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.

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

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.

DNA replication produces a second DNA molecule with the same base sequence as the original, apart from rare errors.

chromosome DNA → replication during interphase → two DNA copies → cell division → one complete genome enters each daughter cell

A cell copies its DNA before dividing so each daughter cell receives a DNA copy.

Biological context	Why replication is required
reproduction	genetic information passes to new cells or offspring
growth	cell number increases without losing the genome
tissue replacement	new cells inherit the instructions needed for their function

Semi-conservative replication gives each daughter DNA molecule one parental strand and one newly synthesized strand.

Exposed template base	New nucleotide added
A	T
T	A
C	G
G	C

The sequence of each original strand determines a complementary new sequence. The two products therefore preserve the original base-pair sequence.

Semi-conservative means that each daughter double helix contains one parental strand and one newly synthesized strand—not that half of every strand is old.

E. coli were first grown with ¹⁵N so their DNA was heavy, then transferred to ¹⁴N medium. DNA from successive generations was separated by density-gradient centrifugation.

Sample	Observed DNA bands	What the result shows
before transfer	one heavy band	all DNA initially contains ¹⁵N
after one replication	one intermediate band	every molecule contains heavy and light material
after two replications	one intermediate + one light band	hybrid molecules persist while fully light molecules appear

The one-generation intermediate band rejects conservative replication, which predicts separate heavy and light bands. The two-generation pattern matches semi-conservative copying.

The isotopes label nitrogen in DNA; they do not make replication radioactive. Both ¹⁵N and ¹⁴N are stable isotopes.

Helicase unwinds the double helix and breaks the hydrogen bonds between complementary bases, exposing two template strands.

Free DNA nucleotides align with exposed bases by complementary base pairing.

DNA polymerase joins adjacent nucleotides by forming phosphodiester bonds in the sugar–phosphate backbone of each new strand.

Enzyme	Bonds affected	Job
helicase	hydrogen bonds between bases	separates the two strands
DNA polymerase	phosphodiester bonds within a new strand	links nucleotides into a backbone

Helicase does not elongate DNA. It opens the template; DNA polymerase performs strand synthesis.

need two genomes → helicase separates the original strands → each old strand acts as a template → complementary nucleotides pair → DNA polymerase links the new backbone → two matching double helices form

Feature to check	Correct outcome
sequence	complementary copying preserves the base-pair sequence
strand origin	each product has one old strand and one new strand
biological result	a complete genome can pass to each daughter cell

A model explains how copying could occur; the Meselson–Stahl band pattern provides experimental evidence that the one-old-one-new strand distribution actually occurs.

The polymerase chain reaction (PCR) amplifies a selected DNA region in vitro, producing enough copies for detection or analysis.

Component	Job
template DNA	contains the target sequence
two DNA primers	bind on opposite strands at the target's ends and provide free 3′ ends
free DNA nucleotides	become the new strands
Taq polymerase	extends primers and remains active after high-temperature heating

Primer sequences determine which interval is copied. The target's flanking sequences must therefore be known well enough to design complementary primers.

PCR combines template DNA, two primers, free nucleotides and Taq polymerase, then uses repeated temperature cycles to make many copies before gel separation.

Denaturation, about 95 °C: hydrogen bonds break and double-stranded DNA separates into templates.

Annealing, a lower temperature: primers bind to complementary sequences flanking the target.

Extension, about 72 °C: Taq polymerase adds nucleotides to each primer's 3′ end, synthesizing new DNA 5′→3′.

Heating begins the next cycle by separating the newly formed double strands. A thermal cycler repeats the temperature sequence automatically.

Cooling does not make DNA polymerase copy the target by itself. Cooling permits primer binding; polymerase extends only after a primer supplies a paired 3′ end.

$N_n=N_0(2^n)$

In the ideal early cycles, each target molecule produces two copies per cycle. Starting from one target, 30 perfect doublings would give about one billion copies.

Taq polymerase comes from the thermophilic bacterium Thermus aquaticus. Its heat stability prevents the enzyme from being destroyed during repeated denaturation.

Limitation	Consequence
primers bind a similar non-target sequence	the wrong interval can be amplified
contamination supplies foreign DNA	contaminant DNA can also be copied
reagents become limiting and products reanneal	later cycles fall below ideal doubling
Taq lacks proofreading	some newly copied molecules contain errors

DNA samples are loaded into wells near the negative electrode of a porous agarose gel immersed in conducting buffer.

Phosphate groups give DNA a net negative charge, so fragments migrate through the gel toward the positive electrode when voltage is applied.

Fragment	Movement through gel pores	Final tendency after the same time
shorter	less hindered; moves faster	farther from the wells
longer	more hindered; moves slower	nearer the wells

DNA is colourless. A stain or labelled probe is needed to reveal separated fragments as bands.

For DNA fragments, charge gives a common migration direction; fragment length is the main variable responsible for separation through the gel.

Observation	Interpretation
band farther from wells	shorter DNA fragments
bands at the same height in two lanes	fragments of the same measured length
more intense band	more stained DNA at that position, not a longer fragment

Compare lanes horizontally, one band position at a time. A profile comparison uses a pattern across many variable markers, not one matching band.

A same-height band supports a shared fragment length; it does not by itself prove that every base in the fragments is identical.

collect and preserve samples → extract DNA → amplify selected STR or VNTR markers by PCR → separate products by electrophoresis → compare band or peak patterns

Result	Supported conclusion	Not established by the profile alone
questioned sample differs at a reliable marker	that reference individual is excluded	who left the sample
all tested markers match	the samples may share a source	when or how the DNA arrived

Match strength increases when more independent, variable markers agree and when the matching pattern is rare in the relevant population.

Contamination, mixed samples, degraded DNA and laboratory error can weaken the inference. Controls and careful sample handling are part of the evidence.

At each tested marker, a child inherits one allele from the biological mother and one from the biological father.

Start with the child's pattern → account for bands or alleles shared with the known parent → compare every remaining child marker with the possible other parent.

Comparison	Inference
a required child marker is absent from the possible parent	exclude that person at that marker, after checking data quality
all non-maternal child markers are present in the possible father	parentage is supported, with strength depending on marker number and frequencies

Close relatives share more DNA than unrelated people. Parentage conclusions use multiple markers and probability, not a visual impression from one or two bands.

Stage	Question answered	Key control
primer design	which region will be copied?	complementary, target-specific flanking sequences
PCR	is there enough target DNA to analyse?	positive and negative controls; contamination prevention
electrophoresis	what fragment lengths are present?	ladder and correct electrode orientation
profile comparison	are patterns consistent with a shared source or inheritance?	enough informative markers and relevant frequencies

PCR changes quantity; electrophoresis changes position; interpretation changes neither. Each step supports a different claim.

A laboratory match is evidence of consistency. The final conclusion must remain limited by sample quality, contamination risk, marker informativeness and the biological question being tested.

The two DNA strands are antiparallel: one runs 5′→3′ while its complement runs 3′→5′.

DNA polymerase forms a phosphodiester bond between the growing strand's free 3′-OH and the incoming nucleotide. New DNA therefore grows only 5′→3′.

Polymerase action	Direction
reads template	3′→5′
synthesizes new strand	5′→3′

DNA polymerase reads a template from 3 prime to 5 prime while adding nucleotides to the new strand's 3 prime end, so the new strand grows 5 prime to 3 prime.

Both templates are exposed at the same fork, but DNA polymerase can extend each new strand only 5′→3′.

On one template, 5′→3′ synthesis follows the advancing fork, so polymerase can remain behind helicase and synthesize continuously.

On the opposite template, 5′→3′ synthesis points away from the fork. Newly exposed template must be copied in short sections that begin repeatedly near the fork.

The lagging strand is discontinuous because of antiparallel geometry and polymerase directionality—not because its polymerase is slower.

A replication fork shows continuous leading-strand synthesis and primer-dependent Okazaki fragments on the lagging strand.

Feature	Leading strand	Lagging strand
relation to fork movement	synthesis proceeds toward the fork	each fragment is synthesized away from the fork
synthesis pattern	continuous	discontinuous
RNA primers at one fork	one initial primer	repeated primers
DNA product before processing	one continuous new strand	multiple Okazaki fragments
ligase requirement	not required to join Okazaki fragments	joins processed fragments

Both new strands are synthesized simultaneously and both grow 5′→3′. The difference is continuous versus fragment-by-fragment synthesis.

The leading strand also needs an RNA primer to begin. Repeated priming is distinctive of the lagging strand.

DNA polymerase III cannot start a strand from unlinked nucleotides; it needs an existing paired chain with a free 3′-OH.

Primase is an RNA polymerase that synthesizes a short RNA primer complementary to the DNA template. It can start without a pre-existing 3′ end.

DNA polymerase III binds at the primer's 3′ end and adds complementary DNA nucleotides, extending the new strand 5′→3′.

New strand	Primase action	DNA polymerase III action
leading	makes one initial primer	extends continuously
lagging	makes a new primer for each fragment	extends each Okazaki fragment

DNA polymerase I removes each RNA primer and replaces its RNA nucleotides with DNA nucleotides.

After replacement, adjacent DNA sections contain the correct nucleotides but remain separated by a nick: one missing phosphodiester bond in the sugar–phosphate backbone.

DNA ligase catalyses formation of the missing phosphodiester bond, joining adjacent DNA fragments into one continuous strand.

Enzyme	Changes nucleotide identity?	Joins a final backbone gap?
DNA polymerase I	yes—RNA is replaced with DNA	leaves a nick between sections
DNA ligase	no	yes—seals the nick

Helicase: separates the parental strands at the replication fork.

Primase: makes an RNA primer that supplies a free 3′-OH.

DNA polymerase III: extends complementary DNA from the primer.

DNA polymerase I: removes the RNA primer and replaces it with DNA.

DNA ligase: seals the remaining nick between DNA sections.

Material present	Next enzyme needed
paired template but no starting chain	primase
RNA primer with free 3′ end	DNA polymerase III
RNA segment embedded before DNA	DNA polymerase I
adjacent DNA sections with a nick	DNA ligase

A newly added nucleotide that does not pair correctly distorts the polymerase–DNA structure and stalls extension.

DNA polymerase III detects a mismatched nucleotide at the new strand's 3 prime end, removes it and adds the correct complementary nucleotide.

DNA polymerase III shifts the 3′ end to its proofreading site, where exonuclease activity removes the mismatched nucleotide.

The corrected 3′ end returns to the polymerizing site; the complementary nucleotide is added and 5′→3′ synthesis resumes.

Proofreading greatly reduces replication errors and therefore mutation rate, but it cannot make replication completely error-free.

Clue at the fork	Consequence
polymerase can add only to a 3′ end	every new segment grows 5′→3′
synthesis can follow helicase	leading strand is continuous
synthesis must restart as template is exposed	lagging strand uses repeated primers and Okazaki fragments
an RNA primer remains	DNA polymerase I replaces it with DNA
two DNA sections meet at a nick	ligase seals the phosphodiester bond
the terminal base is mismatched	DNA polymerase III removes and replaces it

helicase opens → primase starts → DNA polymerase III extends → DNA polymerase I replaces RNA → ligase seals → polymerase III proofreading corrects many terminal mismatches

Directionality creates the leading–lagging difference. Enzyme handoffs solve that geometry, while complementary pairing plus proofreading preserves sequence fidelity.

[ 8 ]

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

[ 3 ]

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

[ 1 ]

What is a function of the enzyme helicase?

[ 4 ]

Describe the polymerase chain reaction (PCR).

[ 4 ]

Outline the process of DNA profiling.

[ 1 ]

How does DNA replicate?

[ 1 ]

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

[ 3 ]

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

[ 4 ]

Explain how mutation is avoided during DNA replication.

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.

Chromosomal DNA is the long-term information store. Protein synthesis instead uses a short-lived messenger RNA (mRNA) copy of a selected gene.

selected gene in DNA → complementary mRNA copy → mRNA reaches a ribosome → nucleotide sequence directs amino-acid order

A working copy solves two problems:

The original DNA sequence remains protected and reusable.
Many mRNA copies can be made, so one gene can support repeated protein production.

mRNA carries information; it is not converted into protein matter. The ribosome reads its sequence while amino acids are joined into a separate molecule.

RNA polymerase binds near the gene and opens a short region of the DNA double helix, exposing the template strand.

RNA polymerase opens a short DNA region, reads the template strand and synthesizes an mRNA strand that leaves the transcription bubble.

Free RNA nucleotides align by complementary base pairing: DNA A pairs with RNA U, DNA T with RNA A, and C with G.

RNA polymerase joins the RNA nucleotides into a sugar–phosphate backbone. The mRNA separates, and the DNA strands pair again.

Strand	Relationship to the new mRNA
DNA template	antiparallel and complementary
DNA coding strand	same base order as mRNA, except DNA has T where RNA has U

DNA template: 3′–TAC CTT GCG–5′

mRNA: 5′–AUG GAA CGC–3′

Codon tables are read from mRNA 5′→3′. Do not look up a DNA template triplet directly.

DNA template bases pair with RNA nucleotides according to A–U and C–G complementarity.

Hydrogen bonds between complementary bases can break and re-form. This lets a short DNA region open temporarily without breaking the covalent sugar–phosphate backbones.

After the mRNA leaves, the original complementary DNA strands pair again. The gene's base sequence is conserved and can be transcribed repeatedly.

Transcription copies one gene region into RNA; it does not duplicate the whole DNA molecule and does not consume the template strand.

DNA opens locally for mRNA synthesis, the mRNA leaves and the original DNA double helix reforms.

Gene expression is the use of gene information to produce a functional RNA or polypeptide. Transcription is a major control point because an untranslated gene cannot supply mRNA to ribosomes.

Transcription state	Immediate consequence	Possible protein outcome
gene active	mRNA is produced	translation can occur
gene inactive	little or no mRNA is produced	little or no corresponding polypeptide is made

Most cells in one organism contain the same genome, but different cell types transcribe different gene sets. Their different protein mixtures produce different structures and functions.

Turning transcription on permits expression; it does not guarantee a fixed protein amount. RNA processing, translation and protein breakdown can also be regulated.

gene selected → DNA opens locally → RNA polymerase reads one template strand → complementary RNA nucleotides are joined → mRNA leaves → DNA re-forms unchanged

Question	Correct check
Which DNA strand determines mRNA?	the template strand
Which base replaces thymine in RNA?	uracil
Which direction is mRNA written?	5′→3′
Why can the gene be reused?	DNA sequence and covalent backbones remain intact

The mRNA now carries codons in an order that a ribosome can translate into an amino-acid sequence.

Translation is the synthesis of a polypeptide whose amino-acid sequence is determined by the codon sequence of an mRNA.

mRNA codons read 5′→3′ → matching tRNAs deliver amino acids → ribosome forms peptide bonds → polypeptide grows from its amino end toward its carboxyl end

The immediate product is a polypeptide: a specific linear amino-acid sequence. Folding and later processing are needed before many polypeptides become functional proteins.

The genetic code determines amino-acid order, not the final three-dimensional shape directly. Shape emerges from interactions within the amino-acid sequence and its environment.

Component	Information or material carried	Translation job
mRNA	ordered codons	supplies the sequence to be read
tRNA	one anticodon and its attached amino acid	matches a codon and delivers the corresponding amino acid
small ribosomal subunit	mRNA-binding site	positions the message for decoding
large ribosomal subunit	A, P and E sites	positions tRNAs and catalyses peptide-bond formation

Translation is accurate only when all three information links agree: the mRNA codon, the tRNA anticodon and the amino acid attached to that tRNA.

The ribosome does not choose amino acids by recognizing their chemical identity. It accepts charged tRNAs whose anticodons pair with the exposed codons.

At the ribosome, a tRNA anticodon pairs antiparallel with a complementary mRNA codon. This selects which tRNA can occupy the decoding site.

Before translation, a specific aminoacyl-tRNA synthetase attaches the correct amino acid to its matching tRNA. This charging step links anticodon identity to amino-acid identity.

Recognition	Example	What it secures
codon–anticodon pairing	mRNA 5′–AUG–3′ pairs with tRNA 3′–UAC–5′	the correct tRNA is selected
enzyme–tRNA/amino-acid recognition	methionine is attached to the initiator tRNA	the selected tRNA carries the correct amino acid

Complementary base pairing alone cannot check which amino acid is attached. Translation fidelity depends on both correct pairing and correct tRNA charging.

The genetic code has several linked properties:

Triplet: three mRNA bases form one codon.
Degenerate: most amino acids are specified by more than one codon.
Unambiguous: each codon specifies only one amino acid or stop signal.
Nearly universal: the same codons usually have the same meanings across organisms, with limited exceptions.
Punctuated: a start codon establishes the reading frame; stop codons end translation.

Degeneracy means some base substitutions are silent, but it does not mean one codon can represent several different amino acids.

Codon type	Function
AUG	usually codes for methionine and can establish the start of translation
UAA, UAG, UGA	stop signals recognized by release factors; they do not add an amino acid

If a DNA template is supplied, write the antiparallel complementary mRNA, replacing DNA A with RNA U.

A DNA template is transcribed into mRNA codons read 5′ to 3′, and a genetic-code table maps those codons to amino acids.

Write the mRNA 5′→3′, locate the stated or biologically relevant start, and divide the sequence into consecutive triplets from that frame.

Look up each mRNA codon, record amino acids in order and stop when a stop codon is reached.

Template 3′–TAC CTT GCG ACT–5′ gives mRNA 5′–AUG GAA CGC UGA–3′, which translates as Met–Glu–Arg–Stop.

A charged tRNA with a complementary anticodon enters the A site beside the tRNA carrying the growing chain in the P site.

A ribosome positions mRNA codons and complementary tRNA anticodons in E, P and A sites while peptide bonds extend a polypeptide.

The ribosome catalyses a peptide bond, transferring the growing polypeptide to the amino acid on the A-site tRNA.

The ribosome moves one codon along the mRNA. The peptidyl-tRNA shifts to P, the empty tRNA shifts to E and exits, and the A site becomes available again.

The entry–bond–movement cycle repeats for successive codons. When a stop codon reaches the A site, no tRNA enters; a release factor releases the completed polypeptide and the ribosome dissociates.

A polysome is one mRNA being translated simultaneously by several ribosomes. Each ribosome starts near the 5′ end and moves toward the 3′ end.

All ribosomes read the same codon sequence, so they produce polypeptides with the same amino-acid order. Simultaneous translation increases the rate of protein production from that mRNA.

The ribosomes do not cooperate to make one long polypeptide. Each ribosome makes a separate copy.

Several ribosomes move in the same 5′ to 3′ direction along one mRNA, each producing a polypeptide.

A mutation can alter a DNA base sequence. If the changed region is transcribed, it can alter an mRNA codon and therefore the amino-acid sequence or the timing of translation.

Codon outcome	Possible protein consequence
codon still specifies the same amino acid	no change to primary structure
codon specifies a different amino acid	altered side-chain interactions, folding or function
codon becomes a stop signal	shortened polypeptide
insertion or deletion shifts the reading frame	many downstream codons may change

Effect size depends on where the mutation occurs and what the affected amino acid does. A substitution at an active site or interaction surface can matter more than one in a tolerant region.

Mutation creates a possibility of changed protein structure; it does not guarantee harm. Some mutations are silent, neutral or beneficial in a particular environment.

In the sickle-cell allele, a β-globin DNA substitution changes an mRNA codon from GAG to GUG, replacing glutamate with valine in the polypeptide.

Glutamate is charged and hydrophilic; valine is non-polar. The replacement creates a hydrophobic patch on deoxygenated haemoglobin S.

Under low oxygen conditions, haemoglobin S molecules associate into long fibres. The fibres distort the red blood cell into a rigid sickle shape.

Rigid cells can block small blood vessels and are removed rapidly from circulation, reducing oxygen delivery and causing anaemia and painful crises.

The mutation does not directly bend the whole cell. It first changes one amino acid, which changes intermolecular interactions, which then changes cell shape under particular oxygen conditions.

DNA template → complementary mRNA → codons read 5′→3′ → charged tRNAs selected → peptide bonds join amino acids → polypeptide primary structure

Checkpoint	Question to ask
transcription	Is the mRNA complementary to the template and written 5′→3′?
code table	Are mRNA codons grouped from the correct start?
tRNA selection	Does the anticodon pair and is the tRNA correctly charged?
elongation	Has the ribosome advanced one codon while the chain remains on the peptidyl-tRNA?
mutation	Did the codon, amino acid or reading frame actually change?

A sequence change matters biologically only through a causal path: altered primary structure → altered interactions or folding → altered protein activity → possible cellular or organism effect.

Process	Molecule being read	Reading direction	Product growth or movement
transcription	DNA template	3′→5′	RNA is synthesized 5′→3′
translation	mRNA	5′→3′	ribosome moves toward the mRNA 3′ end; polypeptide grows N-terminus→C-terminus

Nucleotides are added to a free 3′ hydroxyl group, so every new nucleic-acid strand grows 5′→3′. The template must therefore be read antiparallel to the new strand.

A diagram arrow showing polymerase movement and an arrow showing new-strand growth refer to opposite strands; identify which molecule each arrow belongs to.

RNA polymerase reads a DNA template 3′ to 5′ while synthesizing RNA 5′ to 3′; a ribosome then moves along mRNA 5′ to 3′.

A promoter is a DNA region that positions the transcription machinery near a gene's transcription start site. General transcription factors bind first and help recruit RNA polymerase.

Regulatory input	Binding relationship	Typical effect on initiation
activator transcription factor	binds a regulatory DNA sequence such as an enhancer	stabilizes or recruits the initiation machinery
repressor transcription factor	binds a regulatory sequence or interferes with the complex	reduces assembly or activity of the initiation machinery

Regulatory DNA may be far from the promoter in the linear sequence. DNA looping can bring bound factors into contact with promoter-associated proteins.

The combined transcription-factor state changes how often RNA polymerase initiates, altering the amount of pre-mRNA produced from the gene.

Non-coding DNA is DNA that is not translated into a polypeptide sequence. Non-coding does not mean untranscribed, unused or without function.

Important non-coding categories include:

Regulatory sequences such as promoters and enhancers that affect transcription.
Introns that are transcribed into pre-mRNA and removed during splicing.
Genes for functional RNAs, including rRNA and tRNA, whose products are not translated.
Structural regions such as telomeres and centromeric DNA.
Repeated sequences, some functional and some with no known function.

An exon can include untranslated regions, so exon is not a synonym for protein-coding sequence. Likewise, an intron can influence regulation or splicing even though it is removed from mature mRNA.

Processing change	What happens	Main consequence
5′ cap	a modified nucleotide is added to the 5′ end	protects RNA and supports export and translation initiation
poly-A tail	many adenine nucleotides are added to the 3′ end	increases stability and supports export and translation
splicing	introns are removed and exons joined	creates a continuous mature RNA sequence for translation

pre-mRNA in nucleus → processing completed → mature mRNA exported through a nuclear pore → translation in cytoplasm

The poly-A tail is added after transcription; it is not copied from a long template-DNA sequence of thymine bases.

A eukaryotic pre-mRNA receives a 5′ cap and poly-A tail, has introns removed and exons joined, then is exported from the nucleus as mature mRNA.

In alternative splicing, the same pre-mRNA is spliced in different patterns, so different exon combinations are retained in mature mRNA.

Different mature mRNAs contain different codon sequences and can therefore produce related polypeptides with different amino-acid segments, properties or tissue distributions.

Alternative splicing does not change the DNA sequence. It changes which parts of one RNA transcript remain in the message sent to the ribosome.

One troponin T pre-mRNA is spliced into two mature mRNAs containing different combinations of exons.

The small ribosomal subunit binds the mRNA and positions the start region for scanning or recognition.

A translation initiation complex has AUG paired with an initiator methionine tRNA in the ribosomal P site, with A and E sites labelled.

The initiator tRNA carrying methionine pairs with AUG. This tRNA is placed directly in the P site, establishing the reading frame.

The large subunit joins to complete the ribosome. The A site is now open for the tRNA matching the next codon.

AUG inside an already established reading frame can code for methionine without serving as a new start. Initiation depends on context and assembly, not the triplet alone.

Ribosomal site	Typical tRNA state during elongation
A (aminoacyl)	incoming charged tRNA paired with the next codon
P (peptidyl)	tRNA carrying the growing polypeptide before peptide transfer
E (exit)	uncharged tRNA leaving after translocation

Before peptide transfer: growing chain in P, new amino acid in A.

After peptide transfer and translocation: the chain-bearing tRNA occupies P, the empty tRNA moves to E, and A opens for the next charged tRNA.

The site letters describe functional positions in the ribosome, not permanent labels on a tRNA. A single tRNA can move A → P → E during one cycle.

Translation produces a polypeptide primary structure. Many proteins become functional only after post-translational processing changes their shape, length, location or chemical properties.

Common processing routes include:

Folding into a specific three-dimensional conformation, sometimes assisted by chaperones.
Cleavage to remove a signal peptide or inactive segment.
Covalent modification, such as phosphorylation or glycosylation, which can alter activity, stability or location.
Assembly with other polypeptide subunits or prosthetic groups.

Not every protein undergoes every modification. The required pathway depends on the protein's sequence, destination and function.

Preproinsulin contains an N-terminal signal peptide that directs the growing polypeptide into the endoplasmic reticulum. Removing the signal peptide produces proinsulin.

Preproinsulin loses its signal peptide, folds as proinsulin with disulfide bonds and loses the C-peptide to form mature insulin A and B chains.

Proinsulin folds, and disulfide bonds form between regions that will become the A and B chains.

The connecting C-peptide is removed. The A and B chains remain linked by disulfide bonds, producing mature insulin.

Proteins selected for degradation are commonly marked with a chain of ubiquitin molecules.

A ubiquitin-tagged protein is recognized, unfolded and threaded into a proteasome, which releases short peptides while ubiquitin is recycled.

The proteasome recognizes the tag, removes and recycles ubiquitin, unfolds the target protein and threads it into a proteolytic core.

