What is the role of DNA in the passage of information from parents to offspring?

DNA, or deoxyribonucleic acid, is the hereditary material in humans and almost all other organisms. It carries the genetic instructions used in the growth, development, functioning, and reproduction of all known living organisms. In the context of inheritance, DNA is responsible for passing genetic information from parents to offspring, ensuring that offspring inherit traits from both parents. This process is fundamental to the study of Biology at the Cambridge International A Levels.

How do alleles influence the traits of offspring in inheritance?

Alleles are different forms of a gene that arise by mutation and are found at the same place on a chromosome. They play a crucial role in determining the traits of an organism. During inheritance, offspring receive one allele from each parent for every gene. The combination of these alleles determines the offspring's traits, such as eye color or blood type. Understanding alleles is essential for Cambridge International A Levels Biology students studying inheritance.

What is the difference between dominant and recessive alleles in inheritance?

Dominant alleles are those that express their trait even when only one copy is present, while recessive alleles require two copies to express their trait. In a heterozygous pairing, where one dominant and one recessive allele are present, the dominant allele's trait will be observed. This concept is fundamental in understanding how traits are passed from parents to offspring and is a key topic in Cambridge International A Levels Biology.

How does meiosis contribute to genetic variation in offspring?

Meiosis is a type of cell division that reduces the chromosome number by half, creating four genetically diverse gametes. This process includes crossing over and independent assortment, which shuffle alleles and contribute to genetic variation in offspring. This genetic diversity is crucial for evolution and adaptation, making meiosis a significant topic in the study of inheritance for Cambridge International A Levels Biology students.

What is the significance of Mendel's laws in understanding inheritance?

Mendel's laws, including the Law of Segregation and the Law of Independent Assortment, form the foundation of classical genetics. They explain how alleles are separated and recombined during the formation of gametes and how traits are inherited independently of one another. These principles are essential for understanding the passage of information from parents to offspring and are a critical part of the Cambridge International A Levels Biology curriculum.

Why is "Saying crossing over occurs between sister chromatids" a common mistake in Passage of information from parents to offspring?

Why it happens: Students confuse 'sister chromatids' (identical copies of one chromosome) with 'non-sister chromatids' (one from each of a homologous pair). How to avoid it: Crossing over occurs between **non-sister chromatids** of **homologous chromosomes** (different chromosomes that pair up). Sister chromatids are identical, so exchanging DNA between them would produce no new combinations. Source: 9700 Examiner Reports 2024

Why is "Saying independent assortment occurs at metaphase II" a common mistake in Passage of information from parents to offspring?

Why it happens: Students confuse the two divisions of meiosis. How to avoid it: Independent assortment occurs at **metaphase I** (the random orientation of homologous PAIRS, not sister chromatids). Metaphase II involves sister chromatids of the already-separated single chromosomes. Source: 9700 Examiner Reports 2023

Why is "Forgetting to state the null hypothesis in chi-squared" a common mistake in Passage of information from parents to offspring?

Why it happens: Students focus on the calculation and skip the conceptual step. How to avoid it: **Always** start the χ² test with: 'H₀: there is no significant difference between observed and expected frequencies'. Examiners award a mark for this and may not award the final interpretation mark without it. Source: 9700 Examiner Reports 2024

Why is "Reversing the χ² decision rule" a common mistake in Passage of information from parents to offspring?

Why it happens: Students mistakenly think 'big χ² means good fit'. How to avoid it: Small χ² = small difference = ACCEPT H₀ (good fit). Large χ² (> critical value) = significant difference = REJECT H₀. Mnemonic: 'small chi means same' — small difference means hypothesis holds. Source: 9700 Examiner Reports 2024

Why is "Writing genetic ratios without phenotype labels" a common mistake in Passage of information from parents to offspring?

Why it happens: Students rush to the answer. How to avoid it: Always write '3 purple : 1 white' or '9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green' — not just '3:1' or '9:3:3:1'. Source: 9700 Examiner Reports 2023-2024

Why is "Treating genotype and phenotype as the same" a common mistake in Passage of information from parents to offspring?

Why it happens: The terms sound similar. How to avoid it: Genotype = ALLELES carried (Pp). Phenotype = observable trait (purple flowers). Two different genotypes (PP, Pp) can give the same phenotype. Source: 9700 Examiner Reports 2024

Why is "Using 'gene' when 'allele' is meant (and vice versa)" a common mistake in Passage of information from parents to offspring?

Why it happens: Students learn the terms interchangeably. How to avoid it: Gene = a length of DNA at a locus that codes for a protein (e.g. the 'ABO gene'). Allele = a variant of that gene (e.g. Iᴬ, Iᴮ, i are three alleles of the ABO gene). Everyone has the ABO gene; only some people have allele Iᴬ. Source: 9700 Examiner Reports 2024

InheritanceCambridge International A Levels Biology (9700)

Passage of information from parents to offspring

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Passage of information from parents to offspring — Cambridge International A Level Biology 9700 Study Notes (2025-2027 syllabus)

How meiosis generates genetic variation (crossing over + independent assortment); essential genetic vocabulary (gene, allele, genotype, phenotype, locus, dominance); monohybrid and dihybrid Punnett squares; chi-squared statistical testing of genetic ratios.

What you’ll learn

Mapped to the Cambridge IGCSE 9700 syllabus (2025-2027).

16.1.1 — Describe the behaviour of chromosomes in meiosis, including the pairing of homologous chromosomes, crossing over and the formation of chiasmata.
16.1.2 — Explain how meiosis generates genetic variation through crossing over and the random/independent assortment of homologous chromosomes at metaphase I.
16.1.3 — Define and use the terms: gene, allele, locus, dominant, recessive, codominant, homozygous, heterozygous, genotype and phenotype.
16.1.4 — Use genetic diagrams to predict offspring genotypes and phenotypes in monohybrid and dihybrid crosses with complete dominance.
16.1.5 — Explain how a test cross is used to determine the genotype of an organism showing the dominant phenotype.
16.1.6 — Use the chi-squared (χ²) test to compare observed and expected ratios of phenotypes from genetic crosses.

Meiosis — the engine of genetic variation

Two divisions, four haploid daughter cells; generates variation by crossing over (prophase I) and independent assortment (metaphase I).

Meiosis is a specialised type of nuclear division that:

Produces four daughter cells from one parent cell.
Halves the chromosome number (diploid → haploid), so that fertilisation restores the diploid number.
Generates genetic variation — each of the four daughter cells is genetically different from each other and from the parent.

Meiosis takes place only in cells destined to become gametes (sperm and ova in animals; pollen and embryo sac in plants). It involves two consecutive divisions:

Meiosis I (the reduction division). Separates homologous chromosomes. The daughter cells are haploid, each with a single chromatid pair (still replicated).

