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A-level Biology (7401/7402)Year 1 Comprehension Questions

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Time: 92 Minutes

Marks: 73

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Q1. Read the following passage.

Malaria is a disease so deadly that it has devastated armies and destroyed great civilisations.It has been estimated that in the course of history malaria has been responsible for the deathof one out of every two people who have ever lived. Even today, with all the advantages ofmodern technology, it is still responsible for some three million deaths a year.

5 The first half of the twentieth century was a time of hope for malarial control. The drugs

chloroquine and proguanil had just been discovered and there seemed a real possibility of amalaria-free world. Unfortunately, this honeymoon ended almost as soon as it had started,with the emergence of drug-resistant parasite populations. Scientists now accept that whatevernew drug they come up with, it is likely to have a very limited effective life. As a result, they

10 are increasingly looking at combinations of drugs.

The approach to malaria control which holds the best hope is the production of a vaccine. Oneof these is being developed by a researcher in South America. His vaccine is based on a smallsynthetic polypeptide called SPf66 which is dissolved in a saline solution and given as aninjection. A series of early trials on human volunteers produced confusing results. In one trial

15 the effectiveness of the vaccine was claimed to be 80% while, in others, the results were

statistically insignificant. Not only were the results inconclusive but the methods used werechallenged by other scientists. In particular, the controls were considered inappropriate.

Another, possibly more promising, approach has been the development of a DNA-basedvaccine. In theory, all that is required is to identify the DNA from the parasite which encodes

20 key antigens. Unfortunately, scientists have hit snags. Although they have succeeded in

sequencing the human genome, the genome of the malarial parasite has created majordifficulties. This is partly because of the very high proportion of the bases adenine andthymine. In some places these two bases average 80%, and on chromosomes 2 and 3 nearly100% of the bases present are adenine and thymine. Because of this, it has proved impossible

25 to cut the relevant DNA with the commonly available restriction enzymes into pieces of a

suitable size for analysis.

Use information from the passage and your own knowledge to answer the following questions.

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(a) Explain how a resistant parasite population is likely to arise and limit the life of any new anti-malarial drug (lines 8 - 9).

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(b) A person has a 1 in 500 probability of being infected by a chloroquine-resistant strain of malarial parasite and a 1 in 500 probability of being infected by a proguanil-resistant strain. Use a calculation from these figures to explain why scientists are “increasingly looking at combinations of drugs” (lines 9 - 10).

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(c) (i) Explain why trials of the SPf66 vaccine needed a control.

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(ii) The controls for the SPf66 vaccine trials were considered inappropriate (line 17).

Suggest how the control groups in these trials should have been treated.

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(d) In some of the DNA of a malarial parasite, the proportion of adenine and thymine bases averages 80% (lines 22 - 23). In this DNA what percentage of the nucleotides would you expect to contain

(i) phosphate; ..........................................................................................

(ii) guanine? .............................................................................................(2)

(e) (i) Use your knowledge of enzymes to explain why restriction enzymes only cut DNA at specific restriction sites.

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(ii) Restriction enzymes that can cut the DNA of chromosomes 2 and 3 produce pieces that are too small for analysis. Explain why these restriction enzymes produce small DNA fragments.

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(Total 15 marks)


The plasma membrane plays a vital role in microorganisms. It forms a barrier between the celland its environment, controlling the entry and exit of solutes. This makes bacteria vulnerableto a range of antiseptics and antibiotics

When bacteria are treated with antiseptics, the antiseptics bind to the proteins in the

5 membrane and create tiny holes. Bacteria contain potassium ions at a concentration many

times that outside the cell. Because of the small size of these ions and their concentration inthe cell, the first observable sign of antiseptic damage to the plasma membrane is the leakingof potassium ions from the cell. Some antibiotics damage the plasma membrane in a similarway. One of these is tyrocidin. This is a cyclic polypeptide consisting of a ring of ten

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amino10 acids. Tyrocidin and other polypeptide antibiotics are of little use in medicine.

Other antibiotics also increase the rate of potassium movement from cells. It is thought thatpotassium ions are very important in energy release and protein synthesis, and a loss ofpotassium ions would lead to cell death. Gramicidin A coils to form a permanent pore passingthrough the plasma membrane. This pore enables potassium ions to be conducted from the

15 inside of the cell into the surrounding medium. Vanilomycin also facilitates the passage of

potassium ions from the cell. A molecule of vanilomycin forms a complex with a potassiumion and transports it across the membrane. The potassium ion is released on the outside andthe vanilomycin is free to return and pick up another potassium ion. Vanilomycin depends onthe fluid nature of the plasma membrane in order to function.