The protein is cut into short peptides, which can be hydrolysed to amino acids for new synthesis or metabolism.

Selective degradation removes damaged or misfolded proteins and can rapidly lower the abundance of regulatory proteins. Protein concentration therefore depends on both synthesis and breakdown.

Checkpoint	What can change	Consequence for protein output
transcription initiation	promoter and transcription-factor activity	amount of pre-mRNA made
RNA processing	cap, tail and exon combination	mRNA stability, export and encoded sequence
translation initiation	successful AUG recognition and ribosome assembly	whether and where translation begins
post-translational processing	folding, cleavage, modification and assembly	protein activity or location
proteasomal degradation	ubiquitin tagging and breakdown rate	protein lifetime and abundance

Across these checkpoints, sequence direction stays consistent: RNA is made 5′→3′, the ribosome reads mRNA 5′→3′ and the polypeptide grows from N-terminus to C-terminus.

Gene expression is therefore not a single on/off event. A cell controls which message is made, how it is processed, whether it is translated, how the product is activated and how long the protein remains.

[ 8 ]

Explain the process of transcription in prokaryotes.

[ 1 ]

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

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

What is the sequence of bases on the resulting mRNA?

[ 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.

[ 8 ]

Explain how polypeptides are produced by the process of translation.

[ 7 ]

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

[ 3 ]

Outline how translation depends on complementary base pairing.

[ 5 ]

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

[ 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 $3^{\prime}$

1st base in codon	2nd base U	2nd base C	2nd base A	2nd base G	3rd base in codon
U	Phe	Ser	Tyr	Cys	U
U	Phe	Ser	Tyr	Cys	C
U	Leu	Ser	STOP	STOP	A
U	Leu	Ser	STOP	Trp	G
C	Leu	Pro	His	Arg	U
C	Leu	Pro	His	Arg	C
C	Leu	Pro	Gln	Arg	A
C	Leu	Pro	Gln	Arg	G
A	Ile	Thr	Asn	Ser	U
A	Ile	Thr	Asn	Ser	C
A	Ile	Thr	Lys	Arg	A
A	Met	Thr	Lys	Arg	G
G	Val	Ala	Asp	Gly	U
G	Val	Ala	Asp	Gly	C
G	Val	Ala	Glu	Gly	A
G	Val	Ala	Glu	Gly	G

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-Ile-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

[ 2 ]

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

[ 8 ]

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

[ 1 ]

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

[ 2 ]

Explain the function of a promoter in DNA.

[ 2 ]

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

[ 1 ]

What happens to mRNA after transcription in eukaryotic cells?

[ 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?

[ 4 ]

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

[ 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.

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.

A gene mutation is a change in the order or number of DNA bases within a gene. It creates a new sequence; its later effect depends on where the change occurs.

Substitution: one nucleotide is replaced by another.
Insertion: one or more nucleotides are added.
Deletion: one or more nucleotides are lost.
Duplication: a DNA segment is copied and repeated.

Mutation names describe the DNA change. They do not by themselves tell you whether a protein, cell or organism will be affected.

Aligned DNA sequences show one-base substitution, insertion, deletion and duplication compared with an unchanged sequence.

A base substitution replaces one nucleotide without changing sequence length. In a population, a one-base variant present among individuals is a single-nucleotide polymorphism (SNP).

Coding outcome	Why it happens	Possible protein effect
silent	altered codon still specifies the same amino acid because the code is degenerate	primary structure unchanged
missense	altered codon specifies a different amino acid	folding or function may change
nonsense	altered codon becomes a stop codon	translation ends early

A substitution outside a coding sequence can still matter if it changes a promoter or other regulatory sequence; some substitutions have no detectable effect.

Three codon substitutions illustrate unchanged amino acid, changed amino acid and a new stop codon.

A ribosome reads nucleotides in non-overlapping groups of three. Inserting or deleting a number of bases not divisible by three changes the grouping from that point onward: a frameshift.

A one- or two-base indel usually changes many downstream codons and may create an early stop.
A three-base indel adds or removes one codon without shifting later codon boundaries.
A larger in-frame indel can still disrupt function by adding or removing amino acids.

Do not call every insertion or deletion a frameshift. Test whether the number of bases changed is a multiple of three.

A one-base insertion changes every downstream three-base codon group in a coding sequence.

DNA replication error or DNA damage → proofreading and repair act → unrepaired or misrepaired change persists → later replication copies the altered sequence

Changes can begin inside the cell through:

incorrect nucleotide insertion during DNA replication
failure of proofreading or repair
chromosome breakage or other DNA damage

Mutagens increase mutation frequency by damaging DNA. Important classes include chemical mutagens, ionizing radiation such as X-rays, and ultraviolet radiation.

Exposure causes DNA damage probabilistically; it does not guarantee that a specific gene will mutate. Repair can restore the original sequence.

A mutation does not arise because an organism needs a useful trait. A change can occur before an environmental challenge and later prove harmful, neutral or advantageous.

Mutation probability can still vary with:

base identity and local sequence context
how often a region is copied or transcribed
access to proofreading and repair systems
dose and type of mutagen exposure

Random relative to need is not the same as equally likely at every base. Natural selection is the non-random filter; it does not direct the earlier mutation.

Different DNA variants arise before an environmental filter increases the representation of one favoured variant.

Mutation location	Cells that can inherit it	Biological reach
germ-line lineage	gametes and potentially the offspring formed from them	can pass between generations
somatic cell	mitotic descendants of that body cell	remains within the individual and its cell clone

A somatic mutation can contribute to cancer if it affects control of cell division, DNA repair or cell death. Further mutations and selection among cell clones can then expand the abnormal lineage.

A somatic mutation is not inherited by the person's offspring merely because many body cells carry it; it must enter the germ line to cross generations.

A germ-line mutation can enter a gamete and offspring, whereas a somatic mutation is copied into a local body-cell clone.

Mutation is the original source of new alleles. Recombination can reshuffle existing alleles, but it does not create a new DNA sequence at a locus.

new mutation → new allele → inheritance places the allele in a population → environment affects survival and reproduction → allele frequency may change over generations

Most new mutations are neutral or harmful in their current context; a smaller number can improve reproductive success under particular conditions. The same allele can have different effects in different environments.

Populations evolve because heritable allele frequencies change. An individual does not mutate adaptively and then evolve during its lifetime.

Ask	What the answer predicts
What changed in the DNA?	substitution, insertion, deletion or duplication
Where is the change?	coding, regulatory or other sequence; possible molecular consequence
Which cell lineage carries it?	local somatic clone or possible inheritance through gametes
Does it alter reproductive success?	whether selection may change its frequency

The causal chain can stop at any stage: a DNA change can be repaired, lie in a tolerant region, preserve the amino acid, or make no detectable phenotypic difference.

Mutation generates variants without foresight. Cell lineage controls their reach, and natural selection later filters heritable variation according to environmental conditions.

A gene knockout makes a selected gene non-functional. Researchers compare the knockout phenotype with an otherwise similar control to infer what the missing gene normally contributes.

selected gene active in control → normal gene product present → baseline phenotype

selected gene knocked out → gene product absent or reduced → any reproducible phenotype difference supports a role for that gene

A changed phenotype supports involvement, not a complete one-gene/one-trait explanation. Developmental compensation, background mutations and environmental differences must be controlled.

An active gene produces a normal phenotype, while loss of that gene's function is associated with a changed phenotype in a knockout mouse.

A knockout library contains organisms or cell lines in which different genes have been made non-functional. Researchers can screen the collection for a phenotype and identify candidate genes.

Common model organisms include:

mice for mammalian physiology and disease
Drosophila for development and inheritance
zebrafish for vertebrate development and transparent embryos
Arabidopsis for plant genetics

Model organisms are useful because core genes, the genetic code and many cell processes are shared. They also have short generations, tractable breeding or large existing genetic resources.

A model result is evidence for a testable hypothesis in another species, not automatic proof that the gene has an identical effect in humans.

After viral infection, a bacterium can insert a short fragment of viral DNA as a spacer among CRISPR repeats in its own genome.

The spacer is transcribed into guide RNA. Its base sequence can pair with complementary DNA from the same virus during a later infection.

Guide RNA associates with a Cas nuclease. Complementary pairing positions the nuclease, which cuts the matching viral DNA and prevents successful replication.

Specificity comes from base pairing between guide RNA and target DNA; Cas9 supplies the cutting activity.

Researchers choose a target sequence and design a guide RNA with a complementary region. The guide RNA and Cas9 form a complex that searches DNA for a compatible target beside the required recognition motif.

A guide RNA pairs with target DNA beside a PAM sequence, positioning Cas9 for a double-strand cut before cellular repair.

Guide-target base pairing holds the complex at the selected site. Cas9 then makes a double-strand break in the DNA.

Cas9 makes the cut; the final edit is produced when the cell repairs that break. Similar off-target sequences can sometimes also be cut, so the result must be checked.

Repair context	Likely sequence outcome	Typical use
ends rejoined without a supplied template	small insertions or deletions may disrupt the reading frame	gene disruption or knockout
two sites are cut	intervening segment can be deleted	targeted deletion
repair template is supplied	chosen sequence can be copied into the break	replacement or insertion

define the intended edit → design guide RNA and, if needed, a donor template → deliver editing components → isolate edited cells or organisms → sequence the target → test the phenotype

A desired phenotype alone does not prove the intended edit caused it. Confirm the target sequence, examine plausible off-target sites and compare with suitable controls.

CRISPR-Cas9 can delete, replace, insert or disrupt DNA, but efficiency and precision depend on target sequence, delivery and the cell's repair pathway.

A conserved sequence remains identical or similar among species; a highly conserved sequence has changed little across long evolutionary time.

Read an alignment by columns. A position shared by all sampled species is conserved in that sample; a column containing different bases is variable.

Conservation can occur in protein-coding or regulatory DNA. The species sampled and the time since their divergence affect how strong the pattern appears.

A multi-species DNA alignment highlights positions shared by every species and positions that vary.

Hypothesis	Mechanism	Prediction
functional constraint with purifying selection	harmful variants reduce reproductive success and are removed	disrupting the sequence is likely to impair function
lower mutation rate	fewer sequence changes arise, perhaps because repair is especially effective	conservation can persist even before selection acts

The explanations are not mutually exclusive. An essential sequence may both acquire mutations slowly and lose harmful variants rapidly through purifying selection.

Highly transcribed genes and their transcribed strands can show lower mutation rates, consistent with enhanced access to repair machinery.

Conservation suggests importance, but sequence comparison alone does not reveal the exact function. Knockout or targeted editing can test the hypothesis experimentally.

Evidence route	Core question	Strength	Main caution
knockout	what changes when gene activity is removed?	supports a causal role	other changes and compensation need controls
CRISPR-Cas9 edit	what changes after a chosen sequence is altered?	tests a precise sequence hypothesis	edit and off-target sites must be verified
conservation	which positions persisted across evolution?	prioritizes likely functional regions	correlation does not identify exact function

Conservation can nominate a candidate sequence → targeted editing can change it → phenotype and molecular measurements can test the predicted function → sequencing confirms what was actually altered.

The strongest inference combines comparative evidence with a controlled intervention. Neither an alignment nor an edited phenotype should be interpreted alone.

[ 2 ]

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

[ 4 ]

Outline how a base substitution leads to sickle cell anemia.

[ 2 ]

Explain how chemical substances can cause cancer.

[ 1 ]

What is a feature of mutations?

[ 1 ]

A mutation in which type of cell could be inherited?

[ 1 ]

What causes variation in both sexually and asexually reproducing organisms?

[ 1 ]

What is gene knockout used for?

[ 3 ]

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

[ 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
Corn	QGTLQCSTLLKTKRPAARRWMPCSLLGLHPWEAGRP-SPAV
Daffodil	KSRLSVYSLYTTSKYA-----------CVGFEAENNGKFLI
	* * * * * *
Corn	YSSLPVNPAGEAVVSSEQKVYDVVLKQAALLKRQLRTP--V
Daffodil	RSSLVANPAGEATISSEQKVYDVVLKQAALVKDQTKSSRKS
	* * * * * * * * * * * * * * * * * * * * * * * * * * *
Corn	LDARPQDMDMPRN--GLKEAYDRCGE I CEEYAKTFYLGTML
Daffodil	TDVKP-DIVLPGTVYLLKDAYDRCGEVCAEYAKTFYLGTLL
	* * * * * * * * * * * * * * * * * * * * * * * *
Corn	MTEERRRA I WA I YVWCRRTDELVDGPNANY I TPTALDRWEK
Daffodil	MTPERRRAI WA I YVWCRRTDELVDGHNASHITPSALDRWEA
	** * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Corn	RLEDLFTGRPYDMLDAALSDTISRFPIDIQPFRDMIEGMRS
Daffodil	RLEDLFAGRPYDMFDAALSDTVSRFPVDIQPFMDMVEGMRM
	* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Corn	DLRKTRYNNFDELYMYCYYVAGTVGLMSVPVMGIATESKAT
Daffodil	DLKKSRYKNFDELYLYCYYVAGTVGLMSVPVMGIAPESLAE
	* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Corn	TESVYSAALALGIANQLTNI LRDVGEDARRGRIYLPQDELA
Daffodil	AESVYNAALALGIANQLTNI LRDVGEDARRGRIYLPQDELA
	* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Corn	QAGLSDEDIFKGVVTNRWRNFMKRQIKRARMFFEEAERGVN
Daffodil	EAGLSDEDVFTGKVTDKWRSFMKRQIKRARTFFEQAEKGVT
	* * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Corn	ELSQASRWPVWASLLLYRQI LDEIEANDYNNFTKRAYVGKG
Daffodil	E L SQA SRWP VWASLL LYRQI LDE I EANDYNNF TKRAYVSKV
Corn	KKLLALPVAYGKSLLLPCSLRN---GQT
Daffodil	KRLAALPLA YGKSLLIPLSLRPPSLSKA
	* * * * * * * * * * * * * * * *

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.

Every new cell arises when a pre-existing parent cell divides. The daughter cells inherit cellular material and DNA from that lineage; they are not assembled from non-living parts.

Cell division supplies new cells for:

growth of a multicellular organism
replacement of worn or damaged cells
tissue repair after injury
reproduction in unicellular organisms and some multicellular organisms

Cell division increases cell number. Growth of an individual cell increases cell size; organisms often require both processes.

One parent cell divides by mitosis into daughter cells that contribute to growth, tissue repair, cell replacement or asexual reproduction.

Cytokinesis partitions the cytoplasm after nuclear division so the daughter nuclei become separate cells.

Feature	Animal cell	Plant cell
starting constraint	flexible plasma membrane	rigid existing cell wall
mechanism	actin–myosin contractile ring tightens a cleavage furrow	Golgi-derived vesicles fuse at the equator
result	membrane pinches inward	cell plate becomes new membranes and wall between daughters

Both mechanisms divide one cytoplasm around two daughter nuclei, but they solve the mechanical problem in opposite directions: inward constriction versus construction from the centre outward.

Animal-cell cytokinesis uses a cleavage furrow, whereas plant-cell cytokinesis joins vesicles into a central cell plate.

Cytokinesis often gives daughter cells similar cytoplasmic volumes, but the partition can be unequal when the daughters have different immediate functions.

Oogenesis: one cell retains most cytoplasm, organelles and stored nutrients to become the ovum; small polar bodies mainly remove extra chromosome sets.
Yeast budding: a smaller bud grows from the parent and separates after receiving a nucleus and essential cell components.

Unequal cytokinesis refers to cytoplasmic allocation, not deliberately unequal chromosome inheritance. Normal daughter nuclei still require the appropriate chromosome set.

Equal cytokinesis is contrasted with a large oocyte and polar body and with a small yeast bud on a larger parent cell.

Dimension	Mitosis	Meiosis
nuclear divisions	one	two after one DNA replication
usual nuclei produced	two	four
chromosome number	maintained	halved
genetic relationship	normally identical to parent and each other	varied combinations
main roles	growth, replacement, repair, asexual reproduction	production of cells that develop into gametes

Diploid (2n) nuclei contain two homologous sets of chromosomes. Haploid (n) nuclei contain one set.

Mitosis and meiosis are nuclear divisions. Cytokinesis is the separate division of cytoplasm that usually follows.

A diploid parent produces two genetically similar diploid nuclei by mitosis or four genetically varied haploid nuclei by meiosis.

Before S phase, one chromosome contains one DNA molecule. During S phase that DNA is replicated.

An unreplicated chromosome becomes two sister chromatids joined at a centromere and later separates into daughter chromosomes.

After replication, the chromosome consists of two nearly identical sister chromatids joined at a centromere. The chromosome count has not doubled, but the DNA amount has.

When the sister chromatids separate, each becomes an individual daughter chromosome. Sending one copy to each future nucleus preserves a complete genome.

Do not count chromatids as separate chromosomes while they remain joined at one centromere.

Each chromosome is one very long DNA molecule. Accurate division requires it to be packaged into a compact unit and attached to machinery that can position and separate it.

Mitosis and meiosis share three mechanical features:

DNA wrapped around histones and nucleosomes condenses into chromosomes.
Spindle microtubules attach through centromere-associated structures.
Microtubule dynamics and motor proteins position and move chromosomes.

Condensation does not copy DNA. It changes packaging so long molecules can move with less tangling or breakage.

DNA wraps around histones into chromatin, condenses into a duplicated chromosome and attaches at the centromere region to spindle microtubules.

Prophase: chromatin condenses into visible duplicated chromosomes; the nucleolus disappears, the nuclear envelope breaks down, and the spindle forms.

Metaphase: spindle microtubules attach to the centromere region from opposite poles and position chromosome centromeres at the cell equator.

Identification cues:

condensed chromosomes still scattered or nuclear boundary disappearing → prophase
duplicated chromosomes arranged along one central plane → metaphase

Mitosis is continuous; phase names label dominant visible events rather than pauses between separate processes.

Anaphase: centromere linkage is released and sister chromatids move to opposite poles. Once separated, each chromatid is a daughter chromosome.

Telophase: chromosomes reach the poles and decondense; nuclear envelopes and nucleoli re-form around the two chromosome sets.

Identification cues:

two separating groups, often V-shaped as centromeres lead → anaphase
two groups at opposite ends with new nuclear boundaries → telophase

Because sister chromatids were replicated copies and one moves to each pole, the two daughter nuclei normally contain identical chromosome sets.

Use a fixed observation sequence:

Are individual condensed chromosomes visible?
Are they scattered, aligned at one equator, separating, or already at opposite poles?
Is one nuclear envelope present, absent, or re-forming as two?

Dominant observation	Best stage identification
chromatin diffuse inside an intact nucleus	interphase, not mitosis
condensed chromosomes not yet aligned	prophase
chromosomes aligned at one equator	metaphase
daughter chromosomes moving apart	anaphase
two chromosome groups with re-forming nuclei	telophase

A stained root-tip squash contains many cells in different stages because division is not synchronized. Count only cells whose chromosome evidence is sufficiently clear.

A single still image may lie near a phase boundary. Record the visible cue and acknowledge uncertainty rather than forcing a label from cell outline alone.

DNA replication creates joined sister chromatids → chromosomes condense and attach to the spindle → metaphase alignment checks bipolar attachment → sister chromatids separate → two nuclei re-form → cytokinesis partitions the cytoplasm

Continuity requirement	What supplies it
complete genome in each nucleus	replication followed by equal chromatid separation
controlled chromosome movement	condensation, spindle microtubules and motor proteins
two separate cells	animal cleavage furrow or plant cell plate
suitable daughter-cell contents	equal or functionally unequal cytoplasmic partition

To identify a stage, prioritize chromosome position and nuclear-envelope state. To explain the outcome, connect those observations to what is being separated.

Before meiosis, S phase replicates every chromosome into two sister chromatids. The nucleus is still diploid because both homologues of each pair remain present.

A diploid cell replicates DNA once, separates homologous chromosomes in meiosis I and separates chromatids in meiosis II to form four haploid nuclei.

In meiosis I, homologous maternal and paternal chromosomes separate into different nuclei. Chromosome number is reduced from diploid to haploid, although each chromosome still has two chromatids.

There is no second DNA replication. In meiosis II, sister chromatids separate, producing four haploid nuclei with one chromosome from each homologous pair.

Reduction prevents chromosome number doubling in every generation: haploid gametes fuse at fertilization to restore the diploid number.

A bivalent is a paired set of duplicated homologous chromosomes: one maternal homologue and one paternal homologue, each made of two sister chromatids.

Question	Meiosis I	Meiosis II
what aligns?	bivalents	individual duplicated chromosomes
what separates?	homologous chromosomes	sister chromatids
do centromeres divide?	no	yes
chromosome-number effect	diploid → haploid	remains haploid

Use the object being separated in explanations: homologues move apart in anaphase I; sister chromatids move apart in anaphase II.

Calling both events 'chromosomes separating' hides the reason meiosis I is the reduction division.

A homologous pair can fail to separate in meiosis I, or sister chromatids can fail to separate in meiosis II. This failure is non-disjunction.

Both copies move to one pole while the opposite pole receives none. Later division therefore cannot give every gamete the standard chromosome number.

The affected meiosis produces gametes with an extra chromosome (n + 1) and/or a missing chromosome (n − 1), depending on when the error occurred.

After fusion with a normal haploid gamete, an n + 1 gamete can produce a trisomic zygote; an n − 1 gamete can produce a monosomic zygote.

Non-disjunction can create too many or too few chromosomes. It is not defined only by an extra chromosome.

chromosome 21 non-disjunction during gamete formation → gamete receives two copies of chromosome 21 → fertilization by a normal gamete adds one more copy → zygote has three copies of chromosome 21

The resulting trisomy 21 gives the zygote 47 chromosomes rather than the usual 46 and can cause Down syndrome.

The developmental effects vary among individuals because an extra chromosome changes the dosage of many genes, not just one gene.

Down syndrome is a chromosome-number condition. It is not caused by a point mutation within chromosome 21.

During prophase I, duplicated homologous chromosomes pair closely as a bivalent. Corresponding gene loci align.

Non-sister chromatids break at corresponding positions and rejoin to the other homologue. The visible connection is a chiasma.

Equivalent DNA segments are exchanged, producing recombinant chromatids with new combinations of maternal and paternal alleles.

Crossing over is between non-sister chromatids of homologous chromosomes. It rearranges existing alleles; it does not necessarily create new alleles.

At metaphase I, each bivalent can face either pole independently of the others. The maternal or paternal homologue that enters a gamete is therefore random for each pair.

$2^n$

For humans, $n=23$ , so random orientation alone can produce $2^{23}$ —more than 8 million—chromosome combinations before crossing over is counted.

Random fusion of one genetically varied sperm with one genetically varied egg multiplies the possible allele combinations again.

Crossing over changes allele combinations within chromosomes; random orientation changes which whole maternal and paternal homologues travel together.

Event	What changes	Main consequence
replication before meiosis	each chromosome gains a sister chromatid	DNA prepared for two divisions
meiosis I	homologous chromosomes separate	chromosome number is halved
meiosis II	sister chromatids separate	four haploid products can form
non-disjunction	one required separation fails	aneuploid gametes may form

Variation is layered: crossing over recombines alleles within chromosomes, random orientation assort homologues independently, and random fertilization combines two independently produced gametes.

When explaining any meiotic outcome, name the stage, the chromosome object that should separate, the chromosome number before and after, and whether the result changes number, allele combination or both.

Cell proliferation is an increase in cell number through repeated cell growth and mitotic division. The new cells may remain unspecialized, differentiate or replace cells that were lost.

Early embryo: rapid divisions build cell number before extensive differentiation.
Plant meristems: dividing cells at apical or lateral growth regions supply new plant tissues.
Routine replacement: stem-cell lineages replenish short-lived cells such as epidermal cells.
Wound repair: fibroblasts and other cells proliferate to rebuild tissue and extracellular matrix.

Proliferation describes more cells. Cell enlargement and differentiation can follow, but they are different processes.

After fertilization, repeated mitotic divisions increase cell number from a zygote through two-, four- and eight-cell stages to a blastocyst.

The cell cycle is the repeating sequence in which a cell grows and prepares during interphase, divides its nucleus by mitosis and partitions its cytoplasm by cytokinesis.

G1 growth and biosynthesis → S-phase DNA replication → G2 growth and division preparation → mitosis → cytokinesis → daughter cells enter a new cycle or leave active cycling

A cell is not in mitosis for most of its cycle. Interphase is usually the longest portion, so most cells in a tissue sample are expected to appear in interphase.

A circular cell-cycle diagram orders G1, S and G2 interphase before mitosis and cytokinesis return the lineage to G1.

Interphase stage	Main work	Chromosome state by the end
G1	cell growth, metabolism, protein and organelle production	each chromosome has one chromatid
S	DNA replication	each chromosome has two sister chromatids
G2	further growth, protein synthesis and preparation for spindle formation	replicated chromosomes remain joined

The order matters: the cell first builds capacity, then copies the genome once, then checks and prepares the machinery needed to distribute those copies.

DNA quantity doubles during S phase, but chromosome number does not double because the sister chromatids remain joined at their centromere.

During interphase the cell transcribes genes, synthesizes proteins, grows its cytoplasm, accumulates energy and increases organelles. DNA is dispersed as chromatin and remains accessible for gene expression.

Interphase duration varies with cell fate: rapidly proliferating embryo or meristem cells cycle quickly, many adult cells divide only when signalled, and some differentiated cells remain outside active division for long periods.

A nucleus that looks uniform under a light microscope is not resting. Lack of visible condensed chromosomes is evidence of chromatin state, not metabolic inactivity.

An interphase cell is surrounded by protein synthesis, DNA replication, cell growth and organelle increase.

Cyclin-dependent kinases (CDKs) are present but require particular cyclins to become active. Cyclin concentrations rise and fall, giving the control system time-specific signals.

Cyclin accumulates and binds its CDK → the active complex phosphorylates target proteins → a checkpoint transition can occur once the relevant conditions and threshold are satisfied.

