2.2 Inheritance — Edexcel A Level Biology A (SNAB) 9BI0 Revision Notes & Topical Questions

Section 1 of 8Inheritance

Essential knowledge

DNA

DNA two antiparallel polynucleotide strands with a sugar-phosphate backbone and bases joined by hydrogen bonds between complementary pairs.

Replication

DNA helicase breaks hydrogen bonds to unzip the helix; free nucleotides pair with exposed complementary bases; DNA polymerase joins them by phosphodiester bonds; each new molecule has one old and one new strand.

Transcription

RNA polymerase attaches, DNA unwinds; free RNA nucleotides pair with the template strand; mRNA leaves the nucleus via nuclear pores.

Translation

Translation mRNA binds to a ribosome; tRNA with complementary anticodons bring specific amino acids; peptide bonds form; ribosome moves along to a stop codon.

The code is a non-overlapping, degenerate triplet code; a change of one base may not change the amino acid (silent), may change one amino acid, or may make a stop codon.

Insertions and deletions cause frameshifts, altering every codon downstream, usually severely changing the protein.

Monohybrid inheritance of CF (recessive), and also albinism, thalassaemia (recessive), and haemophilia (X-linked recessive).

Gene therapy for CF

Gene therapy for CF normal CFTR alleles inserted into epithelial cells using viral vectors or liposomes delivered by inhaler; effects are temporary because epithelial cells are replaced (somatic therapy only).

Somatic gene therapy alters body cells (not inherited); germ line therapy alters gametes/embryos (inherited) and is illegal in humans in the UK.

Screening can identify carriers, embryos (PGD) or fetuses (amniocentesis, CVS, NIPT) with alleles for disorders.

Ethical frameworks

Ethical frameworks rights, maximising benefit (utilitarian), informed choice, and weighing false positives/negatives.

CFTR and mucus

The CFTR gene (chromosome 7) codes for a chloride ion channel in epithelial membranes. Faulty channels move fewer Cl⁻ ions out of cells, so less water follows by osmosis and secretions such as mucus become thick and sticky.

CF

CF gas exchange

Cilia cannot shift the thick mucus, so trapped pathogens are not removed and lung infections are frequent. Mucus blocks bronchioles, reducing the surface area available for gas exchange and causing breathlessness; physiotherapy helps loosen it.

CF

CF digestion and fertility

Mucus blocks the pancreatic duct, so fewer digestive enzymes reach the small intestine, and mucus on the gut lining reduces absorption of nutrients. Males may have blocked sperm ducts; in females thick cervical mucus hinders sperm reaching the oviduct.

Meselson-Stahl

Meselson-Stahl second generation

After two replications in ¹⁴N, half the DNA is intermediate (hybrid) density and half is light, again matching semi-conservative replication. Conservative replication would instead give heavy and light bands after the first generation, with no hybrid band.

Duplication and inversion

In a duplication a gene or section is copied, so the spare copy can mutate and gain new functions (important in evolution). In an inversion a section is cut out, turned 180° and reinserted, altering many codons and often giving a non-functional protein.

Incomplete dominance

Neither allele fully masks the other, so heterozygotes show an intermediate third phenotype, e.g. snapdragons: RR red, rr white, Rr pink. A cross between two pink plants gives 1 red : 2 pink : 1 white.

Pedigree diagrams

Squares are males, circles females, shading shows affected individuals; horizontal lines link parents and vertical lines lead to children. Two unaffected parents with an affected child show the allele is recessive and both parents are carriers.

Chi-squared test

Tests whether observed results differ significantly from expected ratios for categorical data. χ² = Σ(O − E)² ÷ E; degrees of freedom = classes − 1. If χ² ≥ critical value at p = 0.05, reject the null hypothesis that there is no significant difference.

Genetic counselling

Counsellors explain the chance of a child inheriting a disorder, interpret test results, and discuss options such as treatment, termination, costs and effects on siblings, so that couples make informed decisions without being directed.

Activated nucleotides

Free nucleotides in the nucleus carry three phosphate groups. DNA polymerase removes two of them and uses the energy released to form the phosphodiester bond with the growing strand.

Section 2 of 8Process & sequence

1

Write parental phenotypes and genotypes, e.g. carrier × carrier: Ff × Ff.

2

List the gametes each parent can produce: F and f from each.

3

Draw a Punnett square combining gametes.

4

List offspring genotypes and ratio: 1 FF : 2 Ff : 1 ff.

5

Convert to phenotypes: 3 unaffected : 1 affected (ff has CF).

6

State probability: 0.25 (25%) chance each child has CF, independently each pregnancy.

From gene to proteinwrite these steps in this order; each step is a marking point.

1

DNA unwinds; RNA polymerase transcribes the template strand into mRNA.

2

mRNA exits through a nuclear pore to a ribosome.

3

tRNA anticodons pair with mRNA codons, delivering specific amino acids.

4

Peptide bonds form between adjacent amino acids.

5

Ribosome reaches a stop codon; polypeptide released and folds into its 3D shape.

