IB Biology HL 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
HL

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.

30% of analysed papers 34 papers · 47 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.

DNA Polymerase Extends Only 5′ to 3′

HL only

DNA strands have chemically different 5′ and 3′ ends, and DNA polymerase synthesizes only in the 5′→3′ direction.

The 5′ end is associated with the phosphate on the sugar's 5′ carbon, while the 3′ end has a free hydroxyl on the 3′ carbon. Polymerase attaches the 5′ phosphate of an incoming DNA nucleotide to the free 3′ end of the growing strand.

New nucleotide's 5′ end + growing strand's 3′ end → phosphodiester bond. Therefore extension occurs only at 3′ and the new strand lengthens 5′→3′.

A polymerase moving along a template 3′→5′ builds the complementary strand 5′→3′, adding every next nucleotide at the new strand's 3′ end.

5′→3′ describes the direction of new-strand synthesis. Polymerase cannot add to the growing strand's 5′ end.

DNA polymerase directionality

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

Explain

What earns marks

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

Representative question

Question 1

[Maximum number: 1]

How does DNA replicate?

A

The deoxyribose of a free nucleotide is linked to the phosphate of the last nucleotide in the chain.

B

The phosphate of a free nucleotide is linked to the deoxyribose of the last nucleotide in the chain.

C

Nucleotides are linked in a 33^{\prime} to 55^{\prime} direction and the new strands are anti-parallel to the template strands.

D

Nucleotides are linked in a 55^{\prime} to 33^{\prime} direction and the new strands are parallel to the template strands.

Leading and Lagging Strands Solve Antiparallel Geometry

HL only

At each replication fork, the leading strand is synthesized continuously and the lagging strand discontinuously as Okazaki fragments.

DNA polymerase can extend only 5′→3′ while the templates are antiparallel. The leading template supports synthesis toward the fork; the lagging template requires repeated synthesis away from the fork as more DNA is exposed.

Leading strand: one RNA primer, then continuous synthesis. Lagging strand: repeated RNA primers, discontinuous Okazaki-fragment synthesis, followed by primer replacement and fragment joining.

As helicase advances, one polymerase follows the fork continuously; on the other template, each newly exposed section receives another primer and becomes a separate Okazaki fragment.

Both new strands are synthesized 5′→3′. Leading/lagging refer to synthesis pattern relative to fork movement, not gene content or biological importance.

Leading vs. lagging strand

HL only

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: Leading versus lagging strand depends on matching each strand, enzyme or laboratory step to its exact function.

Watch for

Saying both strands are copied continuously instead of identifying discontinuous lagging-strand synthesis.

Representative question

Question 1

[Maximum number: 1]

What is a difference between the leading and lagging strands in DNA replication?

A

Fewer Okazaki fragments are produced on the leading strand.

B

Exons are only produced on the lagging strand.

C

More RNA primers are assembled on the lagging strand.

D

DNA nucleotides are linked 55^{\prime} to 33^{\prime} on the leading strand and 33^{\prime} to 55^{\prime} on the lagging strand.

Replication Enzymes Divide the Work

HL only

In prokaryotic DNA replication, primase, DNA polymerase III, DNA polymerase I and DNA ligase perform distinct sequential jobs.

Enzyme Required function
DNA primase Synthesizes short RNA primers that provide a 3′ end
DNA polymerase III Extends from each primer by adding DNA nucleotides 5′→3′; performs most new-strand synthesis
DNA polymerase I Removes RNA primers and replaces them with DNA nucleotides
DNA ligase Seals remaining nicks in the sugar–phosphate backbone, joining adjacent DNA sections

On the lagging strand, primase repeatedly starts fragments, polymerase III extends them, polymerase I replaces each primer, and ligase seals the final backbone gaps.

This named-enzyme sequence is limited to the prokaryotic system. Polymerase I replaces primers; ligase does not synthesize the missing DNA nucleotides.

Functions in replication

HL only

Assessment in practice

1–3 marks
How it is assessed

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

Command terms

Identify / Explain / Outline / Describe

What earns marks

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

Watch for

Saying helicase forms new strands instead of unwinding DNA and breaking hydrogen bonds.

Representative question

Question 1

[Maximum number: 3]

Describe the function of three named enzymes involved in DNA replication.

Proofreading Removes Many Replication Errors

HL only

DNA polymerase III proofreads a newly added base at the growing strand's 3′ terminal and corrects a mismatch before replication continues.

A non-complementary base pair distorts the new DNA. Polymerase III removes the mismatched terminal nucleotide, exposes the 3′ end again and inserts a nucleotide complementary to the template.

Mismatch at 3′ terminal → polymerase III detects it → incorrect nucleotide removed → correct complementary nucleotide added → 5′→3′ extension resumes.

If an incorrect nucleotide is added opposite a template G, proofreading removes it from the 3′ end and replaces it with C before the strand is extended further.

Proofreading greatly improves accuracy but does not eliminate every mutation. This objective is specifically polymerase III correction of a mismatched 3′ terminal.

DNA proofreading

HL only

Assessment in practice

1–4 marks
How it is assessed

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

Command terms

Identify / Explain

What earns marks

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

Watch for

Saying helicase forms new strands instead of unwinding DNA and breaking hydrogen bonds.

Representative question

Question 1

[Maximum number: 4]

Explain how mutation is avoided during DNA replication.

HL Replication Details

HL only

DNA strands have 5' and 3' ends; DNA polymerase adds nucleotides to the 3' end, so new DNA forms 5' to 3'. Leading strand synthesis is continuous; lagging strand synthesis is discontinuous as Okazaki fragments using repeated RNA primers. In the prokaryotic model, primase starts, DNA polymerase III extends, DNA polymerase I replaces primers, and ligase joins fragments. DNA polymerase III removes mismatched nucleotides from the 3' end; proofreading improves copying accuracy and reduces mutations.

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.

50% of analysed papers 56 papers · 79 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.

Nucleic-Acid Synthesis Has a Defined Direction

HL only

Transcription and translation both proceed with respect to the mRNA 5′→3′ direction.

RNA polymerase reads the DNA template 3′→5′ and adds RNA nucleotides to the transcript's 3′ end, so RNA is synthesized 5′→3′.

The ribosome moves along mRNA from its 5′ end toward its 3′ end, reading successive codons in that direction while the polypeptide elongates.

Label both ends before interpreting a diagram: a template shown 3′→5′ supports a transcript drawn antiparallel 5′→3′, and the ribosome follows the transcript toward 3′.

‘Left to right’ is meaningless without strand labels. Transcription direction refers to new RNA synthesis; translation direction refers to ribosome movement along mRNA.

Directionality exam focus

HL only

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: RNA synthesis proceeds in the 5 prime to 3 prime direction.

Watch for

Choosing a start codon without using the 5 prime to 3 prime direction shown in the diagram.

Representative question

Question 1

[Maximum number: 1]

Using the diagram, identify, with a reason, whether X or Y is the start codon.

Promoters Position Transcription Initiation

HL only

A promoter is a DNA region where transcription machinery binds to position RNA polymerase and define where transcription begins.

Promoter sequence and associated factors determine strand choice, start site and transcription level. Opening and early RNA synthesis proceed from that positioned complex.

Name: promoter; binding factor; start site; direction of RNA synthesis.

A promoter mutation can reduce mRNA production without changing the protein-coding sequence. RNA polymerase begins at the promoter and then moves along the template, so promoter position fixes which downstream sequence is copied.

A promoter is regulatory DNA, not the entire gene or translated region.

Initiation of transcription at promoter

HL only

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: Promoters are non-coding DNA sequences linked to transcription start sites.

Representative question

Question 1

[Maximum number: 2]

Explain the function of a promoter in DNA.

Non-Coding DNA Can Regulate the Genome

HL only

Non-coding DNA does not code for polypeptide sequence, but it can have regulatory, processing, structural or functional-RNA roles.

Required eukaryotic example Why it is non-coding
Regulators of gene expression Control when/how strongly genes are transcribed
Introns Transcribed into pre-mRNA but removed before translation
Telomeres Repetitive chromosome-end DNA with a protective structural role
Genes for rRNA and tRNA Produce functional RNAs rather than polypeptides

A tRNA gene is transcribed to a functional tRNA molecule; its RNA product participates in translation but is not itself translated.

Non-coding does not mean non-functional or never transcribed. Limit examples here to the four syllabus categories.

Non-coding sequences in DNA

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

State

What earns marks

Build the answer around this relationship: Non-coding DNA can regulate transcription.

Watch for

Equating non-coding DNA with having no function.

Representative question

Question 1

[Maximum number: 2]

DNA has regions that do not code for proteins. State two functions of these regions.
1.
2.

Post-Transcriptional Processing Makes Mature mRNA

HL only

Eukaryotic pre-mRNA is processed by adding a 5' cap and poly-A tail and removing introns so mature mRNA can be exported and translated.

Processing protects RNA, supports export and helps ribosomes recognize the message. Exons are joined in the order defining the coding sequence.

Trace: pre-mRNA; cap and tail; intron removal; exon joining; mature mRNA export.

Removing an intron prevents its sequence from being read as part of the final translated message.

Processing changes RNA, not the original DNA sequence; different processing choices can alter the final protein.

Post-transcriptional modification

HL only

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: Eukaryotic pre-mRNA can contain both introns and exons.

Watch for

Saying introns are added during processing rather than removed.

Representative question

Question 1

[Maximum number: 1]

What happens to mRNA after transcription in eukaryotic cells?

A

Binding to the large subunit of the ribosome

B

Removal of introns

C

Addition of exons

D

Attachment of an amino acid

Alternative Splicing Produces Multiple Messages

HL only

Alternative splicing joins different combinations of exons from one pre-mRNA, allowing one gene to code for different polypeptides.

After introns are removed, selected exons can be retained or omitted in different mature mRNAs. Each exon combination creates a different codon sequence for translation.

One DNA gene → one pre-mRNA containing exons/introns → introns removed → different exon combinations joined → different mature mRNAs → polypeptide variants.

If mature mRNA A contains exons 1–2–3 and mature mRNA B contains exons 1–3, translation can produce two polypeptides from the same gene.

Alternative splicing changes RNA processing, not the DNA gene. Specific named protein examples or detailed splicing-factor mechanisms are not required.

Alternative splicing

HL only

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: Alternative splicing joins different exon combinations.

Watch for

Assuming one gene can produce only one protein product in all circumstances.

Representative question

Question 1

[Maximum number: 1]

The number of protein-coding genes in the human genome is estimated to be about 20000 , which is much less than the size of the proteome. What is one reason for this?

A

Exons are removed from RNA before translation.

B

There are more types of amino acids than nucleotides.

C

mRNA can be spliced after transcription.

D

Base substitutions occur during transcription.

Translation Initiation Sets the Reading Frame

HL only

Translation initiation assembles the ribosome at the mRNA start codon and establishes the reading frame; A, P and E sites then organize tRNA movement during elongation.

The small ribosomal subunit attaches near the mRNA 5′ terminal and moves to the start codon. An initiator tRNA pairs with that codon; the large subunit attaches, and the next matching tRNA can enter.

