ScienceCBSE Class 10 · Heredity

Dihybrid Cross and Mendel's Laws of Inheritance

Updated 14 October 2026 · Fact-checked

A dihybrid cross follows two traits at once. Crossing round-green with wrinkled-yellow peas gives all round-yellow F1 plants. Their F2 shows four kinds of seeds in about 9 : 3 : 3 : 1, including new combinations. This shows that the genes for seed shape and seed colour are inherited independently of each other.

Understand Dihybrid Cross and Laws of Inheritance

From one trait to two. In a monohybrid cross Mendel followed one pair of traits. He then asked what happens when plants differing in two characters are crossed. In NCERT's example the traits are seed shape (round R dominant over wrinkled r) and seed colour (yellow Y dominant over green y).

The cross.

Parents: RRyy (round, green) × rrYY (wrinkled, yellow).
Gametes: Ry from the first parent, rY from the second.
F1: RrYy, all round and yellow. Round and yellow are the dominant traits.
F1 × F1 (self-pollination): each F1 plant makes four kinds of gametes in equal numbers: RY, Ry, rY, ry. Combining 4 × 4 gives 16 combinations in a checkerboard.

F2 results. Counting the 16 boxes gives four kinds of seeds:

round, yellow: 9
round, green: 3
wrinkled, yellow: 3
wrinkled, green: 1

In NCERT's data, 556 F2 seeds were 315 round yellow, 108 round green, 101 wrinkled yellow and 32 wrinkled green, very close to 9 : 3 : 3 : 1 (expected 312.75 : 104.25 : 104.25 : 34.75).

What it shows. Two of the F2 types, round yellow and wrinkled green, were not present in the parents. These new combinations can only arise if the factors for shape and colour separate and recombine independently when gametes form and fuse. So seed shape and seed colour are inherited independently. The same happens with height and seed shape: crossing tall-round with short-wrinkled plants gives all tall-round F1, but F2 includes new tall-wrinkled and short-round plants.

Why independence works: chromosomes. If each parent passed on one single long thread of all its genes, R and y would always travel together and could not be separated. Instead, each gene set is in separate pieces, the chromosomes. Each cell has two copies of each chromosome, one from each parent, and every germ cell takes one chromosome from each pair, which may be of either maternal or paternal origin. Genes on different chromosomes therefore reach gametes independently.

Mendel's rules of inheritance (as used in Class 10).
1. Dominance: of two different copies of a gene, one (dominant) is expressed and the other (recessive) is hidden.
2. Segregation: the two copies of a gene separate when germ cells form, so each germ cell carries only one copy; fusion restores the pair.
3. Independent inheritance (assortment): genes for different traits, such as shape and colour, are passed on independently, giving new combinations in the offspring.

Dihybrid Cross and Laws of Inheritance: formulas and key facts

Dihybrid parents and F1: $R R y y \times r r Y Y \to F_{1} : R r Y y (round, yellow)$; Gametes Ry and rY.
F1 gametes: $R r Y y \to R Y, R y, r Y, r y (equal numbers)$; 4 × 4 = 16 boxes in the checkerboard.
F2 phenotypic ratio: $9 R_Y_: 3 R_y y : 3 r r Y_: 1 r r y y$; Round yellow : round green : wrinkled yellow : wrinkled green.
Expected numbers: $number = \frac{ratio part}{16} \times total$; For 556 seeds: 312.75, 104.25, 104.25, 34.75.
Each trait alone: $round : wrinkled = 12 : 4 = 3 : 1; yellow : green = 3 : 1$; Each pair still follows the monohybrid ratio.

How to solve Dihybrid Cross and Laws of Inheritance questions

Use this routine for any two-trait cross.

1Assign letters: capital for dominant, small for recessive, one letter per trait (R/r, Y/y).
2Write parent genotypes from their phenotypes (pure-breeding parents have two identical copies of each gene).
3List gametes: each carries one copy of each gene. A double hybrid (RrYy) gives RY, Ry, rY, ry.
4Draw a 4 × 4 checkerboard and fill in the 16 offspring.
5Group by phenotype: at least one R and one Y → round yellow; R but yy → round green; rr with Y → wrinkled yellow; rryy → wrinkled green.
6Write the ratio (9 : 3 : 3 : 1) and, if asked, convert to numbers using $\frac{part}{16} \times$ total.
7Identify new combinations by comparing F2 types with the parents.

