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MCB2021F > Carbohydrates & Lipids > Flashcards

Carbohydrates & Lipids Flashcards

1
Q

What are the biochemical roles of Carbs?

A

Energy
Structural support
Protection
Recognition

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2
Q

3 classes of Carbs

A

Monosaccharides
Disaccharides
Polysaccharides

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3
Q

2 functional groups of Carbs

A

Aldehydes

Ketones

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4
Q

Aldehydes vs Ketones

A

Aldehydes: C=O on C1; name will begin with ‘Aldo’
Ketones: C=O not on C1; name will begin with ‘Keto’

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5
Q

Chiral carbon

A
  • C that is bonded to 4 different groups (H2 doesn’t count)
  • Monosaccharides can have more than 1 chiral carbon
  • Aldoses with >=3C and Ketones with >=4C will have a chiral carbon
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6
Q

Highest Chiral carbon

A

Chiral carbon that is furthest from the functional group

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7
Q

D & L isomers

A
  • Monosaccharide is a D isomer if OH group on highest chiral carbon is on the right
  • Monosaccharide is a L isomer if OH group is on the left of the highest chiral carbon
  • number of stereoisomers = 2^n when n=no. chiral centers
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8
Q

Enantiomer

A

sugars that are mirror images of each other

- e.g. D&L isomers

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9
Q

Diastereoisomers

A

Same formula & number/type of atoms but different structure and they not mirror images of each other

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10
Q

Epimers

A

Diastereoisomers which differ at only 1 chiral center

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11
Q

Hemiacetal

A

Cyclic form of carb formed from reaction of alcohol with aldehyde

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12
Q

Hemiketal

A

Cylic form of carb formed from reaction of alcohol with ketone

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13
Q

Furanose

A

4 Carbon ring structure

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14
Q

Pyranose

A

5 Carbon ring structure

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15
Q

Haworth Projection

A

Graphical representation of a cyclic monosaccharide

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16
Q

How to draw a cyclic structure (Fischer to Pyranose)

A
  1. Number the carbons on the Fischer projection
  2. Rotate Fischer projection 90 degrees clockwise (till its on its side)
  3. Perform bond rotation on C5 (C with OH group that reacts with carbonyl group
  4. Close the ring:
    a. C5-OH attacks C1, breaking C=O bond and forming C1-O
    b. C1-O bond is protonated = C1-OH
    c. All bonds facing down go at the bottom of the ring (OH of C2, C4). All bonds facing up go above the ring
    d. Exception is C1 can be alpha or beta configured
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17
Q

Anomeric Carbon

A

C atoms that carries the carbonyl group (usually C1)

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18
Q

Anomers

A

Isomers of monosaccharides that differ only in their configuration about the anomeric carbon atom

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19
Q

alpha/beta-anomers

A

alpha-anomer: if OH at C1 points in opp direction to CH2OH (i.e. if its below the ring)
beta-anomer: If OH at C1 points in same direction as CH2OH (i.e. above the ring)

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20
Q

Why does the beta-anomer predominate in monossacharides?

A
  • Pyranoses preferential take on a chair configuration
  • in Chair config, groups that point below pyranose ring are axial, groups above ring are equatorial
  • equatorial substituents are lower in energy & therefore more stable
  • B-anomer has OH on anomeric carbon in equatorial config so its more stable than the a-anomer configuration
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21
Q

Biologicaly important derivates of simple sugars

A 
sugar acids
sugar alcohols
deoxy sugars
sugar phosphates
amine derivates
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22
Q

Sugar acids

A
  • Aldoses (ketones can’t be oxidised)
  • Oxidation at aldehyde and/or OH can yield 3 types of sugar acids:
    1. Gluconic acid - oxidation of C=O at C1
    2. Glucuronic acid - oxidation of OH at C6
    3. Glucaric acid - oxidation at C1 AND C6
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23
Q

