Exam Questions (P1)

Question 1

Describe the induced fit model of enzyme action (3 marks)

Answer 1

Substrate binds to the active site to form an enzyme substrate complex.
The active site changes shape so that it is complementary to the substrate.

Question 2

Explain how the counter current principle allows efficient oxygen uptake in the fish gas exchange system
(2 marks)

Answer 2

Blood and water flow in opposite directions.
Diffusion gradient is maintained along whole length of lamella

Question 3

Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation
(3 marks)

Answer 3

tRNA brings specific amino acid to ribosome.
Anticodon on tRNA binds to codon on mRNA.
Amino acid joins by a condensation reaction using ATP to form a peptide bond.

Question 4

Describe how a sample of chloroplasts could be isolated from leaves
(4 marks)

Answer 4

Break open cells and filter out large cell debris.
In a cold, isotonic, buffered solution spin the solution in a centrifuge and remove nuclei.
Spin again at higher speed and remove chloroplasts.

Question 5

Why do iron deficient plants have a reduced growth rate?
(3 marks)

Answer 5

Less thylakoid membrane with few grana.
Small surface area means less chlorophyll.
Less chlorophyll reduces light absorption meaning that there is a slower rate of photosynthesis.

Question 6

Describe the role of spindles fibres and the behaviour of chromosomes during;
1. Prophase
2. Metaphase
3. Anaphase
(5 marks)

Answer 6

Prophase: chromosomes condense and centromeres attach to spindle fibres
Metaphase: chromosomes line up in the centre of the spindle
Anaphase: centromeres divide and chromatid from each pair are pulled to opposite ends of the pole

Question 7

Explain the process of phagocytosis
(3 marks)

Answer 7

Cell engulfs the antibody.
Lysosomes fuse with vesicle.
Lysozymes digest the antibody.

Question 8

Describe how a triglyceride molecule is formed
(3 marks)

Answer 8

One glycerol and three fatty acids.
Condensation reactions and removal of 3 molecules of water.
Ester bonds are formed.

Question 9

Describe the structure of DNA
(5 marks)

Answer 9

Polymer of nucleotides.
Each nucleotide formed from deoxyribose, a phosphate group and an organic nitrogenous base.
There are phosphodiester bonds between nucleotides.
Double helix structure held by hydrogen bonds.
Hydrogen binds between adenine & thymine and cytosine & guanine.

Question 10

Name and describe 5 ways substances can move across the cell surface membrane into a cell
(5 marks)

Answer 10

Simple diffusion of small molecules down a concentration gradient.
Facilitated diffusion down a concentration gradient via protein carriers/channels.
Osmosis of water down a water potential gradient.
Active transport against a concentration gradient via a protein carrier using ATP.
Co-transport of 2 different substances using a carrier protein.

Question 11

Explain the function of ATP hydrolase
(2 marks)

Answer 11

Hydrolyses ATP to release energy.
The energy allows ions to be moved against a concentration gradient.

Question 12

Explain how the movement of sodium ions out of the cell allows the absorption of glucose into the cell lining of the ileum
(2 marks)

Answer 12

Maintains a concentration gradient for sodium ions.
Sodium ions moving in by facilitate diffusion brings glucose with it.

Question 13

Describe the role of micelles in the absorption of fats into the cells lining the ileum
(3 marks)

Answer 13

Micelles include bile salts and fatty acids.
They make the fatty acids more soluble in water.
They bring the fatty acid to the cells lining the ileum.
They maintain a high concentration of fatty acids to the cell.
Fatty acids are absorbed by diffusion.

Question 14

Describe the role of DNA polymerase in the semi-conservative replication of DNA
(2 marks)

Answer 14

Joins nucleotides together.
Catalyses condensation reactions.
Catalyses the formation of phosphodiester bonds between adjacent nucleotides.

Question 15

Describe how an enzyme can be phosphorylated
(2 marks)

Answer 15

Attachment of phosphate to the enzyme which is released from the hydrolysis of ATP.

Question 16

Name 3 difference between DNA in the nucleus of a plant cell and the DNA in a prokaryotic cell
(3 marks)

Answer 16

Plant:
- associated with histones
- linear
- no plasmids
- introns
- longer
Prokaryote
- no histones
- circular
- plasmids
- no introns
- shorter

Question 17

Describe how mRNA is formed by transcription in eukaryotes
(5 marks)

Answer 17

Hydrogen bonds between DNA bases break.
One DNA strand acts as a template.
Free RNA nucleotides align by complementary base pairing.
In RNA uracil base pairs with adenine.
RNA polymerase joins adjacent nucleotides by phosphodiester bonds.
Pre-mRNA is spliced to remove introns and form mRNA.

Question 18

Describe how a polypeptide is formed by the translation of mRNA
(6 marks)

Question 19

Define gene mutation and explain how a gene mutation can have no effect and a positive effect on an individual
(4 marks)

Answer 19

A gene mutation is a change in the base sequence of chromosomes and results in the formation of a new allele.
Gene mutations can have no effect as the genetic code is degenerate. This means that the amino acids are changed but has no effect on tertiary structure.
Gene mutations can have a positive effect because they result in a change in a polypeptide that changes the properties that may result in increased survival chances.

