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Monohybrid inheritance deals with the inheritance of a single gene with two alleles. Mastering monohybrid crosses is essential before tackling more complex patterns such as dihybrid inheritance, codominance, and epistasis. The Edexcel specification requires you to construct and interpret genetic diagrams, use correct genetic terminology, and predict phenotypic and genotypic ratios.
Before constructing any genetic cross, you must be fluent in the following terms:
| Term | Definition |
|---|---|
| Gene | A length of DNA that codes for a polypeptide (or functional RNA) |
| Allele | An alternative form of a gene, found at the same locus on homologous chromosomes |
| Locus | The specific position of a gene on a chromosome |
| Dominant allele | An allele whose effect on the phenotype is expressed in both homozygous and heterozygous individuals |
| Recessive allele | An allele whose effect on the phenotype is only expressed in homozygous individuals |
| Homozygous | Having two identical alleles at a locus (e.g. AA or aa) |
| Heterozygous | Having two different alleles at a locus (e.g. Aa) |
| Genotype | The combination of alleles an organism possesses |
| Phenotype | The observable characteristics of an organism, resulting from genotype and environment |
| Carrier | A heterozygous individual who carries a recessive allele without showing the recessive phenotype |
| Wild type | The most common phenotype in a natural population |
Exam tip: Always define your symbols at the start of a genetic diagram. For example: "Let A = the allele for brown fur (dominant); let a = the allele for white fur (recessive)."
A standard genetic diagram should include:
Fur colour in mice: brown (B) is dominant to white (b).
Cross: Bb × Bb
Parental phenotypes: Brown × Brown
Gametes: Each parent produces B and b gametes in equal proportions.
Punnett Square:
| B | b | |
|---|---|---|
| B | BB | Bb |
| b | Bb | bb |
Offspring genotypes: 1 BB : 2 Bb : 1 bb
Offspring phenotypes: 3 brown : 1 white
This 3:1 ratio is the hallmark of a monohybrid cross between two heterozygous parents and was first described by Gregor Mendel.
A test cross is used to determine the genotype of an individual showing the dominant phenotype. The individual is crossed with a homozygous recessive individual.
Cross: Bb × bb
Gametes: Bb parent produces B and b; bb parent produces only b.
Punnett Square:
| B | b | |
|---|---|---|
| b | Bb | bb |
| b | Bb | bb |
Offspring genotypes: 1 Bb : 1 bb
Offspring phenotypes: 1 brown : 1 white
If the unknown parent were homozygous dominant (BB), all offspring would be brown (Bb). The appearance of any white offspring proves the unknown parent was heterozygous (Bb).
Cystic fibrosis is an autosomal recessive condition. Let F = normal allele, f = cystic fibrosis allele.
Cross: Two carrier parents (Ff × Ff)
Punnett Square:
| F | f | |
|---|---|---|
| F | FF | Ff |
| f | Ff | ff |
Offspring:
Common exam error: Students write "1 in 3 chance of being a carrier" instead of "2 in 3 chance of being a carrier among unaffected offspring." The question wording matters — read it carefully.
This is the classic Mendelian cross. In peas, tall (T) is dominant to dwarf (t).
P generation cross: TT (tall) × tt (dwarf)
| T | T | |
|---|---|---|
| t | Tt | Tt |
| t | Tt | Tt |
F₁ generation: All Tt (all tall). This shows complete dominance.
F₁ × F₁ cross: Tt × Tt
| T | t | |
|---|---|---|
| T | TT | Tt |
| t | Tt | tt |
F₂ generation: 1 TT : 2 Tt : 1 tt → phenotypic ratio 3 tall : 1 dwarf
Mendel observed this 3:1 ratio consistently across many traits, leading him to formulate the law of segregation.
The two alleles of a gene separate into different gametes during meiosis, so each gamete carries only one allele for each gene.
This law is a direct consequence of the behaviour of homologous chromosomes during anaphase I of meiosis, when homologous pairs separate.
