Biological Molecules

1. Monomers joining together.
2. Forming a (named if possible) bond.
3. Releasing a molecule of water.

Note: all three points can be required for a single mark

1. Ester bonds are broken.
2. By the addition of water.

1. It is insoluble so doesn’t affect water potential of cell or osmosis.
2. It is coiled so is compact.
3. It contains many glucose monomers for respiration.
4. It is branched, allowing for fast hydrolysis.
5. It is large so can’t leave the cell.

1. It is a polysaccharide of alpha glucose.Note: You had to state it was ‘alpha’ glucose.
2. The alpha glucose molecules are joined by glycosidic bonds.
3. It is branched and coiled due to its 1,4- and 1,6- glycosidic bonds.

1. It has a helix structure so is compact.
2. It is a polymer of (alpha) glucose so is easily hydrolysed.
3. Its branched endings allow for greater surface area for enzyme action, causing faster hydrolysis.
4. It is insoluble so does not affect the water potential of cells.

1. It has beta glucose molecules that alternate being flipped upside down.
2. The beta glucose molecules are joined by glycosidic bonds.
3. It forms long, unbranched, straight chains.Note: This is because of the alternate flipping of glucose molecules.
4. It forms microfibrils.

1. They hold chains together by forming cross links.
2. They are strong in large numbers so provide strength.

1. It is strong due to its compact structure, caused by its straight and long chains (which are caused by the flipping over of alternate molecules) which are held together by many hydrogen bonds.
2. The bonds are difficult to digest for herbivores.
3. It is insoluble which helps the cell resist osmotic pressure.
4. Its fibres are permeable, allowing water and solutes to freely reach the cell membrane.

1. Energy storage - they are an efficient store of energy due to their long hydrocarbon chains (in their fatty acids) which release a lot of energy when oxidised. Being insoluble, they also don’t affect the water potential of the cells they are stored in.
2. Thermal insulation - they provide a layer of insulation under the skin, reducing heat loss.
3. Buoyancy - fat has a low density, aiding in buoyancy.
4. Nervous transmission - they form the myelin sheath, insulating nerves and speeding up transmission.

1. Both contain ester bonds between glycerol and fatty acids.
2. Both contain glycerol.
3. The fatty acids on both may be saturated or unsaturated.
4. Both are insoluble in water.
5. Both contain C, H and O but phospholipids also contain P.
6. A triglyceride has three fatty acids but a phospholipid has two fatty acids plus a phosphate group.
7. Triglycerides are hydrophobic/non polar but phospholipids have a hydrophilic head and hydrophobic tail.
8. Phospholipids form a bilayer in water but triglycerides don’t.Note: triglycerides form lipid droplets.

1. Its structure is determined by the position of its amino acids.
2. The primary structure is the sequence of amino acids.
3. The secondary structure is formed by hydrogen bonding between amino acids.Note: This is between the amine group on one amino acid and the carboxyl group on another amino acid.
4. The tertiary structure is formed by interactions between R groups.
5. The tertiary structure creates active sites in enzymes and complementary shapes in antibodies, carrier proteins and receptors.Note: this is a fairly random point that you should include in every question on protein structure.
6. The quaternary structure is formed by the interactions between polypeptides.

1. Hydrogen bonds form between the NH group on one amino acid and the C=O group on another amino acid.Note: If you are going to mention where the hydrogen bonds form (which you may as well do), use the chemical structures not names (amine group, carboxyl group).
2. This forms an alpha helix or beta pleated sheet.

1. It has a base sequence.
2. In triplets.
3. This determines the amino acid sequence of the polypeptide.

1. Each amino acid is oriented in the same direction in the chain.
2. This means that when the NH group of one amino acid forms a peptide bond with the COO group of another, there is always a free NH₂ group at one end and a free COOH group at the other.

1. Heat it with Benedict’s solution. If it doesn't contain reducing sugars, it will remain light blue.Note: It is important to mention this to show the absence of reducing sugars.
2. Boil with acid, then neutralise.
3. Heat with Benedict’s solution to 95 degrees C.Note: It is wise to mention the specific temperature.
4. A red precipitate forms.Note: It is important to know it’s a precipitate that causes the colour change.

Ethanol test:

1. Add ethanol and shake, then add water and shake.
2. A white, milky emulsion forms if lipids are present.Note: Remember to clearly describe adding ethanol first THEN water, and also remember to mention shaking at least once.

1. Add biuret reagent.
2. The solution turns from blue to purple if protein is present.

1. Before the substrate binds to the active site to form an enzyme-substrate complex, the active site of the enzyme is not perfectly complementary to the shape of the substrate.
2. When the substrate binds, the active site changes shape so they are perfectly complementary to each other.
3. This distorts, weakens, and/or bends bonds in the substrate.
4. This causes the activation energy to be lowered.
5. When the substrate leaves, the active site returns to its original shape.

1. The wrong molecule is a different shape to the substrate.
2. The enzyme is specific only to the shape of the substrate.
3. Due to the specific shape of the active site due to the specific tertiary structure of the enzyme.
4. This means the wrong molecule won’t be able to fit and bind to the active site.
5. This is because they are not complementary.
6. This means no enzyme-substrate complex forms.

