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Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
72 original questions across 15 selected topic areas. No account or payment required to practise.
Cambridge International AS & A Level Biology (9700), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.
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Original English concept practice addressing selected identified gaps. Not complete syllabus coverage or practical competence.
Selected subtopics: Prokaryotes, eukaryotes and viruses · Chromosome and chromatid counts · Biochemical test interpretation · Cellulose and water
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Question 1Three hypothetical samples are described: A has 70S cytoplasmic ribosomes, a peptidoglycan wall and circular DNA but no membrane-bound nucleus; B has a nucleus, mitochondria and 80S cytoplasmic ribosomes; C has a protein capsid around RNA with no ribosomes. Classify each as a typical bacterium, eukaryotic cell or virus, and use the supplied evidence to explain two important differences.4 indicative marks
Explain using the evidence4 indicative marksNo calculator
Three hypothetical samples are described: A has 70S cytoplasmic ribosomes, a peptidoglycan wall and circular DNA but no membrane-bound nucleus; B has a nucleus, mitochondria and 80S cytoplasmic ribosomes; C has a protein capsid around RNA with no ribosomes. Classify each as a typical bacterium, eukaryotic cell or virus, and use the supplied evidence to explain two important differences.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked bacterial peptidoglycan/nuclear distinction, cytoplasmic versus organelle ribosomes and viral noncellular structure against 1.2; no unsupported plant/animal assignment. This is conceptual placement, not completion of the entire referenced subsection.
9700: 1.2: Eukaryotic, prokaryotic and viral structures
Use combinations of features. Circular DNA alone is not a sufficient bacterial identifier because some eukaryotic organelles also contain circular DNA.
The descriptions do not establish B as specifically a plant or animal cell. C's lack of cellular machinery makes it different from a very small bacterium, not simply a cell lacking a nucleus.
Self-marking points
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A is a typical bacterium: prokaryotic organisation with peptidoglycan and no membrane-bound nucleus.
B is a eukaryotic cell: nucleus and membrane-bound mitochondria distinguish it from A.
C is a virus: a non-cellular capsid/nucleic-acid structure, without its own ribosomes.
Cells contain translation machinery whereas this virus depends on host-cell machinery; 70S versus 80S refers here to cytoplasmic ribosomes, not every ribosome in a eukaryote.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 2A diploid cell has 2n=6 chromosomes before DNA replication. State chromosome and chromatid counts at mitotic metaphase and explain what each daughter nucleus receives after normal mitosis. Distinguish S phase from chromosome separation and explain why uncontrolled cell-cycle progression can produce a tumour.4 indicative marks
Explain using the evidence4 indicative marksNo calculator
A diploid cell has 2n=6 chromosomes before DNA replication. State chromosome and chromatid counts at mitotic metaphase and explain what each daughter nucleus receives after normal mitosis. Distinguish S phase from chromosome separation and explain why uncontrolled cell-cycle progression can produce a tumour.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Independently tracked 6 chromosomes/12 chromatids at metaphase and 6 per daughter, with explicit stage and no confusing mitosis with meiotic reduction. This is conceptual placement, not completion of the entire referenced subsection.
Count duplicated chromosomes by their still-joined centromere units at metaphase. Replication doubles DNA/chromatid amount without changing the cell's metaphase chromosome number from six to twelve.
Mitosis apportions one sister-derived chromosome of each pair to each daughter. The chromosome number can transiently be counted differently after centromeres split within the undivided cell, so specify the requested stage.
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At metaphase there are 6 duplicated chromosomes, each consisting of two sister chromatids: 12 chromatids.
S phase replicates DNA before mitosis; it does not itself separate sister chromatids into daughter nuclei.
At anaphase sisters separate; each daughter nucleus ultimately receives 6 chromosomes with the same genetic information, absent mutation.
Loss of normal cell-cycle control can lead to repeated inappropriate cell division and a growing mass of cells.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 3A supplied report states solution X gives a negative reducing-sugar test before hydrolysis but a positive result after hydrolysis and neutralisation. A separate insoluble fibre contains parallel beta-glucose chains joined by many hydrogen bonds. Interpret X without claiming one unique sugar, identify the fibre polymer and explain its mechanical role. Explain also why water buffers temperature changes. No tests are to be performed.4 indicative marks
Explain using the evidence4 indicative marksNo calculator
A supplied report states solution X gives a negative reducing-sugar test before hydrolysis but a positive result after hydrolysis and neutralisation. A separate insoluble fibre contains parallel beta-glucose chains joined by many hydrogen bonds. Interpret X without claiming one unique sugar, identify the fibre polymer and explain its mechanical role. Explain also why water buffers temperature changes. No tests are to be performed.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked non-reducing-sugar inference limits, beta-cellulose versus alpha storage polymers and hydrogen-bond/high-heat-capacity link; no hazardous test instructions. This is conceptual placement, not completion of the entire referenced subsection.
The hydrolysis stage breaks glycosidic bonds; neutralising before the reported subsequent test avoids attributing a pH artefact to sugar identity. This is interpretation of a report, not a reagent protocol.
Cellulose's beta-linked straight chains differ from the coiled/branched alpha-glucose storage polymers. Water's bulk thermal behaviour is a separate consequence of intermolecular interactions, not breaking its O-H covalent bonds on mild warming.
Self-marking points
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X is consistent with a non-reducing sugar that yields reducing sugars on hydrolysis, such as sucrose; these observations alone do not uniquely identify it.
The fibre is cellulose with beta-glucose chains linked into microfibrils by hydrogen bonding.
Many interchain bonds support tensile strength in plant cell walls.
Water's hydrogen bonding contributes to high specific heat capacity, so a given heat transfer causes a relatively small temperature change.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
