The shape of the lesson
By the end of the lesson, learners can
- The vocabulary and the monohybrid cross: Every genetics answer starts with the same words.
- Two genes at once: 9: 3: 3: 1 and the test cross: Choose 9:3:3:1 and read the PHENOTYPE RATIO for RrYy × RrYy peas (round R, wrinkled r, yellow Y, green y).
- When the ratio breaks: linkage and epistasis: Back in the simulator, choose linkage, then epistasis, and watch the phenotype ratio change.
- Layout: phenotypes, genotypes, gametes, Punnett square, offspring, ratio.
- Codominant and sex-linked alleles are written as superscripts.
- Test cross: with the homozygous recessive; offspring mirror the gametes.
- Linkage: few recombinants; epistasis: one gene masks another.
One picture every idea returns to
A gene is a length of DNA at a fixed position, its locus, coding for a polypeptide; alleles are different versions of it. The genotype is the alleles present (homozygous if the two are the same, heterozygous if different) and the phenotype is the observable feature. A dominant allele is expressed in the heterozygote; a recessive one only when homozygous; codominant alleles are both expressed. In a genetic diagram, write parental phenotypes and genotypes, gametes (one allele of each gene per gamete), a Punnett square, then offspring genotypes, phenotypes and ratio. A monohybrid cross of two heterozygotes gives 3 : 1; a dihybrid cross of two double heterozygotes gives 9 : 3 : 3 : 1 if the genes assort independently. A test cross with a homozygous recessive reveals an unknown genotype. Autosomal linkage (genes on the same chromosome) and epistasis (one gene affecting the expression of another) change the expected ratios.
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, Draw the gametes, fill the square.. A genetic diagram is a prediction machine: write the parents' genotypes, list every type of gamete, combine them in a Punnett square and read off the ratio. 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 vocabulary and the monohybrid cross | Every genetics answer starts with the same words. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Two genes at once: 9: 3: 3: 1 and the test cross | Choose 9:3:3:1 and read the PHENOTYPE RATIO for RrYy × RrYy peas (round R, wrinkled r, yellow Y, green y). Project tab 3 of the lab (Inheritance lab, dihybrid crosses, linkage and epistasis) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: When the ratio breaks: linkage and epistasis | Back in the simulator, choose linkage, then epistasis, and watch the phenotype ratio change. | Mix-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, 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 When the ratio breaks: linkage and epistasis, 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.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
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 |
|---|---|
| Write a full genetic diagram: parental phenotypes, parental genotypes, gametes (one allele of each gene), Punnett square, offspring genotypes and phenotypes, ratio. | For sex-linked genes always attach the allele to the X, and give the Y no allele. |
| Linkage means two genes on the same chromosome; sex linkage means one gene on the X chromosome. | Do not use the words interchangeably. |
Monohybrid and dihybrid crosses: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
Explain the difference between the terms gene and allele, and between genotype and phenotype.
Haemophilia is caused by a recessive allele of the F8 gene on the X chromosome. A woman whose father had haemophilia has children with an unaffected man. Draw a genetic diagram and state the probability that a son will have haemophilia.
In a plant, tall (T) is dominant to short (t) and purple flowers (P) are dominant to white (p); the genes are on different chromosomes. Use a genetic diagram to predict the offspring of TtPp × ttpp.
Explain why a cross between two plants heterozygous for two linked genes (AB/ab) gives mostly two phenotypes instead of four.
In a species of flower, gene E controls pigment production: ee plants cannot make any pigment and are white. Gene R controls pigment colour: R gives red, rr gives yellow. Predict the phenotype ratio from EeRr × EeRr.
Sickle cell anaemia: alleles HbA and HbS are codominant. Two parents with sickle cell trait have a child. Give the possible genotypes and phenotypes of the child, with their probabilities.
Answer key and marking notes
- Gene: a length of DNA at a particular locus that codes for a polypeptide.
- Allele: one of the alternative forms of a gene (differing in base sequence).
- Genotype: the alleles present in an organism; phenotype: its observable characteristics, resulting from the genotype and the environment.
- The woman must be a carrier, XHXh (she received Xh from her father); the man is XHY.
- Gametes: XH, Xh from the mother; XH, Y from the father.
- Offspring: XHXH normal female, XHXh carrier female, XHY normal male, XhY haemophiliac male.
- Probability that a son has haemophilia = 1/2 (0.5).
- Gametes of TtPp: TP, Tp, tP, tp (in equal proportions); gametes of ttpp: all tp.
- Offspring genotypes: TtPp, Ttpp, ttPp, ttpp.
- Phenotypes: tall purple, tall white, short purple, short white.
- Ratio 1 : 1 : 1 : 1 (this is a test cross).
- The two genes are on the same chromosome, so alleles A and B (and a and b) are inherited together / do not assort independently.
