Genetics & Inheritance – A-Level Biology Revision Notes
Complete revision notes on monohybrid and dihybrid crosses, sex linkage, epistasis, codominance, multiple alleles, chi-squared, Hardy-Weinberg and genetic diagrams. Written by a Chartered Biologist and former WJEC/Eduqas & Edexcel examiner who knows exactly how these questions are marked.
Last updated: February 2026
Why Genetics Is the Topic That Separates Grades
Genetics and inheritance is one of the most challenging topics in A-Level Biology, and it is also one of the most heavily examined. Every exam board allocates significant marks to genetic crosses, and the questions are unforgiving – you either apply the logic correctly or you do not. There is no room for vague answers.
Having examined genetics questions for both WJEC/Eduqas and Edexcel, I can tell you that the biggest problem is not that students lack knowledge. Most students can define terms like “genotype” and “phenotype.” The problem is that they do not set out their genetic diagrams in the format that earns marks, they lose precision in their terminology, and they panic when they see chi-squared calculations. These are all fixable problems.
On this page, I am going to take you through every aspect of genetics your exam board requires. More importantly, I am going to show you exactly how examiners mark these questions – what earns credit and what does not. This is information that comes from sitting in standardisation meetings, not from reading a textbook.
Key Terminology – Get These Right Before Anything Else
Before you attempt a single genetic cross, you need to know these definitions precisely. Examiners award and withhold marks based on whether you use the correct term. The most common error across all exam boards, flagged in examiner reports every single year, is confusing “gene” with “allele.”
Gene
A section of DNA at a specific locus (position) on a chromosome that codes for a particular polypeptide (or functional RNA).
Allele
An alternative version of a gene. Different alleles of the same gene have different nucleotide sequences but occupy the same locus on homologous chromosomes.
Genotype
The combination of alleles an organism possesses for a particular gene (e.g., Bb). Written using letters, not words.
Phenotype
The observable characteristics of an organism, resulting from the interaction between genotype and the environment. Written in words (e.g., “brown fur”), not letters.
Dominant allele
An allele whose effect on the phenotype is expressed in both the homozygous (BB) and heterozygous (Bb) condition. Shown as a capital letter.
Recessive allele
An allele whose effect on the phenotype is only expressed when present in the homozygous condition (bb). Shown as a lowercase letter.
Codominance
When both alleles in a heterozygote are expressed simultaneously in the phenotype, so that the heterozygote shows features of both homozygotes. Neither allele is dominant over the other.
Dominant does NOT mean “stronger” or “more common.” This is a deeply embedded misconception. Dominance simply describes which phenotype is expressed in the heterozygote. Huntington’s disease is caused by a dominant allele, yet it affects only 3–7 people per 100,000. Achondroplasia is dominant but extremely rare. Blood group O is caused by a recessive allele, yet it is the most common blood group in the UK. Dominance is about the molecular interaction between alleles, not about frequency in the population.
Monohybrid Crosses – One Gene, Two Alleles
A monohybrid cross involves the inheritance of a single gene. This is where you learn the fundamental technique for setting out genetic diagrams – a technique that carries through to every other type of cross you will encounter.
How to Set Out a Genetic Diagram for Maximum Marks
Having marked thousands of genetic diagram questions, I can tell you exactly what earns credit. Most exam boards award marks in three stages: correct parental genotypes (1 mark), all possible offspring genotypes shown (1 mark), and offspring phenotypes with the correct ratio (1 mark – both elements needed for this single mark). Here is the format that consistently earns full marks:
State the allele key
Define your allele symbols if not given in the question. Use letters where upper and lower case are clearly distinct (B and b are good; C and c are risky because they look similar in handwriting).
Write parental phenotypes and genotypes
Show the phenotypes in words and the genotypes in letters, clearly labelled. For example: Brown fur (Bb) × Brown fur (Bb).
Show the gametes
Circle the gametes. Each gamete contains one allele only. From Bb, the gametes are B and b. Circling them is not strictly required for marks, but it shows the examiner you understand the process.
Draw a Punnett square
Show all possible combinations of gametes. This earns the “offspring genotypes” mark.
State offspring phenotypes and ratio
Write the phenotypic ratio clearly. For Bb × Bb: 3 brown : 1 white. Both the phenotypes and the ratio are needed for the final mark.