Prophase I — chromosomes condense; homologous pairs come together (synapsis) to form bivalents (tetrads); crossing over occurs at chiasmata.
Metaphase I — bivalents line up at the equator with random orientation (independent assortment).
Anaphase I — homologous chromosomes separate (one to each pole); sister chromatids stay together.
Telophase I — two haploid cells form, each containing replicated chromosomes.

Meiosis II (similar to mitosis). Separates sister chromatids. Result: four haploid cells, each with single (unreplicated) chromosomes.

Prophase II — chromosomes recondense in each haploid cell.
Metaphase II — single chromosomes line up at equator.
Anaphase II — sister chromatids separate.
Telophase II — four haploid cells form.

Comparison with mitosis.

Feature	Mitosis	Meiosis
Divisions	1	2
Daughter cells	2	4
Genetically identical to parent?	Yes	No (variation)
Chromosome number	Same (diploid → diploid)	Halved (diploid → haploid)
Homologue pairing	No	Yes (bivalents in prophase I)
Crossing over	No	Yes (prophase I)
Purpose	Growth, repair, asexual reproduction	Gamete formation

Two sources of variation in meiosis (covered in detail next):

Crossing over at prophase I — shuffles alleles within chromosomes.
Independent assortment at metaphase I — shuffles whole chromosomes.

Variation is then amplified by random fertilisation (any gamete with any gamete).

Meiosis: 1 diploid parent → 4 haploid daughter cells.
Two divisions: Meiosis I (separates homologues) + Meiosis II (separates chromatids).
Crossing over: prophase I, between non-sister chromatids of homologous pair, at chiasmata.
Independent assortment: metaphase I, random orientation of bivalents.
Random fertilisation amplifies variation.
Differs from mitosis: 4 cells (not 2), haploid (not diploid), genetically variable (not identical).

Crossing over and chiasmata

Non-sister chromatids of homologous pair exchange DNA at chiasmata in prophase I → recombinant chromatids.

Crossing over is the physical exchange of segments of DNA between the non-sister chromatids of a pair of homologous chromosomes during prophase I of meiosis.

Sequence of events:

Synapsis. As prophase I begins, the two members of each homologous pair come together along their length to form a bivalent (also called a tetrad — four chromatids).
Crossing over. At several points along the bivalent, the non-sister chromatids (one chromatid from the maternal chromosome and one from the paternal chromosome) cross over each other. Enzymes cut and rejoin the DNA at these crossing points, swapping segments of DNA.
Chiasmata. The points of crossing become visible under the microscope as chiasmata (singular: chiasma). Each chiasma is a physical link between non-sister chromatids where DNA has been exchanged.
Recombinant chromatids. After crossing over, each chromatid in the bivalent carries a new combination of alleles — some from the maternal chromosome and some from the paternal chromosome. These are recombinant chromatids.

Sister vs non-sister chromatids — critical distinction:

Sister chromatids are the two identical copies produced by DNA replication of one chromosome. They are joined at the centromere.
Non-sister chromatids are chromatids from different homologous chromosomes (one maternal, one paternal). They are NOT identical — they carry the same genes but possibly different alleles.

Crossing over occurs only between non-sister chromatids; exchanging DNA between sister chromatids would produce no genetic variation because they are identical.

Why this matters. Without crossing over, all the alleles on one chromosome would be inherited as a 'block'. Crossing over allows alleles to be shuffled within a chromosome, so genes that are physically linked on the same chromosome can still end up in different gametes. The frequency of crossing over between two loci depends on the distance between them — this is the basis of genetic linkage mapping.

Quantitative effect. Even a single chiasma per bivalent doubles the number of allele combinations in the resulting gametes. In humans, an average of 1-3 chiasmata occur per bivalent in every meiosis, making the number of possible allele combinations effectively unlimited.

Crossing over: prophase I, between non-sister chromatids of homologous pair.
Chiasmata = visible exchange points; enzymes cut and rejoin DNA.
Result: recombinant chromatids with new allele combinations.
Sister chromatids = identical copies; non-sister = from different homologues.
Frequency of crossing over depends on distance between loci.

Independent assortment

Random orientation of bivalents at metaphase I → 2ⁿ chromosome combinations (n = haploid number).

At metaphase I, the bivalents (paired homologous chromosomes) line up at the equator of the cell. The orientation of each bivalent — which chromosome (maternal or paternal) faces which pole — is random and independent of the orientation of every other bivalent.

When the homologues separate at anaphase I, each daughter cell receives a random mix of maternal and paternal chromosomes. Importantly, the alleles of one gene (on one chromosome) sort independently of the alleles of another gene (on a different chromosome) — this is Mendel's Law of Independent Assortment.

Counting combinations. With $n$ homologous pairs, the random orientation gives $2^n$ possible combinations of chromosomes in the gametes:

Species	$n$ (haploid number)	$2^n$ combinations
Fruit fly (Drosophila)	4	16
Human	23	8 388 608 (~ $8.4 \times 10^6$ )
Wheat (hexaploid)	21	2 097 152

Important caveat — Mendel's law only applies to genes on DIFFERENT chromosomes. Genes on the same chromosome are linked (covered in topic 16.2) and do NOT assort independently — unless crossing over separates them.

Random fertilisation. Each of the millions of possible gametes from the father can fertilise any one of the millions from the mother. For humans, this gives approximately $(8.4 \times 10^6) \times (8.4 \times 10^6) \approx 7 \times 10^{13}$ possible zygotic combinations from just independent assortment + random fertilisation — before crossing over is included. With crossing over added, every offspring (except identical twins) is genetically unique.

Evolutionary significance. This staggering amount of variation is the raw material on which natural selection acts. Different combinations of alleles produce different phenotypes; those better adapted to the environment survive and reproduce; populations evolve. Without meiosis (and the variation it produces), evolution would proceed only by mutation — far slower.

Metaphase I: random orientation of bivalents at equator.
Each bivalent orients independently of every other.
2ⁿ combinations of chromosomes in gametes (n = haploid number).
Humans: n = 23 → ~8.4 million combinations.
Random fertilisation multiplies this further (~10¹³).
Only applies to genes on DIFFERENT chromosomes (Mendel's Law of Independent Assortment).

See the full worked example for passage of information from parents to offspring →

Essential genetic vocabulary

Gene, allele, locus, genotype, phenotype, homozygous, heterozygous, dominant, recessive — the building blocks.

Genetics has a precise vocabulary. Cambridge expects exact definitions — using imprecise language is the most common source of lost marks.

