Advanced Inheritance Patterns Revision Notes | A Level AQA Biology

1. Why Standard Mendelian Ratios Break Down

You already know that a standard dihybrid cross produces a 9:3:3:1 phenotypic ratio — but real organisms rarely cooperate that neatly. When genes interact, sit on the same chromosome, or are carried on sex chromosomes, the ratios shift in predictable ways that examiners love to test.

The key insight is this: every deviation from a standard Mendelian ratio has a biological explanation. Your job in the exam is to identify the pattern, name the mechanism, and use the chi-squared test to decide whether any remaining deviation is just chance. This lesson builds each mechanism in turn, then shows you how to apply the statistical test.

2. Sex-Linkage

Picture the X and Y chromosomes side by side. The X chromosome carries thousands of genes; the Y chromosome carries very few, and most of those are unrelated to the traits on the X. This asymmetry is the engine of sex-linkage.

Because males are hemizygous, a recessive X-linked allele is always expressed in males — there is no second allele to mask it. Females, carrying two X chromosomes, can be homozygous dominant, heterozygous (carrier), or homozygous recessive. This produces the characteristic pattern: affected males are more common than affected females, and the trait appears to skip generations through carrier females.

Notation matters here. Write X-linked alleles as superscripts on the X: $X^H$ for the dominant allele and $X^h$ for the recessive. A carrier female is $X^H X^h$; an affected male is $X^h Y$. Never write the allele on the Y chromosome — the Y carries no corresponding locus.

Worked example: Sex-Linked Cross: Carrier Female × Unaffected Male

Red-green colour blindness is X-linked recessive. A carrier female ($X^N X^n$) has children with an unaffected male ($X^N Y$). Determine the expected phenotypic ratio among offspring.

1. Female: $X^N X^n$ (carrier) Male: $X^N Y$ (unaffected) — The superscript N = normal vision allele (dominant); n = colour-blind allele (recessive). The male has only one X-linked locus.
2. Female gametes: $X^N$ or $X^n$ (each with probability 0.5) Male gametes: $X^N$ or $Y$ (each with probability 0.5) — Each parent produces two types of gamete in equal proportions.
3. | | $X^N$ | $Y$ | |---|---|---| | $X^N$ | $X^N X^N$ | $X^N Y$ | | $X^n$ | $X^N X^n$ | $X^n Y$ | — Four equally probable genotype combinations.
4. $X^N X^N$: unaffected female $X^N X^n$: carrier female (unaffected phenotype) $X^N Y$: unaffected male $X^n Y$: colour-blind male — The recessive allele is only expressed in the hemizygous male. Both female genotypes show normal vision.
5. Females: 2 unaffected: 0 affected Males: 1 unaffected: 1 colour-blind Overall: 3 unaffected: 1 colour-blind (but the affected individual is always male) — The ratio looks like a standard monohybrid 3:1, but the affected phenotype is sex-specific — a key examiner signal.

Answer: Expected ratio: 1 unaffected female: 1 carrier female: 1 unaffected male: 1 colour-blind male. The colour-blind phenotype appears only in males in this cross.

The tell-tale sign of X-linkage in exam data is that the affected phenotype appears predominantly or exclusively in one sex. If you see that pattern in a question, X-linkage should be your first hypothesis to test.

3. Autosomal Linkage

Independent assortment — the basis of the 9:3:3:1 dihybrid ratio — only holds when two genes are on different chromosomes. When two genes sit on the same autosome, they are linked and tend to be inherited together. This breaks the 9:3:3:1 ratio in a characteristic way.

In a dihybrid cross where both genes are fully linked (no crossing over), you get only two phenotypic classes instead of four — the parental combinations. In practice, some crossing over always occurs, so you see mostly parental phenotypes with a smaller proportion of recombinants. The recombination frequency (RF) is the proportion of offspring showing recombinant phenotypes, expressed as a percentage.\n\n**Worked Example: Calculating Recombination Frequency (RF)**\n\nSuppose a cross between two double heterozygotes produces offspring in these phenotypic numbers: 420 AB, 130 Ab, 140 aB, 10 ab (total = 700). The AB and ab phenotypes are parental; Ab and aB are recombinant. To calculate the recombination frequency:\n\nNumber of recombinants = 130 + 140 = 270\nTotal offspring = 700\n\n$$RF = \frac{\text{number of recombinants}}{\text{total offspring}} \times 100\% = \frac{270}{700} \times 100 = 38.6\%$$\n\nThis percentage estimates the distance between the linked genes (in centiMorgans, cM). A higher RF means greater physical distance (more likely crossing over), and a lower RF means close linkage.

