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IGCSE · Cambridge (CIE) · 0973

Trace resistant bacteria into later generations

Surviving bacteria can multiply and pass resistance information to descendants.

Selection · Cambridge IGCSE 0973 · B17.2
Exam scope and assessment notes

Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.

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.

Sources for this lesson

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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Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

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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 Sciences — Co-ordinated (9-1)

Course scope and assessment

This 9–1 pathway maps all 152 detailed subject-content sections in the published 0973 syllabus for exams in 2028 and 2029. The course map follows the same 2028–2029 section codes as Cambridge's A*–G 0654 syllabus, while keeping 0973's grading and entry rules separate. It adapts lessons from the related single-science courses and excludes clear single-science-only topics. A 0973-specific objective-by-objective review, Core/Supplement audit, human subject-teacher review, supervised laboratory experience and cumulative assessment remain in progress. Written practice is self-checked and does not predict an official grade. The separate 2025–2027 syllabus has not been reconciled here. Cambridge lists 0973 only in a limited number of administrative zones: June in zone 3 and centres in Oman and the UAE; November in zone 3 only. Confirm entry with your centre.

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