Gene. A length of DNA that codes for a polypeptide (or a functional RNA molecule). Each gene has a specific position (locus) on a particular chromosome. Different versions of a gene are called alleles. Example: the gene for ABO blood group is on chromosome 9.

Allele. A variant form of a gene. Different alleles encode slightly different versions of the polypeptide, producing phenotypic variation. Example: the ABO gene has three alleles: Iᴬ, Iᴮ and i.

Locus. The specific position of a gene on a chromosome. Homologous chromosomes have the same loci (but possibly different alleles at those loci).

Genotype. The genetic constitution of an organism — the specific alleles it carries at one or more loci. Written as letter pairs: e.g. Pp, Iᴬi, X^h Y. Cannot be observed directly.

Phenotype. The observable characteristics of an organism, resulting from the interaction of the genotype with the environment. Example: genotype Pp → phenotype purple flowers.

Homozygous. Having two identical alleles at a given locus (e.g. PP or pp). 'Homo' = same. A homozygous organism is also called a true-breeding for that gene — when crossed with itself, all offspring have the same genotype.

Heterozygous. Having two different alleles at a given locus (e.g. Pp). 'Hetero' = different.

Dominant allele. An allele that is expressed in the phenotype when present in only one copy (heterozygous). Written with an uppercase letter. Example: the purple-flower allele P is dominant — both PP and Pp give purple flowers.

Recessive allele. An allele that is only expressed in the phenotype when present in two copies (homozygous recessive). Written with a lowercase letter. Example: the white-flower allele p is recessive — only pp gives white flowers.

Codominance (covered more in 16.2). Two alleles both contribute to the phenotype of a heterozygote. Example: ABO blood group — IᴬIᴮ gives blood type AB, with both A and B antigens present.

Test cross. A cross of an organism showing the dominant phenotype (could be PP or Pp) with a homozygous recessive (pp). Used to determine the unknown genotype:

If the test cross gives all dominant offspring → unknown was PP.
If the test cross gives a 1:1 dominant:recessive ratio → unknown was Pp.

A summary table of differences:

Term	Definition	Example
Gene	Length of DNA coding for a polypeptide	ABO gene
Allele	Variant of a gene	Iᴬ, Iᴮ, i
Locus	Position of gene on chromosome	Chromosome 9 (ABO locus)
Genotype	Alleles carried	Iᴬi
Phenotype	Observable trait	Blood type A
Homozygous	Two identical alleles	PP or pp
Heterozygous	Two different alleles	Pp
Dominant	Expressed in heterozygote	P → purple
Recessive	Only expressed in homozygote	p → white (only pp)

Gene = DNA → polypeptide; allele = variant of a gene.
Locus = position; genotype = alleles carried; phenotype = observable.
Homozygous = same alleles (PP, pp); heterozygous = different (Pp).
Dominant = expressed in heterozygote (uppercase); recessive = needs two copies (lowercase).
Test cross with homozygous recessive identifies unknown genotype.

See the full worked example for passage of information from parents to offspring →

Monohybrid crosses — one gene at a time

Pp × Pp gives F₂ 1 PP : 2 Pp : 1 pp genotypes → 3 purple : 1 white phenotype ratio.

A monohybrid cross involves a single gene with two alleles. The classic Mendelian inheritance pattern with complete dominance gives:

Pure-breeding parents (P generation):

Purple PP × White pp

F₁ generation:

All offspring Pp (heterozygous) → all show purple phenotype.
Demonstrates that the purple allele is dominant over white.

F₁ self-cross (Pp × Pp):

	P	p
P	PP	Pp
p	Pp	pp

F₂ genotypes: 1 PP : 2 Pp : 1 pp.

F₂ phenotypes: 3 purple : 1 white — the classic Mendelian monohybrid ratio.

Test cross (Pp × pp):

	p	p
P	Pp	Pp
p	pp	pp

Genotype ratio: 1 Pp : 1 pp. Phenotype ratio: 1 purple : 1 white.

How to set out a genetic cross in an exam:

State parental genotypes clearly (with phenotypes labelled).
State the gametes each parent produces (encircled letters).
Draw the Punnett square with gametes along top and side.
Fill in offspring genotypes in the body of the square.
State the phenotype ratio clearly with labels (not just '3:1').

Mendel's First Law (Law of Segregation). During gamete formation, the two alleles of a gene separate so that each gamete carries only one of the alleles. Each gamete is equally likely to carry either allele. (This is a direct consequence of meiosis — homologues separate in meiosis I.)

Monohybrid cross: one gene, two alleles.
F₁ (Pp × Pp): genotypes 1 PP : 2 Pp : 1 pp; phenotypes 3:1.
Test cross (Pp × pp): 1:1 dominant:recessive — identifies heterozygote.
Always show parental genotypes, gametes, Punnett square, ratio with labels.
Mendel's First Law of Segregation: alleles separate at gamete formation.

See the full worked example for passage of information from parents to offspring →

Dihybrid crosses — two genes at once

RrYy × RrYy: 4 gamete types each → 9:3:3:1 phenotypic ratio in F₂.

A dihybrid cross involves two genes (each with two alleles). Mendel's classic dihybrid experiment used pea seeds:

Round (R) dominant to wrinkled (r) — seed shape.
Yellow (Y) dominant to green (y) — seed colour.

Pure-breeding parents: RRYY (round yellow) × rryy (wrinkled green).

F₁ generation: all RrYy — all round yellow.

F₁ self-cross (RrYy × RrYy):

Each parent can produce four types of gametes by independent assortment, in equal proportion: RY, Ry, rY, ry.

Punnett square (4 × 4 = 16 boxes):

	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRYy	RRyy	RrYy	Rryy
rY	RrYY	RrYy	rrYY	rrYy
ry	RrYy	Rryy	rrYy	rryy

Phenotypes (count by phenotype):

Round, yellow (R_Y_): 9 — includes RRYY, RRYy, RrYY, RrYy genotypes.
Round, green (R_yy): 3 — RRyy, Rryy.
Wrinkled, yellow (rrY_): 3 — rrYY, rrYy.
Wrinkled, green (rryy): 1.

Phenotypic ratio = 9 : 3 : 3 : 1 — round yellow : round green : wrinkled yellow : wrinkled green.

Why 9:3:3:1? Each gene independently shows a 3:1 ratio in F₂:

Round : wrinkled = 3 : 1 (so 3/4 round, 1/4 wrinkled).
Yellow : green = 3 : 1 (so 3/4 yellow, 1/4 green).

