Inquiry Question 4: How can the genetic similarities and differences within and between species be compared?
Investigate the inheritance of patterns including but not limited to: predicting genotypic and phenotypic ratios using Punnett squares and probability rules
A focused answer to the HSC Biology Module 5 dot point on Mendelian inheritance. Mendel's laws, dominant vs recessive alleles, Punnett squares step by step, monohybrid and dihybrid crosses, the standard 3:1 and 9:3:3:1 ratios, and worked HSC past exam questions.
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What this dot point is asking
NESA wants you to use Punnett squares and probability rules to predict the genotypic and phenotypic ratios of offspring from a given parental cross. This is a calculation skill, and Punnett squares appear in almost every HSC Biology exam.
The answer
Gregor Mendel's experiments on pea plants in the 1860s established three core laws of inheritance.
- Law of Segregation. Each parent contributes one of two alleles for each gene to its offspring, randomly.
- Law of Independent Assortment. Alleles for different genes segregate independently (assuming they are on different chromosomes).
- Law of Dominance. When two different alleles are present, the dominant allele determines the phenotype; the recessive allele is masked.
Key terminology
- Gene: a section of DNA that codes for a trait.
- Allele: a version of a gene (e.g. A or a).
- Genotype: the alleles an individual has (e.g. AA, Aa, aa).
- Phenotype: the observed trait (e.g. tall, short).
- Homozygous: two identical alleles (AA or aa).
- Heterozygous: two different alleles (Aa).
- Dominant: the allele expressed when heterozygous (capital letter).
- Recessive: the allele masked when heterozygous (lowercase letter).
Setting up a Punnett square
A Punnett square predicts the possible genotypes and phenotypes of offspring from a given parental cross.
Step 1. Identify the parental genotypes.
Step 2. Write the possible gametes from each parent across the top and down the side.
Step 3. Fill in each cell with the combined genotype.
Step 4. Read off the genotypic ratio.
Step 5. Convert to phenotypic ratio using the dominance rules.
Standard monohybrid cross: Aa × Aa
| A | a | |
|---|---|---|
| A | AA | Aa |
| a | Aa | aa |
- Genotypic ratio: 1 AA : 2 Aa : 1 aa
- Phenotypic ratio: 3 dominant : 1 recessive
This is the canonical 3:1 ratio Mendel observed.
Test cross: Aa × aa
| A | a | |
|---|---|---|
| a | Aa | aa |
| a | Aa | aa |
- Genotypic ratio: 1 Aa : 1 aa
- Phenotypic ratio: 1 dominant : 1 recessive
A test cross with a homozygous recessive individual is used to determine whether a dominant-phenotype individual is homozygous (AA) or heterozygous (Aa).
Dihybrid cross: AaBb × AaBb
When tracking two independent genes, set up a 4 × 4 Punnett square with the four possible gametes from each parent (AB, Ab, aB, ab).
The classic phenotypic ratio is 9:3:3:1 (9 dominant for both : 3 dominant for A only : 3 dominant for B only : 1 recessive for both).
Probability rules
When tracking multiple events:
- Multiplication rule (independent events): P(A AND B) = P(A) × P(B). E.g. probability that two consecutive children are both affected = 1/4 × 1/4 = 1/16.
- Addition rule (mutually exclusive events): P(A OR B) = P(A) + P(B). E.g. probability that a child is either homozygous dominant OR heterozygous = 1/4 + 1/2 = 3/4.
Common Punnett-square traps
- Wrong gametes
- Each parent contributes only ONE allele per gene to each gamete. A heterozygous parent (Aa) produces TWO types of gametes (A and a), not four (AA, Aa, Aa, aa).
- Failing to simplify ratios
- 2:2 is the same as 1:1. 4:2 is the same as 2:1. Markers reward simplified ratios.
- Confusing genotype and phenotype
- Genotype = letters (AA, Aa, aa). Phenotype = trait (purple, white).
- Independent probability for separate children
- Each child is an independent event. The "probability the next child is affected" does not depend on the previous children. 1/4 each time.
- Capitalisation matters
- A means dominant, a means recessive. Be consistent. Markers often use unfamiliar letters (e.g. R/r for red, h/H for hairy); follow the question's convention.
Examples in context
Example 1. Coat colour in NSW DPI Angus cattle. In Angus cattle, the allele for black coat (B) is dominant over the allele for red coat (b). A NSW Department of Primary Industries breeder crosses a heterozygous black bull (Bb) with a herd of red cows (bb). The Punnett square predicts 50 percent Bb (black) and 50 percent bb (red) calves. Over a calving season of 200 calves, the breeder records 96 black and 104 red, close to the expected 100:100 ratio with deviation explained by chance. This test cross also tells the breeder the bull is heterozygous, not homozygous BB, which is information used to plan future matings for breed registration.
Example 2. Pea seed shape and colour, the original Mendel dihybrid cross. Mendel crossed pure-breeding round yellow peas (RRYY) with pure-breeding wrinkled green peas (rryy) to produce a uniform F1 generation of RrYy (round yellow). When he self-pollinated the F1 to produce F2, a 4 by 4 Punnett square of the 16 possible gamete combinations predicted a 9:3:3:1 phenotypic ratio: 9 round yellow, 3 round green, 3 wrinkled yellow, 1 wrinkled green. Mendel counted 556 F2 seeds and recorded 315:108:101:32, almost exactly the predicted ratio. This was the empirical basis for the Law of Independent Assortment.
