(a) Meselson and Stahl carried out an experiment to prove the nature of DNA replication. Recall the experiment

Question

Q. (a) Meselson and Stahl carried out an experiment to prove the nature of DNA replication. Recall the experiment and answer the following questions.

(i) Which two types of nitrogen were used by them in their experiment and why ?

(ii) Why did they take samples of E. coli at definite time intervals for their observation ?

(iii) State the role of caesium chloride density gradient in their experiment.

(iv) Write the conclusions they arrived at.

(OR)

(b)

(i) A true breeding tall pea plant with round seeds is crossed with a recessive dwarf pea plant having wrinkled seeds. Work out the cross up to F2 generation giving the phenotypic ratios of F1 and F2 generation respectively.

(ii) State the Mendelian principle that can be derived only with the help of such a cross.

CBSECBSE Class XII Board 2023Subjective· 5mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Complementary Base Pairing

Imagine you have a zipper. Each tooth on one side fits perfectly only with a specific tooth on the other side — a left tooth always clicks into a right tooth, and they lock together. If you tried to force a left tooth into another left tooth, the zipper would jam. That simple "lock-and-key" fit is the everyday intuition behind complementary base pairing.

In biology, the "zipper" is the DNA molecule, which is made of two long strands twisted together. Each strand is a chain of smaller units called nucleotides. Every nucleotide contains one of four chemical "letters" — Adenine (A), Thymine (T), Guanine (G), and Cytosine (C). The two strands are held together by weak bonds between these letters, but they don't pair randomly. The rule is strict and unchanging:

A (Adenine) always pairs with T (Thymine)
G (Guanine) always pairs with C (Cytosine)

This is the complementary base pairing rule. It means that if you know the sequence of letters on one strand, you can instantly write the sequence on the other strand. For example, if one strand reads A–T–G–C, the opposite strand must read T–A–C–G.

Note

The NCERT Class 12 Biology textbook (Chapter 6, "Molecular Basis of Inheritance") states this rule exactly as: "Adenine pairs with Thymine (A = T) and Guanine pairs with Cytosine (G ≡ C)." The double lines (=) and triple lines (≡) indicate the number of hydrogen bonds — A–T has two bonds, G–C has three bonds. This difference in bond strength matters for how DNA unwinds, but the pairing rule itself is what you need to remember.

Why does this matter? Three big reasons:

Replication: When a cell divides, the two strands of DNA separate. Each strand acts as a template. Using the complementary rule, the cell builds a new partner strand for each old strand. The result is two identical DNA molecules — one for each new cell. Without this rule, copying would be random and life couldn't pass on genetic information accurately.
Transcription: To make proteins, a cell first makes a temporary copy of a gene in the form of mRNA. The mRNA is built using the complementary rule, but with one change: wherever there is an A on the DNA, the mRNA puts a U (Uracil) instead of T. So the pairing becomes A–U, T–A, G–C, C–G. This ensures the genetic message is faithfully transcribed. …

Part (b)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. …

Start your 14-day free trial to unlock the full solution →

Part (a): Meselson & Stahl labelled DNA with heavy ¹⁵N then light ¹⁴N, sampled E. coli each generation and used CsCl density-gradient centrifugation to see one hybrid band after one generation and hybrid + light after two — proving semiconservative replication.

Part (b): TTRR × ttrr → all tall-round F1 (TtRr); F1 selfed → F2 = 9:3:3:1, which establishes Mendel's Law of Independent Assortment.

Part (a) — The Meselson–Stahl experiment

This 1958 experiment settled how DNA replicates, distinguishing the semiconservative model from the conservative and dispersive models.

(i) The two types of nitrogen and why

They used ¹⁵N (heavy, non-radioactive isotope) and ¹⁴N (normal, light isotope) of nitrogen.
Reason: nitrogen is a constituent of the nitrogen bases of DNA. Bacteria grown for many generations in a medium with heavy nitrogen (e.g. ¹⁵NH₄Cl) incorporate ¹⁵N into their DNA, making it denser. On transferring them to a ¹⁴N medium, all newly made DNA uses light nitrogen. The density difference between heavy and light DNA is what allows old and new DNA to be distinguished — without using any radioactivity.

(ii) Why samples were taken at definite time intervals

DNA replication occurs once per cell generation (~20 minutes in E. coli under their conditions).
By taking samples at exactly one, two, etc. generation intervals, they could track the step-by-step change in DNA density after each round of replication. This timed sampling was essential to reveal the characteristic banding pattern that identifies the mode of replication; random sampling would miss the key transitions.

(iii) Role of the caesium chloride density gradient

When a CsCl solution is centrifuged at very high speed, the heavy Cs⁺ ions redistribute to form a continuous density gradient in the tube.
DNA molecules migrate to the point where their own density equals that of the surrounding CsCl and form a band there.
This separates DNA by density: heavy (¹⁵N/¹⁵N) DNA lowest, light (¹⁴N/¹⁴N) DNA highest, and hybrid (¹⁵N/¹⁴N) DNA in between — allowing the different DNA populations to be seen as distinct bands.