Because the genes assort independently (different chromosomes), we multiply the probabilities:

P(round AND yellow) = 3/4 × 3/4 = 9/16
P(round AND green) = 3/4 × 1/4 = 3/16
P(wrinkled AND yellow) = 1/4 × 3/4 = 3/16
P(wrinkled AND green) = 1/4 × 1/4 = 1/16

Ratio: 9 : 3 : 3 : 1.

Mendel's Second Law (Law of Independent Assortment). Alleles of different genes assort independently into gametes — provided the genes are on different chromosomes. (Genes on the same chromosome are linked and do NOT obey this law — see topic 16.2.)

Dihybrid cross: two genes, two alleles each.
F₁ (RrYy × RrYy): four gamete types each → 4×4 Punnett.
F₂ phenotype ratio: 9:3:3:1.
Arises from (3/4) × (3/4) = 9/16 etc. when genes assort independently.
Mendel's Second Law: independent assortment (for unlinked genes).

See the full worked example for passage of information from parents to offspring →

The chi-squared (χ²) test

Compares observed and expected counts. χ² < critical (at p=0.05, df=categories−1) → accept null hypothesis.

Real experimental data rarely gives perfect Mendelian ratios — by chance, the observed counts deviate from the expected ones. The chi-squared (χ²) test is a statistical test that tells us whether the deviation is small enough to be due to chance alone, or whether it is statistically significant.

When to use χ²: comparing observed and expected categorical counts (not measurements, not percentages). Common applications:

Testing whether offspring ratios fit a predicted Mendelian ratio (3:1, 9:3:3:1, 1:2:1, etc.).
Testing whether observed allele frequencies fit Hardy-Weinberg equilibrium.

Step-by-step:

Step 1: State the null hypothesis (H₀).

"There is no significant difference between the observed and expected frequencies; the offspring fit the predicted X:Y:Z ratio."

Step 2: Calculate expected frequencies. From the predicted ratio and the total number of offspring.

For a 9:3:3:1 ratio with $N$ total: expected counts are $9N/16$ , $3N/16$ , $3N/16$ , $N/16$ .

Step 3: Compute (O − E)²/E for each category.

Step 4: Sum all values to find χ²:

$\chi^2 = \sum \frac{(O - E)^2}{E}$

Step 5: Degrees of freedom. $df =$ (number of categories) − 1.

Monohybrid (3:1 or 1:1): 2 categories → df = 1.
Dihybrid (9:3:3:1): 4 categories → df = 3.

Step 6: Compare with the critical value at the chosen significance level (conventionally $p = 0.05$ , meaning we accept up to 5% chance of being wrong).

Critical χ² values at p = 0.05:

df	Critical χ²
1	3.84
2	5.99
3	7.815
4	9.49
5	11.07

Step 7: Interpret.

χ² calculated < critical value → accept H₀. The deviation is not significant; the data fit the predicted ratio.
χ² calculated > critical value → reject H₀. The deviation is significant; the data do NOT fit the predicted ratio.

Worked example. A dihybrid cross gives observed 92 : 28 : 31 : 9 in a 9:3:3:1 expectation (N = 160). Expected: 90 : 30 : 30 : 10.

$\chi^2 = \frac{(92-90)^2}{90} + \frac{(28-30)^2}{30} + \frac{(31-30)^2}{30} + \frac{(9-10)^2}{10}$

$= 0.044 + 0.133 + 0.033 + 0.100 = 0.310$

df = 4 − 1 = 3. Critical at p=0.05 = 7.815. χ² (0.310) << critical (7.815), so accept H₀ — data fit 9:3:3:1.

Biological interpretation if H₀ is rejected. A significant difference suggests the predicted model is wrong. Possible causes:

Genes are linked (on the same chromosome).
Epistasis (one gene masks another).
Lethal alleles (some genotypes die before birth).
Sample bias or small sample size.

Important assumptions of χ². Categories must be mutually exclusive and exhaustive; observations must be independent; expected counts should be ≥5 in every category (otherwise the test is unreliable).

χ² compares observed vs expected counts.
H₀: no significant difference from expected ratio.
Formula: χ² = Σ (O−E)²/E.
df = categories − 1.
Critical values at p=0.05: df 1 → 3.84; df 3 → 7.815.
χ² < critical → accept H₀ (data fit); > critical → reject (data don't fit).
Rejecting H₀ suggests linkage, epistasis, lethal alleles, etc.

See the full worked example for passage of information from parents to offspring →

How it’s examined

Cambridge 9700 examines this topic on Paper 4 (A Level structured questions, often 8-12 marks per part). Frequent questions: (a) describe how meiosis generates variation (5-6 marks); (b) construct a genetic cross showing parents, gametes, Punnett square and ratios (5-7 marks); (c) interpret a dihybrid cross — explain the 9:3:3:1 ratio (5-6 marks); (d) full χ² test on a given dataset (6-8 marks); (e) define key genetic terms with examples (3-4 marks). Paper 4 also asks AO3 application questions requiring candidates to interpret novel pedigrees or experimental data.

Sources: Cambridge International AS & A Level Biology 9700 syllabus (2025-2027); 9700 Examiner Reports 2022-2024; 9700/42 May/Jun 2024 question paper and mark scheme; 9700/42 Oct/Nov 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Passage of information from parents to offspring

Step-by-step solutions to past-paper-style questions on passage of information from parents to offspring, written exactly the way a tutor would explain them at the board.

