Three crosses were carried out in pea plants with respect to flower colour violet/white (V/v) and flower posit

Question

Q. Three crosses were carried out in pea plants with respect to flower colour violet/white (V/v) and flower position axial/terminal (A/a). Study in the table the crosses 'a', 'b' and 'c' where parental phenotypes and their F1 progeny phenotypes are given. Parental plants (Phenotypes) | F1 Progeny (Phenotypes)

(a) Violet, axial × white, axial | 6/16 white, axial; 2/16 white, terminal; 6/16 violet, axial; 2/16 violet, axial

(b) Violet, axial × white, terminal | 1/4 violet, axial; 1/4 violet, terminal; 1/4 white, axial; 1/4 white, terminal

(c) Violet, axial × violet, axial | 3/4 violet, axial; 1/4 white, axial Find the genotypes of each of the parental pairs of crosses 'a', 'b' and 'c'.

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

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Concept understanding — Dihybrid Cross Ratio

Let’s begin with something you already know from everyday life. Think about a family where the parents have two different traits — say, one parent has curly hair and brown eyes, the other has straight hair and blue eyes. Their children might inherit any combination: curly hair with brown eyes, straight hair with blue eyes, curly hair with blue eyes, or straight hair with brown eyes. You can see that traits don’t always travel together; they can mix and match.

That mixing is exactly what a dihybrid cross is about. In biology, a dihybrid cross is a breeding experiment that tracks two different traits at the same time — for example, seed shape (round vs wrinkled) and seed colour (yellow vs green) in pea plants. The “dihybrid cross ratio” is the predictable pattern in which these two traits appear in the offspring when both parents are hybrid (carrying one dominant and one recessive version) for both traits.

The classic result, as stated in the NCERT textbook, is a 9:3:3:1 ratio in the second generation. That means:

9 out of 16 offspring show both dominant traits (e.g., round and yellow)
3 out of 16 show the first dominant trait and the second recessive trait (e.g., round and green)
3 out of 16 show the first recessive trait and the second dominant trait (e.g., wrinkled and yellow)
1 out of 16 shows both recessive traits (e.g., wrinkled and green)

Important

The 9:3:3:1 ratio is not a random outcome. It is the direct consequence of independent assortment — the principle that genes for different traits are inherited independently of one another. This is one of Mendel’s key laws, and the ratio is its visible proof.

Why does this matter for a commerce or humanities student? Because this ratio is a classic example of probability in action. It shows how combinations of independent events produce predictable patterns — the same logic that underlies risk assessment in insurance, portfolio diversification in finance, or even the likelihood of certain combinations in a game of cards. You don’t need to calculate anything; you just need to see that nature follows rules, and those rules can be expressed as simple proportions. …

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Each cross reveals the hidden genotypes of the parents through the ratios seen in the F1 offspring — the key is to work trait by trait, using the classic dihybrid ratios as a guide.

Let’s begin by recalling the basic idea. In pea plants, flower colour and flower position are two independent traits. Violet (V) is dominant over white (v), and axial (A) is dominant over terminal (a). When we cross two plants that differ in both traits, the offspring ratios tell us exactly what combination of alleles each parent carried. The trick is to examine each trait separately first, then combine the results.

Cross (c) — Violet, axial × Violet, axial

This is the simplest place to start. Both parents show the same dominant phenotypes, and the F1 progeny are all axial — no terminal flowers appear at all. That means both parents must be homozygous dominant for flower position (AA), because if either carried a recessive a, some terminal offspring would show up.

Now look at flower colour. The F1 gives 3/4 violet and 1/4 white — a perfect 3:1 monohybrid ratio. That ratio appears only when both parents are heterozygous for colour (Vv). So for cross (c), each parent is Vv AA.

Note

A 3:1 ratio in one trait, combined with no variation in the other trait, is the hallmark of a monohybrid cross for the varying trait and a homozygous dominant for the constant one.

Cross (b) — Violet, axial × white, terminal

Here the white parent is recessive for both traits — its genotype must be vv aa. The violet, axial parent shows both dominant traits, but we need to find out whether it is homozygous or heterozygous for each.