Examiners will give you offspring data showing two phenotypic classes in high frequency and two in low frequency. The high-frequency classes are the parental types; the low-frequency classes are the recombinants. Recognising this pattern is the first step to identifying autosomal linkage.

4. Multiple Alleles

Most GCSE genetics involves just two alleles per locus. At A Level, you need to handle loci with three or more alleles in the population — though any individual still carries at most two alleles (one per homologue).

The ABO blood group system is the classic example. The $I$ locus has three alleles: $I^A$, $I^B$, and $i$. $I^A$ and $I^B$ are both dominant over $i$, but codominant with each other — both are expressed when present together, producing blood group AB. This gives six possible genotypes but only four phenotypes (blood groups A, B, AB, and O).

• $I^A I^A$ or $I^A i$ → Blood group A
• $I^B I^B$ or $I^B i$ → Blood group B
• $I^A I^B$ → Blood group AB (codominance)
• $ii$ → Blood group O

When constructing crosses with multiple alleles, the method is identical to standard Mendelian crosses — write parental genotypes, list gametes, complete the Punnett square, and assign phenotypes. The only additional step is knowing the dominance hierarchy for the alleles at that locus.

5. Epistasis

Epistasis is where the lesson gets genuinely interesting. The word comes from the Greek for 'standing upon' — one gene locus stands upon another, masking its expression. The result is a modified dihybrid ratio that tells you exactly which type of epistasis is operating.

Think of a metabolic pathway: gene A produces enzyme A, which converts a colourless precursor into an intermediate pigment. Gene B produces enzyme B, which converts that intermediate into the final pigment. If gene A is homozygous recessive (aa), no intermediate is made — so even if gene B is present, no final pigment appears. Gene A is epistatic to gene B in this pathway.

The 9:7 ratio from duplicate recessive epistasis is worth examining closely. In a standard 9:3:3:1 dihybrid, the 9 class has at least one dominant allele at both loci. In the 9:7 case, both loci must have at least one dominant allele for the dominant phenotype to appear — if either locus is homozygous recessive, the pathway is blocked. The 3:3:1 classes all collapse into the recessive phenotype, giving 9 dominant: 7 recessive.

Worked example: Recessive Epistasis Cross: 9:3:4 Ratio

In Labrador retrievers, coat colour is controlled by two loci. The B locus controls pigment type (B = black, b = brown). The E locus controls pigment deposition (E = pigment deposited, e = no pigment deposited — yellow coat). A dihybrid cross BbEe × BbEe is performed. Determine the expected phenotypic ratio.

1. ee is epistatic to the B locus. If genotype is ee, coat is yellow regardless of B locus genotype. — The E locus controls whether any pigment is deposited. Without pigment deposition, the B locus alleles are irrelevant — this is recessive epistasis.
2. From BbEe × BbEe, the standard 9:3:3:1 genotypic classes are: 9 B_E_: 3 B_ee: 3 bbE_: 1 bbee — Use the standard dihybrid shorthand. B_ means at least one B allele; E_ means at least one E allele.
3. 9 B_E_ → black (pigment deposited, black type) 3 B_ee → yellow (ee epistatic; no pigment deposited) 3 bbE_ → brown (pigment deposited, brown type) 1 bbee → yellow (ee epistatic; no pigment deposited) — The ee genotype produces yellow regardless of the B locus. The two yellow classes (3 + 1) merge.
4. Black: 9 Brown: 3 Yellow: 3 + 1 = 4 — The two ee classes are phenotypically identical, so their frequencies add together.
5. 9 black: 3 brown: 4 yellow — This 9:3:4 ratio is the hallmark of recessive epistasis. Seeing it in exam data should immediately suggest that one locus is epistatic to the other via a recessive mechanism.

Answer: Expected phenotypic ratio: 9 black: 3 brown: 4 yellow. This is the characteristic ratio for recessive epistasis.

A useful exam strategy: whenever you see a dihybrid cross that does not give 9:3:3:1, add up the total ratio parts. If they sum to 16, you are still dealing with a two-locus system — the classes have just been redistributed by epistasis. The specific ratio tells you which type.

6. The Chi-Squared Test

You have now seen how to predict expected ratios from different inheritance patterns. The chi-squared ($\chi^2$) test answers the next question: when your observed data does not perfectly match the expected ratio, is that difference due to chance, or does it suggest your hypothesis about the inheritance pattern is wrong?

