BiologyCBSE Class 12 · Principles of Inheritance and Variation

Dihybrid Cross and Linkage

Updated 15 October 2026 · Fact-checked

A dihybrid cross follows two characters at once. Round yellow × wrinkled green peas give round yellow F1; F2 shows 9:3:3:1 because each pair segregates independently of the other (law of independent assortment). Morgan found that genes on the same chromosome are linked: parental combinations far exceed recombinants, so the 9:3:3:1 ratio is not seen.

Understand Dihybrid Cross and Linkage

The dihybrid cross. Mendel crossed true-breeding round yellow (RRYY) with wrinkled green (rryy) peas.

Gametes: RY and ry.
F1: all RrYy, round and yellow (round and yellow are dominant).
F1 gametes: RY, Ry, rY, ry, each 1/4.
F2: a 4 × 4 Punnett square (16 boxes) gives 9 round yellow : 3 wrinkled yellow : 3 round green : 1 wrinkled green.

Why 9:3:3:1? Each pair alone gives 3:1 (round:wrinkled and yellow:green). If the two pairs are independent, the combined ratio is the product (3:1) × (3:1) = 9:3:3:1. This also predicts genotypes: 9 genotypes in F2, e.g. RrYy = 1/2 × 1/2 = 4/16.

Law of independent assortment. When two pairs of traits are combined in a hybrid, the segregation of one pair is independent of the other. At meiosis I, the two homologous pairs line up independently, so R can go with Y or with y equally often.

Dihybrid test cross. RrYy × rryy → 1 RrYy : 1 Rryy : 1 rrYy : 1 rryy (1:1:1:1), revealing the four gamete types of the dihybrid in equal numbers.

Chromosomal theory of inheritance. Mendel's work was rediscovered in 1900 by de Vries, Correns and von Tschermak. Microscopy showed chromosomes in pairs, separating at meiosis. Sutton and Boveri noted that chromosome pairing and separation explain the segregation of factors; Sutton united this with Mendel's principles as the chromosomal theory of inheritance: genes are on chromosomes, and chromosomes segregate and assort independently.

Morgan's experiments. Thomas Hunt Morgan used the fruit fly *Drosophila melanogaster*, which is ideal because it grows on simple synthetic medium, completes its life cycle in about two weeks, produces a large number of progeny from a single mating, shows clear differences between the sexes (males are smaller), and has many hereditary variations visible under a low-power microscope.

Linkage and recombination. Morgan made dihybrid crosses with X-linked genes (e.g. yellow-bodied, white-eyed females × brown-bodied, red-eyed males, then intercrossed F1). The genes did not assort independently; the F2 deviated strongly from 9:3:3:1. When two genes are on the same chromosome, the parental combinations are much more frequent than non-parental ones.

Linkage: the physical association of genes on a chromosome (Morgan's term).
Recombination: generation of non-parental gene combinations (by crossing over between homologous chromosomes in meiosis).
Tightly linked genes show low recombination (white and yellow: 1.3 per cent); loosely linked genes show higher recombination (white and miniature wing: 37.2 per cent).
Morgan's student Alfred Sturtevant used recombination frequency as a measure of the distance between genes and mapped their positions on the chromosome (genetic maps).

How to read a test cross for linkage. Cross the double heterozygote with the double recessive. Independent genes give 1:1:1:1. Linked genes give two large parental classes and two small recombinant classes.

recombination frequency (%) = (number of recombinant offspring ÷ total offspring) × 100

Flow chart: dihybrid F1 → meiosis → genes on different chromosomes? → independent assortment → 4 equal gametes → 9:3:3:1 (or 1:1:1:1 in a test cross). Same chromosome? → linkage → mostly parental gametes, few recombinants (from crossing over) → deviation from 9:3:3:1.

Draw and label: independent assortment at meiosis I. Checklist: two arrangements of two homologous pairs at the metaphase plate. Arrangement 1: R and Y on the same side; arrangement 2: R and y on the same side. Below each, the two gamete types. Labels: homologous pair 1 (R/r), homologous pair 2 (Y/y), metaphase plate, gametes RY, ry, Ry, rY.

