IB Biology SL Continuity and Change Concepts

Continuity and Change links inheritance, regulation, reproduction, selection and environmental change to explain stability and transformation in cells, organisms and populations.

Syllabus
First assessment 2025
Section
Level
SL

Exam analysis

Published Concept evidence in Continuity and Change repeatedly connects genetic mechanisms, cell division, inheritance, regulation and selection. The strongest pattern is explaining how information is maintained, altered and acted on across cells, organisms and populations.

Most tested topics

Practice this section

Recent 5 years · Updated 22 Jul 2026

In this section

Topic D1.1

D1.1 DNA replication

DNA replication copies genetic information through template strands, complementary pairing, enzyme action, proofreading, and laboratory amplification or separation techniques used to analyse DNA.

28% of analysed papers 39 papers · 46 questions

Objectives in this topic

DNA Replication Copies the Genome Before Division

DNA replication produces exact copies of DNA with identical base sequences, apart from rare copying errors.

Accurate copies preserve genetic information when cells or organisms reproduce. In multicellular organisms, replication supplies genomes for cell division during growth and replacement of damaged or worn tissues.

Original DNA sequence → replication → two matching DNA molecules → genetic continuity in reproduction, growth and tissue replacement.

Before a skin cell divides to replace lost tissue, its DNA is copied so both daughter cells can inherit the same base sequence.

Replication copies DNA; transcription makes RNA and translation makes polypeptide. ‘Identical’ describes base-sequence information, not two newly synthesized strands without templates.

DNA replication

Assessment in practice

1–8 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Explain.

Command terms

Identify / Explain

What earns marks

Build the answer around this relationship: DNA replication depends on matching each strand, enzyme or laboratory step to its exact function.

Watch for

Saying both parental strands stay together instead of one parental strand entering each daughter molecule.

Representative question

Question 1

[Maximum number: 8]

Growth in living organisms includes replication of DNA. Explain DNA replication.

Semi-Conservative Replication Keeps One Old Strand

Semi-conservative replication produces DNA molecules in which each double helix contains one parental strand and one newly synthesized strand.

When the original strands separate, each acts as a template. Complementary base pairing preserves information while retaining one physical strand from the original molecule in each product.

Identify a product by checking: one old strand; one new strand; complementary pairing between them.

After one round, heavy parental DNA in a density experiment is replaced by two intermediate molecules, each containing one old and one new strand.

Semi-conservative does not mean half the bases are copied randomly; the strand pattern is the key prediction.

Semi-conservative replication

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain / Outline / Distinguish.

Command terms

Explain / Outline / Distinguish / Identify / State

What earns marks

Build the answer around this relationship: Semi-conservative replication depends on matching each strand, enzyme or laboratory step to its exact function.

Watch for

Saying both parental strands stay together instead of one parental strand entering each daughter molecule.

Representative question

Question 1

[Maximum number: 3]

Outline the reason that DNA replication is described as semi-conservative.

Helicase Opens DNA and Polymerase Extends It

Helicase separates the two DNA strands, while DNA polymerase builds complementary DNA strands from the exposed templates.

Helicase unwinds the double helix and breaks hydrogen bonds between complementary bases. DNA polymerase selects complementary DNA nucleotides and joins them into a growing strand.

Helicase: unwind and break inter-strand hydrogen bonds. DNA polymerase: use each original strand as a template and join complementary nucleotides.

At a replication fork, helicase exposes template bases; polymerase then places A opposite T and C opposite G while extending the new DNA.

Helicase does not synthesize DNA, and polymerase does not separate the original strands. Detailed 5′/3′ directionality belongs to the later HL objective.

Role of helicase and DNA polymerase

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, structured response, commonly using Explain / Outline / Identify.

Command terms

Explain / Outline / Identify

What earns marks

Build the answer around this relationship: Role of helicase and DNA polymerase depends on matching each strand, enzyme or laboratory step to its exact function.

Watch for

Saying both parental strands stay together instead of one parental strand entering each daughter molecule.

Representative question

Question 1

[Maximum number: 1]

What is a function of the enzyme helicase?

A

It coils DNA up into a double helical shape.

B

It links DNA nucleotides in a new DNA strand.

C

It breaks hydrogen bonds between the DNA strands.

D

It forms temporary hydrogen bonds to produce messenger RNA.

PCR Amplifies DNA and Electrophoresis Separates It

PCR amplifies a selected DNA region; gel electrophoresis then separates DNA fragments mainly by length.

Primers define the target ends. Each thermal cycle uses high temperature to separate strands, lower temperature for primer binding, and a suitable extension temperature for heat-stable Taq DNA polymerase to synthesize new DNA.

Load amplified fragments into wells. Negatively charged DNA moves toward the positive electrode through the gel; shorter fragments move farther than longer fragments in the same time.

Primers amplify a variable DNA locus; electrophoresis separates the resulting fragments, and a size marker allows their approximate lengths to be compared.

PCR increases the amount of target DNA; electrophoresis separates fragments. Primer specificity, contamination controls and the size marker affect interpretation.

PCR and gel electrophoresis

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, multiple choice, commonly using Explain / Outline / State.

Command terms

Explain / Outline / State / Identify / Determine / Describe / Compare / Deduce / Predict / Suggest

What earns marks

Build the answer around this relationship: PCR and gel electrophoresis depends on matching each strand, enzyme or laboratory step to its exact function.

Watch for

Describing PCR as DNA separation instead of DNA amplification.

Representative question

Question 1

[Maximum number: 4]

Describe the polymerase chain reaction (PCR).

DNA Profiles Compare Variable Fragment Patterns

PCR and gel electrophoresis support DNA profiling by amplifying and comparing variable DNA markers.

In a paternity investigation, a child's marker alleles must be explainable by the biological parents. In a forensic investigation, a crime-scene DNA pattern can be compared with reference samples.

Use several independent markers: each additional matching marker reduces the probability that an unrelated person shares the full pattern by chance. Include positive/negative controls and guard against contamination.

A child's allele not supplied by the known parent must match the candidate parent's allele at each tested marker; one matching marker is weak, whereas a consistent multi-marker pattern is stronger evidence.

A profile supports or excludes a biological relationship/source; it does not alone prove when or how DNA reached a location. More markers reduce false-match probability but do not make laboratory error impossible.

Applications exam focus

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Determine / Explain.

Command terms

Identify / Determine / Explain / Describe / Outline

What earns marks

Build the answer around this relationship: Applications depends on matching each strand, enzyme or laboratory step to its exact function.

Watch for

Confusing DNA polymerase I primer replacement with DNA polymerase III strand elongation.

Representative question

Question 1

[Maximum number: 4]

Outline the process of DNA profiling.

Core DNA Replication

DNA replication produces exact DNA copies before cell division and maintains genetic continuity for reproduction, growth, and tissue replacement. Semi-conservative replication gives each new DNA molecule one original strand and one new strand; complementary base pairing and Meselson-Stahl isotope evidence support the model. Helicase unwinds DNA and breaks hydrogen bonds; DNA polymerase joins complementary nucleotides to build new strands. PCR amplifies selected DNA using primers, temperature cycles, and Taq polymerase; gel electrophoresis separates DNA fragments by size and charge. PCR and gel electrophoresis support DNA profiling for forensic identification and paternity testing.

Topic D1.2

D1.2 Protein synthesis

Protein synthesis links DNA information to functional proteins through transcription, RNA processing, translation, genetic-code reading and post-translational modification in cells.

29% of analysed papers 41 papers · 51 questions

Objectives in this topic

Transcription Copies a Gene into mRNA

Transcription is the synthesis of an RNA strand using one DNA strand as the template.

RNA polymerase binds the selected DNA region, separates a short section, matches RNA nucleotides to exposed template bases and joins the nucleotides into RNA.

DNA template exposed → complementary RNA nucleotides pair → RNA polymerase forms the RNA backbone → RNA separates and DNA re-forms its double helix.

For a protein-coding gene, RNA polymerase produces an mRNA transcript whose base sequence can later be translated.

Transcription synthesizes RNA, not DNA or polypeptide. Promoter/transcription-factor detail belongs to the later HL initiation objective.

Transcription exam focus

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Outline / Explain.

Command terms

State / Outline / Explain / Describe / Distinguish

What earns marks

Build the answer around this relationship: Transcription copies one DNA strand into an RNA molecule.

Watch for

Confusing transcription with translation by naming ribosomes or amino acids instead of RNA polymerase and mRNA.

Representative question

Question 1

[Maximum number: 8]

Explain the process of transcription in prokaryotes.

Hydrogen Bonds Let the Template Open and Close

Temporary hydrogen bonds between complementary bases allow an RNA sequence to be specified from a DNA template during transcription.

DNA hydrogen bonds separate locally while the covalent sugar–phosphate backbones remain intact. RNA nucleotides form complementary hydrogen bonds to exposed template bases before RNA polymerase joins them.

DNA template A pairs with RNA U; DNA template T pairs with RNA A; DNA C pairs with RNA G; DNA G pairs with RNA C.

A DNA template segment 3′-TACG-5′ specifies RNA 5′-AUGC-3′: template adenine is represented by uracil in RNA, not thymine.

Complementary does not mean identical. Hydrogen bonds guide base choice; phosphodiester bonds form the stable RNA backbone.

Hydrogen bonding in transcription

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Determine.

Command terms

Determine

What earns marks

Build the answer around this relationship: RNA uses uracil instead of thymine during base pairing.

Watch for

Writing thymine in an mRNA sequence instead of uracil.

Representative question

Question 1

[Maximum number: 1]

The sequence of bases on a short section of the antisense strand of a gene undergoing transcription is shown:

5 CATG 35^{\prime} \text { CATG } 3^{\prime}

What is the sequence of bases on the resulting mRNA?

A

33^{\prime} CATG 55^{\prime}

B

5' GUAC 3'

C

33^{\prime} GUAC 55^{\prime}

D

3GTAC3^{\prime} \mathrm{GTAC} 5'

DNA Stays Stable While Genes Are Read

A single DNA strand can act repeatedly as a transcription template without its base sequence changing.

RNA polymerase reads the template and joins separate RNA nucleotides; it does not consume or replace the DNA bases. After the transcription bubble passes, the DNA strands pair again.

DNA opens locally → RNA copy is synthesized → RNA leaves → original DNA sequence remains conserved and available for later transcription.

A non-dividing neuron may transcribe the same essential gene many times while conserving that gene's DNA sequence throughout the life of the cell.

Stable does not mean inaccessible: local opening permits transcription. The RNA transcript can change or degrade while the DNA template sequence remains unchanged.

Transcription Turns Gene Information into a Message

Transcription is the first stage of gene expression and a key point at which expression can be switched on or off.

Cells contain many genes but do not express all of them at the same time. If a gene is not transcribed, its mRNA is unavailable for translation; starting transcription makes expression possible.

Regulatory state → transcription on/off → mRNA absent/present → translation possible/not possible. Different cell types express different subsets of the same genome.

A cell can switch on transcription of an enzyme gene when the enzyme is needed while other genes remain untranscribed.

Transcriptional control is a key switch, not the only possible control. More mRNA does not guarantee proportionally more protein if later steps are limiting.

Transcription for gene expression

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Determine / Outline / Compare.

Command terms

Determine / Outline / Compare / Deduce / Evaluate

What earns marks

Build the answer around this relationship: mRNA presence indicates that a gene has been transcribed.

Watch for

Treating a visible mRNA band as protein evidence rather than evidence of transcription.

Representative question

Question 1

[Maximum number: 2]

The scientists concluded that auxin activates the transcription of the GH 3 gene. Using the information on the auxin concentration in the stem base in the graph on page 4 and the Northern blot, evaluate whether this conclusion is supported.

Translation Builds a Polypeptide from mRNA

Translation uses a ribosome to read mRNA codons and join amino acids in the encoded order.

tRNA molecules pair anticodons with codons and deliver specific amino acids. The ribosome forms peptide bonds as it moves from start toward stop.

Trace: start codon; tRNA matching; peptide bond; codon movement; stop codon.

A ribosome reading three codons adds three specified amino acids before a stop signal ends the chain.

Translation reads RNA; it does not copy mRNA back into DNA. The ribosome assembles the chain; later folding and modification determine whether the polypeptide becomes functional.

Translation exam focus

Assessment in practice

1–8 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Explain / Describe.

Command terms

Outline / Explain / Describe

What earns marks

Build the answer around this relationship: Translation occurs on ribosomes in the cytoplasm or on rough endoplasmic reticulum.

Watch for

Naming RNA polymerase as the enzyme for translation instead of locating translation at ribosomes.

Representative question

Question 1

[Maximum number: 8]

Explain how polypeptides are produced by the process of translation.

Ribosome, tRNA and mRNA Divide Translation Work

During translation, mRNA supplies codon order, tRNAs deliver specific amino acids and the ribosome positions the molecules and catalyses peptide-bond formation.

mRNA first binds to the small ribosomal subunit. The large subunit forms the catalytic complex and can hold two tRNAs simultaneously so the amino acid on one can be linked to the growing chain on the other.

mRNA: ordered codons. tRNA: anticodon plus specific amino acid. Small subunit: binds/positions mRNA. Large subunit: binds two tRNAs and supports peptide-bond formation.

One tRNA holds the growing chain while a second matching tRNA places the next amino acid beside it; a peptide bond extends the polypeptide.

No component determines the protein alone: mRNA sets sequence, tRNAs match codons, and the ribosome coordinates assembly.

Roles in translation

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Outline / Explain.

Command terms

Identify / Outline / Explain

What earns marks

Build the answer around this relationship: mRNA supplies codons that are read by the ribosome.

Watch for

Swapping the roles of codons and anticodons between mRNA and tRNA.

Representative question

Question 1

[Maximum number: 7]

Explain the role of RNA in translation, resulting in the formation of polypeptide chains.

Complementary Pairing Transfers Information

A three-base tRNA anticodon forms complementary base pairs with a three-base mRNA codon during translation.

Correct codon–anticodon pairing positions the tRNA carrying the amino acid specified by that codon, transferring mRNA sequence information into polypeptide sequence.

Read the mRNA codon 5′→3′; write the antiparallel complementary anticodon; pair A–U and C–G; identify the amino acid from the mRNA codon, not the anticodon.

mRNA codon 5′-AUG-3′ pairs with tRNA anticodon 3′-UAC-5′, positioning the tRNA carrying methionine.

Codon and anticodon are complementary and antiparallel, not identical. DNA–RNA pairing during transcription is a different Objective.

Complementary base pairing

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Explain.

Command terms

Outline / Explain

What earns marks

Build the answer around this relationship: mRNA codons pair with complementary tRNA anticodons.

Watch for

Using DNA base-pairing letters instead of RNA uracil when writing anticodons.

Representative question

Question 1

[Maximum number: 3]

Outline how translation depends on complementary base pairing.

The Genetic Code Maps Codons to Amino Acids

The genetic code is a triplet, degenerate and nearly universal mapping from mRNA codons to amino acids or stop signals.

Four bases taken two at a time give only 16 combinations, fewer than the 20 amino acids; triplets give 4³ = 64 codons, enough for all amino acids plus stop signals.

Degenerate means more than one codon can specify the same amino acid. Universal means the same codon usually specifies the same amino acid across organisms, with limited exceptions.

UUU and UUC both code for phenylalanine, showing degeneracy; UUU has one defined meaning, so degeneracy is not ambiguity.