Ribosome site Role during elongation
A (aminoacyl) Accepts the incoming charged tRNA
P (peptidyl) Holds the tRNA carrying the growing polypeptide
E (exit) Holds/releases the uncharged tRNA before it leaves

After peptide-bond formation and translocation, tRNAs shift through the sites: the growing-chain tRNA moves to P, the empty tRNA moves to E, and A becomes available for the next tRNA.

The start codon fixes the triplet reading frame. A/P/E describe tRNA positions, not three separate ribosomes.

Translation initiation

HL only

Assessment in practice

1–3 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: The small ribosomal subunit binds mRNA early in initiation.

Watch for

Putting the initiator tRNA into the E site or A site instead of linking initiation to the start codon and P site.

Representative question

Question 1

[Maximum number: 4]

Outline the roles of the different binding sites for tRNA on ribosomes during translation.

Polypeptides Are Modified after Translation

HL only

Many translated polypeptides must be folded, cleaved or chemically modified before becoming functional proteins.

Insulin provides a required two-stage example: removal of the signal peptide converts pre-proinsulin to proinsulin; later removal of the connecting C-peptide converts proinsulin to mature insulin, whose A and B chains remain linked by disulfide bonds.

Pre-proinsulin → signal peptide removed → proinsulin folds/disulfide bonds form → C-peptide removed → functional insulin.

Cleavage changes one precursor polypeptide into the mature hormone structure able to bind its receptor appropriately.

Translation alone does not guarantee a functional protein. The insulin example is processing after translation, not alternative splicing of insulin exons.

Polypeptide modification

HL only

Assessment in practice

2 marks
How it is assessed

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

Command terms

Suggest

What earns marks

Build the answer around this relationship: A newly translated polypeptide may be an inactive precursor.

Representative question

Question 1

[Maximum number: 2]

Insulin is produced by cutting C -peptide from the precursor molecule proinsulin. Suggest why group 1 has a greater level of C-peptide than group 2.

Proteasomes Recycle Unneeded Proteins

HL only

Proteasomes break selected proteins into peptides, allowing amino acids to be recycled for new protein synthesis or metabolism.

Damaged or short-lived proteins are tagged and fed into a proteasome, where peptide bonds are hydrolysed. Released amino acids return to the cellular pool.

Trace: protein tag; proteasome entry; peptide breakdown; amino-acid reuse or catabolism.

A misfolded protein can be removed and its amino acids reused to build a different protein.

Proteasomal degradation is controlled recycling, not random digestion of every protein. Proteasomes do not recycle every protein immediately; tagging and recognition determine which substrates are selected.

HL Protein Synthesis Details

HL only

RNA polymerase reads template DNA 3' to 5' and synthesizes RNA 5' to 3'; ribosomes translate mRNA codons in the 5' to 3' direction. Promoters mark transcription start regions and orientation; transcription factors help RNA polymerase bind and initiate in eukaryotes. Non-coding DNA does not code for polypeptide amino acid sequences and includes introns, regulatory sequences, telomeres, rRNA genes, and tRNA genes. Eukaryotic pre-mRNA is modified before export and translation by adding a 5' cap and poly-A tail and removing introns by splicing. Alternative splicing joins different exon combinations from one pre-mRNA, so one gene can produce multiple protein variants in different cells or stages. Translation initiation assembles ribosomal subunits at the start codon AUG; initiator tRNA enters the P site and A, P, and E sites organize tRNA movement. Newly made polypeptides may be folded, cleaved, or chemically modified; preproinsulin processing to active insulin is a key example. Proteasomes degrade tagged, damaged, or unneeded proteins; amino acid recycling supports new protein synthesis and proteome quality control.

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.

12% of analysed papers 14 papers · 17 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.

Gene Knockout Tests What a Gene Does

HL only

Gene knockout investigates gene function by changing a chosen gene so that it becomes inoperative.

Researchers compare the knockout phenotype with an appropriate control. A consistent difference suggests that the disabled gene contributes to the affected process, while rescue or other controls strengthen the inference.

Libraries of knockout organisms are available for some research species, allowing scientists to study many genes systematically. Common model groups include mice, fruit flies, zebrafish and Arabidopsis plants.

If p53-knockout mice develop tumours and are then used to test anti-cancer treatments, the knockout model connects loss of the tumour-suppressor gene with the cancer phenotype.

Students do not need the technical steps used to create a knockout. A missing phenotype also does not prove that the gene has no function, because another gene may compensate.

Gene knockout

HL only

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: Gene knockout deliberately makes a specific gene inoperative.

Representative question

Question 1

[Maximum number: 1]

What is gene knockout used for?

A

Increasing protein production by editing a gene

B

Investigating the function of a gene by replacing it to make it inoperative

C

Identifying the presence of a gene by editing it to produce a different protein

D

Editing a gene to initiate cell death

CRISPR-Cas9 Targets a Chosen DNA Sequence

HL only

CRISPR-Cas9 gene editing uses a guide RNA to direct the Cas9 enzyme to a complementary DNA target, where Cas9 cuts the DNA so the sequence can be changed.

Changing the guide RNA changes the target sequence. After cutting, gene editing can delete DNA, insert or replace a sequence, or disrupt the gene so that it no longer functions.

Choose a target → design a complementary guide RNA → guide RNA directs Cas9 → Cas9 cuts near the target → the DNA is edited → verify the sequence and biological result.

Successful-use example from an official IB paper: CRISPR-Cas9 repair of a mutated DMD gene restored expression of dystrophin. Depending on the mutation, editing can replace a changed codon, add DNA missing after a deletion, or remove DNA added by an insertion.

Restoring dystrophin expression demonstrates a successful molecular result; it does not by itself prove complete, safe clinical treatment. Delivery, unintended edits and regulation must still be considered.

CRISPR-Cas9 gene editing

HL only

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: Guide RNA directs Cas9 to a complementary DNA target.

Representative question

Question 1

[Maximum number: 3]

Explain ways in which CRISPR-Cas9 gene editing could be used to change the mutated dystrophin protein produced.

Conserved Sequences Reveal Important Functions

HL only

Conserved gene sequences remain identical or similar across species; highly conserved sequences remain similar across long evolutionary periods.

Hypothesis Why similarity persists
Functional requirement Many sequence changes reduce the gene product's function, so purifying selection removes them
Slower mutation rate The sequence accumulates new mutations less often than less-conserved regions

If the same coding region is very similar in distantly related species, researchers can hypothesize that its amino-acid sequence is strongly constrained by the protein's function.

Conservation supports hypotheses; it does not by itself identify the exact function or distinguish functional constraint from a lower mutation rate. Further evidence is required.

Conserved sequences

HL only

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: Conserved sequences are similar across species or long evolutionary times.

Representative question

Question 1

[Maximum number: 1]

A bioinformatics analysis was performed on the protein PSY transcribed from the gene from corn and from daffodil to obtain the sequence alignment.

On the alignment, identify the longest part of the sequence where the consecutive amino acids are the same.

Corn---MAI I LVRAASP-------GLSAAD---------SISH-
Daffodil---MVVAILRVVSAIEIPIRLGFSEANWRFSSPKYDNLGRK
CornQGTLQCSTLLKTKRPAARRWMPCSLLGLHPWEAGRP-SPAV
DaffodilKSRLSVYSLYTTSKYA-----------CVGFEAENNGKFLI
* * * * * *
CornYSSLPVNPAGEAVVSSEQKVYDVVLKQAALLKRQLRTP--V
DaffodilRSSLVANPAGEATISSEQKVYDVVLKQAALVKDQTKSSRKS
* * * * * * * * * * * * * * * * * * * * * * * * * * *
CornLDARPQDMDMPRN--GLKEAYDRCGE I CEEYAKTFYLGTML
DaffodilTDVKP-DIVLPGTVYLLKDAYDRCGEVCAEYAKTFYLGTLL
* * * * * * * * * * * * * * * * * * * * * * * *
CornMTEERRRA I WA I YVWCRRTDELVDGPNANY I TPTALDRWEK
DaffodilMTPERRRAI WA I YVWCRRTDELVDGHNASHITPSALDRWEA
** * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
CornRLEDLFTGRPYDMLDAALSDTISRFPIDIQPFRDMIEGMRS
DaffodilRLEDLFAGRPYDMFDAALSDTVSRFPVDIQPFMDMVEGMRM
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
CornDLRKTRYNNFDELYMYCYYVAGTVGLMSVPVMGIATESKAT
DaffodilDLKKSRYKNFDELYLYCYYVAGTVGLMSVPVMGIAPESLAE
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
CornTESVYSAALALGIANQLTNI LRDVGEDARRGRIYLPQDELA
DaffodilAESVYNAALALGIANQLTNI LRDVGEDARRGRIYLPQDELA
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
CornQAGLSDEDIFKGVVTNRWRNFMKRQIKRARMFFEEAERGVN
DaffodilEAGLSDEDVFTGKVTDKWRSFMKRQIKRARTFFEQAEKGVT
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
CornELSQASRWPVWASLLLYRQI LDEIEANDYNNFTKRAYVGKG
DaffodilE L SQA SRWP VWASLL LYRQI LDE I EANDYNNF TKRAYVSKV
CornKKLLALPVAYGKSLLLPCSLRN---GQT
DaffodilKRLAALPLA YGKSLLIPLSLRPPSLSKA
* * * * * * * * * * * * * * * *

HL Gene Editing Evidence

HL only

Gene knockout makes a specific gene non-functional to investigate phenotype; model organisms such as mice, Drosophila, zebrafish, and Arabidopsis support KO libraries. Guide RNA directs Cas9 to a complementary DNA target sequence; Cas9 cutting enables deletion, replacement, insertion, or gene disruption. Conserved sequences remain similar across species or long evolutionary times; conservation suggests essential function, lower mutation rate, or strong purifying 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.

45% of analysed papers 51 papers · 83 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.

Cell Proliferation Expands Cell Number

HL only

Cell proliferation is repeated passage through growth and mitosis that increases cell number for growth, routine replacement or tissue repair.

Biological need Syllabus example
Growth Repeated division in plant meristems and early animal embryos
Routine replacement New skin cells replace cells continually lost from the surface
Tissue repair Cells near a skin wound proliferate to replace damaged tissue

Proliferation increases cell number only when new cells are produced faster than cells are lost. Signals regulate which cells enter and continue through the cell cycle.

Cell growth increases the size or contents of one cell; cell proliferation increases the number of cells. Uncontrolled proliferation can disrupt tissue organization.

Cell proliferation

HL only

Assessment in practice

1–2 marks
How it is assessed

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

Command terms

Distinguish / Deduce / Calculate / Describe / Compare / Evaluate / Outline / Identify / Suggest

What earns marks

Build the answer around this relationship: Cell proliferation increases cell number by repeated division.

Watch for

Using raw numbers without calculating the fraction of mitotic cells.

Representative question

Question 1

[Maximum number: 3]

Based on the data, evaluate the evidence for leptin promoting regeneration of liver tissue.

The Cell Cycle Coordinates Growth and Division

HL only

The cell cycle is the ordered sequence G1 → S → G2 → mitosis → cytokinesis that enables cell proliferation.

Stage Main event
G1 Cell grows and synthesizes components
S DNA is replicated
G2 Further growth and preparation for nuclear division
Mitosis The nucleus divides and chromosomes are segregated
Cytokinesis The cytoplasm splits to form daughter cells

G1, S and G2 together form interphase; mitosis follows interphase, and cytokinesis completes production of separate daughter cells.

