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IGCSE · Cambridge (CIE) · 0970

Trace resistant bacteria into later generations

Surviving bacteria can multiply and pass resistance information to descendants.

Section 18.3 · Selection

Selection pathway · 0970 (9–1) · 18.3
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.

Trace resistant bacteria into later generationsOriginal schematic for this worked context. Trait, population and selection claims use only the supplied assumptions; stage labels are not measurements of every individual.Trace resistant bacteria into latergenerations1Step 1Separate immediate survivors from descendants.2Step 2Use only the stated multiplication and exposureassumptions.3Step 3Equal growth changes counts but preserves thegroup ratio.
Original schematic for this worked context. Trait, population and selection claims use only the supplied assumptions; stage labels are not measurements of every individual.
Selection evidence key
Stage or recordMeaning
After selection8 resistant; 9 susceptible
Five equal doublingsMultiply each by 32
Resistant total256
Susceptible total288; 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?

Put the idea to work

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
  1. Each group is multiplied by 2 five times: factor 2⁵ = 32.
  2. 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.

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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.

1 · Guided

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
  1. Resistant: 5×8=40; susceptible: 15×8=120.
  2. 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
2 · Independent

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
  1. 40 is an absolute resistant count.
  2. 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
3 · Transfer

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
  1. Both groups are multiplied by the same factor.
  2. 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
Recall without your notes

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.

After trying it yourself, choose your next review. This is your self-assessment.

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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.

Keep learning Biology (9-1)

Course scope and assessment

780 sequenced lessons provide 2340 guided, independent and application problems in characteristics, classification, cells, organisation, specimen size, membrane transport, biological molecules, enzymes, plant and human nutrition, plant transport, circulatory systems, the heart, blood vessels, blood, disease transmission, body defences, immunity, gas exchange in humans, respiration energy uses and investigations, aerobic respiration, anaerobic respiration and recovery, excretion in humans, coordination and response, sense organs, hormones, homeostasis, tropic responses, drugs, asexual reproduction, sexual reproduction, sexual reproduction in plants, sexual reproduction in humans, sex hormones in humans, sexually transmitted infections, chromosomes, genes and proteins, mitosis, meiosis, monohybrid inheritance, variation, adaptive features, selection, energy flow, food chains and webs, nutrient cycles, population, community, ecosystem, population-growth phases, food supply, habitat destruction, pollution, conservation, biotechnology and genetic modification. 61 of 61 syllabus sections have linked teaching and staged practice. The earlier inheritance overview remains introductory, and the monohybrid pathway develops the relevant Core and Supplement requirements. This is not full programme completion or guaranteed exam readiness. Written practice is self-checked. This is the separate 9–1 qualification 0970; Cambridge states that its 2026–2028 subject content is otherwise the same as 0610. 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. The 2029 syllabus is separately published at https://www.cambridgeinternational.org/Images/764352-2029-syllabus.pdf and is not yet mapped here. This course supports exams in 2026, 2027 and 2028 only.

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