BiologyICSE Class 10 · Genetics

Dihybrid Cross and the Law of Independent Assortment (ICSE Class 10 Biology)

Updated 15 October 2026 · Fact-checked

A dihybrid cross follows two pairs of contrasting characters together. Crossing pure round yellow (RRYY) with wrinkled green (rryy) peas gives all round yellow F1 (RrYy). Selfing F1 gives F2 phenotypes 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green, showing the law of independent assortment.

Understand Dihybrid Cross and Law of Independent Assortment

Two characters at once. After studying one character at a time, Mendel crossed plants differing in two characters to see whether they are inherited together or separately. The ICSE syllabus asks for the dihybrid cross: only phenotype (no genotypic ratio), and the law of independent assortment.

Mendel's dihybrid cross (seed shape and seed colour). Round (R) is dominant over wrinkled (r); yellow (Y) is dominant over green (y).
1. P: pure round yellow RRYY × pure wrinkled green rryy.
2. Gametes: RY (from the first) and ry (from the second).
3. F1: all RrYy, all round yellow (both dominant characters show).
4. F1 gametes: each F1 plant forms four kinds of gametes in equal numbers: RY, Ry, rY, ry. Each gamete carries one allele of each pair, and any shape allele can go with any colour allele.
5. F2: a 4 × 4 Punnett square of 16 equally likely combinations.

F2 phenotypes (out of 16).

Phenotype	Number of boxes	Type
Round yellow	9	parental
Round green	3	new combination
Wrinkled yellow	3	new combination
Wrinkled green	1	parental

Phenotypic ratio 9 : 3 : 3 : 1. Two new combinations (round green and wrinkled yellow), present in neither grandparent, appear in 6 of every 16 plants.

Each character still follows 3 : 1. Count seed shape alone: round 9 + 3 = 12, wrinkled 3 + 1 = 4, so 12 : 4 = 3 : 1. Seed colour alone: yellow 12 : green 4 = 3 : 1. The 9 : 3 : 3 : 1 is simply (3 : 1) × (3 : 1): 3/4 × 3/4 = 9/16 both dominant; 3/4 × 1/4 = 3/16 for each mixed class; 1/4 × 1/4 = 1/16 both recessive.

Law of independent assortment. When two (or more) pairs of characters are considered together, the alleles of one pair separate independently of the alleles of the other pair during gamete formation. Because of this, all four gamete types form in equal numbers and new combinations appear in F2. (The law holds for genes on different chromosome pairs.)

Why it happens (cell basis). The gene for seed shape and the gene for seed colour lie on different pairs of homologous chromosomes. In meiosis each pair lines up and separates independently of the other, so R may go with Y or with y equally often.

Mendel's three laws together.

Law	Shown by	Key idea
Dominance	F1 of monohybrid or dihybrid	one allele masks the other
Segregation	F2 of monohybrid	alleles of one pair separate into gametes
Independent assortment	F2 of dihybrid	alleles of different pairs separate independently

Flow chart. P: RRYY (round yellow) × rryy (wrinkled green) → gametes RY, ry → F1: RrYy (round yellow) → selfed → gametes RY, Ry, rY, ry (each 1/4) → F2: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

Draw-and-label checklist: dihybrid cross chart. P line with phenotypes in words and genotypes; circled gametes RY and ry; F1 RrYy (round yellow); '⊗ selfing'; four circled gametes RY, Ry, rY, ry along the top and side of a 4 × 4 grid; shade or mark the boxes by phenotype; a key; the final ratio in words with phenotypes in the order 9 dominant-dominant : 3 dominant-recessive : 3 recessive-dominant : 1 recessive-recessive.

Dihybrid Cross and Law of Independent Assortment: key facts

Dihybrid F2 phenotypes: Round yellow 9
Round green 3
Wrinkled yellow 3
Wrinkled green 1; Out of 16; two new combinations.
Ratio from probabilities: $\frac{9}{16} = \frac{3}{4} \times \frac{3}{4}, \frac{3}{16} = \frac{3}{4} \times \frac{1}{4}, \frac{1}{16} = \frac{1}{4} \times \frac{1}{4}$; Two independent 3 : 1 ratios multiplied.
Gametes of the F1 (RrYy): $R Y ∣ R y ∣ r Y ∣ r y$; Four kinds, in equal numbers (one quarter each).
Law of independent assortment: Statement: alleles of different characters separate independently during gamete formation | Evidence: new combinations in F2, ratio 9 : 3 : 3 : 1; Not the law of segregation.

