AQA AS Level Biology Paper 2, June 2024: Question 8

9 marks · Medium difficulty · Extended Answer

Explain how an indirect ELISA test for Lyme disease works and evaluate clinical trial data comparing symptoms between PTLDS patients and a control group.

Practise this question

Question

Question 8 outlines an ELISA test for detecting antibodies to the bacterium Borrelia burgdorferi. Figure 5 depicts a 4-step flowchart: Step 1 attaches B. burgdorferi antigens to a well; Step 2 adds a blood sample where specific antibodies bind; Step 3 washes the well and adds a second antibody conjugated to an enzyme; Step 4 washes again and adds a substrate converted to a coloured product. Subquestions 08.1 to 08.3 ask why poor washing causes false positives, why syphilis causes false positives, and why testing within 2 weeks causes false negatives. Question 08.4 provides Table 3 detailing the percentage of PTLDS patients versus a control group experiencing fatigue, joint pain, depression, fever, and muscle pain, along with P values for symptom intensity, and asks to evaluate the scientists' conclusions.
Question text

08 Lyme disease is most frequently caused by the bacterium Borrelia burgdorferi.

Lyme disease can be difficult to diagnose.

Figure 5 shows an ELISA test that is used to find out if a person has antibodies to

B. burgdorferi.

Figure 5

A false positive in this test is a result which incorrectly indicates that antibodies to

B. burgdorferi are present.

08.1 Failure to thoroughly wash the well in Step 4 can result in a false positive.

Explain why.

[2 marks]

08.2 A false positive can be produced if a person has been infected by another bacterium

that causes a disease called syphilis.

*19* Suggest why.

[1 mark]

08.3 A false negative in this test is often produced if a person is tested within 2 weeks of

being infected with B. burgdorferi.

Explain why.

[2 marks]

08.4 Sometimes, symptoms of Lyme disease can persist for 6 months following antibiotic

treatment. This condition is known as Post-Treatment Lyme Disease Syndrome

(PTLDS).

Scientists investigated the symptoms experienced by a large number of PTLDS

patients and a control group. During a 2-week period, they asked all the participants:

• if they had experienced symptoms of PTLDS

• to record the intensity of these symptoms.

The scientists used a statistical test to determine if there was a difference in the21

intensity of symptoms of PTLDS between these two groups.

Table 3 shows some of the scientists’ results, including the probability (P) values

obtained using the statistical test.

Table 3

Percentage of Percentage of P value for

PTLDS group control group difference in the

Symptom

experiencing experiencing intensity of

symptom symptom symptoms

Fatigue 100 57 <0.001

Joint pain 96 32 <0.001

Depression 40 4 <0.005

Fever 35 3 <0.005

Muscle pain 86 62 <0.001

The scientists concluded that more PTLDS patients than the control group

experienced:

• symptoms

• greater intensity of symptoms.

Evaluate the scientists’ conclusions.

[4 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 8. Part 08.1 awards 2 marks for enzyme-linked second antibody remaining and substrate being converted to coloured product. Part 08.2 awards 1 mark for antibodies to syphilis having similar structure/complementary binding to B. burgdorferi antigens, or similar antigens between the bacteria. Part 08.3 awards 2 marks for low antibody concentration/plasma cells not yet formed during the primary response. Part 08.4 awards up to 4 marks from 5 points: all symptoms have a higher percentage in PTLDS; significant differences in symptom intensity with P < 0.05; fatigue/joint pain/muscle pain show greatest significance (P < 0.001); study limited to only 2 weeks; and difficulty/subjectivity in measuring symptom intensity.

Question Marking Guidance Mark Comments

1. (Second antibody with) enzyme remains; 1. Accept any

08.1 – 2 – –

description of

2. (So substrate converted to) coloured product; antibody with enzyme

(2 x remaining eg ‘not

AO2) washed out’.