Sugar alcohols

A

Aldehyde or ketone group is reduced from C=O to C-OH

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24
Q

Deoxy sugars

A

1 or more O atoms in hydroxyl group is removed leaving just an H

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25
Q

Sugar Phosphates

A

Glucose molecule is attached to phosphate group

26
Q

Amino Sugars

A

1 or more OH groups replaced by amino group

27
Q

Reducing Sugar

A

Sugar that reduces another compound and is itself oxidised (have to have an anomeric carbon)

28
Q

How to locate the anomeric carbon

A
  1. locate O inside the ring
  2. Look at C on either side of O. one will be attached to CH2OH group, this isn’t the anomeric carbon, the other one is
  3. Other C is bonded to OH, that’s the anomeric C. When it opens to linear it will revert to carbonyl group and be able to react
29
Q

Benedicts Test

A

Benedicts reagent contains a copper compound. The reducing sugar (carbonyl group) is oxidized to a carboxylic acid by the cupric ions (Cu2+) reducing CU2+ to cuprous (CU+), which then forms copper oxide (Cu2O) and precipitates out of solution as a brick-red colored compound. This change in color of the blue Benedict’s reagent is indicative of a reducing sugar.

30
Q

Maltose

A
  • Dissacharide
  • 2 mono’s linked by O-glycosidic bond in dehydration reaction
  • a-glucose bonded to a/b-glucose
  • Joined at C1 and (C4 of 2nd glucose)
31
Q

Maltase

A

Enzyme that breaks down Maltose

32
Q

Lactose

A
  • b-galactose bonded to a/b-glucose by dehydration reaction
33
Q

Lactase

A

Enzyme that digests Lactose

34
Q

Sucrose

A
  • a-glucose bonded to b-fructose by dehydration reaction

- C1 of glucose bonds to C2 of fructose

35
Q

Sucrase

A

Enzyme that digests sucrose

36
Q

O-glycosidic Bonds

A
  • Anomeric Carbon of 1 sugar with OH of another sugar
  • naming keeps the a/b configuration of C1 of the first sugar
  • a/b(1-4(Or 2 in sucrose)) glycosidic bond
37
Q

Reducing Dissaccharides

A
  • Maltose
  • the reducing end is the mol with a free anomeric carbon (with OH) that isn’t involved in a glycosidic bond.
  • Lactose
38
Q

Non-reducing disaccharides

A

Sucrose

39
Q

Functions of Polysaccharides

A
  • energy storage
  • Structure & protection
  • Recognition
40
Q

Starch

A
  • Energy storage in plants
  • Amylose & Amylopectin are 2 forms
  • Amylose & Amylopectin for helical structures (a(1-4) linkages are naturally angled in chair config, resulting in bends in polymer hence helical structure)
41
Q

Amylose

A
  • linear chain of monomers
  • glucose units (homopolysaccharide)
  • a(1-4) glycosidic bonds
  • 1 reducing end
42
Q

Amylopectin

A
  • Linear chain with branches every 12-30 residues
  • 70-90% of all starch
  • glucose units
  • many non-reducing ends, 1 reducing end
  • a(1-4) glycosidic bonds BUT a(1-6) glycosidic bonds where branches occur
  • Purpose of branching = non-reducing ends provide regions where molecules can be hydroluzed
43
Q

Glycogen

A
  • energy storage in animals and humans
  • glucose units linked by a(1-4) glycosidic bonds with a(1-6) branches every 8-12 residues
  • hydrolysis by a&b-amylases and glycogen phosphorylase
44
Q

Structural polysaccharides

A
  • Cellulose

- CHitin

45
Q

Glycosaminoglycans (GAG)

A

Linear polysaccharides with repeating disaccharide units (acetylated amino sugars + negatively charged group/s) – acidic

46
Q

Peptidoglycan

A
  • Polysaccharide of bacterial cells walls
  • In Gram+ & Gram- bacteria.
  • Thick in Gram+ and surrounds the membrane.
  • Thin in Gram- and sandwiched between plasma and outer membrane.
  • NAM (N-acetyl-muramic acid) & NEG (N-acetly-glucosamine)in b(1,4) glycosidic linkage
47
Q

Glycoproteins vs Proteoglycans

A

Glycoproteins:

