Biological Molecules

Question 1

How do hydrogen bonds form between water molecules?

Answer 1

Water is polar: O more electronegative than H, so
attracts electron density in covalent bond more strongly. Forms O 𝛿- (slightly negative) & H 𝛿+ (slightly positive).
There are intermolecular forces of attraction between a lone pair on O 𝛿- of one molecule & H 𝛿+ on an adjacent molecule.

Question 2

State 7 biologically important properties of water.

Answer 2

● reaches maximum density at 4℃
● high surface tension
● incompressible
● metabolite/ solvent for chemical reactions in the body
● high specific heat capacity
● high latent heat of vaporisation
● cohesion between molecules

Question 3

Why is the incompressible nature of water important for organisms?

Answer 3

Provides turgidity to plant cells.
Provides hydrostatic skeleton for some small animals e.g. earthworms.

Question 4

Explain why ice floats on water. Why is this important for organisms?

Answer 4

Ice is less dense than water because H-bonds hold molecules in fixed positions further away from each other.
Insulates water in arctic climates so aquatic organisms can survive. Water acts as a habitat.

Question 5

Why is the high surface tension of water important for organisms?

Answer 5

Slows water loss due to transpiration in plants.
Water rises unusually high in narrow tubes, lowering demand on root pressure.
Some insects can ‘skim’ across the surface of water.

Question 6

Why is water an important solvent for organisms?

Answer 6

Polar universal solvent dissolves & transports charged particles involved in intra & extracellular reactions e.g. PO43-
for DNA synthesis

Question 7

Why are the high specific heat capacity and latent of vapourisation of water important for organisms?

Answer 7

Acts as a temperature buffer which enables endotherms to resist fluctuations in core
temperature to maintain optimum enzyme activity.
Cooling effect when water evaporates from skin surface as sweat/ from mouth when panting

Question 8

Define monomer and polymer. Give some examples.

Answer 8

monomer: smaller units that join together to form larger molecules
● monosaccharides (glucose, fructose, galactose, ribose)
● amino acids
● nucleotides

polymer: molecules
formed when many monomers join together
● polysaccharides
● proteins
● DNA/ RNA

Question 9

What happens in condensation and
hydrolysis reactions?

Answer 9

Condensation: chemical bond forms between 2 molecules & a molecule of water is produced.

Hydrolysis: a water molecule is used to break a chemical bond between 2 molecules e.g. peptide bonds in proteins, ester bonds between fatty acids & glycerol in lipids.

Question 10

Name the elements found in carbohydrates, lipids, proteins and
nucleic acids.

Answer 10

carbohydrates & lipids: C, H, O
proteins: C, H, O, N, S
nucleic acids: C, H, O, N, P

Question 11

Describe the properties of alpha glucose

Answer 11

● Small & water soluble = easily transported in bloodstream.
● Complementary shape to antiport for co-transport for absorption in gut.
● Complementary shape to enzymes for glycolysis = respiratory substrate

Question 12

What type of bond forms when monosaccharides react?

Answer 12

(1,4 or 1,6) glycosidic bond
● 2 monomers = 1 chemical bond = disaccharide.
● Multiple monomers = many chemical bonds
= polysaccharide

Question 13

Name 3 disaccharides. Describe how they form.

Answer 13

Condensation reaction forms glycosidic bond
between 2 monosaccharides.
● maltose: glucose + glucose
● sucrose: glucose + fructose
● lactose: glucose + galactose
all have molecular formula C12H22O11

Question 14

Describe the structure and functions of starch.

Answer 14

Storage polymer of 𝛼-glucose in plant cells:
● insoluble = no osmotic effect on cells
● large = does not diffuse out of cells

made from amylose:
● 1,4 glycosidic bonds
● helix with intermolecular
H-bonds = compact

and amylopectin:
● 1,4 & 1,6 glycosidic bonds
● branched = many terminal ends
for hydrolysis into glucose

Question 15

Describe the structure and functions of glycogen.

Answer 15

Main storage polymer of 𝛼-glucose in animal cells (but also found in plant cells):
● 1,4 & 1,6 glycosidic bonds.
● Branched = many terminal ends for hydrolysis.
● Insoluble = no osmotic effect & does not diffuse
out of cells.
● Compact.

