← Unit 4: Heredity and continuity of life
Topic 2: Inheritance
Apply Mendel's laws of segregation and independent assortment to predict the outcomes of monohybrid and dihybrid crosses using Punnett squares, and explain the purpose of a test cross
A focused answer to the QCE Biology Unit 4 dot point on Mendelian genetics. Defines genotype, phenotype, allele, homozygous and heterozygous, applies the laws of segregation and independent assortment to monohybrid and dihybrid Punnett squares (3:1 and 9:3:3:1 ratios), and explains how a test cross with a homozygous recessive parent reveals the genotype of an unknown dominant phenotype.
Have a quick question? Jump to the Q&A page
What this dot point is asking
QCAA wants you to use Punnett squares to predict the outcomes of monohybrid and dihybrid crosses, state and apply Mendel's two laws, and design and interpret a test cross. Genetics calculation problems are core marks every year.
The answer
Gregor Mendel worked out the basic rules of inheritance in pea plants in the 1860s without knowing about DNA, chromosomes or genes. His laws still describe single-gene inheritance accurately and form the foundation of every genetics problem.
Vocabulary
- Gene. A heritable factor (a stretch of DNA) that codes for a particular trait.
- Allele. An alternative form of a gene (for example T for tall, t for short).
- Genotype. The combination of alleles an individual carries (TT, Tt or tt).
- Phenotype. The observable trait (tall or short).
- Homozygous. Both alleles the same (TT or tt).
- Heterozygous. Two different alleles (Tt).
- Dominant. An allele that is expressed in the phenotype of a heterozygote. Written with a capital letter.
- Recessive. An allele whose phenotype is only seen in the homozygote. Written with a lowercase letter.
Mendel's laws
Law of segregation. Each individual has two alleles for each gene; the two alleles separate during meiosis so that each gamete carries only one. This is the chromosomal behaviour of homologous chromosomes separating in anaphase I.
Law of independent assortment. Alleles for different genes segregate into gametes independently of one another, provided the genes are on different chromosomes (or far enough apart on the same chromosome to recombine freely). This is the chromosomal behaviour of independent metaphase I alignment.
Monohybrid crosses
A monohybrid cross follows one gene with two alleles.
Heterozygous x heterozygous. Tt x Tt.
| T | t | |
|---|---|---|
| T | TT | Tt |
| t | Tt | tt |
Genotype ratio: 1 TT to 2 Tt to 1 tt.
Phenotype ratio: 3 tall to 1 short.
Heterozygous x homozygous recessive. Tt x tt.
| T | t | |
|---|---|---|
| t | Tt | tt |
| t | Tt | tt |
Phenotype ratio: 1 tall to 1 short.
Dihybrid crosses
A dihybrid cross follows two genes simultaneously. The dihybrid ratio is the product of the two monohybrid ratios.
Heterozygous at both loci x heterozygous at both loci. TtYy x TtYy.
Each parent produces four gamete types in equal proportions: TY, Ty, tY, ty. A 4 by 4 Punnett square gives 16 boxes, and the phenotype ratio is:
- 9 with both dominant traits (T_Y_, tall yellow)
- 3 with one dominant and one recessive (T_yy, tall green)
- 3 with the other one dominant and one recessive (ttY_, short yellow)
- 1 with both recessive (ttyy, short green)
The classic 9:3:3:1 ratio.
The mathematical shortcut: probability of being tall (3/4) times probability of being yellow (3/4) = 9/16 tall yellow, and so on. Use the product rule when the genes are independent and the sum rule when combining mutually exclusive outcomes.
The test cross
When an organism shows the dominant phenotype, its genotype could be either homozygous dominant (TT) or heterozygous (Tt). A test cross distinguishes them.
Method. Cross the unknown with a homozygous recessive individual (tt). The recessive partner contributes only t gametes, so any t in the offspring must come from the unknown parent.
Possible outcomes.
- If the unknown is TT: all offspring are Tt, and all show the dominant phenotype.
- If the unknown is Tt: half the offspring are Tt (dominant) and half are tt (recessive). A roughly 1:1 phenotype ratio.
A single recessive offspring is enough to conclude the unknown is heterozygous. The more offspring observed, the more confident you can be that an apparent "all dominant" result is genuine.
Test crosses are still used in plant and animal breeding to confirm whether a desirable individual is homozygous (will breed true) or heterozygous.
Pedigree and probability work
A Punnett square gives probabilities for each offspring. For two independent events use the product rule (multiply probabilities); for either of two mutually exclusive outcomes use the sum rule (add probabilities). For example, the probability that a TtYy x TtYy cross gives an offspring that is both homozygous recessive at both loci is 1/4 x 1/4 = 1/16. The probability that it shows at least one recessive trait is 1 minus the probability of showing both dominants = 1 minus 9/16 = 7/16.
Common traps
Not separating alleles into gametes. A TtYy parent produces four gamete types (TY, Ty, tY, ty), not the four alleles directly. Gametes are haploid.
Mixing genotype and phenotype ratios. Tt x Tt gives a 1:2:1 genotype ratio but a 3:1 phenotype ratio.
Forgetting independent assortment requires unlinked genes. Genes on the same chromosome do not always assort independently; they may be linked and produce ratios that deviate from 9:3:3:1.