At the G2/M transition, cyclin–CDK activity forms mitosis-promoting factor (MPF), triggering chromosome condensation, nuclear-envelope breakdown and spindle assembly.

Cyclin is then degraded while CDK can be reused. Falling complex activity helps make the transition directional and resets control for a later cycle.

Cyclin level is part of checkpoint control, not the only input. DNA integrity, replication completion, spindle attachment and external signals also influence progression.

Gene class	Normal role	Cancer-promoting change	Analogy
proto-oncogene	stimulates division when an appropriate signal is present	gain-of-function mutation creates an oncogene with excessive or constant activity	accelerator stuck on
tumour-suppressor gene	stops the cycle, repairs damage or promotes apoptosis	loss-of-function mutation removes restraint	brake lost

A mutant Ras oncogene can keep growth-signalling pathways active without the normal extracellular signal. One activated allele may be sufficient to increase signalling.

Loss of p53 function can prevent arrest or apoptosis after DNA damage. Because a remaining functional copy can still supply p53, both copies often need to be lost in the affected cell lineage.

Cancer usually develops through accumulated mutations and clonal selection, not one mutation that instantly creates a fully malignant tumour.

Property	Benign tumour	Malignant tumour
local growth	often slower and bounded	often faster and poorly controlled
invasion	does not invade neighbouring tissue	penetrates and damages neighbouring tissue
metastasis	absent	cells may detach, travel and form secondary tumours

Metastasis requires several steps: cells detach from the primary tumour, enter blood or lymph, survive transport, leave the vessel and establish a secondary tumour elsewhere.

A tumour is an abnormal mass produced by excess cell division. Not every tumour is cancerous; malignant invasion and metastatic capacity define cancerous behaviour.

A local benign tumour is contrasted with a malignant tumour invading neighbouring tissue and with cells spreading through a vessel to form a secondary tumour.

$MI=\frac{M}{T}\times 100\%$

Let M be the number of cells observed in mitosis and T the total number of cells observed. If 10 of 95 visible cells are in mitosis,

$\frac{10}{95}\times 100\%=10.5\%$

A higher mitotic index means a larger fraction of sampled cells was in mitosis at the observation time. It can support comparison of tissue proliferation or response to treatment.

Mitotic index is not a unique cancer diagnosis. It depends on sampling, stage-identification accuracy and how long cells remain in mitosis; replicate fields and other evidence are required.

Growth or repair signal → G1 growth → S-phase genome replication → G2 preparation and checkpoint control → cyclin–CDK activation → mitosis and cytokinesis → daughter cells differentiate, replace tissue or re-enter the cycle

Oncogene activation can increase positive growth signalling, while tumour-suppressor loss can remove arrest, repair or apoptosis. Accumulated changes allow an abnormal clone to proliferate.

Evidence scale	Question it answers
cell-cycle proteins and genes	why is checkpoint control altered?
mitotic index	what fraction of sampled cells is dividing now?
tissue boundary	is the tumour local or invading?
secondary growths	has metastatic spread occurred?

Rapid division alone is not the whole definition of malignancy. A strong explanation connects molecular control failure to clonal growth, tissue invasion and possible metastasis.

[ 1 ]

Which process contributes to growth of a multicellular body?

[ 8 ]

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

[ 1 ]

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

[ 5 ]

Distinguish between the processes of meiosis and mitosis.

[ 1 ]

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

[ 1 ]

What occurs in cell division during both mitosis and meiosis?

[ 9 ]

Describe the events that occur during mitosis.

[ 1 ]

The following shows a micrograph.

How many cells are in metaphase?

[ 5 ]

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

[ 4 ]

Describe how non-disjunction can cause Down syndrome.

[ 7 ]

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

[ 3 ]

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

[ 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.

[ 4 ]

Outline the processes occurring during interphase in the cell cycle.

[ 4 ]

Explain how the cell cycle is controlled.

[ 4 ]

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

[ 1 ]

Which processes occur during the development of secondary tumours?

I. Cytokinesis
II. Metastasis
III. Mitosis

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.

Gene expression is the use of information in a gene to produce a functional RNA or protein whose activity can affect cell structure, cell behaviour and phenotype.

For a protein-coding gene:

DNA base sequence → transcription → mRNA sequence → translation → amino-acid sequence → folded protein and cellular function → contribution to phenotype

A gene does not act on phenotype by itself. Its effect depends on whether it is expressed, how much product is made, what the product does and the cellular or environmental context.

A gene is transcribed into mRNA, translated into protein, and the protein contributes to phenotype.

The amount of a protein present at one time reflects a balance between its production and removal, not a single on/off switch.

Control point	Immediate effect	Consequence for protein output
transcription initiation	changes how many mRNA molecules are made	changes the number of templates available
mRNA processing and export	changes which mature transcripts reach ribosomes	changes which messages can be translated
mRNA degradation	changes transcript lifetime	changes how many translation rounds can occur
translation rate	changes ribosome use of each mRNA	changes protein production per transcript
protein modification or degradation	changes activity or lifetime of the product	changes functional protein present

More mRNA often permits more translation, but mRNA abundance alone does not determine phenotype: translation efficiency, protein activation and protein lifetime also matter.

A promoter is a non-coding DNA region near the start of a gene where general transcription factors and RNA polymerase assemble to initiate transcription.

Specific DNA sequences are recognized by proteins → a transcription-initiation complex assembles at the promoter → RNA polymerase is positioned on the template strand → transcription begins at the correct site and direction.

Proteins that stabilize or recruit the initiation complex increase transcription. Proteins that block assembly, polymerase recruitment or promoter access decrease transcription.

The promoter is DNA; a transcription factor is a protein that binds DNA. Neither should be called a gene that codes for the protein being regulated.

Sequence-specific transcription factors bind regulatory DNA only when their DNA-binding region matches the target base sequence.

Regulatory input	Binding site	Usual effect
activator	enhancer	helps recruit or stabilize the transcription machinery, increasing transcription
repressor	silencer or other regulatory sequence	interferes with activation, promoter access or initiation, decreasing transcription

An enhancer may lie thousands of base pairs upstream or downstream of its gene. DNA looping can bring enhancer-bound activators into contact with proteins at the promoter.

A gene's transcription rate reflects the combination of regulatory proteins present in that cell. One enhancer does not make a gene permanently active in every tissue.

An mRNA can be translated repeatedly while it remains intact and accessible to ribosomes. Regulating its degradation therefore changes how long protein production can continue.

A long poly-A tail supports repeated translation, whereas tail shortening is followed by mRNA breakdown and less protein production.

poly-A tail shortens → the transcript becomes less stable → protective structures are removed → nucleases degrade the mRNA → fewer further rounds of translation occur

A short mRNA lifetime allows a rapid change in protein production when conditions change. A stable mRNA can sustain output even after new transcription has decreased.

regulatory proteins and chromatin access → transcription initiation → mature mRNA abundance and lifetime → translation → protein activity and lifetime → cellular effect

Observation	First control question to ask
little or no mRNA	is transcription initiation repressed or is mRNA rapidly degraded?
abundant mRNA but little protein	is translation inefficient or is protein rapidly removed?
protein present but no expected effect	is the protein inactive, misplaced or missing a required partner?

Gene expression is not simply gene present versus gene absent. Explain a change by naming the regulated stage and tracing its consequence to functional protein and phenotype.

Epigenesis is the development of organized patterns of cell differentiation from an initially undifferentiated zygote.

one zygote → repeated mitosis → cells receive different positional or signalling inputs → different genes remain active or inactive → different proteins accumulate → cells acquire specialized structures and functions

A pancreatic β cell expresses proteins for insulin production and secretion.
A skeletal-muscle fibre expresses abundant contractile proteins.
A neuron expresses receptors, ion channels and synaptic proteins suited to signalling.

Differentiation usually changes which genes are expressed, not which genes the cell possesses. Most somatic cell types retain essentially the same genome.

Layer	What it contains	How it changes
genome	the complete DNA information of a cell	relatively stable across most somatic cells
transcriptome	all RNA transcripts present in a cell or tissue at a stated time	changes with cell type, signals and developmental stage
proteome	all proteins present in a cell, tissue or organism at a stated time	changes through translation, modification, location and degradation

The genome supplies possible information. The transcriptome records which regions are being expressed as RNA now. The proteome records the protein products actually present after further regulation.

The transcriptome and proteome must be tied to a sample and time. Neither is a permanent list of everything the genome could produce.

Cells produced from the same zygote usually share the same genome, but regulatory states make different subsets of that genome accessible and active.

Same underlying genome	Different active output	Functional consequence
insulin gene present	abundant insulin mRNA and protein in pancreatic β cells	blood-glucose regulation
collagen genes present	high collagen expression in many connective-tissue cells	extracellular support
ion-channel genes present	neuron-specific channel and receptor profile	electrical signalling

regulatory state → selected transcriptome → selected proteome → specialized structure and function

A cell type does not normally lose every gene it does not use. Inactivity and absence are different claims.

DNA methylation adds methyl groups to particular cytosines without changing the DNA base sequence. Dense methylation in a promoter is usually associated with reduced transcription of the downstream gene.

promoter cytosines become methylated → proteins that recognize methylated DNA can bind and recruit chromatin-compacting machinery → transcription factors and RNA polymerase gain less access → transcription decreases

Because methyl groups can be added, maintained or removed by enzymes, promoter methylation is potentially reversible even when the underlying gene sequence is unchanged.

The regulatory effect of methylation depends on its location. A methyl group anywhere in a gene does not automatically switch that gene off.

DNA wraps around histone proteins in nucleosomes. Chemical modification of histone tails changes which proteins associate with chromatin and how accessible nearby DNA is to transcription machinery.

Histone modification	Typical interpretation	Important condition
acetylation	often loosens chromatin and supports transcription	the effect reflects recruitment of acetyl-binding and chromatin-remodelling proteins
deacetylation	often supports compact, less accessible chromatin	reduced access can lower transcription
methylation	may activate or repress transcription	the effect depends on the amino acid and methylation state

Promoter DNA methylation is usually taught as repressive, but histone methylation is not automatically repressive. Its meaning depends on the modified histone site.

Epigenetic inheritance transmits a gene-expression state without transmitting a new DNA base sequence.

During DNA replication, each daughter DNA molecule initially contains one parental strand carrying the previous local methylation pattern and one newly synthesized strand.

Maintenance enzymes recognize those local cues and restore methylation or chromatin features on the new material, so daughter cells inherit a similar regulatory state.

This cellular memory helps a dividing liver cell lineage remain different from a dividing skin or muscle cell lineage even though their genomes are very similar.

Persistence through mitosis is not the same as transmission to offspring. Gamete development and early embryos erase many epigenetic marks.

An external factor changes gene expression only through a biological sensing and regulatory pathway: signal or exposure → receptor, metabolic sensor or cellular damage response → regulatory proteins or chromatin change → altered transcription or translation.

Examples include:

light cycle altering transcription in photosynthetic organisms or seasonal tissues
oxygen concentration activating hypoxia-responsive transcription factors
steroid hormones binding intracellular receptors
nutrient availability regulating bacterial operons
diet, drugs, temperature or pollutants altering signalling and sometimes epigenetic tags

The molecular change may alter a cell's proteome; coordinated changes across many cells can alter physiology or phenotype.

The environment does not rewrite genes because an organism needs a trait. It changes regulatory inputs, and some exposures may also damage DNA through separate mutagenic mechanisms.

Traffic-related pollution includes particulate matter, ozone, carbon monoxide and nitrogen oxides. Exposure has been associated with altered DNA or histone methylation patterns and with respiratory inflammation or disease.

pollutant exposure → oxidative stress or inflammatory signalling in lung tissue → altered regulatory or epigenetic state → changed expression of inflammatory and repair genes → possible contribution to chronic tissue damage

Evidence	What it supports	What it does not prove alone
exposure correlates with methylation difference	pollution may influence regulation	that methylation caused the disease
methylation correlates with altered expression	the tag may mark a regulatory change	that no other pathway contributed
disease risk rises with exposure	exposure is a risk factor	that every exposed individual will develop disease

A strong conclusion separates observation from mechanism: pollution-related epigenetic change is a plausible contributor, not automatically the only cause of a lung condition.

During germ-cell development and early embryonic development, extensive epigenetic reprogramming removes most existing tags so the new organism can establish appropriate developmental expression patterns.

Most environmentally acquired or tissue-specific states are not transmitted unchanged.
Early embryonic genes can be activated rather than remaining locked in a parental tissue pattern.
A small set of parent-of-origin imprints escapes complete resetting or is re-established during gamete formation.

Evidence that an epigenetic mark persists in a person's somatic cells does not by itself show that the mark will pass through gametes to descendants.

In genomic imprinting, an epigenetic mark causes one parental allele to be expressed differently from the other. The active copy depends on whether the allele came from the mother or father.

Inheritance at an imprinted locus	Expression after fertilization
maternal allele carries the silencing imprint	paternal allele is the working copy
paternal allele carries the silencing imprint	maternal allele is the working copy

Because only one allele may be active, deletion, mutation or abnormal imprinting of that active parental copy can have a phenotype even when the other DNA copy is present but epigenetically silent.

Imprinting does not mean the maternal and paternal alleles have different base sequences. It describes parent-specific regulation of expression.

shared genome → cell-specific regulatory and epigenetic state → cell-specific transcriptome → cell-specific proteome → differentiated structure and function

Process	What remains continuous?	What can change?
mitosis in a differentiated lineage	much of the expression state	reversible tags and regulatory inputs
environmental response	DNA base sequence usually remains	signalling, transcription and sometimes epigenetic marks
gamete and early-embryo reprogramming	genome and selected imprints	most parental epigenetic patterns are reset

For any epigenetic claim, ask where the tag is, how it changes access or protein binding, whether it persists through mitosis or gametes, and whether the evidence shows correlation or causation.

Monozygotic twins develop when one zygote splits and therefore begin with nearly identical genomes. Dizygotic twins develop from two separate fertilizations and share, on average, about half of their segregating genetic variation.

Comparison	Main use
monozygotic twins reared together	holds genome similar while observing outcomes within a largely shared environment
monozygotic twins reared apart	holds genome similar while increasing environmental differences
monozygotic versus dizygotic twin similarity	asks whether greater genetic similarity is associated with greater trait similarity

Monozygotic twins are not biologically identical in every respect. Developmental chance, somatic mutations, epigenetic divergence and non-shared experiences can produce differences.

Monozygotic twins arise when one zygote splits, whereas dizygotic twins arise from two separate eggs and sperm and are genetically distinct.

Pattern in a sufficiently large study	Supported inference
monozygotic pairs are more similar than dizygotic pairs	genetic differences probably contribute to variation
twins reared together are more similar than comparable twins reared apart	shared environment probably contributes
older or differently exposed monozygotic twins show epigenetic divergence	lifetime environment or stochastic development may alter regulation

Check sample size, how the trait was measured, whether environments were truly different, whether twins were selected representatively, and whether age, sex or socioeconomic conditions confound the comparison.

A trait can reflect gene–environment interaction. High similarity does not mean genes act independently of environment, and a difference within a monozygotic pair does not identify one environmental cause by itself.

A lipid-soluble steroid hormone can cross the plasma membrane and bind an intracellular receptor. Ligand binding changes the receptor into an active transcription-regulating complex.

The hormone enters a target cell and binds its specific receptor. The receptor changes conformation, may dimerize and becomes able to regulate transcription.

The activated complex enters or acts in the nucleus, binds a hormone-response DNA sequence and recruits coactivators or corepressors, changing target-gene transcription.

Only cells with the required receptor and accessible target regulatory sequences respond directly. The same hormone can therefore cause different expression changes in different tissues.

The hormone is the signal, not the transcription factor's DNA target. The activated receptor complex regulates transcription by binding specific DNA sequences with other regulatory proteins.

An operon is a group of adjacent bacterial structural genes controlled together and transcribed from one promoter into a shared mRNA.

Component	Function
regulator gene	encodes a regulatory protein such as a repressor; it may lie outside the operon
promoter	RNA polymerase binding and transcription-start region
operator	DNA site where a regulatory protein can block or permit transcription
structural genes	encode related proteins, often enzymes in one pathway
terminator	signals the end of transcription

Coordinated transcription allows a bacterium to make a complete pathway when it is useful and avoid spending energy and materials on several unnecessary enzymes.

The operator is a DNA binding site, not a protein. The repressor is the protein that binds it. Operons are characteristic of prokaryotic gene organization, not the usual organization of eukaryotic genes.

Without lactose, the lac repressor has a shape that binds the operator. This obstructs productive RNA-polymerase transcription of the lactose-use structural genes, so the operon is off or expressed only at very low basal levels.

When lactose is available, its inducer form binds the repressor allosterically → repressor shape changes → the operator is released → RNA polymerase transcribes the structural genes → enzymes for lactose uptake and metabolism are produced.

The lac operon is inducible: its default repressed state is relieved by substrate availability. When the inducer is depleted, the repressor can bind the operator again.

When tryptophan is scarce, the trp repressor is inactive and has little affinity for the operator. RNA polymerase transcribes the structural genes, producing enzymes that synthesize tryptophan.

When tryptophan accumulates, it binds the repressor as a corepressor → repressor conformation changes → the active complex binds the operator → transcription of tryptophan-synthesis genes is inhibited.

This is negative feedback: the pathway's product helps switch off further production of the enzymes that make it.

The trp operon is repressible: it is normally available for transcription but can be switched off by abundant product. Tryptophan does not induce the operon.

Dimension	lac operon	trp operon
pathway role	lactose uptake and breakdown	tryptophan synthesis
default regulatory state	repressor active; operon off	repressor inactive; operon available
small effector	lactose-derived inducer	tryptophan corepressor
effector action	inactivates repressor	activates repressor
useful outcome	enzymes made when substrate is present	enzymes withheld when product is abundant

To predict the state, ask two questions: does the small molecule activate or inactivate the repressor, and does the resulting repressor bind the operator?

Both systems conserve resources, but they do so with opposite default states. Do not memorize only that a nutrient is present; trace its effect on repressor shape and operator occupancy.

external molecule → binds a receptor or regulatory protein → protein conformation, location or DNA binding changes → transcription changes → mRNA and protein abundance change → cell response

System	Signal sensor	DNA-control event	Expression result
steroid-hormone target cell	intracellular receptor	activated receptor complex binds response sequence	selected eukaryotic genes increase or decrease transcription
lac operon	repressor binds lactose-derived inducer	repressor leaves operator	lactose-use genes are transcribed
trp operon	repressor binds tryptophan corepressor	active repressor occupies operator	tryptophan-synthesis genes are repressed

Gene regulation is specific because the responding cell must contain the matching receptor or regulator and the target DNA sequence. The same external condition therefore does not activate every gene or every cell.

[ 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.

[ 7 ]

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

[ 1 ]

Which statement correctly describes genome and proteome?

[ 2 ]

Explain how methylation of nucleosomes affects DNA transcription.

[ 3 ]

Arabidopsis is a small flowering plant in the mustard family (Brassicaceae) that is widely used in basic research. It has a short life cycle, flowers quickly producing a large number of seeds and is easy to cultivate. It forms a circle of leaves known as a rosette that lies close to the soil. Flowers form at the end of short stems.

A study was carried out of differences in development between Arabidopsis plants grown in long days ( 16 hours light, 8 hours dark) or short days ( 8 hours light, 16 hours dark). The sixth leaf (L6) to emerge in the rosette of each plant was used in all investigations.

New leaves are initiated by the meristem and go through four stages as they develop.
- Stage 1 (S1) - rapid cell division
- Stage 2 (S2) - cell division has ceased, cell expansion continues
- Stage 3 (S3) - decreasing cell expansion rate
- Stage 4 (S4) - leaf growth complete

The start of each stage of leaf development for plants grown in long days and short days is shown above the first graph.

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.

[ 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?

[ 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.

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.

Solvation is the interaction of a solvent with dissolved particles. In water, it produces an organized shell of water molecules around an ion or polar molecule.

Around a positive ion, the partially negative oxygen ends point inward.
Around a negative ion, the partially positive hydrogen ends point inward.
Around a polar molecule, water forms hydrogen bonds with exposed partial charges.

These attractions separate and stabilize solute particles, allowing them to remain dispersed in solution.

Polar water molecules orient around an ion and a polar solute to form hydration shells.

Hydrogen bonding and ion–dipole attraction hold some water molecules in hydration shells, reducing their freedom to move compared with water in a more dilute solution.

Solution	Solute particles	Water associated with solute	Tendency of water to leave
more dilute	fewer	smaller fraction	greater
more concentrated	more	larger fraction	lower

This molecular difference explains why water tends to move from a dilute solution toward a more concentrated solution when the two are connected through a membrane permeable to water.

The solute does not pull every water molecule across. Net movement results from unequal random movement in the two directions.

Osmosis is the net movement of water across a selectively permeable membrane from the side with lower solute concentration to the side with higher solute concentration.

The definition requires all three ideas:

the moving substance is water
the two sides differ in solute concentration
the membrane allows water to cross but restricts at least some solute

Say that water moves by osmosis. Osmosis is the process, not a verb performed by water.

Water crosses a partially permeable membrane from a hypotonic solution toward a hypertonic solution.

Tonicity is relative: a solution is described as hypotonic, isotonic or hypertonic to a particular cell or reference solution.

External solution relative to cytoplasm	External solute concentration	Net water movement
hypotonic	lower	into the cell
isotonic	equal	none overall
hypertonic	higher	out of the cell

To predict direction, compare solute concentration on the two sides first; then point the net water movement toward the side with the higher solute concentration.

Hypotonic does not mean that the cell contains little water. It describes the external solution's solute concentration relative to the cell.

In an isotonic condition, water crosses the membrane in both directions at equal rates, so there is no net osmosis and no sustained change in cell volume.

Molecular level	Whole-cell result
water molecules continue moving randomly both ways	inflow equals outflow
individual crossings still occur	average cell volume remains stable

“No net movement” does not mean “no movement.” It means that the two opposing rates balance.

more dissolved particles → more water associated in hydration shells → lower freedom and movement tendency of water → net water movement across a selective membrane toward that side

For any cell diagram:

identify inside and outside
compare solute concentration
confirm that water can cross the membrane
draw net water movement toward the higher solute concentration
if concentrations are equal, state dynamic equilibrium

Use	Avoid
water moves by osmosis	water osmosises
no net osmosis	no water movement
external solution is hypertonic to the cell	the cell is hypertonic, without a reference

A plant-tissue osmosis investigation asks how bathing-solution concentration changes tissue mass or length while other influential variables are controlled.

Prepare: make a range of accurately measured sucrose concentrations; cut tissue from the same plant source to equal starting dimensions; record each sample's initial mass or length.

Expose: immerse replicate samples at every concentration for the same time and at the same temperature.

Measure: remove each sample, blot it with the same pressure and duration, then record final mass or length.

Control or design choice	Why it matters
same tissue source and starting dimensions	limits biological and surface-area variation
same solution volume, time and temperature	keeps exposure and movement rate comparable
uniform blotting	prevents surface liquid from being counted as tissue mass
replicates at every concentration	reveals variation and supports a reliable mean

percentage change = ((final value − initial value) ÷ initial value) × 100

A potato cylinder changes from 4.00 g to 4.36 g.

change = 4.36 − 4.00 = 0.36 g

percentage change = (0.36 ÷ 4.00) × 100 = +9.0%

Result	Interpretation
positive percentage	tissue gained water overall
zero percentage	no detected net change
negative percentage	tissue lost water overall

Plot mean percentage change on the y-axis against sucrose concentration on the x-axis. Percentage change normalizes samples that did not begin with exactly the same mass or length.

The sucrose concentration where the fitted relationship crosses 0% change estimates the concentration isotonic with the tissue: water still moves, but there is no net osmosis.

Judge the estimate using the data, not only the drawn line:

use replicate means rather than one sample
inspect spread or error bars
interpolate only within the measured concentration range
report appropriate precision rather than extra decimal places

A zero mean can hide variation among samples. It estimates the tissue's isotonic point under these conditions; it is not proof that every cell had identical concentration.

Mean percentage change in plant-tissue mass falls as sucrose concentration increases and crosses zero at the estimated isotonic concentration.

External condition	Net water movement	Animal-cell outcome
hypotonic	into cell	swelling; possible lysis if membrane tension becomes too high
isotonic	balanced	stable average volume
hypertonic	out of cell	shrinking and crenation

The plasma membrane can stretch only a limited amount and provides little resistance to internal hydrostatic pressure. Without a rigid wall, continued water entry can rupture it.

A wall-less cell swells and may lyse in a hypotonic solution, crenates in a hypertonic solution, while a freshwater protist expels excess water.

Freshwater is usually hypotonic to a protist's cytoplasm, so water enters continuously by osmosis even when the organism is functioning normally.

Entry: water moves into the cell by osmosis because freshwater has higher water potential than the cytoplasm.

Collection: excess water enters the contractile-vacuole system and the vacuole fills.

Expulsion and reset: the filled vacuole connects to the cell surface and contracts, releasing water outside; the collection cycle then begins again.

Osmotic entry is passive, but vacuole filling and expulsion require cellular work. The vacuole does not stop osmosis; it balances its consequence by active water removal.

External condition	Water and pressure change	Cell state
hypotonic	water enters; vacuole expands; wall pressure rises	turgid
isotonic or mildly water-losing	little turgor pressure	flaccid
strongly hypertonic	water leaves; protoplast shrinks from wall	plasmolysed

The cellulose wall limits expansion. Its resistance creates turgor pressure, protects the membrane from bursting and helps support non-woody plant tissue.

During plasmolysis the plasma membrane and cytoplasm withdraw from the wall; the cell wall itself does not shrink away.

Plant cells are shown as turgid, flaccid and plasmolysed, with changes in vacuole volume and membrane position.

