Biological molecules

Question 1

Describe and explain 3 features of glycogen that make it well suited to its role as an energy store (6 marks)

Answer 1

• It is poorly soluble which means that it won’t affect water potential in cells
• It is compact which means it can fit into small storage granules
• It is easily synthesised and broken down which means that it rapidly responds to the hormones insulin and glucagon

Question 2

Describe the structural differences between amylose and amylopectin and explain their effects on the starch molecule (4 marks)

Answer 2

Amylose contain 1,4 glycosidic bonds between monomers. These bonds cause amylose to form a compact coiled shape.
Amylopectin contains a mixture of 1,4 and 1,6 glycosidic bonds between monomers. The presence of these bonds results in a branched structure that allows folding for even more compactness.

Question 3

Outline one structural similarity and one structural difference between amylopectin and glycogen (2 marks)

Answer 3

Similarity: they both have a combination of 1,4 and 1,6 glycosidic bonds
Differences: Glycogen has a much greater proportion of 1,6 glycosidic bonds compared to amylopectin.
Glycogen has a more compact structure

Question 4

Give one similarity and one difference between the structures of glucose and fructose
(2 marks)

Answer 4

SIMILARITIES
- both are hexose sugars
- they have the same molecular formula
DIFFERENCES
- glucose is a 6 atom ring whereas fructose is a 5 atom ring

Question 5

Describe 2 experimental methods by which the glycosidic bond in a disaccharide can be broken
(4 marks)

Answer 5

METHOD 1
Enzyme hydrolysis
Mix the disaccharide with the enzyme at room temperature
METHOD 2
Acid hydrolysis
Boil a sample with dilute hydrochloric acid

Question 6

Explain the advantages of storing food as starch, rather than as sugars, in plant cells
(3 marks)

Answer 6

Starch is compact and highly branch
Starch is insoluble so does not have an effect of the water potential of the cell
Starch is easily broken down when needed

Question 7

Describe a biochemical test for lipids
(2 marks)

Answer 7

Mix the sample with ethanol followed by water.
A milky white emulsion will form

Question 8

Describe and explain the role played by bile and its component in the processing of lipids
(3 marks)

Answer 8

Bile emulsifies lipids by bile slats attaching to lipids and turning them into smaller droplets.
Bile salts form micelles with the the hydrolysis product of triglycerides. Micelles transports monoglycerides to the outer surface of the intestinal epithelium for absorption

Question 9

Describe the differences between phospholipids and triglycerides
(2 marks)

Answer 9

Phospholipids have 2 fatty acids and are also bonded to a phosphate group

Question 10

Explain how the secondary structure of a protein contributes to its overall 3D shape
(4 marks)

Answer 10

Due the hydrogen bonding between the amine and carboxyl groups of separate amino acids, every 4 amino acids along the chain the protein forms a alpha helix. A beta pleated sheet forms by folding the chain back on itself. Both structures provide bulk to the protein.

Question 11

Describe the quaternary structure of haemoglobin and explain how its structure relates to its function in mammals
(6 marks)

Answer 11

STRUCTURE
A quaternary structure is 4 separate polypeptide chains bonded together. Each polypeptide chain is bonded to a haem group which contains an iron ion.
BENEFITS TO FUCNTION
The function of haemoglobin is to bind to oxygen in the lungs and release oxygen at respiring tissue. The binding of oxygen to one haem group increases the affinity of haemoglobin for oxygen. By changing the shape of the whole protein haemoglobin can deliver oxygen readily to the tissues that require it the most.

Question 12

Describe the biochemical test for proteins
(2 marks)

Answer 12

Mix the solution with Biuret reagent. A positive result will produce a purple solvent.

Question 13

Explain the induced fit hypothesis
(3 marks)

Answer 13

In the induced fit hypothesis:
- enzyme structure is not rigid
- the active site and substrate are not a perfect 3-D fit
- the active site changes shape to fit the substrate
- this can stress the substrate and weaken its bonds which lower the activation energy

Question 14

Describe and explain 2 ways in which an enzyme can become denatured
(4 marks)

Answer 14

High temperatures
Extremes of pH
Both of these conditions cause the weak bonds in the tertiary structure to break whcih changes the shape of the active site.