Dihybrid Cross and Laws of Inheritance in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Independent inheritance and expected numbers

Suppose a seed company wants a new pea variety with round green seeds, but its two pure-breeding lines are round-yellow (RRYY) and wrinkled-green (rryy).

Show how a cross could produce round green seeds, and find how many round green seeds to expect among 1600 F2 seeds. What fraction of these round green seeds would be pure-breeding?

Try it first, then show the solutionSolution

Cross RRYY × rryy: gametes RY and ry; F1 all RrYy (round yellow).
Self F1: 16 combinations; round green (R_yy) makes up 3 of 16 boxes: RRyy (1) and Rryy (2).
Expected round green = $\frac{3}{16} \times 1600 = 300$ .
Pure-breeding round green is RRyy, 1 of the 3 boxes, so $\frac{1}{3}$ of them: about 100 seeds.
Round green is a new combination not present in either parent; it appears because shape and colour genes are inherited independently.

Answer: Cross the two lines and self the F1. About 300 of 1600 F2 seeds will be round green, and about one-third of these (about 100) will be pure-breeding RRyy.

Memory hook: Mix two traits, shuffle the genes, and new combinations come out in 3 of every 16.

Tricks to remember Dihybrid Cross and Laws of Inheritance

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

9-3-3-1 by 'both, one, other, none'
9 with both dominants, 3 with the first only, 3 with the second only, 1 with none.
Use it when: Writing F2 phenotypes in order.
FOIL the gametes
For RrYy: First (RY), Outer (Ry), Inner (rY), Last (ry).
Use it when: Listing the four gametes quickly.
Each trait still 3 : 1
In a 9 : 3 : 3 : 1 cross, round : wrinkled = 12 : 4 = 3 : 1.
Use it when: Checking your checkerboard.

Dihybrid Cross and Laws of Inheritance: solved examples

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

Example 1: Independent inheritance · 3 marks · Short answerBoard level

How do Mendel's experiments show that traits are inherited independently? Use the cross between tall plants with round seeds and short plants with wrinkled seeds.

Try it first, then show the solutionSolution

F1 plants are all tall with round seeds, so tallness and roundness are dominant (1 mark).
In F2, besides tall-round and short-wrinkled plants, new combinations appear: tall-wrinkled and short-round (1 mark).
New combinations can arise only if the height trait and seed-shape trait are passed on separately; so they are inherited independently (1 mark).

Answer: New combinations (tall-wrinkled, short-round) in F2 show height and seed shape are inherited independently.

Example 2: Full dihybrid cross · 5 marks · Long answerHarder board level

A pure round-green pea plant (RRyy) is crossed with a pure wrinkled-yellow plant (rrYY). Show the cross up to F2 and give the F2 phenotypic ratio. Which F2 types are new combinations?

Try it first, then show the solutionSolution

Parents RRyy × rrYY; gametes Ry and rY (1 mark).
F1: RrYy, all round yellow (1 mark).
F1 gametes: RY, Ry, rY, ry from each parent (1 mark).
16-box checkerboard gives round yellow 9, round green 3, wrinkled yellow 3, wrinkled green 1 (1 mark).
New combinations compared with the parents: round yellow and wrinkled green (1 mark).

Answer: F1 RrYy (round yellow); F2 9 : 3 : 3 : 1; new combinations round yellow and wrinkled green.

Example 3: Checking Mendel's data · 3 marks · Short answerHarder board level

In one experiment, 556 F2 seeds were 315 round yellow, 108 round green, 101 wrinkled yellow and 32 wrinkled green. (a) Calculate the expected numbers for a 9 : 3 : 3 : 1 ratio. (b) Find the observed ratio of round to wrinkled seeds and compare it with 3 : 1.