Section 3 of 8Compare

DNA vs RNA
Feature	DNA	mRNA	tRNA
Sugar	Deoxyribose	Ribose	Ribose
Bases	A, T, C, G	A, U, C, G	A, U, C, G
Strands	Double helix	Single, linear	Single, folded clover-leaf
Length	Very long	Shorter (one gene)	About 80 nucleotides
Role	Stores genetic information	Carries code to ribosome	Carries amino acid; anticodon

Genetic screening: ethical issues
Test	Benefits	Concerns
Carrier testing	Informed reproductive choices	Stigma; insurance/employment discrimination
Pre-implantation genetic diagnosis (PGD)	Only unaffected embryos implanted	Destroying embryos; designer babies
Amniocentesis (15–20 weeks)	Diagnosis before birth	About 1% miscarriage risk; later termination
Chorionic villus sampling (10–13 weeks)	Earlier result	About 1–2% miscarriage risk
Non-invasive prenatal testing (fetal DNA in mother's blood)	No miscarriage risk	May need confirming by invasive test; false positives

Section 4 of 8Worked examples

Worked examples

A DNA sample has 22% adenine. Calculate the percentage of each other base.

T = A = 22%. C + G = 100 − 44 = 56%, so C = G = 28%.

0.5625 ≈ 0.56 (56%)

Two carriers of CF have children. Probability that their first two children are both unaffected?

P(unaffected) = 3/4 = 0.75. Both = 0.75 × 0.75 = 0.5625 ≈ 0.56 (56%).

411 amino acids

An mRNA has 1236 nucleotides between and including the start codon and the stop codon. How many amino acids in the polypeptide?

Codons = 1236 ÷ 3 = 412; subtract the stop codon: 411 amino acids.

Section 5 of 8Keep going

Inheritance Edexcel A Level Biology SNAB: what to do next

These Inheritance Edexcel A Level Biology SNAB notes covered the key ideas, worked examples and exam technique for this topic. Check your Inheritance Edexcel A Level Biology SNAB understanding with the practice questions below, then reread any Inheritance Edexcel A Level Biology SNAB point you were unsure about before moving on.

Inheritance Edexcel A Level Biology SNAB: revision checklist

1

Revise the main Inheritance Edexcel A Level Biology SNAB definitions until you can write them from memory.

2

Practise Inheritance Edexcel A Level Biology SNAB exam questions under timed conditions.

3

Link each Inheritance Edexcel A Level Biology SNAB idea to a diagram or a worked example.

4

List the common Inheritance Edexcel A Level Biology SNAB mistakes and how to avoid them.

5

Explain one Inheritance Edexcel A Level Biology SNAB idea out loud as if teaching a friend.

6

Summarise Inheritance Edexcel A Level Biology SNAB in five key points on a revision card.

7

Check which command words appear most often in Inheritance Edexcel A Level Biology SNAB questions.

1

Previous lesson: 1.1 The Circulatory System — 9BI0 Notes

2

Next lesson: 1.2 Diet & Health — 9BI0 Notes

3

Course: Edexcel A Level Biology A (snab) 9BI₀ — all units

4

Official syllabus: Pearson Edexcel A Levels

Section 6 of 8Key terms

Section 7 of 8Command terms

Command terms

Give / State / Name

Recall a fact; one idea per mark.

Define

Give the exact meaning using the specification keywords.

Describe

Give an account of what happens / the pattern in data, quoting values; no reasons needed.

Explain

Give reasons with linked statements (cause → effect). Each linked step is usually a separate marking point.

Suggest

Use knowledge in a new context; more than one answer may be accepted.

Compare

Similarities and differences; each point must refer to both items.

Calculate

Show working; include units; check significant figures. Method marks are given even if the final answer is wrong.

Discuss

Explore the issue from different angles and link the points.

Evaluate / Assess

Weigh strengths and weaknesses and reach a supported judgement.

Deduce

Draw a conclusion from the information given.

Justify

Support a statement/choice with evidence.

Section 8 of 8Exam technique

Keywords the mark scheme looks for in this lesson

Gene · Allele · Genotype · Phenotype · Dominant allele · Recessive allele · Homozygous · Heterozygous · Codominance · Gene mutation · Triplet code · Semi-conservative replication · Transcription · Translation

Learn each definition in the Key words tab exactly; a missing keyword in a definition usually loses the mark.

How to get full marks (examiner guidance)

1

Use specification keywords exactly (e.g. enzyme–substrate complex, active site, complementary).

2Link ideas

Link ideas for “explain” each marking point is usually a cause → consequence step; unlinked facts often score nothing.

3Use the information in the question

“discuss/evaluate” answers must use the data provided and match the number of marks.

4

Convert units carefully (mg → g, cm³ → dm³) and show working for calculations.

5Avoid vague words

“it”, “stuff”, “goes up”, “energy is made”. Name the substance and the process.

6

Extended (6-mark) responses are marked on content AND a logical, linked line of reasoning.

Words to avoid

1

a gene codes for a characteristica gene codes for the sequence of amino acids in a polypeptide.

2

the mutation changes the amino acid (only)changes base sequence, so codon, amino acid sequence, folding and 3D shape of the protein.

3

the recessive allele is weakerthe recessive allele is only expressed in the homozygous condition.

4

1 in 4 children will have CFeach child has a probability of 0.25 of having CF.

5

gene therapy cures CF permanentlysomatic therapy is temporary and not passed to offspring.