Question 15

Describe and explain the similarities and differences between competitive and non-competitive inhibitors
(6 marks)

Answer 15

SIMILARITIES
- the inhibitor binds to the enzyme’s surface
- there is a reduction in the enzyme’s rate
- the rate of initial reaction is inversely proportional to inhibitor concentration
DIFFERENCES
- competitive inhibitors bind to the active site and non-competitive inhibitors bind away from the active site
- in competitive inhibition the active site is blocked and in non-competitive inhibition the active site changes shape
- increasing substrate concentration will decrease the effect of competitive inhibition but will not affect non-competitive inhibition

Question 16

Define the term ‘specificity’ in the context of enzymes
(2 marks)

Answer 16

One enzyme has a small number of substrates due to the complementary shape between the substrate and active site

Question 17

Describe the term ‘frameshift mutation’ and evaluate the likely impact of this kind of mutation on the phenotype of the organisms
(6 marks)

Answer 17

A frameshift mutation is a deletion of a base that occurs during DNA replication. It causes all downstream triples to shift by one base.
It results in a different protein being synthesised and therefore results in a loss of function for that gene. If a mutation occurs in an intron no effect will occur.

Question 18

The ends of a DNA strand are referred to as the 3’ (“3-prime”) end and the 5’ (“5-prime”) end. Describe the aspects of DNA structure that give rise to this naming system
(3 marks)

Answer 18

Carbons 3 and 5 are numbered carbon atoms.
C5 of one nucleotide is bonded by phosphodiester bonds to C3 of an adjacent nucleotide.
There will always be one free C3 and one free C5 at each end of the strand

Question 19

Define semi-conservative replication and outline the steps in Meselson and Stahl’s experiment that provided this proof.
(6 marks)

Answer 19

The semi-conservative replication is where one strand of the original DNA molecule is contained in the new DNA molecule.
1. bacteria were cultured on a heavy nitrogen medium.
2. the bacteria took up this nitrogen and used it to synthesis DNA bases.
3. the bacteria is then transferred to a light nitrogen medium
4. after a period of time the DNA was extracted and centrifuged and it settles in an intermediate position.

Question 20

Compare and contrast the structures of DNA and RNA
(6 marks)

Answer 20

SIMILARITIES
- both are polynucleotides
- both contain A,C and G bases
- both contain a pentose sugar and a phosphate group
- both are linked by phosphodiester bonds between 3’ and 5’ carbon atoms
DIFFERENCES
- DNA is double stranded whereas RNA is single stranded
- DNA is long and RNA is short
- DNA contains deoxyribose sugar and RNA contains ribose sugar
- DNA contains thymine and RNA contains uracil

Question 21

Identify and explain the structural differences between a single strand of DNA and an RNA strand that makes DNA a better storage molecule than RNA
(2 marks)

Answer 21

DNA is a better storage molecule than RNA because:
- ribose sugar contains a hydroxyl group in the 2’ position which makes ribose more unstable.

Question 22

Describe and explain the structure and function of a ribosome
(4 marks)

Answer 22

A ribosome is a small organelle that is either free or attached to rough ER.
It is made from RNA and protein and contains ribosomal RNA which catalyses peptide bond formation between amino acids.
Ribosomes are the site of protein synthesis.

Question 23

Describe and explain three structural features of DNA
(6 marks)

Answer 23

Double helix with base pairs on the inside and a sugar phosphate backbone on the outside which gives the DNA excellent stability.
The strength of the hydrogen bonds between the base pairs are strong enough to hold strand together but weak enough to be split when need for replication.
Supercoiled around histones provides a compact structure.

Question 24

Suggest why most mutations in eukaryotic cells are regarded as neutral
(3 marks)

Answer 24

The mutation has no effect on the phenotype of the organism.
They occur in genes that are not expressed and also occur in introns.

Question 25

Define the term ‘polar’ and describe how this property manifests itself in the structure of a water molecule
(3 marks)

Answer 25

Polar is the uneven spread of electronic charge.
Electrons are more strongly attracted towards oxygen.
This causes an asymmetrical molecular shape.

Question 26

Define latent heat of vaporisation and explain how water’s high value benefits organisms.
(4 marks)

Answer 26

Latent heat of vaporisation is the amount of heat energy required to evaporate without causing an increase in temperature.
This benefits organisms as water can carry away of a lot of heat energy so it cools the organism down with a minimal loss of water to organisms.

Question 27

Describe the role of an inorganic ion required for the absorption of important food molecules in the small intestines. Name the substance concerned.
(5 marks)

Answer 27

The inorganic ion required for the absorption of important food molecules in the small intestines is sodium ions.
Co-transport.
Sodium ions are actively transported out of the intestinal epithelial cells into the lumen.
Sodium ions re-enters the intestinal epithelial cells by facilitate diffusion via co-transport proteins.
Glucose and amino acids enter at the same time.