1Sources of genetic variation in meiosis (5 marks)
Extended• Adapted from 9700/42 May/Jun 2024• meiosis, variation, Paper 4
Question
Describe two ways in which meiosis generates genetic variation in the gametes. (5 marks)
Step-by-step solution
1. Step 1
  Crossing over (prophase I). During prophase I of meiosis, homologous chromosomes pair up to form bivalents. The non-sister chromatids of the homologous pair physically exchange segments of DNA at points called chiasmata (singular: chiasma).
2. Step 2
  Effect of crossing over. This exchanges alleles between maternal and paternal chromosomes, producing recombinant chromatids with new combinations of alleles that were not present in either parental chromosome. Each chromatid in the resulting gametes is therefore a unique mix of maternal and paternal alleles.
3. Step 3
  Independent assortment (metaphase I). At metaphase I, the bivalents line up at the equator of the cell. The orientation of each bivalent — which chromosome (maternal or paternal) faces which pole — is random and independent of the orientation of every other bivalent.
4. Step 4
  Effect of independent assortment. With $n$ homologous pairs, this generates $2^n$ possible combinations of chromosomes in the gametes. For humans ( $n=23$ ), this gives $2^{23} \approx 8.4$ million combinations from independent assortment alone, before crossing over is even considered.
5. Step 5
  Random fertilisation amplifies variation. Any one of the millions of possible gametes from one parent can fertilise any one of the millions from the other — so each zygote is one of about $8.4 \times 10^6 \times 8.4 \times 10^6 \approx 7 \times 10^{13}$ unique combinations (before crossing over). This is the source of genetic variation upon which natural selection acts.
Answer
Crossing over (prophase I): non-sister chromatids of homologous pair exchange DNA at chiasmata → recombinant chromatids with new allele combinations. Independent assortment (metaphase I): bivalents orient randomly at equator → 2ⁿ combinations of chromosomes in gametes (n=23 for humans → ~8.4 million). Random fertilisation further amplifies variation.
Examiner tip
Mark scheme: (1) crossing over named + prophase I; (2) chiasmata / exchange between non-sister chromatids; (3) independent assortment named + metaphase I; (4) random orientation of bivalents / 2ⁿ; (5) random fertilisation. Examiner Reports note candidates often confuse 'homologous pair' with 'sister chromatids' — crossing over is between NON-SISTER chromatids of a HOMOLOGOUS PAIR.
2Monohybrid cross with Punnett square (4 marks)
Extended• monohybrid, Punnett, Paper 4
Question
In pea plants, the allele for purple flowers (P) is dominant over the allele for white flowers (p). A heterozygous purple-flowered plant is crossed with a homozygous white-flowered plant. Determine the expected genotype and phenotype ratios in the offspring. (4 marks)
Step-by-step solution
1. Step 1
  Parental genotypes. Heterozygous purple = Pp; homozygous white = pp.
2. Step 2
  Gametes. Pp parent produces gametes containing either P or p in a 1:1 ratio. pp parent produces gametes all containing p.
3. Step 3
  Punnett square.
  
  p p
  P Pp Pp
  p pp pp
4. Step 4
  Ratios. Genotype: 1 Pp : 1 pp (i.e. 50% heterozygous, 50% homozygous recessive). Phenotype: 1 purple : 1 white (50:50). This is a test cross — used to determine whether an unknown purple plant is homozygous PP or heterozygous Pp.
Answer
Pp × pp → gametes P/p × p only. Punnett square gives Pp:pp = 1:1. Phenotype ratio = 1 purple : 1 white (50:50). This is a test cross.
Examiner tip
Mark scheme: (1) parental genotypes correct; (2) gametes correctly identified; (3) Punnett square drawn correctly; (4) ratios stated (genotype + phenotype). Examiner Reports note candidates lose marks by writing '1:1 purple to white' without explicitly stating phenotype/genotype labels.
3Dihybrid cross — F₂ 9:3:3:1 ratio (6 marks)
Extended• Adapted from 9700/42 Oct/Nov 2024• dihybrid, Punnett, Paper 4
Question
In Mendel's pea experiments, round (R) is dominant to wrinkled (r) and yellow (Y) is dominant to green (y). A double-heterozygous plant (RrYy) is self-pollinated. Predict the phenotypic ratios in the F₂ generation. (6 marks)
Step-by-step solution
1. Step 1
  Parental genotype. Both parents in the self-cross are RrYy (double heterozygous, round-yellow phenotype).
2. Step 2
  Gamete types. Each parent can produce four types of gametes by independent assortment: RY, Ry, rY, ry — each in 1/4 frequency.
3. Step 3
  4 × 4 Punnett square.
  
  RY Ry rY ry
  RY RRYY RRYy RrYY RrYy
  Ry RRYy RRyy RrYy Rryy
  rY RrYY RrYy rrYY rrYy
  ry RrYy Rryy rrYy rryy
4. Step 4
  Phenotypes. Count by phenotype:
  
  Round, yellow (at least one R AND at least one Y): 9 boxes
  
  Round, green (R_yy): 3 boxes
  
  Wrinkled, yellow (rrY_): 3 boxes
  
  Wrinkled, green (rryy): 1 box
5. Step 5
  Phenotypic ratio = 9 : 3 : 3 : 1. Round Yellow : Round Green : Wrinkled Yellow : Wrinkled Green = 9 : 3 : 3 : 1. This is Mendel's classic dihybrid F₂ ratio.
6. Step 6
  Why the 9:3:3:1 ratio arises. Each character independently shows a 3:1 ratio in F₂ (round:wrinkled = 3:1; yellow:green = 3:1). Because they assort independently (different chromosomes), the combined probability is 3/4 × 3/4 = 9/16 for both dominant, 3/16 for each one dominant + one recessive, 1/16 for both recessive — giving 9:3:3:1.
Answer
RrYy × RrYy → 4 gamete types each (RY, Ry, rY, ry). 4×4 Punnett square gives phenotypic ratio 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green = 9:3:3:1. Arises from independent assortment (3/4 × 3/4 = 9/16 etc.).
Examiner tip
Mark scheme: (1) parents correct; (2) gametes correct (4 types); (3) Punnett square set up correctly; (4) phenotypes counted; (5) ratio 9:3:3:1; (6) reason — independent assortment / 3/4 × 3/4. Examiner Reports note candidates lose marks by writing ratios without phenotype labels — 'round yellow:round green:wrinkled yellow:wrinkled green' must be specified.
4Chi-squared test on dihybrid ratio (7 marks)
Extended• Adapted from 9700/42 May/Jun 2024• chi-squared, statistics, Paper 4
Question
A geneticist crosses two double-heterozygous fruit flies (GgRr × GgRr) and observes 160 offspring: 92 grey-red, 28 grey-white, 31 black-red, 9 black-white. Use a chi-squared test to determine whether the observed ratios fit the expected 9:3:3:1 ratio (p = 0.05). (7 marks)
Step-by-step solution
1. Step 1
  Null hypothesis. $H_0$ : there is no significant difference between the observed and expected ratios; the offspring fit the expected 9:3:3:1 ratio.
2. Step 2
  Expected frequencies. Total = 160. Expected: 9/16 × 160 = 90; 3/16 × 160 = 30; 3/16 × 160 = 30; 1/16 × 160 = 10.
3. Step 3
  Calculate (O − E)² / E for each class.
  
