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How are characteristics inherited, and how do we predict offspring ratios?

Genes, alleles and genotypes, monohybrid and dihybrid inheritance, sex linkage, codominance, and the use of genetic diagrams and the chi-squared test.

A CCEA A-Level Biology answer on genes, alleles and genotypes, monohybrid and dihybrid inheritance, sex linkage and codominance, and the use of genetic diagrams and the chi-squared test.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Key genetic terms
Monohybrid and dihybrid crosses
Sex linkage, codominance and chi-squared
Examples in context
Try this

What this dot point is asking

CCEA wants you to define genes, alleles and genotypes, work out monohybrid and dihybrid crosses, explain sex linkage and codominance, and use genetic diagrams and the chi-squared test to test inheritance ratios.

Key genetic terms

Monohybrid and dihybrid crosses

A test cross (crossing an unknown dominant phenotype with a homozygous recessive) reveals whether the unknown is homozygous or heterozygous: any recessive offspring show the unknown carried a recessive allele. Codominance and multiple alleles can change the expected ratios, as in human ABO blood groups where alleles I-A and I-B are codominant and both dominant over the recessive i.

Sex linkage, codominance and chi-squared

In sex linkage, a gene is carried on the X chromosome. Because males (XY) have only one X, recessive conditions such as haemophilia and red-green colour blindness appear more often in males, who cannot be carriers; a female (XX) needs two copies to be affected but can be a carrier. In codominance both alleles are expressed together, as in the AB blood group or roan coat colour in cattle. The chi-squared test compares observed numbers with the expected ratio: a small calculated value (below the critical value at $p = 0.05$ ) means any difference is likely due to chance, so the data fit the prediction.

Working a sex-linkage cross

A carrier woman for haemophilia ( $X^H X^h$ ) has children with an unaffected man ( $X^H Y$ ). Predict the proportion of children who will be affected.

step Write the parental genotypes and gametes

Mother $X^H X^h$ produces gametes $X^H$ and $X^h$ . Father $X^H Y$ produces gametes $X^H$ and $Y$ .

step Combine the gametes in a Punnett square

The offspring are $X^H X^H$ , $X^H X^h$ , $X^H Y$ and $X^h Y$ , in equal proportions of one quarter each.

step Identify the phenotypes

$X^H X^H$ is an unaffected daughter, $X^H X^h$ is a carrier daughter, $X^H Y$ is an unaffected son, and $X^h Y$ is an affected son (haemophiliac).

step State the answer

One quarter of all children (half of the sons) are affected. No daughters are affected because each inherits a healthy $X^H$ from the father.

Examples in context

Example 1. Colour blindness running through a family. Red-green colour blindness is X-linked recessive. A colour-blind father passes his $X^b$ only to his daughters, who become carriers (not affected) because they also receive a normal X from their mother. Those carrier daughters then have a one in two chance of passing $X^b$ to each son, who would be colour blind. This "skipping a generation through the female line" is the signature of sex linkage in a pedigree.

Example 2. Predicting blood groups for a transfusion clinic. A mother of group A ( $I^A i$ ) and a father of group B ( $I^B i$ ) can have children of group A, B, AB or O in a 1 to 1 to 1 to 1 ratio, because the alleles combine as $I^A I^B$ , $I^A i$ , $I^B i$ and $ii$ . This shows codominance ( $I^A$ and $I^B$ together give AB) and recessiveness ( $ii$ gives O), and is exactly the kind of multiple-allele cross CCEA asks you to work through.

Try this

Q1. Two heterozygous tall pea plants (Tt) are crossed. State the expected phenotype ratio of the offspring. [1 mark]

Cue. Three tall to one short (3 to 1).

Q2. Explain why haemophilia is more common in males than females. [2 marks]

Cue. It is X-linked recessive; males have only one X, so a single recessive allele is expressed.

Q3. A cross predicted a 3 to 1 ratio in 80 offspring. The observed numbers were 54 and 26. Calculate the chi-squared value. [2 marks]

Cue. Expected are 60 and 20; $\chi^2 = \frac{(54-60)^2}{60} + \frac{(26-20)^2}{20} = 0.6 + 1.8 = 2.4$ .

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20196 marksIn fruit flies, grey body (G) is dominant to black body (g). A heterozygous grey fly is crossed with a black fly. Draw a genetic diagram to predict the genotype and phenotype ratios of the offspring, and explain why the ratio may differ from the prediction in a real cross.

Show worked answer →

A 6-mark answer needs a complete labelled genetic diagram plus reasoning about real ratios.

Parents: grey heterozygous (Gg) crossed with black (gg).

Gametes: Gg gives G and g; gg gives g and g.

Offspring genotypes: combining gives Gg and gg in equal numbers.

Genotype ratio: 1 Gg to 1 gg. Phenotype ratio: 1 grey to 1 black.

Why real ratios differ: predicted ratios are probabilities for large samples. With small numbers of offspring, chance (random fertilisation of which gamete meets which) causes the actual numbers to deviate from the exact ratio, just as 10 coin tosses rarely give exactly 5 heads.

Markers reward parental genotypes, gametes, offspring genotypes, the 1 to 1 ratios, and a correct explanation of chance deviation.

CCEA 20215 marksA geneticist predicted a 9 to 3 to 3 to 1 ratio from a dihybrid cross of 320 offspring. The observed numbers were 170, 65, 60 and 25. Carry out a chi-squared test and state what it tells you about the prediction.

Show worked answer →

A 5-mark answer needs expected values, the chi-squared calculation, and a conclusion against the critical value.

Expected (from 9:3:3:1 of 320): $\frac{9}{16} \times 320 = 180$ , $\frac{3}{16} \times 320 = 60$ , $60$ , and $\frac{1}{16} \times 320 = 20$ .

Chi-squared is the sum of (observed minus expected) squared divided by expected:

\chi^2 = \frac{(170-180)^2}{180} + \frac{(65-60)^2}{60} + \frac{(60-60)^2}{60} + \frac{(25-20)^2}{20}

\chi^2 = 0.56 + 0.42 + 0 + 1.25 = 2.23

Degrees of freedom = categories minus 1 = 3. The critical value at p = 0.05 is 7.82. Since 2.23 is less than 7.82, the difference is not significant, so the results fit the predicted 9 to 3 to 3 to 1 ratio (any difference is due to chance).

Markers reward correct expected values, the chi-squared formula with working, degrees of freedom, and the comparison with the critical value.

Related dot points

Sources & how we know this

CCEA GCE Biology specification — CCEA (2016)