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Alevel Biology > Biomolecules > Flashcards

Biomolecules Flashcards

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1
Q

What’s the difference between an aldo group and keto group?

A

Aldo: C=O at beginning of C skeleton
Keto: C=O within C skeleton

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2
Q

What’s the difference between alpha and beta glucose?

A

α-glucose: -OH grp attached to C1 below plane of the ring

β-glucose: -OH grp attached to C1 above plane of the ring

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3
Q

Describe the structure and properties of glucose.

A
  1. Small size + many polar hydroxyl grps (-OH) which forms H bonds w water⇒ readily soluble in water (easily transported)
  2. Linear forms has a free carbonyl grp (C=O) ⇒all reducing sugars
  3. Pentose and hexose exists as rings⇒ rings are stable building blocks for larger mlcs (not impt pt)
  4. Ring struc exhibit α & β isomerism⇒ ↑ diversity of
    monosaccharides, building blocks for diff mlcs
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4
Q

Describe how maltose –> glucose + glucose

A
  1. maltose–> 2 α glucose
  2. maltase
  3. α(1-4) glycosidic bond
  4. hydrolysis
    • H2O
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5
Q

Describe the procedure for the Benedict’s test.

A

Equal vol of sample + benedict’s solution
Shake
Boiling water bath for 3-4 min

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6
Q

Describe the procedure for the test for non-reducing sugars.

A
  1. -ve benedict’s test
  2. Boil equal vol of test soln + dilute HCl for abt 1 min→ hydrolyses disaccharide to monosaccharide
  3. Cool contents of tube
  4. Neutralise acidic content w sodium bicarbonate soln
  5. Carry out Benedict’s test for reducing sugar
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7
Q

Compare the structure and function of starch (amylose, amylopectin), glycogen and cellulose: monomers + no. of units

A
  1. (all) Large mlc–> insoluble, won’t affect wp of cell
  2. starch and glycogen made up of a-glucose, while cellulose is made up of b-glucose
  3. starch and and glycogen is made up of thousands of glucose mlcs–> large energy store–> hydrolysed ultimately to glucose for aerobic respiration for ATP
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8
Q

Compare the structure and function of starch (amylose, amylopectin), glycogen and cellulose: bond

A

α(1-4) glycosidic bonds: allow α-glucose monomers to be packed into a helical coil within a branch
(amylopectin & glycogen) α(1-6) glycosidic bonds: links the α-glucose monomer at branch points→ branched helical structure
Bonds are all between α-glucose
- Enzymes that hydrolyse these bonds commonly available. Glucose readily released for respiration
- helical–> compact molecule, more glucose units packed per unit volume

Cellulose has 𝛽(1-4) glycosidic bonds between 𝛽-glucose–> Cellulase found in v few organisms → cellulose X hydrolysed and used as a respiratory substrate

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9
Q

Compare the structure and function of starch (amylose, amylopectin), glycogen and cellulose: monomer orientation & shape

A

All with same orientation.
Helical coil: α(1-4) glycosidic bonds, each residue bent in one direction w respect to the previous residue
- Intramolecular bonding by OH grps & OH grps project into core of helix formation→ relatively fewer OH groups available H bonding with water, insoluble in water & will not affect wp
- compact mlc, pack many glucose units per unit volume for storage

Alternate 𝛽-glucose monomers rotated 180° wrt each other→ forms long, linear, unbranched mlc w OH grps projecting out in both directions, which can form intermolecular H bond w OH grps of other adjacent cellulose mlcs lying // to it→ form microfibrils with high tensile strength

Only sf of microfibril exposed to water + most OH grps involved in interchain H bonding w other OH grps→ relatively fewer OH grps available for H bonding w water + large molecule
- insoluble

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10
Q

Compare the structure and function of starch (amylose, amylopectin), glycogen and cellulose: branching

A

amylose: unbranched
amylopectin: branched
glycogen: extensively branched
- compact
- more branch ends to allow multiple hydrolytic enzymes to work on at the same time→ more glucose mlcs released rapidly at the same time→ ↑ ATP generation by respiration per unit time

cellulose: unbranched

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11
Q

Why is cellulose synthesised in the cell sf membrane?

