Home / A-Level Biology 9700 / Topic 16 / Revision notes
Download PDFQuestion bank →Topic 16 lessons

Inheritance

A-Level Biology 9700 · Topic 16 revision notes

Everything from the 5 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.

  1. Meiosis and genetic variation
  2. Monohybrid and dihybrid crosses
  3. The chi-squared test
  4. Genes, proteins and phenotype
  5. Gene control

1Meiosis and genetic variation

Syllabus 16.1

A diploid (2n) cell has two complete sets of chromosomes, one set from each parent; a haploid (n) cell has one set. The two chromosomes of a homologous pair have the same genes at the same loci (but possibly different alleles), the same length and the same centromere position. Gametes must be haploid so that fertilisation restores the diploid number instead of doubling it each generation, so they are made by meiosis, a reduction division. In meiosis I homologous chromosomes pair up and are separated; in meiosis II sister chromatids are separated, giving four haploid cells. Gametes differ genetically because of crossing over between non-sister chromatids in prophase I, independent assortment (random orientation) of homologous pairs at metaphase I and of sister chromatids at metaphase II, and then random fertilisation.

diploid, 2n = 4 DNA replicated meiosis I haploid, n = 2 meiosis II 4 haploid cells (gametes), n = 2, one chromatid of each chromosome red: from mother blue: from father each gamete has a different mix meiosis I separates homologous chromosomes; meiosis II separates sister chromatids
Fig. 1: meiosis in a cell with two homologous pairs (2n = 4) makes four genetically different haploid cells

Pairs, then chromatids.

  • Meiosis I separates homologous chromosomes: 2n becomes n.
  • Meiosis II separates sister chromatids, as in mitosis.
  • Crossing over: non-sister chromatids, prophase I, chiasmata.
  • Combinations: 2n gametes; random fertilisation multiplies them.

Watch out!

Do not mix up the two anaphases. In anaphase I whole homologous chromosomes separate (each still has two chromatids); in anaphase II the sister chromatids separate, just as in mitosis. The chromosome number halves in meiosis I, not meiosis II.

Crossing over happens between non-sister chromatids of a homologous pair, in prophase I. Sister chromatids are identical copies, so swapping between them would change nothing.

2Monohybrid and dihybrid crosses

Syllabus 16.2

A gene is a length of DNA at a fixed position, its locus, coding for a polypeptide; alleles are different versions of it. The genotype is the alleles present (homozygous if the two are the same, heterozygous if different) and the phenotype is the observable feature. A dominant allele is expressed in the heterozygote; a recessive one only when homozygous; codominant alleles are both expressed. In a genetic diagram, write parental phenotypes and genotypes, gametes (one allele of each gene per gamete), a Punnett square, then offspring genotypes, phenotypes and ratio. A monohybrid cross of two heterozygotes gives 3 : 1; a dihybrid cross of two double heterozygotes gives 9 : 3 : 3 : 1 if the genes assort independently. A test cross with a homozygous recessive reveals an unknown genotype. Autosomal linkage (genes on the same chromosome) and epistasis (one gene affecting the expression of another) change the expected ratios.

parents: carrier mother XH Xh × unaffected father XH Y F8 gene on the X chromosome; H normal, h haemophilia (recessive) XH Xh female gametes XH Y male gametes XH XH normal female XH Xh carrier female XH Y normal male Xh Y haemophiliac male each son has a 1 in 2 chance of haemophilia; each daughter has a 1 in 2 chance of being a carrier
Fig. 1: a sex-linked cross for haemophilia; alleles are written as superscripts on the X chromosome

One allele per gene per gamete.

  • Layout: phenotypes, genotypes, gametes, Punnett square, offspring, ratio.
  • Codominant and sex-linked alleles are written as superscripts.
  • Test cross: with the homozygous recessive; offspring mirror the gametes.
  • Linkage: few recombinants; epistasis: one gene masks another.

Watch out!

Write a full genetic diagram: parental phenotypes, parental genotypes, gametes (one allele of each gene), Punnett square, offspring genotypes and phenotypes, ratio. For sex-linked genes always attach the allele to the X, and give the Y no allele.

Linkage means two genes on the same chromosome; sex linkage means one gene on the X chromosome. Do not use the words interchangeably.

3The chi-squared test

Syllabus 16.2

The chi-squared (χ2) test compares observed (O) numbers in each category with the expected (E) numbers predicted by a genetic ratio. Start with a null hypothesis: there is no significant difference between observed and expected results, and any difference is due to chance. Calculate E from the total and the ratio, then χ2 = Σ (O − E)2 / E (the formula is provided). Find the degrees of freedom = number of categories − 1, and compare χ2 with the critical value at p = 0.05 from the table. If χ2 is less than the critical value, the difference is not significant and the null hypothesis is accepted; if it is greater, the difference is significant (probability less than 0.05 of arising by chance), so the null hypothesis is rejected and something else, such as linkage or epistasis, may be operating.

test cross AaBb × aabb, 200 offspring, expected 1 : 1 : 1 : 1 phenotypeOEO − E(O − E)² / E AB (dominant both)5650+60.72 Ab4450−60.72 aB4750−30.18 ab (recessive both)5350+30.18 χ² = Σ (O − E)² / E 1.80 degrees of freedom = 4 − 1 = 3; critical value at p = 0.05 is 7.82 1.80 is below 7.82: not significant, the data fit 1 : 1 : 1 : 1
Fig. 1: a worked chi-squared test for a dihybrid test cross

Five steps, every time.

