Inheritance
Nine key genetics terms precisely defined; monohybrid Punnett square crosses with genotype and phenotype ratios; test cross method; Extended codominance with ABO blood groups; sex-linked inheritance with X-chromosome notation; carrier females and affected males; Punnett squares for sex-linked traits; and pedigree chart analysis.
Inheritance
CORE EXTENDEDKey Terminology
| Term | Definition | Example |
|---|---|---|
| Gene | A length of DNA that codes for a specific protein (and therefore a specific characteristic) | The gene for eye colour; the gene for ABO blood group |
| Allele | One of the different forms (variants) of a gene | The allele for blue eyes; the allele for brown eyes |
| Genotype | The genetic makeup of an organism — the alleles it actually carries | BB, Bb, or bb for coat colour |
| Phenotype | The observable characteristics of an organism — what it looks and functions like | Brown coat; blue eyes; blood group A |
| Dominant | An allele whose effect is expressed in the phenotype even when only one copy is present (heterozygous) | B (brown coat) is dominant over b (white coat) |
| Recessive | An allele whose effect is only expressed when two copies are present (homozygous recessive) | b (white coat) is recessive — only expressed as bb |
| Homozygous | Having two identical alleles for a gene (can be homozygous dominant or homozygous recessive) | BB (homozygous dominant); bb (homozygous recessive) |
| Heterozygous | Having two different alleles for a gene | Bb (heterozygous) |
| Codominance | Both alleles are expressed equally in the phenotype — neither is dominant or recessive | ABO blood group: Iᵃ and Iᵇ are codominant; A and B both expressed → blood group AB |
Always use capital letters for dominant alleles and lower case for recessive alleles. Use the same letter for both alleles of the same gene (e.g. B/b for coat colour; T/t for height). Write genotypes as two letters (diploid organisms carry two alleles for each gene): BB, Bb, or bb.
Monohybrid Crosses
A monohybrid cross examines the inheritance of a single gene with two alleles. A Punnett square is used to predict the probability of each genotype and phenotype in the offspring.
Cross: Bb × Bb (both parents heterozygous)
| B (from parent 1) | b (from parent 1) | |
|---|---|---|
| B (from parent 2) | BB | Bb |
| b (from parent 2) | Bb | bb |
Genotype ratio: 1 BB : 2 Bb : 1 bb
Phenotype ratio: 3 black : 1 yellow
Probability of yellow offspring = 1 in 4 = 25%
① Write parents’ phenotypes and genotypes. ② Show gametes (separate the two alleles). ③ Draw the Punnett square. ④ State genotype ratio. ⑤ State phenotype ratio. In exam answers, always show all five steps — marks are awarded at each stage.
In pea plants, tall (T) is dominant over short (t). A homozygous tall plant (TT) is crossed with a short plant (tt). State the genotype and phenotype of all F1 offspring. Then cross two F1 plants together and give the expected phenotype ratio of F2 offspring. [4 marks]
- F1 cross: TT × tt → all offspring Tt [1 mark]; all F1 are tall (phenotype) [1 mark]
- F2 cross: Tt × Tt → Punnett square gives TT : 2Tt : tt [1 mark]
- F2 phenotype ratio: 3 tall : 1 short [1 mark]
Test Cross
A test cross is used to determine whether an individual showing the dominant phenotype is homozygous dominant (BB) or heterozygous (Bb) — they look the same but have different genotypes.
Method: Cross the unknown individual with a homozygous recessive individual (bb).
Result 1: If ALL offspring show the dominant phenotype → unknown parent is BB (homozygous dominant).
Result 2: If 50% show dominant, 50% show recessive phenotype → unknown parent is Bb (heterozygous).
Codominance — Extended
In codominance, both alleles are expressed equally in the phenotype — neither masks the other. Codominant alleles are written with a superscript notation.