Interphase is not a stage of mitosis. Checkpoint control belongs to the later cyclin Objective; this card establishes the cycle's sequence.

Cell cycle phases

HL only

Assessment in practice

1–2 marks
How it is assessed

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

Command terms

Identify / Describe / Outline / Compare / Deduce

What earns marks

Build the answer around this relationship: The cell cycle sequence is G1, S, G2, mitosis and cytokinesis.

Watch for

Putting S phase after G2 or after mitosis instead of between G1 and G2.

Representative question

Question 1

[Maximum number: 7]

Following germination of seeds, plants undergo a rapid increase in the number of cells. Describe stages in the cell cycle that result in this increase of cells.

Interphase Growth Prepares a Cell to Divide

HL only

Interphase is a metabolically active period in which a cell grows and biosynthesizes the components needed for division.

Proteins and other cell materials are synthesized, DNA is replicated in S phase, and cytoplasm increases so the future daughter cells inherit sufficient cellular contents.

Mitochondria—and chloroplasts in photosynthetic cells—increase in number by growth and division of pre-existing organelles. Their increase is therefore part of cell growth, not manufacture from nothing.

Before a plant cell divides, it can synthesize proteins, replicate nuclear DNA and increase its chloroplast and mitochondrial populations during interphase.

Interphase is not a resting period and does not include nuclear division; active biosynthesis and preparation make division possible.

Cell growth during interphase

HL only

Assessment in practice

1–3 marks
How it is assessed

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

Command terms

Deduce / State / Outline

What earns marks

Build the answer around this relationship: Interphase includes active growth and metabolism.

Watch for

Listing G1, S and G2 without linking them to actual processes.

Representative question

Question 1

[Maximum number: 4]

Outline the processes occurring during interphase in the cell cycle.

Cyclin Thresholds Control Cell-Cycle Checkpoints

HL only

Different cyclins rise and fall in concentration during the cell cycle, and a cell passes a checkpoint only when the relevant cyclin reaches its threshold level.

Cyclin synthesis raises concentration before a transition; cyclin breakdown lowers it afterwards. This repeating pattern coordinates each checkpoint with the correct stage of the cycle.

Cyclin below threshold → checkpoint is not passed. Cyclin reaches threshold → transition can be triggered. Cyclin is degraded → its signal falls before the next cycle stage.

A graph may show one cyclin accumulating before mitosis, crossing a threshold as mitosis begins, then falling rapidly after it is degraded.

The syllabus requires changing cyclin concentrations and threshold control, not names or detailed roles of specific cyclins.

Cell cycle control by cyclins

HL only

Assessment in practice

1–4 marks
How it is assessed

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

Command terms

Compare / Explain / State

What earns marks

Build the answer around this relationship: Cyclin concentrations fluctuate through the cell cycle.

Watch for

Saying cyclins are constant instead of rising and falling during stages.

Representative question

Question 1

[Maximum number: 4]

Explain how the cell cycle is controlled.

Cell-Cycle Gene Mutations Can Cause Uncontrolled Division

HL only

Mutations can cause uncontrolled cell division by converting proto-oncogenes into oncogenes or by disabling tumour-suppressor genes.

Gene class Normal role Cancer-promoting mutation
Proto-oncogene Promotes division only when appropriate Gain of function produces an oncogene that sends excessive growth signals
Tumour-suppressor gene Slows the cycle, enforces checkpoints or prevents damaged cells dividing Loss of function removes a brake on division

A clone carrying one control mutation can continue dividing and acquire further mutations, progressively weakening the systems that restrain proliferation.

An activated oncogene can keep a growth pathway switched on while loss of a tumour suppressor allows damaged cells to pass a checkpoint; together they strongly favour tumour growth.

One mutation does not guarantee cancer. Several independent controls normally limit division, so tumour development often involves accumulated mutations.

Mutations in cell cycle genes

HL only

Assessment in practice

1–2 marks
How it is assessed

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

Command terms

State / Outline / Explain / Predict

What earns marks

Build the answer around this relationship: Tumours result from uncontrolled cell division.

Watch for

Describing cancer as ordinary growth without linking it to uncontrolled cell division.

Representative question

Question 1

[Maximum number: 4]

Explain how changes to the cell cycle can result in tumour formation.

Distinguish Tumours and Measure Mitotic Index

HL only

Tumours differ in division and growth rate, invasion of neighbouring tissue and capacity for metastasis; these features determine whether they cause cancer.

Term Meaning
Benign tumour Remains localized and does not invade or metastasize; it is non-cancerous
Malignant tumour Invades neighbouring tissue and can metastasize; it is cancerous
Primary tumour Original site at which the tumour developed
Secondary tumour New tumour formed elsewhere after malignant cells spread

Metastasis occurs when cells leave a primary tumour, travel through the body and establish one or more secondary tumours.

\text{mitotic index}=\frac{\text{number of cells observed in mitosis}}{\text{total number of cells observed}}

If 24 of 200 observed tumour cells are in mitosis, the mitotic index is 24 ÷ 200 = 0.12, or 12%. Under comparable sampling conditions, a higher index indicates a larger proportion actively dividing.

Tumour size alone does not establish malignancy. Invasion and metastasis distinguish malignant cancer, while mitotic index estimates division activity only for the sampled population and time.

Tumour differences

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

Suggest

What earns marks

Build the answer around this relationship: Benign tumours grow locally and do not invade distant tissues.

Watch for

Assuming every tumour is malignant or metastatic.

Representative question

Question 1

[Maximum number: 1]

Which processes occur during the development of secondary tumours?

I. Cytokinesis
II. Metastasis
III. Mitosis

A

I and II only

B

II and III only

C

I and III only

D

I, II and III

HL Cell Cycle and Cancer

HL only

Cell proliferation increases cell number by repeated mitosis, as in plant meristems, early embryos, skin replacement, and wound healing. The cell cycle includes interphase, mitosis, and cytokinesis; interphase has G1 growth, S-phase DNA replication, and G2 preparation. Interphase is metabolically active, not resting; cells synthesize proteins, replicate DNA, grow cytoplasm, and increase organelles. Cyclin concentrations rise and fall to activate cyclin-dependent kinases; CDK-cyclin complexes such as MPF control checkpoints and mitosis entry. Proto-oncogene activation and tumour suppressor loss disrupt checkpoints; accumulated mutations can cause uncontrolled proliferation and cancer. Benign tumours grow locally, malignant tumours invade neighbouring tissues, and metastasis spreads cancer cells to form secondary tumours.

Topic D2.2

D2.2 Gene expression [HL only]

Gene expression connects DNA, transcription regulation, epigenetic control, transcript stability, proteomes and environmental signals to phenotype in living cells and tissues.

19% of analysed papers 21 papers · 25 questions

Objectives in this topic

Gene Expression Converts DNA Information into Output

HL only

Gene expression is the mechanism by which information in a gene affects phenotype, most commonly through production and function of a protein.

The DNA base sequence is transcribed into mRNA, the mRNA sequence is translated into a polypeptide, and the folded protein performs a function such as catalysing a reaction. That function contributes to the cell's traits.

Gene information → transcription → mRNA → translation → protein → cellular function → phenotype.

Expression of a gene for a digestive enzyme produces mRNA, then enzyme protein; the enzyme's catalytic activity contributes to the digestive phenotype of that cell.

Possessing a gene does not mean it is expressed in every cell. Regulation can change the amount of mRNA and protein without changing the DNA sequence.

Gene expression mechanism

HL only

Assessment in practice

1 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: DNA information affects phenotype through gene products.

Representative question

Question 1

[Maximum number: 1]

One important chemical in the mobilization of stem cells is a protein, CXCL12, which maintains the stem cells inside the bone marrow. The breakdown of CXCL12 causes the mobilization of stem cells to the blood vessels.

The graph below shows the mobilization of stem cells and the production of mRNA for CXCL12 when the bone marrow is treated with two different chemicals (isoprenaline and clenbuterol).

Explain how the amount of mRNA for CXCL12 gives an indication of the amount of protein CXCL12 produced.

Transcriptional Regulation Sets mRNA Production

HL only

Transcription factors regulate transcription by binding specific DNA sequences such as promoters and enhancers.

Regulatory element Role
Promoter Region near the gene where RNA polymerase and transcription factors assemble to begin transcription
Enhancer Regulatory sequence that increases transcription when an activator transcription factor binds
Repressor-bound sequence Binding can reduce polymerase recruitment or activity and lower transcription

Different cell types contain different combinations or activities of transcription factors, so the same genome can produce different amounts of a gene's mRNA.

An activator bound to an enhancer can contact the promoter complex and increase how often RNA polymerase initiates transcription.

A transcription factor recognizes a particular DNA sequence; it does not bind every promoter or enhancer indiscriminately.

Regulation of transcription

HL only

Assessment in practice

1–7 marks
How it is assessed

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

Command terms

Suggest / Explain

What earns marks

Promoters mark where RNA polymerase can bind to begin transcription, while transcription factors bind specific DNA sequences to activate or repress transcription.

Watch for

Treating promoters as translation start sites instead of RNA polymerase binding regions.

Representative question

Question 1

[Maximum number: 7]

Explain how gene expression can be regulated during transcription to determine an organism's phenotype.

mRNA Stability Controls How Long a Message Is Available

HL only

Regulating mRNA degradation controls how long an mRNA remains available for translation and therefore how much protein can be produced.

Human mRNAs can persist from minutes to days. Shortening of the protective 3′ poly-A tail helps initiate degradation, after which nucleases break down the transcript.

Longer mRNA lifetime → more opportunities for ribosomes to translate it. Faster degradation → fewer translation rounds and a shorter-lived protein-production response.

If two cells transcribe equal numbers of mRNA molecules but one transcript is degraded within minutes while the other persists for hours, the longer-lived transcript can usually support more translation.

mRNA abundance reflects both its rate of transcription and its rate of degradation; high transcription alone does not guarantee high protein output.

Epigenesis Builds Differentiated Cell Patterns

HL only

Epigenesis is the development of patterns of cell differentiation in a multicellular organism from an initially undifferentiated zygote.

Cells descended from the zygote usually contain the same DNA sequence, but different genes become active or repressed. Epigenetic changes such as DNA methylation or histone modification alter how genes are read without altering their base sequences.

One zygote → repeated cell division → different epigenetic states and gene-expression patterns → different proteins → specialized cell phenotypes.

A developing muscle cell activates genes for contractile proteins while maintaining repression of genes needed only in neurons, producing differentiation without changing genotype.

Epigenetic change can alter phenotype but not genotype because the nucleotide sequence remains unchanged. It is therefore not a gene mutation.

Genome, Transcriptome and Proteome Describe Different Layers

HL only

A cell's genome is its complete genetic information, its transcriptome is the set of mRNA transcripts present at a particular time, and its proteome is the set of proteins present at that time.

Layer What it contains How it varies
Genome Complete DNA information Usually shared by an individual's somatic cells and relatively stable
Transcriptome mRNAs produced in a cell or tissue at that time Changes with cell type, development and environment
Proteome Proteins present in a cell, tissue or organism at that time Dynamic because translation, processing and degradation change protein abundance

A liver cell and muscle cell usually share a genome but express different genes, so their transcriptomes and proteomes differ and support different functions.