  Phenotype O E O − E (O − E)² (O − E)²/E
  Grey-red 92 90 2 4 0.044
  Grey-white 28 30 −2 4 0.133
  Black-red 31 30 1 1 0.033
  Black-white 9 10 −1 1 0.100
4. Step 4
  Sum to find χ². $\chi^2 = 0.044 + 0.133 + 0.033 + 0.100 = 0.310$ .
5. Step 5
  Degrees of freedom. $df =$ number of categories − 1 = 4 − 1 = 3.
6. Step 6
  Critical value. From chi-squared tables, at $p = 0.05$ and $df = 3$ , the critical value is 7.815.
7. Step 7
  Conclusion. Calculated χ² (0.310) is much less than the critical value (7.815). Therefore we accept the null hypothesis: the observed ratios are not significantly different from 9:3:3:1. The data are consistent with independent assortment of two unlinked genes with complete dominance.
Answer
H₀: no significant difference from 9:3:3:1. Expected: 90, 30, 30, 10. χ² = (2²/90) + (−2)²/30 + 1²/30 + (−1)²/10 = 0.310. df = 3. Critical value at p=0.05 = 7.815. χ² < critical → accept H₀. Data fit 9:3:3:1.
Examiner tip
Mark scheme: (1) null hypothesis stated; (2) expected values calculated; (3) (O-E)²/E for each class; (4) χ² total; (5) df = 3; (6) critical value cited; (7) correct conclusion (accept H₀). Examiner Reports note candidates often forget to state the null hypothesis explicitly — this is a key A-Level skill.
5Genotype vs phenotype (3 marks)
Extended• definitions, Paper 4
Question
Distinguish between genotype and phenotype, using a named example. (3 marks)
Step-by-step solution
1. Step 1
  Genotype. The genetic constitution of an organism — the specific alleles it carries at one or more loci. Written using letters (uppercase for dominant alleles, lowercase for recessive). For ABO blood group: Iᴬi is a genotype.
2. Step 2
  Phenotype. The observable characteristics of an organism — the result of the interaction of the genotype with the environment. For ABO blood group, the phenotype of Iᴬi is blood type A (red blood cells display A antigen). Phenotype is what you see; genotype is what you cannot see directly.
3. Step 3
  Same phenotype, different genotypes. Two different genotypes can give the same phenotype if the dominant allele is masking. E.g. PP and Pp both give purple flowers — they share a phenotype but differ in genotype. A test cross with pp is needed to distinguish them.
Answer
Genotype = genetic make-up / alleles carried (e.g. Iᴬi, Pp). Phenotype = observable characteristics / how genotype is expressed (e.g. blood type A, purple flowers). Different genotypes can give same phenotype (PP and Pp both purple).
Examiner tip
Mark scheme: (1) genotype defined (alleles / genetic make-up); (2) phenotype defined (observable / appearance); (3) named example showing they can differ. Use ABO blood groups OR Mendel's peas as standard examples.

Phenotype	O	E	O − E	(O − E)²	(O − E)²/E
Grey-red	92	90	2	4	0.044
Grey-white	28	30	−2	4	0.133
Black-red	31	30	1	1	0.033
Black-white	9	10	−1	1	0.100

Model Answers — Passage of information from parents to offspring

High-scoring sample answers for passage of information from parents to offspring on the Cambridge International A Level 9700 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question
9700/42 May/Jun 2024 Q5 (adapted)5 marks
Q (5 marks). Describe two ways in which meiosis produces genetic variation.
Model answer
Meiosis generates genetic variation in the gametes by two main mechanisms: crossing over and independent assortment.

1. Crossing over (prophase I). Early in meiosis I, homologous chromosomes pair up to form bivalents (the process is called synapsis). Within each bivalent, the non-sister chromatids (one from each homologous chromosome) physically swap segments of DNA at structures called chiasmata (the singular is chiasma). Each chromosome consists of maternal- and paternal-derived stretches, with the chiasma being the visible crossing point.

The exchange produces recombinant chromatids with new combinations of alleles — combinations not present in either of the parental chromosomes. This creates allele combinations that did not exist before fertilisation and means every chromatid in the resulting gametes is a unique mosaic. The frequency of crossing over depends on the distance between gene loci: genes far apart on the same chromosome cross over more often than genes close together. This is the basis of genetic mapping.

2. Independent assortment (metaphase I). When the bivalents line up at the cell equator in metaphase I, the orientation of each bivalent is random — which homologue (maternal or paternal) ends up facing which pole is independent of every other bivalent. With $n$ homologous pairs, this random orientation gives $2^n$ possible combinations of whole chromosomes in the gametes.

For humans, $n = 23$ , so $2^{23} \approx 8.4 \times 10^6$ chromosomal combinations are possible from independent assortment alone — and that is before any crossing over is added in.

Random fertilisation amplifies variation further. Each of these millions of possible male gametes can fuse with any one of the millions of possible female gametes. So the total number of possible zygote combinations from a single human couple is approximately $(8.4 \times 10^6)^2 \approx 7 \times 10^{13}$ , before crossing over. With crossing over, the figure becomes effectively unlimited — every offspring is genetically unique (except for identical twins).

This variation is the raw material on which natural selection acts. Without sexual reproduction (and the variation generated by meiosis), evolution would proceed only by mutation — far more slowly.
Why this scores
Why this scores 5/5. Mark scheme: (1) crossing over named + prophase I; (2) chiasmata + non-sister chromatids; (3) independent assortment named + metaphase I; (4) random orientation / 2ⁿ; (5) random fertilisation amplifies. 9700 Examiner Reports note candidates often muddle 'homologous chromosomes' (different chromosomes that pair) with 'sister chromatids' (identical copies of one chromosome). Crossing over is between non-sister chromatids of a homologous pair.
Question
9700/42 Oct/Nov 2024 Q5(b) (adapted)4 marks
Q (4 marks). Explain why crossing over at prophase I increases genetic variation in the offspring.
Model answer
Crossing over is the exchange of segments of DNA between the non-sister chromatids of homologous chromosomes during prophase I of meiosis. The points of physical exchange are called chiasmata.

How variation arises. Each chromosome carries many gene loci. Before crossing over, every chromatid carries the parental combination of alleles at all loci. After crossing over, each chromatid carries a new combination of alleles: some alleles from the maternal chromosome and some from the paternal chromosome on the same chromatid. These are recombinant chromatids — they were not present in either parental gamete.

Why this increases variation. Independent assortment alone shuffles whole chromosomes (giving $2^n$ combinations from $n$ homologous pairs). Crossing over additionally shuffles the alleles within each chromosome, so even genes that are linked on the same chromosome can end up in new combinations in the gametes. Without crossing over, all genes on one chromosome would always be inherited together.

Quantitative effect. The number of possible allele combinations becomes effectively unlimited. For two linked genes with one chiasma between them in 50% of meioses, the recombination frequency is 25% (each recombinant type appears in 25% of gametes); without crossing over, recombinant gametes would never form.

