Year 12 Biology Module 5 · IQ4 ⏱ ~40 min Practice bank · 2 Short Answer Lesson 14b of 19

Dihybrid Crosses and Independent Assortment

When two genes are tracked at once, a 4x4 Punnett square sorts 16 equally likely combinations into four phenotypic classes in a 9:3:3:1 ratio, a direct consequence of Mendel's second law of independent assortment.

Today's hook: A student crosses two pea plants, each heterozygous for seed colour and seed shape, and predicts just two kinds of offspring. Mendel counted 556 F2 seeds and found four distinct classes in a 9:3:3:1 ratio. Where do the extra classes come from, and why do they appear in that exact proportion?
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Orient yourself

Make a prediction, scan the goals and the key terms, then set up the question this lesson answers.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Think First
warm-up

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?

Learning Intentions
goals

Know

  • How to set up and complete a 4x4 Punnett square for a dihybrid cross.
  • The expected 9:3:3:1 phenotypic ratio and what each class represents.

Understand

  • Why independent assortment arises from random bivalent orientation in meiosis I.
  • Why linked genes do not produce a 9:3:3:1 ratio.

Be Able To

  • Predict phenotypic and genotypic outcomes from any AaBb x AaBb cross.
  • Calculate expected numbers of offspring from a given ratio and total.
Scan these before reading
vocab
Dihybrid crossA genetic cross that tracks the inheritance of two different genes simultaneously.
Independent assortmentMendel's second law, genes on different chromosomes segregate independently during meiosis.
9:3:3:1 ratioThe expected phenotypic ratio from an AaBb x AaBb cross when both genes assort independently.
Linked genesGenes located on the same chromosome that tend to be inherited together, violating independent assortment.
BivalentA paired homologous chromosome structure formed during meiosis I; its random orientation produces independent assortment.
Phenotypic classA group of offspring sharing the same observable combination of traits.

A dihybrid cross tracks two genes

See why crossing two AaBb parents needs four gamete types from each side.

1
A dihybrid cross tracks two genes simultaneously
+5 XP

Foundation

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.

P generation

AABB x aabb all gametes from the first parent are AB; all from the second are ab.

F1 generation

All offspring are AaBb heterozygous for both genes. All show the dominant phenotype for both traits.

F2 cross

AaBb x AaBb each parent produces four gamete types: AB, Ab, aB, ab.

Key point
When 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.

An AaBb parent produces _____ genetically different gamete types when the two genes assort independently.

The 4x4 Punnett square

Fill 16 boxes, then group them into the four phenotypic classes.

2
The 4x4 Punnett square reveals four phenotypic groups
+5 XP

The grid

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.

ABAbaBab
ABAABBAABbAaBBAaBb
AbAABbAAbbAaBbAabb
aBAaBBAaBbaaBBaaBb
abAaBbAabbaaBbaabb

F2 Punnett square for AaBb x AaBb, 16 equally probable genotype combinations.

Grouping by phenotype (using A_ for any genotype with at least one A, and B_ for any with at least one B):

A_B_ (9 boxes)

At least one dominant allele for each gene. Shows both dominant phenotypes.

A_bb (3 boxes)

Dominant phenotype for gene A only; homozygous recessive for gene B.

aaB_ (3 boxes)

Dominant phenotype for gene B only; homozygous recessive for gene A.

aabb (1 box)

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.

Pause, copy the highlighted ratio into your book before moving on.

How many of the 16 boxes fall into the A_B_ class (both dominant phenotypes)?

Independent assortment

Trace the 9:3:3:1 ratio back to random bivalent orientation in meiosis I.

3
Independent assortment explains the 9:3:3:1 ratio
+5 XP

Mendel's second law

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.

What independent assortment means

  • Each gamete receives one allele from each gene pair independently.
  • All four gamete types (AB, Ab, aB, ab) are equally likely from an AaBb parent.
  • Alleles on different chromosomes do not travel together.

The meiotic mechanism

  • During meiosis I, bivalents line up randomly at the metaphase plate.
  • Either homologue can face either pole for each bivalent.
  • This random alignment generates all possible allele combinations.
HSC link
Independent assortment only applies to genes on different chromosomes. It is the random orientation of bivalents in meiosis I that produces the four equally frequent gamete types from a dihybrid parent.

What is the cellular basis of independent assortment?

Worked example: Mendel's peas

Follow a full dihybrid cross from P generation to the F2 counts Mendel actually recorded.

4
Mendel's pea plant dihybrid cross
+5 XP

Worked example

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).

