The shape of the lesson
By the end of the lesson, learners can
- Reading a pedigree: This family tree follows albinism, caused by a recessive allele (a).
- Codominance and blood groups: Extended.
- Sex linkage: colour blindness: Extended.
- Pedigree: affected people first (aa for a recessive feature), then their parents and children.
- Extended codominance: both alleles contribute to the phenotype; IAIB = group AB, Io is recessive.
- Extended sex linkage: write alleles on the X: XB, Xb; the Y carries none.
- Extended: sons get their X from their mother; colour blindness is more common in males.
One picture every idea returns to
In a pedigree diagram, two unaffected parents with an affected child show that the allele is recessive and both parents are carriers. Extended: in codominance both alleles in a heterozygote contribute to the phenotype (IAIB is blood group AB). A sex-linked gene is on a sex chromosome, so the feature is more common in one sex, as with red-green colour blindness.
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, Family trees, blood groups and colour vision.. A pedigree diagram lets you track a feature through a family and work out who must carry a hidden allele. 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: Reading a pedigree | This family tree follows albinism, caused by a recessive allele (a). Project tab 3 of the lab (A pedigree diagram for albinism) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Codominance and blood groups | Extended. Project tab 5 of the lab (ABO blood groups) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Sex linkage: colour blindness | Extended. Project tab 6 of the lab (Red-green colour blindness) and let a student drive it while the class predicts. | 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, 21 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 Sex linkage: colour blindness, 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 |
|---|---|
| An unaffected person is not automatically AA. | Look at their parents and children: an aa child or an aa parent proves they carry a. |
| A person can only have TWO of the three alleles. | Group A is not always I A I A: it can be I A I o, which is how two group A parents can have a group O child. |
| A son can never inherit colour blindness from his father: a father passes his Y, not his X, to every son. | Return to Sex linkage: colour blindness: ask a pair to state the correct version and show it on the lab before moving on. |
Pedigrees, codominance and sex linkage: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
A pedigree shows a feature caused by a recessive allele, d. Two unaffected parents, Ana and Ben, have three children: two unaffected and one affected. (a) State the genotype of the affected child. (b) Deduce the genotypes of Ana and Ben, explaining your answer. (c) State the possible genotypes of the unaffected children.
Extended. Describe what is meant by codominance, using the ABO blood groups as an example.
Extended. A man with blood group A and a woman with blood group B have a child with blood group O. (a) State the genotype of the child. (b) Draw a genetic diagram to work out the genotypes of the parents and all the possible blood groups of their children, with the ratio.
Extended. Describe what is meant by a sex-linked characteristic and name one example.
Extended. A woman with normal colour vision whose father was colour blind has children with a man with normal colour vision. Use a genetic diagram to find the probability that a son of theirs is colour blind.
Extended. Explain why a colour-blind man cannot pass colour blindness to his sons, but all his daughters will carry the allele.
Answer key and marking notes
- (a) dd.
- (b) Both Dd: the affected child received d from each parent.
- Both parents are unaffected, so each must also carry the dominant allele D.
- (c) DD or Dd.
- Codominance is when both alleles in a heterozygous organism contribute to the phenotype.
- For example IAIB gives blood group AB, with both antigen A and antigen B on the red blood cells.
- (a) IoIo.
- (b) The child received Io from each parent, so the parents are IAIo (group A) and IBIo (group B).
- Gametes: IA or Io from the man; IB or Io from the woman.
- Offspring: IAIB, IAIo, IBIo, IoIo.
- Blood groups AB, A, B and O in the ratio 1 : 1 : 1 : 1.
- A feature in which the gene responsible is located on a sex chromosome (the X), which makes it more common in one sex than in the other.
- Example: red-green colour blindness.
- The woman received Xb from her colour-blind father, and she has normal vision, so she is XBXb. The man is XBY.
- Gametes: XB or Xb from the mother; XB or Y from the father.
- Offspring: XBXB, XBXb, XBY, XbY.
- Sons are XBY (normal) or XbY (colour blind).
