Revision Notes
OxfordAQA International GCSE Biology: Genetic variation, inheritance, DNA and disorders (9201) – Revision Notes
Condensed revision notes on variation, XX and XY, monohybrid crosses, family trees, DNA, polydactyly, cystic fibrosis, sickle cell and Down's syndrome.
- Subject
- Biology
- Level
- IGCSE
- Topic
- Genetic variation, monohybrid inheritance, DNA and genetic disorders
- Author
- Marlbridge Academic Team
- Updated
Aligned to OxfordAQA IGCSE Biology (9201), Version 5.2, for exams May/June 2018 onwards. Official specification .
Syllabus page (what it covers and how it is assessed): OxfordAQA IGCSE Biology.
Syllabus points this page covers
9201
- 5 Inheritance (whole topic)
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These notes condense sections 3.5.3 (Genetic variation) and 3.5.4 (Genetic disorders) of the OxfordAQA International GCSE Biology (9201) specification, Version 5.2, for exams May/June 2018 onwards. The qualification is untiered. For full explanations read the study guide, then test yourself with the practice questions. Course pages: the OxfordAQA IGCSE Biology hub, the checklist and the free 10-minute diagnostics.
Causes of variation (3.5.3 a)
| Cause | Example |
|---|---|
| Genetic | Inherited blood group |
| Environmental | Less light, so a shorter plant |
| Both | Height, body mass |
Clones and identical twins have the same genes. Differences between them are environmental.
Genes, chromosomes, gametes (3.5.3 b, c)
- Genes are carried on chromosomes in the nucleus.
- Chromosomes normally come in pairs.
- Genes pass to offspring in the gametes.
- Each gamete receives just one chromosome of every pair, and fertilisation puts the pairs back.
Sex determination (3.5.3 d)
Human body cells have 23 pairs of chromosomes. One pair decides sex: female XX, male XY.
| Parent | Gametes |
|---|---|
| Mother XX | all X |
| Father XY | half X, half Y |
Offspring: 1 XX (girl) : 1 XY (boy). Probability of a boy is 1/2 every time.
Worked reminder: a couple have four children. The probability that all four are boys is (1/2)⁴ = 1/16. The probability that the next child is a boy is still 1/2, whatever the earlier children were, because each fertilisation is independent.
Key terms (3.5.3 e, f, g)
| Term | Meaning |
|---|---|
| Gene | A short section of DNA that codes for a protein |
| Allele | A different form of a gene |
| Homozygous | Same two alleles (RR or rr) |
| Heterozygous | Two different alleles (Rr) |
| Dominant | Shows in the phenotype when present on one chromosome only |
| Recessive | Shows only when no dominant allele is present |
| Genotype | The alleles present |
| Phenotype | The observable characteristic |
Key distinction: a carrier is a heterozygous person who has a recessive disorder allele but does not show the disorder.
Method for a monohybrid cross
- Choose a letter. State which allele is dominant (capital) and which is recessive (lower case).
- Write the parents’ genotypes.
- Write each parent’s gametes.
- Fill in a Punnett square.
- State genotypes, phenotypes, the ratio and the probability.
The four crosses to know
| Cross | Offspring genotypes | Phenotypes |
|---|---|---|
| Rr × Rr | 1 RR : 2 Rr : 1 rr | 3 dominant : 1 recessive |
| Rr × rr | 1 Rr : 1 rr | 1 dominant : 1 recessive |
| RR × rr | all Rr | all dominant |
| RR × Rr | 1 RR : 1 Rr | all dominant |
Worked reminder
In sweet peas a tall stem (R) is dominant to a short stem (r). Two heterozygous plants give 80 seedlings.
- Expected tall: 80 × 3/4 = 60. Expected short: 80 × 1/4 = 20.
- Predicted counts are not guaranteed. A result of 57 tall and 23 short is consistent with the model because fertilisation is random.
Unknown genotype
Cross the dominant-looking individual with a homozygous recessive one (rr). A recessive offspring proves the unknown parent was Rr. If all offspring show the dominant phenotype, the parent was probably RR.
Genetic diagrams are models. They predict probabilities. They do not say what happens in one family.
Mendel
- Used garden peas with clear either-or characteristics.
- Began with pure-breeding lines and studied one characteristic at a time.
- Controlled which plants were crossed.
- Counted large numbers of offspring.
- Found first-generation offspring all showed one form, and the next generation was about 3 : 1 (for seed shape, 5474 round and 1850 wrinkled).
- His “inherited factors” were linked with chromosomes later, by other scientists.
Family trees
| Pattern | Conclusion |
|---|---|
| Neither parent shows it, a child does | Recessive; the parents must both be carriers |
| Both parents show it, a child does not | Dominant; the parents must both be Dd-type heterozygotes |
Method:
- Find a family where parents share a phenotype and a child differs.
- Decide dominant or recessive from the table above.
- Give genotypes: recessive phenotype is homozygous recessive; dominant phenotype has at least one dominant allele.