Mendel observed a 3:1 ratio in the F₂ generation because:
Pedigree diagrams show the inheritance of a trait through generations of a family. You need to be able to:
| Clue | Autosomal dominant | Autosomal recessive | X-linked recessive |
|---|---|---|---|
| Affected in every generation? | Usually yes | Often skips generations | Can skip generations |
| Unaffected parents → affected child? | No (unless new mutation) | Yes (both carriers) | Yes (mother is carrier) |
| Sex bias? | No | No | Males mainly affected |
| Father-to-son transmission? | Yes (possible) | Yes (possible) | Never |
When calculating the probability of genetic outcomes:
Two parents are both carriers for cystic fibrosis (Ff). What is the probability that their first two children will both be unaffected?
What is the probability that their first child is an affected boy?
What is the probability of having three children who are all carriers (Ff)?
Important: Each conception is an independent event. Previous children's genotypes do not influence the probabilities for the next child.
| Genetic condition | Inheritance | Alleles | Carrier frequency (approx.) |
|---|---|---|---|
| Cystic fibrosis | Autosomal recessive | CFTR gene | 1 in 25 (European descent) |
| Sickle cell anaemia | Autosomal recessive | HBB gene | 1 in 12 (African descent) |
| Huntington's disease | Autosomal dominant | HTT gene | ~1 in 10,000 |
| Phenylketonuria (PKU) | Autosomal recessive | PAH gene | 1 in 50 (European descent) |
| Achondroplasia | Autosomal dominant | FGFR3 gene | ~1 in 25,000 (mostly new mutations) |
| Tay-Sachs disease | Autosomal recessive | HEXA gene | 1 in 30 (Ashkenazi Jewish descent) |
Monohybrid inheritance follows Mendel's first law of segregation: alleles separate during gamete formation and recombine randomly at fertilisation. A cross between two heterozygotes produces a 3:1 phenotypic ratio. Test crosses with homozygous recessive individuals reveal unknown genotypes. Pedigree analysis and probability rules extend these principles to real-world genetics. Always show full working in genetic diagrams: parental genotypes, gametes, Punnett square, and offspring ratios.
This material sits in Edexcel 9BI0 Topic 8 (Grey Matter — Coordination, Response and Gene Technology) for the inheritance strand, where candidates are expected to construct and interpret monohybrid genetic diagrams for a single gene with two alleles, define and use gene, allele, locus, dominant, recessive, homozygous, heterozygous, genotype, phenotype and carrier, deploy a Punnett square to derive genotypic (1 : 2 : 1 from heterozygote × heterozygote) and phenotypic (3 : 1 under complete dominance) ratios, and apply the test cross (unknown × homozygous recessive) to distinguish homozygous dominant from heterozygous parents. The lesson also covers the cytological basis of Mendel's law of segregation in the separation of homologous chromosomes at anaphase I of meiosis, and the modifying patterns of incomplete dominance (heterozygote shows an intermediate phenotype, generating a 1 : 2 : 1 phenotypic ratio that equals the genotypic ratio) and codominance (heterozygote shows both parental phenotypes simultaneously, also 1 : 2 : 1). Synoptic links run backwards to lessons 1 and 2 (mutation) for the origin of the allelic variation that monohybrid crosses track, and to lesson 3 (meiosis) for the cytological mechanism of segregation; forwards to lesson 5 (dihybrid inheritance), which extends monohybrid logic to two genes (9 : 3 : 3 : 1 under independent assortment), to lesson 6 (codominance and multiple alleles) which deepens the modifying-ratio cases (e.g. ABO blood groups), to lesson 7 (sex-linkage and autosomal linkage) which explains the systematic deviations from Mendelian ratios when loci are physically linked, and to lesson 8 (chi-squared test) which quantifies whether observed offspring counts deviate significantly from the expected Mendelian ratios; outwards to Topic 4 (Biodiversity and Natural Resources) for the Hardy–Weinberg principle, which uses monohybrid allele frequencies to predict population-level genotype frequencies and the action of natural selection. Refer to the official Pearson Edexcel 9BI0 specification document for exact wording.
