Cambridge International AS & A Level Biology 9700: Inheritance – Revision Notes

These notes condense topic 16, Inheritance, of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027: sections 16.1, 16.2 and 16.3. It is A Level content, examined on Paper 4, with the chi-squared test also possible on Paper 5. For full explanations and worked examples, use the inheritance study guide.

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16.1 Meiosis in one page

Haploid (n): one set of chromosomes. Diploid (2n): two sets, one from each parent. Homologous pair: one maternal and one paternal chromosome with the same genes at the same loci; alleles may differ. Why reduce? Gametes must be haploid so that fertilisation restores 2n instead of doubling the chromosome number each generation.

Stage	Chromosomes	Nuclear envelope, membrane, spindle
Prophase I	Condense; homologues pair (bivalents); crossing over at chiasmata	Envelope breaks down; spindle forms
Metaphase I	Bivalents on the equator, random orientation	Spindle fibres attach at centromeres
Anaphase I	Homologues to opposite poles; centromeres intact	Spindle fibres shorten
Telophase I	Two haploid groups	Envelopes may re-form; membrane pinches in (animal) or cell plate (plant)
Prophase II	Re-condense	Envelope breaks down; new spindles
Metaphase II	Single chromosomes on the equator	Spindle fibres attach
Anaphase II	Centromeres divide; chromatids to poles	Spindle fibres shorten
Telophase II	Four haploid groups	Envelopes re-form; cytokinesis gives four cells

Spotting stages in a photomicrograph: pairs → meiosis I; two cells each dividing → meiosis II; four nuclei → telophase II.

Three sources of variation

1. Crossing over (prophase I): non-sister chromatids exchange sections → new allele combinations.
2. Random orientation / independent assortment of homologous pairs (metaphase I) and of sister chromatids (metaphase II) → 2ⁿ chromosome combinations, where n = number of pairs.
3. Random fertilisation: any gamete can fuse with any other.

16.2 Genetics toolkit

Must-know distinctions

Pair	Difference
Gene vs allele	Gene = length of DNA at a locus coding for a polypeptide; allele = one version of that gene
Dominant vs codominant	Dominant masks the other allele; codominant alleles are both expressed in the heterozygote
Multiple alleles vs polygenes	One gene with more than two alleles vs many genes affecting one feature
Autosomal linkage vs sex linkage	Two genes on the same autosome vs a gene on the X chromosome
Linkage vs epistasis	Genes on one chromosome inherited together vs one gene affecting expression of another
Test cross vs F2 cross	Unknown × homozygous recessive vs F1 × F1

Expected ratios to recognise

Cross	Expected phenotype ratio
Monohybrid, Aa × Aa (dominance)	3 : 1
Monohybrid, codominance	1 : 2 : 1
Monohybrid test cross, Aa × aa	1 : 1
Dihybrid, AaBb × AaBb, unlinked	9 : 3 : 3 : 1
Dihybrid test cross, AaBb × aabb, unlinked	1 : 1 : 1 : 1
Dihybrid with autosomal linkage	Parental classes in excess; few recombinants
Epistasis	Modified ratio (you work it out; you need not learn it)

Method: any genetic diagram

1. Define symbols (a key). Use superscripts for codominant or multiple alleles (Iᴬ, Iᴮ, Iᴼ) and for sex linkage (Xᴴ, Xʰ, Y).
2. Parental phenotypes → parental genotypes.
3. Gametes: one allele of each gene per gamete, each type listed once.
4. Punnett square.
5. Offspring genotypes, each linked to its phenotype.
6. Ratio, in the same order as the phenotypes.

Method: chi-squared test

1. State the null hypothesis: no significant difference between observed and expected.
2. Calculate expected numbers from the ratio and the total.
3. For each class calculate (O − E)² / E, then add: χ² = Σ (O − E)² / E (formula given).
4. Degrees of freedom = number of classes − 1 (not given).
5. Compare with the critical value at p = 0.05 (1 df: 3.84; 2 df: 5.99; 3 df: 7.82).
6. χ² < critical value → difference not significant, due to chance. χ² > critical value → significant; something other than chance (for example linkage) is acting.

Worked reminder. 480 offspring, expected 9 : 3 : 3 : 1, observed 279, 97, 83, 21. E = 270, 90, 90, 30. χ² = 0.300 + 0.544 + 0.544 + 2.700 = 4.09. 3 df, critical value 7.82, so not significant.

