B32: Genetic Inheritance
How genetic characteristics are inherited, Punnett squares and genetic disorders
Key Definitions
| Term | Definition |
|---|---|
| Dominant allele | An allele that is expressed (shown in the phenotype) even if only one copy is present; represented by a capital letter (e.g. B) |
| Recessive allele | An allele that is only expressed if two copies are present (homozygous); represented by a lowercase letter (e.g. b) |
| Homozygous | Having two identical alleles for a gene (e.g. BB or bb) |
| Heterozygous | Having two different alleles for a gene (e.g. Bb) |
| Genotype | The combination of alleles an organism has for a gene (e.g. BB, Bb, bb) |
| Phenotype | The observable characteristic produced by the genotype (e.g. brown eyes, blue eyes) |
| Carrier | A person with one recessive allele for a genetic disorder — they do not show the disorder but can pass it on |
A dominant allele only needs one copy to be expressed in the phenotype. A recessive allele needs two copies (homozygous) to be expressed.
Punnett Squares
A Punnett square is a diagram used to predict the possible genotypes and phenotypes of offspring from a genetic cross. It shows all possible combinations of alleles from the two parents.
Worked Example 1: Heterozygous Cross (Bb × Bb)
Eye colour: B = brown (dominant), b = blue (recessive). Both parents are heterozygous (Bb — brown eyes but carry the blue allele).
| B | b | |
|---|---|---|
| B | BB | Bb |
| b | Bb | bb |
Genotype ratio: 1 BB : 2 Bb : 1 bb
Phenotype ratio: 3 brown eyes : 1 blue eyes
Probability of blue eyes: 25% (1 in 4)
Worked Example 2: Homozygous Dominant × Homozygous Recessive (BB × bb)
Parent 1: BB (brown eyes). Parent 2: bb (blue eyes).
| B | B | |
|---|---|---|
| b | Bb | Bb |
| b | Bb | Bb |
All offspring: Bb — 100% brown eyes (all carriers of the blue allele)
Worked Example 3: Heterozygous × Homozygous Recessive (Bb × bb)
Parent 1: Bb (brown eyes, carrier). Parent 2: bb (blue eyes).
| B | b | |
|---|---|---|
| b | Bb | bb |
| b | Bb | bb |
Genotype ratio: 1 Bb : 1 bb
Phenotype ratio: 1 brown : 1 blue (50% each)
Sex Determination
Human sex is determined by the X and Y chromosomes. Females are XX, males are XY. The father determines the sex of the offspring because he can pass on either an X or a Y chromosome. The mother always passes on an X.
Worked Example 4: Sex Determination Cross (XX × XY)
| X | Y | |
|---|---|---|
| X | XX | XY |
| X | XX | XY |
Ratio: 1 XX (female) : 1 XY (male) — so there is a 50% chance of each sex.
Genetic Disorders
Polydactyly
Polydactyly is a condition where a person has extra fingers or toes. It is caused by a dominant allele (D). Only one copy is needed to have the condition. Because it is dominant, it cannot be “carried” unnoticed — anyone with the allele shows the condition.
- DD or Dd = has polydactyly
- dd = does not have polydactyly (normal)
Worked Example 5: Polydactyly Cross (Dd × dd)
One parent is heterozygous for polydactyly (Dd — has condition), the other is homozygous recessive (dd — normal).
| D | d | |
|---|---|---|
| d | Dd | dd |
| d | Dd | dd |
Phenotype ratio: 1 polydactyly : 1 normal (50% each)
Cystic Fibrosis
Cystic fibrosis is a condition causing thick, sticky mucus in the lungs and digestive system. It is caused by a recessive allele (f). A person must be homozygous recessive (ff) to have the disorder. Carriers (Ff) are healthy but can pass on the allele.
Worked Example 6: Cystic Fibrosis — Two Carrier Parents (Ff × Ff)
| F | f | |
|---|---|---|
| F | FF | Ff |
| f | Ff | ff |
Genotype ratio: 1 FF : 2 Ff : 1 ff
Phenotype ratio: 3 normal : 1 cystic fibrosis
There is a 75% chance each child is normal and a 25% chance of having cystic fibrosis. 2 out of 3 of the normal children are carriers (Ff).
Comparison: Polydactyly vs Cystic Fibrosis
| Feature | Polydactyly | Cystic Fibrosis |
|---|---|---|
| Caused by | Dominant allele (D) | Recessive allele (f) |
| Genotype for disorder | DD or Dd | ff only |
| Carriers exist? | No — dominant allele always expressed | Yes — Ff individuals are carriers |
| Can two unaffected parents have an affected child? | No — at least one parent must have the allele | Yes — if both parents are carriers (Ff) |
| Effect | Extra fingers or toes | Thick mucus in lungs and digestive system |
Practice Questions
Q1: Foundation Define the terms genotype and phenotype.