A
  1. Cellulose is a macromolecule (too large) + insoluble in hydrophobic core of p.lipid bilayer→ can’t pass through cell sf membrane if it needs to be transported to the cell exterior
  2. Cellulose is a macromolecule found outside the cell as part of the cell wall→ easier to just deposit it there (closer proximity)
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12
Q

Mesh work of criss-crossing microfibrils form cellulose cell wall. What properties of the cell wall does this result in?

A

Porous struc: gaps between microfibrils→ freely permeable to water and solutes; allows free movement of substances in and out of the cell

Strong & rigid substance: meshwork distributes stresses in all directions→ enclose plant cells, protecting them from physical damage and bursting due to osmotic stress

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13
Q

What’s the difference between saturated and unsaturated fatty acids?

A
  • saturated: only C-C single bonds & C-H bonds; linear chain

- unsaturated: 1 or more C=C cis double bonds → kink in HC chain

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14
Q

Describe the formation of an ester bond.

A
  • between -OH and -COOH

- condensation rxn, one water mlc is removed for each fatty acid joined to the glycerol

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15
Q

Describe the structure and property of triglycerides.

A

Structure: 3 long, non-polar, hydrophobic HC chains/ fatty acid tails joined to a glycerol backbone via ester linkages
Property: non-polar

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16
Q

Describe the role of triglycerides in living organisms

A
  1. Suitable for energy storage
    - Cannot form H bonds w water→ insoluble, does not affect wp & won’t be easily transported out of the cell
    - For an equivalent mass of carbs, triglycerides release x2 as much energy→ compact energy store
    > Less oxidised than carb→ yield more energy upon oxidation
    > Higher proportion of C and H atoms & C-H bonds, which releases energy in the form of ATP during oxidation→ undergo more oxidation events (need more O2 per gram of for oxidation)
  2. Highly reduced, produce more metabolic water per unit mass (compared to carbs) when oxidised during respiration
  3. Protective layer around delicate internal organs; act as shock absorber, protect from mechanical damage
  4. Thermal insulation by fat beneath layer of skin, for mammals in cooler climates
  5. Less dense than water→ Improve buoyancy in marine mammals
  6. Reservoir for storage of fat soluble vitamins eg A, D
17
Q

Describe the structure of phospholipids.

A

glycerol backbone + 2 long, non-polar, hydrophobic HC tails via ester linkages + 1 -vely charged phosphate grp via phosphoester linkage. Additional small mlcs, e.g. nitrogen containing choline, may be present

18
Q

Describe the ethanol emulsion test for solid and liquid samples.

A

Solution: 2cm3 ethanol + 2 drops sample–> mix & stand for 2 min–> decant ethanol into 2cm3 water

Solid: 2cm3 ethanol + grounded sample–> mix well to dissolve any lipids if its present. Stand for 2 min–> Decant ethanol into 2cm3 of water

19
Q

Describe the structure of amino acids.

A

α-carbon atom bonded to 4 groups→ H atom, amino group (NH2), carboxyl group (COOH), variable R group

20
Q

Describe properties of amino acids

A
  1. Classified according to their R groups as hydrophobic/non-polar or hydrophilic
  2. Exist as zwitterions: carry both +ve and -ve charges
    - NH2 receives H+ → +vely charged NH3+
    - COOH dissociates, releasing H+ → -vely charged -COO–
  3. Act as buffers, due to being amphoteric: can donate or accept H+ to act as acid or base
    - Basic: NH3+ loses H+, which neutralises OH– to form H2O
    - Acidic: COO– accept H+ & becomes COOH
    - OH- and H+ removed from solution→ minimise changes in pH of surrounding when a small amt of acid or alkali is added; buffer
21
Q

Describe the formation of a peptide bond

A

condensation, remove water mlc

between aa

22
Q

Define: primary structure

A

no. and sequence of amino acids in a single polypeptide chain

23
Q

What bonds maintain primary struc?