Read codons in a fixed triplet frame. Universality is a broad biological pattern, not a claim that no exceptions exist.

Genetic code features

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Outline / Suggest.

Command terms

Identify / Outline / Suggest / Describe

What earns marks

Build the answer around this relationship: A codon is a triplet of bases on mRNA.

Watch for

Counting amino acids as if one base rather than three bases codes for each residue.

Representative question

Question 1

[Maximum number: 5]

Describe the genetic code and its relationship to polypeptides and proteins.

A Code Table Converts Codons into a Sequence

A genetic-code table is used by reading mRNA codons 5' to 3' and matching each to its amino acid.

The first, second and third bases select positions in the table. Reading-frame errors shift every later codon, so strand labels and grouping matter.

Use: locate start; split triplets; read each codon; stop at a stop codon; report amino-acid order.

For 5'-AUG-GCU-UAA-3', AUG starts, GCU adds alanine and UAA terminates. If one base is omitted before AUG, every later triplet may be regrouped, so the table must be used from the correct start frame.

Do not use the DNA template directly as if it were mRNA; transcribe first.

Using genetic code table

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Determine.

Command terms

Determine

What earns marks

Build the answer around this relationship: Genetic code tables use mRNA codons.

Watch for

Using DNA triplets directly in the mRNA code table without converting thymine to uracil.

Representative question

Question 1

[Maximum number: 1]

Which sequence of mRNA bases and amino acids could be produced by transcription and translation of the DNA molecule shown?

3' AAAGTGGCACGTATATTT 5'
5' TTTCACCGTGCATATAAA 33^{\prime}

\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline \multirow{6}{*}{} & \multicolumn{6}{|c|}{2nd base in codon} & \multirow{14}{*}{3rd base in codon} \\
\hline & & U & C & A & G & & \\
\hline & \multirow{4}{*}{U} & Phe & Ser & Tyr & Cys & U & \\
\hline & & Phe & Ser & Tyr & Cys & C & \\
\hline & & Leu & Ser & STOP & STOP & A & \\
\hline & & Leu & Ser & STOP & Trp & G & \\
\hline \multirow[t]{8}{*}{} & C & Leu & Pro & His & Arg & U & \\
\hline & \multirow{3}{*}{A} & lle & Thr & Asn & Ser & U & \\
\hline & & lle & Thr & Lys & Arg & A & \\
\hline & & Met & Thr & Lys & Arg & G & \\
\hline & \multirow{4}{*}{G} & Val & Ala & Asp & Gly & U & \\
\hline & & Val & Ala & Asp & Gly & C & \\
\hline & & Val & Ala & Glu & Gly & A & \\
\hline & & Val & Ala & Glu & Gly & G & \\
\hline
\end{tabular}

Sequence of mRNA bases

Sequence of amino acids

UUU-GAG-GCU-CGA-UAU-UUU

Phe-Glu-Ala-Arg-Tyr-Phe

AAA-CUC-CGA-GCU-AUA-UUU

Lys-Leu-Arg-Ala-lle-Phe

UUU-CAC-CGU-GCA-UAU-AAA

Phe-His-Arg-Ala-Tyr-Lys

AAA-GUG-GCA-CGU-AUA-UUU

Lys-Val-Ser-Arg-Ile-Phe

Elongation Repeats a Three-Step Ribosome Cycle

During elongation, a matching tRNA enters, a peptide bond forms and the ribosome moves to the next codon.

Repeating this cycle extends the chain in a fixed direction while mRNA advances through the ribosome. Energy from charged tRNAs and factors supports the process.

Track each cycle: codon recognition; chain transfer; translocation; empty tRNA exit.

After one cycle the chain has one additional amino acid and the next codon is exposed.

Elongation is not random amino-acid addition; codon recognition controls the order. The cycle stops at a termination signal rather than continuing until the physical end of the mRNA.

Elongation of polypeptide

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain.

Command terms

Explain

What earns marks

Build the answer around this relationship: Elongation lengthens a polypeptide one amino acid at a time.

Watch for

Describing several ribosomes on one mRNA as several genes rather than one transcript being translated many times.

Representative question

Question 1

[Maximum number: 2]

Six polypeptides are shown in the diagram. Explain the different lengths of these polypeptides.

A Mutation Can Change Protein Structure

A point mutation can change one mRNA codon, replace one amino acid and alter how a polypeptide folds or functions.

The effect depends on code degeneracy, mutation position and the chemical properties of the original and replacement amino acids. Some substitutions are synonymous; others alter interactions stabilizing protein structure.

In sickle-cell disease, a point substitution changes a β-globin codon from GAG to GUG in mRNA, replacing glutamic acid with valine. The hydrophobic replacement promotes abnormal haemoglobin association and changes red-cell shape under low oxygen.

DNA base substitution → changed mRNA codon → possible amino-acid replacement → altered side-chain interactions → altered protein structure/function.

A mutation is not automatically harmful or structure-changing: degeneracy can make it synonymous, and some amino-acid replacements have little effect.

Mutations changing protein structure

Assessment in practice

1–6 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Explain.

Command terms

Outline / Explain

What earns marks

Build the answer around this relationship: A base substitution can change one mRNA codon.

Watch for

Stopping at the DNA mutation without tracing the change through mRNA codon and amino acid sequence.

Representative question

Question 1

[Maximum number: 8]

Explain the cause of sickle cell anemia and how this disease affects humans.

Core Protein Synthesis

  • Transcription: RNA polymerase builds complementary mRNA from the DNA template; A pairs with U and C with G.
  • Translation: ribosomes read mRNA codons 5′ to 3′ while tRNA anticodons deliver specific amino acids for peptide-bond formation.
  • Genetic code: codons are triplets; the code is degenerate and almost universal, with start and stop signals. Use mRNA—not DNA—when reading a code table.
  • Information flow: codon order determines amino-acid sequence, which determines protein folding and function.
  • Expression and variation: cells regulate which genes are transcribed. A mutation may change a codon, primary structure and phenotype, as in sickle-cell haemoglobin.

Topic D1.3

D1.3 Mutation and gene editing

Mutation and gene editing explain how DNA sequence changes arise, affect proteins, create variation and can be studied or altered deliberately.

14% of analysed papers 20 papers · 24 questions

Objectives in this topic

A Gene Mutation Changes a DNA Sequence

A gene mutation is a structural change in the base sequence of DNA within a gene.

Mutation Sequence change
Substitution One base is replaced by another
Insertion One or more bases are added
Deletion One or more bases are removed
Duplication A DNA section is copied, producing an extra copy

Changing 5′-ACT-3′ to 5′-AGT-3′ is a substitution; changing it to 5′-ACCT-3′ is an insertion.

A mutation is the DNA sequence change itself. Its effect on a codon, protein or phenotype is a possible consequence, not part of the definition.

Gene mutations

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Define / Identify / Distinguish.

Command terms

Define / Identify / Distinguish / Compare

What earns marks

Build the answer around this relationship: Gene mutations are changes in DNA nucleotide sequence.

Watch for

Naming a disease such as sickle-cell anemia instead of naming a mutation type.

Representative question

Question 1

[Maximum number: 2]

Mutations may increase variation within a species. Compare and contrast substitution and insertion mutations.

A Base Substitution Can Be Silent, Missense or Nonsense

A single-nucleotide polymorphism (SNP) results from a base substitution, but the substitution may or may not change one amino acid in a polypeptide.

Codon outcome Polypeptide consequence
Silent The new codon specifies the same amino acid because the code is degenerate
Missense The new codon specifies a different amino acid
Nonsense The new codon is a stop codon, so translation ends early

An mRNA codon change from GAA to GAG is silent because both specify glutamate; a change to a stop codon can shorten the polypeptide.

A substitution does not automatically change protein function. First identify the new codon and its amino-acid or stop outcome.

Base substitution consequences

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, multiple choice, commonly using Describe / Explain / Outline.

Command terms

Describe / Explain / Outline

What earns marks

Build the answer around this relationship: A substitution changes one base in a DNA sequence.

Watch for

Stopping at the DNA substitution without tracing the codon and amino acid consequence.

Representative question

Question 1

[Maximum number: 4]

Outline how a base substitution leads to sickle cell anemia.

Insertions and Deletions Can Shift the Reading Frame

Insertions and deletions are likely to stop a polypeptide functioning when they shift its reading frame or change a large section of its sequence.

Ribosomes read mRNA in triplets. Adding or removing a number of bases that is not a multiple of three regroups every downstream codon, often changing many amino acids and creating an early stop codon.

Not a multiple of three → frameshift and changed downstream codons. Multiple of three → no frameshift, but amino acids are added or removed. A major insertion or deletion can still disrupt structure and function even without a frameshift.

Deleting one base near the start of a coding sequence shifts the triplet grouping for most of the remaining mRNA and is therefore likely to produce a non-functional polypeptide.

A three-base insertion or deletion avoids a frameshift, but it is not automatically harmless because the added or missing amino acid may be important.

Mutations Arise from Replication Errors and Mutagens

Gene mutations can result from errors in DNA replication or repair and from DNA damage caused by mutagens.

Cause Example or route to mutation
Replication error An incorrect nucleotide escapes proofreading
Repair error Damaged DNA is repaired with an altered base sequence
Chemical mutagen NNK in tobacco smoke can increase DNA base-sequence changes
Ultraviolet radiation UV can create abnormal links between adjacent bases
Ionizing radiation X-rays or gamma rays can damage DNA, including strand breaks

Damage is not yet a permanent mutation if accurate repair restores the original sequence. It becomes a mutation when the altered sequence remains and is copied.

A mutagen increases mutation probability; it does not produce the same mutation in every exposed cell.

Causes of mutation

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Explain / Evaluate.

Command terms

State / Explain / Evaluate / Identify

What earns marks

Build the answer around this relationship: Mutagens increase the frequency of DNA sequence changes.

Watch for

Giving vague environmental factors without identifying radiation, chemicals or carcinogens.

Representative question

Question 1

[Maximum number: 2]

Explain how chemical substances can cause cancer.

Mutation Occurs Randomly Relative to Need

Mutations occur randomly with respect to an organism's need: no known natural mechanism deliberately changes a particular base in order to create a useful trait.

A mutation can occur anywhere in the genome before its consequence is tested by the environment. Natural selection later changes variant frequencies because some carriers reproduce more successfully.

Random relative to need does not mean uniform probability. Base identity and sequence context, DNA repair, gene activity and mutagen exposure can make some sites or cells more likely to mutate than others.

Antibiotic exposure does not instruct bacteria to make a resistance mutation. A resistant variant may already exist, then increase in frequency when susceptible cells die.

Mutation bias can make some changes more frequent without making them purposeful or directed toward advantage.

Randomness in mutation

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice.

What earns marks

Build the answer around this relationship: Mutations are not directed by an organism’s needs.

Representative question

Question 1

[Maximum number: 1]

What is a feature of mutations?

A

They occur randomly.

B

They only occur in germ cells.

C

The frequency cannot be increased by external factors.

D

They only occur in certain base sequences of the genome.

Germline and Somatic Mutations Have Different Reach

The consequence of a mutation depends on whether it occurs in the germ line or in a somatic cell lineage.

Location Who can receive the mutation? Important consequence
Germ-line cell or gamete Offspring, if the mutated gene is transmitted at fertilization The mutation can be inherited and enter the descendant's cell lineages
Somatic cell Descendant body cells produced by mitosis A clone of altered cells can form; mutations affecting growth control can contribute to cancer

A mutation in a sperm cell may be inherited by a child, whereas a mutation acquired in one skin cell can spread through a local clone but is not normally passed to offspring.

Somatic does not mean harmless: a non-inherited mutation can still cause cancer or other serious effects in the individual.

Consequences in germ vs. somatic cells

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Distinguish.

Command terms

Distinguish

What earns marks

Build the answer around this relationship: Only germ-line mutations can normally be passed to offspring.

Watch for

Saying any mutation can automatically be inherited regardless of cell type.

Representative question

Question 1

[Maximum number: 1]

A mutation in which type of cell could be inherited?

A

Beta cell in the pancreas

B

T-cell in the lymph

C

Sperm cell in the testis

D

Skeletal muscle cell in the diaphragm

Mutation Supplies Variation for Natural Selection

Gene mutation is the original source of new alleles and therefore of all genetic variation.

Most mutations are neutral or harmful to an individual, but a population needs heritable variants for natural selection to act on. Selection changes allele frequencies; it does not create the initial DNA differences.

Mutation creates a new allele → inheritance can place it in a population → environmental conditions affect reproductive success → natural selection can change its frequency over generations.

A new allele that improves drought survival may spread when carriers leave more offspring in dry conditions, while the same allele may provide no advantage in another environment.

Mutation alone is not adaptation. The variant must be heritable and influence reproductive success in the relevant environment.

Mutation as source of variation

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice.

What earns marks

Build the answer around this relationship: Mutation produces new alleles.

Representative question

Question 1

[Maximum number: 1]

What causes variation in both sexually and asexually reproducing organisms?

A

Mutations

B

Polygenic inheritance

C

Crossing over

D

Independent assortment

Core Mutation Effects

Gene mutations are changes in the base sequence of DNA; main types are substitution, insertion, deletion, and duplication. Base substitutions can create SNPs and change codons; degeneracy can make substitutions silent, missense, or nonsense. Insertions or deletions not in multiples of three cause frameshifts that alter downstream codons and often disrupt protein function. Mutations can arise from replication errors, repair errors, or chromosome damage; mutagens include chemicals, ionizing radiation, and ultraviolet radiation. Mutations occur randomly with respect to organism need or advantage; mutation rate varies with DNA sequence, gene expression, repair, and mutagen exposure. Germ-line mutations can be inherited by offspring; somatic mutations affect only descendant body cells and can contribute to cancer. Mutation is the original source of new alleles and genetic variation; many are neutral or harmful, but variation supplies material for natural selection.

Topic D2.1

D2.1 Cell and nuclear division

Cell and nuclear division coordinate DNA replication, chromosome movement, cytokinesis, meiosis, cell-cycle control and proliferation to produce new cells in organisms.

48% of analysed papers 67 papers · 117 questions

Objectives in this topic

Cell Division Makes New Cells

In every living organism, cell division generates new cells when one parent cell divides to produce two daughter cells.

Producing daughter cells allows an organism to grow, replace worn cells, repair damaged tissue or reproduce asexually. The genetic material must be distributed before the cytoplasm separates so both daughters can function.

A skin cell can divide into two daughter cells that replace cells lost from the surface, while a single-celled organism can divide to produce two organisms.

The parent cell is sometimes called a mother cell, but this does not imply sex or fertilization. Cell division means generation of daughter cells, not growth of one cell alone.

Generation of new cells

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Cells arise from pre-existing cells by cell division.

Representative question

Question 1

[Maximum number: 1]

Which process contributes to growth of a multicellular body?

A

Exocytosis

B

Meiosis

C

Mitosis

D

Osmosis

Cytokinesis Separates the Cytoplasm

Cytokinesis splits the cytoplasm of a parent cell between its daughter cells, but animal and plant cells achieve the split differently.

Cell type Cytokinesis mechanism
Animal A contractile ring of actin and myosin tightens, pulling the plasma membrane inward to form a cleavage furrow
Plant Vesicles fuse at the centre to build new membrane; their contents contribute to a cell plate and new cell wall

An animal cell pinches from its outer edge toward the centre, whereas a plant cell builds the separating plate from the centre outward.