The 3:1 Ratio and Test Crosses
When two heterozygous parents are crossed (Bb × Bb), the expected offspring ratio is 3:1 – three showing the dominant phenotype to one showing the recessive phenotype. The underlying genotypic ratio is 1 BB : 2 Bb : 1 bb.
A test cross is used to determine whether an organism showing the dominant phenotype is homozygous dominant (BB) or heterozygous (Bb). You cross the unknown individual with a homozygous recessive individual (bb). The tester must be homozygous recessive because it can only contribute recessive alleles – meaning the offspring phenotypes directly reveal what allele the unknown parent contributed. If any offspring show the recessive phenotype, the unknown parent must be heterozygous.
Dihybrid Crosses – Two Genes, Independent Assortment
A dihybrid cross follows the inheritance of two genes simultaneously. This is where many students start to struggle, but the underlying logic is identical to a monohybrid cross – you just have more gamete combinations to track.
The critical concept is independent assortment. During meiosis, the alleles of one gene segregate independently of the alleles of another gene (provided the genes are on different chromosomes). This means the inheritance of one trait does not affect the inheritance of another.
Gamete Formation – Where Most Errors Happen
From a parent with genotype AaBb, the possible gametes are: AB, Ab, aB, ab. Each gamete takes one allele from each gene. The most common error is writing “Aa” or “Bb” as a gamete – these are genotypes, not gametes. A gamete can never contain both alleles of the same gene.
When you cross AaBb × AaBb, each parent produces four types of gamete. The Punnett square has 16 boxes, and the expected phenotypic ratio is 9:3:3:1. This ratio applies only when both genes show complete dominance and assort independently.
The Dihybrid Test Cross
Crossing AaBb × aabb produces a 1:1:1:1 ratio of all four phenotypes. This is fundamentally different from the 9:3:3:1 ratio and students regularly confuse the two. The test cross ratio is 1:1:1:1 because the homozygous recessive parent can only contribute ab gametes, so each offspring phenotype directly reflects the gamete contributed by the AaBb parent.
Codominance and Multiple Alleles
Codominance
In codominance, both alleles are expressed simultaneously in the heterozygote. Neither allele is dominant over the other. The classic example is the snapdragon flower: homozygous red (CRCR) crossed with homozygous white (CWCW) produces heterozygous pink (CRCW) offspring. The pink phenotype shows both alleles being expressed, not a “blend” – at the molecular level, both alleles are producing their respective pigments, and the visual result is an intermediate colour because both pigments are present.
For allele notation in codominance, you cannot use capital and lowercase letters (since neither is dominant). Instead, use a common letter with superscripts: CR and CW, or IA and IB.
Multiple Alleles – ABO Blood Groups
The ABO blood group system is the most important example of multiple alleles at A-Level. The gene has three alleles: IA, IB, and IO. IA and IB are codominant with each other (so the heterozygote IAIB has blood group AB). Both IA and IB are dominant over IO.
This means there are six possible genotypes but only four phenotypes:
- Group A: IAIA or IAIO
- Group B: IBIB or IBIO
- Group AB: IAIB
- Group O: IOIO
The ABO system demonstrates that dominance is not a fixed property of an allele. IA is dominant over IO, but codominant with IB. The same allele can show different dominance relationships depending on which other allele it is paired with.
Sex Linkage – Genes on the X Chromosome
A sex-linked gene is one located on the X chromosome, in a region that has no corresponding locus on the Y chromosome. The Y chromosome is much smaller than the X and carries very few genes, so most genes on the X have no Y-linked counterpart. This has profound consequences for inheritance.
Why Males Are More Commonly Affected
Males are hemizygous for X-linked genes – they have only one copy because they have only one X chromosome (XY). This means a single recessive allele on the X chromosome will be expressed in males because there is no second allele to mask it. Females have two X chromosomes (XX), so they need to be homozygous recessive to express the condition – they must inherit the recessive allele from both parents.
This is why X-linked recessive conditions like haemophilia and red-green colour blindness are far more common in males. A female can be a carrier – heterozygous, with one normal allele and one recessive allele, phenotypically unaffected but able to pass the recessive allele to offspring. Males cannot be carriers of X-linked recessive conditions. They either have the condition or they do not.