1. Enzyme concentration is now limiting the factor.
2. Because all of the active sites are occupied.

1. Boil to denature the enzymes.
2. Add a strong acid/ alkali to denature the enzymes.
3. Put it in ice to lower the temperature and kinetic energy so no enzyme-substrate complexes form.
4. Add a high concentration of an inhibitor to prevent enzyme-substrate complexes forming.

1. The inhibitor is a similar shape to the substrate.
2. This causes it to fit in and bind to the active site as they are complementary.
3. This prevents enzyme-substrate complexes from forming.

1. The inhibitor attaches to the enzyme at a site other than the active site (also called the allosteric site)
2. This changes the tertiary structure of the enzyme, changing the shape of the active site.
3. This means the active site and substrate are no longer complementary so substrates can no longer bind to form enzyme-substrate complexes.

1. The binding alters the tertiary structure of the enzyme.
2. This causes the shape of the active site to change.
3. This causes enzyme-substrate complexes to form more quickly, further lowering activation energy.Note: basically it works like a noncompetitive inhibitor but increases the rate of reaction instead of decreasing it.

1. pH.Note: control this using a buffer solution.
2. Enzyme concentration.
3. Initial substrate concentrations.

1. It gives quantitative data.
2. Colour change can be subjective, but numerical readings aren’t.
3. This means a meter provides greater accuracy.

1. Phosphodiester bonds.Note: not just ‘ester bonds’ - always say ‘phosphodiester’.
2. Hydrogen bonds.Note: these occur between complementary bases and hold the two strands together.
3. Glycosidic bonds.Note: these form between the deoxyribose sugars and the bases.

1. They form in condensation reactions.
2. Catalysed by DNA polymerase.
3. Between phosphate and deoxyribose.

1. The sugar-phosphate backbone provides strength to the molecule and protects the bases.
2. It is a large molecule, meaning it can store lots of information.
3. It is a helix/ coiled so it is compact.
4. The base sequence codes for amino acids.
5. It is double stranded so replication can occur semi-conservatively.
6. Its (fairly) weak hydrogen bonds allow for replication.

1. The nucleotides have the same structure.
2. The nucleotides are joined by phosphodiester bonds.
3. The DNA in mitochondria/ chloroplasts is very similar to the DNA in prokaryotes (circular, no introns, not bound to histones)

1. DNA is shorter and contains fewer genes.
2. DNA is circular not linear.
3. DNA is not associated with histones
4. DNA contains no intronsNote: Essentially the DNA in chloroplasts is the same as the DNA found in bacteria, because chloroplasts and mitochondria evolved from bacteria.

1. DNA helicase breaks the hydrogen bonds between strands, causing the strands to separate.
2. Each strand from the original molecule acts as a template.
3. Nucleotides line up in complementary pairs due to complementary base pairing.
4. A with T, G with C.
5. Hydrogen bonds reform between the complementary bases.
6. DNA polymerase joins adjacent nucleotides.
7. Forming phosphodiester bonds.
8. Each DNA molecule consists of one template/ parent strand and one new strand.

1. Joining adjacent nucleotides.
2. Catalysing condensation reactions.
3. Forming phosphodiester bonds.

1. To act as a template for the new strand.
2. To determine the order of nucleotides.

1. To form complementary base pairs.
2. To form the complementary strand.

1. Polypeptides/ proteins.
2. Functional RNA (tRNA, rRNA)

1. The mutated triplet codes for the same amino acid.Note: this is possible because DNA is degenerate.
2. The mutation occurs in an intron/ non-coding region, which is removed during splicing.Note: a change in primary structure not causing a change tertiary structure was not an acceptable answer - I’m guessing because the question said 'no change' (at all) in structure.

1. Releasing and providing energy for processes such as muscle contraction or active transport.Note: It's important you name a process in these types of questions.
2. Increasing the reactivity of molecules by phosphorylating themNote: these two should be your go-to.

1. It can easily and quickly be reformed.
2. It phosphorylates substances when it is broken down by releasing a Pi, making them more reactive.
3. It is soluble.
4. It doesn’t leave cells.

1. ATP is hydrolysed into ADP + Pi and the phosphate associates with the enzyme.
2. This changes the shape of the active site.
3. This causes the active site to become complementary to the substrate, activating the enzyme.

1. It contains ribose, adenine and 3 phosphates.
2. It is broken down in the hydrolysis reaction: ATP -> ADP + Pi by ATP hydrolase.
3. It is resynthesised in the condensation reaction: ADP + Pi -> ATP by ATP synthase.

1. It allows it transport dissolved substances around organisms.
2. It allows metabolic reactions to occur at a faster rate because they occur faster in solution.

1. It allows columns of water can form.Note: this is important for transpiration in xylem.
2. It allows surface tension to occur on the top of bodies of water, supporting organisms.

1. They are charged.
2. This allows them to associate with O₂ to form oxyhaemoglobinNote: this is a good reminder that if the question doesn’t state the role of an ion, you should do it in your answer, even if it seems too obvious to mention.

1. Phosphorylating other substances to make them more reactive/ change their shape.
2. To form DNA/ RNA/ phospholipids. I’d always go with phosphorylating substances to make them more reactive, not activating enzymes.