Try this
Q1. In labrador retrievers, black coat (B) is dominant over chocolate coat (b). A breeder crosses two heterozygous black labradors. From a litter of eight puppies, predict the most likely number of chocolate puppies and identify the relevant probability rule. [3 marks]
- Cue. Bb x Bb gives 3:1 black to chocolate, so the expected number of chocolate puppies is one quarter of eight, equal to two, with chance variation expected around this mean.
Q2. A heterozygous individual (Aa) for a recessive disease allele has three children with another heterozygote. Calculate the probability that (a) exactly one child is affected, and (b) at least one child is affected. [2+2 marks]
- Cue. (a) Use the binomial form: three ways to choose the affected child times (1/4)(3/4)(3/4). (b) Use the complement: 1 minus the probability that none are affected, which is 1 minus (3/4) cubed.
Q3. In a dihybrid cross between two AaBb pea plants (round yellow), a student observes the following F2 phenotypes among 320 seeds: 175 round yellow, 65 round green, 60 wrinkled yellow, 20 wrinkled green. (a) State the expected ratio under Mendelian inheritance. (b) Calculate the expected numbers and compare them with the observed values. (c) Suggest a reason for any deviation. [1+3+1 marks]
- Cue. (a) 9:3:3:1. (b) Expected: 180, 60, 60, 20; observed is close. (c) Chance variation in a finite sample, or possibly linkage if very different.
Exam-style practice questions
Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 HSC3 marksTwo healthy parents, heterozygous for cystic fibrosis, have a child that does not have cystic fibrosis. They are planning to have a second child. Using a Punnett square, determine the probability of their second child being born with the condition. Use 'R' for the normal CFTR allele, and 'r' for the faulty CFTR allele. [Cystic fibrosis is recessive; affected individuals have two faulty alleles.]Show worked answer →
Full marks (3) require the correct parental genotypes with a suitable Punnett square AND the correct probability. 2 marks for the correct probability with some working, or correct genotypes with working; 1 for some relevant information.
Sample answer (marking guidelines): Cross Rr x Rr:
| R | r | |
|---|---|---|
| R | RR | Rr |
| r | Rr | rr |
There is a 25% chance of the second child having cystic fibrosis (one rr in four outcomes).
Note the second child is an independent event, so the unaffected first child does not change the 1 in 4 probability. Markers flagged inconsistent probability formats (e.g. writing both 1/4 and 33%).
2021 HSC3 marksIn a population of rabbits, black fur colour is dominant over white fur. A black rabbit, whose mother has white fur, mates with a white rabbit. Predict the phenotypic ratio for the offspring of this cross. Show your working.Show worked answer →
3 marks for the correct phenotypic ratio with parental genotypes and suitable working; 2 for the ratio with some working (or correct genotypes with working); 1 for some relevant information.
Sample answer (marking guidelines): Because the black rabbit's mother had white fur (bb), the black rabbit must have inherited a recessive b allele, so it is heterozygous (Bb). Cross Bb x bb:
| b | b | |
|---|---|---|
| B | Bb | Bb |
| b | bb | bb |
Phenotypic ratio Black : White = 1 : 1.
Markers stressed deducing that the black rabbit is heterozygous (using the white-furred mother) and presenting a phenotypic, not genotypic, ratio.
2020 HSC2 marksUse the pedigree chart to explain why the yellow allele is recessive. [A pedigree shows the inheritance of yellow vs orange colour in a fish; two orange parents have yellow offspring.]Show worked answer →
2 marks for using the pedigree to explain that the yellow allele is recessive; 1 mark for some relevant information.
Sample answer (marking guidelines): The inheritance of yellow colour is recessive since both parents are orange but have yellow offspring. The yellow allele must be present in both parents but it is not expressed.
The key reasoning is that a trait appearing in offspring but not in either parent must be recessive (both parents are heterozygous carriers). Markers noted weak terminology and misreading of pedigree relationships.
2019 HSC2 marksComplete the tables, showing the TWO alleles the patient inherited from each parent. [The patient is heterozygous for Huntington's (Hh) and Stargardt disease (Rr); his father's family has cases of both, and his mother is homozygous unaffected for both genes (Huntington's H = dominant disease allele; Stargardt R = dominant healthy allele).]Show worked answer →
2 marks for identifying suitable alleles for both parents; 1 mark for some relevant information.
Sample answer (marking guidelines): Alleles from father = H, r; Alleles from mother = h, R.
Reasoning: Huntington's is autosomal dominant, so the disease allele H came from the affected father's side; the mother is homozygous unaffected (hh), contributing h. Stargardt is autosomal recessive, so the patient (Rr) must have received the recessive disease allele r from the father and the dominant healthy allele R from the homozygous-unaffected mother. Markers advised distinguishing allele from genotype and using all the information in the stem.
Related dot points
- Model the processes involved in cell replication, including but not limited to: mitosis and meiosis, the role of meiosis and gamete formation in maintaining the chromosome number across generations
A focused answer to the HSC Biology Module 5 dot point on meiosis. The two divisions, crossing over and independent assortment as sources of genetic variation, comparison with mitosis, and how gamete formation maintains chromosome number across generations.
- Investigate the inheritance patterns including but not limited to: sex-linkage, codominance, incomplete dominance, multiple alleles
A focused answer to the HSC Biology Module 5 dot point on sex-linked (X-linked) inheritance. Why X-linked recessive disorders affect males more than females, the standard worked Punnett squares for carrier mothers, named examples (haemophilia, colour blindness, Duchenne muscular dystrophy), and worked HSC past exam questions.