Cambridge O Level Biology 5090: Inheritance (5090)

This study guide teaches Topic 17, Inheritance, for Cambridge O Level Biology 5090. It is aligned to the Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028 (version 4), and covers every learning outcome in sections 17.1 Variation, 17.2 DNA, 17.3 Inheritance and 17.4 Selection. The syllabus is not tiered, so everyone studies all of it. One practical outcome is listed: investigating continuous and discontinuous variation.

Use it with the revision notes and the practice questions. The course hub lists every topic and the printable checklist lets you tick off each outcome. Chromosomes, haploid and diploid, and meiosis are taught in Development of organisms and continuity of life; DNA as a molecule made of nucleotides is in Biological molecules. To find weak spots quickly, try the free 10-minute diagnostics.

What this topic covers

Syllabus ref	What you must be able to do
17.1.1–17.1.5	Define variation; continuous v discontinuous; causes; investigate examples
17.2.1	Describe the structure of DNA
17.2.2–17.2.5	Gene; DNA controls protein production; base sequence → amino acid sequence → protein shape
17.3.1–17.3.6	Inheritance terms; genetic diagrams and Punnett squares; 1 : 1 and 3 : 1 ratios; pure-breeding
17.3.7–17.3.8	Codominance (ABO blood groups); sex determination
17.3.9–17.3.12	Gene and chromosome mutation; sources of variation; mutagens
17.4.1–17.4.3	Natural selection; evolution; antibiotic-resistant bacteria
17.4.4–17.4.5	Artificial selection and its economic role

17.1 Variation

Variation is the differences between individuals of the same species.

• Continuous variation gives a range of phenotypes between two extremes, for example body length and body mass. It is caused by genes and the environment. A mass of 61.3 kg is possible, so any value in the range can occur.
• Discontinuous variation gives a limited number of phenotypes with no intermediates, for example ABO blood groups, and seed shape (round or wrinkled) and seed colour in peas. It is usually caused by genes only.

Investigating variation (17.1.5)

1. Continuous: measure one feature of many individuals of one species, such as the length of one leaf from each of 24 plants of one species, using a ruler in mm. Group the values into equal classes and draw a histogram (bars touching), because the data are continuous.
2. Discontinuous: count individuals in each category, such as round and wrinkled seeds from a sample of pea pods. Draw a bar chart (bars of equal width, not touching), because the data are in categories.
3. Use a large sample, and measure the same way each time.

Worked example 1: which chart?

Leaf lengths (mm) were sorted into classes: 35–44: 2 leaves; 45–54: 7; 55–64: 10; 65–74: 5.

• These are lengths, so any value is possible: continuous variation. Plot a histogram with class on the x-axis and number of leaves on the y-axis.
• The data peak in the middle class and tail off at both ends. That shape is typical of continuous variation.
• Like other continuous features, leaf length is affected by genes and by the environment (light, water, soil).

17.2 DNA

Structure of DNA (17.2.1)

• Two strands coiled together to form a double helix.
• Each strand is a chain of nucleotides.
• Each nucleotide contains a base: A, T, C or G (full names are not required).
• Bonds between pairs of bases hold the two strands together.
• The bases always pair in the same way: A with T and C with G.

Genes and proteins (17.2.2–17.2.5)

• A gene is a length of DNA that codes for a protein.
• DNA controls cell function by controlling the production of proteins, including enzymes. Enzymes control the reactions in a cell, so the genes a cell uses decide what it does.
• The sequence of bases in a gene determines the sequence of amino acids needed to make a specific protein.
• Different sequences of amino acids give different shapes to protein molecules. Shape decides function: the active site of an enzyme and the binding site of an antibody both depend on it (see Enzymes).

Worked example 2: base pairing

(a) One strand reads ATCGGA. Write the other strand. Pair each base: A–T, T–A, C–G, G–C, G–C, A–T. Answer: TAGCCT.

(b) In a DNA molecule, 30% of the bases are A. Find the percentage of C. A pairs with T, so T = 30%. A + T = 60%, leaving 40% for C + G. C pairs with G, so they are equal: C = 20%.

17.3 Inheritance

Inheritance is the transmission of genetic information from generation to generation.

Key terms (17.3.2–17.3.3)

Term	Meaning
Allele	An alternative form of a gene
Genotype	The alleles an organism has, e.g. Rr
Phenotype	The observable features, e.g. round seeds
Homozygous	Two identical alleles of a gene, e.g. RR or rr
Heterozygous	Two different alleles of a gene, e.g. Rr
Dominant	An allele expressed in the phenotype when one copy is present (capital letter)
Recessive	An allele expressed only when no dominant allele is present (small letter)

Genetic diagrams and Punnett squares (17.3.4)

Lay out every cross the same way:

1. Choose a letter; capital for dominant, small for recessive.
2. Write the parental phenotypes and genotypes.
3. Write the gametes: one allele of the gene in each gamete.
4. Combine gametes in a Punnett square.
5. List offspring genotypes, phenotypes and the ratio.

Worked example 3: a 3 : 1 cross

In peas, the allele for round seeds (R) is dominant to the allele for wrinkled seeds (r). Two heterozygous plants are crossed.

Parental phenotypes:  round   x   round
Parental genotypes:    Rr     x    Rr
Gametes:             R or r      R or r

          |  R  |  r
      R   | RR  | Rr
      r   | Rr  | rr

Offspring genotypes: 1 RR : 2 Rr : 1 rr. RR and Rr are both round, so the phenotypic ratio is 3 round : 1 wrinkled.