- Probability that a son is colour blind = 1 in 2 (0.5).
- The allele is on the X chromosome; the Y carries no copy of the gene.
- A father passes his Y chromosome to his sons, so his Xb never reaches them; a son's X comes from his mother.
- A father passes his only X (Xb) to every daughter, so every daughter carries the allele.
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.
The pedigree shows a condition caused by a recessive allele, n. What is the genotype of person 1?
- ANn
- BNN
- Cnn
- DNN or Nn
Use the same pedigree. Which statement about persons 4 and 5 is correct?
- ABoth must be Nn, because their son 6 is affected
- BBoth must be NN, because they are unaffected
- CPerson 4 is Nn and person 5 must be NN
- DPerson 5 must be nn, because her son is affected
In a pedigree for a recessive condition, which observation proves that two unaffected parents are both carriers?
- AThey have an affected child
- BThey have only unaffected children
- COne of their grandparents is unaffected
- DThey have both sons and daughters
Which describes codominance?
- Aboth alleles in a heterozygote contribute to the phenotype
- Bthe dominant allele masks the recessive allele in a heterozygote
- Cthe gene is carried on a sex chromosome
- Dneither allele is expressed in a heterozygote
A woman with blood group AB has children with a man with blood group O. Which blood groups can their children have?
- AA or B only
- BAB or O only
- CA, B, AB or O
- DO only
Two parents both have blood group A. Their child has blood group O. What are the parents' genotypes?
- AIAIo and IAIo
- BIAIA and IAIo
- CIAIA and IAIA
- DIAIB and IoIo
A man with red-green colour blindness (XbY) has children with a woman who is homozygous for normal vision (XBXB). What will their children be?
- Aall sons normal; all daughters carriers
- Ball sons colour blind; all daughters normal
- Chalf the sons colour blind; daughters normal
- Dall children colour blind
A carrier woman (XBXb) and a man with normal vision (XBY) are expecting a child. What is the probability that the child will be a colour-blind boy?
- A0.25
- B0.5
- C0.75
- D0
Quiz key
Person 3 is affected, so she is nn and received one n from each parent; person 1 is unaffected, so he must also have N: Nn. NN is impossible because he passed on an n. nn would make him affected. "NN or Nn" ignores the evidence from his affected daughter.
Person 6 is nn, so he received n from both 4 and 5; both are unaffected, so each is Nn. Being unaffected does not mean NN. If person 5 were NN, son 6 would have received an N and been unaffected. Person 5 is unaffected, so she cannot be nn.
An affected child must be homozygous recessive, so each unaffected parent gave it a recessive allele and is a carrier. Having only unaffected children is consistent with either parent being homozygous dominant. Unaffected grandparents can be carriers or not, so they prove nothing. The sexes of the children are irrelevant for a condition that is not sex-linked.
In codominance both alleles in a heterozygous organism contribute to the phenotype, as in IAIB (group AB). One allele masking the other is ordinary dominance. A gene on a sex chromosome describes sex linkage. If neither allele were expressed the heterozygote would show neither feature, which does not happen with codominance.
The mother gives IA or IB and the father only Io, so the children are IAIo (group A) or IBIo (group B). No child can be AB, because the father has no IA or IB to give; and no child can be O, because every child receives IA or IB from the mother.
The child is IoIo, so each parent gave an Io; both are group A, so both are IAIo. An IAIA parent could not give Io, which rules out the second and third options. IAIB is group AB and IoIo is group O, not group A.
Every son gets XB from his mother and Y from his father, so all sons have normal vision; every daughter gets Xb from her father and XB from her mother, so all are carriers. Sons cannot inherit the father's X, so none is colour blind. Half the sons being colour blind would need a carrier mother. No child receives two recessive alleles.
The square gives XBXB, XBXb, XBY and XbY, so 1 of 4 children is a colour-blind boy: 0.25. 0.5 is the chance that a son is colour blind, which only applies once you know the child is a boy. 0.75 is the chance of normal vision. 0 forgets that the mother can pass on Xb.
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.