Biological Molecules

Question 1

Water

Answer 1

• polar molecule
• water molecules are attracted to each other and form ‘hydrogen bonds’
• high specific heat capacity
• thermal E. weakens H bonds instead of increasing particles’ eK
• a buffer against rapid temp. changes which is good for enzymes

Question 2

Water 2

Answer 2

• high latent heat of vaporisation
• ^ allows organisms to cool without losing a large vol. of water e.g. sweating
• a good solvent as it transports dissolved substances through xylem
• water in blood plasma is vital in transferring heat around the body
• the ‘universal solvent’ due to its polarity

Question 3

Water 3

Answer 3

• contains dissolved oxygen for aquatic organisms to carry out respiration
• cohesion causes surface tension which also acts as a habitat for insects
• cohesion allows long columns of water to travel in xylem tubes
• less dense when a solid (ice)
• reactant/product in metabolic reactions
  e.g. photosynthesis/aerobic respiration
  hydrolysis/condensation

Question 4

Carbohydrates - Monosaccharides

Answer 4

• examples; glucose, galactose & fructose
• soluble in water due to number of hydroxyl groups (OH)
• ^ hydrophilic molecules
• ‘ribose’ = pentose sugar
  ‘glucose’ = hexose sugar
• αlpha & βeta are ‘isomers’ of glucose
• C1 hydroxyl BELOW = ALPHA
  C1 hydroxyl ABOVE = BETA

Question 5

Disaccharides

Answer 5

• examples; maltose, sucrose & lactose
• produces a water molecule as H atom from one mono. bonds with hydroxyl group from the other (condensation)
• glucose molecules are chemically bonded by glycosidic bond
• maltose forms a 1, 4 glycosidic bond
• adding water breaks glycosidic bond (hydrolysis)
• sucrose = glucose + fructose
  lactose = glucose + galactose

Question 6

Polysaccharides

Answer 6

• glucose = store of chemical energy
• glucose is hydrophilic so it causes water to move into cell if cell contains a lot of it by osmosis
• ^ why plants store glucose as starch
• starch = amylose + amylopectin
• polymers are too large to diffuse through plasma membrane
• enzymes are used to break glycosidic bones in starch

Question 7

Amylose & Amylopectin

Answer 7

• amylose = thousands of α glucose molecules joined by 1, 4 glyco. bonds
• amylose twists into a compact helix with H bonds forming between neighbouring chains
• hydrolysis takes place to release glucose if cell needs it
• amylopectin branches after 25-30 glucose molecules
• branch forms 1, 6 glycosidic bond joined to main chain
• enzymes are at ends of branches & work rapidly due to excessive amount

Question 8

Glycogen

Answer 8

• the glucose storage molecule in animals (liver + muscle)
• insoluble; so cannot diffuse out of cell
• similar to amylopectin, however, it’s more branched it’s so more compact
• many free ends allow enzymes to convert glycogen back to glucose rapidly
• ^ good for animals when they rapidly require energy for respiration e.g. predator chase

Question 9

Cellulose

Answer 9

• polymer of beta glucose
• unbranched polysaccharide
• forms straight chains
• hydroxyl groups points in different directions when side-by-side
• every 2nd β glucose molecule flips to form 1, 4 glycosidic bonds
• H bonds between neighbouring chains give cellulose strength

Question 10

Cellulose 2

Answer 10

• cellulose chains grouped together are called microfibril
• many microfibril = macrofibril
• many macrofibril = cellulose fibre
• c. fibres form plant cell wall

Question 11

3.5 Lipids

Answer 11

• molecules in fats/oils
• major source of energy in human diet
• stores energy e.g. ‘adipose tissue’ under skin for insulation & around internal organs for protection
• used for waterproofing and membrane structure

Question 12

Triglycerides

Answer 12

• non-polar/hydrophobic
• 1 glycerol molecule + 3 fatty acids (3x)
• fatty acid = carboxylic group bonded to hyrdrocarbons
• saturated fatty acids contain single covalent bonds between carbon (C)
• unsaturated fatty acids contain at least one double covalent bond between C
• polyunsaturated = more than one double carbon bond

Question 13

Triglycerides 2

Answer 13

• glycerol = 3 hydroxyl groups + carbon + hydrogen (at ends of carbon)
• glycerol + fatty acid = ester bond (esterification aka. condensation)
• lipase needs 3 water molecules to break ester bonds (hydrolysis)
• a lot of energy can be released from triglycerides

Question 14

Phospolipids

Answer 14

• polar/hydrophilic
• 1 glycerol molecule bonded to 2 fatty acids + phosphate (negative)
• heads are hydrophilic
  tails are hydrophobic
• ^ these create a phospholipid bilayer