Question (8 marks):
(a) In garden peas, the allele for tall stems (T) is completely dominant to the allele for dwarf stems (t). A homozygous tall plant is crossed with a homozygous dwarf plant, and the F₁ offspring are then self-pollinated to produce an F₂ generation. Construct a full genetic diagram for both crosses, and state the expected F₂ genotypic and phenotypic ratios. (4)
(b) In a separate cross, two parents heterozygous for the cystic-fibrosis allele (genotype Cc, with C dominant for the normal CFTR phenotype and c recessive for cystic fibrosis) are planning a child. Calculate the probability that (i) any one child is affected; (ii) any one unaffected child is a carrier. Show your reasoning. (2)
(c) Identify the cytological event in meiosis that is the physical basis of Mendel's law of segregation, and state which fundamental property of meiosis (and not of mitosis) makes the law work. (2)
Solution with mark scheme:
(a) M1 (AO1) — symbol definition and parental cross. Let T = tall (dominant) and t = dwarf (recessive). Parental cross: TT (tall) × tt (dwarf). Gametes: TT parent produces only T gametes; tt parent produces only t gametes.
A1 (AO2) — F₁ Punnett square and outcome.
| T | T | |
|---|---|---|
| t | Tt | Tt |
| t | Tt | Tt |
All F₁ are Tt, all phenotypically tall. This is the expected uniformity of the F₁ under complete dominance.
M1 (AO2) — F₁ × F₁ self-pollination. Cross: Tt × Tt. Each parent produces T and t gametes in equal proportion.
| T | t | |
|---|---|---|
| T | TT | Tt |
| t | Tt | tt |
A1 (AO2) — F₂ ratios. F₂ genotypic ratio = 1 TT : 2 Tt : 1 tt. F₂ phenotypic ratio = 3 tall : 1 dwarf, the classic Mendelian monohybrid ratio.
(b) M1 (AO2) — affected child probability. Cross Cc × Cc. Punnett square gives 1 CC : 2 Cc : 1 cc. P(affected, genotype cc) = 1/4 (25%).
A1 (AO2) — carrier-given-unaffected probability. Among the 3 unaffected outcomes (1 CC + 2 Cc), 2 are carriers (Cc). P(carrier | unaffected) = 2/3 (~67%). Common error trap: P(carrier among all four outcomes) = 2/4 = 1/2, but the question conditions on unaffected, so the correct answer is 2/3.
(c) M1 (AO1) — cytological event. Mendel's law of segregation reflects the separation of homologous chromosomes at anaphase I of meiosis: each gamete receives only one of each homologous pair, hence only one of the two alleles of a heterozygote.
A1 (AO3.1) — meiosis-specific property. Mitosis does not pair homologues, so there is no segregation step; meiosis I, with its synapsis (bivalent formation) and homologue separation, is the unique cytological basis of Mendelian segregation. Without meiosis I, gametes could not carry one allele each, and the 3 : 1 ratio would not arise.
Total: 8 marks (M3 A5).
Question (6 marks): A student writes that "the 3 : 1 phenotypic ratio is what you always get in a monohybrid cross between two heterozygous parents." Evaluate this claim, with reference to (i) the genotypic ratio that always sits beneath the 3 : 1, (ii) the conditions under which the phenotypic ratio differs from 3 : 1, and (iii) the role of chance in observed counts. Include at least one worked example with non-3 : 1 phenotypes.