Try it first, then show the solutionSolution

(a) $\frac{556}{16} = 34.75$ ; expected: $9 \times 34.75 = 312.75$ , $3 \times 34.75 = 104.25$ (twice), $1 \times 34.75 = 34.75$ (1½ marks).
(b) Round = 315 + 108 = 423; wrinkled = 101 + 32 = 133 (½ mark).
$\frac{423}{133} \approx 3.18$ , close to 3 : 1, so the seed-shape gene still follows the monohybrid ratio (1 mark).

Answer: (a) About 313 : 104 : 104 : 35. (b) Round : wrinkled = 423 : 133 ≈ 3.2 : 1, close to 3 : 1.

Example 4: Why chromosomes matter · 3 marks · Short answerHOTS / competency

If each parent passed on all its genes as one single long thread of DNA, could the F2 of the RRyy × rrYY cross show round-yellow and wrinkled-green seeds? Explain how the chromosome arrangement allows them.

Try it first, then show the solutionSolution

With one single thread, R would always travel with y, and r with Y, from the original parents (1 mark).
Then only the parental combinations (round green, wrinkled yellow) could appear; round yellow and wrinkled green could not (1 mark).
Genes are carried on separate chromosomes; each germ cell takes one chromosome from each pair, of either maternal or paternal origin, so R and Y (or r and y) can end up together, making new combinations possible (1 mark).

Answer: No; linked genes would keep parental combinations. Separate chromosomes sorted independently into germ cells allow new combinations.

Example 5: Probability from the checkerboard · 3 marks · Short answerHarder board level

From the F1 RrYy self-cross, what fraction of F2 seeds is expected to be (a) wrinkled green, (b) pure-breeding round yellow (RRYY), (c) round (any colour)? How many different genotypes appear in total?

Try it first, then show the solutionSolution

(a) rryy: 1 box of 16, so $\frac{1}{16}$ (½ mark).
(b) RRYY: 1 box of 16, so $\frac{1}{16}$ (½ mark).
(c) Round = 9 + 3 = 12 boxes, so $\frac{12}{16} = \frac{3}{4}$ (1 mark).
Genotypes: for each gene there are 3 kinds (RR, Rr, rr and YY, Yy, yy), so $3 \times 3 = 9$ genotypes (1 mark).

Answer: (a) 1/16 (b) 1/16 (c) 3/4; 9 genotypes.

Where students lose marks in Dihybrid Cross and Laws of Inheritance

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

Mistake: Wrong gametes of a dihybrid
Students write RY and ry only, or put two copies of a gene in one gamete.
Wrong working
Gametes of RrYy: Rr, Yy.
Right working
Gametes: RY, Ry, rY, ry.
Do this: One letter of each pair in every gamete; four combinations.
Usually costs: 2 marks in a 5-mark cross.
Mistake: Misidentifying new combinations
Students always call round green and wrinkled yellow 'new'.
Wrong working
In RRyy × rrYY, the new combinations are round green and wrinkled yellow.
Right working
Those are the parental types here; the new combinations are round yellow and wrinkled green.
Do this: Compare F2 types with the actual parents in the question.
Usually costs: 1 mark.
Mistake: Ratio in the wrong order
Phenotypes listed randomly.
Wrong working
Round green 9, round yellow 3...
Right working
Round yellow 9 : round green 3 : wrinkled yellow 3 : wrinkled green 1.
Do this: Both dominants first, double recessive last.
Usually costs: ½ to 1 mark.
Mistake: Forgetting to show the checkerboard
Students jump to 9 : 3 : 3 : 1.
Wrong working
F2 = 9 : 3 : 3 : 1 (no working).
Right working
Gametes listed and a 4 × 4 checkerboard filled before the ratio.
Do this: Show the working; marks are split across steps.
Usually costs: 2-3 marks in long answers.

Exam tips

Use the same letter for both forms of a trait and different letters for different traits.
Label the checkerboard rows and columns with gametes from each F1 parent.
When asked 'what does this show?', write 'traits are inherited independently' in those words.
For number questions, divide the total by 16 first, then multiply by 9, 3, 3 and 1.
Name Mendel's rules as dominance, segregation and independent inheritance, with one line each.