Round yellow	9
Round green	3
Wrinkled yellow	3
Wrinkled green	1

How to answer Dihybrid Cross and Law of Independent Assortment questions

ICSE gives a dihybrid situation (peas, tomatoes, guinea pigs) and asks for the F1 phenotype, F2 phenotypes and ratio, numbers out of a total, and the law shown.

1Assign letters for both characters and state which is dominant: 'T tall, t dwarf; O round, o oval'.
2Write P genotypes, gametes (one letter of each pair) and F1 genotype and phenotype.
3List the four F1 gametes. Either draw the 16-box Punnett square or use 3/4 × 3/4 reasoning.
4Write the F2 phenotypes with their ratio in the order 9 : 3 : 3 : 1, naming each phenotype in words.
5For numbers, divide the total by 16 and multiply by 9, 3, 3 and 1.
6Name the law: independent assortment (not segregation), and state it fully.

Dihybrid Cross and Law of Independent Assortment in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

New combinations for plant breeders

Suppose a farmer has one pure tomato variety that is tall with round fruits and another that is dwarf with oval fruits. Tall (T) is dominant over dwarf (t) and round fruit (O) over oval (o). She wants a dwarf plant with round fruits, which is easier to harvest.

Can crossing the two varieties give her what she wants? In what fraction of the F2 will it appear?

Try it first, then show the solutionSolution

P: TTOO (tall round) × ttoo (dwarf oval); F1: TtOo, all tall round.
Selfing F1 gives F2 in the ratio 9 tall round : 3 tall oval : 3 dwarf round : 1 dwarf oval.
Dwarf round is a new combination, possible because height and fruit-shape alleles assort independently.
It appears in 3 of every 16 F2 plants.

Answer: Yes: dwarf round-fruited plants appear as a new combination in 3/16 of the F2, thanks to independent assortment.

Memory hook: Independent assortment lets breeders mix and match.

Tricks to remember Dihybrid Cross and Law of Independent Assortment

Short memory aids. Say each one aloud once, then use it on a question the same day.

9-3-3-1 adds to 16
Big-both, mixed, mixed, small-both: 9 both dominant, 3 and 3 one of each, 1 both recessive.
Use it when: Writing the ratio in the right order.
Two coins tossed together
Seed shape and seed colour are like two separate coins; how one lands does not affect the other.
Use it when: Understanding independent assortment.
Di = two = independent
Mono shows segregation; di shows independent assortment.
Use it when: Naming the right law.

Dihybrid Cross and Law of Independent Assortment: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: F1 and F2 phenotypes · 2 marks · Short answerBoard level

A pure round yellow-seeded pea plant is crossed with a wrinkled green-seeded plant. (a) What is the phenotype of the F1? (b) Give the F2 phenotypes and their ratio.

Try it first, then show the solutionSolution

(a) All round yellow (½ mark).
(b) Round yellow, round green, wrinkled yellow, wrinkled green in the ratio 9 : 3 : 3 : 1 (1½ marks).

Answer: (a) Round yellow. (b) 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

Example 2: The law and its statement · 2 marks · Short answerBoard level

Name and state the law of inheritance that can be explained only by a dihybrid cross. What is the evidence for it in the F2?

Try it first, then show the solutionSolution

Law of independent assortment: when two pairs of characters are inherited together, the alleles of each pair separate independently of the other pair during gamete formation (1 mark).
Evidence: new combinations (round green, wrinkled yellow) appear in F2 in the ratio 9 : 3 : 3 : 1 (1 mark).

Answer: Law of independent assortment; new combinations in a 9 : 3 : 3 : 1 ratio.

Example 3: Numbers out of a total · 3 marks · Short answerHarder board level

In a dihybrid cross between pure tall violet-flowered and pure dwarf white-flowered pea plants, 640 F2 plants are obtained. How many would you expect in each phenotypic class?