Evolutionary significance. The new allele combinations created by crossing over may produce phenotypes that are better adapted to the environment. Natural selection can act on this variation, leading to evolution.
Why this scores
Why this scores 4/4. Mark scheme: (1) crossing over between non-sister chromatids at chiasmata; (2) new allele combinations on each chromatid; (3) increases variation beyond independent assortment alone; (4) significance for evolution / natural selection. Examiner Reports flag candidates who write 'sister chromatids exchange DNA' — this is WRONG; sister chromatids are identical, so exchanging produces no new combinations.
Question
Practice question — recurrent Paper 4 style4 marks
Q (4 marks). A heterozygous black guinea pig (Bb) is crossed with a homozygous white guinea pig (bb). State the expected genotype and phenotype ratios in the F₁ generation.
Model answer
Parental cross: Bb × bb.

Gametes produced:

Bb parent → ½ B : ½ b

bb parent → all b

Punnett square:

b b
B Bb Bb
b bb bb

Genotype ratio: 1 Bb : 1 bb (50% heterozygous, 50% homozygous recessive).

Phenotype ratio: 1 black : 1 white (because B is dominant, so Bb shows black; bb shows white).

Significance. This 1:1 ratio is the expected outcome of a test cross — crossing an organism showing the dominant phenotype with a homozygous recessive. A test cross is used in genetics to determine whether an organism showing the dominant phenotype is homozygous (BB) or heterozygous (Bb). If the test cross gives all dominant offspring, the unknown parent is BB. If it gives a 1:1 ratio (as here), the unknown parent is Bb. This was the method Mendel used to confirm his hypothesis of factor (now allele) segregation.
Why this scores
Why this scores 4/4. Mark scheme: (1) parental gametes correct; (2) Punnett square correct; (3) genotype ratio 1 Bb : 1 bb; (4) phenotype ratio 1 black : 1 white / significance as test cross. Examiner Reports note candidates lose marks by writing '50:50' without labelling phenotype vs genotype, or by not naming this as a test cross.
Question
9700/42 May/Jun 2024 Q5(c) (adapted)6 marks
Q (6 marks). Explain how to use the chi-squared (χ²) test to determine whether observed offspring ratios fit a predicted 9:3:3:1 dihybrid ratio. Include the formula, degrees of freedom, and the interpretation of the result.
Model answer
The chi-squared (χ²) test is a statistical test for comparing observed with expected frequencies in categorical data. In genetics, it is used to test whether observed offspring ratios are consistent with a predicted Mendelian ratio.

Step 1 — State the null hypothesis. $H_0$ : there is no significant difference between the observed and expected frequencies; the offspring fit the expected ratio.

Step 2 — Calculate expected frequencies. For a predicted 9:3:3:1 ratio with total $N$ offspring, expected counts are $9N/16$ , $3N/16$ , $3N/16$ , $N/16$ .

Step 3 — Calculate χ². Use the formula:

$\chi^2 = \sum \frac{(O - E)^2}{E}$

where $O$ = observed count, $E$ = expected count, and the sum is over all categories.

Step 4 — Find the degrees of freedom. $df =$ (number of categories) − 1. For a dihybrid cross with 4 phenotype categories, $df = 3$ .

Step 5 — Compare with the critical value. Read the critical χ² from tables at the chosen significance level (conventionally p = 0.05) and the appropriate df. For $df = 3$ and $p = 0.05$ , the critical value is 7.815.

Step 6 — Decide.

If χ² calculated < critical value: accept $H_0$ — there is no significant difference; the data fit the expected ratio.

If χ² calculated > critical value: reject $H_0$ — there IS a significant difference; the data do NOT fit the expected ratio. The genes may be linked, or one of the assumptions (dominance, independent assortment, viability) may be wrong.

Step 7 — Interpret biologically. Acceptance of $H_0$ supports the hypothesis of independent assortment of two unlinked genes with complete dominance. Rejection prompts investigation of possible causes — linkage, epistasis, gene-environment interaction, or non-Mendelian inheritance.

Important caveats. The chi-squared test requires:

Categorical data (counts, not measurements or percentages).

Independent observations.

Expected counts $E \geq 5$ in every category (otherwise the test is unreliable).

Always state the null hypothesis explicitly at the start of the analysis — examiners regularly award a mark for this alone.
Why this scores
Why this scores 6/6. Mark scheme: (1) null hypothesis; (2) calculate expected from ratio; (3) χ² formula with (O-E)²/E; (4) df = categories − 1 = 3; (5) compare with critical value at p=0.05; (6) interpret — accept/reject H₀ and biological meaning. 9700 Examiner Reports note candidates routinely lose marks by writing χ² < critical → 'reject H₀' (wrong direction) — small χ² means CLOSE to expected, so ACCEPT H₀.
Question
9700/42 Oct/Nov 2024 Q5(a) (adapted)3 marks
Q (3 marks). Distinguish between genotype and phenotype, using a named example.
Model answer
Genotype is the genetic constitution of an organism — the specific combination of alleles it carries at one or more gene loci. Genotypes are usually written as letter pairs (e.g. Pp, Iᴬi, X^h Y), with uppercase for dominant alleles and lowercase for recessive.

Phenotype is the observable, expressed characteristic of an organism — what we can see (or measure) as the result of the interaction between the genotype and the environment.

Example — pea flower colour. In Mendel's pea plants, the allele for purple flowers (P) is dominant over the allele for white (p). Plants with genotype PP or Pp both show purple flowers (the phenotype). Plants with genotype pp show white flowers. So:

Genotype PP → phenotype purple

Genotype Pp → phenotype purple

Genotype pp → phenotype white

Key point. Two different genotypes (PP and Pp) can give the same phenotype (purple). To distinguish them, a test cross with the homozygous recessive (pp) is performed: PP × pp gives all purple offspring; Pp × pp gives a 1:1 ratio of purple : white.

Environmental influence. Phenotype is not always solely determined by genotype — environment matters too. For example, the Himalayan rabbit has the genotype for dark fur only at low temperatures (ears, paws), so its phenotype depends on environmental temperature even with constant genotype.
Why this scores
Why this scores 3/3. Mark scheme: (1) genotype defined (alleles / genetic make-up); (2) phenotype defined (observable / appearance / result of genotype + environment); (3) named example showing they can differ (PP and Pp both purple). Examiner Reports note candidates lose marks by writing 'phenotype = what you look like' without mentioning the role of the genotype and environment.

	b	b
B	Bb	Bb
b	bb	bb

Key Definitions and Keywords — Passage of information from parents to offspring

Definitions to memorise and the exact keywords mark schemes credit for passage of information from parents to offspring answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9700 sitting.