Question 20

Describe how a non-competitive inhibitor can reduce the rate of an enzyme controlled reaction
(3 marks)

Answer 20

A non-competitive inhibitor attaches to the enzyme at a site other than the active site. It changes the shape of the active site which means that it is no longer complementary to the active site so less substrate can bind.

Question 21

How would you ensure that an index of diversity was representative of each habitat?
(2 marks)

Answer 21

Random samples
Large number of samples

Question 22

Describe the structure of human immunodeficiency virus (HIV)
(4 marks)

Answer 22

RNA and reverse transcriptase is located in the core of HIV.
It is surrounded by a protein capsid which is found inside a lipid envelope made from membranes of previous host cells.
On the outside of the envelope are attachment proteins.

Question 23

Using knowledge of surface area to volume ration explain the higher metabolic rate of a mouse compared to a horse.
(3 marks)

Answer 23

A mouse is smaller so has a larger surface area to volume ration. This means that more heat is lost and results in a faster rate of respiration.

Question 24

Explain 5 properties that make water important for organisms.
(5 marks)

Answer 24

• it is a metabolite in condensation and hydrolysis reactions
• it is a solvent so it allows reactions to occur and allows the transport of substances
• it has a high heat capacity so can buffer changes in temperature
• it has a large latent heat of vaporisation so provides a cooling effect
• strong cohesion between water molecules supports columns of water in plans and produces surface tension supporting small organisms

Question 25

Describe the biochemical tests you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample.
(5 marks)

Answer 25

LIPID
Add ethanol and water to the sample and shake.
If a lipid is present the emulsion turns milky.
NON-REDUCING SUGAR
Carry out Benedict’s test on sample. It will stay blue (negative result for a reducing sugar).
Boil sample with acid and then neutralise with alkali.
Heat with Benedict’s and a positive result will cause the mixture to form a red/orange precipitate.
AMYLASE
Add biuret reagent, resulting in the mixture becoming purple.
Add starch and then test for a reducing sugar using the Benedict’s test.

Question 26

Describe the chemical reactions involved in the conversion of polymers to monomers and monomer to polymers. Give examples
(5 marks)

Answer 26

A condensation reaction joins monomers together and forms a chemical bond and releases water.
A hydrolysis reaction breaks a chemical bond between monomers and uses water.

Question 27

Give 2 ways in which the arrangement of prokaryotic DNA is different from the arrangement of human DNA.
(2 marks)

Answer 27

PROKARYOTIC DNA
- circular
- not associated with histones
- only one molecule of DNA
HUMAN DNA
- linear
- associated with histones

Question 28

What variables should be controlled when investigation the effect of surface area on osmosis in cubes of potatoes?
(3 marks)

Answer 28

• control temperature by keeping samples at room temperature
• dry cubes before measuring
• measure mass of cubes at stated time intervals
• ensure all cut surfaces of the eight cubes are exposed to the sucrose solution

Question 29

Explain how the formation of an enzyme substrate complex increases the rate of reaction.
(2 marks)

Answer 29

Reduces activation energy due to bending bonds

Question 30

Explain how the treatment with antivenom works and why it is essential to use passive immunity rather than active immunity.
(2 marks)

Answer 30

Antivenom is an example of passive immunity; antibodies bind to the toxin and cause its destruction.
Active immunity would be too slow.

Question 31

Describe the role of two named enzymes in the process of semi-conservative replication of DNA.
(3 marks)

Answer 31

DNA helicase causes the breaking of hydrogen bonds between DNA strands.
DNA polymerase joins the DNA nucleotides together and forms phosphodiester bonds.

Question 32

Describe the gross structure of the human gas exchange system and how we breathe in and out
(6 marks)

Answer 32

Air flows through the trachea into the bronchi, then into the bronchioles and finally into the alveoli.
During inspiration the diaphragm and external intercostal muscles contract which causes a volume increase and an pressure decrease in the thoracic cavity. This creates a pressure gradient and results in air moving in.
During expiration the diaphragm relaxes and the internal intercostals contract which causes a volume decrease and a pressure increase in the thoracic cavity. This creates a pressure gradient and results in air moving out.

Question 33

Compare and contrast the structure and properties of triglycerides and phospholipids
(5 marks)

Answer 33

SIMILARITIES
- both contain ester binds between glycerol and fatty acids
- both contain glycerol
- fatty acids on both may be saturated or unsaturated
- both are insoluble in water
- both contain C, H , O
DIFFERENCES
- triglycerides have 3 fatty acids and phospholipids have 2 fatty acids plus a phosphate group
- triglycerides are hydrophobic and phospholipids have hydrophilic and hydrophobic regions
- phospholipids form a monolayer but triglycerides don’t

Question 34

Describe how lactose is formed and where in the cell it would be attached to a polypeptide to form a glycoprotein
(4 marks)

Answer 34

Glucose and galactose are joined together by a condensation reaction and a glycosidic bond is formed to hold them together.
This is added to a polypeptide in the Golgi apparatus