Q2: Foundation Explain the difference between homozygous and heterozygous.
Q3: Foundation Two carrier parents for cystic fibrosis (Ff) have a child. Use a Punnett square to find the probability the child will have cystic fibrosis.
Q4: Higher A heterozygous parent with polydactyly (Dd) and a normal parent (dd) have four children. Explain why not all four children will necessarily have polydactyly.
Q5: Higher Explain why a person with polydactyly cannot be a “carrier” of the condition.
Q6: Foundation Explain why the father determines the sex of the offspring in humans.
Answers
- Genotype: the combination of alleles an organism has for a gene (e.g. Bb). Phenotype: the observable characteristic produced by the genotype (e.g. brown eyes).
- Homozygous: having two identical alleles for a gene (e.g. BB or bb). Heterozygous: having two different alleles for a gene (e.g. Bb).
- Punnett square shows Ff × Ff → 1 FF : 2 Ff : 1 ff. Probability of cystic fibrosis (ff) = 1/4 = 25%.
- Each birth is an independent event. The Punnett square shows a 50% chance for each child to have polydactyly (Dd) and 50% to be normal (dd). Over four children, probability does not guarantee exactly half will be affected — each child has an independent 50% chance.
- Polydactyly is caused by a dominant allele. A dominant allele is always expressed in the phenotype when present, so anyone with the allele (Dd or DD) will show the condition. A carrier is someone who carries a recessive allele without showing the trait — this is not possible with a dominant condition.
- The mother has two X chromosomes (XX) and can only pass on an X chromosome. The father has one X and one Y chromosome (XY) and can pass on either X or Y. If the father passes on X, the child is XX (female); if he passes on Y, the child is XY (male). Therefore the father’s sperm determines the sex.
Exam Tips
- Always use capital letters for dominant alleles and lowercase for recessive — and use the same letter (e.g. B/b, not B/g)
- When drawing Punnett squares, label the parents and their gametes clearly
- Give both genotype ratio AND phenotype ratio for full marks
- Polydactyly = dominant, cystic fibrosis = recessive — memorise this
- Probabilities are predictions, not certainties. If the probability is 25%, it doesn’t mean exactly 1 in 4 children will be affected
- For sex determination, always show XX and XY in the Punnett square, not just the letters
🔢 Maths Skills
Mathematical Skills
Punnett square probability: outcomes can be expressed as fractions, percentages or ratios. For Ff × Ff: genotype ratio 1 FF : 2 Ff : 1 ff; phenotype ratio 3 normal : 1 affected; probability of being a carrier = 2/4 = 1/2 = 50%.
Converting between forms: fraction → decimal → percentage. E.g. 1/4 = 0.25 = 25%. Ratio to fraction: in a 3:1 ratio, the probability of the recessive phenotype is 1/(3+1) = 1/4.
⚠️ Common Misconceptions
Watch Out!
Students often think dominant alleles are more common in the population. Wrong: Dominant alleles are always more common Correct: Dominance is about expression, not frequency — the allele for polydactyly is dominant but rare
Students often think being a carrier means having the disease. Wrong: Carriers have the disease Correct: Carriers have one recessive allele but no symptoms because the dominant allele masks it
✍️ 6-Mark Question
Extended Answer
6 marks: Explain genetic inheritance using a Punnett square for two carriers of cystic fibrosis having a child. Give the genotype and phenotype ratios and the probability of having an affected child.
Both parents are carriers with genotype Ff. The Punnett square shows: gametes F and f from each parent. Offspring genotypes: FF (normal, not a carrier), Ff (normal, carrier), Ff (normal, carrier), ff (has cystic fibrosis). Genotype ratio: 1 FF : 2 Ff : 1 ff. Phenotype ratio: 3 normal : 1 cystic fibrosis. Probability of having a child with cystic fibrosis = 1/4 = 25%. There is also a 50% (2/4) chance the child will be a carrier.
Mark scheme: 1 mark for correct Punnett square, 1 mark for genotype ratio, 1 mark for phenotype ratio, 1 mark for probability stated, 1 mark for carrier probability, 1 mark for clear explanation
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A family pedigree shows: both parents are unaffected by cystic fibrosis, but they have one child with the condition and two unaffected children. Use this information to determine the carrier status of the parents. Explain your reasoning. If they have another child, what is the probability it will have cystic fibrosis?
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