Notation for Sex-Linked Crosses
Use superscripts on the X chromosome to show the alleles. For haemophilia:
- XHXH = normal female
- XHXh = carrier female
- XhXh = affected female (haemophiliac)
- XHY = normal male
- XhY = affected male (haemophiliac)
The Y chromosome is written without a superscript allele because it does not carry this gene.
Epistasis – Gene Interactions That Modify Ratios
Epistasis occurs when one gene affects the expression of a different gene at a separate locus. It is NOT the same as dominance, which describes interactions between alleles at the same locus. The key to understanding epistasis is thinking about biochemical pathways – if gene A controls a step that must occur before gene B can have its effect, then gene A is epistatic to gene B.
All Epistasis Ratios Come from 9:3:3:1
The crucial insight is that every epistasis ratio is a modification of the standard 9:3:3:1 dihybrid ratio, produced by combining phenotypic classes. All the ratios sum to 16, confirming they arise from a dihybrid cross.
Recessive epistasis – 9:3:4
The classic example is Labrador coat colour. Two genes are involved: gene E controls pigment deposition, and gene B controls pigment colour. The homozygous recessive genotype (ee) prevents any pigment from being deposited, regardless of what the B gene says. So ee dogs are always yellow, whether their B genotype is BB, Bb, or bb. Dogs with at least one E allele can deposit pigment, and then the B gene determines colour: B_ gives black, bb gives chocolate (brown). The 3:1 “brown” class from the normal dihybrid ratio merges with the 1 “double recessive” class because both produce the same phenotype (yellow) – giving 9 black : 3 chocolate : 4 yellow.
Dominant epistasis – 12:3:1
Here, a dominant allele at one locus prevents expression of the second gene. The 9 and 3 classes merge because the dominant epistatic allele masks the effect of the second gene in both. For example, in squash colour, a dominant allele W produces white fruit regardless of the second gene; only ww allows the second gene (Y/y) to determine whether the fruit is yellow or green. Result: 12 white : 3 yellow : 1 green.
Complementary epistasis – 9:7
Both genes must contribute a dominant allele for one phenotype; any other combination produces a different phenotype. The 3, 3, and 1 classes all merge. Example: in sweet pea flower colour, both genes A and B must be present (at least one dominant allele each) for coloured flowers. A_B_ = purple (9), all others = white (7).
The Chi-Squared Test – Are Your Results Due to Chance?
The chi-squared (χ²) test is a statistical test that determines whether the difference between observed results and expected results is significant or simply due to chance. In genetics, you use it to test whether your observed offspring ratios match a predicted Mendelian ratio.
This is the question where I have seen the most marks thrown away, year after year. AQA’s examiner report revealed that only 22% of students could complete a chi-squared calculation correctly in context. The errors are predictable and entirely avoidable.
Step-by-Step Method
State the null hypothesis
“There is no significant difference between the observed and expected results.” Note the word difference – this word is essential and its absence loses the mark.
Calculate expected values
From the total number of offspring, calculate what you would expect based on the predicted ratio. For a 3:1 ratio with 120 offspring: expected = 90 and 30.
Apply the formula
χ² = Σ(O – E)² / E. For each category, subtract expected from observed, square the result, then divide by expected. Sum all categories.
Calculate degrees of freedom
Degrees of freedom = number of categories − 1. For a 3:1 ratio with 2 phenotypic classes, df = 1. For 9:3:3:1 with 4 classes, df = 3.
Compare to the critical value at p = 0.05
Look up the critical value in the chi-squared table for your degrees of freedom at the 0.05 significance level.
Draw your conclusion
If χ² is less than the critical value: accept the null hypothesis. The difference is not significant – your observed results are consistent with the expected ratio. If χ² is greater than the critical value: reject the null hypothesis. The difference IS significant – something other than chance is affecting the results.
The Hardy-Weinberg Principle – Predicting Allele Frequencies
The Hardy-Weinberg principle states that in a large population with random mating and no evolutionary forces acting, allele frequencies remain constant from generation to generation. It provides two equations that allow you to calculate allele and genotype frequencies.
p + q = 1 where p = frequency of the dominant allele and q = frequency of the recessive allele.
p² + 2pq + q² = 1 where p² = frequency of homozygous dominant, 2pq = frequency of heterozygous, and q² = frequency of homozygous recessive.