20 Polyene antibiotics have flattened ring-shaped molecules. The two sides of the ring differ from

each other. One side consists of an unsaturated carbon chain. This part is stronglyhydrophobic and rigid. The opposite side is a flexible, strongly hydrophilic region. It has beenshown that polyene antibiotics bind only to sterols. Sterols are lipids found in the membranesof eukaryotes but not in the membranes of prokaryotic organisms. It is thought that several

25 sterol-polyene complexes come together. The plasma membranes of eukaryotic cells treated

with these polyene antibiotics lose the ability to act as selective barriers and small ions andmolecules rapidly leak out

Use information in the passage and your own knowledge to answer the questions.

(a) By what process do potassium ions normally enter a bacterial cell? Explain the evidence for your answer.

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(b) (i) Draw a peptide bond showing how the COOH group of one amino acid joins to the NH2 group of another.

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(1)

(ii) How many peptide bonds are there in a molecule of tyrocidin (lines 9 - 10)?

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(c) Experiments have shown that vanilomycin is unable to transport potassium ions across a membrane when it is cooled. Gramicidin A continues to facilitate the movement of potassium ions at these low temperatures. Explain these results.

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(d) Draw a simple diagram of one of the phospholipid layers to show how polyene antibiotics allow small ions and molecules to leak rapidly through a plasma membrane. Use the following symbols to represent the different molecules.

Note that the zigzag line on the symbol for the polyene antibiotic represents its hydrophobic region.

(2)

(Total 9 marks)


An anti-gal antibody is a type of antibody that helps to fight infections causedby bacteria. If a person has a bacterial infection, for example Salmonella, anti-galantibodies bind to antigens on the surface of the Salmonella. Not all theanti-gal antibodies are used to fight the infection. Even after the infection, anti-gal

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antibodies remain in the blood. 5

Scientists have made adaptor molecules to try to use the anti-gal antibodiesagainst viruses such as HIV. The adaptor molecules are proteins. Each adaptormolecule had a receptor site to which the HIV binds. This receptor site wassimilar to the receptor site on human cells to which the HIV binds. Theadaptor molecule has another site to which an anti-gal antibody will bind. 10

The scientists then investigated whether adding adaptor molecules and anti-galantibodies can prevent HIV entering cells. They added adaptor moleculesand anti-gal antibodies to a culture of human cells. They then added HIVto the culture. Their results showed that 90% of the virus particles failedto infect cells. 15

The scientists are hoping to develop a different type of adaptor molecule to use against MRSA.

(a) (i) What is an antigen? (line 3)

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(ii) Explain why antibodies against Salmonella do not normally bind to HIV.

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(iii) Explain how the adaptor molecule allows anti-gal antibodies to associate with HIV.

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(b) Describe how humans produce antibodies against a pathogen such as Salmonella.

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(c) (i) HIV infects some human cells, such as T-cells, but not others. Suggest why.

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(ii) Antibiotics are not used to treat viral infections, such as HIV. Explain why.

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(d) (i) When HIV, anti-gal and the adaptor molecule were added to a culture of human cells, 90% of the virus did not infect human cells. (lines 12-15). Explain why.

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(ii) Explain why a different type of adaptor molecule will have to be made to use against MRSA. (lines 16-17)

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(Total 20 marks)

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Q4. Vitamin D deficiency reduces the uptake of calcium ions by epithelial cells lining the small intestine. The diagrams show how calcium ions are transported through normal epithelial cells and those deficient in vitamin D.

(i) Use the information in the diagrams to explain how vitamin D deficiency reduces calcium ion uptake through gut epithelial cells.

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S (ii) Membrane proteins A and B transport calcium ions through cell surface membranes. Explain how each type of membrane protein transports calcium ions.

Protein A ......................................................................................................

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Protein B ......................................................................................................

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(Total 6 marks)

Q5. Read the following passage. 5

10

Campylobacter jejuni is a bacterium. It is one of the commonest causes of diarrhoea in humans. The illness that it causes does not usually last very long and many sufferers do not even go to the doctor. The only treatment required is the use of oral rehydration solutions to replace the water lost by diarrhoea. In 1998, laboratory tests confirmed 60 000 cases of diarrhoea caused by this bacterium in the UK. The bacterium was more frequently found in males than in females with a ratio of 1.5 : 1.