  • State the null hypothesis: no significant difference.
  • Expected numbers = total × fraction for each class.
  • Degrees of freedom = number of classes − 1.
  • Compare with the critical value at p = 0.05, then conclude.

Watch out!

Always use the actual numbers of offspring for O and E, never percentages or ratios. And E does not have to be a whole number: 100 × 3/4 = 75, but 50 × 3/4 = 37.5 is fine.

A non-significant result does not prove the hypothesis is true; it means the data are consistent with it. Say "the difference is not significant" rather than "there is no difference".

4Genes, proteins and phenotype

Syllabus 16.2

A gene's base sequence determines the amino acid sequence of a polypeptide, which determines how it folds and so its function; the protein's activity then produces the phenotype. The TYR gene codes for tyrosinase, which converts tyrosine to melanin; a recessive allele gives a non-functional enzyme and albinism. The HBB gene codes for β-globin; a single base substitution replaces glutamic acid with valine, giving HbS, which forms fibres at low oxygen and causes sickle cell anaemia. The F8 gene on the X chromosome codes for clotting factor VIII; a recessive allele causes haemophilia. The HTT gene codes for huntingtin; a dominant allele with too many CAG repeats causes Huntington's disease. In peas, the dominant allele Le codes for a functional enzyme that makes active gibberellin (GA1), which causes stem elongation; lele plants make a non-functional enzyme and are dwarf.

geneprotein mademutant allelephenotype TYRtyrosinaserecessive, noworking enzymealbinism(no melanin) HBBβ-globin ofhaemoglobincodominant; one basechange, glu → valsickle cellanaemia F8factor VIII(clotting)recessive, on theX chromosomehaemophilia(slow clotting) HTThuntingtin(in neurones)dominant, extraCAG repeatsHuntington’sdisease gene → amino acid sequence → protein shape → function → phenotype
Fig. 1: four human genes, the proteins they code for and the phenotype of the mutant allele

Follow the chain.

  • TYR tyrosinase, albinism; HBB β-globin, sickle cell.
  • F8 factor VIII, haemophilia (X-linked); HTT huntingtin, dominant.
  • Le codes for the enzyme GA20 → GA1; lele plants are dwarf.
  • Every answer: bases, amino acids, shape, function, phenotype.

Watch out!

Link every step: gene → base sequence → amino acid sequence (primary structure) → folding and shape → protein function → phenotype. Marks are lost when an answer jumps straight from "mutation" to "disease".

5Gene control

Syllabus 16.3

Structural genes code for proteins with a function in the cell, such as enzymes; regulatory genes code for proteins, such as repressors and transcription factors, that control the expression of other genes. An inducible enzyme is made only when its substrate is present (β-galactosidase in E. coli); a repressible enzyme is made until its product builds up and switches its synthesis off (the enzymes that make tryptophan). In the lac operon, the regulatory gene lacI codes for a repressor that binds the operator, so RNA polymerase cannot bind the promoter and the structural genes lacZ, lacY and lacA are not transcribed. Lactose binds the repressor, changing its shape so it leaves the operator, and the genes are transcribed. In eukaryotes, transcription factors are proteins that bind to DNA and increase or decrease the rate of transcription. Gibberellin binds a receptor and causes DELLA proteins to be broken down; DELLA proteins normally inhibit transcription factors, so their removal lets transcription of growth genes begin.

no lactose: genes off lacIPOlacZlacYlacA regulatory genepromoteroperatorstructural genes repressor bound to operator RNA polymerase blocked lactose present: genes on lacIPOlacZlacYlacA transcription lactose (blue) binds the repressor, which leaves the operator; RNA polymerase binds the promoter and transcribes lacZ, lacY, lacA β-galactosidase and lactose permease are made: an inducible system
Fig. 1: the lac operon of E. coli with and without lactose

Who binds what.

  • Repressor binds the operator; RNA polymerase binds the promoter.
  • Lactose binds the repressor: an inducible system.
  • Transcription factors are proteins that raise or lower transcription rate.
  • Gibberellin causes DELLA breakdown, freeing transcription factors.

Watch out!

The repressor binds the operator, not the promoter, and lactose binds the repressor, not the DNA. The repressor is coded for by the regulatory gene lacI, which is not one of the structural genes.

Gibberellin does not bind to DNA and is not a transcription factor itself. It acts indirectly: it causes the breakdown of DELLA, and that frees the transcription factors.

Back to A-Level Biology 9700 contents · (c) 2026 TheLucidSTEM · thelucidstem.com