Cytokinesis divides cytoplasm; mitosis or meiosis divides the nucleus. The events can overlap in time but are not the same process.

Cytokinesis exam focus

Assessment in practice

1–8 marks
How it is assessed

This objective is assessed through structured response, commonly using Compare / Explain.

Command terms

Compare / Explain

What earns marks

Build the answer around this relationship: Cytokinesis separates cytoplasm to complete cell division.

Watch for

Confusing cytokinesis with mitosis instead of separating cytoplasmic division from nuclear division.

Representative question

Question 1

[Maximum number: 8]

Compare and contrast the processes of mitosis and cytokinesis in animal and plant cells.

Cytokinesis Can Partition Cell Contents Unequally

Cytokinesis is usually equal, but unequal cytokinesis can give daughter cells very different amounts of cytoplasm.

Whatever their final sizes, both daughters must receive at least one mitochondrion and any other organelle that can arise only by growth and division of a pre-existing organelle.

Example Partitioning result
Typical equal cytokinesis Daughters receive similar shares of cytoplasm
Human oogenesis One large ovum retains most cytoplasm; small polar bodies receive little
Yeast budding A smaller bud separates from the larger parent cell

Unequal cytoplasm does not mean unequal nuclear DNA: chromosomes can still be segregated correctly before asymmetric cytokinesis.

Equal and unequal cytokinesis

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Equal cytokinesis gives daughter cells similar cytoplasmic volumes.

Representative question

Question 1

[Maximum number: 1]

Daughter cells usually receive equal amounts of cytoplasm as parent cells undergo cytokinesis. Which of the following is an exception?

A

Asexual reproduction by budding in yeast

B

Bacterial cell division

C

Cloning of lymphocytes during an immune response

D

Formation of a zygote during fertilization

Mitosis Preserves Cells; Meiosis Makes Gametes

Eukaryotic cells use mitosis to maintain chromosome number and genome, whereas meiosis halves chromosome number and generates genetic diversity.

Nuclear division must occur before cell division so each daughter receives a nucleus rather than becoming an anucleate cell.

Feature Mitosis Meiosis
Main roles Growth, repair, replacement, asexual reproduction Production of cells for sexual reproduction
Nuclear divisions One Two
Chromosome number Maintained Halved
Genetic outcome Genome normally maintained New allele combinations generated

Meiosis is not simply mitosis twice: homologous chromosomes pair and separate in its first division, creating the reduction in chromosome number.

Roles of mitosis and meiosis

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Distinguish / Describe.

Command terms

State / Distinguish / Describe / Explain / Outline

What earns marks

Build the answer around this relationship: Mitosis produces two genetically identical daughter nuclei or cells.

Watch for

Failing to state both sides of a mitosis-versus-meiosis comparison.

Representative question

Question 1

[Maximum number: 5]

Distinguish between the processes of meiosis and mitosis.

DNA Replication Must Precede Nuclear Division

DNA replication occurs before nuclear division so each daughter nucleus can receive a complete chromosome set.

During S phase, each chromosome becomes two sister chromatids joined at a centromere. Division then separates chromatids or homologues according to the process.

Trace: replication; duplicated chromosome; spindle attachment; chromosome separation; daughter nuclei.

If a cell entered mitosis without replicating DNA, one daughter could receive too little genetic material.

Replication doubles DNA amount, not chromosome number in the usual chromosome-counting convention.

DNA replication prerequisite

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Distinguish.

Command terms

Distinguish

What earns marks

Build the answer around this relationship: DNA replication occurs in S phase of interphase.

Watch for

Counting chromosomes and chromatids as the same thing after DNA replication.

Representative question

Question 1

[Maximum number: 1]

Distinguish between the quantity of DNA of the cell at G1 and G2.

Mitosis and Meiosis Share a Controlled Division Logic

Mitosis and meiosis both condense chromosomes and move them accurately between new nuclei.

Histones organize DNA into nucleosomes, and further supercoiling condenses the long chromatin fibres into compact chromosomes that can be moved without tangling.

Spindle microtubules attach to chromosomes, and microtubule motors plus microtubule shortening or growth generate directed movement. Alignment and separation differ between divisions, but the same general machinery organizes chromosome distribution.

In both mitosis and meiosis II, sister chromatids move toward opposite poles; in meiosis I, homologous chromosomes move apart while sister chromatids remain together.

Shared condensation and movement mechanisms do not make the outcomes identical: the chromosome partners separated and the number of divisions differ.

Shared features

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Both mitosis and meiosis use spindle microtubules to move chromosomes.

Representative question

Question 1

[Maximum number: 1]

What occurs in cell division during both mitosis and meiosis?

A

Condensation of DNA by supercoiling in telophase

B

Movement of microtubules to move chromatids in anaphase

C

Pairing of homologous chromosomes in prophase

D

Crossing over between chromosomes in metaphase

Mitosis Moves Chromosomes Through Four Main Stages

Mitosis proceeds through prophase, metaphase, anaphase and telophase, each solving a different chromosome-distribution problem.

Chromosomes condense and attach to spindle fibers, align at the equator, separate sister chromatids and re-form nuclei at opposite poles. Cytokinesis follows.

Use the sequence: condense; align; separate chromatids; rebuild nuclei.

A metaphase cell has chromosomes aligned at the equator; anaphase begins when sister chromatids move apart.

Interphase is not a mitosis stage, although it prepares the cell by growing and replicating DNA.

Phases of mitosis

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Identify / Outline.

Command terms

State / Identify / Outline / Describe / Deduce / Evaluate / Suggest / Label

What earns marks

Build the answer around this relationship: Prophase condenses chromosomes and begins nuclear-envelope breakdown.

Watch for

Confusing metaphase alignment with anaphase separation in micrographs.

Representative question

Question 1

[Maximum number: 9]

Describe the events that occur during mitosis.

Chromosome Features Identify Mitosis Stages

Identify a mitosis phase from the position and appearance of chromosomes, using several features rather than cell shape alone.

Phase Reliable visible cues
Prophase Chromosomes condense; the nuclear envelope begins to disappear
Metaphase Condensed chromosomes align at the cell equator
Anaphase Sister chromatids separate and move toward opposite poles
Telophase Chromosomes reach the poles and new nuclear envelopes form

Two groups of V-shaped chromatids moving away from the equator indicate anaphase, whether seen in a diagram, a prepared root-tip cell or a micrograph.

A dark stain or rounded cell outline is not enough to identify a phase. Confirm chromosome condensation, alignment or separation and nuclear-envelope state.

Identification of mitosis phases

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Condensed unaligned chromosomes suggest prophase.

Representative question

Question 1

[Maximum number: 1]

The following shows a micrograph.

How many cells are in metaphase?

A

2

B

3

C

5

D

7

Meiosis Reduces Chromosome Number

Meiosis is a reduction division: one diploid nucleus undergoes two nuclear divisions after one DNA replication to produce four haploid nuclei.

Diploid nuclei contain two homologous chromosome sets; haploid nuclei contain one. Homologous chromosomes separate in meiosis I, halving the number of sets, and sister chromatids separate in meiosis II without another round of replication.

Diploid nucleus → DNA replication → homologous pairs segregate in meiosis I → two haploid nuclei with duplicated chromosomes → chromatids segregate in meiosis II → four haploid nuclei.

In a sexual life cycle, haploid gametes produced by meiosis fuse at fertilization, restoring the diploid chromosome number instead of doubling it in every generation.

Chromosome number is reduced in meiosis I, even though each chromosome still has two chromatids until meiosis II.

Meiosis as reduction division

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Identify / Outline.

Command terms

State / Identify / Outline

What earns marks

Build the answer around this relationship: Meiosis halves chromosome number to produce haploid nuclei.

Watch for

Saying sister chromatids separate in meiosis I instead of homologous chromosomes.

Representative question

Question 1

[Maximum number: 5]

Outline what occurs in cells in the first division of meiosis.

Nondisjunction Can Produce Trisomy

Nondisjunction is failure of homologous chromosomes or sister chromatids to separate, producing gametes with abnormal chromosome numbers.

If an extra chromosome enters a gamete, fertilization can create a zygote with three copies of one chromosome. The phenotype depends on chromosome and gene dosage.

Trace: division error; abnormal gamete; fertilization; chromosome count; developmental consequence.

A gamete containing two copies of chromosome 21 can combine with a normal gamete to produce trisomy 21.

Nondisjunction is a chromosome-segregation error, not a point mutation in one gene.

Down syndrome

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain / Outline / Determine.

Command terms

Explain / Outline / Determine / State / Identify / Describe

What earns marks

Build the answer around this relationship: Non-disjunction is failed chromosome separation during meiosis.

Watch for

Calling Down syndrome a gene mutation instead of a chromosome-number abnormality.

Representative question

Question 1

[Maximum number: 4]

Describe how non-disjunction can cause Down syndrome.

Meiosis Creates Variation through Pairing and Recombination

Meiosis generates genetic diversity through random orientation of bivalents and crossing over between non-sister chromatids.

At metaphase I, each bivalent can face either pole independently. The maternal and paternal homologues therefore segregate into many possible whole-chromosome combinations.

During prophase I, non-sister chromatids of homologous chromosomes exchange corresponding DNA at chiasmata, producing recombinant chromatids with new combinations of linked alleles.

One gamete can receive a maternal chromosome carrying a short paternal segment after crossing over, plus a different random mixture of the remaining maternal and paternal homologues.

Random fertilization adds further variation but is not a meiotic process. Meiosis does not direct combinations toward future advantage.

Meiosis generates variation

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using State / Explain / Describe.

Command terms

State / Explain / Describe / Draw / Outline / Identify

What earns marks

Build the answer around this relationship: Crossing over exchanges DNA between non-sister chromatids of homologous chromosomes.

Watch for

Saying crossing over occurs between sister chromatids instead of non-sister chromatids of homologous chromosomes.

Representative question

Question 1

[Maximum number: 7]

Explain the stages and processes of meiosis leading to genetic variation.

Core Cell Division

  • Cell division produces daughter cells for growth, repair or reproduction; cytokinesis divides cytoplasm by a contractile ring in animals or a cell plate in plants.
  • DNA replication creates sister chromatids joined at centromeres before nuclear division.
  • Mitosis preserves chromosome number: chromosomes condense, align, sister chromatids separate and nuclei reform, producing genetically identical nuclei.
  • Meiosis follows one replication with two divisions: homologous chromosomes separate in meiosis I and sister chromatids in meiosis II, producing haploid cells.
  • Crossing over, independent orientation and random fertilization generate allele combinations.
  • Non-disjunction is failed chromosome separation and can produce aneuploid cells, including trisomy 21.
  • Identify stages in micrographs from chromosome condensation, equatorial alignment, separation and nuclear-envelope cues.

Topic D2.3

D2.3 Water potential

Water potential explains osmosis, solute effects, plant tissue changes, cell swelling, plasmolysis and isotonic medical conditions in living systems and cells.

13% of analysed papers 18 papers · 26 questions

Objectives in this topic

Water Surrounds Solutes during Solvation

Solvation occurs when polar water molecules surround and interact with dissolved ions or polar solute molecules.

Water's partially negative oxygen is attracted to positive ions, while its partially positive hydrogens are attracted to negative ions. Polar solutes can also form hydrogen bonds with water.

These attractions form hydration shells, separate solute particles and keep them dispersed. Water molecules engaged around solutes have less freedom of movement than in pure water.

When sodium chloride dissolves, oxygen ends of water face Na⁺ and hydrogen ends face Cl⁻, producing oriented hydration shells.

Water does not form hydrogen bonds with every solute: ion–dipole attraction hydrates ions, while hydrogen bonding requires suitable polar groups.

Water Moves from Hypotonic to Hypertonic Solution

Across a partially permeable membrane, net water movement is from the less concentrated solution toward the more concentrated solution.

Comparison term Solute concentration relative to the other solution Expected net water movement
Hypotonic Lower Away from this solution
Hypertonic Higher Toward this solution
Isotonic Equal effective concentration No net movement

If solution A is 0.10 mol dm⁻³ sucrose and solution B is 0.40 mol dm⁻³, A is hypotonic to B and net water movement is from A to B if water can cross but sucrose cannot.

At SL, express the comparison using solute concentration—not 'high water concentration'. Tonicity is relative and depends on solutes that do not freely cross the membrane.

Water movement

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, multiple choice, commonly using Explain / Outline.

Command terms

Explain / Outline

What earns marks

Build the answer around this relationship: Osmosis requires a partially permeable membrane.

Watch for

Saying solute moves by osmosis instead of water.

Representative question

Question 1

[Maximum number: 2]

Outline the conditions necessary for osmosis to occur.

Use Tonicity to Predict Osmosis in Cells

Osmosis is the net movement of water across a partially permeable membrane, and the cell's environment determines its direction.

External environment Relative external solute concentration Net water movement
Hypotonic Lower than inside Into the cell
Hypertonic Higher than inside Out of the cell
Isotonic Equal effective concentration No net movement

In an isotonic environment, water molecules continue crossing in both directions at equal rates. This is dynamic equilibrium, not an absence of molecular movement.

A cell placed in hypertonic solution loses water and decreases in volume because more water leaves than enters.

Always state the solution relative to the cell. 'Hypotonic' or 'hypertonic' without a comparison has incomplete meaning.

Osmosis into/out of cells

Assessment in practice

3 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain.

Command terms

Explain

What earns marks

Build the answer around this relationship: Hypotonic external solutions cause net water entry into cells.

Representative question

Question 1

[Maximum number: 3]

Explain the reason that animal cells and tissues under investigation must be maintained in solutions with the same osmolarity.

Estimate Isotonic Concentration with Plant Tissue

Changes in plant-tissue mass or length across a series of sucrose concentrations can be used to estimate the isotonic concentration.

Prepare equal tissue pieces, record initial mass or length, incubate them for the same time in known sucrose solutions, blot consistently, record final values and calculate change.

%\text{ change}=\frac{\text{final value}-\text{initial value}}{\text{initial value}}\times100

Plot mean percentage change against sucrose concentration. The x-intercept, where change is 0%, estimates the isotonic concentration. Replicates allow standard deviation to compare spread and standard error/error bars to compare uncertainty in means.

Positive mass change at 0.20 mol dm⁻³ and negative change at 0.30 mol dm⁻³ place the isotonic estimate between those concentrations; interpolate from the fitted graph rather than choosing the nearest raw point.

Zero mean change estimates isotonic conditions; it does not mean water molecules stopped moving. Consistent blotting, initial size and incubation time are needed for a fair comparison.

Changes in plant tissue

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through experimental design, commonly using Identify / Describe / Explain.

Command terms

Identify / Describe / Explain / Suggest / Evaluate / Outline

What earns marks

Build the answer around this relationship: Percentage change in mass shows relative water gain or loss.

Watch for

Using raw mass change instead of percentage change when initial masses differ.

Representative question

Question 1

[Maximum number: 2]

Student osmosis experiments often involve putting plant tissue such as potato cylinders in several salt solutions of different concentrations and measuring the mass before and after immersion. Outline how data collected from such an experiment could be used to estimate the osmolarity of the plant tissues.

Cells without Walls Can Swell or Shrink

Cells without a wall can burst in hypotonic solution or shrink in hypertonic solution because their plasma membrane cannot resist large volume changes.