Dihybrid Cross and Linkage: key facts

Dihybrid F2: $R r Y y \times R r Y y \to 9 R_Y_: 3 r r Y_: 3 R_y y : 1 r r y y$; Product of two 3:1 ratios.
Dihybrid test cross: $R r Y y \times r r y y \to 1 : 1 : 1 : 1$; Linked genes: two big parental classes, two small recombinant classes.
Gamete types: $2^{n} kinds of gametes for n heterozygous gene pairs$; RrYy: 4 kinds; RrYyTt: 8 kinds.
Probability of a combined genotype: $P (R r Y y) = P (R r) \times P (Y y) = \frac{1}{2} \times \frac{1}{2} = \frac{4}{16}$; Multiply for independent genes.
Recombination frequency: $R F (%) = \frac{recombinants}{total progeny} \times 100$; 1% RF is taken as 1 map unit; white-yellow 1.3%, white-miniature 37.2%.

How to answer Dihybrid Cross and Linkage questions

Treat each gene separately, then combine. This avoids drawing a 16-box square when only one class is asked.

1Write symbols for both genes and the parents' genotypes; note which traits are dominant.
2List gametes: for each parent combine one allele of each gene (RrYy gives RY, Ry, rY, ry).
3For the whole F2, draw a 4 × 4 Punnett square or use (3:1) × (3:1).
4For one class only, multiply single-gene probabilities, e.g. P(round green) = 3/4 × 1/4 = 3/16.
5Convert to numbers by multiplying by the total; check that the classes add up.
6If the data show two large and two small classes (or deviate strongly from 9:3:3:1), suspect linkage; the large classes are the parental combinations.
7Compute RF = recombinants/total × 100; a smaller RF means genes are closer (tightly linked).

Dihybrid Cross and Linkage 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.

Independent assortment in breeding

Suppose a plant breeder crosses a pea variety with round yellow seeds (RRYY) with one having wrinkled green seeds (rryy), selfs the F1, and harvests 1600 F2 seeds. She wants seeds that are round and green and will breed true.

How many F2 seeds will be round green, and how many of those are expected to breed true (RRyy)?

Try it first, then show the solutionSolution

Round green = R_yy: P = 3/4 × 1/4 = 3/16; number = 1600 × 3/16 = 300.
True-breeding round green = RRyy: P = 1/4 × 1/4 = 1/16; number = 1600 × 1/16 = 100.
So 100 of the 300 round green seeds (one-third) breed true; she must self each round green plant to identify them.

Answer: 300 round green seeds, of which about 100 are RRyy (true-breeding).

Memory hook: New combinations appear because genes shuffle independently.

Tricks to remember Dihybrid Cross and Linkage

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

9-3-3-1 = (3+1)^2
Square the monohybrid: (3:1)(3:1) = 9:3:3:1, sum 16.
Use it when: Deriving the F2 ratio quickly.
Linked = luggage strapped together
Linked genes travel together like strapped bags; only crossing over cuts the strap.
Use it when: Why parental types dominate.
Low RF, close friends
Smaller recombination frequency = genes sit closer.
Use it when: Ordering genes on a map.

Dihybrid Cross and Linkage: solved examples

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

Example 1: Gametes of RrYy · 2 marks · Very short answerBoard level

Write the gametes formed by a pea plant of genotype RrYy and state the law that explains their equal proportions.

Try it first, then show the solutionSolution

RY, Ry, rY, ry, each 1/4 (1 mark).
Law of independent assortment: the R/r pair segregates independently of the Y/y pair (1 mark).

Answer: RY, Ry, rY, ry in equal numbers; independent assortment.

Example 2: Dihybrid cross to F2 · 5 marks · Long answerHarder board level

A true-breeding tall pea plant with round seeds (TTRR) is crossed with a dwarf plant with wrinkled seeds (ttrr). Work out the cross to F2 and give the phenotypic ratio of F1 and F2. State the law shown.

Try it first, then show the solutionSolution

Parents TTRR × ttrr; gametes TR and tr; F1 TtRr, all tall round: F1 phenotype 100% tall round (1 mark).
F1 gametes: TR, Tr, tR, tr, each 1/4 (1 mark).
4 × 4 Punnett square of F1 × F1 (1 mark).
F2: 9 tall round : 3 tall wrinkled : 3 dwarf round : 1 dwarf wrinkled (1 mark).
Law of independent assortment: the height pair segregates independently of the seed-shape pair, giving new combinations (tall wrinkled, dwarf round) (1 mark).

Answer: F1 all tall round; F2 9:3:3:1; independent assortment.