Question 16

Describe the structure and functions of cellulose

Answer 16

Polymer of 𝛽-glucose gives rigidity to plant cell walls (prevents bursting under turgor pressure, holds stem up).
● 1,4 glycosidic bonds.
● Straight-chain, unbranched molecule.
● Alternate glucose molecules are rotated 180°.
● H-bond crosslinks between parallel strands form microfibrils = high tensile strength.

Question 17

How do triglycerides form?

Answer 17

Condensation reaction between 1 molecule of glycerol & 3 fatty acids which forms ester bonds

Question 18

Condensation reaction between 1 molecule of glycerol & 3 fatty acids which forms ester bonds

Answer 18

Saturated:
● contain only single bonds
● straight-chain molecules
have many contact points
● higher melting point = solid at room temperature
● found in animal fats

Unsaturated:
● contain C=C double bonds
● ‘kinked’ molecules have
fewer contact points
● lower melting point = liquid at room temperature
● found in plant oils

Question 19

Relate the structure of triglycerides to their functions.

Answer 19

● High energy:mass ratio = high calorific value from
oxidation (energy storage).
● Insoluble hydrocarbon chain = no effect on water potential of cells & used for waterproofing.
● Slow conductor of heat = thermal insulation e.g. adipose tissue.
● Less dense than water = buoyancy of aquatic animals.

Question 20

Describe the structure and function of phospholipids.

Answer 20

Amphipathic: glycerol backbone attached to 2
hydrophobic fatty acid tails & 1 hydrophilic polar
phosphate head.
● Forms phospholipid bilayer in water = component of membranes.
● Tails can splay outwards = waterproofing e.g. for skin.

Question 21

Are phospholipids and triglycerides polymers?

Answer 21

No; they are not made from a small repeating unit. They are macromolecules

Question 22

Describe the structure and function of cholesterol.

Answer 22

Steroid structure of 4 hydrocarbon rings.
Hydrocarbon tail on one side, hydroxyl group
(-OH) on the other side.
Adds stability to cell surface phospholipid bilayer by connecting molecules & reducing
fluidity

Question 23

What is the general structure of an amino acid?

Answer 23

-COOH carboxyl / carboxylic acid group.
-R variable side group consists of carbon
chain & may include other functional groups e.g. benzene ring or -OH
(alcohol).
-NH2 amine/ amino group.

Question 24

How do polypeptides form?

Answer 24

Condensation reactions
between amino acids
form peptide bonds
(-CONH-).
There are 4 levels of
protein structure.

Question 25

Define primary and secondary structure of a protein.

Answer 25

Primary: sequence, number & type of amino acids in the polypeptide, determined by sequence of codons on mRNA.
Secondary: hydrogen bonds form between O 𝛿- attached to ‒C=O & H 𝛿+ attached to ‒NH.

Question 26

Describe the 2 types of secondary protein structure.

Answer 26

α-helix:
● All N-H bonds on same side of protein chain.
● Spiral shape.
● H-bonds parallel to helical axis.

β-pleated sheet:
● N-H & C=O groups alternate from one side to the other.

Question 27

Define ‘tertiary structure’ of a protein. Describe the bonds present.

Answer 27

3D structure formed by further folding
● Disulfide bridges: strong covalent S-S bonds between molecules of the amino acid cysteine.
● Ionic bonds: relatively strong bonds between charged R groups (pH changes cause these bonds to break).
● Hydrogen bonds: numerous & easily broken

Question 28

Define ‘quaternary structure’ of a protein.

Answer 28

● Functional proteins may consist of more than one polypeptide.
● Precise 3D structure held together by the same types of bond as tertiary structure.
● May involve addition of prosthetic groups e.g
metal ions or phosphate groups.

Question 29

Describe the structure and function of globular proteins.

Answer 29

● Spherical & compact.
● Hydrophilic R groups face outwards & hydrophobic R groups face inwards = usually water-soluble.
● Involved in metabolic processes e.g. enzymes such as amylase, insulin (2 polypeptide chains linked by 2 disulfide bonds), haemoglobin.

Question 30

Describe the structure of haemoglobin.

Answer 30

● Globular conjugated protein with prosthetic group.
● 2 𝛼-chains, 2 𝛽-chains, 4 prosthetic haem groups.
● Water-soluble so dissolves in plasma.
● Fe2+ haem group forms coordinate bond with O2.
● Tertiary structure changes so it is easier for subsequent O2 molecules to bind (cooperative binding).