Environment Net water movement Wall-less cell response
Hypotonic Into cell Swelling; excessive entry may cause lysis or haemolysis
Hypertonic Out of cell Shrinkage; animal cells such as red blood cells become crenated
Isotonic Balanced Stable average volume

Freshwater unicellular organisms continually gain water from their hypotonic environment, so contractile vacuoles collect and expel excess water. Multicellular animals instead maintain near-isotonic tissue fluid around cells.

A plasma membrane can deform but does not provide the rigid mechanical restraint of a cell wall; active water removal is an adaptation, not a reversal of osmosis.

Effects on cells without wall

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, multiple choice, commonly using Deduce / Outline / Predict.

Command terms

Deduce / Outline / Predict / Explain / Describe / Suggest

What earns marks

Build the answer around this relationship: Hypotonic solutions can cause animal cells to swell or lyse.

Watch for

Calling crenated animal cells turgid, a term that applies to walled plant cells.

Representative question

Question 1

[Maximum number: 3]

Explain the effect of placing red blood cells in distilled water (0.000 M NaCl).

Cell Walls Limit Swelling and Create Turgor

A cell wall resists expansion when water enters, converting osmotic water uptake into turgor pressure.

The wall’s rigidity balances the inward tendency of water. If water leaves, pressure falls and the membrane can pull away from the wall, producing plasmolysis.

Distinguish wall restraint from membrane transport: water direction first, then pressure and shape.

A plant cell in dilute solution becomes turgid rather than bursting because the wall pushes back as the vacuole expands.

A wall prevents unlimited swelling but does not stop osmosis or guarantee that a severely dehydrated cell survives.

Effects on cells with wall

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / State / Explain.

Command terms

Outline / State / Explain

What earns marks

Build the answer around this relationship: Water entry can make plant cells turgid.

Watch for

Saying plant cells burst in hypotonic solution ignores the protective cell wall.

Representative question

Question 1

[Maximum number: 7]

Explain the process of osmosis with reference to its effects on plant cells.

Water Potential Guides Medical Fluid Choices

Medical fluids are made isotonic with body tissues to avoid harmful net water gain or loss by cells.

Application Why isotonic conditions matter
Intravenous fluid Prevents red blood cells and other body cells swelling, lysing, shrinking or crenating while fluid/solutes are delivered
Organ awaiting transplantation An isotonic bathing solution limits osmotic damage to the organ's cells before implantation

An isotonic saline infusion replaces extracellular fluid without causing appreciable net movement of water into or out of red blood cells.

Isotonic means matched effective osmotic concentration relative to the tissue; it does not mean the solution has the same chemical composition as cytoplasm.

Medical applications

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, multiple choice, commonly using Explain.

Command terms

Explain

What earns marks

Build the answer around this relationship: Isotonic fluids prevent net water movement into or out of body cells.

Representative question

Question 1

[Maximum number: 4]

Explain the need for isotonic conditions in human blood plasma and tissue fluid.

Core Osmosis Effects

Water forms hydration shells around ions and polar solutes; hydrogen bonding and charge attraction reduce free water movement. Water moves by osmosis across partially permeable membranes from hypotonic/lower solute solutions toward hypertonic/higher solute solutions. Osmosis direction depends on internal and external solute concentration; isotonic conditions have dynamic water movement but no net osmosis. Plant tissue changes mass or length in sucrose solutions; percentage change graphs estimate isotonic or osmotic concentration. Animal cells can lyse in hypotonic solutions and crenate in hypertonic solutions; freshwater protists use contractile vacuoles to expel excess water. Plant cells become turgid in hypotonic solutions as vacuoles swell; hypertonic solutions cause flaccidity and plasmolysis from water loss. Isotonic saline prevents harmful water gain or loss in body cells; IV fluids and transplant organ baths must match tissue osmotic concentration.

Topic D3.1

D3.1 Reproduction

Reproduction covers cloning, human and plant reproductive anatomy, cycles, fertilization, pregnancy, seed development and hormonal coordination across sexual life cycles.

24% of analysed papers 34 papers · 43 questions

Objectives in this topic

Sexual and Asexual Reproduction Differ in Parentage

Asexual reproduction uses one parent without gamete fusion and normally produces genetically identical offspring; sexual reproduction uses meiosis and fertilization to produce new allele combinations.

Mode Relative advantage Relative limitation
Asexual Rapidly preserves a successful genotype when a parent is already adapted to a stable environment Little new genetic variation makes a changed environment risky for many offspring
Sexual Variation among offspring increases the chance that some are suited to changed conditions Requires production and fusion of gametes and does not preserve one genotype exactly

A strawberry runner produces a clone suited to the parent's current habitat, whereas a seed formed after fertilization carries a new allele combination.

Asexual offspring can still differ after mutation or environmental effects; 'clone' refers to their inherited genome, not guaranteed identical phenotype.

Sexual vs. asexual reproduction

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Outline / Describe.

Command terms

Identify / Outline / Describe

What earns marks

Build the answer around this relationship: A clone is genetically identical to the single parent or source cell that produced it.

Representative question

Question 1

[Maximum number: 3]

Outline natural methods of cloning in some eukaryotes.

Meiosis and Gamete Fusion Restore the Life-Cycle Number

Meiosis halves chromosome number in gametes, and fusion of two gametes restores the diploid number in the zygote.

The alternation prevents chromosome number doubling every generation. Independent assortment and crossing over also create combinations before fusion adds another random combination.

Track chromosome number through meiosis; then track the fusion event and the first embryo cell.

A diploid human cell with 46 chromosomes produces gametes with 23; fusion returns the zygote to 46.

Meiosis does not simply make ‘smaller’ cells; its defining outcome is reduced chromosome number plus variation.

Role of meiosis and gamete fusion

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Explain.

Command terms

Identify / Explain

What earns marks

Build the answer around this relationship: Meiosis produces haploid gametes by halving chromosome number.

Representative question

Question 1

[Maximum number: 4]

Explain the need for both fusion of gametes and meiosis in a sexual life cycle.

Sexes Are Defined by the Gametes Produced

In anisogamous species, the male produces smaller motile gametes and the female produces larger nutrient-rich gametes.

The distinction is based on gamete type, not on every secondary trait or an individual’s identity. Different species organise reproductive roles around these gametes in different ways.

Identify the gametes first, then infer the biological sex category used in the syllabus model.

In humans, sperm are small and motile while ova are large and non-motile, so the model labels their producers male and female.

Gamete definitions do not justify assumptions about behaviour, gender, or all reproductive biology.

Map Human Reproductive Structures to Their Functions

Human reproductive systems link gamete production, transport, fertilization, implantation and birth through specialized structures.

Male-typical structure Main function
Testis Produces sperm and testosterone
Epididymis Stores and matures sperm
Sperm duct (vas deferens) Carries sperm toward the urethra
Seminal vesicles/prostate Add fluid to form semen
Urethra and penis Conduct and deliver semen outside the body
Female-typical structure Main function
Ovary Produces oocytes and ovarian hormones
Oviduct Transports the oocyte; usual site of fertilization
Uterus/endometrium Supports implantation and development
Cervix Muscular opening between uterus and vagina
Vagina/vulva Receives semen; vagina forms the birth canal and vulva is the external region

A labelled diagram must show position and connections as well as names. Fertilization normally occurs in an oviduct; implantation occurs later in the endometrium.

Human reproductive system anatomy

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Draw.

Command terms

Identify / Draw

What earns marks

Build the answer around this relationship: The epididymis is where sperm complete maturation and become motile.

Watch for

Misidentifying epididymis, sperm duct, prostate and seminal vesicles on male diagrams.

Representative question

Question 1

[Maximum number: 6]

Draw a labeled diagram of the female reproductive system.

Four Hormones Coordinate the Menstrual Cycle

The menstrual cycle combines ovarian and uterine cycles controlled by FSH, LH, oestradiol and progesterone through negative and positive feedback.

Stage Hormonal control and linked event
Follicular phase FSH promotes follicle growth; the follicle secretes oestradiol, which rebuilds the endometrium and usually inhibits FSH
Ovulation Sustained high oestradiol produces positive feedback, causing an LH surge that triggers ovulation
Luteal phase LH supports the corpus luteum; progesterone maintains the endometrium and inhibits FSH/LH
Menstruation if no pregnancy Corpus luteum breaks down; progesterone and oestradiol fall, so the endometrium is shed and inhibition is removed

A sharp LH peak follows the high-oestradiol positive-feedback switch and occurs just before ovulation.

Feedback direction changes with hormone concentration and cycle stage; oestradiol is not always a positive-feedback signal.

Ovarian and uterine cycles

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Sketch / Identify / Outline.

Command terms

Sketch / Identify / Outline / Explain

What earns marks

Build the answer around this relationship: FSH promotes follicle development and estrogen secretion.

Watch for

Confusing LH with FSH or progesterone when identifying the hormone that triggers ovulation.

Representative question

Question 1

[Maximum number: 8]

Explain the roles of specific hormones in the menstrual cycle, including positive and negative feedback mechanisms.

Human Fertilization Joins Parental Chromosomes

Human fertilization begins in the oviduct when sperm and egg cell membranes fuse and ends with paternal and maternal chromosomes sharing the first zygotic mitosis.

Sperm membrane fuses with egg membrane → sperm nucleus enters while its tail and mitochondria are destroyed → sperm and egg nuclear membranes dissolve → both condensed chromosome sets attach to one mitotic spindle → chromosomes segregate to form two diploid nuclei.

This sequence brings one haploid paternal and one haploid maternal chromosome set into a diploid zygote genome while preventing paternal sperm mitochondria becoming part of the embryo.

The 23 paternal and 23 maternal chromosomes participate together in the first mitosis, so each of the first two embryonic nuclei receives a diploid set.

Fertilization is not implantation: nuclear union begins in the oviduct, while attachment to the endometrium happens later.

Fertilization in humans

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Fertilization is a cellular process involving sperm and egg nuclei.

Representative question

Question 1

[Maximum number: 6]

Describe the process of fertilization in humans.

IVF Moves Key Reproductive Steps into a Controlled Setting

IVF treatment temporarily takes control of normal reproductive hormone signalling so artificial hormone doses can induce superovulation.

Normal pituitary hormone secretion is first suppressed to prevent an uncontrolled ovulation. Carefully timed FSH-like stimulation matures several follicles, and an LH-like trigger completes egg maturation before collection.

Suppress normal cycle → stimulate multiple follicles → trigger maturation → collect oocytes → fertilize outside the body → culture embryo(s) → transfer selected embryo(s) to uterus.

Producing several mature oocytes in one controlled cycle gives more opportunities for fertilization and embryo selection than the usual release of one oocyte.

Superovulation increases the number of available oocytes but does not guarantee fertilization, implantation or live birth.

In vitro fertilization (IVF)

Assessment in practice

1–8 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline.

Command terms

Outline

What earns marks

Build the answer around this relationship: FSH stimulation is used to produce more eggs than in a normal cycle.

Watch for

Listing IVF steps but omitting either hormone stimulation, egg collection, external fertilization or embryo transfer.

Representative question

Question 1

[Maximum number: 9]

Embryos that are produced by in vitro fertilization can be screened for genetic disease. Outline the process of in vitro fertilization, including one example of a situation when it is used.

Flowering-Plant Sexual Reproduction Uses Pollen and Ovules

Flowering-plant reproduction is sexual because male and female gametes fuse, even when one hermaphroditic flower produces both pollen and ovules.

Male gametes develop inside pollen grains in anthers; female gametes develop inside ovules in the ovary. Pollination transfers pollen to a stigma, the pollen grain develops a tube, and male nuclei travel to the ovule for fertilization.

Fusion produces a diploid zygote that develops into an embryo; the ovule develops into a seed that contains the embryo.

Pollen carried by an insect reaches a compatible stigma, grows a tube down the style and delivers a male nucleus to the egg cell in an ovule.

Pollination is transfer, not fertilization. A hermaphroditic flower still reproduces sexually when gamete nuclei fuse.

Sexual reproduction in flowering plants

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Identify / State.

Command terms

Outline / Identify / State / Distinguish / Define

What earns marks

Build the answer around this relationship: Pollination is transfer of pollen from anther to stigma.

Watch for

Confusing pollination with fertilization or seed dispersal.

Representative question

Question 1

[Maximum number: 4]

Outline pollination, fertilization and seed dispersal.

Insect-Pollinated Flowers Advertise and Deliver Pollen

An insect-pollinated flower attracts a pollinator and positions its reproductive structures so pollen is picked up and later deposited on a stigma.

Structure/feature Function in insect pollination
Coloured or scented petals Attract and guide insects
Nectary Rewards feeding visits
Anthers held inside flower Brush sticky/rough pollen onto the insect
Sticky stigma inside flower Receives pollen carried on the insect
Ovary with ovules Contains female gametes that may be fertilized after pollen-tube growth

As a bee reaches nectar, the flower's anthers brush pollen onto its body; a later visit places some pollen on another flower's stigma.

For a diagram, annotate each named structure with its function; colour alone does not establish insect pollination.

Insect-pollinated flower features

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Draw.

Command terms

Identify / Draw

What earns marks

Build the answer around this relationship: Nectar attracts animals that can transfer pollen between flowers.

Representative question

Question 1

[Maximum number: 4]

Draw a half-view of an animal-pollinated flower.

Plants Promote Cross-Pollination to Mix Pollen Sources

Plants promote cross-pollination by separating pollen and receptive female structures in time, space or among different plants, then using animals or wind as transfer vectors.

Method How it reduces self-pollination
Different maturation times Pollen is released when the same flower's stigma is not receptive, or vice versa
Separate male/female flowers Anthers and stigmas are physically separated on one plant
Separate male/female plants (dioecy) Pollen must travel between plants
Animal or wind transfer Carries pollen from anthers of one plant to stigmas of another

If pollen matures before the stigma of the same flower, pollen arriving later from another plant is more likely to fertilize its ovules.

Cross-pollination increases new gene combinations but is not guaranteed on every visit; self-incompatibility is a separate genetic recognition mechanism.

Promoting cross-pollination

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline.

Command terms

Outline

What earns marks

Build the answer around this relationship: Different maturation times of anthers and stigmas can reduce self-pollination.

Representative question

Question 1

[Maximum number: 2]

Outline how cross-pollination can be promoted by flowering plants.

Self-Incompatibility Rejects Genetically Similar Pollen

Self-incompatibility is a genetic recognition system that prevents self-pollen from fertilizing ovules and thereby promotes cross-fertilization.

Matching incompatibility alleles in pollen and stigma can block pollen germination or pollen-tube growth. Compatible pollen from another plant can continue to the ovule.

Self-pollination increases inbreeding, which reduces genetic diversity and can reduce vigour by increasing expression of harmful recessive alleles. Rejecting self-pollen helps maintain variation within the species.

Pollen sharing the stigma's incompatibility class is rejected, while pollen carrying a different compatible class grows a tube and can fertilize the ovule.

Self-incompatibility is not pollen sterility or physical separation; the same pollen may function normally on a genetically compatible plant.

Self-incompatibility mechanisms

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice.

What earns marks

Build the answer around this relationship: Self-incompatibility prevents inbreeding rather than decreasing variation.

Representative question

Question 1

[Maximum number: 1]

Cherry trees (Prunus avium) have two self-incompatibility alleles. What benefit do self-incompatibility alleles have?

A

They decrease genetic variation.

B

They prevent inbreeding.

C

They decrease the chances of mutations taking place within the gametes.

D

They prevent the plant from releasing pollen at certain times of the year.