Worked example 4: a 1 : 1 cross

The same heterozygous plant (Rr) is crossed with a wrinkled plant (rr). Gametes: R or r from the first parent, only r from the second.

          |  R  |  r
      r   | Rr  | rr

Offspring: 1 Rr : 1 rr, so 1 round : 1 wrinkled.

Observed v expected ratios (17.3.5)

Each fertilisation is a chance event: which gamete meets which is random. A predicted ratio is what you expect on average. With small numbers of offspring, chance has a big effect, so the observed ratio often differs from the expected one.

Worked example 5

The cross in Worked example 3 gave just 8 seeds: 5 round and 3 wrinkled. Expected: 8 × 3/4 = 6 round and 8 × 1/4 = 2 wrinkled. The observed ratio is 5 : 3, about 1.67 : 1, not 3 : 1. The sample is small, and fertilisation is random, so this is not surprising. With 2400 seeds you would expect very close to 1800 round and 600 wrinkled.

Pure-breeding (17.3.6)

Two identical homozygous individuals that breed together will be pure-breeding: RR × RR gives only RR; rr × rr gives only rr. Every offspring has the same phenotype as the parents.

Codominance: ABO blood groups (17.3.7)

In codominance, both alleles in a heterozygote are expressed in the phenotype. The ABO gene has three alleles: Iᴬ, Iᴮ and Iᵒ. Iᴬ and Iᴮ are codominant; both are dominant to Iᵒ.

Genotype	Blood group (phenotype)
IᴬIᴬ or IᴬIᵒ	A
IᴮIᴮ or IᴮIᵒ	B
IᴬIᴮ	AB (both alleles expressed)
IᵒIᵒ	O

Worked example 6

A man of group AB (IᴬIᴮ) and a woman of group O (IᵒIᵒ) have children. Gametes: Iᴬ or Iᴮ from the man; only Iᵒ from the woman.

          |  Iᴬ   |  Iᴮ
     Iᵒ   | IᴬIᵒ  | IᴮIᵒ

Children: 1 group A : 1 group B. None can be AB or O.

Sex determination (17.3.8)

Human females are XX and males are XY. Every egg carries an X. Half the sperm carry X and half carry Y.

          |  X  |  Y      (sperm)
  egg X   | XX  | XY

The ratio is 1 female : 1 male, so each child has an equal chance of being either sex. The sperm decides the sex.

Mutation (17.3.9–17.3.12)

• A gene mutation is a random change in the base sequence of DNA. Example: sickle cell anaemia. A change in the base sequence of the gene for haemoglobin changes the amino acid sequence, so the haemoglobin has a different shape. Red blood cells can become sickle-shaped. This links straight back to 17.2.4–17.2.5.
• A chromosome mutation is a change in chromosome number or structure. Example: Down’s syndrome, where a person has 47 chromosomes instead of 46.
• Sources of genetic variation in populations: mutation, meiosis, random mating and random fertilisation.
• Ionising radiation (for example X-rays and gamma rays) and some chemicals increase the rate of mutation.

17.4 Selection

Natural selection (17.4.1–17.4.2)

1. There is variation within a population.
2. Organisms produce many offspring, more than the environment can support.
3. There is a struggle for survival, including competition for resources such as food and mates.
4. Individuals better adapted to the environment are more likely to survive and reproduce.
5. They pass on their alleles to the next generation.

Repeated over many generations, the alleles for the useful feature become more common. In this way the inherited features of a population evolve over time.

Antibiotic-resistant bacteria (17.4.3)

Within a population of bacteria, a few may be resistant to an antibiotic because of a mutation. The antibiotic kills the others; the resistant bacteria survive, reproduce and pass on the resistance allele. Repeated use builds up a resistant strain. MRSA (methicillin-resistant Staphylococcus aureus) is such a strain, resistant to many commonly used antibiotics. Limiting antibiotic use is covered in Disease and immunity.

Artificial selection (17.4.4–17.4.5)

Artificial selection (selective breeding):

1. Humans select animals or plants with desirable features.
2. They cross these to produce the next generation.
3. They select the offspring showing the desirable features.
4. They repeat this over many generations.

It is used to produce economically important plants and animals: crops with high yield or disease resistance, cattle that give more milk, and sheep with more wool.

	Natural selection	Artificial selection
Who or what selects	The environment	Humans
Features favoured	Those that help survival and reproduction	Those useful to humans

Common errors

• Writing “gene” when you mean “allele”. A gene for seed shape has two alleles, R and r.
• Putting both alleles in one gamete. A gamete carries one allele of each gene.
• Expecting exactly 3 : 1 in a small sample. Ratios are chances, not fixed numbers.
• Pairing A with C or G with T. Only A–T and C–G.
• Saying the mother decides the sex of a child. Every egg carries X; the sperm decides.
• Writing AB as “IAB” or “incomplete dominance”. Use IᴬIᴮ and the term codominance.
• Saying the antibiotic “causes” the mutation. Resistant bacteria were already there; the antibiotic selects them.
• Calling a continuous-variation feature “caused by genes only”.

Next steps

Condense this into the revision notes, then test yourself with the practice questions. Track your progress on the printable checklist, see other topics on the course hub, and check your gaps with the free 10-minute diagnostics.

Official syllabus

Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028 (version 4), Topic 17: Inheritance (17.1 Variation; 17.2 DNA; 17.3 Inheritance; 17.4 Selection), published by Cambridge Assessment International Education.