The shape of the lesson
By the end of the lesson, learners can
- Four genes, four proteins: Choose TYR, HBB, F8 and HTT in turn.
- Gibberellin and the Le gene: Plant height is also controlled by an enzyme.
- TYR tyrosinase, albinism; HBB β-globin, sickle cell.
- F8 factor VIII, haemophilia (X-linked); HTT huntingtin, dominant.
- Le codes for the enzyme GA20 → GA1; lele plants are dwarf.
- Every answer: bases, amino acids, shape, function, phenotype.
One picture every idea returns to
A gene's base sequence determines the amino acid sequence of a polypeptide, which determines how it folds and so its function; the protein's activity then produces the phenotype. The TYR gene codes for tyrosinase, which converts tyrosine to melanin; a recessive allele gives a non-functional enzyme and albinism. The HBB gene codes for β-globin; a single base substitution replaces glutamic acid with valine, giving HbS, which forms fibres at low oxygen and causes sickle cell anaemia. The F8 gene on the X chromosome codes for clotting factor VIII; a recessive allele causes haemophilia. The HTT gene codes for huntingtin; a dominant allele with too many CAG repeats causes Huntington's disease. In peas, the dominant allele Le codes for a functional enzyme that makes active gibberellin (GA1), which causes stem elongation; lele plants make a non-functional enzyme and are dwarf.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, From one base to a whole-body trait.. Albinism, sickle cell anaemia, haemophilia and Huntington's disease look nothing alike, but each follows the same chain: a change in the base sequence of one gene changes a protein, and the faulty protein changes the phenotype. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Four genes, four proteins | Choose TYR, HBB, F8 and HTT in turn. Project tab 5 of the lab (Inheritance lab, genes, proteins and phenotype) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Gibberellin and the Le gene | Plant height is also controlled by an enzyme. | Corners |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 20 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Gibberellin and the Le gene, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Link every step: gene → base sequence → amino acid sequence (primary structure) → folding and shape → protein function → phenotype. | Marks are lost when an answer jumps straight from "mutation" to "disease". |
Genes, proteins and phenotype: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
Explain how a mutation in the TYR gene can result in albinism.
Explain how a single base substitution in the HBB gene leads to the symptoms of sickle cell anaemia.
State the protein coded for by the F8 gene and explain why haemophilia is described as sex-linked.
Explain how the allele causing Huntington's disease differs from the normal HTT allele and why the disease is dominant.
Explain the role of the Le gene in determining the height of pea plants.
Dwarf lele pea plants and tall LeLe plants were grown; half of each group was sprayed with GA1. Mean heights after 4 weeks: LeLe unsprayed 62 cm, LeLe sprayed 66 cm, lele unsprayed 18 cm, lele sprayed 58 cm. Explain these results.
Answer key and marking notes
- A change in the base sequence of TYR changes the sequence of codons / triplets.
- This changes the amino acid sequence (primary structure) of tyrosinase, so it folds differently and the active site changes shape.
- The enzyme is non-functional, so tyrosine is not converted to melanin.
- Without melanin the skin, hair and eyes lack pigment (albinism); only seen in homozygous recessive individuals.
- The substitution changes one codon so that valine replaces glutamic acid in β-globin.
- Valine is non-polar / hydrophobic, so at low oxygen concentration HbS molecules stick together and form long fibres.
- Red blood cells become sickle shaped, carry less oxygen and are destroyed more quickly (anaemia).
- Sickle cells block capillaries, reducing blood flow and causing pain and organ damage.
- F8 codes for factor VIII, a blood-clotting protein.
- The F8 gene is on the X chromosome (no equivalent locus on the Y), so the inheritance pattern differs between males and females.
- The mutant allele has an increased number of CAG repeats (more than about 36, compared with fewer than about 27).
- Huntingtin therefore contains a long chain of glutamine, so it misfolds and forms clumps that kill neurones in the brain.
- The faulty protein causes damage even when a normal allele is present, so one copy causes the disease.
- The Le gene codes for an enzyme in the gibberellin synthesis pathway that converts inactive GA20 to active GA1.
- The dominant allele Le codes for a functional enzyme, so LeLe and Lele plants make GA1.
- GA1 causes stem elongation (cell division and elongation in the internodes), so these plants are tall.
- The recessive allele le codes for a non-functional enzyme (one amino acid different), so lele plants make little GA1 and are dwarf.