Seeds Disperse, Then Germinate when Conditions Permit

Seed dispersal separates offspring from the parent, and germination begins when water, oxygen and a suitable temperature allow metabolism and growth.

Dispersal reduces crowding and competition. During germination, water activates enzymes, oxygen supports respiration, and the embryo uses stored food until photosynthesis begins.

Check each condition before deciding whether a seed can germinate.

A bean seed kept dry does not germinate; after water and warmth are supplied, respiration rises and the radicle emerges.

A seed can be viable but remain dormant; failure to germinate does not prove it is dead.

Seed dispersal and germination

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Outline.

Command terms

Identify / Outline

What earns marks

Build the answer around this relationship: Water uptake is the first step that reactivates metabolism in a dry seed.

Watch for

Listing water, oxygen and temperature without explaining their biological roles.

Representative question

Question 1

[Maximum number: 6]

Outline the metabolic processes that occur in starchy seeds during germination.

Retrieve the Core Reproduction Route

Core D3.1 route: reproduction creates offspring, gametes or pollen move, fertilization or germination follows, and the consequence is variation, embryo formation, seed production, or successful early growth.

  • mitosis makes clones; meiosis and fertilization create variation
  • hormones, anatomy, fertilization, and IVF support gamete fusion and embryo development
  • pollination and pollen-tube growth bring gametes together inside ovules
  • dispersal reduces competition and germination starts with water, enzymes, and reserves

Core Reproduction

Core D3.1 exam questions usually combine reproduction strategy with gamete formation, fertilization, human cycles, IVF, plant pollination, or seed germination. Treat each answer as a route: name the process, say what moves or changes, then give the biological consequence.

  • Compare asexual and sexual reproduction by mechanism and genetic outcome.
  • Link meiosis, fertilization, reproductive anatomy, and hormonal cycles to successful reproduction.
  • Explain plant pollination and seed stages by connecting structures to transfer, fertilization, dispersal, and germination.

Topic D3.2

D3.2 Inheritance

Inheritance explains how alleles, chromosomes, meiosis, pedigrees, linkage, variation and statistical tests predict genetic outcomes across generations in inheritance problems.

48% of analysed papers 67 papers · 96 questions

Objectives in this topic

Gamete Fusion Restores Diploidy

In the sexual life cycle common to eukaryotes, meiosis makes haploid gametes and fertilization fuses two gametes to form a diploid zygote.

Meiosis halves chromosome number from 2n to n. Fusion combines one maternal and one paternal set, restoring 2n without chromosome number doubling in every generation.

A diploid individual normally carries two copies of each autosomal gene, one on each homologous chromosome. Each gamete carries one copy, and the zygote receives two again.

A human sperm and ovum each contain 23 chromosomes; after nuclear fusion the zygote contains 46.

Fertilization combines two haploid sets; it does not copy one gamete or make the zygote genetically identical to either parent.

Haploid gametes + fusion = diploid

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Gametes are haploid so fusion can restore the diploid number.

Representative question

Question 1

[Maximum number: 1]

For what reason do gametes contain only one allele of each gene?

A

To prevent inbreeding in a population

B

Haploid cells contain only one set of chromosomes

C

The two alleles of a gene are separated during mitosis

D

Crossing over will always produce one allele of a gene

Use Flowering Plants to Perform Genetic Crosses

A flowering-plant cross transfers pollen carrying male gametes to a stigma so fertilization can combine known parental alleles.

Generation What to do and record
P Choose parents with known contrasting traits and control which pollen reaches the stigma
F1 Grow the first filial offspring and record their phenotype(s)
F2 Cross or self-pollinate suitable F1 plants, then compare observed offspring with a Punnett-grid prediction

Pollen is the practical source of male gametes; female gametes are inside ovules in the ovary. Pea flowers can self-pollinate, which helps maintain pure-breeding lines and produce controlled F2 generations.

Cross two pure-breeding parents with contrasting traits, record a uniform F1, then self-pollinate the F1 and compare the F2 counts with the predicted genotype and phenotype ratios.

A Punnett grid predicts probabilities, not exact counts. Controlled crosses are used in crop and ornamental breeding, but pollination is transfer of pollen, not fertilization itself.

Genetic crosses in flowering plants

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Parental genotypes determine the gametes available in a cross.

Representative question

Question 1

[Maximum number: 3]

L. purpureus can have purple or white flowers. Two pure-breeding varieties were crossed: HA 4 with white flowers and GL 424 with purple flowers. All of the F1F_{1} plants had purple flowers. The F1F_{1} plants were self-pollinated to produce an F2F_{2} generation. There were 97 plants with purple flowers and 38 plants with white flowers in the F2F_{2} generation.

Using a Punnett grid, explain the results of this cross.

Genotype Records Alleles at a Locus

A gene is a DNA sequence affecting a characteristic; an allele is one version of that gene; a genotype is the allele combination carried at one or more loci.

Term Meaning Example at an A/a locus
Homozygous Two identical alleles AA or aa
Heterozygous Two different alleles Aa

Writing Aa identifies the genotype at one locus; it does not by itself name the visible phenotype until the allele relationship is known.

Do not use gene, allele, genotype and phenotype as synonyms. An individual has two alleles at an autosomal locus, while a population may contain more than two.

Genotype exam focus

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Distinguish / Define.

Command terms

Identify / Distinguish / Define

What earns marks

Build the answer around this relationship: A genotype records the alleles an organism carries.

Representative question

Question 1

[Maximum number: 1]

Define the term genotype.

Phenotype Results from Genotype and Environment

Phenotype is an observable or measurable characteristic produced by genotype, environment, or an interaction between them.

Main influence Example and explanation
Genotype ABO blood group follows the inherited ABO alleles
Environment An acquired scar depends on injury rather than an inherited allele
Interaction Human height is influenced by many genes and by conditions such as nutrition

Genotype sets biological possibilities, while environmental conditions can alter gene expression, development or physiology; the size of each contribution depends on the trait.

A phenotype is not always visible, and an environmental effect does not necessarily change DNA sequence or become inherited.

Phenotype exam focus

Assessment in practice

2 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Phenotype means the expressed or observable characteristic.

Representative question

Question 1

[Maximum number: 2]

Identify the phenotypes of each part of the phenotypic ratio.

RatioPhenotypes
9
3
3
1

Dominant and Recessive Describe an Allele Relationship

For a complete-dominance locus, the dominant allele determines the heterozygous phenotype; the recessive phenotype appears only when no dominant allele is present.

One dominant allele may produce enough functional product for the dominant phenotype, so AA and Aa look alike in this model, whereas aa lacks that contribution.

Genotype Phenotype in a complete-dominance model
AA Dominant
Aa Dominant
aa Recessive

In Aa × Aa, the expected genotypes are 1 AA : 2 Aa : 1 aa but the expected phenotypes are 3 dominant : 1 recessive.

Dominant does not mean common, beneficial or stronger. Dominance describes the phenotype of a heterozygote.

Dominant and recessive alleles

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Explain / Deduce.

Command terms

Identify / Explain / Deduce

What earns marks

Build the answer around this relationship: Dominant alleles are expressed in heterozygotes.

Representative question

Question 1

[Maximum number: 4]

Many genetic diseases are due to recessive alleles of autosomal genes that code for an enzyme. Using a Punnett grid, explain how parents who do not show signs of such a disease can produce a child with the disease.

Phenotypic Plasticity Changes Expression, Not Genotype

Phenotypic plasticity is the capacity of one genotype to produce different phenotypes under different environmental conditions.

Environmental signals can change which genes are expressed and how much product is made, altering physiology or form without changing the DNA sequence.

Many plastic responses can reverse during an individual's lifetime if the environment changes again; the inherited genotype remains the same.

The same plant genotype may form broader leaves in shade and smaller leaves in bright, dry conditions because development responds to the local environment.

Plasticity is not mutation and does not guarantee that the acquired phenotype is inherited by offspring.

Phenotypic plasticity

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice.

What earns marks

Build the answer around this relationship: One genotype can produce different phenotypes in different environments.

Representative question

Question 1

[Maximum number: 1]

Scientists incubated larvae of the moth Utetheisa ornatrix at either 15C15^{\circ} \mathrm{C} or 22C22^{\circ} \mathrm{C} until they hatched. They found the hatched moths had different wing colour patterns due to phenotypic plasticity.

Moth from larvae incubated at \(15^{\circ

Moth from larvae incubated at \(22^{\circ

Which of the following explains the observed differences in wing colour?

A

Colder temperatures induce mutations in genes for wing colour.

B

The expression of genes for wing colour is affected by temperature.

C

A mutation makes moths less visible to predators in cold climates.

D

Wing colour is the result of polygenic inheritance.

PKU Connects a Recessive Allele to Metabolism

Phenylketonuria (PKU) is an autosomal recessive disorder in which mutation reduces the enzyme that converts phenylalanine to tyrosine.

With insufficient enzyme activity, phenylalanine accumulates and tyrosine production is reduced. Two recessive disease alleles are normally required for the affected phenotype.

Genotype → reduced phenylalanine-hydroxylase activity → disrupted phenylalanine-to-tyrosine conversion → altered metabolite concentrations and phenotype.

Restricting dietary phenylalanine lowers the substrate entering the blocked pathway and can reduce the severity of the phenotype.

Diet can change the phenotype but does not remove or rewrite the inherited PKU alleles.

Phenylketonuria (PKU)

Assessment in practice

2–3 marks
How it is assessed

This objective is assessed through essay response, commonly using Explain / Outline.

Command terms

Explain / Outline

What earns marks

Build the answer around this relationship: PKU is usually autosomal recessive, so carriers can be unaffected.

Representative question

Question 1

[Maximum number: 4]

Discuss the causes and treatments of phenylketonuria.

SNPs and Multiple Alleles Create Variation

A SNP is a common one-base DNA difference; multiple alleles are variants of one locus present in a population.

A base change may affect coding, regulation or nothing observable; a diploid person carries at most two alleles even when a population has many. Trace the allele combination through the stated biological mechanism before predicting the result.

Separate population allele variety from the two alleles in one individual.; compare the stated alleles and outcome

The ABO locus has three common alleles, while one person may carry only IA and IB. This gives a concrete prediction from the stated parental information.

A SNP is not automatically harmful or visible. Interpret the result within the stated inheritance model and its sample or environmental limits.

SNPs and multiple alleles

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: An SNP is variation at a single nucleotide position.

Representative question

Question 1

[Maximum number: 1]

Which statement defines alleles?

A

They are the different forms of a gene that have the same effect on the phenotype.

B

They are the similar forms of a gene in different positions of a chromosome.

C

They are the various forms of a gene with slight differences in their base sequences.

D

They are the different forms of a gene coding for identical polypeptide chains.

ABO Blood Groups Use Codominance

ABO phenotype is determined by IA, IB and i: IA and IB are codominant, while i is recessive to either.

IA makes A antigen; IB makes B; i makes neither; IAIB therefore displays both antigens. Trace the allele combination through the stated biological mechanism before predicting the result.

List the two alleles; apply dominance; identify antigens and phenotype.; compare the stated alleles and outcome

IAi × IBi can produce AB, A, B or O offspring. This gives a concrete prediction from the stated parental information.

Blood type requires alleles from both parents. Interpret the result within the stated inheritance model and its sample or environmental limits.

ABO blood groups

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Describe / State / Identify.

Command terms

Describe / State / Identify / Outline

What earns marks

Build the answer around this relationship: IA and IB are codominant in blood group AB.

Representative question

Question 1

[Maximum number: 9]

Describe the inheritance of ABO blood groups.

Separate Incomplete Dominance from Codominance

In incomplete dominance the heterozygote has an intermediate phenotype; in codominance both allele products are detectably expressed.

Pattern Heterozygote IB example
Incomplete dominance Intermediate phenotype Red × white four-o'clock flower (Mirabilis jalapa) can produce pink F1 flowers
Codominance Both products expressed IᴬIᴮ produces both A and B antigens

Self-crossing two pink Mirabilis F1 plants predicts a 1 red : 2 pink : 1 white phenotype ratio when the two alleles show incomplete dominance.

Codominance is not blending: both products remain present. Incomplete dominance does not make either allele 'partly dominant' in every genotype.

Incomplete dominance and codominance

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Identify / Describe.

Command terms

Identify / Describe

What earns marks

Build the answer around this relationship: Incomplete dominance produces an intermediate heterozygote phenotype.

Representative question

Question 1

[Maximum number: 1]

A Mirabilis jalapa plant with red flowers was crossed with one with white flowers. All plants in the F1 generation had pink flowers. What phenotype ratio would be expected in the F2 generation?

A

100 % pink

B

50 % red and 50 % white

C

25 % white, 50 % pink and 25 % red

D

75 % red and 25 % white

The Sperm Sex Chromosome Determines the Typical XX/XY Outcome

In the simplified human model, eggs carry X, whereas sperm carry X or Y; therefore the sperm's sex chromosome determines whether the zygote is typically XX or XY.

Egg Sperm Typical zygote outcome
X X XX, typical female sex characteristics
X Y XY, typical male sex characteristics

The X chromosome carries far more genes than the Y chromosome. Sex-chromosome inheritance therefore also affects many genes unrelated to sex determination.

An X-bearing and a Y-bearing sperm are expected in roughly equal proportions, so each fertilization has approximately equal model probabilities of XX and XY.

XX/XY is a simplified model of typical development; chromosome variation and differences in gene function can produce other biological outcomes.

Sex determination

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Explain.

Command terms

Identify / Explain

What earns marks

Build the answer around this relationship: Eggs normally contribute an X chromosome.

Representative question

Question 1

[Maximum number: 4]

Distinguish between autosomes and sex chromosomes in humans.

Haemophilia Shows X-Linked Recessive Inheritance

Haemophilia alleles on the X chromosome reduce a clotting factor; the recessive pattern makes affected XY individuals more common.

An XY individual has one X allele; an XX individual may have a second functional allele; a carrier mother can pass the allele to sons or daughters. Trace the allele combination through the stated biological mechanism before predicting the result.

Write X-linked genotypes and track which parent supplies each X.; compare the stated alleles and outcome

Carrier mother XH Xh and unaffected father XH Y can have an affected son Xh Y. This gives a concrete prediction from the stated parental information.

Probabilities describe a model, not one guaranteed child. Interpret the result within the stated inheritance model and its sample or environmental limits.

Haemophilia exam focus

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Deduce / Explain.

Command terms

Identify / Deduce / Explain / Outline / State / Predict

What earns marks

Build the answer around this relationship: Males express an X-linked recessive allele if it is on their single X chromosome.

Representative question

Question 1

[Maximum number: 8]

Explain how males inherit hemophilia and how females can become carriers for the condition.

Use Pedigrees to Test Inheritance Hypotheses

A pedigree records phenotype and family relationships across generations so inheritance patterns and possible genotypes can be deduced.

Step Reasoning
Read symbols and relationships Identify affected/unaffected individuals, sex, partners and offspring
Look for a pattern Recessive traits may skip generations; sex linkage and dominance give different parent-offspring constraints
Assign only forced genotypes Use each mating and offspring to test the hypothesis; leave uncertain alleles unknown

Inductive reasoning proposes a pattern from the observed family data; deductive reasoning predicts who could be affected or carry an allele if that pattern is correct.

Two unaffected parents with an affected child support a recessive hypothesis; if the trait is autosomal recessive, both parents must carry the allele.

Consanguineous partners are more likely to share a rare ancestral recessive allele, but relatedness does not guarantee an affected child. Small pedigrees may fit more than one model.