External condition	Cell without wall	Cell with wall
hypotonic	swells; may lyse	becomes turgid as wall resists expansion
isotonic	stable average volume	commonly flaccid because little pressure develops
hypertonic	shrinks or crenates	loses turgor; may plasmolyse

The direction of net water movement follows the same concentration difference in both cell types. The difference is the mechanical response after water moves.

Use lysis and crenation for wall-less animal cells; use turgid, flaccid and plasmolysed for plant cells.

Application	Why isotonicity matters
intravenous replacement fluid	avoids widespread red-cell swelling, lysis or crenation
solution carrying an injected medicine	prevents the solvent from imposing a harmful osmotic gradient
transplant-organ bath	limits cell-volume damage while the organ is outside the body

The solution should match the tissue's effective osmotic concentration, not reproduce every molecule in cytoplasm. Sterile 0.9% sodium chloride is commonly used because it is approximately isotonic with human extracellular fluid.

Isotonic does not mean chemically identical. It means that the solution does not cause harmful sustained net water gain or loss by the cells.

Red blood cells swell, remain stable or crenate in hypotonic, isotonic or hypertonic fluids, and a transplant organ is protected in an isotonic bath.

solution concentration → net water movement → change in tissue mass, length, cell volume or pressure → observable cell or tissue outcome

Evidence	What it supports
positive plant-tissue percentage change	bathing solution was hypotonic to the tissue
zero-change graph intercept	estimated isotonic concentration
animal-cell lysis or crenation	large water gain or loss without wall resistance
plant-cell turgor or plasmolysis	water gain or loss modified by wall mechanics
stable cells in a medical bath	solution is effectively isotonic with the tissue

First determine where water moves; then use cell structure to predict the consequence. The same osmotic rule explains a graph intercept, a red blood cell, a plant cell and the choice of a medical fluid.

Water potential (Ψw) is the potential energy of water per unit volume, expressed in pressure units, usually kilopascals (kPa). It indicates the tendency of water to move.

Pure water at standard atmospheric pressure and 20°C is assigned Ψw = 0 kPa. Dissolved solute lowers water potential, so an unpressurized solution normally has a negative value.

A value closer to zero is higher. For example, −100 kPa is higher water potential than −500 kPa.

Pure water is placed at zero kilopascals and increasingly concentrated solutions at progressively lower water potentials.

When water can cross between two regions, net movement is from the region with higher water potential to the region with lower water potential.

Case	Side A	Side B	Higher Ψw	Net water movement
1	0 kPa	−300 kPa	A	A → B
2	−100 kPa	−500 kPa	A	A → B
3	−700 kPa	−250 kPa	B	B → A
4	−400 kPa	−400 kPa	equal	no net movement

Compare the signed numbers on a number line. The value farther to the right—closer to zero—is higher, even when both values are negative.

Solutes lower Ψw because water–solute interactions reduce the potential energy and movement freedom of water relative to pure water.

Ψw = Ψs + Ψp

water potential = solute potential + pressure potential

Component	Physical cause	Usual sign	Effect when magnitude increases
solute potential, Ψs	dissolved particles interacting with water	0 or negative	more solute makes Ψs more negative
pressure potential, Ψp	hydrostatic pressure on water	often positive in a turgid walled cell	more pressure raises Ψw

Pressure potential is not always positive: tension in xylem can create negative Ψp. Use the stated biological context rather than memorizing one sign for every system.

The water-potential equation is paired with solute lowering potential and wall pressure raising potential in a plant cell.

A plant cell has Ψs = −700 kPa and Ψp = +300 kPa. It is placed in a solution with Ψs = −250 kPa and Ψp = 0 kPa. Predict the initial net movement of water.

Cell: Ψw = −700 + 300 = −400 kPa.
Solution: Ψw = −250 + 0 = −250 kPa.
−250 kPa is higher than −400 kPa.
Water therefore moves from the solution into the cell.

A complete answer shows both totals with units, compares the signed values and states the direction. Comparing Ψs alone would ignore the cell's pressure potential and can give the wrong conclusion.

In a hypotonic external solution:

water enters → vacuole expands → cell contents press harder on the wall → Ψp rises; solutes are diluted → Ψs becomes less negative; total Ψw rises

In a hypertonic external solution:

water leaves → cell pressure falls → Ψp decreases; remaining solutes become more concentrated → Ψs becomes more negative; total Ψw falls

Water movement changes solute concentration, not necessarily the number of solute particles. The amount of water in the cell has changed.

Water entry expands a plant-cell vacuole and raises pressure, while water exit lowers pressure and concentrates solutes.

Net water movement changes the receiving or losing cell in ways that reduce the original water-potential difference.

Initial condition	Changes inside plant cell	How the gradient is reduced
external Ψw is higher	water enters; Ψp rises and Ψs becomes less negative	cell Ψw rises toward external Ψw
external Ψw is lower	water leaves; Ψp falls and Ψs becomes more negative	cell Ψw falls toward external Ψw

A turgid cell can reach equilibrium because positive wall pressure raises its Ψw.
In a strongly hypertonic solution, pressure can fall to approximately zero and the protoplast may plasmolyse.
At equal Ψw, water still crosses both ways but net movement is zero.

This is why total Ψw—not solute concentration alone—is needed for walled cells: pressure develops as part of the response and changes the final equilibrium.

solute interactions and pressure set Ψw → water crosses a selective membrane from higher Ψw to lower Ψw → water movement changes cell volume, Ψs and Ψp → the gradient decreases toward equilibrium

For an HL prediction:

calculate or identify total Ψw on both sides
compare signed values correctly
move water from higher Ψw to lower Ψw
predict how cell Ψs and Ψp will change
state the structural outcome and the equilibrium condition

Qualitative language	Water-potential language
hypotonic external solution	external Ψw is higher than cell Ψw
hypertonic external solution	external Ψw is lower than cell Ψw
isotonic / no net osmosis	total Ψw is equal across the membrane

Solute concentration predicts many simple cases, but total water potential is the general rule. In walled cells, pressure is part of both the response and the final balance.

[ 2 ]

An experiment was carried out to determine the change in mass due to osmosis. Potato cubes were soaked in different concentrations of sodium chloride for 12 hours. The table shows the mean data for the mass of potato before and after the 12-hour period.

	Mean mass / g ( $\boldsymbol{\pm} \mathbf{0 . 0 1 ~ g} \boldsymbol{)}$
Sodium chloride concentration / mol dm ${ }^{-3}$	Original	Final
0.00	1.02	1.37
0.10	1.00	1.25
0.20	1.01	1.08
0.30	1.04	1.05
0.40	0.98	0.80
0.50	1.03	0.60
0.60	1.00	0.34

Outline the conditions necessary for osmosis to occur.

[ 3 ]

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

[ 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.

[ 3 ]

The table shows the percentage of red blood cells that have ruptured (hemolysis) when placed in sodium chloride solutions of different molar concentrations.

NaCl molarity / M	Hemolysis / \%
0.000	100
0.075	67
0.150	2
0.460	2

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

[ 7 ]

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

[ 4 ]

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

[ 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?

\Psi_{\mathrm{w}}=\Psi_{\mathrm{s}}+\Psi_{\mathrm{p}}

D3.1 Reproduction

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

Feature	Asexual reproduction	Sexual reproduction
gametes and fertilization	absent	present
main nuclear divisions	mitosis	meiosis makes gametes; mitosis grows offspring
parentage	usually one parent	genetic contribution from two gametes
genetic outcome	clones, except for new mutations	new allele combinations

Asexual: binary fission in Amoeba, budding in yeast or Hydra, runners in strawberry, bulbs and tubers.
Sexual: fusion of sperm and egg in animals, or male and female gametes delivered through pollen and ovules in flowering plants.

Asexual reproduction can increase numbers rapidly when a successful genotype suits stable conditions. Sexual reproduction is slower and depends on gamete transfer, but variation increases the chance that some offspring suit changed conditions.

Clones are not guaranteed to remain absolutely identical: mutation and environmental effects can still create differences.

diploid germ cell (2n) → meiosis → haploid gametes (n) → fertilization → diploid zygote (2n) → mitosis → multicellular diploid organism

Meiosis halves the chromosome-set number before gametes fuse. Fertilization then adds one haploid set from each gamete, restoring diploidy rather than doubling the species' chromosome number every generation.

Haploid means one chromosome set and diploid means two sets. The human counts 23 and 46 are examples, not definitions for every species.

A diploid organism makes haploid gametes by meiosis, two gametes fuse into a diploid zygote, and mitosis produces the next diploid organism.

Event	What is randomized	Result
crossing over in meiosis I	DNA segments exchanged between homologous non-sister chromatids	recombinant chromatids
independent assortment	maternal and paternal homologues sent to gametes in different combinations	many chromosome combinations
random fertilization	which genetically distinct sperm and egg fuse	a unique zygote combination

Meiosis creates genetically varied haploid gametes; fertilization combines two of those independently varied genomes. Mutation can add new alleles, but recombination mainly rearranges alleles already present in the parents.

Variation is a population advantage under changing conditions, not a promise that every individual offspring will be better adapted.

In anisogamous species, the sex producing small gametes is defined as male and the sex producing large gametes as female.

Gamete feature	Sperm / male gamete	Egg / female gamete
size	small	large
number produced	very many	relatively few
usual movement	motile or carried to the egg	usually non-motile
cytoplasm and reserves	little	abundant, supporting early development
shared requirement	haploid nucleus capable of fusion	haploid nucleus capable of fusion

Producing many small gametes increases the chance that at least one reaches an egg. Investing more cytoplasm and resources in each egg supports the zygote before it can obtain resources independently.

Sex is defined here by gamete type, not by stereotypes about whole-organism behaviour or parental care.

seminiferous tubules in testes → epididymis → sperm duct → fluids added by seminal vesicles and prostate → urethra → penis

Structure	Essential function
testes in scrotum	make sperm and testosterone at a temperature below body core
epididymis	completes sperm maturation and stores sperm
sperm duct	transports sperm during ejaculation
seminal vesicles and prostate	add nutrient-containing, alkaline seminal fluid
urethra and penis	conduct and deliver semen outside the body

Semen is sperm plus glandular fluid. It is not produced entirely in the testes, and semen production is not the same process as spermatogenesis.

Paired male and female human reproductive-system diagrams label the main organs and connecting ducts.

ovary releases secondary oocyte → oviduct receives and transports it; fertilization usually occurs in the upper oviduct → embryo reaches uterus → implantation occurs in endometrium → cervix and vagina form the exit route at birth

Structure	Essential function
ovary	produces oocytes, oestradiol and progesterone
oviduct	transports oocyte or embryo; usual fertilization site
uterus	muscular organ supporting embryo and foetus
endometrium	vascular lining for implantation and early support
cervix	muscular neck between uterus and vagina
vagina and vulva	canal and external structures for intercourse and birth

Draw the oviduct opening into the uterine cavity, not ending inside the uterine wall. The ovary lies near the funnel-shaped end but is not directly joined by a sealed tube.

Approximate time	Ovary	Dominant signal	Endometrium
days 1–5	new follicles begin developing	FSH rises	menstruation
days 5–13	one follicle becomes dominant	rising oestradiol	rebuilds
around day 14	ovulation	LH surge	remains prepared
days 15–28	corpus luteum active, then degenerates if no pregnancy	progesterone rises, then falls	maintained, then shed

Twenty-eight days and day 14 are useful averages, not fixed dates for every person or every cycle. Interpret event order and hormone relationships before relying on a calendar number.

A 28-day timeline aligns FSH, LH, oestradiol and progesterone curves with follicle growth, ovulation, corpus luteum and endometrial change.

FSH from anterior pituitary → follicle growth → follicle secretes oestradiol → endometrium proliferates

Oestradiol condition	Feedback effect	Functional consequence
rising during follicle development	supports further follicle responsiveness and oestradiol production	one follicle becomes dominant
high before mid-cycle	suppresses FSH but stimulates strong LH release	limits additional follicles and prepares ovulation

FSH primarily stimulates follicle development. LH is not a substitute name for FSH; its surge triggers the later ovulation transition.

high oestradiol → LH surge → ovulation → ruptured follicle becomes corpus luteum

corpus luteum → progesterone plus oestradiol → endometrium maintained and FSH/LH inhibited

If pregnancy does not begin: corpus luteum degenerates → progesterone and oestradiol fall → endometrium is no longer maintained → menstruation; loss of inhibition allows FSH to rise for a new cycle.

Ovulation is caused by the LH surge. Menstruation is caused later by loss of ovarian steroid support; the two events are not simultaneous.

Human fertilization normally begins in the upper oviduct when a sperm reaches the secondary oocyte and their plasma membranes fuse.

sperm reaches and penetrates oocyte coverings → sperm and oocyte membranes fuse → sperm nucleus enters while most sperm mitochondria and tail remain outside → oocyte completes meiosis II → male and female haploid nuclear material forms the diploid zygote genome

The zygote receives one homolog of each chromosome pair from each parent. Nearly all cytoplasmic organelles, including mitochondria, come from the egg.

Fertilization occurs in the oviduct; implantation occurs later in the uterus. These are separate events at different locations.

Control and stimulation: suppress the natural cycle when the protocol requires it → inject FSH to develop multiple follicles → give an hCG trigger to coordinate final oocyte maturation.

Collection and culture: retrieve oocytes → combine with prepared sperm or inject one sperm → culture fertilized embryos and assess early development.

Transfer and support: place selected embryo or embryos in the uterus → provide progesterone support to maintain a receptive endometrium → test for pregnancy.

Transferring more embryos can increase the chance that one implants but also increases multiple-pregnancy risk. Modern protocols balance success with maternal and foetal safety.

meiosis halves chromosome-set number and reshuffles alleles → specialized haploid gametes form → random fertilization restores diploidy → mitosis grows the genetically unique zygote

Transition	Main location	Key control or structure
sperm production and maturation	testis → epididymis	FSH/testosterone; seminiferous tubules
follicle growth and ovulation	ovary	FSH, oestradiol and LH surge
fertilization	upper oviduct	gamete encounter and membrane fusion
implantation preparation	endometrium	oestradiol and progesterone
IVF intervention	clinic/laboratory → uterus	controlled stimulation, retrieval, culture and transfer

When explaining a reproductive event, name the chromosome state, the anatomical location and the controlling signal. Mixing these three levels is the main source of fragmented answers.

pollen transferred from anther to receptive stigma (pollination) → compatible pollen germinates → pollen tube grows through style → male nuclei enter ovule through micropyle → one male nucleus fuses with egg nucleus (fertilization) → diploid zygote develops into embryo

Before fertilization	After fertilization
ovule contains female gamete	ovule develops into seed
ovary contains one or more ovules	ovary develops into fruit
egg nucleus is haploid	zygote nucleus is diploid

Pollination ends at the stigma. Pollen does not travel intact to the ovary; a pollen tube carries male nuclei from the stigma through the style.

A pollen grain germinates on the stigma and grows a pollen tube through the style into an ovule where male and female nuclei fuse.

Whorl or organ	Parts	Reproductive function
sepals	calyx	protect the developing flower bud
petals	corolla	advertise to animal pollinators
stamen	anther + filament	produce and present pollen
carpel	stigma + style + ovary	receive pollen, guide tube and contain ovules

Anther produces pollen → filament presents anther → stigma receives pollen → style provides pollen-tube path → ovary contains ovules where fertilization occurs.

A labelled insect-pollinated flower shows petals, nectary, anther, filament, stigma, style, ovary and ovules along a pollinator's contact route.

Feature	Effect on pollination
conspicuous petals and scent	attract insects from a distance
nectar	rewards entry and repeated visits
anthers placed on the route	dust visiting insect with pollen
sticky or sculptured pollen	adheres to animal body
stigma placed on the route	contacts pollen carried from another flower
sticky stigma	retains deposited pollen

The features work as a system: attraction brings the insect, reward draws it into a predictable position, and flower geometry makes its body contact anthers or stigmas.

The plant gains gamete transfer; the pollinator gains nectar or pollen food. This mutualism increases transfer efficiency but does not guarantee that every visit causes cross-pollination.

Cross-pollination transfers pollen from an anther on one plant to a stigma on a different plant of the same species, increasing opportunities to combine different parental alleles.

Mechanism	How self-pollination is reduced
anthers and stigma mature at different times	self-pollen is absent when the stigma is receptive
anthers and stigma are at different heights	a visitor contacts them on different body regions
separate male and female flowers or plants	one flower or plant cannot supply both gametes
pollinator movement or wind	carries pollen between individuals

Cross-pollination can increase genetic variation, but it also depends on pollen reaching a compatible plant. The mechanism reduces selfing; it does not ensure successful fertilization.

Self-incompatibility is a genetically controlled recognition system in which a carpel rejects pollen carrying an incompatible self identity.

pollen lands on stigma → pollen and carpel recognition alleles or proteins are compared → incompatible self pollen fails to hydrate, germinate or extend its tube → male nucleus cannot reach ovule → self-fertilization is prevented

Rejecting genetically similar pollen promotes outcrossing, reduces inbreeding and lowers the chance that identical harmful recessive alleles meet in offspring.

Self-incompatibility acts after pollen arrives. It is different from temporal or spatial separation, which reduces the chance that self pollen reaches the stigma.

Flower structure	Mature product
zygote	embryo root, shoot and cotyledon or cotyledons
fertilized ovule	seed containing embryo and food reserve
ovule wall	protective seed coat or testa
ovary	fruit surrounding or carrying seeds

Wind, water, animals or explosive fruit mechanisms carry seeds away from the parent. Separation reduces competition for light, water, minerals and space and can colonize new sites.

Pollination moves pollen before fertilization. Seed dispersal moves a fertilized, developing offspring after seed formation.

Seeds move away from a parent plant and later take up water, mobilize food reserves and emerge as seedlings.

Dormancy is an internal block to growth in a viable seed; germination is the resumed growth of the embryo using stored food.

Dormancy may persist because the embryo is immature, inhibitory abscisic acid remains, the testa is impermeable, or a species-specific chilling period has not occurred.

External condition	Why germination needs it
water	rehydrates cells, softens testa and enables metabolic reactions
oxygen	supports aerobic respiration and ATP production
suitable temperature	permits enzyme-catalysed reserve mobilization and growth

Light is required for germination in some species but not all. Water, oxygen and a suitable temperature are the general conditions emphasized here.

water uptake activates embryo → embryo releases gibberellin → signal reaches food-storage tissue → hydrolytic enzymes such as amylase are produced → starch becomes maltose and then glucose; proteins become amino acids → soluble products move to growing root and shoot

Mobilized product	Immediate use
glucose	respiration for ATP and carbon skeletons for biosynthesis
amino acids	synthesis of enzymes and structural proteins

The seedling depends on stored reserves until leaves expand and photosynthesis supplies enough organic material for independent growth.

pollen production → transfer to stigma → compatibility check → pollen-tube delivery → gamete fusion in ovule → seed and fruit formation → dispersal → dormancy release → germination and reserve mobilization

Transition problem	Adaptation
pollen must reach a conspecific stigma	pollinator attraction, wind transfer and flower geometry
selfing can reduce variation	timing, spatial separation, separate sexes and self-incompatibility
offspring compete with parent	seed dispersal
growth begins in poor conditions	dormancy and environmental germination requirements

Pollination is gamete delivery, fertilization is nuclear fusion, and dispersal moves the resulting seed. Keep those transitions separate while explaining how each supports successful outcrossing and establishment.

hypothalamus increases pulsatile GnRH → anterior pituitary releases FSH and LH → gonads increase gametogenesis and steroid-hormone secretion → reproductive organs mature and secondary sexual characteristics develop

Shared level	Testis-dominant response	Ovary-dominant response
pituitary gonadotropins	FSH supports sperm production; LH stimulates testosterone secretion	FSH supports follicle development; LH supports ovulation and ovarian steroid secretion
gonadal steroids	testosterone supports reproductive maturation and male-typical secondary traits	oestradiol and progesterone support reproductive maturation and female-typical cyclic changes

Sex steroids also influence bone growth, muscle protein synthesis and changes in body composition. Puberty is therefore a coordinated endocrine transition, not a single anatomical event.

FSH and LH operate in all sexes; their names reflect how they were first studied, not hormones found only in ovaries.

Spermatogenesis begins at puberty in the germinal epithelium of seminiferous tubules and continues toward the tubule lumen with support from Sertoli cells.

Cell stage	Division or change	Ploidy / product
spermatogonium	mitosis and growth	primary spermatocyte, 2n
primary spermatocyte	meiosis I	two secondary spermatocytes, n
secondary spermatocytes	meiosis II	four spermatids, n
spermatids	differentiation, no further division	four mature sperm, n

Sertoli cells nourish and organize developing germ cells. Interstitial cells between tubules secrete testosterone; the epididymis completes sperm maturation and stores the cells after they leave the testis.

Spermatids are the products of meiosis; sperm are the differentiated cells. Adding seminal-vesicle and prostate fluid later produces semen, not more sperm.

Oogenesis begins before birth. Primary oocytes remain arrested in follicles until cycles after puberty recruit a small group for further development.

oogonium divides by mitosis and grows → primary oocyte begins meiosis → meiosis I completes with unequal cytokinesis → large secondary oocyte plus first polar body → secondary oocyte begins meiosis II and is ovulated while arrested → sperm entry triggers completion → ovum plus second polar body

Unequal cytokinesis keeps almost all cytoplasm, organelles and stored materials in one cell while polar bodies remove extra chromosome sets. The products are genetically haploid but not equal in developmental capacity.

The cell released at ovulation is normally a secondary oocyte, not yet a completed ovum. Meiosis II finishes only if fertilization begins.

Dimension	Spermatogenesis	Oogenesis
begins	puberty	before birth, resumes after puberty
production pattern	continuous after puberty	cyclic recruitment from finite primary-oocyte pool
cytokinesis	approximately equal	highly unequal
products per primary cell	four functional sperm	one functional ovum plus polar bodies
meiosis II	completes before sperm differentiation	completes only after sperm entry

Both begin with diploid germ cells, include a multiplication and growth phase, use meiosis I to separate homologues and meiosis II to separate sister chromatids, and produce haploid nuclei.

The contrast explains anisogamy: sperm production maximizes the number of motile gametes, whereas oogenesis concentrates resources into one large cell capable of supporting the earliest embryo.

Parallel spermatogenesis and oogenesis routes track mitosis, meiosis I, meiosis II, four sperm and one ovum with polar bodies.

GnRH → pituitary FSH and LH → gonadal steroid secretion plus gametogenesis → sexually mature reproductive systems

Question	Sperm route	Oocyte route
where?	seminiferous tubules	ovarian follicles
when?	continuous from puberty	begins before birth; resumes cyclically
how many functional products?	four per primary spermatocyte	one per primary oocyte
why different?	maximize delivery probability	concentrate cytoplasm and reserves

Track cell name, chromosome-set number, cytoplasm and timing together. Those four dimensions distinguish the routes more reliably than memorizing isolated stage lists.

sperm passes between follicle cells → sperm surface interacts with zona pellucida → acrosomal membrane releases hydrolytic enzymes → enzymes digest a local path through zona → sperm reaches and fuses with oocyte plasma membrane

The acrosome is a membrane-bound enzyme compartment at the sperm head. Its reaction is localized: it enables one sperm to cross the extracellular coat rather than dissolving the entire zona pellucida.

The acrosome reaction permits penetration. It does not itself prevent polyspermy; the block is triggered after sperm–oocyte membrane fusion.

A sperm acrosome releases enzymes to cross the zona pellucida, then sperm–oocyte fusion triggers cortical granule release that blocks other sperm.

Fusion of the first sperm and oocyte plasma membranes triggers a rapid calcium-mediated release of cortical-granule contents by exocytosis.

first membrane fusion → cortical granules release enzymes outside oocyte → zona-pellucida proteins are modified and hardened → additional sperm can no longer bind or penetrate effectively → only one paternal nucleus enters

If two sperm nuclei fused with the oocyte nucleus, the zygote would receive an abnormal extra chromosome set. Blocking later sperm therefore protects the diploid chromosome complement required for normal development.

The block must follow the first successful fusion quickly. A barrier acting before any fusion would prevent fertilization altogether.

zygote → 2-cell and 4-cell cleavage stages → compact morula → fluid-filled blastocyst

During cleavage, repeated mitoses partition the original zygote cytoplasm into progressively smaller blastomeres. Cell number rises, but the embryo has not yet gained substantial mass.

Cilia and smooth-muscle activity move the dividing embryo down the oviduct so that blastocyst formation and uterine arrival are coordinated.

A zygote divides through cleavage and morula stages to a blastocyst with inner cell mass and outer trophoblast before implantation.

Blastocyst region	Early role
inner cell mass	gives rise to the embryo proper and later foetus
outer trophoblast cells	attach to and invade endometrium; contribute to placenta
fluid-filled cavity	creates the blastocyst organization rather than a solid morula

about day 7: blastocyst contacts receptive endometrium → trophoblast attaches and invades → embryo embeds over the following days → maternal tissue and embryonic membranes begin forming the placenta

Implantation anchors the embryo and establishes access to endometrial nutrients before the mature placental exchange system is complete.

The blastocyst is not simply a larger morula. Its cells have reorganized into an inner mass, outer layer and cavity with different developmental roles.

Without a pregnancy signal, the corpus luteum degenerates, progesterone falls and the endometrium is shed. An implanted embryo must interrupt that normal reset.

trophoblast / early placenta secretes hCG → hCG maintains corpus luteum → corpus luteum continues progesterone secretion → endometrium remains thick and vascular → early pregnancy is sustained

As the placenta develops, it becomes an endocrine organ and increasingly supplies progesterone and oestradiol itself, so pregnancy no longer depends on the corpus luteum indefinitely.

The signal must be produced early because the non-pregnant cycle is already programmed to reduce luteal hormone support.

