Botany · Principles of Inheritance and Variation

Dihybrid Cross & Law of Independent Assortment

The dihybrid cross extends Mendel's work from one gene to two, and the 9:3:3:1 ratio it generates is the single most tested numerical result in NEET genetics. This subtopic dissects how RRYY x rryy produces that ratio, what independent assortment means at the chromosome level, the 4x4 Punnett square, the 1:1:1:1 dihybrid test cross, and the linkage exception that breaks the rule.

NCERT grounding

NCERT Class 12 Biology, Chapter 4, Section 4.3 (Inheritance of Two Genes) opens the dihybrid story: Mendel "crossed pea plants that differed in two characters, as is seen in the cross between a pea plant that has seeds with yellow colour and round shape and one that had seeds of green colour and wrinkled shape." The F1 seeds were all yellow and round, identifying yellow as dominant over green and round as dominant over wrinkled. On selfing the F1, the F2 showed the four phenotypes in the proportion 9:3:3:1. Section 4.3.1 then states the Law of Independent Assortment and Section 4.3.3 introduces the linkage exception found by Morgan in Drosophila.

"When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters."

NCERT Class 12 Biology — Mendel's Law of Independent Assortment

NIOS Biology, Chapter 22, reinforces the same numbers: it defines a dihybrid cross as one in which the parents differ in two pairs of contrasting characters, names the F2 result the "Mendelian dihybrid ratio (9:3:3:1)", and contrasts it with the monohybrid 3:1. It also records that Bateson and Punnett, working with sweet pea, obtained a ratio that departed from 9:3:3:1, the first documented case of linkage. These two sources fix every fact used on this page.

The dihybrid cross and the 9:3:3:1 ratio

A dihybrid cross tracks two genes at once. Mendel began with two true-breeding parents: a plant with round, yellow seeds (genotype RRYY) and a plant with wrinkled, green seeds (genotype rryy). R is the dominant allele for round seed shape, r the recessive for wrinkled; Y is dominant for yellow seed colour, y recessive for green. Because each parent is homozygous at both loci, RRYY makes only RY gametes and rryy makes only ry gametes. Fertilisation therefore yields a single F1 genotype: the dihybrid RrYy, phenotypically round and yellow, resembling the dominant parent in both characters.

The interesting biology begins when the F1 is self-pollinated. The F1 RrYy is heterozygous at both loci, and the question is what kinds of gametes it can make. This is where the dihybrid cross diverges sharply from the monohybrid cross, and where the 9:3:3:1 ratio is actually born.

Why the F1 dihybrid makes four gamete types

Consider the R/r gene pair alone. During meiosis the two homologous chromosomes carrying R and r separate, so 50 per cent of gametes carry R and 50 per cent carry r — exactly as in a monohybrid cross. Now consider the Y/y pair. It also segregates 50:50. The decisive point is that the two pairs lie on different chromosomes, so the way the R/r pair orients at the metaphase plate has no effect on how the Y/y pair orients. Each combination is equally likely.

NCERT puts it precisely: "the segregation of 50 per cent R and 50 per cent r is independent from the segregation of 50 per cent Y and 50 per cent y. Therefore, 50 per cent of the r bearing gametes has Y and the other 50 per cent has y." Multiplying the two independent choices gives four equally frequent gamete genotypes.

4

Gamete types from RrYy

The F1 dihybrid produces RY, Ry, rY and ry gametes, each at a frequency of 1/4 (25 per cent). A diploid heterozygous for n loci produces 2ⁿ gamete types — so two loci give 4, three loci give 8, four loci give 16.

Figure 1 Independent assortment generates four gamete types F1 dihybrid RrYy → four gamete types R r Y y R (1/2) r (1/2) RY Ry rY ry 1/4 1/4 1/4 1/4 Each gene segregates 1/2 : 1/2; the two splits are independent, giving 2 × 2 = 4 equal classes

Figure 1. The R/r pair splits 1/2 R : 1/2 r, and the Y/y pair splits 1/2 Y : 1/2 y. Because the two splits are independent, the four combinations RY, Ry, rY and ry each occur at 1/4 frequency.

The 4x4 Punnett square

With four egg types and four pollen types, the Punnett square for the F1 self-cross has 16 boxes. Reginald Punnett's grid is, in NCERT's words, "a graphical representation to calculate the probability of all possible genotypes of offspring in a genetic cross." Writing RY, Ry, rY and ry along the top and side and filling every cell produces all 16 zygote genotypes. Counting phenotypes — not genotypes — across those 16 boxes gives the famous ratio.

