When two genes are tracked simultaneously, the 4x4 Punnett square reveals four phenotypic classes in a 9:3:3:1 ratio — a direct consequence of Mendel's second law, independent assortment.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A student crosses two pea plants that are each heterozygous for two genes: seed colour (Yy) and seed shape (Rr). The student predicts there will be two phenotypic classes among the offspring.
Before reading on, explain whether you think the student is correct. How many phenotypic classes would you predict, and why?
Monohybrid crosses track one gene. A dihybrid cross tracks two genes at once, asking: what combination of phenotypes appears in offspring, and in what proportions?
The classic setup uses two true-breeding lines: AABB (homozygous dominant for both traits) crossed with aabb (homozygous recessive for both traits). This cross is called the P generation.
AABB x aabb — all gametes from the first parent are AB; all from the second are ab.
All offspring are AaBb — heterozygous for both genes. All show the dominant phenotype for both traits.
AaBb x AaBb — each parent produces four gamete types: AB, Ab, aB, ab.
AaBb parents are crossed, each parent can produce four different gamete types because each gene independently contributes one allele. This is what makes the 4x4 Punnett square necessary.Because each AaBb parent produces four gamete types, the Punnett square has 4 columns and 4 rows — 16 boxes total.
The gametes from each parent are: AB, Ab, aB, ab. Each combination in the grid gives a two-gene genotype.
| AB | Ab | aB | ab | |
|---|---|---|---|---|
| AB | AABB | AABb | AaBB | AaBb |
| Ab | AABb | AAbb | AaBb | Aabb |
| aB | AaBB | AaBb | aaBB | aaBb |
| ab | AaBb | Aabb | aaBb | aabb |
Grouping by phenotype (using A_ for any genotype with at least one A, and B_ for any with at least one B):
At least one dominant allele for each gene. Shows both dominant phenotypes.
Dominant phenotype for gene A only; homozygous recessive for gene B.
Dominant phenotype for gene B only; homozygous recessive for gene A.
Homozygous recessive for both genes. Shows both recessive phenotypes.
Phenotypic ratio: 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb = 9:3:3:1
Mendel's law of independent assortment states that alleles of different genes segregate independently of each other during gamete formation.
The cellular mechanism is the random orientation of bivalents (paired homologous chromosomes) on the metaphase plate during meiosis I. When two gene pairs are on different chromosomes, the way one pair aligns has no effect on how the other pair aligns.
Mendel's original dihybrid experiment crossed pea plants differing in seed colour (yellow Y vs green y) and seed shape (round R vs wrinkled r).
Yellow round YYRR x green wrinkled yyrr
All offspring are YyRr — all yellow and round (both dominant phenotypes)
YyRr x YyRr — produces four gamete types: YR, Yr, yR, yr
9 yellow round : 3 yellow wrinkled : 3 green round : 1 green wrinkled
In Mendel's actual experiment with 556 F2 seeds, he observed 315 yellow round, 101 yellow wrinkled, 108 green round, and 32 green wrinkled — very close to the expected 9:3:3:1 ratio.
Mendel's law of independent assortment holds only for genes on different chromosomes. Genes on the same chromosome are said to be linked.
Linked genes tend to be inherited together because they travel on the same chromosome during meiosis. They do not assort independently, so the four gamete types from a dihybrid parent are not equally frequent. Instead, parental combinations (the original allele pairings) appear more often than recombinant combinations.
At HSC level, you are expected to know that linked genes exist and that they explain deviations from 9:3:3:1, but detailed recombination frequency calculations are not required.
Tracks two genes simultaneously. AaBb x AaBb produces four gamete types (AB, Ab, aB, ab) and requires a 4x4 Punnett square with 16 boxes.
The expected phenotypic ratio from AaBb x AaBb: 9 with both dominant phenotypes, 3 with only dominant A, 3 with only dominant B, 1 with both recessive phenotypes.
Mendel's second law. Genes on different chromosomes segregate independently due to random bivalent orientation in meiosis I.
Genes on the same chromosome do not assort independently and produce ratios that deviate from 9:3:3:1.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Activities
Activities
1. Two pea plants YyRr are crossed. What fraction of the offspring would be expected to be yellow AND wrinkled (Y_rr)?
2. From an AaBb x AaBb cross, how many phenotypically distinct classes are produced (assuming simple dominance for each gene)?
3. Which genotype ratio arises from an AaBb x AaBb cross?
4. The 9:3:3:1 phenotypic ratio from a dihybrid cross assumes which of the following?
5. A dihybrid cross produces offspring in a 9:3:3:1 ratio. If 160 offspring are produced, how many would be expected to show both recessive phenotypes (aabb)?
6. A student crosses two plants that are both PpTt (heterozygous for pod colour P and plant height T). Construct a complete 4x4 Punnett square for this cross and determine the expected phenotypic ratios among the offspring.
5 marks
7. Explain how Mendel's law of independent assortment is supported by the 9:3:3:1 phenotypic ratio, and identify the cellular mechanism from meiosis that produces independent assortment.
4 marks
Result of AaBb x AaBb when genes assort independently. The ratio is 9 both dominant : 3 A only : 3 B only : 1 both recessive.
An AaBb parent produces AB, Ab, aB, and ab gametes in equal frequency because each gene contributes one allele independently.
Random orientation of bivalents on the metaphase plate in meiosis I produces independent assortment for genes on different chromosomes.
Genes on the same chromosome do not assort independently and will produce ratios that deviate from 9:3:3:1.
Return to your Think First response. How many phenotypic classes did you predict? Rewrite a corrected, more complete answer using precise biological language.
1. B — Y_rr represents 3/16 of offspring from YyRr x YyRr (3/4 chance of Y_ multiplied by 1/4 chance of rr = 3/16).
2. C — Four phenotypic classes: both dominant, A dominant only, B dominant only, both recessive.
3. C — The 9 genotype classes from AaBb x AaBb in full ratio form are 1:2:2:4:1:2:1:2:1 (AABB:AABb:AaBB:AaBb:AAbb:Aabb:aaBB:aaBb:aabb).
4. C — The 9:3:3:1 ratio requires that the two genes assort independently (are on different chromosomes).
5. B — aabb is 1/16 of offspring. 1/16 x 160 = 10.
Gametes from PpTt: PT, Pt, pT, pt (four gamete types in equal frequency).
The 4x4 Punnett square produces 16 boxes. Phenotypic grouping: P_T_ = 9, P_tt = 3, ppT_ = 3, pptt = 1. Phenotypic ratio = 9 : 3 : 3 : 1.
Mark allocation: 1 mark for correct gametes listed, 2 marks for correctly completed Punnett square (allow one error), 1 mark for identifying four phenotypic classes, 1 mark for correct 9:3:3:1 ratio.
The 9:3:3:1 ratio supports independent assortment because it can only arise if all four gamete types (AB, Ab, aB, ab) are produced with equal frequency from an AaBb parent. If the genes were linked, the parental combinations would appear more frequently than recombinant combinations, and the ratio would deviate from 9:3:3:1.
The cellular mechanism is the random orientation of bivalents (paired homologous chromosomes) on the metaphase plate during meiosis I. When homologue pairs for two different genes are on different chromosome pairs, the way each pair aligns is independent of the other. This random alignment means that any combination of maternal and paternal alleles from the two genes can end up in the same gamete.
Tick this once you can construct a 4x4 Punnett square, predict the 9:3:3:1 ratio, and explain the meiotic mechanism of independent assortment.
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