Antibodies (produced against syphilis and

B. burgdorferi) are similar

OR

Antibodies (to syphilis) are complementary to

08.2 (1 x

antigens (of B. burgdorferi)

AO2)

OR

Antigen(s) (of syphilis and B. burgdorferi) are similar

(in structure);

1. Low concentration of antibodies (against

B. burgdorferi)

OR

No/little antibody produced (during first two

weeks);

2. (Only) primary response 2

08.3 OR (2 x

AO2)

By plasma cells

OR

Plasma cells not produced

OR

– – –

B cells not divided/cloned;

1. For all symptoms higher percentage in PTLDS

group;

2. Significant difference/increase (in intensity of all 2. and 3. Reject

symptoms) in PTLDS group; ‘results are significant’

3. Most significant difference/increase (in intensity) 2. Accept

in fatigue/joint pain/muscle pain symptoms in ‘difference/increase in

PTLDS group (all) results is

significant’.

OR

2. and 3. If neither mp

Less significant difference/increase (in intensity) 4 max

2 or mp 3 is awarded

in depression/fever symptoms in PTLDS group;

08.4 (4 x accept for one mark

4. Only (investigated) for 2 weeks AO3) ‘there is less than a

5% or less than 0.05

OR 15

probability of

Short time period (for investigation); difference (in intensity

of symptoms) being

due to chance’.

5. Difficult to determine intensity of symptoms

5. Accept any

OR description of

Determining intensity of symptoms is subjective; determining

symptoms eg, judging

symptoms.

How to answer it

Indirect ELISA Testing & Evaluating PTLDS Clinical Data

WHAT THIS QUESTION TESTS

This question assesses your mastery of immunological methods, humoral response kinetics, and critical statistical evaluation:

  • Principles of Indirect ELISA: Understanding the molecular role of each washing step, secondary antibodies, and enzyme-substrate colour changes.
  • Antibody Specificity & Cross-Reactivity: How structural similarities in antigens or variable regions cause false-positive test results.
  • Primary Immune Response Timeline: Clonal selection, clonal expansion, plasma cell differentiation, and the lag time required for detectable antibody titres.
  • Data Evaluation & Statistical Significance: Interpreting P-values (P < 0.05, P < 0.005, P < 0.001) relative to chance, and identifying study design limitations (subjectivity, trial duration).
Question 08.1

Step 4 Washing Failure & False Positives

Explain why failure to thoroughly wash the well in Step 4 can result in a false positive [2 marks]

✅ Correct Answer

  • Unbound secondary antibody with enzyme remains (is not washed away). [1 mark]
  • Enzyme reacts with substrate, converting it to a coloured product (even without target antibodies present). [1 mark]
Marking Guidance: Accept any clear phrasing for "second antibody with enzyme remains", such as "enzyme-linked antibody not washed out". Both the presence of the enzyme and the formation of coloured product are needed for full marks.

💡 Key Knowledge

The ELISA in Figure 5 is an indirect ELISA:

  • Step 3: Unbound primary antibodies are washed away. Second antibody (carrying enzyme) binds to primary antibody.
  • Step 4: Essential to wash away any unbound secondary enzyme-linked antibodies.
  • If unbound enzyme stays in the dish, it will hydrolyse the substrate, causing a colorimetric signal regardless of whether the patient actually has B. burgdorferi antibodies.

🧠 Exam Technique

Always state which antibody carries the enzyme. Don't simply write "the antibody remains". Specify that it is the second antibody or the antibody with enzyme attached.

❌ Common Errors

Confusing the Step 3 wash with the Step 4 wash. In Step 4, the primary antibody has already bound; the wash is solely to clear free, unbound enzyme-linked secondary antibodies.

Question 08.2

Cross-Reactivity & Syphilis Infection

Suggest why infection by another bacterium (syphilis) can cause a false positive [1 mark]

✅ Correct Answer

Any one of the following:

  • Antibodies produced against syphilis and B. burgdorferi have a similar structure / similar antigen-binding sites. [1 mark]
  • Antibodies produced against syphilis are complementary to the antigens of B. burgdorferi. [1 mark]
  • Antigens of the syphilis bacterium and B. burgdorferi are similar in structure. [1 mark]

💡 Key Knowledge

Antibody-antigen binding depends entirely on tertiary structure complementarity.

Both Borrelia burgdorferi and Treponema pallidum (syphilis) are spirochaete bacteria. They share evolutionary history and may express surface antigens with very similar tertiary epitopes. Syphilis antibodies can therefore bind to the immobilised B. burgdorferi antigens via molecular cross-reactivity.