Figure 2 The 4x4 dihybrid Punnett square RrYy × RrYy — 16-box Punnett square RY Ry rY ry RY Ry rY ry RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy 9 Round Yellow 3 Round Green 3 Wrinkled Yellow 1 Wrinkled Green

Figure 2. All 16 boxes of the dihybrid Punnett square. Nine boxes carry at least one R and one Y (round yellow), three are round green, three are wrinkled yellow, and one box — rryy — is wrinkled green. Phenotype ratio 9:3:3:1.

Reading the 9:3:3:1 out of the 16 boxes

A box shows a dominant phenotype for a gene whenever it carries at least one dominant allele. For round seed shape that means any genotype with an R; for yellow colour, any genotype with a Y. Counting across the 16 boxes: 9 boxes have both an R and a Y (round, yellow); 3 have an R but are yy (round, green); 3 are rr but have a Y (wrinkled, yellow); and exactly 1 box, rryy, is homozygous recessive for both (wrinkled, green). The phenotypic ratio is therefore 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

NCERT also asks students to confirm that the F2 contains nine different genotypes. The genotypic ratio is 1:2:2:4:1:2:1:2:1 across those nine classes — quite different from the simple 9:3:3:1 phenotypic ratio. Keeping genotype counts and phenotype counts separate is essential.

The 9:3:3:1 ratio is two 3:1 ratios multiplied

The deepest insight in this subtopic is that you never need to draw the 16-box grid at all. Because each gene assorts independently, each character segregates in its own monohybrid 3:1 ratio: round to wrinkled is 3:1, and yellow to green is 3:1. The dihybrid ratio is simply the product of these two separate ratios. NCERT writes the derivation explicitly:

Derivation. (3 Round : 1 Wrinkled) × (3 Yellow : 1 Green) = 9 Round Yellow : 3 Round Green : 3 Wrinkled Yellow : 1 Wrinkled Green. The cross-multiplication 3×3, 3×1, 1×3, 1×1 gives 9, 3, 3, 1 directly.

Seed shape

3 : 1

Round : Wrinkled

R is dominant; 3/4 of F2 carry at least one R.

Seed colour

3 : 1

Yellow : Green

Y is dominant; 3/4 of F2 carry at least one Y.

Combined

9 : 3 : 3 : 1

Product of the two 3:1 ratios

Independent assortment makes the probabilities multiply.

This multiplicative behaviour is the whole content of independent assortment. The probability of a round-yellow seed is P(round) × P(yellow) = 3/4 × 3/4 = 9/16. A wrinkled-green seed is 1/4 × 1/4 = 1/16. Two events multiply only when they are independent — which is exactly why a deviation from 9:3:3:1 is the signature that two genes are not assorting independently.

Independent assortment at the chromosome level

Mendel inferred independent assortment from ratios alone; he had no idea what a chromosome was. The physical explanation came only with the Chromosomal Theory of Inheritance. After Mendel's work was rediscovered in 1900 by de Vries, Correns and von Tschermak, improved microscopy let cytologists watch chromosomes double and divide. By 1902 chromosome movement in meiosis had been described, and Walter Sutton and Theodore Boveri noticed that chromosomes behaved exactly as Mendel's "factors" should — they occur in pairs, segregate at gamete formation so each gamete gets one of each pair, and different pairs segregate independently of one another. Sutton united chromosomal segregation with Mendelian principles and called it the chromosomal theory of inheritance.

The mechanism sits in metaphase I of meiosis. The R/r homologous pair and the Y/y homologous pair are separate bivalents. Each bivalent aligns at the metaphase plate independently — the pole that R faces is unrelated to the pole that Y faces. As NCERT's Figure 4.9 shows, a cell may send R and Y to the same pole in one meiocyte and R with y to the same pole in another. This random orientation of non-homologous bivalents is the cellular event that produces the four equally frequent gamete types. Thomas Hunt Morgan and his colleagues later supplied the experimental verification of the chromosomal theory using Drosophila.

Figure 3 Independent alignment of bivalents at metaphase I Metaphase I — two equally likely alignments Orientation 1 R r Y y Top pole → R Y    Bottom pole → r y Gametes: RY and ry Orientation 2 R r y Y Top pole → R y    Bottom pole → r Y Gametes: Ry and rY Both orientations occur with equal probability → all four gamete types at 1/4 each

Figure 3. Two non-homologous bivalents orient at the metaphase I plate at random. Orientation 1 yields RY and ry gametes; orientation 2 yields Ry and rY. Equal probability of the two orientations is independent assortment.