  • more protein than carb
  • short glycol chains
  • branched
  • no repeating disaccharides

Proteoglycans:

  • More carb than protein
  • Long glycan chain
  • linear
  • repeating disaccharide units
48
Q

Glycosylation

A
  • formation of glycoproteins
    1. O-linked glycosylation:
  • glycan chain (carb) attached to O group or serine/threonine residues of protein
  1. N-linked glycosylation:
    - glycan chain attached to amide nitrogen of asparagine residues within the motif
    Asp-X-Ser/Thr [X = any amino residue except Proline. MUST be followed by Ser/Thr]
49
Q

Glycoprotein Functions

A
  1. Give rise to blood groups
  2. All immune System antibodies are glycoproteins
  3. Mucin in respiratory tracts
  4. Anti-freeze glycoproteins in fish in the arctic
  5. Cell signalling
50
Q

Proteoglycans structure

A

Core protein attached to linear glycan chains either by O-linked or N-linked glycosylation.

51
Q

Proteoglycans in the ECM of cartilage

A
  • Cartilage = soft tissue between bones and joints. Comprised of collagen and proteoglycan called aggrecan.
  • Aggrecan is negatively charged (from GAG) – associates strongly with H20. Acts as a lubricant and ‘shock-absorber’.
52
Q

Structure & Chemistry of Fatty Acids

A
  • Carboxylic acid group is the head of the molecule.

- Technically amphipathic molecules (polar head, non-polar tail)

53
Q

Melting temp in Fatty Acids

A
  • length & degree of saturation determines melting temperature
  • higher length has higher melting temp because there are more C’s that need to be broken
  • Unsaturated has lower melting temp than saturated
54
Q

Location of double bonds in Fatty Acids (naming)

A

∆ - energy change because double bonds take energy to break.
i.e. 18:2∆9,12 or 18:2(9,12)

55
Q

Effect of Unsaturated Fatty Acids on lipid fluidity

A

o Double bond produces bend in hydrocarbon tail.

o Produces flexible, fluid aggregates and weaker bonds (easier for heat to access)

56
Q

Essential fatty Acids

A

polyunsaturated fatty acids that mammals cannot synthesise

57
Q

Elongation of Fatty Acids

A
  • palmitic acid is primary product of FA bio-synthesis (synthesis usually stops at 16C) BUT eukaryotic cells generate longer fatty acids

o These elongation reactions add 2 C units sequentially to carboxyl ends of both saturated and unsaturated FA substrates. Always even number of C’s.
o In mitochondria and ER, where source of Carbon is acetyl-CoA or malonyl CO-A, respectively.

58
Q

Triacylglycerols (TAG)

A
  • 3 fatty acid chains joined to glycerol molecule from dehydration reaction
  • can be simple: same type of fatty acid
  • or can be mixed: 3 different fatty acids
  • can be saturated or unsaturated
59
Q

Triacylglyerol Location, Importance & Function

A
  • In adipocytes – pH stability (keeps away from higher pH of cytoplasm)
  • Very efficient energy source
    o 1g of protein/carbs ->17 KJ energy but 1g TG’s -> 38 KJ energy (From structure of C-H bonds)
    o Hydrophobic and anhydrous (functions without water) in contrast to hydrophilic proteins and carbs. This means that protein and carbs need an equal mass of water to hydrate them.
    o Metabolic water can be produced via FA oxidation (yields CO2 and H2O in TCA cycle).
  • Provide insulation. Don’t conduct heat.
60
Q

Saponification

A
  • Reaction between TGs and NaOH or KOH yields free FA salts and glycerol.
    o If NaOH is used = hard soap
    o If KOH yield = soft soaps.
  • The FA salt is an emulsifying agent and surfactant (amphiphilic).
  • Hydrophilic head attracted to water – hydrophobic tail attracted to fat.
  • In aqueous solution the FA salt self-assembles to form a micelle.
  • Grease molecules are secured by the lipophilic tails of the soap molecules crowding in the center of the micelle.

MCB2021F (5 decks)

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