Question 31

Describe the structure and function of fibrous proteins

Answer 31

● Can form long chains or fibres.
● Insoluble in water.
● Useful for structure and support e.g. collagen in skin.

Question 32

List the functions of collagen, elastin and keratin

Answer 32

Collagen: component of bones, cartilage, connective tissue, tendons.

Elastin: provides elasticity to connective tissue, arteries, skin, lungs, cartilage, ligaments.

Keratin: structural component of hair, nails, hooves/ claws, horns, epithelial cells of outer layer of skin.

Question 33

Describe how to test for proteins in a sample.

Answer 33

Biuret test confirms presence of peptide bond
1. Add equal volume of sodium hydroxide to sample at room temperature.
2. Add drops of dilute copper (II) sulfate solution.
Swirl to mix. (steps 1 & 2 make Biuret reagent).
3. Positive result: colour changes from blue to purple. Negative result: solution remains blue

Question 34

Describe how to test for lipids in a sample.

Answer 34

1. Dissolve solid samples in ethanol.
2. Add an equal volume of water and shake.
3. Positive result: milky white emulsion forms

Question 35

Describe how to test for reducing sugars.

Answer 35

1. Add an equal volume of Benedict’s reagent to a sample.
2. Heat the mixture in an electric water bath at 100℃ for 5 mins.
3. Positive result: colour changes from blue to orange & brick-red precipitate forms.
   Or use test strip coated in a reagent that changes colour if reducing sugar is present.

Question 36

Describe the Benedict’s test for non-reducing sugars.

Answer 36

1. Negative result: Benedict’s reagent remains blue.
2. Hydrolyse non-reducing sugars e.g. sucrose into their monomers by adding 1cm3 of HCl. Heat in a boiling water bath for 5 mins.
3. Neutralise the mixture using sodium carbonate solution.
4. Proceed with the Benedict’s test as usual.

Question 37

Describe the test for starch.

Answer 37

1. Add iodine solution.
2. Positive result: colour changes from orange to blue-black.

Question 38

State the role and chemical symbol of nitrates and ammonium

Answer 38

NO3- is used to make DNA, amino acids, NADP
for photosynthesis & NAD for respiration.

NH4+ can be converted to NO3-by saprobionts
during nitrogen cycle. Produced by deamination
of amino acids during ornithine cycle in liver.

Question 39

State the role and chemical symbol of hydroxide and phosphate ions.

Answer 39

OH- ions affect pH & can interact with bonds
in 3° protein structure to cause denaturation.
PO43- is a component of ATP/ ADP for energy release & NADP.

Question 40

State the role and chemical symbol of sodium, potassium and chloride ions.

Answer 40

Na+ & K+ are involved in maintenance of resting
potential of neurons/ generation of action potentials. Na+ is also involved in co-transport mechanisms.
Cl- is involved in inhibitory synapses to cause
hyperpolarisation.

Question 41

State the role and chemical symbol of hydrogen and hydrogencarbonate ions.

Answer 41

H+ & HCO3- form in organisms when CO2 dissolves in water.
H+ regulates pH & can interact with bonds in 3°
protein structure to cause denaturation. H+ pump is
involved in chemiosmosis & active loading in translocation.

Question 42

State the role and chemical symbol of calcium ions.

Answer 42

Ca2+ is used to make calcium pectate to add stability to middle lamella of plant cell walls. Regulates exocytosis of neurotransmitter. Binds to troponin to stimulate muscle contraction.

Question 43

How can the concentration of a solution be measured quantitatively?

Answer 43

● Use colorimetry to measure absorbance/
%transmission. Interpolate a calibration curve from
solutions of known concentration.
● Use biosensors. A bioreceptor detects the
presence of a chemical. A transducer converts the
response into a detectable electrical signal.

Question 44

Outline the principles and process of paper/ thin-layer chromatography.

Answer 44

1. Use capillary tube to spot solution onto pencil ‘start line’ (origin) 1 cm above bottom of paper.
2. Place chromatography paper in solvent. (origin should be above solvent level).
3. Allow solvent to run until it almost touches other end of the paper. Molecules in mixture move different distances based on relative solubility in solvent/attraction to paper.

Question 45

What are Rf values? How can they be calculated?

Answer 45

Ratios that allow comparison of how far molecules have moved in chromatograms.
Rf value = distance between origin and centre of pigment spot / distance between origin and solvent front