NCEA Level 2 Biology · Learner resource

Cell Biology Learner Evidence Pack

Use the numbered sheet named in each lesson. All numerical datasets are clearly labelled classroom models; no value is presented as research or clinical data.

Ingoa | Name

Akomanga | Class

Ngā Whāinga Ako | Learning Intentions

Use diagrams, observations, and numerical evidence to explain cellular life processes.
Connect cell structure, process, and consequence in a supplied context.
Distinguish sourced knowledge from inference and from teacher-authored model data.

Paearu Angitu | Success Criteria

I label the evidence I use rather than relying on a memorised paragraph.
I give biological reasons for how or why a result occurs.
I link ideas when the question asks me to justify, compare, analyse, or evaluate.

← Return to the Cell Biology unit

Lesson 1 Read the cell backwards

For each evidence profile, infer the most likely cell type or job. Name the two observations that carry the most weight, then explain how each structure supports that function.

Profile A

Abundant rough endoplasmic reticulum; a large Golgi apparatus; many membrane-bound secretory vesicles; several mitochondria.

Profile B

Long, cylindrical cell; many mitochondria between bundles of contractile protein; more than one nucleus.

Profile C

Rigid cell wall; large central vacuole; many chloroplasts positioned near the cell edge; tightly packed with neighbouring cells.

Profile D

One long, thin extension; no chloroplasts; many mitochondria; large membrane area in contact with soil water.

Profile	Cell type or job	Observation 1 → function	Observation 2 → function
A
B
C
D

He ara mātai | Observe with authority and care

Harakeke is not a disposable generic lab specimen. The Science Learning Hub advises schools to seek local iwi knowledge about protocols before harvesting or using harakeke. If that guidance and permission are not in place, use teacher-provided material, an existing slide, or a digital image instead.

Read Observing harakeke — Science Learning Hub and Harakeke — Te Papa.
Make two headings: what our microscopy shows and what the named Māori source teaches.
Place each claim under its actual source. Do not turn whakapapa, tikanga, or rongoā into a metaphor for organelles.
Write one action your class will take before collecting any living material.

Lesson 2 Membrane decisions and tonicity

A. Membrane decision grid

Substance	Key property	Main route or routes: direct, protein-mediated, or energy-requiring?	Biological reason
O₂	Small, non-polar molecule
CO₂	Small, non-polar molecule
Glucose	Large, polar molecule
Na⁺	Charged ion
Water	Small, polar molecule; direct bilayer crossing is slow

B. Six tonicity scenarios

In each row, draw an arrow showing net water movement and predict the cell outcome.

Cell	Solute concentration outside relative to inside	Net water movement	Predicted outcome and reason
Animal	Lower outside
Animal	Equal
Animal	Higher outside
Plant	Lower outside
Plant	Equal
Plant	Higher outside

Extension: Explain why facilitated diffusion is passive even though a membrane protein is involved.

Lesson 3 SA:V and agar diffusion evidence

Data status: teacher-authored model data for practising analysis. Four cubes were placed in the same solution for 10 minutes. The indicator changed colour where solution penetrated.

Cube side length (cm)	Surface area (cm²)	Volume (cm³)	SA:V	Mean penetration from each face (cm)	Calculated percentage of cube reached
1				0.45
2				0.45
3				0.45
4				0.45

Calculate surface area using 6s² and volume using s³. Express each SA:V as a ratio to 1.
Plot side length against SA:V.
For each cube, calculate the unpenetrated core side: side length − (2 × penetration). Use that to find the percentage reached.
Use the numbers to explain why folded exchange surfaces help lungs, gills, and root systems.

Lesson 4 Two-phase photosynthesis map

Process	Exact chloroplast location	Inputs	Outputs	Link to the other process
Light-dependent processes
Light-independent processes

Fictional classroom dataset · light intensity and photosynthesis rate: Plot the supplied series, describe where the response begins to plateau, and explain why more light no longer increases the rate once another factor becomes limiting.

Light intensity (relative units)	Photosynthesis rate (relative units)
0	0
20	4
40	8
60	11
80	12
100	12

Atom trace: Use a different colour for carbon, oxygen, and hydrogen. Mark which molecule supplies the released oxygen and which supplies the carbon in carbohydrate.

Lesson 5 Respiration comparison

Level 2 Biology requires comparison of aerobic and anaerobic respiration. Use substantially more ATP for aerobic respiration unless a question supplies a particular yield; fixed totals vary by model and cell conditions.

Feature	Aerobic respiration	Anaerobic respiration in animal cells	Anaerobic respiration in yeast
Oxygen required?
Cell location(s)
Products
Relative ATP yield
Advantage in context
Limitation in context

Explain: Why can a rapidly contracting muscle use a low-yield pathway temporarily?

Lesson 6 Enzyme inhibitor datasets

Data status: all values below are simulated classroom data for pattern interpretation. Dataset A and Dataset B each include the same enzyme plus a different inhibitor.

