Cambridge International AS & A Level Biology 9700: Inheritance – Study Guide

This guide teaches topic 16, Inheritance, of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027. It covers sections 16.1 Passage of information from parents to offspring, 16.2 The roles of genes in determining the phenotype and 16.3 Gene control. This is A Level content, not AS: it is examined on Paper 4 (A Level Structured Questions), and the chi-squared test can also appear on Paper 5 (Planning, Analysis and Evaluation).

Quick links: course hub · printable checklist · revision notes · practice questions · A Level diagnostic

What this topic covers

Section	What you must be able to do	Paper
16.1 (1–3)	Explain haploid and diploid, homologous pairs, and why gametes need a reduction division	4
16.1 (4–5)	Describe chromosome, nuclear envelope, membrane and spindle behaviour in each stage of meiosis; identify stages in photomicrographs	4
16.1 (6–7)	Explain how crossing over, independent assortment and random fertilisation produce variation	4
16.2 (1–4)	Use genetic terms; construct and interpret monohybrid, dihybrid and test crosses, including codominance, multiple alleles, sex linkage, autosomal linkage and epistasis	4
16.2 (5)	Use the chi-squared test (formula provided)	4, 5
16.2 (6–7)	Link gene → protein → phenotype for TYR, HBB, F8 and HTT; explain Le/le and gibberellin	4
16.3 (1–4)	Structural vs regulatory genes; repressible vs inducible enzymes; the lac operon; transcription factors; gibberellin and DELLA	4

16.1 Passage of information from parents to offspring

Haploid, diploid and homologous pairs

A diploid (2n) cell has two sets of chromosomes, one from each parent; a haploid (n) cell has one set. Human body cells: 2n = 46; gametes: n = 23.

Homologous chromosomes are a maternal and a paternal chromosome with the same genes at the same loci (same length and centromere position), possibly carrying different alleles.

Why a reduction division is needed

If gametes were diploid, the zygote would be 4n and chromosome number would double every generation. Meiosis halves it so fertilisation restores 2n.

The stages of meiosis

You need the names of the eight stages, not the sub-stages of prophase I.

Stage	What happens
Prophase I	Chromosomes condense; homologous chromosomes pair (bivalents); crossing over at chiasmata; nuclear envelope breaks down; spindle forms
Metaphase I	Bivalents line up on the equator; each pair’s orientation is random
Anaphase I	Homologous chromosomes are pulled to opposite poles; centromeres do not divide
Telophase I	Nuclear envelopes may re-form; cell surface membrane pinches in (animal) or a cell plate forms (plant): two haploid cells
Prophase II	Nuclear envelope breaks down again; new spindles form, often at right angles to the first
Metaphase II	Individual chromosomes line up on the equator
Anaphase II	Centromeres divide; sister chromatids are pulled to opposite poles
Telophase II	Nuclear envelopes re-form; cytokinesis gives four haploid cells

Animal cells have centrioles at the spindle poles; plant cells do not.

Identifying stages from photomicrographs. Pairs on the equator: metaphase I. Single chromosomes on the equator of one of two cells: metaphase II. Whole chromosomes moving to the poles: anaphase I; single chromatids moving: anaphase II.

Sources of genetic variation

• Crossing over in prophase I swaps sections between non-sister chromatids of a homologous pair. This makes new combinations of alleles on each chromatid.
• Random orientation (independent assortment) of homologous pairs at metaphase I gives each gamete a random mix of maternal and paternal chromosomes. Crossing over makes sister chromatids different, so their random orientation at metaphase II adds more variation.
• Random fertilisation: any male gamete can fuse with any female gamete.

Worked example. How many chromosome combinations can one human make by independent assortment alone?

number of homologous pairs = 23
combinations = 2^23 = 8 388 608
two parents: 8 388 608 × 8 388 608 ≈ 7.04 × 10^13 possible zygotes

Crossing over adds far more variation.