Try it first, then show the solutionSolution

F2 ratio: 9 tall violet : 3 tall white : 3 dwarf violet : 1 dwarf white (1 mark).
640 ÷ 16 = 40 (1 mark).
Tall violet 9 × 40 = 360; tall white 120; dwarf violet 120; dwarf white 40 (1 mark).

Answer: 360 tall violet, 120 tall white, 120 dwarf violet, 40 dwarf white.

Example 4: Each character alone · 3 marks · Short answerHarder board level

Show that each character in a dihybrid F2 (round/wrinkled, yellow/green) still follows a 3 : 1 ratio. What does this tell you?

Try it first, then show the solutionSolution

Shape: round 9 + 3 = 12; wrinkled 3 + 1 = 4; 12 : 4 = 3 : 1 (1 mark).
Colour: yellow 9 + 3 = 12; green 3 + 1 = 4; 3 : 1 (1 mark).
Each pair segregates as in a monohybrid cross, unaffected by the other pair: the two characters are inherited independently (1 mark).

Answer: Both give 12 : 4 = 3 : 1, so each pair segregates normally and independently.

Example 5: Pod breeding case · 3 marks · Case-basedHOTS / competency

Suppose a breeder crosses a pure pea plant with inflated green pods (IIGG) with a pure plant with constricted yellow pods (iigg). (a) Give the F1 genotype and phenotype, and justify why the F1 is described as a hybrid. (b) Give the F2 phenotypic ratio in the correct order. (c) What fraction of the F2 will be the new combinations, and which law explains them? (The 2025 paper set a dihybrid question of this type.)

Try it first, then show the solutionSolution

(a) IiGg, inflated green pods; it is a hybrid because it is heterozygous for both pairs, having received one dominant and one recessive allele of each from different parents (1 mark).
(b) 9 inflated green : 3 inflated yellow : 3 constricted green : 1 constricted yellow (1 mark).
(c) Inflated yellow and constricted green: (3 + 3)/16 = 6/16 = 3/8; law of independent assortment (1 mark).

Answer: (a) IiGg, inflated green, heterozygous for both. (b) 9 : 3 : 3 : 1 as above. (c) 3/8; independent assortment.

Example 6: A dihybrid test cross · 3 marks · Short answerHOTS / competency

Suppose an F1 round yellow plant (RrYy) is crossed with a wrinkled green plant (rryy). Predict the phenotypes and ratio of the offspring. How does this result support independent assortment?

Try it first, then show the solutionSolution

RrYy forms RY, Ry, rY, ry equally; rryy forms only ry (1 mark).
Offspring: round yellow, round green, wrinkled yellow, wrinkled green in the ratio 1 : 1 : 1 : 1 (1 mark).
Equal numbers of the four gamete types from the F1 are seen directly, which is what independent assortment predicts (1 mark).

Answer: 1 round yellow : 1 round green : 1 wrinkled yellow : 1 wrinkled green; four equal gamete types show independent assortment.

Where students lose marks in Dihybrid Cross and Law of Independent Assortment

The slips examiners see most often, and what to write instead.

Mistake: Wrong ratio, wrong order or wrong law
CISCE examiners' comments (2025): in a dihybrid tomato question, candidates wrote only 'tall red' without justifying the hybrid nature, gave a wrong phenotypic ratio, reversed the order or gave the monohybrid ratio, and stated the law of segregation or dominance, or paraphrased the law inaccurately, for independent assortment.
Wrong working
F2: 3 tall round : 1 dwarf oval; law of segregation.
Right working
F2: 9 tall round : 3 tall oval : 3 dwarf round : 1 dwarf oval; law of independent assortment.
Do this: Write all four phenotypes in the order both dominant, dominant–recessive, recessive–dominant, both recessive.
Usually costs: 1-3 marks.
Mistake: Giving a dihybrid genotypic ratio
The syllabus requires only the phenotype for a dihybrid cross; a long genotypic working wastes time and invites errors.
Do this: Give phenotypes and the 9 : 3 : 3 : 1 ratio unless genotypes are clearly asked.
Usually costs: Time.
Mistake: Wrong gametes from the F1
Students write gametes like 'Rr' or 'YY', which contain two alleles of the same gene.
Do this: Each gamete has one allele of every gene: RY, Ry, rY, ry.
Usually costs: 1 mark.