Practice questions on Dihybrid Cross and Laws of Inheritance

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
In a dihybrid cross between pure round-yellow and pure wrinkled-green pea plants, the F2 phenotypic ratio is:
(a) 3 : 1 (b) 1 : 2 : 1 (c) 9 : 3 : 3 : 1 (d) 1 : 1 : 1 : 1
Hint
Two traits, each 3 : 1.
Answer
(c) 9 : 3 : 3 : 1.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): New combinations of traits appear in the F2 of a dihybrid cross.
Reason (R): The factors for different traits are inherited independently of each other.
(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
What makes recombination possible?
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
List the four types of gametes formed by an RrYy plant and explain why they form in equal numbers.
Hint
Each gamete gets one copy of each gene.
Answer
RY, Ry, rY, ry. Each gamete gets one of R or r and one of Y or y; because the two genes are passed on independently, each combination is equally likely.
Q4 · 4 marks · Case-basedHOTS / competency
Read the passage and answer.
Suppose a student crosses a pure tall plant with violet flowers (TTVV) with a pure short plant with white flowers (ttvv), where tall and violet are dominant. She then self-pollinates the F1 and raises 800 F2 plants.
(a) Write the genotype and phenotype of F1. (1 mark)
(b) How many F2 plants are expected to be short with violet flowers? (1 mark)
(c) Which F2 types are new combinations, and how many plants of each are expected in total? (2 marks)
Hint
Short violet is the 'second only' class: 3 of 16.
Answer
(a) TtVv, tall with violet flowers. (b) $\frac{3}{16} \times 800 = 150$ . (c) New combinations are tall-white and short-violet (the parents were tall-violet and short-white): $\frac{3}{16} \times 800 = 150$ each, so 300 in total.
Q5 · 3 marks · Short answerHOTS / competency
A plant with round yellow seeds of unknown genotype is crossed with a wrinkled green plant (rryy). The offspring are round yellow, round green, wrinkled yellow and wrinkled green in roughly equal numbers. What is the genotype of the round yellow parent? Explain.
Hint
rryy gives only ry gametes, so offspring show exactly what the other parent's gametes carry.
Answer
RrYy. The rryy parent gives only ry gametes, so each offspring's appearance shows the other parent's gamete. Four kinds of offspring in equal numbers mean the round yellow parent made RY, Ry, rY and ry gametes equally, which needs both genes in the hybrid form, Rr and Yy.
Q6 · 3 marks · Short answerHarder board level
State Mendel's rule of segregation and explain how it accounts for the reappearance of the recessive trait in F2.
Hint
Two copies separate into different germ cells.
Answer
The two copies of a gene separate when germ cells form, so each germ cell carries only one copy. An F1 hybrid (Tt) makes T and t germ cells in equal numbers; when two t germ cells fuse, the offspring is tt and shows the recessive trait. This is why one quarter of the F2 show it.

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.F1 of RRyy × rrYY?Show answer
RrYy, round yellow.
2.How many boxes in a dihybrid checkerboard?Show answer
16.
3.F2 phenotypic ratio?Show answer
9 : 3 : 3 : 1.
4.What fraction of F2 is double recessive?Show answer
1/16.
5.What does the dihybrid cross prove?Show answer
Different traits are inherited independently.
6.Why can genes for different traits separate?Show answer
They are on separate chromosomes, and each germ cell takes one chromosome from each pair.

Dihybrid Cross and Laws of Inheritance: frequently asked questions

What is a dihybrid cross?

A cross that follows two pairs of contrasting traits at once, such as seed shape and seed colour in peas.

Why is the F2 ratio 9 : 3 : 3 : 1?

Each F1 plant forms four kinds of gametes in equal numbers. Their 16 combinations give 9 with both dominant traits, 3 with one, 3 with the other and 1 with both recessive traits.

What are Mendel's laws of inheritance?

Dominance (one form of a trait is expressed over the other), segregation (the two copies of a gene separate into different germ cells), and independent inheritance (genes for different traits are passed on independently).

What are new combinations in a dihybrid cross?

F2 types not seen in either parent. They show that the genes for the two traits were inherited independently.

Does each trait still show a 3 : 1 ratio in a dihybrid cross?

Yes. For example, round to wrinkled is 12 : 4, which is 3 : 1.