The shape of the lesson
By the end of the lesson, learners can
- The language of inheritance: In the sim, tall (T) is dominant to dwarf (t) in pea plants.
- Pure-breeding and the 3: 1 ratio: Press Next generation twice: pure-breeding parents, then F1, then F2.
- The test cross: Extended.
- Parents: phenotype and genotype of each (choose a letter, capital for dominant).
- Gametes: one allele per gamete, circled.
- Punnett square: every combination of the gametes gives the offspring genotypes.
- Offspring: phenotypes and the ratio (3 : 1 from Aa × Aa; 1 : 1 from Aa × aa). Extended: a test cross uses aa.
One picture every idea returns to
Genotype is the genetic make-up of an organism in terms of its alleles; phenotype is its observable features. A dominant allele is expressed if it is present; a recessive allele only when no dominant allele is present. Crossing two heterozygotes gives a 3 : 1 ratio; crossing a heterozygote with a homozygous recessive gives 1 : 1.
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, One gene, two alleles, predictable ratios.. Inheritance is the passing of genetic information from one generation to the next. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The language of inheritance | In the sim, tall (T) is dominant to dwarf (t) in pea plants. Project tab 1 of the lab (Punnett squares for pea plants) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Pure-breeding and the 3: 1 ratio | Press Next generation twice: pure-breeding parents, then F1, then F2. Project tab 2 of the lab (P, F1 and F2 generations) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: The test cross | Extended. Project tab 4 of the lab (Test cross with a mystery guinea pig) and let a student drive it while the class predicts. | Think-Pair-Share |
| 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, 22 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 The test cross, 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.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
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 |
|---|---|
| Each gamete carries only ONE allele of the gene. | Write the gametes as T and t, never as Tt, and circle each one. |
| All-black offspring do not prove the parent is BB: a Bb parent could, by chance, give only black pups. | A large number of offspring makes the conclusion much more reliable. |
Monohybrid crosses: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Define (a) genotype, (b) homozygous, (c) heterozygous and (d) recessive allele.
In tomato plants, red fruit (R) is dominant to yellow fruit (r). Two heterozygous plants are crossed. Draw a complete genetic diagram and state the expected ratio of the offspring phenotypes.
A heterozygous red-fruited tomato plant (Rr) is crossed with a yellow-fruited plant (rr). 120 seeds grow into plants. (a) State the expected phenotype ratio. (b) Calculate the expected number of yellow-fruited plants. (c) State the probability that one particular seed grows into a red-fruited plant.
Explain why a plant that is homozygous for an allele is pure-breeding, but a heterozygous plant is not.
Extended. Purple flowers (P) are dominant to white flowers (p). Explain how a test cross could show whether a purple-flowered plant is PP or Pp.
Two pure-breeding pea plants, one with green pods and one with yellow pods, were crossed. All the F1 plants had green pods. Two F1 plants were then crossed, giving 580 plants with green pods and 202 with yellow pods. (a) State which allele is dominant. (b) Calculate the ratio green : yellow to one decimal place. (c) State the genotype of the F1 plants, using G and g.
Answer key and marking notes
- (a) The genetic make-up of an organism, in terms of the alleles present.
- (b) Having two identical alleles of a particular gene.
- (c) Having two different alleles of a particular gene.
- (d) An allele that is only expressed when there is no dominant allele of the gene present in the genotype.
- Parents: red Rr × red Rr.
- Gametes: R and r from each parent.
- Punnett square: RR, Rr, Rr, rr.
- Phenotypes: RR, Rr, Rr red; rr yellow.
- Ratio 3 red : 1 yellow.
- (a) Gametes R or r × r only, giving Rr and rr: 1 red : 1 yellow.
- (b) 120 ÷ 2 = 60 yellow-fruited plants.
- (c) 1 in 2 (0.5, or 50%).
- A homozygous plant (for example TT) can only make gametes carrying one type of allele (T).
- Bred with an identical homozygous plant, all offspring have the same genotype and phenotype as the parents, so it is pure-breeding.
- A heterozygous plant (Tt) makes gametes carrying T or t, so its offspring can have different genotypes and show either phenotype: it is not pure-breeding.
- Cross the purple plant with a white-flowered plant, which must be homozygous recessive (pp).
- The pp plant only gives p, so the offspring phenotypes show the allele received from the purple plant.