No cell expresses all its genes. Detecting an mRNA also does not guarantee that its protein is abundant or functional.

Genome, transcriptome, proteome

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

State / Calculate / Explain

What earns marks

Build the answer around this relationship: The genome is the complete genetic information.

Watch for

Assuming the genome and proteome are identical in all tissues.

Representative question

Question 1

[Maximum number: 1]

Which statement correctly describes genome and proteome?

A

Only the genome but not the proteome can be analysed using gel electrophoresis.

B

The genome and the proteome are the same in all tissues in an organism.

C

In cells of different tissues, the genome is the same while the proteome varies.

D

Only mutations in the proteome but not in the genome cause any variability.

Epigenetic Tags Alter Access to DNA

HL only

Methyl groups added to promoter DNA or to histones act as epigenetic tags that alter transcription without changing DNA base sequence.

Tag location Typical expression effect
Cytosine in a DNA promoter Methylation represses transcription of the downstream gene
Amino acids in histone proteins Methylation can repress or activate transcription, depending on the site and context

The tags change which regulatory proteins can associate with chromatin and how accessible the promoter is to transcription machinery.

Extensive methylation of promoter cytosines can make a gene less accessible, reducing its mRNA and protein production.

Do not apply one universal rule to all methylation: promoter DNA methylation is generally repressive, while histone methylation may activate or repress.

Epigenetic tags

HL only

Assessment in practice

1–2 marks
How it is assessed

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

Command terms

Outline / Suggest / Explain

What earns marks

Build the answer around this relationship: Promoter DNA methylation usually represses transcription.

Watch for

Describing histone methylation when the question specifically asks about DNA methylation.

Representative question

Question 1

[Maximum number: 2]

Explain how methylation of nucleosomes affects DNA transcription.

Some Epigenetic States Can Persist through Cell Division

HL only

Epigenetic inheritance passes a change in gene expression to daughter cells or offspring without changing the nucleotide sequence of DNA.

If DNA methylation or histone modifications remain in place or are re-established during chromosome replication and division, the inherited chromatin state can keep a gene active or repressed.

During mitosis, persistent tags can transmit a cell-lineage expression pattern to daughter cells. If tags remain through gamete formation and meiosis, an expression state may influence offspring.

A repressive mark maintained through repeated mitoses can help daughter cells retain the same differentiated identity even though every cell division copies the same DNA sequence.

Epigenetic inheritance is inheritance of expression state, not a change in allele sequence, and most tags are not guaranteed to persist through gamete formation.

Air Pollution Can Alter Epigenetic Gene Regulation

HL only

Environmental exposure can alter gene expression by changing epigenetic tags; air pollution can modify DNA and histone methylation patterns.

Pollutants reaching lung tissue can trigger cellular responses associated with altered methylation. Changed access to regulated genes can affect inflammatory pathways and may contribute to lung disease without changing the DNA sequence.

Air-pollution exposure → altered DNA or histone methylation → altered transcription pattern → changed cell response → possible contribution to inflammation or respiratory disease.

Comparing exposed and less-exposed lung-cell samples may reveal different methylation and expression patterns associated with inflammatory responses.

An association between pollution, methylation and disease does not by itself prove that one methylation change caused the disease; exposure, dose and alternative causes must be evaluated.

Environmental effects on expression

HL only

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through essay response, multiple choice, commonly using Compare / Suggest / Discuss.

Command terms

Compare / Suggest / Discuss

What earns marks

Build the answer around this relationship: Environmental conditions can alter gene expression.

Watch for

Treating altered expression as a necessary DNA base-sequence mutation.

Representative question

Question 1

[Maximum number: 3]

Using the data in the bar chart, discuss the evidence for Arabidopsis plants adapting to different daylight regimes by changing the pattern of gene expression.

Gamete Resetting Leaves Some Parental Imprints

HL only

During human egg and sperm development, most epigenetic tags are removed, but retained imprints can make only the maternal or paternal copy of a gene active in offspring.

Resetting prevents most acquired expression states being passed between generations. For an imprinted gene, a retained parent-specific tag silences one allele, so phenotype depends on whether the active copy came from the mother or father.

Hybrid cross Epigenetic growth outcome
Male tiger × female lion → tigon Tiger paternal genes lack the lion's strong growth promotion, while the lioness contributes anti-growth imprinting; the hybrid is about parental size or smaller
Male lion × female tiger → liger Lion paternal growth promotion is not opposed by the tigress's imprints in the same way; the hybrid can grow larger than either parent

The hybrid pattern supports a parent-of-origin epigenetic explanation, but the exact imprinted genes making the largest growth difference are not established in the approved local textbook.

Epigenetic tag removal

HL only

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through multiple choice.

What earns marks

Some marks, such as imprints, may be retained in specific cases and can affect offspring phenotypes.

Representative question

Question 1

[Maximum number: 3]

Very soon after fertilization, parental epigenetic methylation is reversed in the DNA. Later, tissue-specific epigenetic modifications are made to the embryonic DNA. The graph follows the degree of methylation from different sources during embryonic development.

According to the graph, what are the changes in DNA methylation during embryonic development?

A

Only the paternal DNA becomes demethylated.

B

The maternal DNA becomes demethylated first.

C

The methylation patterns of the parents' DNA are erased before fertilization.

D

The methylation patterns of both parents are erased after fertilization.

Monozygotic Twins Separate Genetic and Environmental Effects

HL only

Studies of monozygotic twins compare genetically similar individuals to estimate how environment and epigenetic differences contribute to traits.

If twins differ despite near-identical DNA, differing environments, developmental history or epigenetic states are possible explanations. Concordance and study design determine the strength of inference.

Evaluate a twin result by checking: shared genes; shared environment; age and exposure; trait concordance; alternative causes.

Twins may both inherit risk alleles but develop different symptoms after different exposures, suggesting environment modifies expression or phenotype.

Twin differences do not prove a purely environmental cause; measurement error and non-shared biology also matter.

Monozygotic twin studies

HL only

Assessment in practice

1–3 marks
How it is assessed

This objective is assessed through data analysis, multiple choice, commonly using Identify / Compare / Analyse.

Command terms

Identify / Compare / Analyse

What earns marks

Build the answer around this relationship: Monozygotic twins share essentially the same genome.

Watch for

Using only numerical values without comparing identical and non-identical twins.

Representative question

Question 1

[Maximum number: 3]

Analyse the data to find whether it supports the hypothesis that genetic factors cause some people to have a much higher chance of cocaine dependence than others.

Hormones and Lactose Can Switch Gene Expression

HL only

External chemical signals can change gene expression in both eukaryotes and bacteria, but the regulatory mechanism depends on the cell type.

External factor Mechanism and expression outcome
Steroid hormone in a eukaryotic cell The hormone binds an intracellular receptor; the activated receptor acts with transcriptional regulators at DNA and changes transcription
Lactose in E. coli Without lactose, a repressor binds the lac operator and blocks transcription. Lactose binds and inactivates the repressor, allowing transcription of genes needed for lactose metabolism

Both mechanisms connect an external condition to selective protein production, so the cell makes a response only when the signal or substrate is present.

The lac operon is an inducible bacterial system and is not the mechanism used by steroid hormones in eukaryotic cells. One biochemical example is sufficient for the syllabus scope.

HL Gene Expression Control

HL only
  • Transcription factors, promoters, enhancers, activators and repressors control RNA polymerase activity.
  • mRNA lifetime limits translation; poly-A shortening and nucleases help remove transcripts.
  • DNA methylation and histone modification alter chromatin access without changing base sequence, creating epigenetic patterns during differentiation.
  • The genome is all genetic information; the transcriptome and proteome vary with cell type, time and environment.
  • Some epigenetic marks persist through cell division or inheritance, although most are reset during gamete formation; genomic imprinting is an exception.
  • Hormones regulate eukaryotic transcription through receptors and transcription factors; lac and trp operons illustrate bacterial control.
  • Twin studies and environmental exposures help separate genetic, epigenetic and environmental effects on phenotype.

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.

17% of analysed papers 19 papers · 24 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.

Water Potential Is Relative Potential Energy per Volume

HL only

Water potential (ψw) is the potential energy of water per unit volume and is usually measured in kilopascals (kPa).

Absolute potential energy cannot be measured, so water potentials are expressed relative to pure water at atmospheric pressure and 20°C, which is assigned ψw = 0 kPa.

Adding solute makes water potential negative relative to pure water. Applying positive pressure can raise water potential. The numerical value predicts which region's water can release more potential energy by moving.

A solution at −300 kPa has lower water potential than pure water at 0 kPa; water has a tendency to move from the pure water toward the solution if a pathway exists.

Water potential is not solute concentration alone and is not an absolute store of energy. State the reference conditions and units.

Water Moves from Higher to Lower Total Potential

HL only

The direction of net water movement is set by total water potential, not by one component in isolation.

A higher Ψ means water has a greater tendency to leave. Comparing total values incorporates both solute and pressure effects and explains why a concentrated but pressurized cell may not lose water.

Write both total Ψ values, compare them numerically, then state the direction and the condition for equilibrium.

Water moves from −0.2 MPa to −0.5 MPa, but not from a region at −0.8 MPa to −0.5 MPa; the latter direction is reversed.

‘Higher’ means numerically less negative; movement stops when the relevant potentials are equal or other forces intervene.

Solute and Pressure Potential Explain Cell Water Status

HL only

In cells with walls, total water potential is the sum of solute potential and pressure potential.

\psi_w=\psi_s+\psi_p

Component Sign and meaning
Solute potential, ψs 0 for pure water and increasingly negative as dissolved solute lowers water's potential
Pressure potential, ψp Usually positive inside turgid walled cells; can be negative in xylem sap under tension

If ψs = −600 kPa and ψp = +100 kPa, then ψw = −600 + 100 = −500 kPa. Water tends to enter from a neighbouring region at −400 kPa.

Keep signs and units through the calculation. Pressure potential is not always zero or positive: tension in functioning xylem gives a negative value.

Solute and pressure potential

HL only

Assessment in practice

1–3 marks
How it is assessed

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

Command terms

Explain

What earns marks

Build the answer around this relationship: Water potential equals solute potential plus pressure potential.

Representative question

Question 1

[Maximum number: 1]

The water potential of a plant cell is -0.24 kPa . If the pressure potential of the cell is 0.46 kPa , what is the solute potential of that cell?

Ψw=Ψs+Ψp\Psi_{\mathrm{w}}=\Psi_{\mathrm{s}}+\Psi_{\mathrm{p}}
A

0.22 kPa

B

-0.22 kPa

C

0.70 kPa

D

-0.70 kPa

Bathing Solutions Change Both Potentials in Plant Cells

HL only

When plant tissue is bathed in hypotonic or hypertonic solution, water movement changes both solute potential and pressure potential until equilibrium or plasmolysis is reached.

External solution Initial water movement Change inside the plant cell Result
Hypotonic (higher ψw outside) Water enters Cell sap is diluted, so ψs becomes less negative; expanding contents raise positive ψp Cell becomes turgid and rising ψp opposes further entry
Hypertonic (lower ψw outside) Water leaves Cell sap becomes more concentrated, so ψs becomes more negative; ψp falls toward zero Cell becomes flaccid and may plasmolyse

A cell initially at ψs = −600 kPa and ψp = +300 kPa has ψw = −300 kPa. In an external solution below −300 kPa, water leaves, turgor falls and the internal potentials change.

