Answer key: Punnett Square Activity

For teachers. This page is hidden from search engines and not linked from student pages, but it is not password-protected, so please don’t share the link with students. Student activity: Punnett Square Activity.

Goal: Predict genotype and phenotype ratios for monohybrid, dihybrid and blood-type crosses, and explain why a Punnett square gives probabilities rather than guarantees.

Connects to NGSS MS-LS3-2 (Model why asexual reproduction gives genetically identical offspring and sexual reproduction gives genetic variation.); HS-LS3-3 (Apply concepts of statistics and probability to explain the variation and distribution of expressed traits in a population.)

Completed results table

Every value is the tool’s own readout for that setting, calculated with the same code as the Punnett square calculator.

Answers

1. Two Aa parents have three children who all show the dominant trait. What is the chance that their fourth child shows the recessive trait? Explain.

   25%, the same as for every child. Each fertilization is independent, so earlier children do not change the odds: 1 box out of 4 in the Punnett square is aa.

2. Why does the test cross Aa × aa give a 1 : 1 ratio, and how could a breeder use a test cross to tell an AA parent from an Aa parent?

   The aa parent makes only a gametes, so each child shows which allele came from the other parent: half dominant, half recessive (1 : 1). A dominant-looking parent that produces any recessive offspring in a test cross must be Aa; an AA parent gives only dominant offspring (AA × aa gives all Aa).

3. For AaBb × AaBb, what fraction of the 16 boxes shows both recessive traits? Write it as a fraction and as a percentage, and give the phenotype ratio.

   Only 1 of the 16 boxes (aabb), so 1/16 = 6.25%. The phenotype ratio is 9 : 3 : 3 : 1.

4. Check your dihybrid answer with the rule of multiplication: what is the chance of aa, of bb, and of both together?

   Each gene on its own is an Aa × Aa cross, so P(aa) = 1/4 and P(bb) = 1/4. Because the genes assort independently, P(aabb) = 1/4 × 1/4 = 1/16 = 6.25%.

5. A type A parent (AO) and a type B parent (BO) have children. Which blood types can the children have, and with what probability?

   All four: A (AO), B (BO), AB and O (OO), each with a 25% chance, a 1 : 1 : 1 : 1 ratio. Neither parent is type O, but both carry the recessive O allele, so a type O child is possible.

6. Challenge: from Aa × Aa crosses Mendel counted 705 purple and 224 white flowers. Is that close to 3 : 1? Check it with the chi-square calculator.

   Yes. Expected counts are 929 × 3/4 = 696.75 purple and 929 × 1/4 = 232.25 white. The chi-square test gives χ² = 0.391, below the critical value 3.841, so the counts fit a 3 : 1 ratio.