- Each parent makes mainly AB and ab gametes, giving offspring mostly with the dominant phenotype for both genes or the recessive phenotype for both (about 3 : 1).
- Crossing over between the loci produces a few Ab and aB recombinant gametes, so a small number of the other two phenotypes appear.
- Of 16 offspring, 9 have E_R_ (red) and 3 have E_rr (yellow).
- 4 are ee (eeR_ and eerr), which are white whatever the R alleles, because gene E is epistatic to gene R.
- Ratio 9 red : 3 yellow : 4 white.
- Parents HbAHbS × HbAHbS; gametes HbA and HbS from each.
- HbAHbA unaffected 0.25; HbAHbS sickle cell trait 0.5; HbSHbS sickle cell anaemia 0.25.
- The heterozygote has its own phenotype (trait) because both alleles are expressed: codominance.
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 is meant by the locus of a gene?
- Athe position of the gene on a particular chromosome
- Bone of the alternative forms of the gene
- Cthe sequence of bases that makes up the gene
- Dthe observable feature that the gene controls
In cattle, red coat (CR) and white coat (CW) are codominant; heterozygotes are roan. What proportion of the offspring of a roan bull and a white cow are expected to be roan?
- A1/2
- B1/4
- C3/4
- Dall of them
A man of blood group O and a woman of blood group AB have a child. Which blood groups are possible for the child?
- AA or B only
- BA, B or AB
- CO or AB only
- DA, B, AB or O
The pedigree shows a family with haemophilia, a sex-linked recessive condition. What is the genotype of female 2?
- AXHXh
- BXHXH
- CXhXh
- DXhY
A plant with genotype AaBb is crossed with an aabb plant. The genes are on different chromosomes. What fraction of the offspring are expected to show both recessive phenotypes?
- A1/4
- B1/16
- C1/2
- D9/16
In a plant, genes for flower colour (P purple, p white) and stem length (T tall, t short) are linked. A PpTt plant with P and T on one chromosome is test crossed. Which result shows that crossing over has occurred?
- Aa small number of purple short and white tall offspring
- Bequal numbers of all four phenotypes
- Conly purple tall and white short offspring
- Dmainly purple short and white tall offspring
In a squash plant, the dominant allele W stops any colour forming, so W_ fruit are white. In ww plants, G gives yellow fruit and gg gives green. What is the phenotype ratio from WwGg × WwGg?
- A12 white : 3 yellow : 1 green
- B9 white : 3 yellow : 4 green
- C9 white : 3 yellow : 3 green : 1 white
- D3 white : 1 green
A woman who is a carrier of haemophilia and has blood group AB (IAIB) has children with an unaffected man of blood group O. What is the probability that their first child is a boy with haemophilia and blood group A?
- A1/8
- B1/4
- C1/16
- D1/2
Quiz key
A locus is the fixed position of a gene on a chromosome. An alternative form is an allele. The base sequence is the gene itself. The observable feature is the phenotype.
CRCW × CWCW gives CRCW and CWCW in equal numbers: 1/2 roan, 1/2 white. 1/4 and 3/4 come from treating it as a cross of two heterozygotes. All roan would need a red parent crossed with a white one.
IoIo × IAIB gives IAIo (group A) and IBIo (group B). AB is impossible because the father cannot give IA or IB. O is impossible because the mother has no Io allele. So the child can be neither AB nor O.
Son 4 is XhY and received his Xh from his mother. She is unaffected, so she must be a carrier, XHXh. XHXH could not pass on Xh. XhXh would make her affected. XhY is a male genotype.
AaBb makes AB, Ab, aB and ab equally; aabb makes only ab, so 1/4 of offspring are aabb. 1/16 is the fraction for AaBb × AaBb. 1/2 counts only one gene. 9/16 is the dominant-dominant class in a dihybrid F2.
Recombinant gametes Pt and pT, made by crossing over, give the few purple short and white tall offspring. Equal numbers would mean independent assortment. Only purple tall and white short means no crossing over at all. Mainly purple short and white tall would mean P and t were linked in the parent, the opposite of the arrangement given.
12 of the 16 combinations have W (9 W_G_ + 3 W_gg), so they are white. Of the 4 ww, 3 are wwG_ (yellow) and 1 is wwgg (green). 9 : 3 : 4 is the recessive-epistasis pattern, which does not fit a dominant W. 9 : 3 : 3 : 1 ignores the epistasis. 3 : 1 treats it as a single gene.
Probability of XhY = 1/4 (1/2 a boy × 1/2 receiving Xh). Probability of group A = 1/2 (IAIo or IBIo). 1/4 × 1/2 = 1/8. 1/4 forgets the blood group. 1/16 uses 1/4 for blood group A. 1/2 uses only the chance of a son being affected.
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.