Meiosis
Examiner keyword
A type of nuclear division that produces four genetically different haploid daughter cells (gametes) from one diploid parent cell. Involves two divisions (meiosis I + II) and generates variation through crossing over and independent assortment.
Homologous chromosomes
Examiner keyword
A pair of chromosomes that carry the same genes at the same loci but possibly different alleles. One member is maternal and one paternal. They pair up during prophase I of meiosis.
Bivalent
A pair of homologous chromosomes joined together during prophase I of meiosis. Each homologue is already replicated, so a bivalent contains four chromatids (a tetrad).
Crossing over
Examiner keyword
The exchange of segments of DNA between the non-sister chromatids of homologous chromosomes at chiasmata, occurring during prophase I of meiosis. Produces recombinant chromatids with new allele combinations.
Chiasma (plural: chiasmata)
Examiner keyword
The visible point of contact between non-sister chromatids of homologous chromosomes during prophase I of meiosis, where crossing over occurs.
Independent assortment
Examiner keyword
The random orientation of homologous pairs (bivalents) at the equator during metaphase I of meiosis, so that the alleles of one gene segregate independently of the alleles of another (if on different chromosomes). Produces 2ⁿ combinations of chromosomes in the gametes.
Gene
Examiner keyword
A length of DNA that codes for a polypeptide or a functional RNA. Found at a specific position (locus) on a chromosome.
Allele
Examiner keyword
A variant form of a gene. Different alleles produce slightly different polypeptides (or different amounts of the same polypeptide), giving rise to phenotypic variation.
Locus (plural: loci)
Examiner keyword
The specific position of a gene on a chromosome. Homologous chromosomes carry the same loci (though possibly different alleles).
Genotype
Examiner keyword
The genetic constitution of an organism — the specific combination of alleles at one or more loci. Written as letter pairs (e.g. Pp, Iᴬi). Not directly observable.
Phenotype
Examiner keyword
The observable characteristics of an organism, resulting from the interaction of the genotype with the environment.
Homozygous
Examiner keyword
Having two identical alleles at a given locus (e.g. PP or pp). 'Homo' = same.
Heterozygous
Examiner keyword
Having two different alleles at a given locus (e.g. Pp). 'Hetero' = different.
Dominant allele
Examiner keyword
An allele that is expressed in the phenotype when present in only one copy (heterozygous). Conventionally written with an uppercase letter.
Recessive allele
Examiner keyword
An allele that is only expressed in the phenotype when present in two copies (homozygous recessive). Conventionally written with a lowercase letter.
Test cross
Examiner keyword
A cross between an individual showing the dominant phenotype (genotype unknown — could be PP or Pp) with a homozygous recessive individual (pp). Used to determine the unknown genotype.
Monohybrid cross
A genetic cross involving only one gene with two alleles. F₂ phenotype ratio is 3:1 (dominant : recessive).
Dihybrid cross
A genetic cross involving two genes (each with two alleles). F₂ phenotype ratio with independent assortment is 9:3:3:1.
Chi-squared (χ²) test
Examiner keyword
A statistical test that compares observed and expected frequencies of categorical data. Used in genetics to evaluate whether observed offspring ratios fit a predicted Mendelian ratio. Formula: $\chi^2 = \sum (O-E)^2/E$ .

Common Mistakes and Misconceptions — Passage of information from parents to offspring

The traps other students keep falling into on passage of information from parents to offspring questions — taken from recent Cambridge International A Level 9700 examiner reports and mark schemes — and how to avoid them.

✕Saying crossing over occurs between sister chromatids
9700 Examiner Reports 2024
Why it happens
Students confuse 'sister chromatids' (identical copies of one chromosome) with 'non-sister chromatids' (one from each of a homologous pair).
How to avoid it
Crossing over occurs between non-sister chromatids of homologous chromosomes (different chromosomes that pair up). Sister chromatids are identical, so exchanging DNA between them would produce no new combinations.
✕Saying independent assortment occurs at metaphase II
9700 Examiner Reports 2023
Why it happens
Students confuse the two divisions of meiosis.
How to avoid it
Independent assortment occurs at metaphase I (the random orientation of homologous PAIRS, not sister chromatids). Metaphase II involves sister chromatids of the already-separated single chromosomes.
✕Forgetting to state the null hypothesis in chi-squared
9700 Examiner Reports 2024
Why it happens
Students focus on the calculation and skip the conceptual step.
How to avoid it
Always start the χ² test with: 'H₀: there is no significant difference between observed and expected frequencies'. Examiners award a mark for this and may not award the final interpretation mark without it.
✕Reversing the χ² decision rule
9700 Examiner Reports 2024
Why it happens
Students mistakenly think 'big χ² means good fit'.
How to avoid it
Small χ² = small difference = ACCEPT H₀ (good fit). Large χ² (> critical value) = significant difference = REJECT H₀. Mnemonic: 'small chi means same' — small difference means hypothesis holds.
✕Writing genetic ratios without phenotype labels
9700 Examiner Reports 2023-2024
Why it happens
Students rush to the answer.
How to avoid it
Always write '3 purple : 1 white' or '9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green' — not just '3:1' or '9:3:3:1'.
✕Treating genotype and phenotype as the same
9700 Examiner Reports 2024
Why it happens
The terms sound similar.
How to avoid it
Genotype = ALLELES carried (Pp). Phenotype = observable trait (purple flowers). Two different genotypes (PP, Pp) can give the same phenotype.
✕Using 'gene' when 'allele' is meant (and vice versa)
9700 Examiner Reports 2024
Why it happens
Students learn the terms interchangeably.
How to avoid it
Gene = a length of DNA at a locus that codes for a protein (e.g. the 'ABO gene'). Allele = a variant of that gene (e.g. Iᴬ, Iᴮ, i are three alleles of the ABO gene). Everyone has the ABO gene; only some people have allele Iᴬ.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Passage of information from parents to offspring — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Passage of information from parents to offspring

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Sources of genetic variation in meiosis (5 marks)

2Monohybrid cross with Punnett square (4 marks)

3Dihybrid cross — F₂ 9:3:3:1 ratio (6 marks)

4Chi-squared test on dihybrid ratio (7 marks)

5Genotype vs phenotype (3 marks)

Meiosis

Homologous chromosomes

Bivalent

Crossing over

Chiasma (plural: chiasmata)

Independent assortment

Gene

Allele

Locus (plural: loci)

Genotype

Phenotype

Homozygous

Heterozygous

Dominant allele

Recessive allele

Test cross

Monohybrid cross

Dihybrid cross

Chi-squared (χ²) test

✕Saying crossing over occurs between sister chromatids

✕Saying independent assortment occurs at metaphase II

✕Forgetting to state the null hypothesis in chi-squared

✕Reversing the χ² decision rule

✕Writing genetic ratios without phenotype labels

✕Treating genotype and phenotype as the same

✕Using 'gene' when 'allele' is meant (and vice versa)

Practice questions

Past paper style quiz

Assess your understanding

Passage of information from parents to offspring — frequently asked questions