- If the purple plant is PP, all the offspring are purple (Pp).
- If it is Pp, about half the offspring are white (pp): a 1 purple : 1 white ratio. Use a large number of offspring.
- (a) Green, because all the F1 had green pods even though they carried both alleles.
- (b) 580 ÷ 202 = 2.9, so about 2.9 : 1 (close to 3 : 1).
- (c) Gg (heterozygous).
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.
What does the term heterozygous mean?
- Ahaving two different alleles of a particular gene
- Bhaving two identical alleles of a particular gene
- Chaving an allele that is expressed whenever it is present
- Dhaving observable features that differ from the parents
Which of these is a phenotype?
- Around seeds
- BRr
- CR
- Dhomozygous
In humans, the allele for free earlobes (E) is dominant to the allele for attached earlobes (e). A man with attached earlobes and a woman who is heterozygous have children. What is the expected ratio of free : attached earlobes?
- A1 : 1
- B3 : 1
- C1 : 3
- Dall free
Two pea plants were crossed. The bar chart shows the pod colour of their offspring. Green (G) is dominant to yellow (g). What were the genotypes of the parents?
- AGg and Gg
- BGg and gg
- CGG and gg
- DGG and Gg
A Punnett square for the cross Aa x Aa is drawn. What fraction of the offspring are expected to be heterozygous?
- A1/2
- B1/4
- C3/4
- D1/3
Which pair of rabbits will always produce offspring identical in coat colour to themselves (pure-breeding)? Grey (G) is dominant to white (g).
- Agg x gg
- BGg x Gg
- CGg x gg
- DGG x gg
A breeder wants to know whether a tall pea plant is TT or Tt. Which plant should it be crossed with?
- Aa dwarf plant, tt
- Ba tall plant, TT
- Ca tall plant, Tt
- Danother plant of unknown genotype
A black bull is crossed with several red cows (bb). Black (B) is dominant. They have 12 calves, all black. Which is the best conclusion?
- AThe bull is very likely BB, but Bb is not ruled out
- BThe bull must be Bb
- CThe bull must be bb
- DThe cows must be carrying the B allele
Quiz key
Heterozygous means having two different alleles of a gene, such as Tt. Two identical alleles is homozygous. An allele expressed whenever present is a dominant allele, which describes an allele rather than a genotype. Differing from the parents describes phenotype and variation, not heterozygosity.
Round seeds is an observable feature, so it is a phenotype. Rr is a genotype, the alleles present. R is a single allele. Homozygous describes a type of genotype (two identical alleles), not a feature you can see.
ee gives only e; Ee gives E or e. The offspring are Ee (free) and ee (attached) in equal numbers, 1 : 1. 3 : 1 comes from crossing two heterozygotes (Ee x Ee), and 1 : 3 reverses that ratio. All free would need one parent to be EE.
74 : 26 is about 3 : 1, the ratio from crossing two heterozygotes, so the parents were Gg and Gg. Gg x gg gives about 1 : 1. GG x gg and GG x Gg give all green offspring, because every offspring receives a dominant G from the GG parent.
The square gives AA, Aa, aA and aa, so 2 of the 4 boxes are heterozygous: 1/2. 1/4 is the fraction that is AA (or aa). 3/4 is the fraction showing the dominant phenotype, which includes AA as well. 1/3 is the fraction of dominant-phenotype offspring that are AA, not a fraction of all offspring.
Two identical homozygous individuals are pure-breeding: gg x gg can only give gg, white like the parents. Gg x Gg can give white offspring from grey parents. Gg x gg gives a mixture of grey and white. GG x gg parents are not identical, and all their offspring are grey Gg, unlike the white parent.
A test cross uses a homozygous recessive individual (tt), which only gives t, so the offspring phenotypes reveal the allele each received from the tall plant. A TT partner gives only T, so all offspring are tall whatever the unknown is. A Tt partner adds its own t alleles and muddles the result. A plant of unknown genotype cannot reveal anything.
If the bull were Bb, each calf would have a 1 in 2 chance of being red, so 12 black calves in a row would be very unlikely; he is very probably BB, but a test cross can only make this more certain, not prove it. Any red calf would have shown Bb. The bull is black, so he cannot be bb. Red cows are bb, so they have no B allele.
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.