Exam tips

State 'phenotypic ratio' and name each class in words.
Justify a hybrid: 'heterozygous for both characters'.
Quote the law in full: alleles of different characters, independently, during gamete formation.
For numbers, show 'total ÷ 16' as a working line.

Practice questions on Dihybrid Cross and Law of Independent Assortment

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The F2 phenotypic ratio of a dihybrid cross is:
(a) 3 : 1 (b) 1 : 2 : 1 (c) 9 : 3 : 3 : 1 (d) 1 : 1 : 1 : 1
Hint
Adds to 16.
Answer
(c) 9 : 3 : 3 : 1.
Q2 · 1 mark · Very short answerBoard level
How many kinds of gametes does a plant of genotype RrYy produce? Name them.
Hint
One allele from each pair.
Answer
Four: RY, Ry, rY, ry.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): New combinations of characters appear in the F2 of a dihybrid cross.
Reason (R): The alleles of the two characters separate independently during gamete formation.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Independent assortment.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
In a dihybrid cross of pure round yellow and wrinkled green peas, 1600 F2 seeds are obtained. How many seeds are expected to be (a) wrinkled green, (b) round green?
Hint
1600 ÷ 16.
Answer
1600 ÷ 16 = 100. (a) 1 × 100 = 100. (b) 3 × 100 = 300.
Q5 · 2 marks · Short answerHarder board level
Which two phenotypes in the F2 of Mendel's dihybrid cross are new combinations? What fraction of the F2 do they make up together?
Hint
Not seen in P.
Answer
Round green and wrinkled yellow; 3/16 + 3/16 = 6/16 = 3/8.
Q6 · 2 marks · Short answerHarder board level
Distinguish between the law of segregation and the law of independent assortment.
Hint
One pair versus two pairs.
Answer
Segregation: the two alleles of one character separate so each gamete gets only one of them (seen in a monohybrid cross). Independent assortment: alleles of different characters separate independently of each other, giving all combinations (seen in a dihybrid cross).
Q7 · 2 marks · Short answerHOTS / competency
Suppose in guinea pigs black fur (B) is dominant to white (b) and rough coat (R) to smooth (r). A pure black rough guinea pig is crossed with a white smooth one, and the F1 are interbred to give 160 young. How many young are expected to be white and rough?
Hint
One dominant, one recessive class.
Answer
White rough is a 3/16 class: 160 ÷ 16 × 3 = 30.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a plant RRYy (round seeds; yellow, heterozygous) is crossed with a wrinkled green plant (rryy).
(a) What gametes does each parent make? (1)
(b) What phenotypes appear in the offspring, and in what ratio? (1)
(c) Why are no wrinkled seeds produced? (1)
Hint
RR can give only R.
Answer
(a) RRYy: RY and Ry; rryy: ry. (b) RrYy round yellow and Rryy round green, 1 : 1. (c) Every offspring receives R from the RR parent, and round is dominant.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the F1 phenotype of RRYY × rryy?Show answer
Round yellow.
2.How many boxes are there in a dihybrid Punnett square?Show answer
16.
3.What fraction of a dihybrid F2 shows both recessive characters?Show answer
1/16.
4.Out of 320 F2 plants, how many are expected to show both dominant characters?Show answer
180.
5.Which law is illustrated only by a dihybrid cross?Show answer
The law of independent assortment.

Dihybrid Cross and Law of Independent Assortment: frequently asked questions

What is a dihybrid cross?

A cross between parents that differ in two pairs of contrasting characters, such as seed shape and seed colour, followed to the F2.

Why is the dihybrid ratio 9 : 3 : 3 : 1?

Because each character gives 3 : 1 independently, and multiplying (3 : 1) × (3 : 1) gives 9 : 3 : 3 : 1 over 16 combinations.

Do I need the genotypic ratio of a dihybrid cross for ICSE?

No. The 2027 syllabus asks for the dihybrid cross with phenotype only.

What is the law of independent assortment?

When two pairs of characters are inherited together, the alleles of each pair separate independently of the other pair during gamete formation.