Question 15

Cholesterol

Answer 15

• sterols are not fats/oils but complex alcohol molecules
• has a hydrophilic hydroxyl group which interacts with phospholipids heads
• rest of molecule is hydrophobic which interacts with fatty acid tail
• helps control fluidity of plasma membrane
• produces bile in liver
• makes vitamin D & steroid hormones
• ^ these hormones can pass through plasma membranes

Question 16

3.6 Protein Structure
Amino Acids

Answer 16

• (amino group (NH₂)) + “HCRgroup” + carboxyl group) = general amino acid
• Rgroup differentiates amino acids
• dipeptide = 2 amino acids chemically bonded (peptide bond)
• ^ condensation reaction that takes place in ribosomes
• polypeptide = 3 or more amino acids
• polypeptides can be hydrolysed
• a polyP has to be folded into a 3D shape to become a functional protein
• proteins usually contain different polyP

Question 17

Primary Structure

Answer 17

• specific order of amino acids in a polyP
• structure is determined by DNA sequence of gene that encodes polyP
• changing one amino acid can affect a protein’s final structure and function

Question 18

Secondary Structure

Answer 18

• H atoms in amino group bond with O atoms in carboxyl group
• H bonds form between amino acids all along polyP chain causing it to twist/fold
• 2 types called ‘alpha helix’ & ‘beta sheet’
• H bonds hold structure in place
• type of structure formed depends on primary structure

Question 19

Tertiary Structure

Answer 19

• further folding of polyP due to bonds forming between R groups of a. acids
• overall 3D shape of polyP shape
• contains both alpha helix & beta sheet
• critical for protein function
• specific bonding depends on R group of amino acids

Question 20

Quaternary Structure

Answer 20

• multiple polypeptide chains
• e.g. haemoglobin (4 polyP chains)
• haemoglobin has 2 subunits that are arranged to form the Q structure
• ‘prosthetic groups’ = non-proteins that bind to protein to help carry out its role
• ^ forms part of the structure
• ‘haem’ is the prosthetic in haemoglobin
• each subunit has 1 haem
• ^ these are called ‘conjugated proteins’

Question 21

Types of Bonding

Answer 21

1. Hydrogen bonding
   - occurs when a polypeptide has 2 amino acids with a hydroxyl group
   - bonds are easily broken by high temperatures or pH changes
2. Hydrophobic/Hydrophilic interactions
   - amino acids with hydrophobic R groups cluster together to avoid water molecules (vice versa)
   - bonds are weak

Question 22

Types of Bonding 2

Answer 22

3. Ionic bonding
   - occurs between amino acids with charged R groups
   - stronger than H bonds but are broken by pH changes
4. Disulphide bridges
   - occurs between cysteine R groups
   - cysteine contains sulphur atoms which form a covalent bond
   - strongest bonds; not broken by temp/pH chages

Question 23

Globular Proteins
(Haemoglobin)

Answer 23

• have amino acids with hydrophilic R groups on its surface so they’re soluble
• compact & spherical
• have functional purposes
• forms during the tertiary structure
• e.g. insulin, haemoglobin & catalase
• each ‘haem’ group in haemoglobin contains a FE 2+ ion
• ^ oxygen binds to this ion which slightly changes its quaternary structure

Question 24

Insulin

Answer 24

• regulates blood glucose concentration
• carries out function by binding to receptor molecules found on plasma membrane of target cells
• precise globular shape allows it to fit into its receptor
• PP chains are linked by disulfide bonds
• soluble with no subunits

Question 25

Catalase

Answer 25

- contains a ‘haem’ prosthetic group in each of its 4 subunits - FE 2+ ions in heam group allow catalase to interact with hydrogen peroxide - ^ this speeds up its break down as hydrogen peroxide is damaging to cells - h. perox. is a by-product of metabolism

Question 26

Fibrous Proteins

Answer 26

- have large proportions of a. acids with H. phobic R groups so they’re insoluble - have structural purposes - not folded so not compact - long/narrow shape - unreactive; less sensitive to internal/external changes - examples; collagen, keratin & elastin

Question 27

Collagen

Answer 27

- the connective tissue in skin, tendons, ligaments & nervous system - 3 polypeptides chains wound tightly together in a triple helix rope structure - polyP chain can wrap around tightly due to every 3rd a. acid being ’glycine’ - ^ glycine has the smallest R group - H bonds form as polyP chains wrap around each other - molecule are staggered; no weak spots

Question 28

Keratin

Answer 28

- makes up outer layer skin, hair & nails - insoluble in water - strong, inflexible molecule due to many disulphide bonds - nails contain more d. bonds than hair making hair “more flexible”

Question 29

Elastin

Answer 29

- provides elastic recoil (flexibility) in alveoli & blood vessels - a quaternary protein made from many stretchy molecules called ‘tropoelastin’ - contains hydrophobic regions that re-associate strands after being stretched