In rare cases, the nervous system may be affected. Scientists are now beginning to understand the cause of this. Sugars in the antigens on the surface of the bacteria are identical to some of the sugars on the surface of nerve cells. Antibodies produced against the bacteria may therefore attack the body’s nerve cells. There can be serious problems if this leads to paralysis of the diaphragm. Breathing difficulties result and the patient may die.

Use information in the passage and your own knowledge to answer the following questions.

(a) (i) The number of cases of diarrhoea confirmed as being caused by Campylobacter jejuni in the UK in 1998 was 60 000 (lines 4–5). Explain why the true number of cases is thought to be more than this.

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(ii) Calculate the number of cases of diarrhoea confirmed as being caused by Campylobacter jejuni in men in 1998.

Answer ..........................................(1)

(b) Explain why antibodies produced against Campylobacter jejuni also attack nerve cells (lines 9 –10).

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(c) Explain how paralysis of the diaphragm leads to breathing difficulties (line 11).

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(Total 7 marks)

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Q6. S The figure below shows the processes involved in absorbing amino acids into a capillary from the small intestine.

(i) Name processes A, B and C. In each case, give the evidence for your answer.

A Process ...............................................................................................

Evidence .............................................................................................

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Evidence .............................................................................................

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(ii) Explain how process B creates the conditions for process A to occur.

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(Total 5 marks)

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Q7. Courtship and mating in fruitflies can occur equally well in the light or dark.

The diagrams show the courtship sequence of males from two closely related species of fruitfly (species A and species B). The numbers show the probability of one courtship element following from another.

(a) Once a male of species A has orientated to the female, what is the probability that he will perform each courtship element once only and then attempt to mate? Show your working.

Probability ........................................(2)

(b) Suggest how the courtship sequences provide evidence to support the claim that the two species are

(i) closely related;

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(ii) separate species.

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(c) During courtship, vibration of the wings creates a sound. The sound is different in the two species of fruitfly. Explain how this prevents mating between members of different species.

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Q8.The Amish are a group of people who live in America. This group was founded by 30 Swiss people, who moved to America many years ago. The Amish do not usually marry people from outside their own group.

One of the 30 Swiss founders had a genetic disorder called Ellis-van Creveld syndrome. People with this disorder have heart defects, are short and have extra fingers and toes. Ellis-van Creveld syndrome is caused by a faulty allele.

In America today, about 1 in 200 Amish people are born with Ellis-van Creveld syndrome. This disorder is very rare in people in America who are not Amish.

(a) In America today, there are approximately 1250 Amish people who have Ellis-van Creveld syndrome. Use the information provided to calculate the current Amish population of America.

Amish population .....................................(1)

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(b) The faulty allele that causes Ellis-van Creveld syndrome is the result of a mutation of a gene called EVC. This mutation leads to the production of a protein that has one amino acid missing.

(i) Suggest how a mutation can lead to the production of a protein that has one amino acid missing.

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(ii) Suggest how the production of a protein with one amino acid missing may lead to a genetic disorder such as Ellis-van Creveld syndrome.

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(Total 5 marks)

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M1. (a) Presence of resistant and non-resistant varieties / mutation produces resistant variety;Resistant ones survive / non-resistant ones killed by treatment;These will reproduce and produce more resistant parasites / pass on resistance allele;

3

(b) Likelihood of being infected (by strain resistant to both drugs) is less;1/500 × 1/500/1/250 000;Drug has longer effective life;

max 2

(c) (i) As comparison / to show that nothing else in the treatment was responsible;

1

(ii) Given injections of saline / injection without SPf66;(otherwise) treated the same as experimental group;

2

(d) (i) 100%;1

(ii) 10%;1

(e) (i) Different lengths of DNA have different base sequences / cut at specific sequence;Results in different shape / different shape of active site;Therefore (specific sequence) will only fit active site of enzyme;

3

(ii) Recognition sites contain only AT pairs;Which would occur very frequently;

2[15]

M2. (a) Rate of movement / diffusion proportional to concentration gradient / difference in concentration;

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High concentration of potassium ions inside cell compared to outside;Must mention high concentration. Ignore reference to other factors ifreasoning is appropriate.