Do not describe only solute concentration: in a walled cell, increasing pressure potential can stop net entry even while the cell sap remains more concentrated than the outside solution.

HL Water Potential

HL only

Water potential is potential energy of water per unit volume, measured in kPa; pure water at standard conditions has water potential of 0 kPa. Water moves from higher water potential to lower water potential; solutes lower water potential by restricting water molecule movement. Water potential equals solute potential plus pressure potential; solute potential is zero or negative and pressure potential is often positive in walled cells. Water entering plant cells increases pressure potential and dilutes solutes; water leaving plant cells lowers pressure potential and makes solute potential more negative.

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.

52% of analysed papers 59 papers · 116 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.

Puberty Begins when Hormonal Signals Activate the Gonads

HL only

Puberty begins when increased hypothalamic GnRH release stimulates the pituitary to release more LH and FSH.

LH and FSH act on the gonads, increasing gamete production and secretion of steroid sex hormones such as testosterone and oestradiol. These hormones drive primary reproductive maturation and secondary sexual characteristics.

Hypothalamus: GnRH ↑ → pituitary: LH/FSH ↑ → ovaries/testes: gametogenesis and sex-steroid secretion ↑ → developmental changes of puberty.

In a typical male pathway, rising LH supports testicular testosterone secretion while FSH contributes to sperm production; testosterone promotes reproductive maturation and secondary traits.

Puberty timing and visible changes vary among individuals; the defining control is the endocrine signal chain, not one external trait.

Puberty control

HL only

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: The hypothalamus initiates puberty control through increased GnRH release.

Representative question

Question 1

[Maximum number: 1]

What controls the developmental changes during puberty?

I. Increased release of gonadotropin-releasing hormone (GnRH) by the hypothalamus
II. Luteinizing hormone (LH) leading to increased sex hormone production
III. Gonadotropin-releasing hormone (GnRH) triggering the onset of increased luteinizing hormone (LH)

A

I and II only

B

I and III only

C

II and III only

D

I, II and III

Gametogenesis Produces Haploid Sperm or Ova

HL only

Human gametogenesis combines mitosis, cell growth, two meiotic divisions and differentiation to produce haploid sperm or eggs.

Stage/outcome Spermatogenesis Oogenesis
Starting-cell supply Germ cells divide by mitosis Germ cells divide by mitosis before birth
Growth Primary spermatocyte grows Primary oocyte grows and stores extensive cytoplasm
Meiosis Two equal divisions Two highly unequal divisions
Products Four haploid cells differentiate into sperm One large ovum plus polar bodies
Differentiation Small motile sperm with specialized structures Large non-motile egg retains resources

One primary spermatocyte can yield four sperm, whereas unequal cytokinesis directs most cytoplasm from one primary oocyte into one ovum.

Both pathways halve chromosome number; the different product numbers arise mainly from equal versus unequal cytokinesis, not a different number of meiotic divisions.

Gametogenesis exam focus

HL only

Assessment in practice

1–4 marks
How it is assessed

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

Command terms

Identify / Compare / Describe / Outline

What earns marks

Build the answer around this relationship: Spermatogenesis occurs in seminiferous tubules of the testes.

Watch for

Confusing spermatogonia, spermatocytes, spermatids and spermatozoa in the sequence.

Representative question

Question 1

[Maximum number: 8]

Compare and contrast the processes of spermatogenesis and oogenesis.

Polyspermy Is Blocked after the First Sperm Fuses

HL only

Two linked reactions allow one sperm to enter an egg and then prevent additional sperm from causing polyspermy.

Reaction Trigger and effect
Acrosome reaction Enzymes released from the sperm acrosome digest a path through the zona pellucida, allowing penetration
Cortical reaction Fusion of the first sperm triggers cortical-granule release; the zona pellucida changes so other sperm cannot pass through

Blocking additional sperm preserves the normal paternal chromosome contribution and prevents an abnormal polyploid zygote.

The acrosome reaction enables penetration; the cortical reaction creates the later block. Reversing these roles gives the wrong mechanism.

Preventing polyspermy

HL only

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: The acrosome reaction helps one sperm penetrate the zona pellucida.

Watch for

Mixing up the acrosome reaction that allows sperm entry with the cortical reaction that blocks other sperm.

Representative question

Question 1

[Maximum number: 2]

Explain the mechanism that prevents polyspermy during fertilization.

The Blastocyst Must Implant in the Uterine Lining

HL only

After early cleavage, the embryo forms a blastocyst whose outer cells attach to and invade the uterine lining during implantation.

The inner cell mass forms the embryo; the outer trophoblast helps attachment and later placenta formation. Implantation connects embryonic development to maternal support.

Distinguish cleavage; blastocyst formation; implantation; and later organ development in sequence.

A blastocyst reaches the uterus and its trophoblast attaches to the endometrium, allowing implantation to begin.

Fertilization in the oviduct is not implantation; implantation requires later attachment to the uterine lining.

Blastocyst and implantation

HL only

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: A blastocyst is an embryo, not an unfertilized egg.

Representative question

Question 1

[Maximum number: 1]

What is a blastocyst?

A

An unfertilized egg surrounded by follicle cells

B

An unfertilized egg cell expelled by menstruation

C

The follicle when it has swelled up with fluid

D

The embryo when it has become a hollow ball of cells

Pregnancy Tests Detect a Hormone after Implantation

HL only

Pregnancy tests detect human chorionic gonadotropin (hCG) using monoclonal antibodies that bind specifically to the hormone.

The early embryo or developing placenta secretes hCG after implantation. hCG maintains the corpus luteum, allowing progesterone secretion to continue, and enters maternal blood and urine.

Urine moves along the test strip → labelled anti-hCG antibody binds hCG if present → the complex is captured by another anti-hCG antibody at the test line → a control line confirms flow and reagent function.

A test taken before urine hCG reaches the detection threshold can be negative even if implantation later produces a detectable concentration.

The test detects hCG, not the embryo directly. A control line does not indicate pregnancy; it indicates that the strip operated correctly.

Pregnancy testing

HL only

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: HCG is secreted by the embryo during early pregnancy.

Watch for

Placing HCG production in the pituitary, corpus luteum or ovary instead of the embryo.

Representative question

Question 1

[Maximum number: 1]

What is a function of human chorionic gonadotropin (HCG)?

A

To stimulate the corpus luteum to produce progesterone during early pregnancy

B

To stimulate contraction of uterine muscles at the onset of birth

C

To inhibit the production of progesterone and prevent menstruation during pregnancy

D

To prevent polyspermy during fertilization in the fallopian tubes

The Placenta Exchanges Materials between Mother and Fetus

HL only

The placenta supports foetal development by exchanging materials across placental villi while maternal and foetal blood normally remain separate.

Numerous villi provide a large surface area and a short exchange path. Oxygen, nutrients and maternal antibodies move toward foetal blood; carbon dioxide, urea and other wastes move toward maternal blood; placental hormones help maintain pregnancy.

The placenta allows the foetus to remain and develop inside the uterus to a later stage than in mammals without a placenta, while maternal physiology supplies continuous exchange.

Oxygen diffuses down its concentration gradient from maternal blood across a villus into foetal capillaries, while carbon dioxide moves in the opposite direction.

Large surface area does not mean the two blood supplies mix. The placental barrier is selective but not complete; some drugs and pathogens can cross.

Placenta role

HL only

Assessment in practice

1–8 marks
How it is assessed

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

Command terms

Explain / Identify

What earns marks

Build the answer around this relationship: Placental villi increase surface area for maternal-fetal exchange.

Representative question

Question 1

[Maximum number: 8]

Explain how the structure and functions of the placenta maintain pregnancy.

Progesterone Maintains Pregnancy; Oxytocin Drives Birth

HL only

Progesterone maintains pregnancy, whereas its decrease near childbirth permits an oxytocin-driven positive-feedback loop of uterine contractions.

After implantation, hCG maintains the corpus luteum so it continues secreting progesterone. Later the placenta becomes the main progesterone source, maintaining the endometrium and reducing uterine contractions.

Near childbirth, progesterone levels fall. Cervical stretch promotes oxytocin release; oxytocin strengthens uterine contractions, which increase cervical stretch and cause still more oxytocin release.

Corpus luteum progesterone → placental progesterone → progesterone falls → contractions/stretch → oxytocin ↑ → stronger contractions → more stretch.

Positive feedback applies to the escalating childbirth loop; pregnancy maintenance is continuity through progesterone, not an LH-surge mechanism from the ovarian cycle.

Hormonal control

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

Identify / Describe

What earns marks

Build the answer around this relationship: Oxytocin uses positive feedback to intensify uterine contractions during birth.

Watch for

Confusing positive oxytocin feedback in childbirth with negative feedback in the menstrual cycle.

Representative question

Question 1

[Maximum number: 7]

Describe the hormone feedback mechanisms that help to prepare a woman's body for pregnancy, sustain the pregnancy and then give birth.

HRT and CHD Show Why Correlation Is Not Causation

HL only

Evidence about hormone replacement therapy (HRT) and coronary heart disease (CHD) changed when randomized controlled trials tested a correlation reported by observational studies.

Evidence source Finding Best interpretation
Early epidemiological studies HRT users had lower CHD incidence Association only; users differed from non-users in other ways
Later randomized controlled trials HRT caused a small increase in CHD risk Random assignment better isolates the causal effect of HRT

HRT users in the observational studies tended to have higher socioeconomic status. Socioeconomic status itself is causally associated with lower CHD risk, so it confounded the apparent protective association.

If lower CHD is caused by healthcare access, diet or other factors linked to socioeconomic status, comparing self-selected HRT users with non-users can wrongly attribute that difference to HRT.

A strong correlation can still be non-causal. The syllabus conclusion is not that HRT prevents CHD: randomized trials found a small increase in risk.

Retrieve the HL Reproduction Route

HL only

HL D3.1 is about ordered mechanisms and evidence judgment: endocrine control, gametogenesis contrast, one-sperm fertilization, early embryo stages, hCG detection, placental exchange, birth feedback, and HRT evaluation.

  • GnRH drives pituitary FSH/LH and sex-hormone production
  • spermatogenesis gives four sperm; oogenesis gives one ovum and polar bodies
  • acrosome entry, cortical block, cleavage, blastocyst, endometrium
  • hCG, placenta, childbirth feedback, and HRT risk-benefit evaluation

HL Human Reproduction

HL only

HL D3.1 moves from reproductive events into control and evidence. The strongest answers keep sequences in order: endocrine axis at puberty, gametogenesis outcomes, fertilization blocks, blastocyst implantation, hCG testing, placental exchange, childbirth feedback, and HRT evaluation.

  • Explain HL mechanisms in order, especially hormone pathways and early-development stages.
  • Compare gametogenesis and fertilization mechanisms using precise structural terms.
  • Evaluate HRT by weighing symptom benefits against risks and evidence quality.

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.

57% of analysed papers 64 papers · 118 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.

Meiosis Produces Segregation and Independent Assortment

HL only

Segregation separates the two alleles of a gene into different gametes; independent assortment gives unlinked gene pairs independent gamete combinations.