Example 3: Guinea pig probability · 3 marks · Short answerHarder board level

In guinea pigs, black coat (G) is dominant over white (g) and brown eyes (B) over blue (b); the genes are not linked. Two GgBb animals are mated. What is the probability of an offspring with black coat and blue eyes? Of a white, blue-eyed one?

Try it first, then show the solutionSolution

Each gene separately: P(black) = 3/4, P(white) = 1/4; P(brown eyes) = 3/4, P(blue) = 1/4 (1 mark).
Black, blue: 3/4 × 1/4 = 3/16 (1 mark).
White, blue: 1/4 × 1/4 = 1/16 (1 mark).

Answer: 3/16 black blue-eyed; 1/16 white blue-eyed.

Example 4: Spotting linkage in a test cross · 3 marks · Short answerHOTS / competency

Suppose a plant heterozygous for two genes (AaBb, received AB from one parent and ab from the other) is test-crossed with aabb. The 1000 offspring are: AaBb 446, aabb 454, Aabb 52, aaBb 48. (a) Are the genes linked? (b) Calculate the recombination frequency. (c) What does this value tell you?

Try it first, then show the solutionSolution

(a) Independent genes would give about 250 of each class (1:1:1:1). Here two classes (AaBb, aabb) are much larger: these are the parental combinations, so the genes are linked (1 mark).
(b) Recombinants = Aabb + aaBb = 52 + 48 = 100; RF = 100/1000 × 100 = 10% (1 mark).
(c) The genes are on the same chromosome and about 10 map units apart; crossing over between them occurred in a minority of meioses (1 mark).

Answer: (a) Linked (b) 10% (c) about 10 map units apart.

Example 5: Morgan's choice of Drosophila · 2 marks · Very short answerBoard level

Give four reasons why Drosophila was suitable for Morgan's genetic experiments.

Try it first, then show the solutionSolution

Grows on simple synthetic medium in the lab; completes its life cycle in about two weeks (1 mark).
A single mating gives a large number of progeny; sexes are easily told apart and many hereditary variations are visible under a low-power microscope (1 mark).

Answer: Easy culture, two-week life cycle, many progeny, clear sexes and visible variations.

Example 6: Mapping three genes · 3 marks · Short answerHOTS / competency

Suppose the recombination frequencies between three linked genes are: P-Q 12%, Q-R 5%, P-R 7%. Arrange the genes in order on the chromosome and justify. Whose idea are you using?

Try it first, then show the solutionSolution

The largest RF (P-Q, 12%) is between the two outer genes (1 mark).
P-R (7%) + R-Q (5%) = 12% = P-Q, so R lies between P and Q: order P-R-Q (1 mark).
Sturtevant's idea: recombination frequency is a measure of the distance between genes on a chromosome (1 mark).

Answer: P-R-Q (P to R 7 units, R to Q 5 units); Sturtevant's gene mapping.

Where students lose marks in Dihybrid Cross and Linkage

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

Mistake: Gametes with two alleles of the same gene
Students write RrY or RR.
Wrong working
Gametes of RrYy: Rr, Yy.
Right working
Gametes: RY, Ry, rY, ry (one allele of each gene).
Do this: Each gamete gets one letter of each gene.
Usually costs: 1-2 marks.
Mistake: Mislabelling F2 classes
The 3:3 classes are swapped.
Wrong working
9 round yellow : 3 round yellow : 3 wrinkled green : 1 round green.
Right working
9 round yellow : 3 wrinkled yellow : 3 round green : 1 wrinkled green.
Do this: Write the phenotype next to each number.
Usually costs: 1 mark.
Mistake: Calling the large classes recombinants
Linkage data are read backwards.
Wrong working
Recombinants = 446 + 454.
Right working
Large classes = parental; small classes = recombinants (52 + 48).
Do this: Linked genes stay together, so parental types are the majority.
Usually costs: 1-2 marks.
Mistake: Confusing linkage with crossing over
Both deal with genes on a chromosome.
Wrong working
Linkage produces recombinants.
Right working
Linkage keeps parental combinations together; crossing over (recombination) produces non-parental ones.
Do this: Linkage = together; recombination = new.
Usually costs: 1 mark.

Exam tips

When only one F2 class is asked, multiply single-gene probabilities instead of drawing the full square.
In 5-mark crosses, draw the 4 × 4 square neatly with gametes along both edges.
Quote Morgan's numbers (1.3% and 37.2%) when asked for tightly and loosely linked genes.
State the law in its exact form: segregation of one pair of characters is independent of the other pair.
Name the scientists correctly: Sutton and Boveri (chromosomal theory), Morgan (linkage), Sturtevant (gene mapping).