A. Temperature factor dataset

Temperature (°C)	Relative reaction rate
10	12
20	29
30	53
40	76
50	39
60	8

B. pH factor dataset

pH	Relative reaction rate
3	7
5	33
7	68
8	81
10	24

C. Substrate concentration and inhibitor datasets

Substrate concentration (mmol L⁻¹)	No inhibitor (relative rate)	Dataset A (relative rate)	Dataset B (relative rate)
1	12	5	6
2	22	11	11
4	38	24	19
8	56	44	28
16	67	61	34

Plot and describe the temperature, pH and no-inhibitor substrate patterns.
Plot the no-inhibitor, Dataset A and Dataset B substrate series on the same axes.
Identify the likely competitive inhibitor and the likely non-competitive inhibitor.
Justify both decisions from what happens as substrate concentration rises and from where each inhibitor binds.
Explain why heat beyond an enzyme's optimum cannot be overcome simply by adding more substrate.
Predict what could happen if this enzyme also required a cofactor or coenzyme that was absent.

Lesson 7 Cell-cycle sequencing and DNA replication

A. Cut and sequence the evidence cards

Card T

Chromosomes condense and become visible. The nuclear envelope begins to break down.

Card K

Chromosomes line up across the cell equator. Spindle fibres attach at centromeres.

Card A

Sister chromatids separate and move towards opposite poles.

Card R

Chromosomes reach the poles and new nuclear envelopes form.

Card I

DNA is uncondensed. The cell grows and replicates its DNA before division.

Card C

The cytoplasm divides; a cleavage furrow or cell plate separates the daughter cells.

B. Semi-conservative replication

The two strands below are antiparallel: they run in opposite 5′ to 3′ directions. Add one complementary new strand beside each original strand.

Original strand 1:  5′—A T G C C A—3′
Original strand 2:  3′—T A C G G T—5′

Label helicase, free nucleotides, and DNA polymerase on your replication diagram.
Show complementary base pairing.
Explain why each daughter DNA molecule contains one original strand and one newly synthesised strand.

Lesson 8 Tissue turnover and division signals

Data status: simulated cell counts for practising mitotic-index calculation. They are not clinical incidence or treatment data.

Tissue sample	Total cells counted	Cells visibly in mitosis	Functional clue
Small-intestine lining	160	32	Constant abrasion and replacement
Bone marrow	180	34	Continuous blood-cell production
Hair-follicle matrix	150	24	Continuous hair growth
Basal skin layer	200	26	Replaces surface cells
Liver	180	3	Usually slow; can increase after damage
Mature neurons	200	0	Most remain outside the active cell cycle

Calculate: mitotic index = cells visibly in mitosis ÷ total cells counted. Rank the six samples, cite the calculated values, and state one limitation of the index.

Direct resource or indirect control?

Sort each factor into direct resource availability or an indirect influence on division. Then explain one from each group through resource supply, enzyme activity or cell-cycle control.

Glucose availability · oxygen availability · amino-acid availability · temperature · pH · growth-factor signalling · DNA-damage checkpoint activity · mutagen exposure

Lesson 9 Moderation practice responses

Response status: these are teacher-authored practice responses, not NZQA candidate scripts. Grade them against the official criteria and the question, not by length or keywords.

Practice question

A palisade cell is exposed to increasing light intensity. Its photosynthetic rate rises and then reaches a plateau. Explain the pattern. In your answer, connect chloroplast structure, the two photosynthesis processes, and at least one limiting factor.

Response A

Palisade cells contain many chloroplasts. Photosynthesis uses light energy to make glucose. When there is more light, photosynthesis is faster, but eventually it cannot get any faster.

Response B

At low light intensity, light is the limiting factor. More light absorbed by chlorophyll allows the light-dependent processes on the thylakoid membranes to occur faster, producing more ATP and NADPH. These products support the light-independent processes in the stroma, so carbohydrate production increases. At high light intensity, another factor such as carbon dioxide concentration or temperature limits the overall rate, causing the plateau.

Response C

Stacked thylakoid membranes provide a large surface area for chlorophyll and the reactions that capture light energy. As light intensity rises, ATP and NADPH production rises, so the stroma reactions can fix carbon dioxide faster. Once light is no longer limiting, extra light cannot raise the rate if carbon dioxide supply is insufficient or enzyme activity is limited by temperature. The two processes are linked: slowing carbon fixation also reduces how quickly ATP and NADPH can be used and regenerated. The plateau therefore reflects the whole chloroplast system rather than the cell “running out of space”.

Response	Provisional grade	Exact evidence that meets the criterion	One missing or uncertain link
A
B
C

Lesson 10 Integrated synthesis question

Data status: simulated classroom dataset. Values are relative indices designed to support causal reasoning.

A greenhouse tomato plant is moved from 22 °C to 8 °C. Light intensity stays constant, and the soil solution becomes more concentrated after fertiliser is added.

Measure	Before: 22 °C	After: 8 °C
Net photosynthesis (relative units)	18	6
Oxygen uptake in darkness (relative units)	10	4
Nitrate uptake by roots (relative units)	100	38
Leaf-cell turgor index (relative units)	100	74

Write one integrated response that:

explains how lower temperature changes enzyme-mediated photosynthesis and respiration;
links reduced respiration to ATP supply and active nitrate transport;
uses osmosis and water potential to explain the turgor change after the soil solution becomes more concentrated;
connects at least two of these effects to the functioning of the whole plant cell.

Self-check: box one description, underline two biological reasons, and draw arrows between at least two linked ideas.