Homologous chromosomes separate in meiosis I, so their alleles segregate. Each bivalent orients independently at metaphase I, so maternal and paternal homologues enter daughter cells in different combinations.

For AaBb with unlinked loci, segregation and random bivalent orientation predict AB, Ab, aB and ab gametes in equal proportions.

The probability of an A gamete is 1/2 and of a B gamete is 1/2, so the unlinked-model probability of AB is 1/2 × 1/2 = 1/4.

Independent assortment applies to unlinked genes; closely linked loci can produce unequal parental and recombinant gamete frequencies.

Segregation and independent assortment

HL only

Assessment in practice

1–6 marks
How it is assessed

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

Command terms

Outline / Determine / Identify

What earns marks

Build the answer around this relationship: Segregation gives each gamete one allele from each pair.

Representative question

Question 1

[Maximum number: 6]

Outline the relationship between Mendel's law of independent assortment and meiosis.

Derive Ratios for Unlinked Dihybrid Crosses

HL only

A dihybrid cross follows two loci at once; for unlinked autosomal genes, the four gamete types from AaBb are expected equally.

Cross under complete dominance Expected phenotypic ratio
AaBb × AaBb 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb
AaBb × aabb (test cross) 1 A_B_ : 1 A_bb : 1 aaB_ : 1 aabb

Multiply the independent 3:1 monohybrid phenotype probabilities to obtain 9:3:3:1; in a test cross, each offspring directly reveals one of the heterozygote's four gamete types.

For AaBb × AaBb, P(A_B_) = 3/4 × 3/4 = 9/16, while P(aabb) = 1/4 × 1/4 = 1/16.

These ratios assume unlinked loci, complete dominance, equal gamete viability and a sufficiently large sample. Genes far apart on one chromosome may also approach 50% recombination.

Dihybrid crosses

HL only

Assessment in practice

1–2 marks
How it is assessed

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

Command terms

Determine / Identify / State / Explain / Predict

What earns marks

Build the answer around this relationship: A dihybrid cross follows two genes simultaneously.

Representative question

Question 1

[Maximum number: 3]

The expected ratio of phenotypes in the offspring of a cross between a plant with narrow, yellow leaves and a plant heterozygous for the genes for leaf width and colour is 1: 1: 1: 1.

Justify this expected ratio using a Punnett grid or other diagram.

Investigate Human Gene Loci and Polypeptide Products

HL only

A locus is a gene's specific chromosomal position. A human-gene database can connect a gene symbol to its genomic location, summary and polypeptide product.

Database step Evidence to record
Search a gene and select Homo sapiens Correct gene record and symbol
Read genomic context Chromosome and locus coordinates
Read the summary/product fields Named polypeptide and its function
Compare a second record Pair on different chromosomes, or zoom in to find a nearby pair on the same chromosome

The mapped local textbook suggests TDF and CFTR as candidate searches. Record what the database reports for each gene rather than guessing its locus or product from the name.

Genes on different chromosomes can assort independently, whereas genes in close proximity on the same chromosome are likely to be linked and inherited together more often.

A database record is evidence only for its stated genome assembly and species. Do not treat two alleles at one locus as two nearby genes, and do not invent coordinates when the database has not been checked.

Linked Autosomal Genes Do Not Assort Independently

HL only

Autosomal linkage occurs when two genes lie on the same autosome and are therefore inherited together more often than expected by independent assortment.

The chromosome, not each allele, moves as a unit during meiosis unless crossing over occurs between the loci. Closer loci have a lower chance of being separated by a chiasma.

Show linked alleles beside two vertical lines representing homologous chromosomes—for example AB/ab for coupling or Ab/aB for repulsion—rather than writing only AaBb.

A test cross producing many AB and ab offspring but few Ab and aB offspring supports linkage in the AB/ab heterozygote.

Gene linkage means loci share a chromosome; sex linkage specifically means a locus lies on a sex chromosome. They are not synonyms.

Autosomal gene linkage

HL only

Assessment in practice

1–2 marks
How it is assessed

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

Command terms

Identify / Define / Compare / Explain / Distinguish / Outline

What earns marks

Build the answer around this relationship: Linked autosomal genes are on the same non-sex chromosome.

Representative question

Question 1

[Maximum number: 3]

Outline how it can be shown that the genes for shell base colour (Cc) and presence or absence of bands (Bb) are linked.

Use a Test Cross to Identify Recombinants

HL only

A recombinant carries a new allele combination relative to the parental chromosome combinations, produced by crossing over for linked genes or reassortment for unlinked genes.

Cross an individual heterozygous at both loci with an individual homozygous recessive at both. The recessive parent contributes only ab, so each offspring genotype and phenotype reveals the gamete made by the heterozygote.

Heterozygote gamete Test-cross offspring genotype Classification for AB/ab parent
AB AaBb Parental
ab aabb Parental
Ab Aabb Recombinant
aB aaBb Recombinant

If parental classes greatly exceed recombinant classes, the loci are linked; approximately 1:1:1:1 supports an unlinked or effectively 50%-recombining model.

Identify parental combinations from the cross, not from which phenotype looks 'normal'. Recombination frequency cannot exceed 50%.

Recombinants exam focus

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

Deduce / Identify

What earns marks

Build the answer around this relationship: Recombinants differ from parental allele combinations.

Representative question

Question 1

[Maximum number: 1]

An individual is heterozygous for two linked genes ABab\frac{\mathrm{AB}}{\overline{\mathrm{ab}}}.

To investigate the frequency of crossing over, a test cross is carried out between the individual and another that is homozygous recessive for both genes. What are the possible recombinants in the offspring of this cross?

A

Abab\frac{\mathrm{Ab}}{\mathrm{ab}} and Abab\frac{\mathrm{Ab}}{\mathrm{ab}}

B

ABab\frac{\mathrm{AB}}{\mathrm{ab}} and AbaB\frac{\mathrm{Ab}}{\mathrm{aB}}

C

Abab\frac{\mathrm{Ab}}{\mathrm{ab}} and aBab\frac{\mathrm{aB}}{\mathrm{ab}}

D

AAaa\frac{\mathrm{AA}}{\mathrm{aa}} and BBbb\frac{\mathrm{BB}}{\mathrm{bb}}

Use Chi-Squared to Test a Dihybrid Ratio

HL only

A chi-squared goodness-of-fit test asks whether differences between observed and expected dihybrid counts are larger than expected from chance sampling.

χ2=Σ((OE)2/E),whereOiseachobservedcountandEisitsexpectedcount.χ2hasnounit;degreesoffreedom=numberofcategories1.χ² = Σ((O − E)² / E), where O is each observed count and E is its expected count. χ² has no unit; degrees of freedom = number of categories − 1.

State H₀: observed counts fit the stated Mendelian ratio; H₁: they do not. Convert the ratio to expected counts, calculate and sum each contribution, then compare χ² with the critical value at p = 0.05.

In local Question Bank record 10900, a four-category 1:1:1:1 model gives E = 575 in each category and df = 3. The supplied χ² = 1002.6 exceeds the p = 0.05 critical value 7.815, so reject H₀: the observed ratio differs significantly from expectation.

Failing to reject H₀ does not prove the model; rejecting it does not by itself identify linkage or another cause. Check assumptions, expected counts and experimental design.

Chi-squared test

HL only

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through structured response.

What earns marks

Build the answer around this relationship: Chi-squared compares observed counts with expected counts.

Representative question

Question 1

[Maximum number: 2]

The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio differed significantly from the expected Mendelian ratio.

c2=Σ( Observed  Expected )2 Expected =1002.6c^{2}=\Sigma \frac{(\text { Observed }- \text { Expected })^{2}}{\text { Expected }}=1002.6
Probability
Degrees of freedom0.9950.9750.200.100.050.0250.020.010.0050.0020.001
10.000040.0011.6422.7063.8415.0245.4126.6357.8799.55010.828
20.0100.0513.2194.6055.9917.3787.8249.21010.59712.42913.816
30.0720.2164.6426.2517.8159.3489.83711.34512.83814.79616.266
40.2070.4845.9897.7799.48811.14311.66813.27714.86016.92418.467
50.4120.8317.2899.23611.07012.83313.38815.08616.75018.90720.515
60.6761.2378.55810.64512.59214.44915.03316.81218.54820.79122.458
70.9891.6909.80312.01714.06716.01316.62218.47520.27822.60124.322

Retrieve the HL Inheritance Route

HL only

HL D3.2 is secure when chromosome behaviour explains the ratios: segregation and independent assortment produce unlinked dihybrid expectations, gene loci explain linkage, recombinants reveal crossing over, and chi-squared decides whether observed counts fit the expected model.

  • homologous chromosomes separate and random bivalent orientation assort unlinked genes
  • unlinked autosomal genes can produce 9:3:3:1 or 1:1:1:1 ratios
  • linked genes give more parental types and fewer recombinants after crossing over
  • observed counts are compared with expected ratios using df and p = 0.05

Solve HL Linkage and Chi-Squared Questions

HL only

HL inheritance transfer is about deciding whether the expected ratio should be Mendelian or linked, then testing the evidence. Start from meiosis and gene location, predict gametes or ratios, identify parental and recombinant classes, and use chi-squared when observed counts need a statistical conclusion.

  • Explain segregation and independent assortment from meiosis before using dihybrid ratios.
  • Use gene loci, linkage, crossing over, and recombinant frequency to interpret offspring classes.
  • Apply chi-squared with observed/expected values, degrees of freedom, p = 0.05, and a null-hypothesis 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.

51% of analysed papers 58 papers · 83 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.

Kidneys Perform Excretion and Osmoregulation

HL only

Kidneys regulate blood composition by excreting metabolic wastes and by osmoregulating water and dissolved ions.

Process Meaning Kidney outcome
Excretion Removal of metabolic waste and unwanted substances Urea and other unwanted solutes leave in urine
Osmoregulation Regulation of osmotic concentration Nephrons adjust water and ion reabsorption, changing urine volume and concentration

Osmoticconcentrationisexpressedinosmolesperlitre(osmolL1).Osmotic concentration is expressed in osmoles per litre (osmol L⁻¹).

When body water is scarce, increased water reabsorption produces a smaller volume of more concentrated urine while urea is still excreted.

Excretion and egestion are different: kidneys remove substances from blood, whereas egestion removes undigested material from the gut.

Kidney role

HL only

Assessment in practice

1–7 marks
How it is assessed

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

Command terms

Explain / State / Deduce / Outline / Identify

What earns marks

Build the answer around this relationship: Kidneys remove wastes while conserving useful substances.

Representative question

Question 1

[Maximum number: 8]

Explain the role of the kidney in osmoregulation.

Ultrafiltration Is Followed by Selective Reabsorption

HL only

The glomerulus and Bowman's capsule form filtrate by ultrafiltration; the proximal convoluted tubule (PCT) then returns useful substances to blood.

Site Mechanism and outcome
Glomerulus → Bowman's capsule High hydrostatic pressure forces water and small solutes through fenestrations, basement membrane and podocyte slits; cells and most plasma proteins remain in blood
PCT Membrane transport selectively reabsorbs all normal glucose and amino acids, much Na⁺ and other ions; water follows by osmosis into surrounding capillaries

Ultrafiltration is mainly size/pressure based, so it cannot distinguish useful small solutes from wastes. Selective transport in the PCT performs that recovery.