Mark scheme decomposition by AO:
| Mark | AO | Earned by |
|---|---|---|
| 1 | AO1.1 | Stating that the genotypic ratio beneath every Cc × Cc cross is 1 : 2 : 1 (homozygous dominant : heterozygous : homozygous recessive), regardless of dominance pattern |
| 2 | AO1.2 | Identifying complete dominance as the condition under which 1 : 2 : 1 collapses to 3 : 1 phenotypically (heterozygote indistinguishable from homozygous dominant) |
| 3 | AO2.1 | Recognising incomplete dominance as a case where heterozygotes show an intermediate phenotype (e.g. Andalusian fowl plumage — black BB, white bb, blue Bb), giving a 1 : 2 : 1 phenotypic ratio identical to the genotypic ratio |
| 4 | AO2.7 | Recognising codominance as a case where heterozygotes show both parental phenotypes simultaneously (e.g. MN blood groups, where MN heterozygotes carry both M and N antigens), again giving 1 : 2 : 1 |
| 5 | AO3.1 | Stating that even under complete dominance, the 3 : 1 ratio is the expected ratio — actual offspring counts deviate from it by chance, especially for small samples; the chi-squared test (lesson 8) is used to decide whether deviations are significant |
| 6 | AO3.2 | Concluding that the student is wrong twice over: 3 : 1 is conditional on complete dominance, and even then it is the expected rather than the observed ratio; the underlying 1 : 2 : 1 genotypic ratio is the more general invariant |
Total: 6 marks (AO1 = 2, AO2 = 2, AO3 = 2). Specimen question modelled on the Edexcel 9BI0 paper format. Edexcel reliably tests monohybrid inheritance through "evaluate / explain the ratio" prompts; candidates who recite 3 : 1 without distinguishing complete dominance, incomplete dominance and codominance, and without flagging chance variation, lose AO3 marks.
Lessons 1 and 2 (mutation) — origin of allelic variation. Monohybrid crosses track alleles but do not explain where alleles come from. Gene mutation (lesson 1: substitutions, insertions, deletions) creates new alleles at a locus; chromosome mutation (lesson 2) can duplicate or delete whole loci. The dominant T and recessive t alleles for pea height descend from a single ancestral allele in which a mutation at some point produced an alternative form. A* candidates frame inheritance as a layered system: mutation creates alleles; meiosis shuffles them; monohybrid analysis tracks the resulting transmission pattern.
Lesson 3 (meiosis) — the cytological basis of segregation. Mendel's law of segregation is the genetic statement of anaphase I: homologous chromosomes carrying the two alleles separate to opposite poles, so each gamete receives only one. The 1 : 1 gamete ratio from a heterozygote is therefore not a probabilistic assumption — it is a direct consequence of meiotic mechanics. Independent assortment (which generates the dihybrid 9 : 3 : 3 : 1 in lesson 5) is meanwhile the cytological signature of metaphase I, where bivalents orient independently. A* candidates pin every Mendelian ratio to its cytological cause.
Lesson 5 (dihybrid inheritance) — the same logic, applied to two genes. Mendel's second law (independent assortment) says that the alleles at one locus segregate independently of the alleles at another, provided the two loci are on different chromosomes (or far apart on the same chromosome). Multiplying two monohybrid 3 : 1 ratios gives the dihybrid (3 + 1)² = 9 + 3 + 3 + 1. The 9 : 3 : 3 : 1 ratio is the monohybrid ratio squared — a direct conceptual extension. A* candidates derive the dihybrid ratio from monohybrid logic rather than memorising it.
Lesson 6 (codominance and multiple alleles) — modifying the standard ratios. Complete dominance gives 3 : 1 phenotypically. Incomplete dominance (Andalusian fowl, snapdragon flower colour) and codominance (ABO blood antigens, MN blood antigens, sickle-cell heterozygote phenotype) preserve the underlying 1 : 2 : 1 genotypic ratio but make the heterozygote phenotypically distinct, giving a 1 : 2 : 1 phenotypic ratio. Multiple alleles (e.g. ABO has three alleles Iᴬ, Iᴮ, i) expand the gene pool but each individual still carries only two alleles per locus — six possible genotypes (IᴬIᴬ, IᴬIᴮ, Iᴬi, IᴮIᴮ, Iᴮi, ii) and four phenotypes (A, B, AB, O). A* candidates state explicitly that multiple alleles are a population-level concept, not an individual-level one.
Lesson 7 (linkage) — when Mendelian ratios fail. Mendel's second law assumes independent assortment, which fails when two loci sit close together on the same chromosome — they are linked and tend to be inherited together. Linkage breaks the 9 : 3 : 3 : 1 dihybrid ratio (parental types become more common than recombinant types in proportion to crossover distance). The monohybrid 3 : 1, however, is unaffected by linkage because it concerns a single locus. A* candidates flag that monohybrid ratios are robust to linkage; dihybrid ratios are not.
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