Look at the F1: all four possible combinations appear in equal 1/4 proportions — violet axial, violet terminal, white axial, white terminal. That is a perfect 1:1:1:1 ratio, which is the classic test cross ratio. A test cross happens when a double heterozygote (Vv Aa) is crossed with a double recessive (vv aa). So the violet, axial parent must be Vv Aa.

Important

A 1:1:1:1 ratio in a dihybrid cross always means one parent is heterozygous for both traits and the other is homozygous recessive for both — no other combination gives that exact distribution.

Cross (a) — Violet, axial × white, axial

Now we have a white parent that is axial — so it is vv for colour, but for position it could be AA or Aa. The violet, axial parent is dominant for both, so its genotype is unknown. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set 57/1/11 markMCQ
$Q. In the following figure, two ways of pairing of two homologous pairs of chromosomes are shown. Which of the following phenomena is expressed ? (A) Linkage of genes (B) Independent assortment of genes (C) Multiple alleles (D) Incomplete dominance$
›Reveal solutionSolution
The figure illustrates the independent assortment of genes, a fundamental principle where different pairs of chromosomes align and separate randomly during meiosis, leading to diverse combinations of traits in offspring.

Mendel's experiments with pea plants laid the foundation for our understanding of heredity. While his monohybrid crosses helped formulate the Law of Segregation, his dihybrid crosses, involving two pairs of contrasting traits simultaneously, led to the formulation of the Law of Independent Assortment. This law explains how different genes are inherited relative to each other.

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. In simpler terms, the alleles for different genes assort independently of one another during the formation of gametes. This means that the inheritance of one trait does not influence the inheritance of another trait, provided the genes are located on different chromosomes or are far apart on the same chromosome.

The figure you described, showing "two ways of pairing of two homologous pairs of chromosomes," directly depicts the chromosomal basis of this law. During meiosis, specifically in Metaphase I, homologous chromosomes pair up and align at the metaphase plate. For two different pairs of homologous chromosomes, there are two possible orientations:

Orientation 1: The maternal chromosome of one pair and the maternal chromosome of the other pair might align on the same side of the metaphase plate, with their paternal counterparts on the opposite side.

Orientation 2: The maternal chromosome of one pair might align with the paternal chromosome of the other pair on one side, and vice versa on the opposite side.

These orientations are entirely random. Because of this random alignment and subsequent separation of homologous chromosomes into daughter cells, different combinations of chromosomes (and thus the genes located on them) are distributed into the gametes. This random distribution of non-homologous chromosomes is precisely what leads to the independent assortment of the genes carried on those chromosomes.

Important

The random orientation of homologous chromosome pairs at the metaphase plate during Meiosis I is the physical basis for the Law of Independent Assortment.

Let's briefly consider why the other options are not expressed by the figure: …
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CBSE 2026Set 57/2/11 markMCQ
$Q. In a dihybrid cross, 2400 individuals are produced in F2 generation. Approximately how many will be phenotypically similar to parents ? (A) 2000 (B) 1500 (C) 580 (D) 450$
›Reveal solutionSolution
In a dihybrid cross, 9 out of every 16 F₂ individuals show at least one dominant trait from each gene pair, making them phenotypically similar to the double-dominant parent; with 2400 individuals, approximately 1350 match this parental phenotype.

When Mendel crossed pea plants differing in two traits simultaneously—say, seed shape (round vs wrinkled) and seed color (yellow vs green)—he performed what we now call a dihybrid cross. The parental generation consisted of plants that were homozygous dominant for both traits (RRYY, round yellow seeds) crossed with plants homozygous recessive for both (rryy, wrinkled green seeds). All F₁ offspring were heterozygous (RrYy) and displayed the dominant phenotype: round and yellow.