16.2 The roles of genes in determining the phenotype

Terms you must define

Term	Meaning
Gene	A length of DNA that codes for a polypeptide (or an RNA)
Locus	The position of a gene on a chromosome
Allele	One of two or more alternative forms of a gene
Dominant	An allele expressed in the phenotype in a heterozygote
Recessive	An allele expressed only when homozygous (no dominant allele present)
Codominant	Both alleles are expressed in the heterozygote
Linkage	Genes on the same chromosome, tending to be inherited together
Test cross	Crossing an individual with a homozygous recessive to find its genotype
F1, F2	First and second filial generations
Genotype / phenotype	The alleles present / the observable features
Homozygous / heterozygous	Two identical alleles / two different alleles

Setting out a genetic diagram

Give parental phenotypes and genotypes, gametes, a Punnett square, offspring genotypes linked to phenotypes, and the ratio. Define your symbols; use superscripts for codominant, multiple and sex-linked alleles.

Codominance and multiple alleles

The ABO blood group gene has three alleles: Iᴬ and Iᴮ are codominant; Iᴼ is recessive to both. A person can carry only two of them.

Worked example. Parents IᴬIᴼ (group A) × IᴮIᴼ (group B).

gametes:        Iᴬ, Iᴼ         Iᴮ, Iᴼ
                 Iᴮ        Iᴼ
      Iᴬ       IᴬIᴮ (AB)  IᴬIᴼ (A)
      Iᴼ       IᴮIᴼ (B)   IᴼIᴼ (O)
ratio: 1 AB : 1 A : 1 B : 1 O

Sex linkage

Genes on the non-homologous part of the X chromosome are sex-linked. Males (XY) have one copy, so a single recessive allele shows. Haemophilia A is an example.

Parents:    carrier female XᴴXʰ  ×  unaffected male XᴴY
gametes:    Xᴴ, Xʰ                  Xᴴ, Y
offspring:  XᴴXᴴ unaffected female, XᴴXʰ carrier female,
            XᴴY unaffected male,   XʰY haemophiliac male

Half the sons are expected to have haemophilia; no daughters are affected.

Dihybrid crosses

Genes on different chromosomes assort independently. AaBb × AaBb gives four gamete types from each parent (AB, Ab, aB, ab) and a 9 : 3 : 3 : 1 phenotype ratio.

Autosomal linkage

Alleles of genes on the same autosome tend to be inherited together. A test cross of AaBb (AB on one chromosome, ab on the other) × aabb gives 1 : 1 : 1 : 1 if unlinked; with linkage the parental combinations are much more common.

Worked example. A test cross gives 412 AaBb, 398 aabb, 93 Aabb and 97 aaBb offspring (total 1000). The two large classes are the parental combinations. The recombinants (Aabb and aaBb) make up 190 out of 1000, or 19%, well below the 50% expected without linkage. They are produced when crossing over occurs between the two loci in prophase I.

Epistasis

Epistasis is when one gene affects the expression of another gene at a different locus. It often happens when two genes code for enzymes in the same pathway. You must be able to construct the diagram; you do not need to learn the ratios.

Worked example. In a flower, a colourless precursor → yellow pigment (enzyme coded by allele A) → red pigment (enzyme coded by allele B). The genes are on different chromosomes.

• aa: no yellow pigment, so white whatever the B genotype (gene A is epistatic to gene B). A_bb: yellow. A_B_: red.

AaBb × AaBb: of the 16 boxes in the Punnett square, 9 are A_B_ (red), 3 are A_bb (yellow) and 4 are aa (white). Ratio 9 red : 3 yellow : 4 white.

The chi-squared test

The test shows whether the difference between observed (O) and expected (E) numbers is significant or due to chance. The formula is provided:

χ² = Σ (O − E)² / E        degrees of freedom = number of classes − 1

Work out the degrees of freedom yourself and compare χ² with the critical value at p = 0.05. Below it: not significant.

Worked example. A dihybrid cross expected to give 9 : 3 : 3 : 1 produces 480 offspring: 279, 97, 83 and 21.

E = 480 × 9/16, 3/16, 3/16, 1/16 = 270, 90, 90, 30
(O − E)²/E = 81/270 + 49/90 + 49/90 + 81/30
           = 0.300 + 0.544 + 0.544 + 2.700 = 4.09
degrees of freedom = 4 − 1 = 3; critical value at p = 0.05 = 7.82
4.09 < 7.82

Not significant (p > 0.05): the results fit 9 : 3 : 3 : 1, so the genes are probably unlinked.