- lele plants are short because they cannot make enough active GA1.
- Spraying lele plants with GA1 increases height greatly (18 to 58 cm), showing they can respond to gibberellin; only its synthesis is faulty.
- LeLe plants already make GA1, so extra GA1 has only a small effect (62 to 66 cm).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which protein is coded for by the HTT gene?
- Ahuntingtin
- Btyrosinase
- Cfactor VIII
- Dβ-globin
Which description of albinism is correct?
- Aa recessive TYR allele gives a non-functional enzyme, so tyrosine is not converted to melanin
- Ba dominant TYR allele gives too much tyrosinase, which destroys melanin
- Ca recessive TYR allele on the X chromosome stops melanin being transported
- Da mutation in HBB stops melanin being made in red blood cells
Why does a person with sickle cell anaemia have symptoms mainly when their blood oxygen concentration is low?
- Adeoxygenated HbS molecules stick together into fibres that distort red blood cells
- BHbS cannot bind oxygen at all, so cells starve at low oxygen
- Clow oxygen causes more base substitutions in the HBB gene
- Dred blood cells lose their nuclei only at low oxygen
A man with haemophilia has children with a woman who is homozygous for the normal F8 allele. What is the probability that a son has haemophilia?
- A0
- B1/4
- C1/2
- D1
What is the function of the enzyme coded for by the dominant Le allele in pea plants?
- Ait converts inactive GA20 into active gibberellin GA1
- Bit breaks down gibberellin so that stems stop growing
- Cit binds to DNA as a transcription factor for growth genes
- Dit converts GA1 into inactive GA20
The bar chart shows the heights of pea plants of two genotypes with and without gibberellin spray. Which conclusion is supported?
- Alele plants respond to GA1 but cannot make enough of it themselves
- Blele plants lack a receptor for gibberellin, so spraying has no effect
- CLele plants are tall because they are more sensitive to gibberellin
- Dthe Le allele codes for gibberellin itself rather than an enzyme
Huntington's disease is dominant, but its symptoms usually begin after the age of 35. Why is the allele passed on so often?
- Aaffected people have usually had children before symptoms appear
- Bthe allele is recessive in people under 35
- Ccarriers of the allele never develop the disease
- Dthe extra CAG repeats are removed during meiosis
Glutamic acid in normal β-globin is coded for by the mRNA codon GAG; in HbS it is replaced by valine (GUG). What change occurs in the template strand of the HBB gene?
- ACTC becomes CAC
- BGAG becomes GTG
- CCUC becomes CAC
- DCTC becomes CTT
Quiz key
HTT codes for huntingtin, needed in neurones. Tyrosinase is coded for by TYR, factor VIII by F8 and β-globin by HBB.
Albinism results from a recessive TYR allele coding for non-functional tyrosinase, so no melanin forms. The mutant allele is not dominant and does not destroy melanin. TYR is autosomal, not on the X chromosome. HBB codes for β-globin and is unrelated to melanin.
Deoxygenated HbS molecules bind to each other through the hydrophobic valine, forming long fibres that make cells sickle. HbS does bind oxygen. Mutations do not depend on oxygen level. Mature red blood cells have no nucleus anyway.
A son receives Y from his father and a normal XH from his mother, so none of the sons are affected. 1/4 and 1/2 would need the mother to be a carrier. 1 confuses sons with daughters, all of whom will be carriers.
The Le enzyme catalyses GA20 to GA1, the active gibberellin. Breaking down gibberellin would make plants dwarf. It is an enzyme, not a transcription factor. Converting GA1 to GA20 is the reverse direction.
Spraying raises lele plants from about 10 cm to about 35 cm, close to Lele plants, so they can respond and only lack GA1. A missing receptor would mean no response, but there is a large one. Lele plants barely respond to extra GA, so sensitivity does not explain their height. Le codes for an enzyme in GA synthesis, not for gibberellin, which is not a protein.
Late onset means most people with the allele have already had children, each with a 1 in 2 chance of inheriting it. Dominance does not change with age. Anyone with the allele (with a large repeat number) develops the disease; there are no unaffected carriers as with recessive alleles. Repeats are not removed in meiosis and can even increase.
The template is complementary to the mRNA: GAG pairs with CTC, GUG with CAC, so CTC becomes CAC. GAG to GTG is the coding strand. CUC contains uracil, which is not in DNA. CTC to CTT is complementary to GAA, which still codes for glutamic acid.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.