The logic of the test is straightforward: if observed and expected values are identical, $(O - E)^2 = 0$ for every class, and $\chi^2 = 0$. The larger the discrepancy between observed and expected, the larger the $\chi^2$ value. You then compare your calculated $\chi^2$ to a critical value from a table, using the appropriate degrees of freedom.

Degrees of freedom (df) = number of phenotypic classes minus 1. For a 9:3:4 ratio (three classes), df = 2. For a 9:3:3:1 ratio (four classes), df = 3. The critical value at $p = 0.05$ is the threshold: if your $\chi^2$ exceeds it, the probability of getting that deviation by chance alone is less than 5%, and you reject the null hypothesis.

Worked example: Chi-Squared Test: Testing a 9:3:4 Ratio

A cross between two black Labrador retrievers (BbEe × BbEe) produces 87 black, 31 brown, and 42 yellow offspring (total = 160). Test whether these results are consistent with the expected 9:3:4 ratio at the 5% significance level.

1. $H_0$: There is no significant difference between the observed and expected phenotypic frequencies. The results are consistent with a 9:3:4 ratio. — The null hypothesis always states that any deviation is due to chance. This is what the test either supports or rejects.
2. Total offspring = 160 Ratio parts: 9 + 3 + 4 = 16 Expected black: $\frac{9}{16} \times 160 = 90$ Expected brown: $\frac{3}{16} \times 160 = 30$ Expected yellow: $\frac{4}{16} \times 160 = 40$ — Divide the total by the sum of ratio parts to find the value of one part, then multiply by each ratio number. Check: 90 + 30 + 40 = 160. Correct.
3. Black: $O = 87$, $E = 90$ $(87 - 90)^2 = (-3)^2 = 9$ $9 \div 90 = 0.100$ Brown: $O = 31$, $E = 30$ $(31 - 30)^2 = (1)^2 = 1$ $1 \div 30 = 0.033$ Yellow: $O = 42$, $E = 40$ $(42 - 40)^2 = (2)^2 = 4$ $4 \div 40 = 0.100$ — Square the difference first (this removes negative signs), then divide by the expected value. Keep at least three decimal places to avoid rounding errors in the sum.
4. $\chi^2 = 0.100 + 0.033 + 0.100 = 0.233$ — Add the three contributions. A small $\chi^2$ value suggests the observed data fits the expected ratio well.
5. df = number of classes $- 1 = 3 - 1 = 2$ Critical value at $p = 0.05$, df $= 2$: $5.991$ — The critical value at p = 0.05 with 2 degrees of freedom is 5.991. This is the threshold your calculated value must exceed to reject the null hypothesis.
6. $\chi^2_{calculated} = 0.233 < \chi^2_{critical} = 5.991$ — Because the calculated value is less than the critical value, we do not reject the null hypothesis. The deviation from the expected 9:3:4 ratio is not statistically significant and is consistent with chance variation.

Answer: $\chi^2 = 0.233$. Since 0.233 < 5.991 (critical value at p = 0.05, df = 2), we fail to reject the null hypothesis. The observed results are consistent with the expected 9:3:4 ratio, supporting the hypothesis of recessive epistasis.

Notice the language of the conclusion: you never 'prove' the null hypothesis — you either reject it or fail to reject it. Examiners award marks for this precision. Also note that a large sample size is important: with very small samples, even large deviations may not reach significance, and the test becomes unreliable.

7. Combining Patterns and Exam Technique

Exam questions often combine two patterns in a single scenario — for example, an X-linked gene that also shows epistasis with an autosomal gene, or a multiple-allele system where you must also apply chi-squared. The approach is always the same: identify each pattern separately, then integrate them into your cross.

When a question asks you to 'suggest' an inheritance pattern from data, work through this diagnostic sequence: first check whether the affected phenotype is sex-specific (sex-linkage signal); then check whether the ratio parts sum to 16 and whether any classes are missing or merged (epistasis signal); then check whether two phenotypic classes dominate with two rare classes (linkage signal). State your reasoning explicitly — examiners award marks for the logic, not just the conclusion.

Bringing this together: sex-linkage, autosomal linkage, multiple alleles, and epistasis are all mechanisms that modify the ratios you would predict from simple Mendelian genetics. Each has a diagnostic signature in the data. The chi-squared test then gives you the statistical language to say, with confidence, whether your observed results support your genetic hypothesis — and that combination of biological reasoning and statistical evaluation is precisely what AQA examiners are looking for at A Level.