Trace resistant bacteria into later generations
Surviving bacteria can multiply and pass resistance information to descendants.
Exam scope and assessment notes
Extended only (Supplement) · Biology 0970, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked; human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0970 has the same subject content as IGCSE 0610, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- This is an invented conditional calculation, not a bacterial growth protocol or a prediction of treatment outcomes.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why state what happens after resistant bacteria survive?
What you will learn
- Explain how survivors contribute to later bacterial populations
- Keep stipulated reproduction separate from survival counts
Original selection schematic. Population counts and scenarios are supplied teaching models; they are not field observations.
| Stage or record | Meaning |
|---|---|
| After selection | 8 resistant; 9 susceptible |
| Five equal doublings | Multiply each by 32 |
| Resistant total | 256 |
| Susceptible total | 288; resistant share stays 8/17 |
Explain how survivors contribute to later bacterial populations
Natural selection changes the make-up of a population when variants differ in survival and reproduction in a particular environment. After an antibiotic selection event, surviving resistant bacteria can multiply. If resistance is inherited, descendants can carry the resistance trait and form a larger share of later generations.
Survival and reproduction are separate stages. A survivor count immediately after exposure does not yet tell us how many descendants will be produced. A follow-up model must state the reproduction assumptions, observation period and whether the environment continues to include the antibiotic.
Keep stipulated reproduction separate from survival counts
For a simple invented arithmetic model, suppose 8 resistant and 9 susceptible survivors each double for five stated rounds, with no further selection or other changes. The totals become 256 resistant and 288 susceptible. Their resistant fraction remains 8/17 because both groups were stipulated to multiply at the same rate after the selection event.
This calculation is not a prediction of a real bacterial culture. Real populations can change through differing growth, resource limits, further exposure, gene transfer and other processes. The syllabus explanation is that resistant bacterial strains develop through natural selection; no laboratory activity or clinical recommendation follows from this numerical illustration.
Pause and explain
If two groups multiply by the same factor after selection, what happens to their ratio?
Worked example
After the stipulated selection counts, 8 resistant and 9 susceptible survivors double equally for five rounds. Calculate both totals and the resistant fraction.
Show the worked solution
- Each group is multiplied by 2 five times: factor 2⁵ = 32.
- Resistant = 8×32=256; susceptible = 9×32=288; compare 256 with the total 544.
Answer The stipulated totals are 256 resistant and 288 susceptible. The resistant fraction is 256/544 = 8/17 ≈ 47.1%, unchanged because equal multiplication was assumed after selection.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Five resistant survivors and 15 susceptible survivors double equally for three stipulated rounds. Find the group totals and resistant share.
Give me a hint
Three doublings multiply each group by 8.
Compare my reasoning
- Resistant: 5×8=40; susceptible: 15×8=120.
- The total is 160; resistant share is 40/160 = 25%.
The totals are 40 and 120; the resistant share stays 25% under equal multiplication.
Look for these in your work
- Power-of-two factor
- Share remains unchanged
A later survey after multiplication shows 40 resistant bacteria but gives no susceptible count. What cannot be inferred?
Give me a hint
The after-population denominator is absent.
Compare my reasoning
- 40 is an absolute resistant count.
- The resistant fraction requires all counted bacteria.
The fraction and whether its share rose cannot be calculated without the total or susceptible count.
Look for these in your work
- Absolute count identified
- Missing denominator named
A model assumes 8 resistant and 9 susceptible survivors double equally, then claims resistance increases every round. Find the inconsistency.
Give me a hint
Compare both groups’ multiplication factors.
Compare my reasoning
- Both groups are multiplied by the same factor.
- Equal multiplication preserves 8/17 as the resistant fraction.
The claim conflicts with the assumption: both counts rise, but the relative fraction remains 8/17 unless another differential process is added.
Look for these in your work
- Absolute and relative changes separated
- No invented selection
Common mistakes
- Applying an unstated second selection event
- Treating a conditional model as real culture growth
- Saying the resistant fraction rises when both groups multiply equally
What can you explain now?
If two groups multiply by the same factor after selection, what happens to their ratio?
Compare with the explanation
It remains the same under that model
Multiplying both numerator and denominator by the same factor preserves the ratio.
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Make it stick
Without your notes, explain how survivors contribute to later bacterial populations and keep stipulated reproduction separate from survival counts. Explain the mechanism, use the given population or breeding records and state what the evidence does not establish.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.