A lateral-flow pregnancy test detects hCG in urine with specific monoclonal antibodies and produces separate test and control lines.

urine moves by capillary action → hCG binds mobile coloured anti-hCG antibody → complex is captured by immobilized antibody at test line → coloured test line appears

excess mobile coloured antibody continues along strip → a different immobilized antibody captures it at control line → coloured control line confirms that liquid flowed and reagents functioned

Lines visible	Interpretation
control + test	hCG detected
control only	no hCG detected above test threshold
no control line	invalid test, regardless of test-line appearance

Placental feature	Exchange advantage
many branching villi and microvilli	large surface area
thin barrier between maternal blood space and foetal capillaries	short diffusion distance
continuous maternal and foetal blood flow	maintains concentration gradients
extensive foetal capillary network	rapidly carries exchanged substances away or toward surface

Maternal blood bathes the outside of villi; foetal blood remains inside villus capillaries connected to the umbilical cord. Substances cross the placental barrier, but the blood cells normally do not mix.

Separation reduces immune and pressure incompatibilities while still permitting efficient molecular exchange across a small distance.

Placental villi containing foetal vessels project into maternal blood spaces while the two blood supplies remain separated by a thin exchange barrier.

Material	Main direction	Typical mechanism or significance
oxygen	mother → foetus	diffusion down maintained gradient
carbon dioxide	foetus → mother	diffusion
glucose	mother → foetus	facilitated diffusion
amino acids and some ions	mother → foetus	active transport contributes
water	both directions as gradients require	osmosis
urea and other wastes	foetus → mother	diffusion / transport for maternal excretion
maternal antibodies	mother → foetus	receptor-mediated transfer gives passive immunity

The placenta is a selective exchange organ, not a perfect shield. Many bacteria are excluded, but some viruses, drugs, alcohol and other small molecules can cross and affect development.

The foetus does not breathe air or eat directly. Maternal respiratory, digestive and excretory systems supply and remove materials through the placental interface.

Early pregnancy: embryonic hCG maintains corpus luteum → corpus luteum secretes progesterone and oestradiol → endometrium remains intact.

Later pregnancy: placenta becomes main source of progesterone and oestradiol → uterine lining and pregnancy conditions continue after corpus luteum declines.

Hormonal effect	Reproductive consequence
progesterone maintains endometrium and suppresses uterine contraction	implantation site remains supportive
progesterone and oestradiol inhibit FSH and LH	further follicle development and ovulation are suppressed

The source changes from ovary to placenta, but hormone output overlaps so support is not interrupted during the handover.

Near birth, reduced progesterone influence removes inhibition of uterine muscle, allowing oxytocin and prostaglandins to promote coordinated contractions.

uterine contractions push foetal head against cervix → cervix stretches → sensory signals reach hypothalamus → posterior pituitary releases more oxytocin → contractions become stronger and more frequent → cervix stretches further

The loop is positive because each response strengthens the original stimulus. Birth removes the foetal head from the cervix, ending stretch and therefore breaking the loop.

Positive feedback is useful here because labour needs a decisive endpoint. It is not the usual stabilizing negative feedback used to maintain a physiological variable near a set point.

During menopause, ovarian responses and secretion of oestradiol and progesterone decline, menstrual cycles cease and loss of steroid-hormone effects can produce symptoms.

Hormone replacement therapy supplies oestradiol, with a progestogen when protection of an intact endometrium is required, to reduce symptoms such as hot flushes and other effects of hormone loss.

Benefit and risk depend on the individual, hormone formulation, dose, route, timing and duration. The biology course uses HRT to examine evidence, not to prescribe one treatment for everyone.

Replacing hormones can reduce symptoms without restoring fertility or reversing ovarian ageing.

Study design	Observation	Main inference limit
early observational studies	HRT users had lower CHD incidence	users differed in socioeconomic status, health-service use and other risk factors
randomized controlled trials	treatment groups showed a small CHD-risk increase in the studied populations	result depends on participants, regimen and timing but better isolates treatment effect

A confounding variable is associated with both the exposure and the outcome. Higher socioeconomic status could increase HRT uptake while independently lowering coronary risk, creating a protective-looking correlation.

To claim causation, ask whether groups were comparable before treatment, whether allocation was randomized, whether outcome measurement was blinded, and whether the effect is consistent with biological mechanism and other trials.

Correlation can motivate a hypothesis, but it cannot by itself show that HRT caused the lower—or higher—disease rate.

acrosome opens a local path → cortical reaction closes entry to later sperm → cleavage builds cell number → blastocyst opens an interface with endometrium → hCG keeps hormonal support open → placenta maintains selective exchange → oxytocin feedback drives birth to completion

Transition	Enabling mechanism	Protective limit
fertilization	acrosomal penetration and membrane fusion	cortical polyspermy block
implantation	trophoblast attachment and invasion	organized endometrial interface
foetal exchange	thin, vascular placental villi	maternal and foetal blood remain separate
pregnancy to birth	progesterone support followed by oxytocin amplification	loop ends when cervical stretch ends
hormone therapy	replacement of lost signalling	evidence-based benefit–risk selection

Successful reproduction is not one event. Each stage depends on a signal or structure that enables the next transition while preventing an incorrect place, time, cell or causal conclusion.

[ 3 ]

Outline natural methods of cloning in some eukaryotes.

[ 4 ]

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

[ 6 ]

Draw a labeled diagram of the female reproductive system.

[ 8 ]

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

[ 6 ]

Describe the process of fertilization in humans.

[ 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.

[ 4 ]

Outline pollination, fertilization and seed dispersal.

[ 4 ]

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

[ 2 ]

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

[ 1 ]

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

[ 6 ]

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

[ 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)

[ 8 ]

Compare and contrast the processes of spermatogenesis and oogenesis.

[ 2 ]

Explain the mechanism that prevents polyspermy during fertilization.

[ 1 ]

What is a blastocyst?

[ 1 ]

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

[ 8 ]

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

[ 7 ]

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

D3.2 Inheritance

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

diploid parent (two homologous copies of each autosome) → meiosis → haploid gametes (one copy) → fertilization → diploid zygote (two copies again)

Cell	Chromosome sets	Alleles possible at one autosomal locus
body cell in a diploid parent	2n	two alleles, one on each homolog
gamete	n	one allele
zygote	2n	one allele from each parent

Two alleles does not mean two different alleles. A diploid genotype may be homozygous, with matching alleles, or heterozygous, with different alleles.

Term	Meaning	Example
gene	DNA sequence with a functional product or role	a gene affecting pea height
locus	the gene's position on a chromosome	the height-gene position
allele	one sequence version at that locus	T or t
genotype	allele combination carried	TT, Tt or tt

homozygous: two matching alleles, such as TT or tt
heterozygous: two different alleles, such as Tt

Use genotype for the allele combination, not for the visible trait. A genotype can be written for one locus, several loci or the whole genome, so state the scope when it matters.

A phenotype is an observable or measurable characteristic produced by the genotype, the environment, or an interaction between them.

Main cause	Example	Why
genotype	ABO blood group	inherited alleles determine A and B antigens
environment	an acquired scar	injury changes tissue without changing inherited alleles
genotype × environment	adult height	growth potential interacts with nutrition and health

The same phenotype can come from different genotypes, and one genotype can produce different phenotypes in different environments. Phenotype alone therefore does not always identify genotype.

Under complete dominance, a dominant allele determines the heterozygous phenotype. A recessive phenotype appears only when no dominant allele is present.

Genotype	Allele state	Phenotype when T is dominant
TT	homozygous dominant	tall
Tt	heterozygous	tall
tt	homozygous recessive	dwarf

Dominant does not mean better, stronger or more common. It only describes expression in a heterozygote; a harmful or rare allele can be dominant.

choose parents with known, contrasting phenotypes and preferably true-breeding genotypes
remove immature anthers from the recipient flower before they release pollen
transfer pollen from the chosen donor to the receptive stigma
prevent stray pollen from reaching the flower, then label the cross
collect seeds and score enough offspring to compare with a predicted ratio

Emasculation and isolation prevent self-pollination, so the experimenter knows the parental sources of both gametes. Reciprocal crosses can also test whether switching the pollen and ovule parents changes the result.

A controlled flowering-plant cross removes self-pollinating anthers, transfers pollen from a chosen parent, grows F1 offspring and crosses the F1 to obtain an F2 generation.

P generation: true-breeding tall TT × true-breeding dwarf tt → every gamete carries T or t, respectively.

F1 generation: every offspring is Tt and tall. The t allele has not blended away; its phenotypic effect is masked in the heterozygote.

F1 self-cross: Tt × Tt → F2 genotypes 1 TT : 2 Tt : 1 tt → phenotypes 3 tall : 1 dwarf in a large sample.

Reappearance of dwarf offspring shows that hereditary factors remain discrete across generations. The observed ratio approaches the probability prediction only when many offspring are counted.

define allele symbols and the dominance relationship
write both parental genotypes
derive the allele carried by each possible gamete
place one parent's gametes across the grid and the other's down the side
combine alleles in each cell, then convert genotypes to phenotypes

Tt × Tt	T gamete	t gamete
T gamete	TT, tall	Tt, tall
t gamete	Tt, tall	tt, dwarf

Each cell has probability 1/4 because each parent produces T and t gametes with probability 1/2. The grid predicts probabilities for each fertilization, not a guarantee that every four offspring will contain one of each cell.

A tall plant could be TT or Tt because both contain the dominant T allele. Cross the unknown plant with a homozygous recessive tester, tt, whose gametes reveal what the unknown parent contributes.

Unknown parent	Cross with tt	Expected offspring evidence
TT	TT × tt	all Tt, all tall
Tt	Tt × tt	about 1 Tt tall : 1 tt dwarf

A small all-tall sample does not prove the parent is TT because a heterozygote can produce tall offspring repeatedly by chance. Confidence increases as more offspring are scored.

Phenotypic plasticity is the capacity of one genotype to develop different phenotypes in different environments by changing patterns of gene expression.

environmental cue → signalling pathway → transcription or translation changes → different proteins, cell activity or growth → phenotype better matched to the experienced environment

Plastic response	Genetic evolution
genotype is unchanged within the individual	allele frequencies change across generations
can be reversible during a lifetime	inherited population change is not reversed by one individual's environment
response range itself can be inherited	selection acts on heritable differences

A suntan or training response is acquired plasticity; it does not mean the altered phenotype is encoded as a new allele and passed directly to offspring.

two loss-of-function PAH alleles → little or no phenylalanine hydroxylase → phenylalanine is not converted efficiently to tyrosine → phenylalanine and harmful derivatives accumulate → untreated neural development is damaged

Intervention	Why it helps
newborn screening	detects the metabolic problem before severe symptoms appear
early low-phenylalanine diet	limits substrate accumulation while still supplying controlled essential phenylalanine
continued monitoring	keeps blood concentrations within a safer range as diet and growth change

The genotype does not change with treatment, but the environment does. Dietary control can greatly change the phenotype, making PKU a clear genotype–environment interaction.

Genotype	PKU status	Reproductive meaning
PP	unaffected, not a carrier	passes P
Pp	unaffected carrier	can pass P or p
pp	affected	passes p

Pp × Pp	P gamete	p gamete
P gamete	PP unaffected	Pp carrier
p gamete	Pp carrier	pp affected

For each pregnancy: 1/4 affected, 1/2 unaffected carrier, 1/4 unaffected non-carrier. The probability resets for every fertilization; previous children do not change the next gamete combination.

Carrier and affected are not synonyms. A heterozygous carrier has a working dominant allele and usually does not show PKU, but can transmit the recessive allele.

A single-nucleotide polymorphism (SNP) is a common difference at one nucleotide position among genomes. A SNP within or near a gene can create or mark a different allele.

Scale	What may exist at one locus
one haploid gamete	one allele
one diploid individual	at most two alleles
a population gene pool	two, three or many alternative alleles

Multiple alleles means several versions of one gene in the population. It does not mean that one ordinary diploid person carries every version.

Phenotype	Possible genotype(s)	Red-cell antigen(s)
A	IᴬIᴬ or Iᴬi	A
B	IᴮIᴮ or Iᴮi	B
AB	IᴬIᴮ	A and B
O	ii	neither A nor B

Iᴬ and Iᴮ are codominant because both antigen products appear in IᴬIᴮ. Each is dominant to i, which produces neither A nor B antigen.

Blood group is a phenotype; Iᴬ, Iᴮ and i are alleles. A person with group A or B may hide an i allele, so phenotype does not always identify genotype.

Red blood cells show A antigen for blood group A, B antigen for group B, both antigens for AB and neither antigen for O, alongside the corresponding ABO genotypes.

A group-A parent could be IᴬIᴬ or Iᴬi, and a group-B parent could be IᴮIᴮ or Iᴮi. State the genotype assumption before predicting children.

Iᴬi × Iᴮi	Iᴮ gamete	i gamete
Iᴬ gamete	IᴬIᴮ, group AB	Iᴬi, group A
i gamete	Iᴮi, group B	ii, group O

With these heterozygous parents, A, B, AB and O phenotypes each have probability 1/4. If either parent is homozygous, the outcome changes.

ABO data can sometimes exclude a proposed parent–child relationship, but common blood groups cannot uniquely prove parentage because many people share the same phenotype and genotype.

Allele relationship	Heterozygote phenotype	Example
complete dominance	matches the dominant homozygote	Tt pea is tall
codominance	both allele products are detectably expressed	IᴬIᴮ has A and B antigens
incomplete dominance	intermediate between the two homozygotes	red × white Mirabilis gives pink F1

Heterozygote × heterozygote	Genotype ratio	Phenotype ratio
complete dominance	1:2:1	3:1
codominance or incomplete dominance	1:2:1	1:2:1 when all three genotypes are distinguishable

Codominance is not blending: both products remain identifiable. In incomplete dominance, the heterozygote is intermediate rather than showing two separate products side by side.

P generation: FᴿFᴿ red × FᵂFᵂ white → all FᴿFᵂ pink F1 offspring.

FᴿFᵂ × FᴿFᵂ	Fᴿ gamete	Fᵂ gamete
Fᴿ gamete	FᴿFᴿ red	FᴿFᵂ pink
Fᵂ gamete	FᴿFᵂ pink	FᵂFᵂ white

F2 genotype ratio = phenotype ratio = 1 red : 2 pink : 1 white because the heterozygote has its own recognizable phenotype.

Segregation has not changed: each heterozygote still makes the two gamete types equally. The different phenotypic ratio comes from how the heterozygous genotype is expressed.

homologous chromosomes carry two alleles in a diploid parent → meiosis separates one allele into each gamete → fertilization restores a two-allele genotype → allele relationship and environment shape phenotype → offspring data test the prediction

For any genetic cross:

define allele symbols and their relationship
write parental genotypes, not only phenotypes
derive gametes before drawing the grid
combine gametes, then translate genotype to phenotype
state probabilities and the assumptions behind them

If the simple 3:1 expectation fails	First question to ask
heterozygote has a third phenotype	incomplete dominance?
both products appear	codominance?
phenotype changes with conditions	environmental or plastic response?
more than two alleles occur in the population	multiple-allele locus such as ABO?

Gamete contribution	X-bearing sperm	Y-bearing sperm
X-bearing egg	XX zygote	XY zygote
expected probability	about 1/2	about 1/2

Y chromosome with functional SRY → testis-determining factor (TDF) is produced → embryonic gonads develop as testes → testicular hormones direct typical male reproductive development

Without a functional SRY signal, the undifferentiated gonads normally follow ovarian development and typical female reproductive structures develop.

XX/XY is the core human chromosomal model, but chromosome number, SRY location or function, hormone synthesis and tissue response can vary. Chromosomal pattern alone does not describe every aspect of sex or gender.

Most genes in the non-homologous region of X have no matching allele on Y. An XY individual is therefore hemizygous for those X-linked genes: the single allele on X is expressed even if it is recessive.

Genotype	Typical status for an X-linked recessive allele h
XᴴXᴴ	unaffected XX
XᴴXʰ	unaffected carrier XX
XʰXʰ	affected XX
XᴴY	unaffected XY
XʰY	affected XY

a son receives his X chromosome from his mother and Y from his father
a father passes his X chromosome to every daughter and to no son
an affected father cannot pass his X-linked allele directly to a son

Write the allele as a superscript on X, such as Xᴴ or Xʰ. Do not place the X-linked allele on Y when the gene is absent from the relevant Y region.

XᴴXʰ × XᴴY	Xᴴ sperm	Y sperm
Xᴴ egg	XᴴXᴴ unaffected daughter	XᴴY unaffected son
Xʰ egg	XᴴXʰ carrier daughter	XʰY affected son

Among daughters: 1/2 carriers and 1/2 non-carriers; none affected in this cross. Among sons: 1/2 affected and 1/2 unaffected. Across all offspring, each grid outcome has probability 1/4.

Haemophilia A and B result from deficient clotting factors. The inheritance pattern reflects an X-linked recessive allele; treatment supplies or supports the missing clotting function but does not alter the inherited allele.

A carrier female XH Xh crossed with an unaffected male XH Y produces unaffected daughters, carrier daughters, unaffected sons and sons with haemophilia.

Pedigree mark	Meaning
square / circle	male / female in the standard convention
filled symbol	individual shows the tracked phenotype
horizontal partner line	mating or reproductive partnership
vertical descent line	offspring connection
shared sibship line	siblings from the same parents
I, II, III …	generations
1, 2, 3 …	individuals within a generation

identify the phenotype being tracked and whether filled means affected
trace parents, siblings and generations accurately
mark genotypes that are forced by phenotype or parent–child transmission
keep uncertain alleles as alternatives until more evidence removes them

A pedigree records observed family evidence, not every conception or every relative. Small families and missing records can make more than one inheritance model possible.

Observation	Strong inference
two unaffected parents have an affected child	supports recessive inheritance; contradicts simple complete-dominant inheritance
affected person has an affected parent in every generation	supports dominance, but does not prove it alone
affected father has an affected son	contradicts X-linked transmission from father to son
affected sons arise from unaffected mothers	consistent with X-linked recessive carrier mothers
males and females affected similarly	supports autosomal inheritance

Start with forced genotypes. For an autosomal recessive trait, affected people are aa; each unaffected parent of an affected child must carry a. For an X-linked recessive trait, an affected son XʰY proves that his mother supplied Xʰ.

Treat each inheritance mode as a hypothesis. Predict transmissions that must or cannot occur, compare them with the pedigree, and reject a model only when a relationship contradicts it.

Skipping a generation suggests recessiveness but is not a proof. Chance, incomplete records, late onset and variable expression can obscure the apparent pattern.

Feature	Continuous variation	Discrete variation
possible values	many intermediate measurements	separate categories
common cause	polygenic effects plus environment	often one or few major loci
examples	height, body mass, skin pigmentation	ABO group, biological sex chromosome category
useful display	histogram or box plot	bar chart or counts table

Ask whether the variable is measured on a scale or counted in named categories. Height remains continuous even if a researcher later groups measurements into artificial height bands.

Continuous does not mean every imaginable value must occur in one sample, and discrete does not mean the categories are controlled by only one allele.

In polygenic inheritance, several loci contribute to one characteristic. Different combinations of small allele effects generate many genotypic values rather than a few Mendelian phenotype classes.

many contributing genotypes + nutrition, health, temperature or other environmental effects + genotype-specific responses to those conditions → overlapping phenotypes across a continuous range

Trait	Genetic contribution	Environmental contribution
height	many growth-related loci	nutrition, disease and developmental conditions
body mass	metabolism and appetite loci	diet, activity and health
skin pigmentation	several melanin-related loci	ultraviolet exposure changes melanin production

Do not confuse multiple genes contributing to one trait with multiple alleles at one gene locus. These are different levels of genetic variation.

order the measurements from lowest to highest
find the median
find Q1 as the median of the lower half and Q3 as the median of the upper half
calculate IQR = Q3 − Q1
draw the box from Q1 to Q3 and the median line inside
extend whiskers to the most extreme non-outlier values and plot outliers separately

A common rule marks values below Q1 − 1.5 × IQR or above Q3 + 1.5 × IQR as outliers. Check the convention used before treating the raw minimum and maximum as whisker ends.

A horizontal box-and-whisker plot labels minimum, lower quartile Q1, median, upper quartile Q3, maximum, interquartile range and an outlier.

Feature	What to compare
median line	typical central value
box length, IQR	spread of the middle 50%
whiskers	spread beyond the quartiles among non-outliers
isolated points	potential outliers requiring biological or measurement review
overlap between boxes	descriptive overlap, not automatically a significance test

If group A has a higher median but strongly overlapping boxes with group B, report both observations. Do not erase the overlap or claim a significant difference without an appropriate inferential test.

A box plot does not show sample size, separate peaks or every raw value unless these are added. Two very different distribution shapes can share the same median and quartiles.

X and Y contributions establish the usual chromosomal starting point → unequal X-linked gene dosage changes inheritance risk → pedigrees reveal transmission across families → polygenes and environment create continuous population variation → box plots summarize centre and spread

Evidence source	Reliable question
X-linked cross	which parent supplies the allele to sons or daughters?
pedigree	which inheritance hypotheses are contradicted by family relationships?
continuous measurements	where is the median and how wide is the middle 50%?
apparent outlier	biological extreme, data error or different process?

Match the claim to the evidence scale. One Punnett grid predicts a fertilization probability, a pedigree constrains family genotypes, and a box plot describes a sample distribution; none alone proves a universal biological rule.

Before meiosis, an Aa diploid cell carries A on one homologous chromosome and a at the same locus on the other homolog.

At anaphase I, homologous chromosomes move to opposite poles. A and a therefore enter different daughter cells; meiosis II separates sister chromatids without reuniting the alleles.

If meiotic products survive equally, an Aa parent makes approximately 1/2 A gametes and 1/2 a gametes. Random fertilization then restores two alleles in the zygote.

Mendel's law of segregation is therefore a chromosome-mechanics statement: paired alleles separate because their homologous chromosomes separate during meiosis.

Each bivalent can face either spindle pole at metaphase I. The orientation of the A/a pair does not determine the orientation of a B/b pair on a different chromosome, so maternal and paternal homologs enter daughter cells in different combinations.

Gene positions	Assortment expectation
on different chromosomes	independent
far apart on the same chromosome with recombination approaching 50%	behaves approximately independent
close together on the same chromosome	linked; parental combinations exceed recombinants

Independent assortment is not the random separation of sister chromatids. It concerns the relative orientation and separation of different homologous chromosome pairs in meiosis I.

Two unlinked homologous chromosome pairs orient in two equally likely ways at meiosis I, producing AB and ab gametes in one orientation or Ab and aB gametes in the other.

Every gamete from AaBb must contain one allele from the A/a locus and one from the B/b locus. For unlinked loci, multiply the independent probabilities.

A-locus choice	B-locus choice	Gamete	Probability
A, 1/2	B, 1/2	AB	1/4
A, 1/2	b, 1/2	Ab	1/4
a, 1/2	B, 1/2	aB	1/4
a, 1/2	b, 1/2	ab	1/4

A valid gamete list has one allele per locus, no duplicate types and probabilities summing to 1. Writing A, a, B and b as four separate gametes loses one gene from each gamete.

The same multiplication rule scales: an individual heterozygous at n independently assorting loci can make 2ⁿ gamete types, before considering linkage or unequal viability.

Assume both genes are autosomal and unlinked, complete dominance acts at each locus, both parents are AaBb, the four gamete types are equally frequent, fertilization is random and offspring classes survive equally.

Phenotype class	Probability calculation	Expected fraction
A_B_	3/4 × 3/4	9/16
A_bb	3/4 × 1/4	3/16
aaB_	1/4 × 3/4	3/16
aabb	1/4 × 1/4	1/16

The ratio is two independent 3:1 monohybrid probabilities multiplied together. A 4 × 4 grid lists all 16 genotype combinations; the product rule reaches the same phenotype expectation more efficiently.

An AaBb by AaBb dihybrid Punnett grid groups offspring into four phenotype classes in a 9:3:3:1 ratio and contrasts an AaBb by aabb test cross with a 1:1:1:1 ratio.

Cross the double heterozygote AaBb with aabb. The tester can contribute only ab, so each offspring genotype reveals which gamete came from the heterozygote.

AaBb gamete	Tester gamete	Offspring	Phenotype class
AB	ab	AaBb	both dominant
Ab	ab	Aabb	A dominant only
aB	ab	aaBb	B dominant only
ab	ab	aabb	both recessive

If A/a and B/b assort independently, AB, Ab, aB and ab gametes each occur with probability 1/4, so the test-cross phenotype expectation is 1:1:1:1.

A strong excess of two reciprocal classes and a shortage of the other two suggests linkage: parental chromosome combinations are being transmitted more often than recombinant combinations.

search an authoritative gene database with the gene name or symbol
confirm the organism is Homo sapiens and select the correct record
record chromosome, cytogenetic band or genomic coordinates from genomic context
record the named RNA and polypeptide product and its supported function
compare a second gene's chromosome and coordinates before calling the pair linked

Database field	Question it answers
gene symbol and stable identifier	which gene record?
chromosome and genomic coordinates	where is the locus?
transcript or product annotation	what is produced?
curated functional summary	what evidence supports its role?

Genes on different chromosomes are unlinked. Genes on the same chromosome are physically linked, but how often crossing over separates them depends on the distance between their loci.

A nearby genomic feature is not automatically the same gene or product. Record the database version, species, coordinates and identifier so the evidence can be checked again.

homologs separate alleles → independently oriented bivalents combine alleles from unlinked genes → AaBb makes AB, Ab, aB and ab gametes equally → random fertilization produces 9:3:3:1 in an F2 or 1:1:1:1 in a test cross

Prediction	Required conditions
1:1 allele segregation	normal meiosis and equal gamete contribution
four equal AaBb gametes	loci unlinked or recombination effectively 50%
9:3:3:1 F2	complete dominance, random fertilization and equal survival
1:1:1:1 test cross	double heterozygote × double recessive tester

A database locates the genes; the cross tests their behaviour. If observed offspring depart systematically from the unlinked expectation, inspect chromosome location, parental phase, crossing over and viability before blaming random sampling alone.

