A level biology module 8 graded A+(exam questions and
answers 2023)
Explain how the structure of DNA is related to its functions (6 marks)
- sugar-phosphate (backbone)/double stranded/helix so provides
strength/stability/protects bases/protects hydrogen bonds
- long/large molecule so can store lots of information
- helix/coiled so compact
- base sequence allows information to be stored/ base sequence codes for amino
acids/protein
- double stranded so replication can occur semi-conservatively/strands can act as
templates/complementary base pairing/A-T and G-C so accurate replication/identical
copies can be made
- (weak) hydrogen bonds for replication/ unzipping/ strand separation/ many hydrogen
bonds so stable/strong
Scientists investigated three genes, C, D and E, involved in controlling cell division.
They studied the effect of mutations in these genes on the risk of developing lung
cancer. The scientists analysed genes C, D and E from healthy people and people with
lung cancer. • If a person had a normal allele for a gene, they used the symbol N. • If a
person had two mutant alleles for a gene, they used the symbol M. They used their data
to calculate the risk of developing lung cancer for people with different combinations of
N and M alleles of the genes. A risk value of 1.00 indicates no increased risk. The
following table shows the scientists' results
What do these data suggest about the relative importance of the mutant alleles of genes
C, D and E on increasing the risk of developing lung cancer? Explain your answer (3
marks)
- (mutation) in E produces highest risk/1.78
- (mutation) in D produces next highest risk/1.45
- (mutation) in C produces least risk/1.30
Chemotherapy is the use of a drug to treat cancer. The drug kills dividing cells. The
figure below shows the number of healthy cells and cancer cells in the blood of a patient
receiving chemotherapy. The arrows labelled F to I show when the drug was given to
the patient.
Calculate the rate at which healthy cells were killed between days 42 and 46. (1 mark)
- 1280-560=720
- 720/4=180
Describe similarities and differences in the response of healthy cells and cancer cells to
the drug between times F and G. (3 marks)
(similarities)
- same/similar pattern/both decrease, stay the same then increase
- number of cells stays the same for same length of time
(differences)
- greater/faster decrease in number of healthy cells/more healthy cells killed/healthy
cells killed faster
, - greater/faster increase in number of healthy cells/more healthy cells
replaced/divide/healthy cells replaced/divide faster
More cancer cells could be destroyed if the drug was given more frequently. Suggest
why the drug was not given more frequently. (2 marks)
- more/too many healthy cells killed
- (so) will take time to place/increase in number
- person may die/have side effects
Mycolic acids are substances that form part of the cell wall of the bacterium that causes
tuberculosis. Mycolic acids are made from fatty acids. Isoniazid is an antibiotic that is
used to treat tuberculosis. The diagram shows how this antibiotic inhibits the production
of mycolic acids in this bacterium.
Treatment with isoniazid leads to the osmotic lysis of this bacterium. Use information in
the diagram to suggest how. (2 marks)
- cell wall not formed/production inhibited
-lower water potential in bacterium
- water enters and causes lysis/expansion/pressure
Human cells also produce fatty acids. Isoniazid does not affect the production of these
fatty acids. Use information in the diagram to suggest one reason why isoniazid does
not affect the production of fatty acids in human cells. (1 mark)
human cells lack enzyme (B)/ have a different enzyme/produce different fatty acids/use
different fatty acids/use different substances
A mutation in the gene coding for enzyme B could lead to the production of a non-
functional enzyme. Explain how. (3 marks)
- change in base sequence (of DNA/gene) leading to change in amino acid
sequence/primary structure (of enzyme)
- change in hydrogen/ionic/disulphide bonds leading to change in the tertiary
structure/active site (of enzyme)
- substrate not complementary/cannot bind (to enzyme/active site)/no enzyme-substrate
complexes form
Phenylketonuria is a disease caused by mutations of the gene coding for the enzyme
PAH. The table shows part of the DNA base sequence coding for PAH. It also shows a
mutation of this sequence which leads to the production of non-functioning PAH.
What is the maximum number of amino acids for which this base sequence could code?
( 1 mark)
4
Explain how this mutation leads to the formation of non-functioning PAH. (3 marks)
- change in amino acid/ (sequence of) amino acids/primary structure
- change in hydrogen/ionic/disulphide bonds leading to change in the tertiary
structure/active site (of enzyme)
- substrate not complementary/cannot bind (to enzyme/active site)/no enzyme-substrate
complexes form
PAH catalyses a reaction at the start of two enzyme-controlled pathways. The diagram
shows these pathways.