2

(b) (i) O || C – N | H;

1

(ii) 10;1

(c) Action of vanilomycin depends on fluidity of membrane;Fluidity reduced / not fluid at low temperatures;Pore formed by gramicidin A remains in place / permanent;

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(d) Pore between sterol molecules lined with polyene antibiotic;Hydrophobic region next to sterol;

2[9]

M3. (a) (i) Molecule/protein/glycoprotein;Stimulates immune response;(That causes) production of antibodies;

2 max

(ii) Antigens on HIV are different (shape);So, antibody will not ‘fit’/not complementary (to antigen);Receptor sites on antibody specific to one antigen;

2 max

(iii) (Has site with) same shape as salmonella antigen so bindsto anti-gal antibodies;(Has site with) same shape as receptor molecule so that HIV will bind;Binds to both molecules;

2 max

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(b) Salmonella pathogen has specific antigen on surface;Salmonella pathogen engulfed by macrophage;T-cells activate B-cells;B-cell with complementary/specific receptor antibody activated/clonal selection;B-cells divide/form clone/clonal expansion;Plasma cells make antibodies;Specific to antigen/bind to salmonella bacterial antigen;

Accept macrophage presents antigen to T/B cells;Accept T-cells release factors;

6 max

(c) (i) HIV binds to specific receptor;Only present on certain cells / T-cells;

2

(ii) Antibiotics stop metabolism, viruses don’t have metabolism;Viruses hide in cells, antibiotics can’t reach;

Two suitable cell components antibiotics work against thatviruses don’t have;e.g. some antibiotics work against ribosomes, that viruses don’t have

2

(d) (i) Adaptor molecule binds to HIV;(This) prevents the HIV binding to the receptor;Therefore few HIV available to infect cells;

2 max

(ii) Would need to be complementary to MRSA (antigens);MRSA has different antigens;But would still need to have binding site for anti-gal;

2 max[20]

M4. (i) less / no calbindin protein;{reject carrier protein)calcium not transported / moved (across the cytoplasm);so diffusion gradient reduced at small intestine interface;

2

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(ii) A is channel / pore protein (for calcium ions);passage by facilitated diffusion down diffusion / concentration gradient;

2

B is carrier protein(for calcium ions);passage by active transport against concentration gradient / requires energy / ATP;

2[6]

M5. (a) (i) Many people do not go to the doctor;1

(ii) 36000;No marks awarded for working here as calculation is very straightforward

1

(b) Same sugars / antigens on bacteria / nerve cells;Do not accept references to same shape as equivalent to complementary.

Bind with antibody / form antigen-antibody complex;Reject react

Have complementary shape / fit binding site;Reject active site

3

(c) Diaphragm will not move down / flatten / contract;Ignore references to breathing out

Thoracic cavity / lung volume not increased so cannot breathe in;2

[7]

M6. (i) In all cases reject ‘energy’ unless qualified

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A – facilitated diffusion as transport protein needed but ATP not needed;B – active transport ‘energy’ unless as (transport protein and) ATP needed;

qualifiedC – (simple) diffusion as neither ATP nor transport protein needed;

(Ignore all references to concentration gradients)3

(ii) creates low concentration of amino acids / Na+ in cell concentration gradient established between lumen and cell (of amino acids or Na+)

2[5]

M7. (a) principle of sequential multiplication (0.9×0.6×0.75×0.67);0.27;

(correct answer 2 marks)2

(b) (i) similar sequence / actions / sign stimuli;1

(ii) additional action in sequence(species A) / scissor wings blockssequence in B;

1

(c) (acts as) sign stimulus;responds only to species-specific sound;

2[6]

M8.(a) 250 000;1

(b) (i) Loss of 3 bases / triplet = 2 marks;;‘Stop codon / code formed’ = 1 mark max unless related to the last amino acid

Loss of base(s) = 1 mark;

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eg triplet for last amino acid is changed to a stop codon / code = 2 marks3 bases / triplet forms an intron = 2 marksAccept: descriptions for ‘intron’ eg non-coding DNA‘Loss of codon’ = 2 marks

2

(ii) 1. Change in tertiary structure / active site;Neutral: change in 3D shape / structure

2. (So) faulty / non-functional protein / enzyme;Accept: reference to examples of loss of function eg fewer E-S complexes formed

2[5]

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