Three alleles control the ABO blood group: Iᵃ (A antigen on red blood cells), Iᵇ (B antigen), and i (no antigen). Iᵃ and Iᵇ are codominant to each other; both are dominant over i.
| Genotype | Blood group (phenotype) |
|---|---|
| IᵃIᵃ or Iᵃi | A |
| IᵇIᵇ or Iᵇi | B |
| IᵃIᵇ | AB (both antigens present — codominance) |
| ii | O (no antigens) |
A person with blood group A (genotype Iᵃi) has a child with a person of blood group B (genotype Iᵇi). Using a Punnett square, determine the possible blood groups of their children and the probability of each. [4 marks]
Parents: Iᵃi × Iᵇi
| Iᵃ | i | |
|---|---|---|
| Iᵇ | IᵃIᵇ (AB) | Iᵇi (B) |
| i | Iᵃi (A) | ii (O) |
- Punnett square correct [2 marks]
- Blood groups: AB : B : A : O in ratio 1:1:1:1 [1 mark]
- Each blood group has 25% probability [1 mark]
Sex-Linked Inheritance — Extended
Sex-linked genes are located on the X chromosome (most common) or, rarely, the Y chromosome. Because males are XY, they only have one copy of X-linked genes — they cannot be heterozygous carriers for X-linked conditions.
Notation for X-linked genes: Write the allele as a superscript on the X: e.g. Xᵒ (normal colour vision), Xⁿ (colour blindness allele, recessive).
| Genotype | Sex | Phenotype |
|---|---|---|
| XᵒXᵒ | Female | Normal colour vision (homozygous) |
| XᵒXⁿ | Female | Normal colour vision (carrier — heterozygous; recessive allele masked) |
| XⁿXⁿ | Female | Colour blind (homozygous recessive) |
| XᵒY | Male | Normal colour vision (only one X allele) |
| XⁿY | Male | Colour blind (only one X allele — no second allele to mask it) |
Males have only one X chromosome. If that X carries the recessive allele (Xⁿ), there is no second X with the dominant allele to mask it → the male is affected. Females need two copies of the recessive allele (XⁿXⁿ) to be affected. This is why colour blindness, haemophilia, and Duchenne muscular dystrophy are much more common in males.
A carrier female (XᵒXⁿ) for colour blindness has children with a male with normal colour vision (XᵒY). Using a Punnett square, determine (i) the probability that any child is colour blind, and (ii) the probability that a son is colour blind. [3 marks]
Parents: XᵒXⁿ (carrier female) × XᵒY (normal male)
| Xᵒ | Y | |
|---|---|---|
| Xᵒ | XᵒXᵒ (normal female) | XᵒY (normal male) |
| Xⁿ | XᵒXⁿ (carrier female) | XⁿY (colour blind male) |
- Punnett square correct [1 mark]
- Any child who is a colour-blind male (XⁿY) = 1 in 4 = 25% [1 mark]
- Probability that a son is colour blind = 1 in 2 = 50% of sons [1 mark]
Pedigree Analysis — Extended
Symbols: □ = male; ○ = female; filled = affected; half-filled = carrier; horizontal line between □ and ○ = mating; vertical line down = offspring.
Determining inheritance pattern:
→ If two unaffected parents have an affected child → the condition is recessive (parents are carriers).
→ If condition appears in every generation → likely dominant.
→ If only males are affected and females are unaffected carriers → likely X-linked recessive.
An organism has the genotype Aa. Which statements about this organism are correct?
I. It is heterozygous.
II. Its phenotype shows the recessive characteristic.
III. It can produce gametes carrying either A or a.
- A. I only
- B. II and III only
- C. I and III only
- D. I, II and III
Comprehensive Practice Questions
Mixed questions across Topic 17.
A plant is heterozygous for flower colour. The dominant allele (R) produces red flowers; the recessive allele (r) produces white flowers. Which statement is correct?
- A. The plant’s genotype is RR and its phenotype is red flowers
- B. The plant’s genotype is Rr and its phenotype is red flowers
- C. The plant’s genotype is Rr and its phenotype is white flowers
- D. The plant’s genotype is rr and its phenotype is red flowers
In guinea pigs, black coat (B) is dominant over white coat (b). Two heterozygous black guinea pigs are crossed.
(a) Using a Punnett square, determine the expected genotype and phenotype ratios of the offspring. [3 marks]
(b) A farmer has a black guinea pig of unknown genotype. Describe how a test cross could be used to determine whether it is BB or Bb. Include the expected results for each case. [3 marks]
- (a) Cross: Bb × Bb; gametes: B and b from each parent [1 mark]; Punnett square giving BB, Bb, Bb, bb [1 mark]; genotype ratio 1 BB : 2 Bb : 1 bb; phenotype ratio 3 black : 1 white [1 mark]
- (b) Cross the unknown black guinea pig with a homozygous white (bb) individual [1 mark]; if all offspring are black → the unknown parent is BB [1 mark]; if 50% are black and 50% are white → the unknown parent is Bb [1 mark]
Haemophilia is an X-linked recessive condition. A pedigree shows that a woman (whose father had haemophilia) marries an unaffected man. They have three sons and two daughters; one son has haemophilia.