A

peptide bonds between successive aa

24
Q

Define: secondary structure

A

regular coiling or pleating of a single polypeptide chain

25
Q

What bonds maintain secondary structure?

A

H bonds between C=O and N-H groups of polypeptide backbone

26
Q

Compare structure of a-helix and b-pleated sheet.

A

Coiled/Spiral shape vs flat-sheet like structure

H bonds formed between C=O group of one aa residue and N-H group of another aa residue:
four aa away VS between adj segments lying side by side of a single polypeptide chain

Coil occurs only in a single direction vs aa in one segment run in same direction or opposite direction relative to another segment

27
Q

Define tertiary structure

A

further extensive folding & bending of a single polypeptide chain, usually forming a compact, globular molecule, giving rise to specific conformation of protein

28
Q

Define quarternary structure

A

association of 2 or more polypeptide chains/subunits into 1 functional protein molecule

29
Q

Describe the 4 interactions that hold tertiary/quaternary structure tgt

A

H bonds between polar R groups w OH, C=O or N-H

ionic bonds: between opp charged R grps of aa

disulfide bonds: between 2 R grps w S (eg cysteine) by oxidation of sulfydryl (-SH) grps

hydrophobic interactions: between non-polar, hydrophobic R grps

30
Q

Compare fibrous and globular proteins.

A

Shape: polypeptides folding/forming long, straight fibres vs roughly spherical shape

Solubility: insoluble (large, extensive H bonds alrdy between aa) vs soluble (hydrophilic R grps/aa on exterior of proteins, form H bond w water)

Variation:

  • less/more variety of aa used to construct protein–> has/dh regular structure w regular repetitive aa seq
  • length of polypeptide and seq of aa varies/identical between 2 samples of the same protein–> protein still functional/else it won’t be functional

Function: structural vs metabolic role

31
Q

Describe the structure and function of haemoglobin.

A
  1. Globular in shape
  2. quarternary struc w 4 subunits: 2 a-globin and 2 b-globin subunits. each subunit has a haem group w Fe2+ that binds reversibly with O2 ==> 1 Hb mlc carries up to 4 O2 at the same time
  3. weak intermlc interactions between R grps of subunits that are easily broken==> Allows movement of subunits wrt each other, change in position that influences affinity for O2→ allow cooperative binding of O2: binding of 1 O2 to 1 subunit→ conformation change in remaining 3 subunits so their affinity for O2 ↑
  4. Each subunit arranged such that aa with hydrophilic R groups found on the exterior of the protein; aa with non-polar, hydrophobic R groups buried in interior of protein, away from aq surrounding==> soluble, transported in blood
32
Q

Describe the structure and function of collagen.

A
  1. 3 helical polypeptide chains wound around each other like a rope to form 1 tropocollagen molecule⇒ high tensile strength
  2. Each chain usually contains a repeating tripeptide seq: glycine-X-Y, where X is usually proline, Y is usually hydroxyproline
    - bulky, inflexible proline and hydroxyproline ⇒ rigidity of molecule
    - Glycine, smallest aa, every third residue→ fit into restricted space in center of triple helix→ form very tight, compact triple helical structure⇒ high tensile strength
  3. Extensive/numerous H bonds between aa residues of adjacent polypeptide chains ⇒high tensile strength; interaction w water molecules are limited→ insoluble
  4. Staggered arrangement of collagen mlcs within fibril (longitudinal displacement) ⇒ minimises point of weakness along length of fibrils
  5. Covalent cross-links between lysine residues at C and N ends of adjacent tropocollagen mlcs to form fibrils ⇒ greatly increase tensile strength
  6. Many collagen fibrils lie in parallel, bundle to form collagen fibres ⇒ high tensile strength; large, insoluble
33
Q

Describe the biuret test for proteins

A

2cm3 test soln + equal vol 5% KOH soln
Shake
Add 2 drops of 1% copper(II) sulphate soln, shaking well after each drop

34
Q

What is denaturation?

A

Loss of 3D conf of protein mlc, due to disruption of interactions that maintain secondary or higher lvl of structure –> lose function

Alevel Biology (16 decks)

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