❌ Common Errors

Writing that "the diseases are similar" or that "the bacteria are the same". Marks require reference to antibodies or antigens having a similar structure, or syphilis antibodies being complementary to B. burgdorferi antigens.

Question 08.3

False Negatives in Early Infection

Explain why a false negative is often produced within 2 weeks of infection [2 marks]

✅ Correct Answer

  • Low concentration of antibodies in the blood / no antibodies yet produced (against B. burgdorferi). [1 mark]
  • Because it is only the primary immune response / plasma cells have not yet developed / B cells have not yet undergone clonal selection and divided into plasma cells. [1 mark]

💡 Key Knowledge

The primary immune response has a pronounced latent period (lag phase):

  1. Antigen presentation by APCs.
  2. Selection and activation of specific T helper and B cells.
  3. Clonal expansion (mitosis) and differentiation into plasma cells.
  4. Synthesis and secretion of antibodies.

Within the first 14 days, the antibody titre is below the minimum detection threshold of the ELISA.

🧠 Exam Technique

Ensure you link the observation (low antibody titre) directly to the cellular mechanism (primary response / plasma cells not yet formed / B cells not cloned). Both elements are needed for the 2 marks.

❌ Common Errors

Claiming "memory cells haven't produced antibodies". Memory cells are responsible for the secondary response; in an initial infection, antibodies are secreted by freshly differentiated plasma cells.

Question 08.4

Evaluating Conclusions on PTLDS Symptoms

Evaluate the scientists' conclusions that more PTLDS patients than controls experienced: (1) symptoms, (2) greater intensity of symptoms [4 marks]

✅ Correct Answer (Award up to 4 marks max)

Evidence Supporting the Conclusions:

  • For symptom occurrence: A higher percentage of PTLDS patients experienced all five symptoms compared to the control group (e.g., fatigue 100% vs 57%, joint pain 96% vs 32%). [1 mark]
  • For symptom intensity: There is a statistically significant difference / increase in intensity for all symptoms in the PTLDS group, because all P values are < 0.05 (less than 5% probability that the difference in intensity is due to chance). [1 mark]
  • Nuance in significance: The difference in intensity is most significant for fatigue, joint pain, and muscle pain (P < 0.001) compared to depression and fever (P < 0.005). [1 mark]

Evidence Questioning / Limiting the Conclusions:

  • Duration limitation: Investigation was only carried out for 2 weeks, which is very short considering PTLDS symptoms persist for 6 months or longer. [1 mark]
  • Subjectivity: Assessing "intensity" of symptoms is subjective / difficult to measure objectively and relies on self-reporting. [1 mark]

🧠 Critical Evaluation Rules

  • Never just say "the results are significant": The mark scheme explicitly states: Reject 'results are significant'. You must say the difference in intensity is significant!
  • Address BOTH parts of the conclusion:
    Part 1: More patients experienced symptoms → Back this up with percentages from Table 3.
    Part 2: Greater intensity → Back this up with P-values and statistical significance.
  • Look for experimental weaknesses: Ask yourself: How long was the trial? (Only 2 weeks). How was the data gathered? (Self-reported rating scale → subjective).

📐 Interpreting the P-Values

Symptom P value Statistical Interpretation
Fatigue, Joint pain, Muscle pain P < 0.001 < 0.1% probability that difference in intensity is due to chance. Highly significant.
Depression, Fever P < 0.005 < 0.5% probability that difference is due to chance. Significant, but less so than the other three.

Since all P < 0.05, we reject the null hypothesis of no difference in intensity between the two groups.

❌ Common Errors

  • Confusing the percentages (which show how many people had the symptom) with the P-value (which evaluates the intensity of the symptom).
  • Stating that "P < 0.05 proves the hypothesis is true". In science, statistics never prove; they show that the probability of differences arising by chance is extremely low.
  • Failing to provide counter-arguments (e.g. failing to mention the 2-week duration or subjectivity). An "Evaluate" question always demands both sides!

Topics

Biology · Practical skills · 3.2 Cells · Data analysis · Experimental design

Question and mark scheme from the AQA AS Level Biology examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.