(a) State the genotype of the mother. [1 mark]
(b) Explain why the father’s haemophilia tells you the mother is a carrier. [2 marks]
(c) Using a Punnett square, determine the probability that their next son will have haemophilia. [3 marks]
- (a) XᵒXʰ (carrier female) [1 mark] — where Xʰ = haemophilia allele
- (b) The mother’s father had haemophilia (XʰY) → he could only pass Xʰ to his daughters [1 mark]; the mother must have inherited one Xʰ from him; since she is unaffected, she must also carry the dominant Xᵒ → she is a carrier XᵒXʰ [1 mark]
- (c) Cross: XᵒXʰ × XᵒY; Punnett square: XᵒXᵒ (normal female), XᵒXʰ (carrier female), XᵒY (normal male), XʰY (haemophilia male) [1 mark for correct Punnett square]; probability of any child having haemophilia = 1 in 4 = 25% [1 mark]; probability of a son having haemophilia = 1 in 2 = 50% of sons [1 mark — these are distinct answers; make clear which is being asked]
High-Frequency Mistakes — Topic 17 Overall
- 🔄Confusing genotype and phenotypeGenotype = the alleles an organism carries (e.g. Bb). Phenotype = the observable characteristic produced (e.g. brown fur). Two organisms can have the same phenotype (brown fur) but different genotypes (BB or Bb). Always state both when answering genetics questions.
- ✏Using wrong notation — lower case for dominant, or different letters for two allelesAlways use a capital letter for the dominant allele and the same letter in lower case for the recessive allele (T/t, not T/s). Never use capital letters for both alleles of a gene — that implies codominance (IᵃIᵇ). Consistent notation is required for full marks.
- 📈Giving phenotype ratios as genotype ratiosA Bb × Bb cross gives a 1:2:1 genotype ratio (BB:Bb:bb) but a 3:1 phenotype ratio (dominant:recessive). Many students state 1:2:1 as the phenotype ratio. Always convert to phenotype ratios unless the question asks for genotype ratios.
- 👶Ext: Saying males can be "carriers" of X-linked recessive conditionsMales have only one X chromosome. If it carries the recessive allele, they are affected — not carriers. Only females (with two X chromosomes) can be unaffected carriers of X-linked recessive conditions (XᵒXⁿ). Males are either affected (XⁿY) or unaffected (XᵒY).
- 🔄Ext: Saying codominance is the same as incomplete dominanceIn codominance, both alleles are fully expressed simultaneously (e.g. blood group AB has both A and B antigens). In incomplete dominance, a blended intermediate phenotype appears (e.g. red × white = pink). The syllabus only requires codominance — do not use the term “incomplete dominance.”
- ⚖Forgetting to show gametes in Punnett square workingA complete genetics answer requires: (1) parents’ genotypes, (2) gametes clearly shown (separate the alleles), (3) Punnett square, (4) offspring genotypes, (5) phenotype ratio. Missing the gamete step or the phenotype ratio typically loses 1 mark each.
- 📈Ext: Confusing probability of a child being affected vs probability of a son being affectedThese are different questions. If a cross gives 1 in 4 chance of XⁿY (colour blind son), then: probability of any child being colour blind = 25%; but probability of a son being colour blind = 50% (only considering the sons). Read the question carefully — “a son” vs “a child.”
Topic 17 is almost always present in Papers 2, 3, and 4. Highest-yield Core items: all nine key terms (must define precisely — especially allele vs gene, genotype vs phenotype, dominant vs recessive, homozygous vs heterozygous); Punnett square layout with all five steps; 3:1 ratio from Aa × Aa cross; test cross method and interpretation. For Extended: codominance with ABO blood groups (genotype → phenotype table); X-linked inheritance notation (Xᵒ/Xⁿ) and why males more affected; sex-linked Punnett squares; pedigree reading (two unaffected parents with affected child = recessive; only males affected = X-linked). The sex-linkage question with a carrier female is a near-certain Paper 4 target every year.