Same AaBb genotype	Alleles on one homolog	Alleles on the other	Name
AB/ab	A with B	a with b	coupling phase
Ab/aB	A with b	a with B	repulsion phase

Writing only AaBb records which alleles are present but loses which combinations share a chromosome. For linked genes, parental phase determines which gametes are common and which require crossing over.

Autosomal linkage means two loci share a non-sex chromosome. Sex linkage is different: it refers to a locus on a sex chromosome, usually X.

A diagram contrasts linked A and B loci on the same chromosome, unlinked loci on different chromosomes and crossing over that creates Ab and aB recombinant chromatids from AB and ab parental chromatids.

Start with a heterozygote in coupling phase: parental homologs carry AB and ab. After DNA replication, each homolog has two sister chromatids.

A chiasma between the A and B loci exchanges matching segments between non-sister chromatids. Two chromatids remain parental, while two become Ab and aB.

Chromatid or gamete class	Allele combination	Origin
parental	AB and ab	no exchange between the loci
recombinant	Ab and aB	crossover between the loci

A crossover elsewhere on the chromosome does not recombine these two loci. The exchange must occur between their positions to separate their parental allele combination.

Cross AB/ab with ab/ab. Because the tester contributes only ab, each offspring's phenotype or genotype identifies the gamete produced by the double heterozygote.

Heterozygote gamete	Offspring	Classification for AB/ab parent
AB	AaBb	parental
ab	aabb	parental
Ab	Aabb	recombinant
aB	aaBb	recombinant

two largest reciprocal classes identify the parental haplotypes
two smaller reciprocal classes identify recombinant gametes
an exact 1:1:1:1 pattern supports independent assortment, not strong linkage

Classify by comparison with the heterozygous parent's chromosome combinations, not by whether an offspring looks common or unusual in the population.

$r=\frac{R}{N}\times 100\%$

Here, R is the number of recombinant offspring and N is the total. If a test cross gives 420 AB, 400 ab, 90 Ab and 90 aB offspring, r = (90 + 90) / 1000 × 100% = 18%.

Observation	Inference
recombinants much below 50%	loci are linked
lower recombinant frequency	loci are usually closer, so fewer chiasmata fall between them
about 50% recombinants	loci behave as unlinked; they may be on different chromosomes or far apart

Recombinant frequency does not exceed 50% in an ordinary two-locus test. Multiple crossovers can restore parental combinations, so frequency underestimates large physical distances.

Statement	Meaning in a dihybrid goodness-of-fit test
null hypothesis, H₀	observed departures from the expected genetic ratio are due to chance sampling
alternative hypothesis, H₁	the departures are too large for that chance model; another explanation is needed

Convert the ratio to expected counts before calculating χ². For a total of 556 offspring and a 9:3:3:1 model: multiply 556 by 9/16, 3/16, 3/16 and 1/16 to obtain 312.75, 104.25, 104.25 and 34.75.

Categories must be mutually exclusive, observations should be independent and expected counts should be large enough for the approximation. The expected ratio must come from a biological model chosen before inspecting the deviations.

Chi-squared tests whether counts fit a specified expectation; it does not identify the true alternative mechanism when they do not fit.

$\chi^2=\sum \frac{(O-E)^2}{E}$

Category	O	E	(O − E)² / E
both dominant	315	312.75	0.016
A dominant only	108	104.25	0.135
B dominant only	101	104.25	0.101
both recessive	32	34.75	0.218
sum	556	556	χ² = 0.470

For four mutually exclusive categories with one fixed total and no parameters estimated from the data, degrees of freedom = categories − 1 = 3.

Use counts, not percentages, inside the formula. Keep expected values unrounded during calculation, and include every category contribution before comparing with a critical value.

A chi-squared table compares observed and expected Drosophila phenotype counts, sums category contributions to 0.47 and compares the result with the p equals 0.05 critical value for three degrees of freedom.

At p = 0.05	Statistical decision	Genetic conclusion
χ² ≤ critical value	fail to reject H₀	deviations are not significant; data are consistent with the expected ratio
χ² > critical value	reject H₀	deviations are significant; the stated ratio does not adequately explain the data

For df = 3, the p = 0.05 critical value is 7.815. Because 0.470 < 7.815, fail to reject H₀: the observed Drosophila counts are consistent with a 9:3:3:1 expectation.

Do not say the test proves H₀ or proves independent assortment. A non-significant result means the observed departure is small enough to be explained by sampling under the model.

A significant result directs further investigation. Possible causes include linkage, differential survival, non-random fertilization, scoring error or an incorrect dominance model; chi-squared alone does not choose among them.

meiosis predicts gamete combinations → chromosome loci determine whether genes assort independently or travel together → crossing over creates reciprocal recombinants → test-cross counts reveal parental phase and recombinant frequency → chi-squared tests whether departures from a chosen ratio exceed chance expectation

Offspring pattern	First model to examine
9:3:3:1 F2	two unlinked genes with complete dominance
1:1:1:1 test cross	four equally frequent gametes from an unlinked double heterozygote
two large and two small reciprocal test-cross classes	linked loci with crossing over
significant χ² departure	assumptions, linkage, viability, fertilization or scoring require review

Keep prediction and evidence separate: define the genetic model first, calculate expected counts from it, compare observations with the expectation, and change the model only when chromosome location or statistical evidence requires it.

[ 1 ]

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

[ 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 $F_{1}$ plants had purple flowers. The $F_{1}$ plants were self-pollinated to produce an $F_{2}$ generation. There were 97 plants with purple flowers and 38 plants with white flowers in the $F_{2}$ generation.

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

[ 1 ]

Define the term genotype.

[ 2 ]

Identify the phenotypes of each part of the phenotypic ratio.

Ratio	Phenotypes
9
3
3
1

[ 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.

[ 1 ]

Scientists incubated larvae of the moth Utetheisa ornatrix at either $15^{\circ} \mathrm{C}$ or $22^{\circ} \mathrm{C}$ 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?

[ 4 ]

Discuss the causes and treatments of phenylketonuria.

[ 1 ]

Which statement defines alleles?

[ 9 ]

Describe the inheritance of ABO blood groups.

[ 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?

[ 4 ]

Distinguish between autosomes and sex chromosomes in humans.

[ 8 ]

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

[ 2 ]

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

[ 7 ]

Explain the reasons for variation in human height.

[ 3 ]

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

[ 6 ]

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

[ 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.

[ 3 ]

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

[ 1 ]

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

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?

[ 2 ]

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

c^{2}=\Sigma \frac{(\text { Observed }- \text { Expected })^{2}}{\text { Expected }}=1002.6

	Probability
Degrees of freedom	0.995	0.975	0.20	0.10	0.05	0.025	0.02	0.01	0.005	0.002	0.001
1	0.00004	0.001	1.642	2.706	3.841	5.024	5.412	6.635	7.879	9.550	10.828
2	0.010	0.051	3.219	4.605	5.991	7.378	7.824	9.210	10.597	12.429	13.816
3	0.072	0.216	4.642	6.251	7.815	9.348	9.837	11.345	12.838	14.796	16.266
4	0.207	0.484	5.989	7.779	9.488	11.143	11.668	13.277	14.860	16.924	18.467
5	0.412	0.831	7.289	9.236	11.070	12.833	13.388	15.086	16.750	18.907	20.515
6	0.676	1.237	8.558	10.645	12.592	14.449	15.033	16.812	18.548	20.791	22.458
7	0.989	1.690	9.803	12.017	14.067	16.013	16.622	18.475	20.278	22.601	24.322

D3.3 Homeostasis

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

Homeostasis maintains selected variables of the internal environment within narrow limits despite internal and external change. The internal environment is blood plasma and tissue fluid—not the world outside the body.

Regulated variable	Why large deviation matters
core temperature	enzyme and membrane processes change rate
blood pH	protein shape and reaction rates change
blood glucose	cells lose a reliable respiratory substrate; osmotic effects rise when excessive
O₂ and CO₂	respiration and acid–base balance are disrupted
ions and osmotic concentration	water movement, membrane potentials and cell volume change

Stable does not mean perfectly constant. A healthy value fluctuates within an acceptable range, and different variables can have different set ranges.

change in a regulated variable → receptor detects the value → coordinator compares it with the set range → signal reaches an effector → effector changes the variable

Component	Question it answers
stimulus / deviation	what moved away from the range?
receptor	what measured the variable?
coordinator	where was information integrated?
effector	what tissue carried out the response?
response	how did the variable change?

Signals may travel as nerve impulses, hormones, or both. A valid explanation names the regulated variable and the direction of correction, not only the organs involved.

Deviation	Corrective response	When correction succeeds
value above the range	response lowers the value	response weakens or switches off
value below the range	response raises the value	response weakens or switches off

The feedback is negative because the response reduces the original deviation. The output feeds back to decrease the stimulus that activated the loop.

Positive feedback amplifies a change and is useful for endpoint processes such as childbirth. By itself it does not restore a homeostatic set range.

State	Immediate problem	Serious consequence if prolonged
hypoglycaemia	too little glucose available, especially to the brain	fainting, convulsions or coma
hyperglycaemia	plasma water potential falls and water is drawn from tissues	dehydration and circulatory damage

Blood glucose rises after carbohydrate absorption. It falls during fasting and prolonged activity as cells remove glucose for respiration.

Glycogen stores buffer these changes. Liver glycogen can support blood glucose; muscle glycogen is primarily a local fuel reserve for muscle activity.

A rise in blood glucose is detected directly by pancreatic beta cells in the islets of Langerhans.
Beta cells secrete insulin into the blood.
Insulin increases glucose uptake by target cells and promotes respiration.
Liver and skeletal muscle convert glucose to glycogen; other excess can be converted to fat.
As blood glucose falls toward the range, insulin secretion decreases.

Insulin is a hormone signal. It does not itself convert glucose into glycogen; it changes target-cell activity so enzymes carry out the conversions.

Two feedback pathways show pancreatic beta cells releasing insulin after blood glucose rises and alpha cells releasing glucagon after blood glucose falls.

A fall in blood glucose is detected directly by pancreatic alpha cells.
Alpha cells secrete glucagon into the blood.
In the liver, glucagon promotes glycogen breakdown and glucose production from non-carbohydrate substrates.
Glucose is released to the blood.
As blood glucose rises toward the range, glucagon secretion decreases.

Feature	Insulin branch	Glucagon branch
source	beta cells	alpha cells
trigger	high blood glucose	low blood glucose
main effect	uptake and storage	liver glucose release
net result	lowers blood glucose	raises blood glucose

Glucagon is a hormone; glycogen is a glucose-storage polymer. Glucagon activates cellular pathways—it is not the enzyme that digests glycogen.

Feature	Type 1 diabetes	Type 2 diabetes
primary failure	autoimmune destruction of beta cells	reduced target-cell response to insulin; beta-cell function may later decline
insulin signal	little or none produced	present but less effective
common management principle	replace insulin and monitor glucose	improve response and glucose balance through individualized diet, activity and medication
association	genetic susceptibility and environmental triggers	age, family history, inactivity and excess body fat are risk factors

A risk factor changes probability; it is not a single sufficient cause. Type 2 diabetes should not be reduced to “eating sugar,” and type 1 is not caused by lifestyle.

Both can produce persistent hyperglycaemia, glucose in urine and long-term damage to blood vessels, kidneys, retina and nerves if control remains poor.

Measure fasting blood glucose.
Give a standard oral glucose dose.
Measure blood glucose at defined times after ingestion.
Compare peak height and recovery toward the fasting range.

Curve feature	Effective regulation	Impaired regulation
starting value	within expected fasting range	may already be elevated
post-dose peak	rises then turns downward	often higher or prolonged
later value	approaches the starting range	remains elevated longer

The curve shows glucose handling under a standardized challenge. Diagnosis uses clinical thresholds and context; one graph alone does not identify the exact cellular cause.

skin thermoreceptors monitor environmental influence + hypothalamic thermoreceptors monitor blood temperature → hypothalamus integrates the signals → skin, skeletal muscle, liver and endocrine pathways alter heat loss or production

Control feature	Core temperature rises	Core temperature falls
corrective direction	increase heat loss and reduce heat gain	reduce heat loss and increase heat production
shut-off condition	response fades as temperature returns toward the range	response fades as temperature returns toward the range

Human core temperature is regulated near 37 °C but is not fixed at exactly 37.0 °C; normal values vary with time and physiological state.

Skin and brain thermoreceptors signal the hypothalamus, which activates cooling above the normal range and warming below it.

Effector response	How it cools
skin arterioles dilate	more warm blood reaches surface capillaries, increasing heat transfer to surroundings
sweat secretion increases	evaporation transfers latent heat from the skin
hair erector muscles relax	hairs lie flat, reducing the insulating air layer
behaviour changes	shade, reduced activity or exposed skin decreases heat gain and increases loss

Sweating cools only when sweat evaporates. Sweat that drips away without evaporating removes far less heat.

A hot-skin cross-section shows active sweat glands, evaporation, dilated skin blood vessels and relaxed hair muscles.

Effector response	How it warms or conserves heat
skin arterioles constrict	less warm blood reaches surface capillaries, reducing heat loss
sweating decreases	less evaporative heat is lost
skeletal muscles shiver	rapid contractions increase respiration and heat production
liver metabolic activity rises	greater respiration releases more heat
brown adipose tissue uncouples respiration	fuel energy is released as heat rather than captured mainly in ATP

Brown fat is especially important in newborns, whose large surface-area-to-volume ratio causes rapid heat loss.

Raised hairs have little insulating effect in humans because human body hair is sparse; the mechanism is more effective in furred mammals.

regulated variable leaves its acceptable range → receptor detects the direction of deviation → coordinator selects an opposing response → effectors alter transfer, storage or metabolism → deviation shrinks → corrective signal declines

System	Sensor / coordinator	High-side response	Low-side response
blood glucose	pancreatic islet cells	beta cells, insulin, uptake and storage	alpha cells, glucagon, liver glucose release
core temperature	thermoreceptors and hypothalamus	vasodilation and sweating	vasoconstriction, shivering and metabolic heat

For any homeostatic explanation, name the variable, direction of deviation, detector, signal, effector action and why that action reverses the deviation.

Kidney job	What is controlled	Example
excretion	removal of metabolic waste from the body	urea leaves in urine
osmoregulation	balance of water and dissolved solutes	urine volume and concentration change with water status

renal artery brings blood → nephrons filter and modify fluid → renal vein carries adjusted blood away; collecting ducts deliver urine → renal pelvis → ureter → bladder → urethra

Excretion is not the same as egestion. Urea is a metabolic waste made inside the body; undigested food leaving the gut was never part of the internal environment.

A kidney-to-nephron overview shows renal blood vessels, ureter, cortex, medulla, Bowman’s capsule, proximal tubule, loop of Henle and collecting duct.

A wide afferent arteriole supplies the glomerulus and a narrower efferent arteriole maintains high hydrostatic pressure.
Water and small solutes cross fenestrated capillary endothelium, the basement membrane and filtration slits between podocyte processes.
Blood cells and most plasma proteins are too large to cross and remain in the blood.
The resulting glomerular filtrate enters Bowman’s capsule and flows into the proximal convoluted tubule.

Ultrafiltration is driven by blood hydrostatic pressure generated by the heart. The filtration step itself is not powered by ATP in podocytes.

A renal corpuscle diagram shows pressure filtration from the glomerulus into Bowman’s capsule while cells and large proteins remain in blood.

Substance	Blood plasma entering glomerulus	Fresh glomerular filtrate	Final urine in health
blood cells	present	absent	absent
most plasma proteins	present	absent or trace	absent or trace
glucose and amino acids	present	present	normally reabsorbed, so absent or trace
water and ions	present	present	variable amounts
urea	present	present	concentrated for excretion

Filtration separates mainly by size; selective reabsorption separates by biological value and regulated need. Urine is therefore not simply “filtered blood.”

Glucose in urine can appear when blood glucose is high enough to exceed tubular reabsorptive capacity. Successful diabetes control can restore urine glucose to absent or trace levels.

Cell feature	Contribution to reabsorption
one-cell-thick epithelium	short transport distance
apical brush-border microvilli	large surface area facing filtrate
many mitochondria	ATP supply for active transport
basolateral membrane infoldings	large area for pumps and carrier proteins
close capillary network	removes reabsorbed substances and maintains gradients

The PCT reabsorbs all normal filtered glucose and amino acids, much sodium and other ions, and much water. Urea is retained more strongly in the tubular fluid.

A proximal convoluted tubule epithelial cell shows apical microvilli, many mitochondria, basolateral membrane folds, sodium–potassium pumps and nearby capillary.

Basolateral sodium–potassium pumps use ATP to move Na⁺ from the PCT cell into tissue fluid.
The resulting low intracellular Na⁺ concentration lets Na⁺ enter from the filtrate down its electrochemical gradient.
Na⁺-dependent cotransport carries glucose or amino acids through the apical membrane.
Glucose, amino acids and ions leave across the basolateral side into tissue fluid and capillaries.
Water follows the reabsorbed solutes by osmosis.

Glucose uptake at the apical membrane is secondary active transport: the cotransporter does not use ATP directly, but it depends on the Na⁺ gradient maintained by an ATP-powered pump.

Reabsorption means movement from nephron filtrate back to blood. Secretion is the opposite transfer—from blood or tubule cells into the tubular fluid.

Limb	Water permeability	NaCl movement	Effect on tubular fluid
descending	high	low solute permeability	water leaves by osmosis; fluid becomes more concentrated
thin ascending	impermeable	NaCl diffuses out	fluid becomes less concentrated
thick ascending	impermeable	Na⁺ and Cl⁻ actively transported out	fluid becomes still more dilute; medulla gains solute

The loop’s main function is to establish the hypertonic medulla that later permits water conservation from collecting ducts; it does not respond directly to ADH by inserting aquaporins.

The loop of Henle shows water leaving the descending limb, sodium chloride leaving the water-impermeable ascending limb and increasing medullary concentration toward the hairpin bend.

Ascending-limb salt transport makes nearby medullary interstitial fluid more concentrated while water stays in that limb.
Water leaves the adjacent descending limb by osmosis until its filtrate approaches the local interstitial concentration.
Continuous flow brings new less-concentrated filtrate into the descending limb and moves concentrated filtrate around the bend.
Repeating the local effect along the loop creates a large cortex-to-inner-medulla gradient.

The vasa recta supplies the loop tissue and carries away reabsorbed water while its countercurrent arrangement limits washout of the gradient.

Longer loops extend farther into the medulla and can support a steeper gradient, increasing the potential to produce concentrated urine.

The loop of Henle makes interstitial osmotic concentration increase down the medulla. Collecting ducts pass through this gradient, so water has a potential route from tubular fluid into the medulla and then blood.

Condition	Collecting-duct permeability	Consequence
few aquaporins in apical membrane	low	most water remains in tubular fluid
many aquaporins in apical membrane	high	water leaves by osmosis down the medullary gradient

The gradient supplies the driving force; ADH changes the permeability. Neither mechanism alone explains regulated water reabsorption.

Water loss or high salt intake raises blood osmotic concentration.
Hypothalamic osmoreceptors detect the change; thirst increases and the posterior pituitary releases more ADH.
ADH binds receptors on collecting-duct cells and promotes insertion of aquaporins into the apical membrane.
Collecting ducts become more permeable, so water leaves by osmosis into the hypertonic medulla and returns to blood.
A small volume of concentrated urine is produced; correction reduces the ADH signal.

ADH is synthesized by hypothalamic neurons and stored and released from the posterior pituitary. The posterior pituitary does not synthesize it.

Low- and high-ADH collecting duct cells show aquaporins stored in vesicles or inserted into the apical membrane and the resulting difference in water movement.

Excess water lowers blood osmotic concentration.
Hypothalamic osmoreceptor stimulation falls, so the posterior pituitary releases less ADH and thirst declines.
Aquaporins are removed from the collecting-duct apical membrane into vesicles.
Collecting ducts become less permeable, so less water leaves the tubular fluid.
A large volume of dilute urine removes excess water; recovery reduces the deviation.

Water status	ADH	Collecting-duct permeability	Urine
dehydrated / salty intake	high	high	small volume, concentrated
excess water	low	low	large volume, dilute

Concentrated urine contains less water per amount of solute; it does not mean the kidney has excreted more total waste in every case.

Physiological state	Flow increases toward	Flow can decrease toward	Reason
vigorous exercise	active skeletal muscle, heart and skin when heat loss is needed	gut and kidneys	oxygen delivery, metabolite removal and thermoregulation take priority
fight or flight	skeletal muscle and heart	much of the gut	epinephrine prepares rapid action
after a meal at rest	digestive organs	relatively inactive muscle	supports secretion, absorption and processing
sleep	selected brain regions vary; diaphragm remains supplied	most skeletal muscle	metabolic demand changes by tissue

Constriction of arterioles raises resistance and reduces flow into a capillary bed; dilation lowers resistance and increases flow. Different vascular beds can respond differently to the same hormone because receptors and local signals differ.

Renal flow may fall during prolonged vigorous exercise, but it cannot remain severely reduced indefinitely because excretion and osmoregulation must continue.

renal blood reaches glomerulus → pressure filters water and small solutes → PCT selectively returns useful substances → loop of Henle builds a medullary solute gradient → collecting-duct permeability sets final water recovery → urine leaves while adjusted blood returns to circulation

If asked about…	Start with…
cells or proteins absent from filtrate	filtration barrier and molecular size
glucose absent from urine	selective reabsorption and transport maximum
salty medulla	opposite loop-limb permeability and countercurrent multiplication
concentrated urine	high ADH plus an existing medullary gradient
organ blood flow during exercise	local arteriole resistance and changing metabolic demand

Do not collapse the nephron into one action. Filtration is pressure-driven and non-selective for small solutes; PCT recovery is selective and transport-dependent; the loop builds the gradient; ADH controls how fully collecting ducts use it.

[ 6 ]

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

[ 4 ]

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

[ 8 ]

Explain the control of blood glucose concentrations in humans.

[ 5 ]

Outline type II diabetes.

[ 8 ]

Explain the control of body temperature in humans.

[ 1 ]

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

[ 8 ]

Explain the role of the kidney in osmoregulation.

[ 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.

[ 1 ]

What is the function of the loop of Henle?

[ 7 ]

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

[ 1 ]

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

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.

Natural selection is differential survival and reproduction caused by heritable differences among individuals. Across generations, alleles associated with greater reproductive contribution can become more frequent.

Individuals in a population vary.
Some variation is heritable.
The environment creates selection pressures.
Heritable variants cause differences in survival or mating success.
Individuals leave different numbers of surviving offspring.
Allele frequencies change across generations.

Individuals are selected; populations evolve. An individual does not genetically adapt because it needs to—the frequency of inherited variants changes in descendants.

Mutation location	What can inherit it?	Evolutionary consequence
germ-line cell or its precursor	gametes and potentially offspring	can add a new allele to the population gene pool
somatic cell	descendant cells within that individual	normally ends when the individual dies

Mutation is random with respect to what the organism needs. Selection is non-random in the sense that existing variants leave different numbers of offspring in a particular environment.

A new allele may be harmful, neutral or advantageous depending on phenotype and environment; mutation does not mean automatically beneficial or harmful.

Process	Source of new combination
crossing over	homologous chromosomes exchange DNA between non-sister chromatids
independent assortment	maternal and paternal homologs enter gametes in varied combinations
random fertilization	one of many possible sperm fuses with one of many possible eggs

Sexual reproduction usually reshuffles existing alleles; mutation is the ultimate source of a completely new allele.

Recombination does not guarantee a better offspring. It creates varied genotypes on which environmental selection pressures can act.

potential offspring production exceeds the number the environment can support → food, space, light, territories or mates become limiting → many individuals die or fail to reproduce → variants differ in representation among the offspring that breed

Competition type	Who competes?	Why it matters for selection
intraspecific	members of the same species	they share a niche and compete for very similar resources and mates
interspecific	members of different species	overlapping resource use can also change survival and reproduction

A roughly stable population size does not mean every offspring survives. High offspring production can be balanced by high pre-reproductive mortality.

Abiotic factor	Variant that could be favoured in one context
drought	lower water loss or deeper roots
salinity	ion exclusion or salt tolerance
temperature	enzymes and membranes that function at local extremes
light	leaf form or pigment investment suited to the light regime
pH	proteins and transport systems stable at local pH

A density-independent pressure can affect survival regardless of how crowded the population is, although its total impact may still differ among places and times.

The environment does not cause organisms to develop the needed inherited trait. It changes which already varying individuals contribute most to the next generation.

Individual outcome	Contribution to evolutionary fitness
survives but produces no surviving offspring	little or none
reproduces early and offspring later breed	high contribution
wins mates but offspring have low survival	mating success may not become high lifetime fitness

Fitness is relative to a particular environment. A phenotype advantageous during drought may be neutral or costly when water is abundant.

Better adapted does not mean perfect. Selection compares variants currently present and is constrained by trade-offs, ancestry and chance.

Feature	Acquired change	Heritable variant
origin	arises during one individual’s lifetime through use, injury or environment	encoded in inherited genetic or transmissible biological information
inheritance	usually not copied into gamete DNA	can be passed through gametes
population effect	ends with the individual	can change frequency across generations

Selection can act on a phenotype, but evolutionary change follows only when differences affecting reproductive success are heritable. Ordinary muscle gain or a scar is not inherited as a new DNA sequence.

Parallel giraffe sequences contrast Lamarckian individual neck lengthening with Darwinian inherited variation and differential reproduction across generations.

The population already contained pale and melanic heritable forms.
Soot darkened resting surfaces and reduced pale lichen cover in industrial areas.
Birds more often detected the less-camouflaged form.
Better-camouflaged moths survived and reproduced at a higher rate.
The allele associated with that form increased locally across generations.

When air pollution later declined and lichens returned, the direction of the camouflage advantage could reverse. Fitness belongs to a phenotype–environment relationship, not permanently to one colour.

Pale and melanic peppered moth forms are shown side by side, with pale associated with non-polluted habitats and melanic with industrially polluted habitats.