The dihybrid test cross — a 1:1:1:1 result

A round, yellow plant could be RRYY, RRYy, RrYY or RrYy — the phenotype alone cannot reveal the genotype. To resolve it, geneticists use a test cross: the organism showing the dominant phenotype is crossed not by selfing but with the double homozygous recessive, here rryy. The recessive parent contributes only ry gametes, so it adds no dominant allele and cannot mask anything. Every offspring phenotype is a direct readout of one gamete from the tested parent.

When the tested plant is the F1 dihybrid RrYy, its four gamete types RY, Ry, rY and ry — each at 1/4 — meet ry from the recessive parent. The offspring are RrYy, Rryy, rrYy and rryy, phenotypically round yellow, round green, wrinkled yellow and wrinkled green in the ratio 1:1:1:1. The test-cross ratio therefore exposes the gamete proportions of the heterozygote without any algebra.

Dihybrid test cross: RrYy × rryy

Expected progeny ratio 1:1:1:1
  1. Step 1

    Dihybrid gametes

    RrYy produces RY, Ry, rY, ry — each 1/4.

  2. Step 2

    Tester gametes

    rryy produces only ry — one type.

  3. Step 3

    Combine

    Offspring RrYy, Rryy, rrYy, rryy.

  4. Step 4

    Phenotype ratio

    1 RY : 1 Ry : 1 rY : 1 ry phenotypes.

A test cross is also the cleanest way to detect linkage. If the two genes assort independently, the test cross gives 1:1:1:1. If they are linked, parental-type combinations far outnumber recombinant types, and the ratio is skewed — precisely the observation NEET 2016 tested.

Linkage and recombination — the exception

Independent assortment is not a universal law. It holds only when the two genes lie on different chromosomes. Genes located on the same chromosome are physically connected and tend to be inherited together. Morgan, crossing yellow-bodied white-eyed female Drosophila with brown-bodied red-eyed males and intercrossing the F1, found the F2 ratio "deviated very significantly from the 9:3:3:1 ratio." Knowing both genes sat on the X chromosome, he coined linkage for the physical association of genes on a chromosome and recombination for the generation of non-parental combinations.

Independent assortment vs Linkage

Independent assortment

9:3:3:1

F2 dihybrid phenotype ratio

  • Two genes on different chromosomes
  • F1 makes four gamete types in equal 1/4 frequency
  • Test cross gives 1:1:1:1
  • Parental and recombinant types equally frequent
VS

Linkage

Skewed

Deviates from 9:3:3:1

  • Two genes on the same chromosome
  • Parental gamete types far exceed recombinants
  • Recombination frequency measures gene distance
  • Tight linkage = low recombination; loose = higher

Recombination arises by crossing over during meiosis. Morgan found that some linked genes were tightly linked with very low recombination, while others were loosely linked with higher recombination — white and yellow showed only 1.3 per cent recombination, white and miniature wing 37.2 per cent. His student Alfred Sturtevant turned recombination frequency into a ruler: the percentage of recombinants measures the distance between two genes on a chromosome, with 1 per cent recombination defined as one map unit, or one centimorgan. This is how the first genetic maps were drawn, and the principle still underpins genome sequencing today.

Worked examples

Worked example 1

In the F2 of the cross RRYY × rryy, what fraction of the offspring are expected to be wrinkled and green?

Wrinkled requires the genotype rr (probability 1/4) and green requires yy (probability 1/4). Because the two genes assort independently, the probabilities multiply: 1/4 × 1/4 = 1/16. This is the smallest class of the 9:3:3:1 ratio and corresponds to the single rryy box of the 16-box Punnett square.

Worked example 2

A diploid organism is heterozygous for four loci. How many types of gametes can it produce, assuming all loci assort independently?

Each heterozygous locus contributes two allele choices to a gamete, and independent assortment makes the choices multiply. For n loci the count is 2ⁿ. With n = 4, the answer is 2⁴ = 16 gamete types. This is the direct extension of the 2² = 4 rule seen in the dihybrid.

Worked example 3

A tall plant with yellow seeds of genotype TtYy is crossed with a tall plant with green seeds, TtYy × Ttyy. What proportion of the offspring are tall and green?

Treat each gene separately. For height, Tt × Tt gives 3/4 tall. For seed colour, Yy × yy gives 1/2 yellow and 1/2 green, so green is 1/2. Multiply the independent probabilities: P(tall and green) = 3/4 × 1/2 = 3/8. The same method gives dwarf and green as 1/4 × 1/2 = 1/8.

Worked example 4

An F1 dihybrid is test-crossed and the progeny show 41 per cent parental types and only 9 per cent of one recombinant class. What does this indicate about the two genes?

Independent assortment predicts a 1:1:1:1 test-cross ratio — each of the four classes at 25 per cent, with parental and recombinant types equally frequent. Here the parental types dominate and recombinants are scarce, so the two genes are linked on the same chromosome. The excess of parental over recombinant types is the diagnostic signature of linkage.