Pedigree charts

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Deduce / Determine.

Command terms

Identify / Deduce / Determine / Draw / Calculate / State / Explain

What earns marks

Build the answer around this relationship: Pedigrees use affected and unaffected relatives to infer hidden genotypes.

Representative question

Question 1

[Maximum number: 2]

Explain how the pedigree chart shows that the dominant allele causing PKD is not on the X chromosome.

Continuous Variation Produces a Measurable Range

Continuous variation has many intermediate values and often results from several genes, environmental factors, or both.

Variation Example Useful description
Continuous Human skin colour or height Distribution, range, mean, median and mode
Discrete ABO blood group Counts or proportions in distinct categories

In polygenic inheritance, many loci each contribute to the phenotype; environmental conditions can add further variation, often producing a broad distribution.

For student heights, the mean uses every value, the median identifies the middle position and the mode identifies the most frequent value or interval.

Continuous does not mean entirely environmental, and discrete does not mean that only one gene is always involved.

Continuous variation

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / State / Outline.

Command terms

Identify / State / Outline / Distinguish / Explain

What earns marks

Build the answer around this relationship: Continuous variation shows a range rather than separate phenotype classes.

Representative question

Question 1

[Maximum number: 7]

Explain the reasons for variation in human height.

Read Box Plots Using Quartiles and the IQR

A box-and-whisker plot summarizes a continuous dataset using the minimum, lower quartile (Q1), median, upper quartile (Q3), maximum and any plotted outliers.

IQR=Q3Q1.Avalueisanoutlierbythe1.5IQRruleifitisbelowQ11.5(IQR)oraboveQ3+1.5(IQR).IQR = Q3 − Q1. A value is an outlier by the 1.5-IQR rule if it is below Q1 − 1.5(IQR) or above Q3 + 1.5(IQR).

The box spans the middle 50% of observations, its internal line is the median, and whiskers show the non-outlying range when outliers are plotted separately.

To compare two student-height samples, compare their medians for centre, their IQRs for spread, their total non-outlying ranges and any outliers.

A longer box means a larger IQR, not necessarily a larger sample. A box plot does not display every observation or prove that two groups differ significantly.

Box-and-whisker plots

Assessment in practice

1 marks
How it is assessed

This objective is assessed through data analysis, commonly using State / Determine / Deduce.

Command terms

State / Determine / Deduce

What earns marks

Build the answer around this relationship: The median is the central line, not the mean.

Representative question

Question 1

[Maximum number: 3]

Using the data, deduce whether the incidence of CHF or the incidence of anemia has a greater effect on the blood hepcidin concentration.

Retrieve the Core Inheritance Route

Core D3.2 is secure when the student can move from allele rules into predictions and evidence: gametes form genotypes, genotypes can produce phenotypes, different dominance patterns need different notation, and pedigrees or plots require evidence-based interpretation.

  • haploid gametes carry one allele and fertilization restores a diploid genotype
  • dominance, codominance, incomplete dominance, environment, and plasticity affect the observed trait
  • PKU, ABO, sex determination, and haemophilia use different inheritance rules and notation
  • pedigrees infer inheritance patterns and box plots summarize continuous variation

Solve Core Inheritance Questions

Core inheritance exam questions reward disciplined reasoning. First identify the inheritance rule, then write the correct notation or evidence, then state the phenotype, ratio, or conclusion. This prevents the common mistake of writing definitions without solving the genetic problem.

  • Use allele and genotype notation correctly for monohybrid, ABO, PKU, haemophilia, and sex-determination contexts.
  • Connect genotype, dominance pattern, environment, or plasticity to phenotype.
  • Use pedigree or box-plot evidence to justify an inheritance or variation conclusion.

Topic D3.3

D3.3 Homeostasis

Homeostasis maintains internal conditions through feedback control of blood pH, glucose, temperature, kidney filtration, osmoregulation and blood flow in human physiology.

30% of analysed papers 42 papers · 65 questions

Objectives in this topic

Homeostasis Keeps the Internal Environment within Limits

Homeostasis maintains variables in an organism's internal environment within preset narrow limits despite external fluctuations.

Human homeostatic variable Why regulation matters
Body temperature Keeps enzyme and membrane processes in a functional range
Blood pH Preserves protein shape and reaction conditions
Blood glucose concentration Maintains a usable respiratory substrate supply
Blood osmotic concentration Limits harmful water movement into or out of cells

Stable tissue fluid lets cells function predictably even when temperature, food intake or water availability outside the body changes.

After a meal raises blood glucose, hormonal regulation brings the concentration back toward its preset range.

Homeostasis is dynamic: values fluctuate around a set point or within limits rather than remaining perfectly constant.

Homeostasis definition

Assessment in practice

1–6 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain / Identify / Outline.

Command terms

Explain / Identify / Outline

What earns marks

Build the answer around this relationship: Homeostasis keeps internal variables within narrow limits.

Representative question

Question 1

[Maximum number: 6]

Explain how the pH of blood is kept constant during exercise.

Negative Regulation Reverses a Deviation

Negative regulation reduces the original change so a regulated variable returns toward its normal range.

The response opposes the disturbance: a rise triggers actions that lower it, and a fall triggers actions that raise it. This stabilizes rather than amplifies the system.

Ask whether the response moves the variable in the opposite direction to the initial deviation.; identify the signal, controller and effector

If body temperature rises, sweating and vasodilation increase heat loss, reducing the rise. This gives a concrete prediction from the stated condition.

Negative means opposing the deviation, not harmful or always below the set point. Interpret the result within the stated biological model and limits.

Negative feedback loops

Assessment in practice

4 marks
How it is assessed

This objective is assessed through essay response, commonly using Discuss.

Command terms

Discuss

What earns marks

Build the answer around this relationship: Negative feedback opposes the original change.

Representative question

Question 1

[Maximum number: 4]

Discuss the use of positive and negative feedback to control levels of variables.

Insulin and Glucagon Regulate Blood Glucose

Pancreatic endocrine cells detect blood glucose: beta cells release insulin when it rises, while alpha cells release glucagon when it falls.

Change Hormone carried in blood Main target effects Result
Glucose above set point Insulin Increased glucose uptake by target cells; glycogen synthesis in liver and muscle Blood glucose falls
Glucose below set point Glucagon Liver glycogen breakdown and glucose release Blood glucose rises

The two opposing hormone responses form negative-feedback loops that reduce the original deviation.

After a carbohydrate-rich meal, rising glucose stimulates beta cells; insulin promotes uptake and storage until secretion falls as the set point is approached.

Glucagon acts mainly on the liver to raise circulating glucose; muscle glycogen is primarily a local fuel store and is not released as blood glucose in response to glucagon.

Blood glucose regulation

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Explain / Describe.

Command terms

Identify / Explain / Describe / Outline / State / Discuss

What earns marks

Build the answer around this relationship: Beta cells secrete insulin when blood glucose is high.

Representative question

Question 1

[Maximum number: 8]

Explain the control of blood glucose concentrations in humans.

Type 1 and Type 2 Diabetes Disrupt Different Parts of Control

Diabetes mellitus causes persistent difficulty controlling blood glucose, but type 1 and type 2 begin with different physiological failures.

Feature Type 1 Type 2
Main physiological change Autoimmune destruction of pancreatic beta cells causes little or no insulin secretion Target cells respond poorly to insulin; beta-cell function may later decline
Important risk pattern Autoimmune susceptibility; not prevented by lifestyle Risk rises with genetic susceptibility, excess body fat and low physical activity
Management Insulin replacement, glucose monitoring and coordinated diet/exercise Activity, diet and healthy body mass can reduce risk and aid control; medication and sometimes insulin may be required

With too little effective insulin signalling, uptake and storage do not adequately reduce blood glucose after a meal, so hyperglycaemia persists.

Lifestyle is a risk modifier for type 2, not a moral diagnosis or the sole cause. A single high reading does not distinguish the two types.

Diabetes exam focus

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Describe / Identify / Explain.

Command terms

Describe / Identify / Explain / State / Discuss / Analyse / Outline

What earns marks

Build the answer around this relationship: Type I diabetes involves insufficient insulin production.

Representative question

Question 1

[Maximum number: 5]

Outline type II diabetes.

Thermoregulation Is a Negative-Feedback Control System

Human thermoregulation detects deviation in core temperature and coordinates effectors that reverse the change.

Control component Role
Peripheral thermoreceptors Detect temperature changes, especially at the skin
Hypothalamus Integrates peripheral and central temperature information
Pituitary/thyroid pathway Alters thyroxin signalling and therefore metabolic heat production
Skeletal muscle Shivering raises respiration and heat production
Brown adipose tissue Uncoupled respiration releases energy as heat

A fall in temperature is detected, the hypothalamus coordinates reduced heat loss and increased muscle/adipose heat production, and the response decreases as core temperature recovers.

The regulated variable is core temperature; skin temperature can change more rapidly and acts partly as an early environmental signal.

Thermoregulation exam focus

Assessment in practice

1–8 marks
How it is assessed

This objective is assessed through structured response, commonly using Describe / Explain / Identify.

Command terms

Describe / Explain / Identify / Outline

What earns marks

Build the answer around this relationship: The hypothalamus coordinates body temperature control.

Representative question

Question 1

[Maximum number: 8]

Explain the control of body temperature in humans.

Human Effectors Alter Heat Loss and Production

Human thermoregulation combines physiological and behavioural responses; each effector changes heat transfer or metabolic heat production.

When hot Effect When cold Effect
Skin vasodilation More warm blood near the surface increases heat loss Skin vasoconstriction Less warm blood near the surface reduces heat loss
Sweating Evaporation removes latent heat Shivering Rapid muscle contraction increases respiration and heat production
Hairs lie flatter Reduces the trapped insulating air layer Hair erection Traps more air, though the effect is small in humans
Behaviour seeks shade/cooling Reduces heat gain or raises loss Brown-fat uncoupled respiration/warmer behaviour Produces or conserves heat

Sweating is most effective when sweat evaporates; high humidity reduces evaporation and therefore reduces cooling.

Vasodilation transfers internal heat toward skin but does not itself remove heat from the body; the environment must accept that heat.

Thermoregulation mechanisms

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Outline.

Command terms

Identify / Outline

What earns marks

Build the answer around this relationship: Evaporation of sweat removes heat from the body.

Representative question

Question 1

[Maximum number: 1]

Outline one change that happens in the human body in response to a rise in body temperature above 36.4C36.4^{\circ} \mathrm{C}.

Retrieve the Core Homeostasis Route

Core D3.3 is secure when every example becomes a feedback route: identify the variable, detect deviation, coordinate a response, activate effectors, and reverse the change. Glucose and temperature are the key worked examples.

  • stable internal environment within narrow limits
  • detects deviation from set point and reverses it
  • insulin lowers high glucose; glucagon raises low glucose
  • hypothalamus coordinates cooling or warming responses

Core Homeostasis

Core homeostasis answers should use a control-loop structure, not a list of responses. The response starts with the variable and set point, then explains how the body detects deviation and activates the response that reverses it. Apply that loop to glucose, diabetes, or temperature.

  • Define homeostasis as maintaining stable internal conditions within narrow limits.
  • Use negative feedback language: receptor, coordinator, effector, set point, and reverse the deviation.
  • Apply the loop to insulin/glucagon, diabetes types, or hot/cold thermoregulation responses.

Topic D4.1

D4.1 Natural selection

Natural selection explains how heritable variation, selection pressures, differential survival, reproduction and allele-frequency changes drive evolutionary adaptation in populations in evolving populations.

30% of analysed papers 42 papers · 52 questions

Objectives in this topic

Natural Selection Drives Evolutionary Change

Natural selection is differential survival and reproduction caused by heritable differences; over generations it changes populations and drives evolution.

Heritable variation exists → an environmental or biological pressure affects individuals differently → some genotypes leave more surviving offspring → their alleles become more frequent → population characteristics change.

Operating continuously over billions of years, repeated selection has contributed to life's biodiversity. It can also be observed over short timescales when pressures are strong.

Darwin supplied a convincing selection mechanism and displaced Lamarckian explanations based on inherited acquired characteristics. Replacing a dominant explanatory framework in this way is a scientific paradigm shift.

Individuals do not evolve because they need to. Selection acts on existing phenotypic differences, while evolutionary change is measured across generations.

Natural selection as mechanism

Assessment in practice

1–7 marks
How it is assessed

This objective is assessed through essay response, commonly using Explain / Identify / Evaluate.

Command terms

Explain / Identify / Evaluate / Outline / State / Describe / Discuss

What earns marks

Build the answer around this relationship: Natural selection acts on individuals but changes populations across generations.

Representative question

Question 1

[Maximum number: 8]

Explain how evolution may happen in response to environmental change with evidence from examples.

Mutation Creates Alleles; Sex Reshuffles Them

Mutation creates new alleles by changing DNA sequence; sexual reproduction creates new combinations of existing alleles.

Source How variation is generated
Mutation A new base sequence can create a new allele
Meiosis Crossing over and independent assortment place existing alleles into new gamete combinations
Random fertilization Combines one gamete from each parent into a new genotype

In a sexually reproducing plant, a mutation may introduce a drought-tolerance allele; meiosis and fertilization then place that allele into varied genetic backgrounds.

Selection does not create a needed mutation. Mutations arise without regard to usefulness, and sexual reproduction reshuffles rather than invents alleles.

Roles of mutation and sexual reproduction

Assessment in practice

1–7 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain / Identify / Outline.

Command terms

Explain / Identify / Outline

What earns marks

Build the answer around this relationship: Mutation can create new alleles.

Representative question

Question 1

[Maximum number: 8]

Explain how sexual reproduction can eventually lead to evolution in offspring.

Overproduction Creates Competition for Limiting Resources

Populations can produce more offspring than the environment can support, so individuals of the same species compete for limiting resources.

Food, water, light, mineral ions, territory, nesting sites or mates can limit carrying capacity. Individuals that obtain these resources more successfully are more likely to survive and reproduce.

Potential offspring exceed available resources → intraspecific competition occurs → heritable differences affect resource access → reproductive success differs.

When food limits a population, individuals whose inherited feeding traits increase food acquisition may leave more offspring than competitors.

Overproduction means reproductive potential exceeds long-term support; it does not require the population to remain permanently above carrying capacity.

Overproduction and competition

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Identify / State.

Command terms

Outline / Identify / State / Explain

What earns marks

Build the answer around this relationship: Overproduction means not all offspring survive to reproduce.

Representative question

Question 1

[Maximum number: 4]

Outline how overpopulation of a species in a given environment may lead to evolution.

Abiotic Factors Apply Density-Independent Selection

Abiotic conditions can act as selection pressures when they affect survival or reproduction differently among heritable phenotypes.

High or low temperature, drought, salinity, pH and other physical conditions may affect individuals regardless of population density, so they are density-independent pressures.

Abiotic condition changes → physiological performance differs → survival or reproduction differs → alleles associated with advantageous traits become more frequent.

During repeated low-temperature events, plants with inherited frost tolerance may survive and set more seed, increasing the frequency of tolerance alleles.

A useful response is evolutionary adaptation only if it is heritable and changes reproductive contribution; temporary acclimatization alone does not alter allele frequency.

Abiotic factors as selection pressures

Assessment in practice

2 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Explain.

Command terms

Outline / Explain

What earns marks

Build the answer around this relationship: Abiotic factors are non-living parts of the environment.