P generation

Yellow round YYRR x green wrinkled yyrr

F1 generation

All offspring are YyRr all yellow and round (both dominant phenotypes)

F2 cross

YyRr x YyRr produces four gamete types: YR, Yr, yR, yr

F2 result

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.

Exam tip
When writing a worked cross, always state: (1) parent genotypes, (2) gamete types from each parent, (3) the Punnett square or combined probability, (4) genotypic and phenotypic ratios.

Mendel's 556 F2 pea seeds appeared in a phenotypic ratio very close to _____ (write it in the form number:number:number:number).

When 9:3:3:1 breaks: linked genes

See why genes on the same chromosome deviate from the expected ratio.

5
Linked genes violate the 9:3:3:1 ratio
+5 XP

Exception

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.

Warning
If a dihybrid cross does not produce a 9:3:3:1 ratio, the most likely explanation is that the two genes are on the same chromosome (linked). The greater the physical distance between linked genes, the more recombination can occur, increasing the frequency of recombinant gametes.

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.

A dihybrid cross gives parental combinations far more often than 9:3:3:1 predicts. What is the most likely explanation?

Copy into your books

Dihybrid cross

  • Tracks two genes simultaneously. AaBb x AaBb produces four gamete types (AB, Ab, aB, ab) and requires a 4x4 Punnett square with 16 boxes.

9:3:3:1 ratio

  • 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.

Independent assortment

  • Mendel's second law. Genes on different chromosomes segregate independently due to random bivalent orientation in meiosis I.

Linked genes

  • Genes on the same chromosome do not assort independently and produce ratios that deviate from 9:3:3:1.

Consolidate and apply

Count the phenotypic classes, predict offspring numbers, then explore meiosis in the interactive tool.

Activity 1
ApplyBand 4

Count the classes

From the 4x4 Punnett square for AaBb x AaBb, count how many of the 16 boxes fall into each of the four phenotypic classes (A_B_, A_bb, aaB_, aabb). Record your counts and verify they add to 16.

Activity 2
AnalyseBand 5

Predict offspring numbers

A genetics experiment crosses two YyRr pea plants and produces 320 offspring. Using the 9:3:3:1 ratio, calculate how many offspring you would expect to be: (a) yellow round, (b) yellow wrinkled, (c) green round, (d) green wrinkled.

Interactive Tool, Meiosis & Mitosis Open fullscreen ↗
After Meiosis I is complete, each daughter cell contains…
Do this next

Practise independently

Attempt every response in your own words first. The model answers are for checking, not for copying.

01
Multiple Choice
+5 XP

A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

02
Short Answer
+5 XP

ApplyBand 5(5 marks) 1. 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.

AnalyseBand 5(4 marks) 2. 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.

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.

Activity 1, Count the classes

A_B_ = 9 boxes, A_bb = 3 boxes, aaB_ = 3 boxes, aabb = 1 box. Total = 9 + 3 + 3 + 1 = 16, confirming the 9:3:3:1 phenotypic ratio.

Activity 2, Predict offspring numbers

(a) Yellow round = 9/16 x 320 = 180. (b) Yellow wrinkled = 3/16 x 320 = 60. (c) Green round = 3/16 x 320 = 60. (d) Green wrinkled = 1/16 x 320 = 20. Check: 180 + 60 + 60 + 20 = 320.

Short Answer 1

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.

Short Answer 2

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.

Finish strong

Retrieve and reflect

Use the Review session first, then compare how your thinking has changed since the opening prediction.

Check what actually stuck
RAPID REVIEW
The big ideas in four tiles

The 9:3:3:1 ratio

Result of AaBb x AaBb when genes assort independently. The ratio is 9 both dominant : 3 A only : 3 B only : 1 both recessive.

Why four gamete types

An AaBb parent produces AB, Ab, aB, and ab gametes in equal frequency because each gene contributes one allele independently.

Meiotic mechanism

Random orientation of bivalents on the metaphase plate in meiosis I produces independent assortment for genes on different chromosomes.

Linkage breaks the rule

Genes on the same chromosome do not assort independently and will produce ratios that deviate from 9:3:3:1.

Test yourself against the clock
boss

Rapid-fire questions on dihybrid crosses, the 9:3:3:1 ratio and independent assortment. Beat the boss to bank a tier, gold (perfect + fast), silver (80%+), or bronze (cleared).

Revisit your thinking

Return to your Think First response. How many phenotypic classes did you predict? A dihybrid cross of two heterozygotes produces four phenotypic classes in a 9:3:3:1 ratio, not two, because each gene assorts independently and contributes one of two alleles to each gamete. Rewrite a corrected, more complete answer using precise biological language.