SBI3U Genetics Practice Questions: Punnett Squares and Explained Answers

SBI3U genetics questions become easier when you follow the same sequence: define the alleles, identify the parents’ genotypes, list their possible gametes, and then calculate the offspring probabilities. The most common mistake is starting a Punnett square before deciding what each letter means.

This original practice set helps Grade 11 biology students review inheritance. It is not an official Ontario test or a complete review of SBI3U. The examples use stated, simplified inheritance models; real traits can involve more genes and environmental influences.

Before you begin: five terms to know

• Allele: a version of a gene.
• Genotype: the allele combination being considered, such as Aa.
• Phenotype: the observable characteristic associated with that genotype in the model.
• Homozygous: having two identical alleles, such as AA or aa.
• Heterozygous: having two different alleles, such as Aa.

Under complete dominance, AA and Aa have the dominant phenotype. Dominant does not mean more common, stronger or better. A gamete receives one allele from each gene pair.

How to use this practice set

Try the questions before reading the solutions. For every cross, write the parent genotypes and possible gametes. Give both the fraction and percentage when asked for a probability. If a question asks for a phenotype, do not stop at a genotype ratio.

Twelve practice questions

• 1. In a model pea plant, purple flowers (P) are dominant to white flowers (p). Identify the genotypes that produce purple flowers and the genotype that produces white flowers.
• 2. Cross two heterozygous plants, Pp × Pp. Find the genotype ratio and the probability of white flowers.
• 3. Cross Pp × pp. What proportion of offspring should have purple flowers?
• 4. A purple-flowered plant has an unknown genotype. Which plant should it be crossed with for a test cross? What does one white-flowered offspring establish about the unknown parent?
• 5. Two Pp parents produce four offspring. Does the expected 3:1 phenotype ratio guarantee three purple and one white offspring? Explain.
• 6. For Pp × Pp, what is the probability that two specified offspring both have white flowers? Assume the outcomes are independent.
• 7. In a flower model with incomplete dominance, RR is red, WW is white and RW is pink. Cross two pink plants. State the phenotype probabilities.
• 8. In a codominance model, CᴿCᴿ cattle have red coats, CᵂCᵂ have white coats, and CᴿCᵂ have roan coats with both red and white hairs. Cross a roan animal with a white animal.
• 9. Assume genes A and B assort independently. What gamete types can an AaBb parent produce, and with what probabilities?
• 10. Under independent assortment and complete dominance at both genes, cross AaBb × aabb. Find the probability of offspring with both dominant phenotypes.
• 11. In a simplified X-linked recessive model, a mother is XᴺXⁿ and a father is XᴺY. What fraction of sons inherit the recessive phenotype? What fraction of all offspring inherit it, assuming equal probabilities of sons and daughters?
• 12. In a simplified autosomal recessive model, two unaffected parents have an affected offspring with genotype aa. What must the parents’ genotypes be? What is the probability that their next offspring is affected?

Worked answers

1. Genotype and phenotype

PP and Pp produce purple flowers. Only pp produces white flowers. The capital letter represents the dominant allele in this model, not the frequency of that allele in a population.

2. A heterozygous cross

Each Pp parent produces P and p gametes, each with probability 1/2. The four equally likely combinations are PP, Pp, pP and pp. Pp and pP represent the same genotype.

The genotype ratio is 1 PP : 2 Pp : 1 pp. The phenotype ratio is 3 purple : 1 white, so the probability of white flowers is 1/4 = 25%.

3. Crossing with a recessive parent

The Pp parent produces P or p. The pp parent produces only p. Half the offspring are Pp and half are pp. The probability of purple flowers is 1/2 = 50%.

4. A test cross

Cross the unknown purple-flowered plant with a pp plant. A white offspring must receive p from both parents, so the unknown parent must be Pp. If all observed offspring are purple, that alone does not prove the unknown parent is PP: a small sample from Pp × pp can also contain only purple offspring.

5. Expected ratios are not guarantees

No. Each offspring has a 3/4 probability of purple flowers and a 1/4 probability of white flowers. A group of four may have different counts. Expected ratios describe probabilities and become more apparent over many offspring; they do not dictate the outcome of each small group.

6. Two independent outcomes

Multiply the probabilities: 1/4 × 1/4 = 1/16 = 6.25%. This answer is for two specified offspring both being white. A question about exactly two white offspring in a larger group would require accounting for the different possible arrangements.

7. Incomplete dominance

RW × RW produces RR, RW, WR and WW. The probabilities are 25% red, 50% pink and 25% white. Here, the heterozygous phenotype is intermediate; do not use the complete-dominance 3:1 phenotype ratio.

8. Codominance

The roan parent supplies Cᴿ or Cᵂ; the white parent supplies only Cᵂ. Offspring are 50% CᴿCᵂ, roan, and 50% CᵂCᵂ, white. Codominance means both allele-associated characteristics are expressed in the heterozygote; it is different from the intermediate phenotype in Question 7.

9. Gametes from two genes

The possible gametes are AB, Ab, aB and ab, each with probability 1/4 under the stated independent-assortment assumption. A gamete contains one allele for A and one for B; it does not contain Aa or Bb.

10. A two-gene test cross

The aabb parent produces only ab gametes. Combining these with the other parent’s four gamete types gives AaBb, Aabb, aaBb and aabb, each with probability 1/4. Only AaBb has both dominant phenotypes, so the probability is 25%.

11. Identify the group in the question

The four equally likely combinations are XᴺXᴺ, XᴺXⁿ, XᴺY and XⁿY. Among sons, one of the two possibilities has the recessive phenotype: 1/2 = 50% of sons. Among all offspring, it is 1/4 = 25%. The denominator changes because “sons” and “all offspring” describe different groups.

12. Inferring parent genotypes

An aa offspring received a from each parent. Because both parents are unaffected under the stated recessive model, each must be Aa. Aa × Aa gives a 25% probability of aa for the next offspring. A previous outcome does not change that probability under the model.

Turn mistakes into a revision plan

If Questions 1–4 were difficult, practise allele notation and single-gene crosses. If Questions 5–6 were difficult, review probability and independence. For Questions 7–8, compare the heterozygote phenotypes. For Questions 9–12, write the gametes and identify the group being counted before calculating.

You can also read our SBI3C vs SBI3U course guide or explore Grade 11 science tutoring.

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Curriculum reference: Ontario Grades 11–12 Science, SBI3U Genetic Processes. Questions and explanations above are original practice material.