Glucose enters Bowman's filtrate because it is small, then is normally reabsorbed in the PCT rather than excreted in urine.

Large proteins or blood cells in urine are not normal products of ultrafiltration and may indicate damage to the filtration barrier.

Glomerulus, Bowman's capsule, PCT

HL only

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: Ultrafiltration occurs from the glomerulus into Bowmans capsule.

Representative question

Question 1

[Maximum number: 8]

Explain the presence of glucose in the urine of a diabetic person and its absence in the urine of a person with type I diabetes that is being successfully treated.

The Loop of Henle Builds a Medullary Gradient

HL only

The loop of Henle uses differing permeabilities and ion transport to create a high solute concentration in the medulla, enabling later water conservation.

The descending limb loses water but has low ion permeability; the ascending limb pumps ions but is water-impermeable. Countercurrent multiplication amplifies the gradient.

For each limb ask whether water can cross and whether solute is transported before predicting fluid concentration.; identify the signal, controller and effector

As filtrate descends, water leaves into the concentrated medulla; as it ascends, ions leave while water stays, making filtrate dilute. This gives a concrete prediction from the stated condition.

The loop creates the gradient; it does not by itself determine the final urine volume. Interpret the result within the stated biological model and limits.

Loop of Henle

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

Identify / State

What earns marks

Build the answer around this relationship: The descending limb allows water to leave the filtrate.

Representative question

Question 1

[Maximum number: 1]

What is the function of the loop of Henle?

A

To reabsorb salt

B

To maintain a hypertonic solution in the medulla

C

To transport liquid from the collecting ducts to the convoluted tubules

D

To reabsorb glucose

ADH Moves Aquaporins to Control Collecting-Duct Water Loss

HL only

Hypothalamic osmoreceptors detect blood osmotic concentration and adjust pituitary ADH release, changing collecting-duct water permeability.

Blood condition ADH and aquaporin location Urine response
Too concentrated / low water More ADH; aquaporins move from intracellular vesicles into collecting-duct cell membranes More water follows the medullary gradient into blood; low-volume concentrated urine
Too dilute / excess water Less ADH; aquaporins are removed from membranes into vesicles Less water is reabsorbed; larger-volume dilute urine

Changing membrane aquaporin number switches permeability rapidly without rebuilding the collecting duct, closing a negative-feedback loop.

During dehydration, increased ADH inserts more aquaporins, so water leaves the collecting duct by osmosis and blood osmotic concentration moves back toward its set point.

ADH changes permeability but does not create the medullary gradient; the loop of Henle establishes the gradient that makes water reabsorption possible.

Osmoregulation by collecting ducts

HL only

Assessment in practice

1–5 marks
How it is assessed

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

Command terms

Describe / Explain / Identify / State / Outline

What earns marks

Build the answer around this relationship: High blood solute concentration stimulates ADH release.

Representative question

Question 1

[Maximum number: 7]

Explain the hormonal control of osmoregulation in the kidney by negative feedback.

Activity Redistributes Blood among Organs

HL only

Arteriolar vasodilation and vasoconstriction redistribute blood so organ supply matches changing metabolic activity while vital functions continue.

State Skeletal muscle Gut Brain Kidneys
Sleep Lower flow to most muscle groups than when awake Depends on digestive activity Total flow changes little, though regions such as hypothalamus/brainstem can rise during REM Maintained for excretion and osmoregulation; lying down can raise renal flow
Vigorous exercise Strongly increased Reduced as blood is redirected Kept relatively stable Reduced during prolonged vigorous exercise, but regulation limits disruption
Wakeful rest Lower than during exercise Increased after a meal for digestion and absorption Kept relatively stable Fairly constant overall; posture can alter flow

Locally active tissues dilate their arterioles, while sympathetic/epinephrine signals can constrict vessels to less immediately required organs and dilate those supplying skeletal muscle.

During vigorous exercise, skeletal-muscle flow rises while gut and renal flow fall; this supplies respiration where demand is greatest.

Redistribution is relative, not complete shut-off. Brain and kidney perfusion must remain sufficient for neural control, excretion and osmoregulation.

Blood supply changes

HL only

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: Exercise increases heat production in muscles.

Representative question

Question 1

[Maximum number: 1]

What is a reason for the changes in blood flow during exercise?

A

Increased blood flow to the kidneys removes waste products from exercise.

B

Blood flow to the brain decreases so that blood is diverted to the kidneys.

C

Increased blood flow to the skin removes heat.

D

Increased blood flow to the digestive system provides more glucose to muscles.

Retrieve the HL Homeostasis Route

HL only

HL D3.3 adds kidney and circulation mechanisms. The route is still feedback logic: nephrons filter and reabsorb, the loop of Henle builds a gradient, ADH changes collecting duct permeability, and blood vessels redistribute flow according to activity.

  • excretion removes urea; osmoregulation adjusts water and ions
  • glomerulus filters and proximal tubule selectively reabsorbs
  • medulla gradient and aquaporins conserve water
  • vasodilation and vasoconstriction redirect flow by activity

HL Kidney and Blood-Flow Control

HL only

HL homeostasis transfer is about structure-function precision. In kidney answers, say where filtration, reabsorption, salt pumping, water movement, ADH, and aquaporins happen. In blood-flow answers, say which vessels dilate or constrict and how activity changes tissue demand.

  • Distinguish excretion from osmoregulation and locate filtration/reabsorption in the nephron.
  • Explain the loop of Henle and ADH using permeability, salt movement, aquaporins, and concentrated urine.
  • Explain blood redistribution using vasoconstriction, vasodilation, exercise, epinephrine, and tissue demand.

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.

42% of analysed papers 48 papers · 65 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.

A Gene Pool Belongs to a Breeding Population

HL only

A gene pool consists of all genes and all their different alleles in an interbreeding population at a particular time.

The next generation receives a sample of this pool through gametes. Comparing allele proportions through time reveals whether the population's genetic composition is changing.

At one diploid locus, ten individuals contribute twenty allele copies to that locus's part of the population gene pool.

A gene pool is a population property, not one organism's genome and not every species living in the same ecosystem.

Gene pool concept

HL only

Assessment in practice

1–3 marks
How it is assessed

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

Command terms

State / Discuss

What earns marks

Build the answer around this relationship: A gene pool includes all alleles in a population.

Representative question

Question 1

[Maximum number: 3]

Discuss how the isolation of populations due to the fragmentation of forests could lead to changes in gene pools.

Compare Geographic Allele Frequencies with Databases

HL only

Allele frequency is the proportion of all gene copies represented by one allele; geographically isolated populations can have different frequencies.

Database step Record before comparing populations
Choose one human variant Gene, rsID and reference/alternate alleles
Select population or subpopulation data Geographic identity and sample size
Read allele counts/frequencies Same allele and same data field in every population
Compare Difference, direction and uncertainty; do not infer a cause from frequency alone

The mapped local textbook uses human LPL variant rs268: dbSNP reports global A = 0.983727 and G = 0.016273 and provides subpopulation frequency data. Use those subpopulation fields to compare geographic groups rather than treating the global value as every population's value.

Isolation reduces gene flow, so drift and different selection pressures can make allele frequencies diverge over generations.

Compare the same variant and compatible datasets. A geographic frequency difference is evidence of population structure, not by itself proof of natural selection.

Allele frequencies

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

State / Deduce

What earns marks

Build the answer around this relationship: Allele frequency is a proportion within the population gene pool.

Representative question

Question 1

[Maximum number: 1]

Define allele frequency.

Natural Selection Changes Allele Frequency

HL only

Neo-Darwinism links Darwin's selection mechanism with genetics: natural selection among individuals changes allele frequencies in the population gene pool.

Individuals differ in heritable traits → the current pressure causes different survival or reproductive success → successful individuals pass associated alleles more often → those allele frequencies rise in the next generation.

Selection acts on phenotypes of individuals, but evolutionary change is recorded as a change in inherited allele frequency across generations.

If an inherited phenotype increases seed production under drought, alleles contributing to that phenotype can become more frequent after repeated drought generations.

A phenotype becoming common is not enough: the trait must be heritable and its carriers must contribute disproportionately to the next generation.

Changes in allele frequency

HL only

Assessment in practice

1–3 marks
How it is assessed

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

Command terms

Compare / Outline

What earns marks

Build the answer around this relationship: Selection changes allele frequencies through differential reproductive success.

Representative question

Question 1

[Maximum number: 4]

If a wild population of cats contained both curled ears and normal ears, explain how the proportion of these two phenotypes could change in the population.

Three Selection Modes Reshape Trait Distributions

HL only

Directional, stabilizing and disruptive selection differ in which phenotypes have highest fitness; all can change allele frequencies.

Mode Favoured phenotype(s) Distribution outcome
Directional One extreme Mean shifts toward that extreme
Stabilizing Intermediate Variation narrows around the existing mean
Disruptive Both extremes Intermediate values decline and the distribution may become bimodal

Antibiotic exposure can cause directional selection toward resistance; selection against very low and very high human birth weights illustrates stabilizing selection.

These labels describe fitness patterns, not the number of genes involved. Disruptive selection does not automatically produce new species.

Types of selection

HL only

Assessment in practice

1 marks
How it is assessed

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

Command terms

Explain / Identify

What earns marks

Build the answer around this relationship: Directional selection shifts the population toward one extreme.

Representative question

Question 1

[Maximum number: 2]

Phenotypic variation allows natural selection within populations. Compare and contrast directional and disruptive selection.

Hardy–Weinberg Relates Allele and Genotype Frequencies

HL only

For two alleles with frequencies p and q, the Hardy–Weinberg model predicts p², 2pq and q² genotype frequencies under specified assumptions.

The equation p+q=1 describes allele frequencies; p²+2pq+q²=1 follows random union of gametes. It provides a null expectation for detecting evolutionary forces.

State p and q; calculate expected genotypes; compare observed values with the model.; separate variation, selection, inheritance and time

If q=0.2, expected aa frequency is q²=0.04 and heterozygote frequency is 2pq=0.32. This gives a concrete prediction from the stated population.

The model is an expectation, not a claim that real populations never evolve. Interpret the result within the stated selection model and evidence limits.

Hardy-Weinberg equation

HL only

Assessment in practice

2–3 marks
How it is assessed

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

Command terms

Outline / Discuss

What earns marks

Build the answer around this relationship: p and q are allele frequencies that add to one.

Representative question

Question 1

[Maximum number: 3]

Discuss the use of the Hardy-Weinberg equation in population genetics studies.

Hardy–Weinberg Needs Restrictive Conditions

HL only

Hardy–Weinberg equilibrium assumes a large population, random mating, no migration, no mutation and no selection.

These conditions prevent systematic changes in allele frequencies or sampling noise. Violating one condition can make observed genotype frequencies depart from expectation.

Check each assumption before using an equilibrium calculation as a biological conclusion.; separate variation, selection, inheritance and time

A small isolated population can deviate because drift acts strongly even if no selection is present. This gives a concrete prediction from the stated population.

A departure does not reveal which assumption failed; independent evidence is needed. Interpret the result within the stated selection model and evidence limits.

Hardy-Weinberg conditions

HL only

Assessment in practice

1–2 marks
How it is assessed

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

Command terms

State / Identify

What earns marks

Build the answer around this relationship: Hardy-Weinberg equilibrium is a null model for no evolution.