Practice questions on Dihybrid Cross and Linkage

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 phenotypic ratio of a dihybrid test cross is:
(a) 9:3:3:1 (b) 3:1 (c) 1:1:1:1 (d) 1:2:1
Hint
Cross with the double recessive.
Answer
(c) 1:1:1:1.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When two genes are on the same chromosome, parental combinations are much more frequent than non-parental ones in a dihybrid cross.
Reason (R): Linked genes do not assort independently.
(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
Is R the cause of A?
Answer
(a).
Q3 · 3 marks · Short answerHarder board level
In tomato, tall stem (T) is dominant over dwarf (t) and red fruit (R) over yellow (r); the genes assort independently. A heterozygous tall, yellow-fruited plant (Ttrr) is crossed with a dwarf, heterozygous red-fruited plant (ttRr). Give the phenotypic ratio of the offspring.
Hint
Each gene alone is a 1:1 test cross.
Answer
Tt × tt → 1/2 tall : 1/2 dwarf; rr × Rr → 1/2 red : 1/2 yellow. Combined: 1 tall red (TtRr) : 1 tall yellow (Ttrr) : 1 dwarf red (ttRr) : 1 dwarf yellow (ttrr) = 1:1:1:1.
Q4 · 2 marks · Very short answerHarder board level
How many kinds of gametes does a plant of genotype AaBBCcDd form if all genes are on different chromosomes? What fraction of them carry abcd?
Hint
BB is homozygous.
Answer
Heterozygous pairs: A, C, D = 3, so 2^3 = 8 kinds. No gamete can carry b (the plant is BB), so the fraction with abcd is 0; the fraction with aBcd is 1/8.
Q5 · 4 marks · Case-basedHarder board level
Read and answer.
Suppose a student test-crosses two F1 plants, each heterozygous for two genes, and plots bar graphs of the progeny. Graph 1 shows four equal bars (25% each). Graph 2 shows two tall bars (about 45% each) and two short bars (about 5% each).
(a) Which graph shows independent assortment? (1)
(b) What does Graph 2 indicate about the location of the genes? (1)
(c) Calculate the recombination frequency in Graph 2 and state whether the genes are tightly or loosely linked compared with white and miniature wing in Drosophila (37.2%). (2)
Hint
Recombinants are the small bars.
Answer
(a) Graph 1 (1:1:1:1). (b) The genes are on the same chromosome (linked); the tall bars are parental combinations. (c) RF = 5% + 5% = 10%; much lower than 37.2%, so these genes are more tightly linked (closer together).
Q6 · 2 marks · Very short answerHOTS / competency
Suppose two genes lie very far apart on the same long chromosome, so far that a crossover happens between them in almost every meiosis. Predict the test-cross result and explain why such genes may look unlinked.
Hint
Recombination frequency cannot exceed about 50%.
Answer
Recombinants approach about half of the progeny, so the four classes become nearly equal (close to 1:1:1:1), the same as independent assortment. Such loosely linked genes behave as if unlinked, which is why high recombination (like 37.2% for white and miniature wing) indicates genes far apart.

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.F2 ratio of a dihybrid cross?Show answer
9 : 3 : 3 : 1.
2.Gametes of RrYy?Show answer
RY, Ry, rY, ry.
3.Who proposed the chromosomal theory of inheritance?Show answer
Sutton and Boveri.
4.Recombination between white and yellow genes in Drosophila?Show answer
1.3 per cent.
5.Who used recombination frequency to map genes?Show answer
Alfred Sturtevant.
6.Probability of rryy in a dihybrid F2?Show answer
1/16.

Dihybrid Cross and Linkage: frequently asked questions

What is the law of independent assortment?

When two pairs of traits are combined in a hybrid, the segregation of one pair is independent of the other, so all combinations of their alleles enter gametes equally.

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

Each gene gives 3:1 on its own; when the two assort independently, the combined ratio is (3:1) × (3:1) = 9:3:3:1.

What is linkage?

The physical association of genes on the same chromosome, so they tend to be inherited together and do not assort independently.

What is recombination?

The appearance of non-parental combinations of genes, produced by crossing over between homologous chromosomes during meiosis.

Why do linked genes not show 9:3:3:1?

They do not assort independently; parental combinations are much more frequent than recombinant ones.