Common confusion & NEET traps

The dihybrid cross packs several look-alike ratios into one topic, and NEET items are usually built around confusing one for another. The 9:3:3:1, 1:1:1:1 and 1:2:1 ratios all live in this chapter, and each belongs to a specific cross. The traps below isolate the most frequent errors.

NEET PYQ Snapshot — Dihybrid Cross & Law of Independent Assortment

Real NEET questions on the 9:3:3:1 ratio, independent assortment and linkage.

NEET 2025

Genes R and Y follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?

  1. Phenotypic ratio — 9 : 7
  2. Phenotypic ratio — 1 : 2 : 1
  3. Phenotypic ratio — 3 : 1
  4. Phenotypic ratio — 9 : 3 : 3 : 1
Answer: (4)

Why: A classical Mendelian dihybrid cross. Yellow is dominant over green and round over wrinkled; selfing the F1 RrYy gives the F2 phenotypic ratio 9:3:3:1.

NEET 2022

Assertion (A): Mendel's law of Independent assortment does not hold good for the genes that are located closely on the same chromosome. Reason (R): Closely located genes assort independently.

  1. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  2. (A) is correct but (R) is not correct
  3. (A) is not correct but (R) is correct
  4. Both (A) and (R) are correct and (R) is the correct explanation of (A)
Answer: (2)

Why: Assertion is true — independent assortment fails for linked genes. The Reason is false: closely located genes do not assort independently, they tend to be inherited together.

NEET 2016

In a test cross involving F1 dihybrid flies, more parental-type offspring were produced than recombinant-type offspring. This indicates —

  1. Chromosomes failed to separate during meiosis
  2. The two genes are linked and present on the same chromosome
  3. Both characters are controlled by more than one gene
  4. The two genes are located on two different chromosomes
Answer: (2)

Why: An excess of parental over recombinant types is the hallmark of incomplete linkage — the two genes lie on the same chromosome. Independent assortment would have given equal classes (1:1:1:1).

NEET 2021

The production of gametes by the parents, formation of zygotes, the F1 and F2 plants, can be understood from a diagram called:

  1. Net square
  2. Bullet square
  3. Punch square
  4. Punnett square
Answer: (4)

Why: The Punnett square, developed by Reginald C. Punnett, graphically represents gamete combinations and zygote genotypes — for a dihybrid cross it is the 4×4, 16-box grid.

FAQs — Dihybrid Cross & Law of Independent Assortment

Quick answers to the most common doubts on dihybrid crosses and independent assortment.

What is the F2 phenotypic ratio of a dihybrid cross?

A dihybrid cross between RRYY and rryy gives an F1 of RrYy, and on selfing the F1 the F2 generation shows four phenotypes in the ratio 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green, i.e. 9:3:3:1. This ratio is the product of two independent 3:1 monohybrid ratios.

What does Mendel's Law of Independent Assortment state?

The Law of Independent Assortment states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters. The R/r alleles separate into gametes without any influence from how the Y/y alleles separate, so the F1 RrYy produces four gamete types (RY, Ry, rY, ry) in equal 1/4 frequency.

Why does the F1 dihybrid RrYy produce four types of gametes?

During meiosis the R/r homologous pair segregates 50 percent R and 50 percent r, and the Y/y pair segregates 50 percent Y and 50 percent y. Because the two pairs lie on different chromosomes, they assort independently, so each gamete combines one allele of each gene in every possible way: RY, Ry, rY and ry, each at 1/4 frequency.

What is the result of a dihybrid test cross?

A dihybrid test cross crosses the F1 RrYy with the double recessive rryy. The recessive parent contributes only ry gametes, so the offspring phenotypes directly mirror the four gamete types of the dihybrid. The expected ratio is 1 round yellow : 1 round green : 1 wrinkled yellow : 1 wrinkled green, i.e. 1:1:1:1.

How does linkage cause a deviation from the 9:3:3:1 ratio?

Independent assortment holds only when two genes lie on different chromosomes. When two genes sit on the same chromosome they are linked and tend to be inherited together. Morgan found that linked genes produce far more parental-type than recombinant-type offspring, so the F2 ratio deviates significantly from 9:3:3:1. The closer the genes, the stronger the linkage and the lower the recombination.

How is the 9:3:3:1 ratio derived from monohybrid ratios?

Because each gene assorts independently, each character segregates in its own 3:1 ratio. Multiplying the two ratios together gives the dihybrid ratio: (3 round : 1 wrinkled) x (3 yellow : 1 green) = 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. The 9:3:3:1 ratio is simply the product of two separate 3:1 ratios.