(iv) Observations and conclusion

Start (0 generation): all DNA heavy — a single heavy band.
After 1 generation in ¹⁴N: a single band of intermediate (hybrid) density. This rules out the conservative model (which would give a heavy + a light band).
After 2 generations: two bands in equal amounts — one hybrid and one light. This rules out the dispersive model.
Conclusion: DNA replication is semiconservative — each daughter DNA molecule keeps one parental (old) strand and has one newly synthesised strand. …

Start your 14-day free trial to unlock the full solution →

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 24 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: …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set F1 markMCQ
$Q. Which of the following is a correct pair?$
(a) G ≡ T
(b) C = G
(c) G ≡ C
(d) A ≡ T
›Reveal solutionSolution

Correct base pairing: G ≡ C (three H-bonds) and A = T (two H-bonds).

In double-stranded DNA the bases pair by complementary hydrogen bonding (Chargaff/Watson-Crick rules). Guanine pairs specifically with Cytosine forming THREE hydrogen bonds (shown as G ≡ C), and Adenine pairs with Thymine forming TWO hydrogen bonds (A = T). Checking the …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. In DNA Double helix Adenine base joins with which base -$
(a) Thymine
(b) Guanine
(c) Cytosine
(d) Uracil
›Reveal solutionSolution

DNA follows Chargaff/Watson-Crick complementary base pairing: A-T (2 H-bonds) and G-C (3 H-bonds).

In the DNA double helix, the purine Adenine always pairs with the pyrimidine Thymine through two hydrogen bonds, while Guanine pairs with Cytosine through three hydrogen bonds. This strict complementarity keeps the diameter of the double helix uniform and is t …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The distance between the two strands of a DNA molecule ............ from one end to another.$
(a) increases
(b) remains the same
(c) decreases
(d) none of these
›Reveal solutionSolution

DNA is a uniform double helix — the inter-strand distance never changes along its length.

According to the Watson–Crick model, DNA is a right-handed double helix made of two antiparallel polynucleotide chains coiled around a common axis. The two strands are held together by hydrogen bonds between complementary bases (A=T, G≡C), and because every base pair is a purine paired with a pyrimidine (which keeps the width of each 'rung' of the ladder essentially identical, about 20 Å), the …

Start your 14-day free trial to see the full solution →
CBSE 2025Set BOTANY1 markMCQ
$Q. Fill in the blank selecting the appropriate one: In a DNA double helix, two polydeoxyribonucleotide molecules are held together by ____ bonds.$
(a) phosphodiester
(b) covalent
(c) ionic
(d) hydrogen
›Reveal solutionSolution

The two strands of DNA are joined by hydrogen bonds between complementary base pairs, not by covalent, ionic or phosphodiester linkages.

Within a single strand, adjacent nucleotides are joined by strong covalent phosphodiester bonds, forming the sugar-phosphate backbone. The two separate strands, however, are held together across the helix by weak hydrogen bonds between complementary nitrogen bases: adenine pairs with thymine through two hydrogen bonds and guanine pairs with cytosine thr …

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which nitrogenous base pairs with guanine nitrogenous base in DNA$
(a) Adenine
(b) Cytosine
(c) Thymine
(d) Uracil
›Reveal solutionSolution

In DNA, the purine guanine always base-pairs with the pyrimidine cytosine (three H-bonds); adenine pairs with thymine (two H-bonds).

DNA's two strands are held together by hydrogen bonding between specific, complementary nitrogenous base pairs (Watson-Crick base-pairing rules): a purine always pairs with a pyrimidine of a complementary shape. Adenine (A) pairs with Thymine (T) throug …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. By how many hydrogen bonds guanine is bonded with cytosine ?$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

G≡C pairing has 3 hydrogen bonds; A=T pairing has 2.

In the double-stranded DNA helix, the purine guanine (G) always pairs with the pyrimidine cytosine (C) through three hydrogen bonds, making the G-C pair relatively more stable than the A-T pair (which is held together by only two hydrogen bonds). This complementary base pairing (Char …

Start your 14-day free trial to see the full solution →

Q.(a) Meselson and Stahl carried out an experiment to prove the nature of DNA replication. Recall the experiment and answer the following questions.

Part (a)Concept understanding — Complementary Base Pairing

Part (b)Concept understanding — Dihybrid Cross Ratio

Part (a) Meselson–Stahl experiment

Part (a) — The Meselson–Stahl experiment

Q.(a) Meselson and Stahl carried out an experiment to prove the nature of DNA replication. Recall the experiment and answer the following questions.

Part (a) — The Meselson–Stahl experiment

🎓Unlock everything free for 14 days

Unlock everything free for 14 days