Representative question

Question 1

[Maximum number: 2]

Explain how natural selection is influenced by changes in the environment.

Fitness Is Relative Reproductive Success

Biological fitness is a genotype's relative contribution of surviving, reproducing offspring to the next generation in a particular environment.

Intraspecific competition exposes differences in adaptation. Survival has evolutionary importance when it increases reproductive opportunity, and reproduction passes the responsible alleles onward.

Compare individuals in the same population: adaptation to current conditions → survival and mating differences → different numbers of fertile offspring → different fitness.

A genotype that survives well but produces no fertile offspring has lower fitness than a competing genotype that leaves many reproducing descendants.

Fitness does not mean strength, health or lifespan in isolation; it is relative, environment-dependent reproductive success.

Differences in adaptation, survival, reproduction

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Outline / Compare / Describe.

Command terms

Outline / Compare / Describe

What earns marks

Build the answer around this relationship: Adaptations are heritable traits that improve performance in context.

Representative question

Question 1

[Maximum number: 1]

In a natural population, what is a feature of individuals that are better adapted?

A

They start to produce offspring at a younger age than less well adapted individuals.

B

They produce identical offspring by cloning that are also better adapted.

C

They tend to produce more offspring during their lifetime than less well adapted individuals.

D

They do not produce more offspring than the environment can support.

Selection Requires Heritable Differences

For selection to produce an evolutionary change, the trait difference associated with reproductive success must be transmitted to offspring.

Environmental effects can change an individual’s phenotype without changing inherited alleles. Only the heritable component can shift population frequencies across generations.

Ask whether offspring resemble parents for the trait before attributing a long-term change to selection.; separate variation, selection, inheritance and time

If dark fur is inherited and dark mice leave more pups, dark alleles rise; if fur darkens only from soot exposure, the population need not evolve. This gives a concrete prediction from the stated population.

A trait can be heritable yet show little response when selection is weak or environments change. Interpret the result within the stated selection model and evidence limits.

Traits must be heritable

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Identify.

Command terms

Identify

What earns marks

Build the answer around this relationship: Natural selection causes evolution only when selected traits are heritable.

Representative question

Question 1

[Maximum number: 1]

What is required for natural selection to occur?

I. Acquired characteristics
II. Advantageous characteristics
III. Genetic variation

A

I only

B

I and III only

C

II and III only

D

I, II and III

Sexual Selection Favors Mating Success

Sexual selection is selection for traits that increase access to mates or fertilization success, even when they carry survival costs.

Mate choice and competition among same-sex individuals change reproductive success. The trait spreads when its mating advantage outweighs its costs.

Separate survival benefit from mating benefit before explaining a conspicuous trait.; separate variation, selection, inheritance and time

Peacock tail feathers may attract mates while making escape harder, so mating success can favour the tail despite predation risk. This gives a concrete prediction from the stated population.

Sexual selection is one component of natural selection, not a guarantee that the trait improves survival. Interpret the result within the stated selection model and evidence limits.

Sexual selection

Assessment in practice

2–3 marks
How it is assessed

This objective is assessed through essay response, commonly using Explain / Suggest / Evaluate.

Command terms

Explain / Suggest / Evaluate

What earns marks

Build the answer around this relationship: Sexual selection acts through mating success.

Representative question

Question 1

[Maximum number: 6]

Explain what is meant by exaggerated traits and how they may develop in males of a species.

Endler Controlled Predation and Background in Guppy Selection

John Endler modelled natural and sexual selection in Trinidadian guppies by experimentally controlling predation pressure and gravel background.

Controlled factor Comparison Selection prediction
Predation No predator, weak predator, dangerous predator Strong predation favours less conspicuous males; low predation allows female choice to favour conspicuous males
Gravel background Coarse versus fine gravel Under predation, spot size that better matches the background improves camouflage

After dangerous predators were introduced, mean spot number decreased; with no or weak predation it continued to increase. Coarse gravel favoured larger spots and fine gravel smaller spots when predators were present.

A field transfer from a dangerous-predator site to a weak-predator site produced more colourful males over 15 generations, consistent with sexual selection becoming stronger relative to predation.

The experiment shows a trade-off: conspicuous colour can improve mating success yet reduce survival. Correlation alone is weaker evidence than Endler's controlled manipulation of selection pressures.

Modelling selection

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through multiple choice, commonly using Describe / Explain.

Command terms

Describe / Explain

What earns marks

Build the answer around this relationship: Models can show how selection changes variant frequencies over generations.

Representative question

Question 1

[Maximum number: 1]

John Endler experimented on populations of guppies (Poecilia reticulata) with different colouration. A male guppy fish is shown with large spots, which makes the fish more attractive to females, but more visible to predators.

The table shows the male colouration of guppy fish and number of predators in three different ponds.

Predator XPredator YMale guppy colouration
Pond 1120Large colourful spots
Pond 2150Medium colourful spots
Pond 3517None/very small drab spots

What can be concluded from the data?

A

There is a positive correlation between numbers of predator X and size of spots.

B

Predator Y has least influence on colouration.

C

There is a negative correlation between number of predators and size of spots.

D

There is no sexual selection.

Retrieve the Core Natural Selection Route

Core D4.1 examples follow the same causal route: heritable variation exists, a selection pressure acts, individuals differ in fitness, and alleles linked to higher reproduction become more common. Endler’s guppies and sexual selection are evidence versions of the same chain.

  • mutation creates alleles; meiosis and fertilization reshuffle combinations
  • overproduction, limited resources, and abiotic factors filter variants
  • passing alleles to offspring in a particular environment
  • Endler controlled predation pressure and guppy colour patterns changed

Core Natural Selection

Core natural-selection exam answers should never stop at “the best adapted survive.” They need the chain: heritable variation exists, a named pressure acts, some individuals have higher fitness, and their alleles become more common over generations. Use this for abiotic pressure, overproduction, sexual selection, and Endler-style data.

  • Explain natural selection using heritable variation, selection pressure, differential survival/reproduction, and population change.
  • Distinguish mutation/recombination as sources of variation from selection as the filtering process.
  • Use examples such as abiotic pressure, sexual selection, or Endler guppy data to support the chain.

Topic D4.2

D4.2 Stability and change

Stability and change in ecosystems depend on sustainable resource use, pollution impacts, keystone species, rewilding, and succession processes over time.

24% of analysed papers 34 papers · 59 questions

Objectives in this topic

Stable Ecosystems Persist while Remaining Dynamic

Ecosystem stability is the capacity to maintain characteristic structure and function over time or recover after disturbance.

Resistance limits the immediate effect of disturbance, while resilience is the capacity to recover. Evidence from forests, deserts and other natural ecosystems shows that some recognizable systems have persisted for millions of years.

A forest can undergo seasonal population changes and recover from storms while retaining its nutrient cycling, food-web structure and dominant vegetation over long periods.

Stability means continuity of key properties, not a frozen species count or absence of all change.

Four Requirements Support Ecosystem Stability

Long-term ecosystem stability requires continuing energy supply, nutrient recycling, genetic diversity and climatic variables within organismal tolerance limits.

Requirement Why it supports stability
Energy supply, usually sunlight Maintains primary production and food-web energy flow
Nutrient recycling Returns finite chemical elements from waste and dead biomass to producers
Genetic diversity Provides variation that can support population survival under disease or change
Climate within tolerance limits Keeps temperature, precipitation and insolation compatible with resident species

If prolonged drought pushes precipitation outside tree tolerances, producer biomass falls and both energy input and habitat complexity decline.

The requirements interact; meeting one cannot compensate indefinitely for failure of another.

Requirements for stability

Assessment in practice

2–4 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline / Describe.

Command terms

Outline / Describe

What earns marks

Build the answer around this relationship: Energy must continually enter ecosystems because it is transferred and lost rather than recycled.

Watch for

Treating sustainability as a list of organisms only, without explaining energy input or nutrient recycling.

Representative question

Question 1

[Maximum number: 4]

Outline the features of ecosystems that make them sustainable.

Amazon Deforestation Can Reinforce a Tipping-Point Shift

A large Amazon forest area is needed to recycle atmospheric water by transpiration, causing cooling, air movement and rainfall that help maintain the forest.

Deforestation lowers transpiration and rainfall, increases drying and fire risk, and fragments habitat; further forest loss can then reinforce the original change. The minimum area needed to maintain these processes remains uncertain.

Percentagechange=((finalforestareainitialforestarea)÷initialforestarea)×100%.Anegativeresultrepresentsforestloss.Percentage change = ((final forest area − initial forest area) ÷ initial forest area) × 100\%. A negative result represents forest loss.

The mapped local textbook reports 3,399,308 km² in 2017 and 3,390,835 km² in 2018: ((3,390,835 − 3,399,308) ÷ 3,399,308) × 100 = −0.25%, so estimated cover fell by 0.25%.

A proposed tipping range is uncertain, so a calculated percentage loss must not be presented as proof that an irreversible threshold has already been crossed.

Use Mesocosms as Controlled Ecosystem Models

A mesocosm is a contained ecosystem model used to test how a controlled variable affects stability.

Design choice Purpose
Sealed glass vessel preferred to an open tank Prevents matter entering or leaving while allowing energy transfer such as light and heat
Aquatic or microbial community More likely than a terrestrial system to function at small contained scale
Replicated control and treatment vessels Separates the manipulated variable from background variation
Repeated abiotic and biotic measurements Tracks stability, disturbance and recovery through time

Replicated sealed aquatic mesocosms can receive different light treatments while temperature, starting organisms and nutrient quantities are held constant.

A mesocosm supports causal inference about its model conditions but does not reproduce every migration, weather event or interaction in a natural ecosystem; it also requires ethical care and maintenance.

Mesocosm model

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through essay response, commonly using Discuss / Outline / State.

Command terms

Discuss / Outline / State / Suggest / Explain

What earns marks

Build the answer around this relationship: A sealed mesocosm restricts matter exchange but can still exchange energy with its surroundings.

Watch for

Assuming a sealed mesocosm exchanges no energy, when light or heat can still pass between the system and surroundings.

Representative question

Question 1

[Maximum number: 3]

Mesocosm experiments using water from Narragansett Bay were completed in the laboratory during a six month period. Discuss advantages and limitations of carrying out mesocosm investigations. be marked.

Keystone Species Have Disproportionate Effects

A keystone species has an effect on community structure much larger than its abundance would suggest.

Its predation, grazing, habitat engineering or other interaction controls competitors or resources. Removing it can trigger a trophic cascade and reduce diversity.

Predict the community change after removal by identifying the interaction the species controls.; separate state, pressure, control loop and time

Removing sea otters can allow sea urchins to increase and overgraze kelp forests, changing habitat for many species. This gives a concrete prediction from the stated ecosystem.

Keystone status is context-dependent; abundance alone does not identify a keystone species. Interpret the result within the stated model and evidence limits.

Keystone species

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through essay response, commonly using Outline / Suggest / Explain.

Command terms

Outline / Suggest / Explain / Define

What earns marks

Build the answer around this relationship: Keystone species have effects on community structure that are disproportionate to their abundance.

Watch for

Equating keystone species with the most abundant species or only with top predators.

Representative question

Question 1

[Maximum number: 6]

Explain how an ecological community structure could be affected by the removal of a keystone species.

Harvest below Replacement to Keep Resources Renewable

A renewable-resource harvest is sustainable only when long-term removal remains below replacement and leaves a viable reproducing population.

Resource Evidence used to assess sustainability
Scots pine (Pinus sylvestris) in managed Finnish forest Compare timber volume removed with regrowth/replanting; survey logged and unlogged forest structure and soil disturbance
Atlantic cod (Gadus morhua) Use stock size, age structure, reproductive rate, juvenile recruitment and a precautionary estimate of maximum sustainable yield

Replacement rates vary with age structure, habitat and climate, so monitoring must update quotas or harvest methods rather than treating one limit as permanent.

A renewable species is not automatically harvested sustainably; incomplete stock data and illegal or unreported removal increase uncertainty.

Sustainable resource harvesting

Assessment in practice

4 marks
How it is assessed

This objective is assessed through essay response, commonly using Discuss.

Command terms

Discuss

What earns marks

Build the answer around this relationship: Harvesting is sustainable only when removal stays at or below the population replacement rate.

Representative question

Question 1

[Maximum number: 4]

Discuss the impact of overfishing in Lake Kariba and how sustainable harvesting of resources can be assessed.

Judge Agriculture across Soil, Inputs, Pollution and Carbon

Sustainable agriculture maintains food production without reducing the soil, water, biodiversity and climate conditions needed by future production.

Factor Sustainability question
Soil erosion Is fertile topsoil being lost faster than it forms?
Nutrient leaching Are soluble nitrates/phosphates leaving soil and polluting water?
Fertilizers and other inputs Can nutrient supply and yield be maintained without growing external dependence?
Agrochemical pollution Are pesticides or fertilizers harming non-target organisms and ecosystems?
Carbon footprint What emissions arise from machinery, fertilizers, livestock, transport and land-use change?

Crop rotation, soil cover and nutrient matching may reduce erosion, fertilizer demand and leaching, but yield and labour trade-offs must still be measured.

No single practice proves a farm sustainable; assessment must include outputs, inputs, pollution and long-term soil condition.

Agriculture sustainability factors

Assessment in practice

2–6 marks
How it is assessed

This objective is assessed through essay response, commonly using Discuss / Distinguish.

Command terms

Discuss / Distinguish

What earns marks

Build the answer around this relationship: Harvesting crops removes nutrients, so agricultural systems need replacement or recycling to maintain production.

Watch for

Treating fertilizer use as only beneficial, without considering phosphate depletion, leaching, and eutrophication.

Representative question

Question 1

[Maximum number: 6]

Discuss the risks and benefits associated with the use of phosphate fertilizers in agriculture.

Fertilizer Leaching Can Raise BOD and Remove Oxygen

Eutrophication occurs when leached nitrogen and phosphate fertilizers enrich aquatic or marine water and stimulate excessive primary production.

Nitrate/phosphate leaching → algal or plant growth → shading and biomass death → decomposer respiration rises → biochemical oxygen demand (BOD) rises → dissolved oxygen falls → hypoxia and organism death.

After fertilizer runoff causes a bloom, bacteria decomposing dead algae consume oxygen; fish may die when oxygen demand exceeds reaeration and photosynthetic supply.

BOD measures oxygen demanded by biological decomposition; it is not the same as dissolved oxygen, and a high BOD predicts stronger oxygen depletion.

Eutrophication exam focus

Assessment in practice

1–6 marks
How it is assessed

This objective is assessed through essay response, commonly using Explain / Discuss.

Command terms

Explain / Discuss

What earns marks

Build the answer around this relationship: Nitrate and phosphate enrichment commonly starts eutrophication in aquatic ecosystems.

Watch for

Saying algae directly use up all oxygen, instead of linking oxygen loss mainly to aerobic decomposition of dead organic matter.

Representative question

Question 1

[Maximum number: 6]

Discuss the causes and consequences of eutrophication.

Persistent Toxins Biomagnify through Food Chains

Biomagnification is increasing tissue concentration of a persistent pollutant in consumers at successively higher trophic levels.

DDT and mercury are retained or eliminated slowly. Predators consume many contaminated prey, so their total intake produces a higher tissue concentration than in organisms below them.

Mercury can be low in water or plankton, higher in fish and highest in fish-eating birds or mammals; DDT similarly reached damaging concentrations in top predators.