The real insight came in the F₂ generation. When Mendel self-crossed these F₁ plants, he observed a characteristic phenotypic ratio of 9:3:3:1. This ratio breaks down as follows:

9 parts show both dominant traits (round and yellow)

3 parts show the first dominant and second recessive (round and green)

3 parts show the first recessive and second dominant (wrinkled and yellow)

1 part shows both recessive traits (wrinkled and green)

Now, the question asks which F₂ individuals are "phenotypically similar to parents." This phrasing requires careful interpretation. The original parents were RRYY (round yellow) and rryy (wrinkled green)—two distinct phenotypes. However, in standard genetics problems of this type, "similar to parents" typically means resembling the dominant parent or the F₁ phenotype, since the F₁ already looks like the dominant parent.

Important

The 9 out of 16 individuals in the F₂ generation that display both dominant traits (round and yellow, in Mendel's case) are phenotypically identical to both the dominant parent and the entire F₁ generation.

With 2400 individuals in the F₂ generation, we calculate how many fall into this "9 parts" category. The fraction is 9/16 of the total:

2400 × (9/16) = 2400 × 0.5625 = 1350 …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The ratio of phenotypes in dihybrid cross is :$
(a) 9:3:3:1
(b) 3:1
(c) 1:2:1
(d) 1:1:1:1
›Reveal solutionSolution

In a dihybrid cross, the F2 phenotypic ratio is the classical Mendelian ratio 9:3:3:1.

A dihybrid cross studies the inheritance of two genes/traits simultaneously (e.g., Mendel's pea cross for seed shape and seed colour, RRYY × rryy). The F1 generation is heterozygous for both traits (RrYy) and shows both dominant phenotypes. When F1 individuals are self-crossed, the gametes segregate independently (Law of Independent Assortment), producing 4 types of gametes (RY, Ry, rY, ry) in equal proportion. A 4×4 …

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CBSE 2025Set 57/6/11 markMCQ
$Q. Which of the following do not follow the law of independent assortment ? (A) Genes on non-homologous chromosomes and absence of linkage (B) Two or more genes on homologous chromosomes (C) Linked genes located on the same chromosomes (D) Two or more distant genes present on the same chromosome$
›Reveal solutionSolution
The Law of Independent Assortment is violated when genes are linked, meaning they are located close together on the same chromosome and tend to be inherited together.

Mendel's Law of Independent Assortment is a fundamental principle of genetics, stating 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. In simpler terms, the alleles of two different genes get sorted into gametes independently of one another. This means that the allele a gamete receives for one gene does not influence the allele received for another gene. This law is typically observed in dihybrid crosses and is crucial for understanding genetic variation.

For the Law of Independent Assortment to hold true, certain conditions are generally met:

Genes on different chromosomes: If genes are located on different, non-homologous chromosomes, they will assort independently because the segregation of one chromosome pair during meiosis does not affect the segregation of another pair.

Genes far apart on the same chromosome: Even if two genes are on the same chromosome, if they are located sufficiently far apart, the probability of crossing over occurring between them is high enough that they effectively behave as if they are on different chromosomes, leading to independent assortment.

Now, let's examine the given options in light of this understanding:

(A) Genes on non-homologous chromosomes and absence of linkage: This scenario perfectly aligns with the conditions for independent assortment. Genes on different chromosomes will segregate independently during gamete formation. The absence of linkage further confirms that their inheritance patterns will not influence each other. Therefore, this option does follow the law.

(B) Two or more genes on homologous chromosomes: This statement is quite general. Genes are always located on chromosomes, and in diploid organisms, these chromosomes exist in homologous pairs. If these genes are on different homologous chromosome pairs, they would assort independently. If they are on the same homologous chromosome, their assortment depends on their distance. This option, by itself, does not definitively state a violation of the law without specifying linkage or distance. …
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CBSE 2025Set 57/6/11 markMCQ
$Q. For Questions number 13 to 16, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : In dihybrid crosses involving sex-linked genes in Drosophila generation of non-parental gene combinations are observed. Reason (R) : Two genes present on different chromosomes show linkage and recombination in Drosophila.$
›Reveal solutionSolution

The Assertion is true but the Reason is false, so the correct choice is (C).