Genes, proteins and phenotype

Gene	Protein	Condition and how it arises
TYR	Tyrosinase, an enzyme that converts tyrosine into melanin (via DOPA)	Albinism. A recessive allele codes for inactive tyrosinase, so homozygous recessive people make little or no melanin: pale skin, hair and eyes
HBB	β-globin polypeptide of haemoglobin	Sickle cell anaemia. A base substitution changes glutamic acid to valine (haemoglobin S). At low oxygen concentration HbS forms fibres, red cells sickle, carry less oxygen and block capillaries. HbA and HbS are codominant
F8	Factor VIII, a blood-clotting protein	Haemophilia A. A recessive allele on the X chromosome codes for non-functional factor VIII, so blood clots slowly. Mostly males are affected
HTT	Huntingtin	Huntington’s disease. A dominant allele has an expanded CAG repeat, so the protein carries an extra-long glutamine chain that damages brain neurones; symptoms usually start in adulthood

Gibberellin and the Le/le alleles

Gibberellins cause stem elongation in internodes. In peas, the Le allele codes for a functional enzyme that converts an inactive gibberellin (GA₂₀) into active GA₁. The recessive le allele has a single base substitution that changes one amino acid near the enzyme’s active site, making the enzyme non-functional. lele plants make little active gibberellin, so they have short internodes and are dwarf. LeLe and Lele plants are tall.

16.3 Gene control

Structural and regulatory genes

• A structural gene codes for a protein with a function in the cell, such as an enzyme or a membrane protein.
• A regulatory gene codes for a protein (for example a repressor) that controls the expression of other genes.

Inducible and repressible enzymes

• An inducible enzyme is made only when its substrate is present, e.g. β-galactosidase in E. coli when lactose is present.
• A repressible enzyme is made until its end product builds up and switches transcription off, e.g. the enzymes that make tryptophan in E. coli.

The lac operon

An operon is a group of structural genes under the control of one promoter and one operator. The lac operon contains:

a promoter (RNA polymerase binds), an operator (the repressor binds) and three structural genes: lacZ (β-galactosidase, hydrolyses lactose to glucose and galactose), lacY (lactose permease) and lacA (transacetylase).

A separate regulatory gene, lacI, with its own promoter, codes for the repressor protein.

Lactose absent: the repressor binds to the operator, so RNA polymerase cannot transcribe the structural genes.

Lactose present: lactose (as allolactose) binds to the repressor and changes its shape, so it cannot bind to the operator. RNA polymerase transcribes lacZ, lacY and lacA as one mRNA. (The role of cAMP is not required.)

Transcription factors in eukaryotes

Transcription factors are proteins that bind to DNA and control gene expression in eukaryotes by increasing or decreasing the rate of transcription.

Gibberellin and DELLA proteins

DELLA proteins are repressors: they bind to transcription factors that would promote transcription of growth genes. Gibberellin binds to a receptor, which leads to the breakdown of DELLA proteins. The transcription factors are released, bind to the promoters of their target genes and transcription starts. In a lele plant there is little active gibberellin, so DELLA proteins persist and growth genes stay off.

Common errors

• Separating sister chromatids in anaphase I. In anaphase I homologous chromosomes separate; chromatids separate in anaphase II.
• Writing gametes as “AaBb”. Each gamete carries one allele of each gene: AB, Ab, aB or ab.
• Writing sex-linked alleles as letters alone (Hh) instead of Xᴴ and Xʰ, or giving the Y chromosome an allele.
• Calling a χ² below the critical value “significant”.
• Describing the lac repressor as binding to the promoter, or lactose binding to the operator.
• Writing that the le allele is “missing” or “codes for no protein”. It codes for a non-functional enzyme.

Next steps

Condense this with the inheritance revision notes, then try the inheritance practice questions. Carry the ideas of alleles, variation and chi-squared into Selection and evolution.

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.