Bioaccumulation is increase within one organism over time; biomagnification is increase between trophic levels. Not every pollutant does either.

Biomagnification exam focus

Assessment in practice

2–3 marks
How it is assessed

This objective is assessed through essay response, commonly using Explain / Define / State.

Command terms

Explain / Define / State / Discuss / Identify / Suggest / Deduce / Outline / Justify

What earns marks

Build the answer around this relationship: Biomagnification requires a pollutant that persists and accumulates in organism tissues.

Watch for

Describing biomagnification as any pollution effect, without explaining increasing concentration at successive trophic levels.

Representative question

Question 1

[Maximum number: 6]

Discuss the use of DDT (dichlorodiphenyltrichloroethane) in the control of the malarial parasite.

Microplastics and Macroplastics Harm Ocean Life

Ocean plastics persist because they are non-biodegradable; large macroplastics and small microplastics expose organisms in different ways.

Plastic scale Example effects on marine life
Macroplastic Entanglement, drowning, injury or blockage after ingestion
Microplastic Ingestion by small organisms, transfer through food webs and exposure to associated chemicals

Weathering fragments plastic but does not mineralize it, while rivers, wind, fishing and currents continually redistribute material.

Clear scientific communication and popular-media coverage changed public perception and helped drive measures to reduce plastic pollution.

Detection alone does not quantify biological effect; particle size, polymer, dose and exposure duration must be evaluated.

Plastic pollution of oceans

Assessment in practice

2 marks
How it is assessed

This objective is assessed through essay response, commonly using State / Outline / Suggest.

Command terms

State / Outline / Suggest / Explain / Describe

What earns marks

Build the answer around this relationship: Macroplastics can kill organisms through entanglement, choking, gut blockage, and starvation.

Watch for

Treating plastic pollution only as litter, without explaining ingestion, entanglement, or digestive blockage.

Representative question

Question 1

[Maximum number: 4]

Explain the consequences of plastic pollution in marine environments.

Rewilding Restores Processes and Habitat Connectivity

Rewilding restores self-sustaining ecosystem processes by reconnecting habitats, reintroducing apex predators or other keystone species, and minimizing human impact through ecological management.

Large connected areas allow movement and gene flow; keystone interactions can restore food-web regulation; reducing intensive intervention lets succession and natural disturbance rebuild habitat complexity.

At Hinewai Reserve in New Zealand, management supports natural regeneration of native forest, removes alien trees and vines, and otherwise uses minimal intervention so endemic flora and fauna can re-establish.

Rewilding is not simply abandoning land. Connectivity, invasive-species control, community effects and monitoring determine whether natural processes can recover safely.

Rewilding exam focus

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response, commonly using Outline.

Command terms

Outline

What earns marks

Build the answer around this relationship: Rewilding often uses keystone or native species to restart ecological interactions.

Representative question

Question 1

[Maximum number: 2]

Outline two methods of restoration of natural processes in ecosystems by rewilding, other than reintroducing a keystone species.
1.
2.

Core Stability and Change

Core D4.2 is secure when students can judge whether a system is being stabilized or pushed toward change. The route is: identify the stability support or disturbance, explain the mechanism, and state the ecosystem consequence using evidence.

  • energy, nutrient cycling, diversity, and tolerance ranges maintain persistence
  • Amazon deforestation and keystone removal can push systems toward instability
  • harvest and agriculture require recovery, soil, nutrients, biodiversity, and monitoring
  • eutrophication, biomagnification, and plastics harm ecosystems through specific mechanisms

Ecosystem Stability and Human Impact

Core transfer questions ask students to explain why an ecosystem remains stable or why a disturbance pushes it toward change. Strong answers do not list threats; they explain mechanisms such as lost rainfall recycling, trophic cascade, overharvest, nutrient enrichment, toxin biomagnification, plastic movement, or restoration through rewilding.

  • Use stability requirements: energy input, nutrient cycling, biodiversity, genetic diversity, and abiotic tolerance ranges.
  • Explain disturbance mechanisms such as Amazon tipping points, trophic cascades, overharvesting, agricultural damage, eutrophication, biomagnification, plastics, or rewilding.
  • Support claims with evidence from controlled models, monitoring, food webs, or pollution pathways.

Topic D4.3

D4.3 Climate change

Climate change affects ecosystems through greenhouse-gas forcing, feedback cycles, habitat shifts, coral stress, phenology changes, and evolutionary responses across many environments.

28% of analysed papers 39 papers · 42 questions

Objectives in this topic

Human CO₂ and Methane Emissions Drive Climate Change

Anthropogenic climate change is driven here by human-caused increases in atmospheric carbon dioxide and methane, which absorb outgoing infrared radiation.

Gas Major anthropogenic sources
Carbon dioxide (CO₂) Fossil-fuel combustion and deforestation/land-use change
Methane (CH₄) Livestock and rice agriculture, waste decomposition, and fossil-fuel extraction or leakage

Higher concentrations strengthen greenhouse forcing, altering Earth's radiative balance and raising long-term mean temperature.

Antarctic ice cores show a long-term positive correlation between CO₂ and temperature, but correlation alone does not establish causal direction. Infrared absorption physics, observations and climate models provide the additional causal evidence.

One weather event cannot demonstrate climate causation; attribution uses long-term patterns and multiple independent evidence sources.

Anthropogenic causes

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain / Outline / Discuss.

Command terms

Explain / Outline / Discuss / Describe / State

What earns marks

Build the answer around this relationship: The greenhouse effect is natural, but human activity enhances it by increasing greenhouse gas concentrations.

Watch for

Confusing the greenhouse effect with ozone-layer depletion or ultraviolet radiation reaching Earth.

Representative question

Question 1

[Maximum number: 7]

Explain the impact of anthropogenic activity on climate change.

Positive Feedback Cycles Amplify Initial Warming

A positive climate feedback produces a change that reinforces the initial warming, making the response larger than the original forcing alone.

Initial warming causes… Reinforcing return to warming
Deep-ocean CO₂ release More atmospheric CO₂ strengthens greenhouse forcing
Snow and ice loss Darker surfaces absorb more solar radiation
Faster peat/permafrost organic-matter decomposition More CO₂ is released
Permafrost melting Methane is released
More drought and forest fire Carbon stores burn and forest uptake falls

Warming melts reflective snow; exposed darker land absorbs more sunlight, causing additional warming and further melt.

Positive means self-reinforcing, not beneficial. Feedback strength and thresholds vary and do not imply one fixed rate of warming.

Positive feedback cycles

Assessment in practice

1 marks
How it is assessed

This objective is assessed through structured response, commonly using Identify / Explain.

Command terms

Identify / Explain

What earns marks

Build the answer around this relationship: Positive feedback reinforces the original warming instead of opposing it.

Representative question

Question 1

[Maximum number: 4]

Explain how positive feedback cycles could increase the rate of warming of the Earth.

Boreal Forests Can Shift from Carbon Sink to Source

A boreal forest tipping point can occur when carbon uptake by growth falls below carbon released by mortality, decomposition and fire.

Warmer temperatures + reduced winter snowfall → drought stress → lower taiga primary production and forest browning → more frequent/intense fires → combustion of living biomass and legacy soil carbon → net carbon loss.

Released carbon strengthens warming, while tree loss reduces future uptake; these feedbacks can make recovery to the former forest state difficult.

Repeated severe fires can burn older stored carbon as well as current vegetation, while drought prevents conifer regeneration from replacing the lost sink.

A tipping point is a risk of persistent state change, not a precisely dated outcome for every boreal region; local moisture, species and management matter.

Boreal forest tipping point

Assessment in practice

2 marks
How it is assessed

This objective is assessed through structured response, commonly using Explain.

Command terms

Explain

What earns marks

Build the answer around this relationship: Boreal warming can increase drought, fires, pests, disease, and tree mortality.

Representative question

Question 1

[Maximum number: 2]

An increase in global temperatures poses a critical threat to boreal forests. Explain the consequences of climate change to this northern ecosystem.

Polar Ice Loss Removes Breeding and Resting Habitat

Climate warming changes polar habitat by causing earlier Antarctic landfast-ice breakout and reducing Arctic sea ice.

Ice change Species-level consequence
Earlier breakup of Antarctic landfast ice Emperor penguins (Aptenodytes forsteri) may lose stable breeding grounds before chicks complete development
Loss of Arctic sea-ice floes Walruses lose resting platforms between feeding dives; calves are especially dependent on the habitat

Ice is physical habitat, not only frozen water: its seasonal timing and position control access to breeding, resting and feeding areas.

Landfast ice is attached to coast, seabed shoals or grounded icebergs; sea-ice extent and ecological effects vary by region and season.

Polar habitat changes

Assessment in practice

1–5 marks
How it is assessed

This objective is assessed through essay response, data analysis, commonly using Outline / State / Distinguish.

Command terms

Outline / State / Distinguish / Describe / Analyse / Discuss

What earns marks

Build the answer around this relationship: Loss of ice habitat can reduce survival and reproduction of ice-dependent species.

Watch for

Assuming all polar sea ice changes have the same direction in Arctic and Antarctic data.

Representative question

Question 1

[Maximum number: 3]

Discuss the use of Adélie penguins in studying the effects of global warming.

Surface Warming Can Suppress Nutrient Upwelling

Warmer surface water strengthens density stratification and can change the timing and extent of ocean upwelling.

Upwelling brings cold nutrient-rich deep water into the sunlit surface. Stronger stratification resists vertical mixing, so fewer nutrients reach phytoplankton.

Surface warming → stronger stratification → reduced/delayed upwelling → lower surface nutrients → lower phytoplankton primary production → less energy entering marine food chains.

If a seasonal upwelling pulse weakens, phytoplankton and then zooplankton production can fall, reducing food available to migrating consumers.

Currents also respond to wind and salinity; one local season is insufficient to attribute a long-term circulation shift.

Ocean current changes

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice.

What earns marks

Build the answer around this relationship: Warmer surface water can strengthen stratification and reduce vertical mixing.

Representative question

Question 1

[Maximum number: 1]

What is a consequence of ocean water having a very high temperature?

A

Decreased bleaching of coral reefs

B

Increased production of oxygen

C

Increase in energy flow through food chains

D

Reduced nutrient upwelling to the surface

Climate Suitability Shifts Ranges Upslope and Poleward

As temperature zones move, species may shift upslope or poleward if dispersal and suitable connected habitat allow them to track their climatic niche.

Evidence case Observed pattern in mapped local textbook
Montane birds, Papua New Guinea Upper range limits shifted upslope by 113 m on Mt Karimui and 152 m on Karkar Island compared with 1960s records
Eastern North American trees Many species showed range contraction or northward spread; Quebec sapling shifts were faster than adult-tree shifts

High-elevation species can run out of cooler habitat, while slow-growing trees may not disperse fast enough to match the speed of climate change.

A shifting observation is not automatically caused only by climate; land use, barriers, competition and survey effort must also be assessed.

Range shifts

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through multiple choice, structured response, commonly using Explain / Compare / Discuss.

Command terms

Explain / Compare / Discuss / Suggest

What earns marks

Build the answer around this relationship: Warming can move suitable climate zones and food supplies northward or upslope.

Representative question

Question 1

[Maximum number: 1]

The data shows how the hardiness zones in part of North America are predicted to change over the next 25 years. A hardiness zone is an area that has a certain average annual minimum temperature, a factor relevant to the survival of many plants. The lower the number, the more cold

resistant the plants must be.

What is a likely consequence of this change for tree species?

A

Tree species will spread northwards as climate changes.

B

Tree species that are not cold-resistant will decline.

C

There will be no change in the distribution of the tree species.

D

Tree species that currently live in the north will outcompete other tree species.

Coral Reefs Face Heat and Carbonate-Chemistry Stress

Coral reefs are threatened when warming causes bleaching and altered seawater chemistry reduces calcification, while pollution and overfishing weaken recovery.

Heat disrupts coral–algal symbiosis; acidification lowers carbonate ion availability; local stressors reduce resilience and recruitment.

Separate direct heat stress, chemistry effects and local pressures before evaluating a reef outcome.; separate driver, mechanism, response and timescale

A marine heatwave expels symbiotic algae, bleaching coral; repeated heat before recovery raises mortality risk. This gives a concrete prediction from the stated climate condition.

Bleaching is a stress response, not immediate death; outcome depends on duration, species and recovery conditions. Interpret the result within the stated evidence and scenario limits.

Coral reef threats

Assessment in practice

1–4 marks
How it is assessed

This objective is assessed through structured response, commonly using Describe / Suggest / Outline.

Command terms

Describe / Suggest / Outline / Discuss / Deduce / Explain

What earns marks

Build the answer around this relationship: Coral bleaching occurs when heat stress disrupts the coral-zooxanthellae symbiosis.

Watch for

Treating bleaching as colour loss only, without explaining zooxanthellae expulsion and reduced nutrient supply.

Representative question

Question 1

[Maximum number: 5]

Outline the reasons that climate change is a threat to coral reefs.

Three Ecosystem Approaches Sequester Carbon

Carbon sequestration transfers atmospheric carbon into a biological store and lowers atmospheric CO₂ only while that storage persists.

Approach Storage mechanism and limitation
Afforestation Establish trees where there was no previous forest; growing biomass and soil store carbon, but plantation species and fire risk matter
Forest regeneration Re-establish forest after harvest, fire, pests or disease; native recovery restores biomass carbon but takes time
Restore peat-forming wetlands Rewet anaerobic soils so decomposition slows and peat accumulates; drainage reverses storage and releases CO₂

There is active debate over non-native plantations versus rewilding with native species: rapid carbon uptake must be weighed against biodiversity, resilience and permanence.

Rewetting drained peat reduces aerobic decomposition and allows long-term soil carbon accumulation to resume.

Sequestration complements emissions reduction; temporary uptake cannot offset continued fossil-carbon release one-for-one.

Carbon sequestration approaches

Assessment in practice

1 marks
How it is assessed

This objective is assessed through multiple choice, structured response, commonly using Explain.

Command terms

Explain

What earns marks

Build the answer around this relationship: Photosynthesis removes carbon dioxide from the atmosphere and stores carbon in organic matter.

Representative question

Question 1

[Maximum number: 1]

Which action will decrease carbon sequestration?

A

Afforestation

B

Primary production

C

Deforestation

D

Rewetting peatlands

Retrieve the SL Climate Chain

Core D4.3 is secure when every climate impact is explained as a chain: human greenhouse-gas sources or feedbacks change climate conditions, which alter habitats, oceans, carbon stores, or species distributions. Carbon sequestration is the mitigation chain that stores atmospheric CO2.

  • human gases and positive feedbacks amplify warming
  • boreal forests, ice habitats, upwelling and reefs shift through specific mechanisms
  • species may move poleward, upslope, contract, or lose ice/reef habitat
  • afforestation, agroforestry, regeneration and peatland rewetting store CO2

Climate Effects on Ecosystems

Core climate-change transfer answers should not list endangered examples. They should identify the climate driver, explain the physical or chemical mechanism, then state the biological consequence. Use this for greenhouse gases, feedbacks, boreal forests, ice-dependent species, upwelling, range shifts, reefs, and carbon sequestration.

  • Link human activities to increased greenhouse gases and enhanced warming.
  • Explain ecosystem impacts using mechanisms such as positive feedback, carbon sink/source shifts, habitat ice loss, reduced upwelling, range shifts, bleaching or acidification.
  • Explain carbon sequestration by naming the storage pathway in biomass, forests, soils or peatlands.