Representative question

Question 1

[Maximum number: 2]

State two assumptions made when using the Hardy-Weinberg equation.
1.

Artificial Selection Uses Deliberate Breeding Choices

HL only

Artificial selection occurs when humans deliberately choose crop plants or domesticated animals with desirable inherited traits to breed.

Choose a heritable target phenotype → select parents showing it → breed them → measure offspring → repeat selection over generations, increasing associated alleles.

Case Selection type and reason
Repeatedly breeding wheat plants with high grain yield Artificial: humans deliberately choose breeders
Repeatedly breeding dairy cattle with high milk yield Artificial: humans deliberately choose breeders
Antibiotic-resistant bacteria becoming common during antibiotic use Natural: the antibiotic is a pressure, but humans do not choose individual bacteria to reproduce

Strong repeated choices can change a population quickly but may narrow genetic diversity or bring correlated undesirable traits.

Artificial selection changes who reproduces; it is not direct DNA editing. An unintended evolutionary consequence of a human action is not automatically artificial selection.

Retrieve the HL Population Genetics Route

HL only

HL D4.1 turns selection into measurable population genetics. A gene pool changes when allele frequencies shift. Hardy-Weinberg gives a no-evolution baseline; selection graphs, isolated populations, artificial selection, and resistance show how forces move populations away from that baseline.

  • all alleles in an interbreeding population
  • directional, stabilizing, or disruptive selection favours different phenotype ranges
  • p and q calculate allele and genotype frequencies in equilibrium
  • selection, mutation, migration, drift, or non-random mating may be acting

HL Population Genetics

HL only

HL population-genetics questions ask students to quantify or model evolution. The answer starts with the gene pool and allele frequencies, then uses the model or selection graph to decide whether the population is at equilibrium or being shifted by selection, mutation, migration, drift, artificial selection, or isolation.

  • Use gene pool and allele frequency language to define evolution quantitatively.
  • Interpret selection graphs and isolated populations as changes in phenotype or allele frequencies.
  • Apply Hardy-Weinberg equations and equilibrium assumptions, then explain what deviations mean.

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 27 papers · 45 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.

Abiotic and Biotic Changes Drive Succession

HL only

Ecological succession is an orderly change in community composition through time, triggered by changes in abiotic conditions, biotic interactions, or both.

Pioneer organisms modify light, soil, nutrients and moisture; competition, facilitation, herbivory and dispersal then change which species can establish next.

In an abandoned quarry, lichens and mosses alter bare substrate and add organic matter, allowing grasses, shrubs and later woodland species to colonize.

Succession has a directional mechanism but not one universal fixed sequence; climate, starting conditions and disturbance can redirect it.

Ecological succession

HL only

Assessment in practice

1–3 marks
How it is assessed

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

Command terms

State / Distinguish / Suggest

What earns marks

Build the answer around this relationship: Succession describes changes in organism communities through time.

Representative question

Question 1

[Maximum number: 3]

Distinguish between primary succession and secondary succession, giving an example of each.

Primary Succession Builds Soil and Ecosystem Complexity

HL only

Primary succession begins on a surface without developed soil, so pioneer organisms must help create the conditions needed by later communities.

From pioneer to later stages General change
Plant size and biomass Increase
Amount of primary production Increase as producer cover develops
Species diversity Increase as more niches form
Food-web complexity Increase as trophic interactions accumulate
Nutrient cycling Becomes larger and more internally recycled as soil and biomass develop

On fresh volcanic rock, lichens and mosses weather substrate and add organic matter; deeper soil later supports herbs, shrubs and trees.

If soil and a biological legacy remain after disturbance, recovery is secondary rather than primary succession.

Changes during primary succession

HL only

Assessment in practice

1–3 marks
How it is assessed

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

Command terms

Describe / Explain / Suggest / State / Outline / Predict

What earns marks

Build the answer around this relationship: Pioneer organisms colonize bare surfaces and begin changing the abiotic environment.

Watch for

Forgetting that primary succession begins on bare substrate with little or no soil.

Representative question

Question 1

[Maximum number: 4]

Outline the ecological changes that will occur on the island of cooled lava.

Cyclical Succession Repeats after Regular Disturbance

HL only

Cyclical succession occurs when recurring disturbance repeatedly shifts a community through a predictable sequence without a single permanent endpoint.

Fire, flooding, grazing or seasonal conditions reset some species while survivors and propagules restart the pathway. Frequency and intensity set the cycle.

Compare disturbance interval with species life histories before predicting the recurring community.; separate state, pressure, control loop and time

A grassland burned every few years returns to fire-tolerant herbs rather than developing into forest. This gives a concrete prediction from the stated ecosystem.

A cycle is not random change; it requires a recurring driver and repeatable response. Interpret the result within the stated model and evidence limits.

Climax Communities Can Be Arrested by Human Pressure

HL only

Under specified environmental conditions, succession tends toward a relatively stable climax community; human activity can arrest it at an earlier or alternative state.

Human influence How succession is arrested
Livestock grazing and associated burning Removes seedlings and prevents woodland regeneration, maintaining grassland or heath
Wetland drainage Changes waterlogged abiotic conditions and permits agriculture, peat extraction, housing or other non-wetland communities

Removing the pressure may allow succession to resume if propagules, soil and hydrology remain recoverable; persistent change can instead hold an alternative stable state.

A climax is conditional on climate, soil and disturbance regime, not a universal permanent endpoint.

Retrieve the HL Succession Route

HL only

HL D4.2 is the long-term change story. Succession begins after new habitat or disturbance, pioneer species modify conditions, primary succession builds soil and complexity, and some systems repeat cycles or are held before climax by disturbance.

  • new habitat or disturbance opens space for community change
  • bare surfaces gain soil, biomass, diversity, food webs, and nutrient cycling
  • recurring disturbance or interactions repeat community changes
  • local stable climax differs from disturbance-maintained earlier stages

HL Succession and Long-Term Change

HL only

HL succession questions focus on community pathways over time. The answer starts from the initial condition, shows how species modify abiotic conditions and interactions, and identifies whether the pathway is primary, cyclical, climax, or arrested by repeated disturbance.

  • Explain succession as community change over time after new habitat or disturbance.
  • Use primary succession features: bare surface, soil formation, increasing biomass, diversity, food-web complexity, and nutrient cycling.
  • Distinguish cyclical succession, climax communities, and arrested succession using the cause of repeated or stopped change.

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.

19% of analysed papers 22 papers · 33 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.

Phenology Tracks the Timing of Seasonal Events

HL only

Phenology is the timing of recurring biological events such as flowering, migration, breeding or leaf-out.

Temperature, day length and resource cues control schedules. Climate warming can advance or delay events, changing overlap among interacting species.

Record the event, cue, date shift and interacting species before inferring ecological impact.; separate driver, mechanism, response and timescale

Warmer springs advance caterpillar emergence; a bird that migrates on a fixed cue may miss the food peak. This gives a concrete prediction from the stated climate condition.

A date shift alone is not a population decline; consequences depend on synchrony and alternative food. Interpret the result within the stated evidence and scenario limits.

Phenology exam focus

HL only

Assessment in practice

1–2 marks
How it is assessed

This objective is assessed through multiple choice, data analysis, commonly using Identify / Analyse.

Command terms

Identify / Analyse

What earns marks

Build the answer around this relationship: Phenology is the study of seasonal timing in biological events.

Representative question

Question 1

[Maximum number: 1]

Which of the following could be a subject of phenological research?

A

Timing of budburst in a tree species each year

B

Changes in allele frequencies in a population over time

C

The relationship between changes in ocean acidification and carbon dioxide concentration

D

The emergence of a new species of migratory bird by divergent evolution

Different Seasonal Cues Can Create Phenological Mismatch

HL only

Phenological synchrony is disrupted when interacting populations shift seasonal events by different amounts because their cues respond differently to climate change.

Interaction Possible mismatch mechanism
Arctic mouse-ear chickweed (Cerastium arcticum) and migrating reindeer (Rangifer tarandus) Temperature advances spring plant growth, while migration timing also depends on snow and other cues; reindeer may arrive after peak forage quality
Great tit (Parus major) breeding and caterpillar peak biomass Warming shifts caterpillar biomass timing and egg-laying responses over different temperature windows, reducing food overlap when chicks need it most

Reduced overlap can lower energy transfer, growth, fledgling mass or reproductive success even if both populations remain present.

Mismatch is about the timing overlap required for an interaction. Species may partly adjust behaviour or evolve, so the outcome is not automatically permanent.

Warming Can Add Spruce Bark Beetle Generations

HL only

Within their tolerance range, warmer seasons accelerate insect development and can increase the number of complete life cycles per year.

Spruce bark beetles such as Ips typographus or Dendroctonus micans can complete additional generations or attack periods when summers are longer and winters milder.

Warmer development period → faster egg–larva–pupa–adult cycle → more reproducing adults in the same year → more attacks on spruce phloem → greater probability that tree defences are overwhelmed.

A second annual generation can attack trees already weakened by the first, allowing fewer beetles to cause lethal phloem damage.

Warming is beneficial only within physiological limits; extreme heat, drought effects on hosts and predators can alter the outcome.

Reduced Snow Cover Selects Tawny-Owl Colour Variants

HL only

Climate change can cause evolution when it changes the relative fitness of heritable variants, shifting their frequencies across generations.

In southern Finland, milder winters and reduced snow changed camouflage conditions for tawny owl (Strix aluco) plumage variants. Dark brown owls gained relative success compared with pale grey owls.

The mapped local textbook reports dark brown owls increasing from about 30% to 50% of the population, while brown owls have higher mortality in severe snowy winters.

Snow cover declines → colour-dependent detection/feeding and survival change → heritable colour variants leave different numbers of offspring → variant frequency changes.

A one-season colour difference or phenotypic plasticity is not evolution; the response must be heritable and demonstrated as a population-frequency change over generations.

Evolution from climate change

HL only

Assessment in practice

2–3 marks
How it is assessed

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

Command terms

Predict / Suggest

What earns marks

Build the answer around this relationship: Climate change can create new selection pressures on survival and reproduction.

Representative question

Question 1

[Maximum number: 3]

Suggest how climate change can influence the natural selection of organisms that live in the Arctic oceans.

HL Timing and Selection

HL only

HL D4.3 adds timing and evolution. Phenology tracks when seasonal events happen; warming can desynchronize interacting species, shorten insect development, add generations, and change selection pressures so phenotype frequencies shift.

  • timing of flowering, migration, breeding, nesting or insect emergence
  • interacting species shift timing by different amounts
  • warmer temperatures can add generations and attack periods
  • milder winters alter selection and phenotype frequencies

HL Phenology and Climate Selection

HL only

HL climate questions require reasoning from timing to ecological or evolutionary consequence. The answer starts with a seasonal event or life-cycle stage, explains how warming shifts timing or selection pressure, and states the effect on interaction success, number of generations, attack periods, survival, or phenotype frequency.

  • Define phenology and identify the seasonal event being shifted.
  • Explain timing mismatch or faster development using named examples such as plant-reindeer, great tit-caterpillar, or spruce bark beetles.
  • Link climate-altered selection pressure to phenotype frequency change using the tawny owl example.