Gene → protein → phenotype

Gene	Protein	Phenotype when the allele is faulty	Inheritance
TYR	Tyrosinase (tyrosine → melanin)	Albinism: little or no melanin	Autosomal recessive
HBB	β-globin of haemoglobin	Sickle cell anaemia: HbS (valine for glutamic acid) polymerises at low O₂, cells sickle	HbA and HbS codominant
F8	Factor VIII (clotting)	Haemophilia: slow clotting	X-linked recessive
HTT	Huntingtin	Huntington’s disease: neurone damage, late onset	Autosomal dominant (expanded CAG repeat)

Gibberellin and Le/le

• Le (dominant) codes for a functional enzyme that makes active gibberellin (GA₁ from GA₂₀).
• le (recessive) codes for a non-functional enzyme (one amino acid changed near the active site).
• lele → little active gibberellin → little stem elongation → dwarf plant.

16.3 Gene control

Term	Meaning
Structural gene	Codes for a protein used in the cell (enzyme, transport protein)
Regulatory gene	Codes for a protein that controls other genes (e.g. the lac repressor)
Inducible enzyme	Made only when its substrate is present (β-galactosidase)
Repressible enzyme	Production stops when its product builds up (tryptophan synthesis enzymes)
Transcription factor	Protein that binds to DNA and increases or decreases the rate of transcription in eukaryotes

The lac operon in steps

Order on the DNA: promoter → operator → lacZ → lacY → lacA. The regulatory gene lacI is separate and has its own promoter.

• No lactose: repressor (from lacI) binds to the operator → RNA polymerase blocked → no transcription of lacZ, lacY, lacA.
• Lactose present: lactose (as allolactose) binds to the repressor → repressor changes shape → leaves the operator → RNA polymerase transcribes all three genes → β-galactosidase, lactose permease and transacetylase made.
• cAMP is not required.

Gibberellin and DELLA in steps

1. DELLA proteins bind to transcription factors (for example PIFs) and stop them promoting transcription of growth genes.
2. Gibberellin binds to its receptor.
3. This causes DELLA proteins to be broken down.
4. Transcription factors are released, bind to promoters and increase transcription.
5. Growth proteins are made; stem cells elongate.

Quick self-test

1. A cell with 2n = 12 undergoes meiosis. How many chromosomes are in each cell at the end of telophase I?
2. How many chromosome combinations can this organism make by independent assortment alone?
3. In which stage do sister chromatids separate?
4. A test cross gives 46 and 54 offspring, expected 1 : 1. Calculate χ² and state whether the difference is significant.
5. Two people who are both heterozygous for the TYR recessive allele have a child. What is the probability that the child is a girl with albinism?
6. A dihybrid cross of two double heterozygotes gives 320 offspring. Give the expected numbers in each class if the genes are unlinked.
7. How many degrees of freedom are there for a chi-squared test with four phenotype classes?
8. A man with haemophilia has children with a woman who is homozygous for the normal allele. What proportion of their sons will have haemophilia?
9. Two people with genotype HbᴬHbˢ have a child. What is the probability that the child is HbˢHbˢ?
10. A person heterozygous for the Huntington’s disease allele has a child with a person who does not carry it. What is the probability the child inherits the allele?
11. A cross expected to give 3 : 1 produces 68 and 32 of 100 offspring. Calculate χ².
12. What binds to the lac operator when lactose is absent?

Answers

1. 6 (each still made of two chromatids).
2. 2⁶ = 64.
3. Anaphase II.
4. χ² = (46 − 50)²/50 + (54 − 50)²/50 = 0.64. 1 df, critical value 3.84; 0.64 < 3.84, so not significant.
5. ¼ (albino) × ½ (girl) = 1/8.
6. 180 : 60 : 60 : 20.
7. 3.
8. None (0): sons get their X from the mother, who has only Xᴴ. All daughters are carriers.
9. 1/4.
10. 1/2 (the allele is dominant, so this is also the probability the child will develop the disease).
11. E = 75 and 25. χ² = 49/75 + 49/25 = 2.61; below 3.84, so not significant.
12. The repressor protein coded by the regulatory gene lacI.

Where marks are usually lost

• Stating that meiosis “halves the DNA” instead of the number of chromosomes, or saying the cells produced are “identical”.
• Mixing up the anaphases: homologues separate in anaphase I; sister chromatids separate in anaphase II.
• Leaving out the nuclear envelope and spindle when a question asks for their behaviour, not just the chromosomes’.
• Writing gametes that contain two alleles of one gene, or listing duplicate gamete types.
• Writing sex-linked genotypes without the X and Y, which loses the mark for the genotype even if the ratio is right.
• Giving χ² without degrees of freedom and the critical value, or concluding the data are “proven”.
• Saying the lac repressor binds to the promoter, or that lactose binds to the operator.
• Describing DELLA as binding directly to DNA. It inhibits the transcription factors that bind to the promoter.
• Saying the le allele is absent or “does not code”. It codes for a non-functional enzyme.

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027 (Version 1), Cambridge University Press & Assessment – topic 16, Inheritance.