Route	Selection event	Example trait
mate choice	one sex preferentially mates with particular partners	bright plumage or complex courtship display
mating competition	individuals compete directly or indirectly for access	body size, weapons, territories or dominance behavior

heritable variation in display or competition trait → unequal mating success → unequal offspring contribution → associated alleles increase

Persistent differences in selected traits between sexes can produce sexual dimorphism, such as bright displaying males and more cryptic females.

Effect of an exaggerated ornament	Possible fitness consequence
more visible or energetically costly	lower survival through predation or resource cost
attractive display or honest condition signal	greater chance of mating
species-specific courtship signal	mating isolation from other populations

The trait can spread when its gain in reproductive success exceeds its survival cost. Fitness integrates the whole route to descendants, not survival alone.

Sexual selection names the pressure—competition for mates or mate choice—rather than competition for food or escape from predators.

Design element	Selection experiment question
independent variable	which pressure is deliberately changed?
dependent variable	which phenotype, survival or mating outcome is measured?
controls	which habitat, starting population and observation conditions stay comparable?
replication	are multiple populations exposed to each treatment?
time	are enough generations observed for frequency change?

Random assignment and controlled treatments make a causal claim stronger; repeated measurements reveal whether a pattern persists rather than reflecting one generation of chance.

A simulation or enclosure simplifies nature. Its value depends on whether the manipulated pressure and measured fitness components represent the biological system.

Treatment	Predator pressure	Prediction for male spots
K	none	female choice can favour conspicuous males
R	weak, Rivulus hartii	sexual selection remains influential
C	dangerous, Crenicichla alta	conspicuous males suffer stronger predation

Fine and coarse gravel were crossed with predator treatments. Under predation, spot size matching the substrate can improve camouflage; without strong predation, contrast can improve visibility to females.

Guppies were allowed to breed, predator treatments were imposed, and spot number and size were measured over later generations in replicated ponds.

Evidence	Interpretation
spots increased before strong predators were added	female choice favored conspicuous males
mean spot number fell with C. alta	dangerous predators selected against conspicuous males
weak-predator field transplants became more colorful over 15 generations	reduced predation allowed sexual selection to dominate
spot size tracked gravel under predation	background matching improved camouflage

Natural and sexual selection acted on the same heritable color traits in opposing directions. The observed phenotype reflected the balance of those pressures, not one universal best color.

Guppy treatments K, R and C show spot number rising with no or weak predation and falling under dangerous C. alta predation.

mutation creates alleles + sexual reproduction reshuffles them → offspring vary → limited resources and abiotic or biotic pressures produce unequal survival and mating → heritable variants reach offspring unequally → population trait and allele frequencies change

Claim to test	Evidence needed
adaptation occurred by selection	heritable variation plus unequal reproductive contribution across generations
sexual selection acted	trait differences predict mating success
environmental pressure caused change	controlled manipulation or strong comparative evidence links pressure to fitness
population evolved	heritable trait or allele frequency changed, not merely one individual

Avoid purpose-driven language. Populations do not create useful mutations on demand; selection filters inherited variation already produced by mutation and recombination.

A gene pool is the complete set of genes and their alleles in an interbreeding population. Individuals carry samples of that pool; gametes sample alleles again for the next generation.

Level	What is counted?
diploid individual	two allele copies at one autosomal locus
population	all allele copies carried by all breeding individuals
allele frequency	proportion of all locus copies that are a specified allele

When allele frequencies change from one generation to another, the gene pool is changing—this is population-level evolutionary change even before a new species appears.

For allele A in a diploid population:

$f(A)=\frac{2N_{AA}+N_{Aa}}{2N_{total}}$

Genotype	Individuals	A copies contributed
AA	30	60
Aa	50	50
aa	20	0
total	100	110 of 200 copies

f(A) = 110/200 = 0.55; therefore f(a) = 0.45. Heterozygotes contribute one copy of each allele, not two copies of the dominant allele.

A barrier reduces or stops gene flow between populations.
Mutation, selection and genetic drift act separately in each gene pool.
Different environments can favor different inherited variants.
Allele frequencies diverge across generations.
Accumulated genetic and reproductive differences may eventually contribute to speciation.

Force	Directional expectation
gene flow	tends to make populations more similar
different local selection	can make populations diverge predictably
drift in small populations	can make frequencies diverge by chance
mutation	introduces new alleles independently

Geographic allele-frequency databases can reveal divergence, but the pattern alone does not identify which force caused it; sampling and population definition must also be checked.

mutation supplies new alleles → meiosis and fertilization create genotypes → genotypes influence heritable phenotypes → selection acts through individual survival and reproduction → allele frequencies change in the population gene pool

Darwinian description	Genetic description
individuals vary	individuals carry different alleles and genotypes
some leave more offspring	alleles are transmitted unequally
population becomes adapted	advantageous inherited variants rise in frequency

Selection is not the only force that changes allele frequency. Mutation, migration and genetic drift also alter gene pools, though not necessarily because one phenotype has higher fitness.

Mode	Highest fitness	Distribution after selection
directional	one extreme	mean shifts toward that extreme
stabilizing	intermediate	variance narrows around the existing mean
disruptive	both extremes	intermediates decline; distribution can become bimodal

The graphs describe phenotype distributions, but evolutionary change ultimately requires associated heritable alleles or allele combinations to change frequency.

Before-and-after phenotype distributions show directional selection favoring one extreme, stabilizing selection favoring intermediates and disruptive selection favoring both extremes.

Resistance variation exists before or arises independently of treatment.
The antibiotic kills susceptible bacteria more effectively.
Resistant survivors reproduce and contribute a larger fraction of the next population.
Resistance alleles rise in frequency; plasmid transfer can spread them between cells.

The antibiotic does not teach individual bacteria to become resistant. It changes the survival difference between pre-existing or newly mutated variants.

A mixed bacterial population is exposed to antibiotic; susceptible cells die, resistant survivors reproduce, and resistance can also move by plasmid transfer.

Birth-weight region	Selective disadvantage can arise from
very low	immature organs, low reserves and greater heat loss
intermediate	lowest mortality under the measured historical conditions
very high	delivery complications and associated maternal or fetal risks

When both extremes leave fewer surviving offspring, intermediate phenotypes contribute more. The mean may remain similar while the distribution narrows.

Human birth weight is affected by genes and environment. Historical hospital data show selection on the phenotype, but do not make every difference genetic or define one universal optimum.

environment contains two contrasting opportunities → individuals near either phenotypic extreme exploit one successfully → intermediate phenotype fits neither well → both extremes contribute more offspring → a bimodal distribution can develop

Yearling male plumage	Territorial outcome proposed in the study
dull brown	may be tolerated as non-threatening and gain territory
bright blue	may be avoided as a strong rival and gain territory
intermediate	attacked more often and gains poorer territory

Disruptive selection can increase divergence, but two peaks do not by themselves prove speciation. Reproductive isolation and continued genetic divergence are additional requirements.

Symbol	Meaning for two alleles A and a
p	frequency of A
q	frequency of a
p²	expected frequency of AA
2pq	expected frequency of Aa
q²	expected frequency of aa

$p+q=1$

$p^2+2pq+q^2=1$

p and q are allele frequencies; p², 2pq and q² are genotype frequencies. The equations do not require p to represent the dominant allele.

Convert the recessive phenotype count to a frequency; under complete dominance this is q².
Calculate q = √q².
Calculate p = 1 − q.
Calculate expected genotype frequencies p², 2pq and q².
Check that the three genotype frequencies sum to 1.

This inference is valid only when the population and locus satisfy the Hardy–Weinberg model and the recessive phenotype identifies homozygous recessives accurately.

A Hardy–Weinberg workflow starts from observed q squared, takes a square root for q, calculates p, then predicts p squared, 2pq and q squared and checks their sum.

Quantity	Calculation	Result
q²	observed non-taster frequency	0.35
q	√0.35	0.592
p	1 − 0.592	0.408
p²	0.408²	0.166
2pq	2 × 0.408 × 0.592	0.483
check	0.166 + 0.483 + 0.350	0.999 ≈ 1

In 200 people, expected counts are approximately 33 homozygous dominant, 97 heterozygous and 70 homozygous recessive after multiplying each frequency by 200.

Keep extra digits until the final answer. Early rounding makes the genotype frequencies fail to sum closely to 1.

Condition	Change prevented in the model
very large population	random genetic drift is negligible
random mating	genotype pairing is not biased
no selection	genotypes contribute equally
no mutation	alleles do not convert into new states
no migration	no alleles enter or leave through gene flow

Under these conditions, random mating produces p²:2pq:q² genotype frequencies and allele frequencies remain stable across generations.

The model is an ideal baseline, not a claim that real populations satisfy every condition perfectly.

Estimate allele frequencies from the population sample.
Use p², 2pq and q² to calculate expected genotype frequencies and counts.
Compare expected with observed genotype counts using an appropriate statistical test.
If the departure exceeds sampling expectation, investigate which assumptions are violated.

Possible violation	Additional evidence to seek
selection	survival or reproduction differs by genotype
non-random mating	mate pairs differ from random expectation
migration	immigrants carry different allele frequencies
drift	small population and random change
mutation	documented new allele conversion

A deviation does not by itself prove natural selection. Several forces, population structure, genotyping error or sampling can produce the pattern.

Choose a target phenotype that has heritable variation.
Select parents showing the desired value.
Breed them and measure offspring.
Retain offspring with the strongest desired phenotype.
Repeat across generations so associated alleles become more frequent.

Rapid improvement can reduce genetic diversity and bring linked harmful alleles or correlated traits along with the selected phenotype.

A maize sequence compares a small wild-type cob with progressively larger cobs produced by repeated selection across generations.

Feature	Artificial selection	Natural selection
source of reproductive advantage	human breeder chooses parents	environmental survival and mating pressures
variation required	heritable variation	heritable variation
population result	chosen alleles change frequency	alleles linked to higher fitness change frequency
guaranteed organismal benefit	no; target serves human goal	no universal benefit; adaptation is environment-specific

Antibiotic resistance is usually natural selection caused by a human-created pressure, not artificial selection: humans do not deliberately choose resistant bacteria as breeding parents.

Both processes reveal the same inheritance principle—repeated unequal reproduction changes a population over generations.

count allele copies in the gene pool → establish p and q → use Hardy–Weinberg as a no-change expectation → compare observed genotype or allele frequencies across generations → use phenotype, fitness and population evidence to identify selection or another evolutionary force

Pattern or action	First model to examine
mean shifts toward one extreme	directional selection
variance narrows around intermediate values	stabilizing selection
both extremes rise and intermediates fall	disruptive selection
observed genotypes differ from p²:2pq:q²	one or more equilibrium assumptions may be violated
humans repeatedly choose parents	artificial selection

Keep description, calculation and cause separate. A changed distribution describes an outcome; allele counts quantify evolution; causal claims require evidence about inheritance, survival, mating, migration, drift or deliberate breeding.

[ 8 ]

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

[ 8 ]

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

[ 4 ]

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

[ 2 ]

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

[ 1 ]

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

[ 1 ]

What is required for natural selection to occur?

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

[ 6 ]

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

[ 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 X	Predator Y	Male guppy colouration
Pond 1	12	0	Large colourful spots
Pond 2	15	0	Medium colourful spots
Pond 3	5	17	None/very small drab spots

What can be concluded from the data?

[ 3 ]

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

[ 1 ]

Define allele frequency.

[ 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.

[ 2 ]

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

[ 3 ]

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

[ 2 ]

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

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.

Ecosystem stability is the ability to maintain characteristic structures and processes continuously over time, despite ordinary fluctuations or disturbance.

Evidence of stability	What may still change?
nutrient cycling continues	population sizes vary seasonally
food-web roles persist	individuals and local patches turn over
dominant community remains recognizable	storms, droughts or fires cause temporary departures

Stable does not mean static. A system can change locally and still be stable if its defining organization persists or recovers.

Ancient rainforests, deserts and long-lived communities show that ecosystems can persist over very long timescales when supporting conditions and ecological processes continue.

Repeated recovery after disturbance supports resilience.
Small fluctuations around a characteristic state support continuity.
A long history does not guarantee survival after a new disturbance of greater magnitude or rate.

Always state the timescale and property being judged. A population can decline while nutrient cycling remains stable, or a forest can persist while its species composition changes.

Requirement	Why stability depends on it
continual energy input	replaces energy dissipated as heat during metabolism
producer biomass	captures light or chemical energy into organic molecules
decomposers and nutrient cycling	return mineral nutrients from waste and dead biomass for reuse
matter inputs and retention	supply elements needed to build biological molecules

Energy is not recycled: it enters, is transferred and is lost as heat. Chemical elements can cycle repeatedly between organisms and the abiotic environment.

Stability support	Mechanism
species diversity	multiple species can occupy different niches or partly replace lost functions
genetic diversity	varied alleles increase the chance that some individuals tolerate disease or environmental change
abiotic conditions within tolerance	key species can survive, reproduce and maintain their ecological roles

Higher diversity often supports resilience, but it is not a guarantee. Stability still depends on interaction strengths, climate, energy supply and the type of disturbance.

Two beetle communities contain the same three species, but the more even community has higher diversity because no single species dominates.

A tipping point is a critical threshold beyond which a small additional change can produce a disproportionately large shift to a different ecological state.

A disturbance weakens a stabilizing process.
The weakened process amplifies the original disturbance through positive feedback.
Recovery becomes progressively harder as the system approaches a threshold.
Beyond the threshold, the system may settle into a different equilibrium.

A tipping point is not simply any large change. The defining feature is a threshold response reinforced by feedback, often with difficult or incomplete reversal.

Deep-rooted trees take up soil water.
Transpiration transfers water vapour to the atmosphere.
Condensation and air movement support regional cloud formation and rainfall.
Rain replenishes soil water and sustains further forest growth and transpiration.

Evapotranspiration cools the regional surface.
Continuous canopy and moisture transport extend rainfall beyond local forest patches.
Large connected forest areas sustain the loop more effectively than fragmented remnants.

Clearing, burning and fragmentation reduce leaf area and transpiration.
Rainfall recycling and regional cooling weaken.
Longer, hotter dry seasons increase drought stress and fire risk.
Fire and tree death remove more forest, reinforcing the original drying.
A forest state can shift toward lower-biomass savanna-like vegetation.

The exact minimum forest area and threshold are uncertain. A possible range is a risk estimate, not a precise prediction that collapse occurs at one universal percentage.

An intact Amazon forest sustains strong transpiration and rainfall recycling, while cleared and burned forest has weaker moisture recycling and moves toward a drier state; an uncertain possible tipping range is marked.

percentage change = (final forest cover − initial forest cover) ÷ initial forest cover × 100

Quantity	Area (km²)
estimated pre-1970 cover	4,100,000
2018 natural forest cover	3,390,835
change	−709,165

(−709,165 ÷ 4,100,000) × 100 = −17.3%. The negative sign indicates loss; the chosen initial year determines what the percentage describes.

A percentage of area lost does not by itself prove a tipping point. The ecological inference also needs evidence about fragmentation, rainfall feedback, fire and recovery.

Mesocosm strength	Mesocosm limitation
isolates one manipulated variable	small scale omits migration and large food webs
allows replication and repeated measurements	container boundaries alter light, heat and movement
reduces damage to a natural ecosystem	short experiments may miss slow feedbacks

A sealed bottle can restrict matter exchange while still receiving light energy and losing heat. Producers, consumers and decomposers must together recycle matter within the model.

A mesocosm can test a mechanism under controlled conditions; transferring its result to nature requires cautious comparison with field evidence.

Design element	Nutrient-enrichment test
independent variable	phosphate added to treatment only
control addition	equal volume of water without phosphate
controlled variables	starting culture, vessel, light, temperature, mixing and sampling
dependent variables	algal cell density and dissolved oxygen over time
reliability	replicate treatment and control vessels

The design must also prevent escape or disposal of organisms and nutrients into natural water, and it must protect organisms from avoidable heat, starvation or oxygen stress.

Matched phosphate-treatment and control mesocosms receive equal light and stirring; only the treatment receives phosphate while algal cell density and oxygen concentration are logged.

A keystone species has a disproportionately large effect on community structure relative to its abundance or biomass.

Observation after removal	Keystone inference
many indirect population changes	the species controlled a trophic or habitat-forming interaction
loss of functions or community reorganization	its role was not readily replaced by other species
effect far exceeds its biomass share	impact is disproportionate, the defining criterion

A common or dominant species can be important without being a keystone. “Keystone” compares ecological effect with abundance; it does not simply mean rare, large or charismatic.

Agoutis open Brazil-nut pods that most animals cannot access.
They carry and bury seeds as food stores.
Some buried seeds are not recovered and germinate away from the parent tree.
Brazil-nut recruitment supports forest structure, nesting sites, food and human livelihoods.
Removing agoutis therefore affects more than one predator–prey pair.

The keystone role here is seed dispersal and recruitment. The prediction follows the controlled interaction: fewer agoutis → fewer successfully dispersed Brazil-nut seeds → long-term decline of trees and dependent species.

The largest effect may be delayed because adult Brazil-nut trees are long-lived. Short-term persistence of mature trees would not show that seed dispersal and future recruitment are secure.

continued energy input + nutrient recycling + diversity + tolerable abiotic conditions → ecological functions persist or recover; disturbance that weakens these supports can trigger reinforcing change and a new state

Question	Evidence to seek
Is the ecosystem stable?	persistence or recovery of named structures and processes over a stated timescale
Is a tipping point plausible?	threshold-like response plus reinforcing feedback
Does one species have a keystone role?	disproportionate community change after its removal
Does a mesocosm support the mechanism?	replicated treatment–control difference with variables held constant

Do not equate stable with unchanged, old with invulnerable, or correlated change with a demonstrated feedback mechanism.

Long-term harvesting is sustainable when removal, natural mortality and habitat damage do not exceed recruitment, growth and recovery of a viable reproducing population.

Estimate stock size and its uncertainty.
Measure recruitment, growth, age or size structure and natural mortality.
Record harvested biomass, by-catch and habitat damage.
Compare removal with replacement across many years.
Reduce harvest when recovery indicators weaken.

Renewable does not mean inexhaustible. A population can replenish itself only while enough breeding individuals and suitable habitat remain.

Stock observation	Likely interpretation
many juveniles reaching reproductive age	recruitment can support future stock, if they survive
few large breeding adults	reproductive capacity may be weakened
catch stays high while fishing effort rises	stock may be declining despite similar landings
by-catch or habitat damage increases	harvest cost extends beyond the target stock

Maximum sustainable yield is an estimated long-term catch near the balance between stock growth and losses. It depends on uncertain, variable recruitment and should be updated with long time-series data.

A single year’s catch cannot establish sustainability. Illegal discarding or poor age data can bias stock estimates and make a quota unsafe.

Measure	Sustainable direction
extracted timber volume	no greater than replacement growth over the management period
replanting and natural regeneration	enough surviving trees of suitable species and age classes
soil disturbance and erosion	minimized so future productivity is retained
habitat structure and carbon storage	maintained across the managed landscape

Seedlings planted today are not immediate replacements for mature trees. Assessment must use the time needed to restore biomass, habitat and carbon, not just a count of stems planted.

High short-term yield can coexist with falling long-term forest capacity. Sustainability prioritizes the continuing system over one harvest cycle.

Vegetation removal exposes soil to rain and wind.
Soil particles and organic matter are detached and transported.
Topsoil depth, structure, water retention and nutrient stores decline.
Crop growth becomes more dependent on irrigation and external fertilizer.
Further poor cover or tillage accelerates the loss.

Practice	Stability benefit
cover crops or residue retention	shields soil and adds organic matter
contouring and reduced tillage	slows runoff and soil detachment
rotations and mixed vegetation	supports soil biota and interrupts pest cycles

Fertile soil can form far more slowly than it is eroded, so it is effectively non-renewable on human farming timescales.

Decision	Potential gain	Sustainability cost to monitor
fertilizer use	replaces limiting nutrients and raises yield	leaching, eutrophication and energy cost
pesticide use	reduces crop loss	non-target effects, resistance and biomagnification if persistent
irrigation	maintains growth during water shortage	aquifer depletion, salinization and energy use
local seasonal supply	can reduce transport	yield, storage and production method still affect total footprint

maintain soil depth, structure and nutrient balance
protect water quality and biodiversity
keep greenhouse-gas emissions and fossil-energy use within long-term limits
judge yield together with external inputs and damage

No single favourable indicator proves sustainability. A high yield can depend on soil depletion, pollution or carbon emissions shifted outside the field boundary.

Nitrate and phosphate inputs remove a nutrient limit on algae or cyanobacteria.
A bloom reduces light penetration, so submerged plants die.
Bloom cells also die, increasing dead organic matter.
Decomposer populations and aerobic respiration increase.
Biochemical oxygen demand rises as dissolved oxygen is consumed.
Hypoxia or anoxia kills or excludes oxygen-dependent animals.

Algae do not directly “use up all the oxygen” simply by growing. The key oxygen-loss step is respiration, especially by decomposers processing dead biomass.

A nutrient-enrichment sequence shows nitrate and phosphate runoff, an algal bloom, shading and plant death, decomposer growth, higher biochemical oxygen demand, falling dissolved oxygen and fish death or migration.

Biochemical oxygen demand (BOD) is the amount of dissolved oxygen used by aerobic biological activity while organic matter in a measured volume of water is broken down.

Measure dissolved oxygen in a known water sample.
Incubate the sealed sample in darkness at 20 °C for five days.
Measure dissolved oxygen again.
Initial oxygen − final oxygen gives the five-day BOD.

Darkness prevents photosynthesis from replacing oxygen during incubation. High BOD indicates much respirable organic matter and/or strong decomposer activity, not high oxygen availability.

Use matched volume, temperature, time and handling. BOD is an indirect pollution measure and does not identify the pollutant by itself.

Ecosystem property	Typical eutrophication change
light penetration	falls as algal biomass and turbidity rise
submerged macrophytes	decline when photosynthesis becomes light-limited
producer diversity	often falls as bloom-forming species dominate
food-chain length	can shorten as habitats and oxygen-sensitive consumers disappear
net production	may rise initially even while stability and diversity fall

Early nutrient enrichment can produce high primary production; later death, decomposition and hypoxia create the ecological damage. Sampling at only one time point can miss this sequence.

High productivity is not the same as a healthy or stable ecosystem. Composition, oxygen regime and food-web structure must also be judged.

Process	Comparison being made
bioaccumulation	toxin concentration in one organism increases over time because uptake exceeds breakdown and excretion
biomagnification	toxin concentration per unit biomass increases from one trophic level to the next

Persistence prevents rapid degradation; fat solubility and poor excretion retain compounds such as DDT; predators consume contaminant loads from many prey.

A food-chain pyramid shows DDT concentration increasing from producers to primary and secondary consumers and a top predator, alongside one fish accumulating more toxin over time.

A dilute persistent pollutant enters producers from water or soil.
Primary consumers eat many contaminated producers.
Predators retain pollutant faster than they eliminate it and consume many contaminated prey.
Concentration per gram of tissue increases at successive trophic levels.
Long-lived top predators receive the greatest dose and may suffer reproductive or neurological harm.

DDT property or effect	Consequence
persistent and fat-soluble	retained in animal tissues
biomagnifies	birds of prey receive high tissue concentrations
disrupts calcium deposition	thin eggshells crack more easily, reducing breeding success

The pollutant does not create extra mass at each trophic level. Concentration rises because predators assimilate contaminants from many prey while biomass and energy are lost between levels.

Plastic class	Operational size and examples	Dominant exposure route
macroplastic	greater than 5 mm; bags, bottles, line and packaging	entanglement, obstruction and ingestion as visible debris
microplastic	less than 5 mm; fragments, fibres and industrial particles	ingestion by small organisms and food-web transfer

Wind, rivers and ships deliver plastic to surface water and coasts.
UV radiation and wave action fragment items but do not mineralize them rapidly.
Biofouling can make particles sink; resuspension can return them to water.
Plastic and associated chemicals enter organisms and food webs.

Fragmentation is not biodegradation. It redistributes the same persistent material into smaller particles with greater surface area and access to smaller organisms.

Route	Example consequence
entanglement	restricted movement, injury, drowning or impaired feeding
mistaken ingestion	gut blockage, false satiation and starvation
transfer to offspring	albatross adults feed floating plastic to chicks that cannot regurgitate it
chemical transport	additives or persistent pollutants associated with plastic can enter tissues and food webs

Sea turtles can mistake plastic bags for jellyfish. Albatrosses skim food from the ocean surface, where floating debris accumulates; mechanism and feeding behavior explain exposure better than a generic pollution claim.

Not every effect requires biomagnification. Entanglement and blockage are direct physical harms; chemical accumulation is a separate pathway that needs pollutant evidence.

reconnect habitat fragments so organisms can disperse and breed
remove or control invasive species that block native recovery
reintroduce missing keystone species or apex predators where risks are understood
reduce intensive harvesting and allow succession, predation and decomposition to operate

The goal is a more self-sustaining system whose interactions maintain diversity and function with less continuous human control.

Rewilding is not simply “leave everything alone.” Reintroductions, invasive removal and corridors require monitoring for animal welfare, conflict, disease and unintended trophic effects.

Formerly cleared land is protected from renewed intensive use.
Alien trees and vines that suppress native regeneration are removed.
Native plants recolonize and succession rebuilds forest structure.
Habitat and food resources permit endemic fauna to re-establish.
Management monitors recovery while allowing natural processes to do most of the work.

Minimal intervention occurs after key pressures are removed. Hinewai therefore links restoration to succession: management creates conditions in which native community change can proceed.