Choosing the wrong test cross partner. A test cross uses a homozygous recessive, not another heterozygote.
Treating small offspring numbers as exact ratios. Real offspring numbers vary around the predicted ratio. A 3:1 ratio in 4 offspring might appear as 4:0 or 2:2 by chance.
In one sentence
Mendel's laws of segregation (one allele per gamete) and independent assortment (different genes assort independently into gametes) produce 3:1 phenotypic ratios in monohybrid heterozygote crosses and 9:3:3:1 in dihybrid heterozygote crosses, with Punnett squares predicting the probabilities and a cross to a homozygous recessive partner (test cross) revealing whether an unknown dominant phenotype is homozygous (all dominant offspring) or heterozygous (roughly 1:1 dominant to recessive offspring).
Past exam questions, worked
Real questions from past QCAA papers on this dot point, with our answer explainer.
2024 QCAA5 marksIn pea plants, tall (T) is dominant to short (t) and yellow seeds (Y) are dominant to green (y). Two plants heterozygous for both traits are crossed. Construct the Punnett square and state the expected phenotypic ratio of the offspring. Explain the underlying laws.Show worked answer →
A 5-mark answer needs the cross, the 16-cell square, the 9:3:3:1 ratio and both laws.
Parents. TtYy x TtYy.
Gametes. Each parent produces TY, Ty, tY, ty in equal proportions (4 gamete types per parent because of independent assortment).
Punnett square (4 by 4). Cross every gamete from one parent with every gamete from the other to fill 16 cells.
Genotypic outcome (counting).
- 1 TTYY
- 2 TTYy
- 1 TTyy
- 2 TtYY
- 4 TtYy
- 2 Ttyy
- 1 ttYY
- 2 ttYy
- 1 ttyy
Phenotypic ratio.
- 9 tall yellow (T_Y_)
- 3 tall green (T_yy)
- 3 short yellow (ttY_)
- 1 short green (ttyy)
So 9 to 3 to 3 to 1.
Underlying laws.
- Law of segregation. Each parent has two alleles for each gene; only one passes to each gamete (paired alleles separate during meiosis I).
- Law of independent assortment. Alleles for different genes assort independently into gametes (provided the genes are on different chromosomes or far apart on the same chromosome). This is why the dihybrid ratio is the product of two independent monohybrid ratios (3:1 x 3:1 = 9:3:3:1).
Markers reward the correct gametes, the 16-cell working, the 9:3:3:1 ratio and both laws by name.
2022 QCAA3 marksA farmer has a black bull (black is dominant to red) but does not know whether it is homozygous or heterozygous. Describe the test cross you would carry out and explain how the offspring phenotypes would reveal the bull's genotype.Show worked answer →
A 3-mark answer needs the cross, the two possible outcomes and the conclusion logic.
The test cross. Cross the black bull (unknown genotype BB or Bb) with several homozygous recessive red cows (genotype bb).
Possible outcomes.
- If the bull is BB: all gametes from the bull are B. Crossed with b from the cows, every calf is Bb (heterozygous black). Phenotype: all black.
- If the bull is Bb: half the bull's gametes are B and half are b. Half the calves are Bb (black) and half are bb (red). Phenotype: roughly 1:1 black to red.
Conclusion logic. Even one red calf is enough to show the bull must carry a b allele and is therefore heterozygous. If a large number of calves are produced and all are black, the bull is almost certainly homozygous dominant.
Why a homozygous recessive partner is used. Their gametes only carry recessive alleles, so any recessive phenotype in the offspring must come from the unknown parent. This makes the bull's genotype directly readable from the offspring.
Markers reward the explicit choice of a homozygous recessive partner, the predicted ratios for each possible bull genotype and the reasoning that a single recessive offspring confirms heterozygosity.
Related dot points
- Describe and apply non-Mendelian patterns of inheritance including codominance, incomplete dominance, multiple alleles, sex linkage and polygenic inheritance
A focused answer to the QCE Biology Unit 4 dot point on non-Mendelian inheritance. Walks through codominance (ABO blood groups, roan cattle), incomplete dominance (snapdragon flower colour), multiple alleles (ABO, coat colour), X-linked inheritance (haemophilia, colour blindness, Punnett squares with sex chromosomes), and polygenic inheritance (skin colour, height) with continuous variation.
- Interpret pedigrees to deduce patterns of inheritance (autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive) and calculate the probability of specified offspring genotypes and phenotypes
A focused answer to the QCE Biology Unit 4 dot point on pedigree analysis. Explains pedigree symbols, generation and individual numbering, the four inheritance patterns and the signature clues for each (skipped generations, sex bias, affected fathers and daughters), and works through probability calculations using the product and sum rules for combined events.
- Describe types of mutation (point, frameshift, chromosomal) and the sources of genetic variation including meiosis, fertilisation and mutation, and explain the consequences of mutations for phenotype and population polymorphism
A focused answer to the QCE Biology Unit 4 dot point on mutations and variation. Covers point mutations (silent, missense, nonsense), frameshift indels, chromosomal mutations (deletion, duplication, inversion, translocation, non-disjunction) and the three sources of variation (independent assortment, crossing over, random fertilisation) plus mutation as the ultimate source.