Let’s unpack this carefully. The Assertion talks about dihybrid crosses involving sex-linked genes in Drosophila (the fruit fly). In such crosses, you do indeed observe non-parental (recombinant) combinations in the offspring. That part is correct — when you cross two flies that differ in two sex-linked traits, the F₂ generation shows new combinations that were not present in either parent. This happens because of crossing over during meiosis in the female (since males have only one X chromosome and do not undergo crossing over for X-linked genes).

Now the Reason claims that two genes present on different chromosomes show linkage and recombination. This is where the error lies. Linkage is the tendency of genes located close together on the same chromosome to be inherited together. If two genes are on different chromosomes, they assort independently — they are not linked. Recombination between them occurs not because of linkage but because of independent assortment. So the Reason mixes up two distinct concepts: linkage (same chromosome) and independent assortment (different chromosomes). …
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CBSE 2024Set 57/1/11 markMCQ
$Q. Assertion (A): Linked genes do not show dihybrid F2 ratio 9 : 3 : 3 : 1. Reason (R): Linked genes do not undergo independent assortment. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
Linked genes, being inherited together, do not assort independently, which prevents them from producing the classic 9:3:3:1 dihybrid F2 phenotypic ratio.

To understand why linked genes deviate from the expected dihybrid ratio, we must first recall the basis of that ratio: Mendel's Law of Independent Assortment. When Mendel conducted his dihybrid crosses, studying the inheritance of two different traits simultaneously (for example, seed colour and seed shape in peas), he observed a consistent pattern in the F2 generation. If he crossed a pure-breeding parent with dominant traits (e.g., yellow, round seeds) with a pure-breeding parent with recessive traits (e.g., green, wrinkled seeds), the F1 generation would all display the dominant traits. However, when these F1 individuals were self-pollinated, the F2 generation showed a specific phenotypic ratio of 9 : 3 : 3 : 1.

This 9:3:3:1 ratio represents:

9 parts showing both dominant traits (e.g., yellow, round)

3 parts showing one dominant and one recessive trait (e.g., yellow, wrinkled)

3 parts showing the other dominant and one recessive trait (e.g., green, round)

1 part showing both recessive traits (e.g., green, wrinkled)

This precise ratio is a direct consequence of Mendel's Law of Independent Assortment, which 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. In simpler terms, the alleles for one gene (like seed colour) separate and distribute into gametes independently of the alleles for another gene (like seed shape). This independent segregation happens because these genes are typically located on different chromosomes, or are very far apart on the same chromosome, allowing for all possible combinations of alleles to form in the gametes with equal probability.

Important

The 9:3:3:1 dihybrid F2 phenotypic ratio is the hallmark outcome when two genes assort independently.

However, not all genes behave this way. The concept of "linked genes" describes genes that are located close together on the same chromosome. Because they are physically close, they tend to be inherited together as a single unit during meiosis, rather than assorting independently. This phenomenon is known as genetic linkage.

When genes are linked, the formation of gametes does not follow the independent assortment pattern. Instead, the parental combinations of alleles are much more likely to be passed on together to the offspring. For instance, if an individual inherited a chromosome carrying alleles for 'yellow' and 'round' from one parent, and 'green' and 'wrinkled' from the other, and these genes are linked, then the gametes produced will predominantly carry either 'yellow' and 'round' together, or 'green' and 'wrinkled' together. The recombinant gametes (e.g., 'yellow' and 'wrinkled', or 'green' and 'round') will be formed much less frequently, primarily through crossing over events, which are less common between closely linked genes. …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. The phenotypic ratio for a dihybrid cross is$
(a) 1 : 2 : 1
(b) 3 : 1
(c) 9 : 4
(d) 9 : 3 : 3 : 1
›Reveal solutionSolution

A dihybrid cross (two independently assorting gene pairs, e.g. RrYy × RrYy) gives four phenotypic classes in the F2 generation in the ratio 9:3:3:1.

Mendel's dihybrid cross used pea plants differing in two traits (seed shape: round R/wrinkled r; seed colour: yellow Y/green y). Crossing a doubly heterozygous RrYy plant with another RrYy gives, in the F2 generation, four phenotype classes: Round-Yellow : Round-green : wrinkled-Yellow : wrinkled-green in the ratio 9 : 3 : 3 : 1. This 9:3:3:1 ratio is the hallmark result demonstrating the Law of Independent …

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CBSE 2023Set ANNUAL1 mark
$Q. What is dihybrid cross?$
›Reveal solutionSolution

A dihybrid cross tracks the inheritance of two different genes/traits at once, revealing whether they assort independently.