The Most Common Calculation Error
The most frequent mistake is treating a phenotypic frequency as an allelic frequency. If a question says “20% of the population has blue eyes (recessive phenotype),” students plug in q = 0.2. This is wrong. The 20% represents q², not q. You must take the square root first: q = √0.2 = 0.447. Then p = 1 – 0.447 = 0.553. The carrier frequency (2pq) = 2 × 0.553 × 0.447 = 0.494, or roughly 49%.
Always start with q². Questions almost always give you the frequency of the recessive phenotype, because this is the only phenotype where genotype can be determined directly (all individuals with the recessive phenotype must be q²). The dominant phenotype includes both p² and 2pq individuals, so you cannot work backwards from it without first calculating q.
The Five Assumptions
Hardy-Weinberg equilibrium only holds when five conditions are met: no mutation, no natural selection, no migration (gene flow), random mating, and a large population size. Any violation of these conditions can cause allele frequencies to change over time – which is, of course, evolution. You need to be able to state these assumptions and explain which real-world factors might violate them.
Autosomal Linkage and Crossing Over
Independent assortment assumes genes are on different chromosomes. When two genes are on the same chromosome, they are linked and tend to be inherited together. This is autosomal linkage (as distinct from sex linkage, which involves the X chromosome).
Linked genes do not follow the expected 9:3:3:1 ratio. Instead, parental combinations appear far more frequently than recombinant combinations. For example, if alleles A and B are on the same chromosome (in cis configuration: AB/ab), the parental phenotypes (AB and ab) will vastly outnumber the recombinant phenotypes (Ab and aB).
Recombinant offspring arise from crossing over during prophase I of meiosis, when homologous chromosomes exchange segments. The frequency of recombinant offspring gives you a measure of the distance between the two genes on the chromosome – the further apart they are, the more likely crossing over will occur between them.
Exam Board Comparison – What Your Board Requires
The genetics specifications differ more between boards than almost any other topic. This table shows exactly what each board requires so you do not waste time on content that will not appear in your exams.
| Content | AQA | OCR A | OCR B | Edexcel A | Edexcel B | WJEC/Eduqas |
|---|---|---|---|---|---|---|
| Dihybrid crosses | Yes | Yes | Yes | No | Yes | Yes |
| Epistasis | Yes | Yes (both types) | No | No | No | Uncertain |
| Hardy-Weinberg | Yes | Yes | Yes | No | Yes | Yes |
| Sex linkage | Yes | Yes | Yes | No | Yes | Yes |
| Autosomal linkage | Yes | Yes | Yes | No | Yes | Yes |
| Chi-squared | Yes | Yes | Yes | General maths skill | Yes | Yes |
| Multiple alleles (ABO) | Yes | Yes | Yes (HLA) | Yes | Yes | Yes |
| Pedigree analysis | Implicit | Implicit | Implicit | Explicit | Explicit | Likely |
| Genetic screening ethics | Minimal | Minimal | Minimal | Major focus | Minimal | Minimal |
| Epigenetics | Basic | Module 6.1.1 | Yes (detailed) | Topic 3 | Yes (detailed) | Yes |
Key takeaway: Edexcel A has by far the least genetics content. If you are on AQA or OCR A, you have the most demanding specification for this topic. WJEC and Eduqas are the only boards requiring both haemophilia and Duchenne muscular dystrophy as sex linkage examples. OCR B uniquely names Drosophila as a required model organism for inheritance studies.
Common Mistakes – Where Students Lose Marks
These errors appear in examiner reports year after year. Learn to avoid them and you will immediately outperform most other candidates.