- Use a Punnett square for any prediction.
Worked reminder: reading a family tree
Two parents are both unaffected by a recessive disorder. They have one affected daughter and one unaffected son.
- The parents show nothing but a child is affected, so the allele is recessive.
- The daughter is homozygous recessive (aa). She received one a from each parent, so both parents are Aa.
- The son is AA or Aa. Out of the three unaffected outcomes from Aa × Aa, two are Aa, so the probability that he is a carrier is 2/3.
- If the parents have another child, the probability it is affected is 1/4.
DNA, genes and proteins (3.5.3 h to k)
| Fact | Detail |
|---|---|
| Chromosomes | Made of large DNA molecules |
| DNA | Holds the coded information for inherited characteristics |
| Gene | Small section of DNA coding for a particular combination of amino acids, to make a specific protein |
| Structure | Very long strands twisted into a double helix |
| Bases | Four different compounds in the strands (names not required) |
| Triplet | Three bases in a row are the code for one amino acid |
| Order | The base sequence decides the sequence in which amino acids are joined |
Calculation: number of bases = 3 × number of amino acids. A protein of 210 amino acids needs at least 630 bases in its coding sequence.
Reading a code: split into threes first. With the invented code XYW = amino acid 1 and WWZ = amino acid 2, the sequence XYWWWZ is two amino acids: 1 then 2.
Worked reminder: from base sequence to protein
A gene fragment has 18 bases. 18 ÷ 3 = 6, so it codes for 6 amino acids. Change the order of the bases and you may change the order of the amino acids, so the protein can fold differently and work differently. This is why a change to one gene can lead to an inherited disorder such as sickle cell anaemia, in which the abnormal allele leads to abnormal haemoglobin.
Chromosome link: chromosomes contain many genes, and each gene is a different section of DNA.
Genetic disorders (3.5.4)
| Disorder | Key facts | Allele |
|---|---|---|
| Polydactyly | Extra fingers or toes | Dominant |
| Cystic fibrosis | Thick sticky mucus affects lungs and digestion | Recessive |
| Sickle cell anaemia | Red blood cells sickle-shaped and can block vessels | Recessive |
| Down’s syndrome | Extra chromosome: 47 in body cells, not alleles | Not allele-based |
Carrier × carrier: 1/4 affected, 1/2 carrier, 1/4 unaffected non-carrier. An unaffected child of two carriers has a 2/3 probability of being a carrier.
Reading disorder data
- State what is measured and the units.
- Compare groups using numbers.
- Compare with the prediction from the cross (for example 25%).
- Comment on sample size.
Worked reminder: 12 children, 4 affected. 4 ÷ 12 × 100 = 33.3%. Prediction 25% means 3 of 12. The difference is small for a small sample.
Down’s syndrome in more detail
- Cause: an extra chromosome. It is a different type of cause from an abnormal allele.
- Counts: the usual number in a human body cell is 46 (23 pairs). In Down’s syndrome it is 47.
- Where it comes from: a gamete with 24 chromosomes joins with a normal gamete with 23, giving 47 in the zygote.
- Exam use: you may be given a table of cases against age or country. Describe the trend with numbers, then calculate any rate or ratio asked for.
Quick self-test
- A plant is grown from a cutting in two different soils and ends up different heights. Name the cause of the height difference.
- Which sex chromosomes does a human male have?
- What is the probability that a baby is a girl?
- Define heterozygous.
- In Tt × tt, state the phenotype ratio if T is dominant.
- A cross of two heterozygous parents gives 120 offspring. How many are expected to show the recessive phenotype?
- Two parents with a dominant characteristic have a child without it. Which type of allele causes the characteristic?
- How many bases code for 90 amino acids?
- How many chromosomes are in a body cell of a person with Down’s syndrome?
- Which of polydactyly, cystic fibrosis and sickle cell anaemia is caused by a dominant allele?
Answers
- Environment (the soils), because cuttings from one plant are genetically identical.
- X and Y.
- 1/2.
- It has two different alleles of the gene.
- 1 dominant : 1 recessive.
- 120 × 1/4 = 30.
- A dominant allele.
- 90 × 3 = 270 bases.
-
- Polydactyly.
Where marks are usually lost
- Writing capital and lower-case letters that look the same (C and c, S and s). Choose clear letters.
- Writing a ratio where a probability is wanted, or the other way round.
- Forgetting to write the gametes, so the Punnett square has no clear source.
- Saying the recessive allele has been lost in a carrier.
- Describing Down’s syndrome as an allele problem.
- Stating “all of the DNA codes for protein”. A gene is only a section of DNA.
- Predicting exact counts when the question asks what is expected.
- Using “the parents have the gene”. Say which allele and whether it is dominant or recessive.
Official syllabus
OxfordAQA International GCSE Biology (9201), Version 5.2, for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.5.3 Genetic variation and 3.5.4 Genetic disorders.
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