose 1-phosphate to glucose 6-phosphate
  
  • Glycogen phosphorylase breaks α-1,4 glycosidic bonds, releasing glucose 1-phosphate from periphery of granule.
    
    • Cannot break α-1,6 bonds and thus stops when it nears outermost branch points.
  • Enzyme is activated by glucagon in liver, so glucose can be provided for rest of the body. In skeletal muscle, it is activated by AMP and epinephrine, which signal that muscle is active and requires more glucose.
    
    • It is inhibited by ATP.
• Debranching enzyme is a two-enzyme complex that deconstructs the branches in glycogen that have been exposed by glycogen phosphorylase.
  
  • Breaks an α-1,4 bond adjacent to branch point and moves the small oligoglucose chain that is released to the exposed end of the other chain.
  • Forms a new α-1,4 bond. Hydrolyzes the α-1,6 bond, releasing the single residue at the branch point as free glucose. This represents the only free glucose produced directly in glycogenolysis.

Question 28

Glycogenesis: synthesis of glycogen granules

• ________________ is the rate-limiting enzyme
• which enzyme is responsible for introducing α-1,6 linked branches?

Answer 28

• Glycogen synthase is the rate-limiting enzyme of glycogen synthesis and forms the α-1,4 glycosidic bond found in the linear glucose chains of the granule. It is stimulated by glucose 6-phosphate and insulin. It is inhibited by epinephrine and glucagon through a protein kinase cascade that phosphorylates and inactivates the enzyme.
• Branching enzyme is responsible for introducing α-1,6 linked branches into the granule as it grows. Hydrolyzes one of the α-1,4 bonds to release a block of oligoglucose (few glucose molecules bound together in a chain), which is then moved and added in a slightly different location
• 1. Glycogen synthase makes a linear α-1,4 linked polyglucose chain. 2. Branching enzyme hydrolyzes an α-1,4 bond. 3. Branching enzyme transfers oligoglucose unit and attaches it with an α-1,6 bond to create a branch. 4. Glycogen synthase extends both branches.

Question 29

Summary: Regulation of glycogen synthesis and breakdown

• Glycogen phosphorylase in ___________ activated by glucagon in ___________. In skeletal muscle, it is activated by _______ and _________. Inhibited by _____________.
• Glycogen synthase in ____________ activated by ___________ & _______________. Inhibited by ___________&__________through a protein kinase cascade that phosphorylates and inactivates the enzyme.

Answer 29

• glycogenolysis, liver, AMP and epinephrine, ATP
• gllycogenesis, glucose 6-phosphate and insulin, epinephrine and glucagon

Question 30

Citric Acid Cycle:

1. Acetyl Co-A production via:

a. glycolysis
b. aerobic decarboxylation
c. fat metabolism
d. protein metabolism

Answer 30

• a. glycolysis: Pyruvate, the product from glycolysis, is transformed into acetyl CoA in the mitochondria.
  
  • Through the enzyme, pyruvate dehydrogenase, which is a multi-enzyme complex, pyruvate loses a carbon to produce a new, 2-carbon molecule called acetyl- CoA. The carbon that is removed takes two oxygens from pyruvate with it and exits the body as carbon dioxide (CO2). CO2 is the waste product that you release when you exhale.
• b. aerobic decarboxylation: (mitochondrial matrix) = convert pyruvate (3 carbons) to an acetyl group (2 carbons)
  
  • 1 NADH made for every pyruvate. Only occurs in the presence of oxygen. Acetyl group attaches to Coenzyme A to make acetyl-CoA.
• c. fat metabolism: a process called beta-oxidation breaks down the fatty-CoA, 2 carbons at a time, to make acetyl-CoA.
  
  • The acetyl-CoA feeds into the Krebs cycle, and the FADH2 and NADH feed into the ETC.
  • ​β-oxidation produces acetyl-CoA and also FADH2 and NADH.
• d. protein metabolism: The carbon in the amino acid is converted to pyruvate or acetyl-CoA, (or other metabolic intermediates such as oxaloacetate), depending on what amino acid it is

Question 31

Citric Acid Cycle:

2. Reactions of the cycle: substrates & products

Location:

what are the substrates & products?

Answer 31

• Location: matrix of mitochondria
• Acetyl-CoA and oxaloacetate are substrates for citrate synthase.
• Acetyl-CoA is the main substrate. Oxaloacetate can be produced by pyruvate carboxylase that converts pyruvate to oxaloacetate, however this is not as energy efficient.
• Major product of this cycle is oxaloacetate, NADH, FADH2, ATP.

Question 32

Citric Acid Cycle:

3. Regualtion of the cycle:

• major form of regulation is through: _____________
• ______________is major substrate so if the body doesn’t have a lot of it, the speed that NADH and FADH2 is produced will ____________
• _____________, under conditions of high ATP, generally shuttles lots of acetyl-CoA into ____________ for fatty acid synthesis.
• When our body breaks down amino acids and enter the TCA cycle, such as conversion to __________. This occurs when body is starving so it wants to be able to produce more. NADH and FADH2 to produce more ATP.
• The main important enzymes are _____________
• ___________ allosterically inhibits all three of these enzymes because it is a product of the overall TCA cycle. Another inhibitor is ________, since this indicates that our body has produced enough energy.
• Products can also inhibit, such as citrate which performs negatively feedback on _____________&_________ which negatively feedbacks onto α-ketoglutarate dehydrogenase, citrate synthase as well.
• First allosteric activator is __________ since this indicates need for energy. __________for a similar reason. Another is ________.
  
  • Remember that muscle cells require an influx of calcium to contract. This is a way for the body, especially in skeletal muscles, to essentially couple muscle contraction with producing more ATP to meet the needs of those contracting muscles. It has been shown to activate_______________

Answer 32

• through allosteric regulation
• Acetyl-CoA , slow down
• Citrate, cytoplasm
• α-ketoglutarate
• citrate synthase, isocitrate dehydrogenase, and α- ketoglutarate dehydrogenase.
• NADH, ATP
• citrate synthase and succinyl-CoA
• ADP NAD+ , calcium
• isocitrate and α-ketoglutarate dehydrogenase

Question 33

Citric Acid Cycle:

Net molecular & energetic results of respiration processes

Answer 33

• Net yield of 2 ATP per glucose molecule (per 2 acetyl CoA)
• Net yield of 6 NADH and 2 FADH₂
• Net yield of 4 CO2 molecules