| # | The Mistake | Why It Costs Marks |
|---|---|---|
| 1 | Writing “gene” instead of “allele” | The single most penalised terminology error across all boards. A gene is the section of DNA; alleles are the different versions. “New combinations of genes” during crossing over is rejected – it must be “new combinations of alleles.” |
| 2 | Writing gametes as Aa or Bb | A gamete contains one allele from each gene, not two alleles from the same gene. From AaBb, gametes are AB, Ab, aB, ab – never Aa, Bb, AaBb, or similar. |
| 3 | Mixing up XX and XY | Flagged explicitly in the Eduqas 2024 examiner report. Females are XX, males are XY. Under exam stress, students swap these. Double-check every time. |
| 4 | Saying “the results are due to chance” for chi-squared | The conclusion must refer to the difference between observed and expected. Saying “the results” are due to chance is not the same thing and loses the mark. |
| 5 | Using q instead of q² for Hardy-Weinberg | The frequency of the recessive phenotype is q², not q. You must take the square root to find q before calculating anything else. |
| 6 | Not showing working in calculations | AQA examiner reports note that students who write only the final (incorrect) answer cannot receive any partial credit. Always show every step. |
| 7 | Adding incorrect extra information | AQA mark schemes apply a contradiction rule: each error cancels a correct response. Volunteering wrong information can reduce your mark to zero even if other parts of your answer were correct. |
| 8 | Saying dominant means “stronger” or “more common” | Dominance describes which phenotype is expressed in the heterozygote. It has nothing to do with allele frequency. Huntington’s is dominant but affects 3–7 per 100,000. |
Need Help With Genetics?
Genetics is the topic where one-to-one tutoring makes the biggest difference. I work through genetic crosses at your pace, for your exam board, focusing on the exact areas where you are losing marks. Many of my students go from struggling with dihybrid crosses to confidently tackling epistasis in just a few sessions.
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Google Frequently Asked Questions About Genetics & Inheritance
Follow this structure: define allele symbols (key), state parental phenotypes and genotypes, show gametes (ideally circled), draw a Punnett square showing all offspring genotypes, then state offspring phenotypes with the correct ratio. Most boards award marks in three stages: correct parental genotypes, correct offspring genotypes, and correct phenotypes with ratio. Showing gametes can rescue marks even if your parental genotypes are wrong – AQA mark schemes have a specific partial-credit rule for this.
Dominance describes the interaction between two alleles at the same gene locus. Epistasis describes the interaction between genes at different loci, where one gene affects the expression of another. All epistasis ratios are modifications of the 9:3:3:1 dihybrid ratio, produced by phenotypic classes merging. Recessive epistasis gives 9:3:4, dominant epistasis gives 12:3:1, and complementary epistasis gives 9:7. Only AQA and OCR A require epistasis.
Males are hemizygous for X-linked genes – they have only one X chromosome, so a single recessive allele on that X will be expressed because there is no second allele to mask it. Females have two X chromosomes and need to be homozygous recessive (inheriting the recessive allele from both parents) to be affected. Females can be carriers (heterozygous), but males cannot – they either express the condition or do not carry the allele.
Compare your calculated χ² value to the critical value at p = 0.05 for the appropriate degrees of freedom. If your value is less than the critical value, the difference between observed and expected is not significant – accept the null hypothesis. If it is greater, the difference is significant – reject the null hypothesis. The essential word in your conclusion is “difference” – you must say the difference is or is not significant, not that “the results” are significant.
Always start with q². Questions typically give you the frequency of the recessive phenotype – this equals q². Take the square root to find q, then calculate p = 1 − q. The carrier (heterozygote) frequency is 2pq. The most common error is plugging the phenotypic frequency directly as q without taking the square root first.
The gametes from AaBb are: AB, Ab, aB, and ab. Each gamete takes one allele from each gene. Never write Aa, Bb, or AaBb as gametes – these are genotypes. A gamete is haploid and cannot contain both alleles of the same gene. This error is one of the most common causes of incorrect dihybrid crosses.
No – the differences are substantial. Edexcel A (Salters-Nuffield) does NOT require dihybrid crosses, epistasis, Hardy-Weinberg, sex linkage, or autosomal linkage. AQA and OCR A have the most demanding specifications. Epistasis is required by AQA and OCR A only. OCR B uniquely requires Drosophila as a model organism. WJEC and Eduqas uniquely require both haemophilia and Duchenne muscular dystrophy for sex linkage. Always check your specification.
Dominance describes which phenotype appears in the heterozygote – it says nothing about how frequently an allele occurs in a population. Huntington’s disease is caused by a dominant allele but affects only 3–7 per 100,000 people. Blood group O is determined by a recessive allele yet is the most common blood group in the UK. Allele frequency in a population is determined by natural selection, genetic drift and gene flow – not by dominance.
Disclaimer: The information provided on this page is intended for educational guidance only. While every effort has been made to ensure accuracy, Biology Education and its author accept no responsibility for individual exam outcomes. Students are advised to consult their own teachers, tutors, and official exam board resources as part of their revision.