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BB10235 > Carbohydrates > Flashcards

Carbohydrates Flashcards

1
Q

Why are carbohydrates important

A
  1. Main source of energy
  2. Structural components (animals, plants, bacteria)
  3. Component of nucleic acids
  4. Contribute to protein structure and function
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2
Q

What are peptidoglycans

A
  1. carbohydrates are involved in cross-linking peptide chains which forms a mesh to protect the cell (prokaryotes - bacteria)
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3
Q

What are proteoglycans

A
  1. form the extracellular matrix in which all our cells are embedded, acts as a molecular sponge for water (eukaryotes - humans)
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4
Q

Why is glucose important

A
  1. major source of metabolic energy
  2. Primary product of photosynthesis
  3. Building block - it is the primary monosaccharide so can form disaccharides and polysaccharides
  4. Makes cellulose, starch and glycogen
  5. Humans can’t digest cellulose as contains B 1-4 glycosidic linkages
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5
Q

Describe Cellulose

A
  1. Source- plant
  2. Subunit- B-glucose
  3. Bonds- 1-4
  4. Branches- no
  5. Each alternate glucose is inverted
  6. Separate strands which form hydrogen bonds between each other
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6
Q

Describe amylose

A
  1. Source- plant
  2. Sub-unit- a-glucose
  3. Bonds 1-4
  4. Branches- no
  5. Glucose forms straight chains- all facing same direction
  6. Coiled structure
  7. forms starch with amylopectin
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7
Q

Describe amylopectin

A
  1. Source- plant
  2. Subunit- a-glucose
  3. bonds- 1-4 and 1-6
  4. Branches- yes
  5. Forms straight chain and branched chains
  6. Forms starch with amylose
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8
Q

Describe glycogen

A
  1. Source- animal
  2. Subunit- a-glucose
  3. Bonds- 1-4 and 1-6
  4. Branches- yes (more than amylopectin)
  5. Very branched structure produced
  6. Main energy store in animals- accumulates in liver and muscles to be broken down when energy needed
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9
Q

Describe sucrose

A
  1. Sucrose is a disaccharide of glucose and fructose
  2. Sweet taste is due to the fructose
  3. Sucrose is the main sugar we eat
  4. Glucose is absorbed by active transporters, therefore there is a maximum rate of glucose absorption when all active transporters are saturated
  5. Fructose is taken up by a different mechanism
  6. More energy is absorbed per unit time when eating sucrose as opposed to starchy foods because sucrose is hydrolysed into fructose and glucose which are then absorbed separately
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10
Q

What are the two types of saccharides

A
  1. Aldose - in the aldehyde state

2. Ketose - in the ketone state

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11
Q

Describe D-glyceraldehyde

A
  1. D-glyceraldehyde (an aldose)
  2. Has one chiral centre
  3. So has two optically isomeric forms (D-glyceraldehyde and L-glyceraldehyde)
  4. Can tell the difference between each enantiomer using optical rotation (plane polarised light)
  5. Glyceraldehyde is the molecule used in the fisher convention
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12
Q

What is the Le Bel-van’t Hoff rule

A
  1. If a molecule has n chiral centres it has 2^n optical isomers
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13
Q

Describe the structure of aldo-tetroses

A
  1. 2 chiral centres so 4 optical stereoisomers
  2. Monosaccharide with 4 carbon atoms. They have either an aldehyde functional group in position 1 or a ketone functional group in position 2.
  3. D (+) Erythrose D (-) threose L(+) erythrose L(-) threose
  4. ‘Erythro’ means substituents on the same side in the Fisher projection
  5. ‘Threo’ means substituents are on opposite sides in the Fisher projection
  6. Enantiomers are named due to the handedness of the most distal chiral centre from the carbonyl group
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14
Q

What is the difference between hemiacetals and hemiketals

A
  1. Hemiacetals- alcohols react with aldehyde

2. Hemiketals- Alcohols react with ketone

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15
Q

How is a ring formed

A
  1. A linear molecule forms a cyclic hemiacetal/hemiketal by the reaction of the aldehyde/ketone group and the alcohol group on carbon 5
  2. React intramolecularly
  3. There are two ways of forming the ring which means that 2 alternate anomers are formed (different conformations of the same molecule)
  4. These ring structures are not as strong as esters meaning they can form and unform (ring formation is reversible)
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16
Q

What type of ring is glucose

A
  1. Hemiacetal = compound formed when an aldehyde reacts with an alcohol
  2. Hemiacetal is less chemically stable than an ester
17
Q

What type of ring is fructose

A
  1. Hemiketal = compound formed when a ketone reacts with an alcohol
  2. Can only form a furanose ring (5C) because the carbonyl group is on carbon 2
  3. Also has beta and alpha anomers
18
Q

What two types of ring can glucose form

A
  1. A six carbon ring (pyranose)

2. A five carbon ring (furanose)

19
Q

Why is each ring either beta or alpha

A
  1. The cyclisation of a monosaccharide results in a pair of diastereomers called anomers
  2. The hemiacetal or hemiketal carbon is referred to as the anomeric carbon
  3. This changes the 3D structure the rings adopt (several 3D forms e.g. boat, chair)
  4. beta anomers are more stable
  5. In free solution of glucose there are more pyranoses than furanoses
20
Q

How do you name a ring beta or alpha

A
  1. Haworth projection
  2. OH group on alpha anomers points down (carbon 1)
  3. OH group on beta anomers points up (carbon 1)
21
Q

What are epimers

A
  1. An epimer = a chirally opposite centre
  2. Galactose is the C4 epimer of glucose
  3. Mannose is the C2 epimer of glucose
  4. Differ only by the configuration about one C atom
22
Q

What determines stability in

A
  1. The repulsion between OH groups determines the stability of the anomers- hinderence etc
  2. The stability of anomers can be measured
  3. Start with a pure form
  4. Anomers will interconvert over a relatively short time period
  5. Measure optical rotation over time to see which anomer is favoured
23
Q

Which 3D structure is the most stable

A
  1. Chair form is the most stable- and the chair form which usually has less crowding among axial substituents is most stable
  2. It can also flip into boat form but more sterically crowded
  3. There are many other different conformers too
24
Q

Describe the stability of phosphate groups in ATP

A
  1. Kinetically stable: doesn’t spontaneously break down - implies a high activation energy.
  2. Thermodynamically unstable: because the phosphate bonds are high in energy
25
Q

Describe the hydrolysation of ATP

A
  1. ATP + H2O –> ADP + HPO4
  2. The negative charge on the phosphate anion can be distributed around all 4 oxygens which gives it more stability than when it is part of ATP
  3. ΔG is driven by entropic stabilisation of phosphate anion.
  4. Lots of energy is released
  5. ATP must be made constantly to maintain metabolic environment in the cell

BB10235 (16 decks)

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