A dihybrid cross involves parents that differ in two pairs of contrasting characters (traits), each controlled by a separate gene, and are heterozygous for both (e.g., RrYy × RrYy). By studying the phenotypic ratios of offspring in such a cross (classically the 9:3:3:1 ratio observed by Mendel in pea plants for seed shape and seed colour), geneticists can determine whether the two genes assort independently of each other during gamete formation, as descr …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. What is the test-cross ratio of a dihybrid?$
(a) 1:1
(b) 7:1:1:7
(c) 1:1:1:1
(d) 1:7:7:1
›Reveal solutionSolution

A dihybrid test cross gives a 1:1:1:1 ratio.

In a test cross, an individual is crossed with the homozygous recessive (aabb). A dihybrid (AaBb) produces four gamete types — AB, Ab, aB, ab — in equal proportion, while aabb produces only ab gametes. The offspring therefore show the four phenotypes …

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CBSE 2021Set TERM11 markMCQ
$Q. The Dihybrid Phenotypic ratio is :$
(a) 1 : 2 : 1
(b) 3 : 1
(c) 9 : 3 : 3 : 1
(d) 9 : 2 : 2 : 1
›Reveal solutionSolution
A standard dihybrid cross (AaBb x AaBb) gives an F2 phenotypic ratio of 9:3:3:1 when the two genes assort independently and show complete dominance.

In Mendel's dihybrid cross (e.g. round-yellow seeds x wrinkled-green seeds), the F1 (RrYy) is selfed. Because the two gene pairs (seed shape and seed colour) assort independently, the F2 generation shows four phenotypic classes in the ratio:

9 (dominant-dominant, e.g. round-yellow)

3 (dominant-recessive, e.g. round-green)

3 (recessive-dominant, e.g. wrinkled-yellow)

1 (recessive-recessive, e.g. wrinkled-green) …
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CBSE 2020Set ANNUAL1 markMCQ
$Q. The ratio of dihybrid cross is$
(a) 3:1
(b) 9:3:3:1
(c) 9:7
(d) 12:3:1
›Reveal solutionSolution
A dihybrid cross (crossing individuals differing in two traits) between two double heterozygotes gives an F2 phenotypic ratio of 9:3:3:1.

In Mendel's dihybrid cross (e.g. seed shape and seed colour in pea plants: round-yellow RRYY x wrinkled-green rryy), the F1 generation is entirely heterozygous (RrYy) and shows both dominant traits. Self-crossing F1 (RrYy x RrYy) produces gametes in a 1:1:1:1 ratio (RY, Ry, rY, ry) which combine via a Punnett square to give an F2 phenotypic ratio of:

9 Round Yellow (dominant-dominant)

3 Round Green (dominant-recessive)

3 Wrinkled Yellow (recessive-dominant)

1 Wrinkled Green (recessive-recessive) …
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CBSE 2019Set ANNUAL1 markMCQ
$Q. Mendel dihybrid ratio is.$
(a) 2:1
(b) 3:1
(c) 9:3:3:1
(d) 9:7
›Reveal solutionSolution

Mendel's dihybrid cross produces an F2 phenotypic ratio of 9:3:3:1.

When Mendel crossed pea plants differing in two traits simultaneously (e.g., seed shape: round/wrinkled, and seed colour: yellow/green), the F1 generation was uniformly round-yellow (showing dominance for both traits). Self-pollinating (selfing) the F1 dihybrids (RrYy x RrYy) produced an F2 generation with four phenotypic classes in the ratio 9 (round, yellow) : 3 (round, green) : 3 (wrinkled, yellow) : 1 (wrinkled, green). This 9:3:3:1 ratio demonstrated that the two genes assort into gametes independently of each other — Mendel's Law of Independent Assortment.

…

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