Success is not measured only by tree cover. Native composition, connectivity, regeneration, fauna and ecological processes are also evidence of recovery.

measure stock and ecosystem condition → keep removal and pollution within recovery capacity → monitor delayed and indirect effects → reduce pressure or restore processes when indicators deteriorate

Pressure	Mechanism to name
excess nutrient input	bloom → dead biomass → decomposer respiration → high BOD → hypoxia
persistent toxin	retention within organisms plus concentration across trophic levels
marine plastic	persistence, fragmentation, entanglement, ingestion and chemical transfer
harvesting	removal relative to recruitment, growth, age structure and habitat condition
degraded habitat	connectivity, species interactions and succession restored through rewilding

A resource being renewable, a crop yield being high or a restoration site becoming greener is not enough. Sustainability claims require evidence that future ecological capacity is retained.

Ecological succession is an orderly change in community composition through time as abiotic conditions and species interactions alter which organisms can establish and compete.

Starting condition	Succession type
new surface without developed soil	primary succession
disturbance removes organisms but leaves soil and some biological legacies	secondary succession

Pioneer species tolerate the starting conditions.
Their growth, death and interactions modify resources and the physical habitat.
New species colonize the changed habitat.
Competition, facilitation, herbivory and further abiotic change replace earlier communities.

Orderly does not mean identical or inevitable. Colonization history, climate and disturbance can redirect the route.

Lichens, mosses, bacteria and other pioneers colonize bare substrate.
Weathering plus dead pioneer biomass begins soil formation.
Small rooted plants stabilize material and add organic matter.
Deeper, wetter and more nutrient-rich soil supports grasses, shrubs and trees.
Animals arrive as plant habitat and food become available.

Nitrogen-fixing organisms can add biologically available nitrogen, while roots reduce erosion and decomposers recycle nutrients. Biotic change therefore improves abiotic conditions for later stages.

A dry-land primary-succession sequence progresses from bare rock to lichens and mosses, grasses and herbs, shrubs and woodland while soil depth and biomass increase.

Property	Early stage	Later stage
soil depth and organic matter	little	greater accumulation and retention
standing biomass	low	high
food-web complexity	few short pathways	more consumers, decomposers and interactions
nutrient source	external inputs and weathering dominate	internal cycling becomes more important
niches and diversity	few harsh-condition specialists	generally more opportunities and species

These are broad trends, not a rule that every variable rises smoothly. Disturbance, invasion, local soil and dominance by a few species can interrupt or reverse them.

NPP = GPP − community respiration

Stage	GPP	Community respiration	NPP and biomass accumulation
pioneer	low producer biomass, so GPP is low	also low	NPP can be a large proportion of GPP; biomass accumulates rapidly
developing	leaf area and GPP rise	respiration rises as biomass and consumers increase	continued positive NPP builds biomass
mature/climax-like	GPP may be high	high maintenance and food-web respiration	NPP is lower as production and respiration approach balance

“Productivity increases throughout succession” is incomplete. GPP and NPP can follow different patterns because community respiration changes.

Cyclical succession is a recurring sequence of community states in the same area rather than progression to one permanently fixed endpoint.

Low grazing allows scrub and young trees to establish.
Woodland structure develops and changes light and forage.
Large herbivores browse seedlings and open vegetation.
Reduced tree regeneration returns patches toward pasture or scrub.
The spatial mosaic repeats as grazing and tree replacement continue.

A cycle requires a repeatable driver and return pathway. Seasonal fluctuation alone is not succession, and one disturbance followed by recovery is not necessarily cyclical.

Control	Effect on the likely mature community
climate	sets temperature, water and seasonality constraints
soil and drainage	determine rooting conditions, oxygen and nutrient availability
species pool and interactions	determine which competitors, consumers and decomposers assemble
disturbance regime	can maintain mosaics or prevent one late stage dominating

Complex food webs, deeper soils, high biomass, nutrient cycling and negative feedback can support relative stability under the prevailing conditions.

Climax is a model of a relatively stable state, not a universal final destination. Climate change, invasion or altered disturbance can move the system again.

Human pressure	Mechanism	Maintained alternative state
repeated grazing	browsers remove seedlings before trees establish	grassland, heath or open wood pasture
controlled burning	kills young woody plants and resets vegetation age	managed heather moorland
wetland drainage	lowers the water table and changes oxygen and soil processes	agricultural land, peat extraction or other dry-land community

Arrested succession or plagioclimax occurs when repeated disturbance prevents the expected local climax and maintains an earlier or alternative stable community.

Removing the pressure may allow succession to resume, but recovery depends on remaining soil, seed sources, hydrology and whether degradation has crossed another threshold.

starting substrate and biological legacies → pioneer establishment → organisms modify soil, light, moisture and nutrients → later species colonize and compete → local controls and repeated disturbance determine whether change approaches a mature state, cycles or remains arrested

Observation	First explanation to test
no soil at the start	primary succession
soil remains after disturbance	secondary succession
GPP high but biomass accumulation slow	community respiration is also high
community states repeatedly return	recurring driver produces cyclical succession
woodland fails under repeated grazing or burning	succession is arrested at a plagioclimax

Succession is directional only relative to current conditions. It is not a universal ladder toward one permanent best community.

[ 4 ]

Outline the features of ecosystems that make them sustainable.

[ 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.

[ 6 ]

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

[ 4 ]

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

[ 6 ]

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

[ 6 ]

Discuss the causes and consequences of eutrophication.

[ 6 ]

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

[ 4 ]

Explain the consequences of plastic pollution in marine environments.

[ 2 ]

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

[ 3 ]

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

[ 4 ]

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

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.

Weather	Climate
atmospheric conditions at a particular time and place	statistical pattern of weather over decades, commonly 20–30 years or longer
one storm, heatwave or cold day	changing averages, ranges, seasonality and frequency of extremes

Climate change is a sustained change in global or regional climate patterns. One unusual day cannot establish it; long time-series evidence can.

Natural climate drivers exist, but the rapid warming since the mid-to-late twentieth century is primarily attributed to human-driven increases in greenhouse gases.

Most incoming shortwave solar radiation passes through the atmosphere.
Land and ocean absorb energy and warm.
The warm surface emits longwave infrared radiation.
Greenhouse gases absorb specific infrared wavelengths.
Re-emission sends energy in all directions, including back toward the surface.

The natural greenhouse effect makes Earth habitable. Enhanced greenhouse effect means that added greenhouse gases increase infrared retention and raise global mean temperature.

Shortwave solar radiation reaches and warms Earth; the surface emits longer-wave infrared radiation, some of which greenhouse gases absorb and re-emit in all directions.

Human activity	Main climate link
fossil-fuel combustion	transfers geologically stored carbon to atmospheric CO₂
cement production	releases CO₂ during fuel use and conversion of limestone
deforestation and land-use change	releases biomass carbon and removes photosynthetic uptake
cattle, rice cultivation and waste	add methane from anaerobic microbial processes
fertilizer and combustion processes	add nitrous oxide and other greenhouse gases

Photosynthesis and other sinks remove much of the emitted CO₂, but not all of it. The remaining annual surplus accumulates in the atmosphere.

Methane absorbs infrared strongly per unit mass but is less abundant and shorter-lived than CO₂. Climate contribution depends on radiative properties, concentration and atmospheric lifetime together.

Evidence	What it establishes
ice-core gas bubbles and oxygen isotopes	past greenhouse-gas concentrations and temperature proxies covaried over long timescales
modern atmospheric measurements	CO₂ now lies above the range recorded across recent glacial cycles and is rising rapidly
carbon sources and budget	fossil fuels, cement and land-use change supply the added carbon
radiative physics and climate models	added greenhouse gases produce the observed direction and scale of warming

Correlation in an ice-core graph alone does not prove causation. The causal conclusion comes from converging physical, chemical, historical and modelling evidence.

Orbital cycles initiated past glacial changes, with ice and greenhouse feedbacks amplifying them. That does not explain the current rapid greenhouse-gas increase from measured human sources.

A positive feedback produces a response that reinforces the original change, moving the system farther from its starting state.

An initial forcing raises temperature.
A temperature-sensitive component changes.
That change increases heat absorption or adds greenhouse gas.
The additional warming strengthens the same component change.

Positive means self-reinforcing, not beneficial. The feedback amplifies a forcing; it need not be the event that began the warming.

Temperature-sensitive change	Reinforcing step
ice melts	darker surface lowers albedo and absorbs more solar energy
permafrost thaws	decomposition releases methane and CO₂
ocean warms	CO₂ solubility falls and carbon release can increase
drought and fire increase	combustion releases CO₂ and removes carbon sinks

Warmer air can also contain more water vapour, itself a greenhouse gas, adding another reinforcing response.

The strength and timing of each feedback differ. Biological uptake and circulation can modify ocean carbon behavior, so a valid explanation states the mechanism without pretending every loop has one certain magnitude.

Four circular pathways connect warming to more warming through ice loss and lower albedo, permafrost thaw and greenhouse-gas release, reduced ocean carbon retention, and drought with wildfire.

Carbon gain	Carbon loss
photosynthesis and tree growth	plant and microbial respiration
carbon stored in wood and litter	decomposition of litter and thawed soil carbon
accumulation in cold or waterlogged soils	combustion during peat and forest fires

net sink: uptake > release → ecosystem carbon store grows
net source: release > uptake → ecosystem carbon store shrinks and atmospheric carbon rises

A forest can remain visibly wooded while switching from sink to source. The classification depends on the net carbon balance, not tree presence alone.

Warmer winters and reduced snowfall alter soil insulation and water supply.
Drought stress and heat reduce growth and cause forest browning.
Insects, tree mortality and fire increase.
Combustion and decomposition release biomass and soil carbon.
Permafrost thaw exposes additional carbon to decomposition.
Extra CO₂ and methane reinforce warming, drought and fire risk.

The tipping point is crossed when repeated carbon loss and impaired recovery maintain a net-source state instead of returning to net accumulation.

The relevant evidence must combine carbon fluxes, tree growth, mortality, fire and soil thaw through time; one photograph of a burned forest cannot establish the budget shift.

Ice habitat	Ecological function	Climate-sensitive change
mobile pack or sea ice	resting platform near offshore feeding grounds	reduced extent and longer ice-free seasons
landfast ice attached to coast or seabed	stable breeding and chick-rearing platform	earlier breakout before breeding is complete
seasonal ice edge	concentrates prey and structures access to food	timing and location shift

Melting floating sea ice does not directly raise sea level like melting land ice, but it can transform habitat, food access and albedo.

Species	Ice dependency	Consequence of earlier or reduced ice
walrus	adults and calves rest on sea ice between benthic feeding dives	longer travel, crowded land haul-outs and reduced safe resting habitat
emperor penguin	colonies breed and rear chicks on stable Antarctic landfast ice	early breakout can remove the platform before chicks are ready to survive at sea

Extent and timing both matter. The same total ice area at the wrong season or in the wrong location may not supply the required habitat function.

An adult female Atlantic walrus and juvenile rest on pack ice surrounded by open water in the Arctic.

Surface water warms and becomes less dense relative to deeper water.
The density difference strengthens stratification.
Vertical mixing and nutrient-rich upwelling weaken or shift in timing.
Fewer nitrate and phosphate ions reach the sunlit layer.
Phytoplankton primary production falls where nutrients become limiting.

Upwelling also depends on winds, coastline and circulation. Warming can alter its timing and extent; it does not impose one identical response in every ocean region.

Normal upwelling brings cold nutrient-rich deep water into the sunlit surface layer; warming strengthens stratification, reduces mixing, lowers surface nutrients and weakens the food chain.

Lower surface nutrients limit phytoplankton growth.
Less producer biomass feeds fewer zooplankton.
Fish recruitment and biomass can fall as prey supply declines.
Seabirds, marine mammals and fisheries receive less energy from the food web.

Phytoplankton also fix CO₂. Changes in production and sinking organic matter can alter biological carbon transfer to the deep ocean, linking food webs to climate feedback.

A predator decline alone does not prove reduced upwelling. Test the chain with current, nutrient, chlorophyll or phytoplankton, zooplankton and consumer data over time.

Range boundary	Response to warming
warm or equatorward/low-elevation edge	local populations decline or disappear as tolerance is exceeded
cool or poleward/upper edge	colonization can extend the range if habitat and dispersal allow
mountaintop or coastline barrier	available climate space contracts because no farther suitable area exists

A range shift is a population-distribution change, not an individual animal choosing to evolve. Movement, establishment and local extinction together redraw the range.

A map and mountain cross-section show a temperate species shifting poleward and a montane bird shifting upslope as its modern climate envelope moves.

Tracking requirement	Possible barrier
dispersal	seeds, juveniles or adults move too slowly
connected habitat	farms, cities or fragmented forest block movement
suitable soil and partners	climate arrives without pollinators, hosts or substrate
population growth	colonists fail to establish at the new edge

Repeated surveys can compare historical and modern upper and lower limits. Consistent upslope or poleward shifts across many species strengthen the climate interpretation, while species-specific exceptions reveal other controls.

Tree saplings may shift faster than mature-tree distributions because adult trees persist long after recruitment conditions change. Range contraction can occur even while old individuals remain.

Coral supplies to zooxanthellae	Zooxanthellae supply to coral
protected, illuminated cells and inorganic nutrients	photosynthetic organic carbon and oxygen
CO₂ and nitrogenous waste from metabolism	much of the energy supporting coral growth and calcification

Reef-building corals live near their upper thermal tolerance. Sustained heat stress disrupts photosynthesis and cellular regulation in the symbiosis.

The coral animal is not a plant. Its photosynthetic partner is the zooxanthella living within coral tissues.

Unusually warm water stresses the coral–algal association.
Damaged photosynthetic processes produce harmful reactive molecules.
Coral expels zooxanthellae or loses their pigments.
Transparent tissue reveals the white calcium carbonate skeleton.
Energy supply falls; prolonged or repeated bleaching increases disease and mortality.

Bleaching is not instant death. Corals can recover if stress ends and compatible symbionts return, but repeated events leave less time to rebuild energy reserves and reproduce.

White colour is the visible symptom; the central mechanism is loss of the mutualistic photosynthetic partner and its energy contribution.

Atmospheric CO₂ dissolves into seawater.
Dissolved CO₂ forms carbonic acid and releases hydrogen ions.
Hydrogen ions combine with carbonate ions.
Carbonate availability and calcium carbonate saturation fall.
Corals must spend more energy calcifying and skeleton growth weakens.

Warming and acidification are different pressures from the same CO₂-driven problem: heat disrupts symbiosis; altered carbonate chemistry suppresses skeleton formation.

Ocean acidification means falling pH, not that seawater necessarily becomes below pH 7. The biologically important change is the shift in carbonate chemistry.

Heat events reduce coral energy and survival.
Acidification slows new skeleton formation and reef repair.
Dead framework erodes faster than living corals rebuild it.
Three-dimensional refuges, feeding sites and nursery habitat disappear.
Fish and invertebrate diversity and food-web functions decline.

Pollution, sedimentation and overfishing can further weaken recovery. Ecosystem collapse arises from interacting pressures and loss of habitat-building coral, not temperature in isolation.

Two pathways show heat stress expelling zooxanthellae and dissolved CO₂ lowering pH and calcium carbonate formation; both converge on degraded reef structure and lower biodiversity.

human greenhouse-gas source → enhanced infrared retention → climate variable changes → feedback may amplify the change → habitat or physiological mechanism alters survival, production or distribution → population and ecosystem consequences

Impact	Mechanism that must appear
boreal sink becomes source	carbon release exceeds uptake through drought, mortality, fire and thaw
ice-dependent species decline	required resting or breeding platform is lost at the critical time
marine productivity falls	stratification reduces nutrient upwelling to phytoplankton
range contracts or shifts	climate space moves but dispersal and habitat constrain tracking
reef degrades	bleaching removes symbiont energy and acidification suppresses calcification

Do not jump directly from “warming” to “species decline.” The biological mechanism and the relevant time or place make the explanation causal.

Carbon sequestration is the capture of atmospheric CO₂ and its storage in a biological or physical pool for a meaningful period.

Term	Starting land and action
afforestation	establish forest where there was not previously forest cover
reforestation or forest regeneration	restore trees and forest processes where forest was lost or damaged
agroforestry	combine trees with crops or livestock on managed land
peatland rewetting	restore waterlogged conditions that slow decomposition

Sequestration is a flow into a store. Existing forest conservation mainly avoids emissions and protects a current store; it is not identical to creating additional uptake.

Approach	Carbon mechanism	Additional benefit or trade-off
afforestation	new tree growth fixes CO₂ into wood, roots and soil	can compete with other land uses or harm native open habitats
forest regeneration	surviving seed sources and succession rebuild native biomass	often supports habitat complexity but takes time
agroforestry	trees add long-lived biomass and reduce soil carbon loss within farms	retains food production but management determines permanence

match species to climate, water and native ecology
protect soil carbon and avoid displacing deforestation elsewhere
count survival and growth, not only seedlings planted
plan for fire, drought, harvest and future climate

A fast-growing monoculture can store carbon yet reduce biodiversity or water availability. Climate benefit must be evaluated with ecosystem consequences.

Waterlogging limits oxygen diffusion into soil.
Anaerobic conditions slow decomposer respiration.
Partly decomposed plant material accumulates as peat.
Drainage introduces oxygen, accelerating decomposition and CO₂ release.
Rewetting raises the water table and slows further carbon loss.

Peatlands hold exceptionally dense soil carbon accumulated over centuries to millennia. Preventing loss of an old store can have a different timescale and value from planting new biomass.

Rewetting can initially affect methane emissions and local land use. Net climate benefit must include multiple gases, water level and long-term avoided CO₂ loss.

Criterion	Question
additionality	would the carbon gain or avoided loss occur without the action?
permanence	could fire, drainage, harvest or drought rapidly release the store?
rate and capacity	how much carbon is stored, and over what time?
leakage	does the action shift deforestation or emissions elsewhere?
biodiversity and people	does it support native habitat, water and livelihoods or create new harm?

Sequestration complements rapid emissions reduction; finite land and reversible biological stores cannot indefinitely compensate for continued fossil-carbon release.

Report the baseline, boundary and monitoring period. Gross tree growth without mortality, soil change and displaced activity can overstate net removal.

reduce fossil and land-use emissions first → protect existing high-carbon ecosystems → add appropriate forest, agroforestry or peatland storage → monitor whole-system net carbon → maintain the store against fire, harvest, drainage and climate stress

Land context	Strong first option
intact high-carbon ecosystem	conserve the existing store and avoid degradation
degraded former forest	support native regeneration
productive farm	integrate suitable trees through agroforestry
drained peat soil	restore waterlogging where feasible
native non-forest habitat	avoid inappropriate tree planting

A tonne captured temporarily is not equivalent to a tonne of fossil carbon left underground permanently. Time, reversal risk and ecosystem fit belong in the comparison.

Event group	Examples
plant development	budburst, flowering, fruiting, leaf fall
animal movement	migration arrival or departure
reproduction	nesting, egg laying, breeding and birth
food availability	insect emergence and caterpillar biomass peak

Phenology is the study of the timing of recurring seasonal or cyclical biological events and the environmental cues associated with them.

A seasonal timeline places budburst, flowering, insect emergence, migration, nesting, autumn leaf change and seed dispersal under photoperiod and temperature cues.

Cue	Climate-change response	Biological implication
temperature	spring thresholds can occur earlier in warmer years	temperature-sensitive development may advance
photoperiod	day length on each calendar date is unchanged	species relying strongly on light may shift less
snow or ice state	melt and accessibility respond to weather and accumulated conditions	migration or feeding habitat can shift nonlinearly
internal state and food	depend on prior stages and resource history	responses can lag behind direct temperature cues

Interacting species remain synchronized only if their event dates shift by similar amounts. Different cue sensitivities create mismatch risk.

An earlier event can be phenotypic plasticity within individuals or evolution across generations. Timing change alone does not distinguish them.

Define the event precisely, such as first flower, median hatch date or peak insect biomass.
Record date and location with a consistent sampling method over many years.
Measure candidate cues such as temperature, photoperiod, snowmelt or rainfall.
Compare trends among interacting species and across sites.
Link timing overlap to survival, reproduction or resource use.

First-event dates are sensitive to observer effort and rare early individuals. Peak or median timing and repeated sampling can better represent a population.

A date shift shows phenological change; ecological harm requires evidence that overlap, food supply or breeding success also changed.

Phenological synchrony occurs when interacting events overlap at the time needed—for example, peak chick food demand coinciding with peak caterpillar biomass.

Interaction	If one event advances farther	Likely consequence
flower–pollinator	flowers open before pollinators are active	reduced pollination and seed set
herbivore–plant	young leaves mature before grazers arrive	lower food quality
chick–caterpillar	prey peak precedes maximum chick demand	slower growth or fewer fledglings

Both species can move earlier and still become mismatched if they shift by different amounts. Compare overlap, not just the direction of change.

Spring warming advances caterpillar development and peak biomass.
Great-tit laying also responds to temperature, but to cues over a different time window.
The prey peak can advance farther than hatching and chick demand.
Parents encounter less food during maximum chick growth.
Fledgling number or mass can fall, reducing reproductive success.

The mechanism is reduced overlap, not simply “warm weather harms birds.” Populations may partly adjust breeding dates, so long-term outcome depends on the pace and limits of that response.

Earlier decades show caterpillar biomass overlapping great-tit chick food need, while recent warm decades show the caterpillar peak shifted earlier and overlap reduced.

Warming and earlier snowmelt advance nutritious young plant growth.
Reindeer migration timing responds to temperature, snow, photoperiod, inherited routes and movement constraints.
If arrival advances less than the plant-growth peak, forage is older and less nutritious.
Reduced maternal condition or calf growth can lower fitness.

Migration can also advance and preserve some synchrony. The correct prediction compares the size of the plant and animal shifts rather than assuming mismatch from warming alone.

Demonstrate mismatch with plant quality and animal arrival data at the same places and years, then connect overlap to feeding or reproductive outcomes.

Temperature effect	Population consequence
faster enzyme-driven development within the viable range	egg-to-adult generation time shortens
longer frost-free season	more time remains for another complete generation
warmer winter	more overwintering stages survive
temperature exceeds tolerance or drought removes host quality	survival can fall despite faster development

Warming does not increase insect generations without limit. Development has an optimum and requires suitable hosts, moisture and enough season length to complete the extra cycle.

Voltinism is the number of generations completed in one year; climate change can shift a population from univoltine to bivoltine conditions.

Adult beetles tunnel through bark and lay eggs beside phloem.
Larvae feed on phloem, disrupting sucrose transport and weakening the tree.
Larvae pupate and new adults emerge.
Warmer seasons allow the cycle to finish earlier.
A second generation attacks again in late summer, when stressed trees may resist fewer beetles.

More adults and two attack periods increase the chance that beetle density exceeds the threshold needed to overwhelm resin and other tree defences.

Drought-stressed trees are easier to kill; dead trees add fuel and reduce carbon uptake, linking insect outbreaks to the boreal carbon-loss feedback.

A trait varies among individuals.
At least part of the variation is heritable.
Climate-linked conditions alter survival or reproductive success among variants.
Trait-associated alleles or heritable phenotypes change frequency across generations.

Observation	Why it is not yet proof of evolution
one animal changes behavior	can be within-lifetime plasticity
population moves north	can be migration without genetic change
phenotype mean changes	environment can alter phenotype directly

Climate change creates a selection pressure; populations evolve only when inherited variants contribute unequally to later generations.

Tawny owls vary heritably from pale grey to brown plumage.
Historically snowy winters favoured pale camouflage; brown birds had higher mortality in severe winters.
Milder winters reduce snow cover and change the camouflage advantage.
Brown birds survive and reproduce relatively more often.
Across decades, the brown phenotype rises from about 30% to about 50% of the population.

The owls do not darken during one winter. Natural selection changes the population frequency of inherited colour variants across generations.

Snowy habitat favours grey tawny-owl camouflage, low-snow forest favours brown camouflage, and frequency bars show the brown phenotype increasing from about 30% to 50%.

climate variable changes → seasonal cue or development rate changes → timing overlap, generations or relative fitness changes → ecological outcome occurs within lifetimes; evolution occurs only if inherited variants change frequency across generations

Observation	First question
flowering or migration date advances	which cue changed, and how was timing measured?
interacting events move earlier	did their overlap increase, persist or decline?
insect outbreak intensifies	did development time, winter survival or generation number change?
phenotype frequency shifts	is the variation heritable and linked to differential reproduction?

Plastic timing responses can occur before genetic evolution and can themselves alter selection. Keep immediate response, ecological interaction and generational change distinct.

[ 7 ]

Explain the impact of anthropogenic activity on climate change.

[ 4 ]

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

[ 2 ]

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

[ 3 ]

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

[ 1 ]

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

[ 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?

[ 5 ]

Outline the reasons that climate change is a threat to coral reefs.

[ 1 ]

Which action will decrease carbon sequestration?

[ 1 ]

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

[ 3 ]

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

Exam analysis

Most tested topics

DNA replication

Semi-conservative replication

Role of helicase and DNA polymerase

PCR and gel electrophoresis

Applications exam focus

DNA polymerase directionality

Leading vs. lagging strand

Functions in replication

DNA proofreading

Transcription exam focus

Hydrogen bonding in transcription

Transcription for gene expression

Translation exam focus

Roles in translation

Complementary base pairing

Genetic code features

Using genetic code table

Elongation of polypeptide

Mutations changing protein structure

Directionality exam focus

Initiation of transcription at promoter

Non-coding sequences in DNA

Post-transcriptional modification

Alternative splicing

Translation initiation

Polypeptide modification

Gene mutations

Base substitution consequences

Causes of mutation

Randomness in mutation

Consequences in germ vs. somatic cells

Mutation as source of variation

Gene knockout

CRISPR-Cas9 gene editing

Conserved sequences

Generation of new cells

Cytokinesis exam focus

Equal and unequal cytokinesis

Roles of mitosis and meiosis

DNA replication prerequisite

Shared features

Phases of mitosis

Identification of mitosis phases

Meiosis as reduction division

Down syndrome

Meiosis generates variation

Cell proliferation

Cell cycle phases

Cell growth during interphase

Cell cycle control by cyclins

Mutations in cell cycle genes

Tumour differences