Use the information in the diagram to give two symptoms you might expect to be visible
in a person who produces non-functioning PAH. (2 marks)
answers 2023)
Explain how the structure of DNA is related to its functions (6 marks)
- sugar-phosphate (backbone)/double stranded/helix so provides
strength/stability/protects bases/protects hydrogen bonds
- long/large molecule so can store lots of information
- helix/coiled so compact
- base sequence allows information to be stored/ base sequence codes for amino
acids/protein
- double stranded so replication can occur semi-conservatively/strands can act as
templates/complementary base pairing/A-T and G-C so accurate replication/identical
copies can be made
- (weak) hydrogen bonds for replication/ unzipping/ strand separation/ many hydrogen
bonds so stable/strong
Scientists investigated three genes, C, D and E, involved in controlling cell division.
They studied the effect of mutations in these genes on the risk of developing lung
cancer. The scientists analysed genes C, D and E from healthy people and people with
lung cancer. • If a person had a normal allele for a gene, they used the symbol N. • If a
person had two mutant alleles for a gene, they used the symbol M. They used their data
to calculate the risk of developing lung cancer for people with different combinations of
N and M alleles of the genes. A risk value of 1.00 indicates no increased risk. The
following table shows the scientists' results
What do these data suggest about the relative importance of the mutant alleles of genes
C, D and E on increasing the risk of developing lung cancer? Explain your answer (3
marks)
- (mutation) in E produces highest risk/1.78
- (mutation) in D produces next highest risk/1.45
- (mutation) in C produces least risk/1.30
Chemotherapy is the use of a drug to treat cancer. The drug kills dividing cells. The
figure below shows the number of healthy cells and cancer cells in the blood of a patient
receiving chemotherapy. The arrows labelled F to I show when the drug was given to
the patient.
Calculate the rate at which healthy cells were killed between days 42 and 46. (1 mark)
- 1280-560=720
- 720/4=180
Describe similarities and differences in the response of healthy cells and cancer cells to
the drug between times F and G. (3 marks)
(similarities)
- same/similar pattern/both decrease, stay the same then increase
- number of cells stays the same for same length of time
(differences)
- greater/faster decrease in number of healthy cells/more healthy cells killed/healthy
cells killed faster
, - greater/faster increase in number of healthy cells/more healthy cells
replaced/divide/healthy cells replaced/divide faster
More cancer cells could be destroyed if the drug was given more frequently. Suggest
why the drug was not given more frequently. (2 marks)
- more/too many healthy cells killed
- (so) will take time to place/increase in number
- person may die/have side effects
Mycolic acids are substances that form part of the cell wall of the bacterium that causes
tuberculosis. Mycolic acids are made from fatty acids. Isoniazid is an antibiotic that is
used to treat tuberculosis. The diagram shows how this antibiotic inhibits the production
of mycolic acids in this bacterium.
Treatment with isoniazid leads to the osmotic lysis of this bacterium. Use information in
the diagram to suggest how. (2 marks)
- cell wall not formed/production inhibited
-lower water potential in bacterium
- water enters and causes lysis/expansion/pressure
Human cells also produce fatty acids. Isoniazid does not affect the production of these
fatty acids. Use information in the diagram to suggest one reason why isoniazid does
not affect the production of fatty acids in human cells. (1 mark)
human cells lack enzyme (B)/ have a different enzyme/produce different fatty acids/use
different fatty acids/use different substances
A mutation in the gene coding for enzyme B could lead to the production of a non-
functional enzyme. Explain how. (3 marks)
- change in base sequence (of DNA/gene) leading to change in amino acid
sequence/primary structure (of enzyme)
- change in hydrogen/ionic/disulphide bonds leading to change in the tertiary
structure/active site (of enzyme)
- substrate not complementary/cannot bind (to enzyme/active site)/no enzyme-substrate
complexes form
Phenylketonuria is a disease caused by mutations of the gene coding for the enzyme
PAH. The table shows part of the DNA base sequence coding for PAH. It also shows a
mutation of this sequence which leads to the production of non-functioning PAH.
What is the maximum number of amino acids for which this base sequence could code?
( 1 mark)
4
Explain how this mutation leads to the formation of non-functioning PAH. (3 marks)
- change in amino acid/ (sequence of) amino acids/primary structure
- change in hydrogen/ionic/disulphide bonds leading to change in the tertiary
structure/active site (of enzyme)
- substrate not complementary/cannot bind (to enzyme/active site)/no enzyme-substrate
complexes form
PAH catalyses a reaction at the start of two enzyme-controlled pathways. The diagram
shows these pathways.
Use the information in the diagram to give two symptoms you might expect to be visible
in a person who produces non-functioning PAH. (2 marks)
