Q.(a) Stability, as one of the properties of genetic material, was very evident in one of the very early experiments in genetics. Name the scientist and describe his experiment. State the conclusion he arrived at.
🔒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 — Nucleic Acid Functions
Nucleic Acid Functions – A First Look
Think of a living organism as a giant, incredibly complex factory. Every second, this factory needs to produce thousands of different products—proteins, enzymes, hormones, structural materials—in exactly the right amounts, at exactly the right places, and at exactly the right times. How does the factory know what to build and when? It needs a master blueprint and a set of working copies that can be carried to the assembly lines.
That master blueprint is DNA (deoxyribonucleic acid). The working copies are RNA (ribonucleic acid). Together, they are the nucleic acids, and their job is to store, transmit, and execute the genetic information that makes every living thing what it is.
The Two Main Functions
1. DNA – The Permanent Blueprint (Storage and Inheritance)
DNA is the long-term, stable repository of genetic information. It is like the original architectural plan for the entire factory, locked in a secure vault. Its functions are:
- Storing genetic information: DNA contains the instructions for building every protein the organism will ever need. These instructions are written in a chemical language using four "letters" (nucleotides: A, T, G, C). The sequence of these letters is the code.
- Replication (making copies): Before a cell divides, DNA makes an exact copy of itself. This ensures that each daughter cell receives a complete set of instructions. This is why children inherit traits from their parents—the DNA blueprint is passed down.
- Transmission to offspring: DNA is the molecule of heredity. It is passed from parents to offspring, carrying the genetic information that determines everything from eye colour to susceptibility to certain diseases.
DNA never leaves the nucleus of a cell. It is too precious and too large to move around. It stays safely inside, like a reference book that cannot be taken out of the library.
2. RNA – The Working Copy (Execution of the Blueprint)
RNA is the temporary, mobile copy of specific parts of the DNA blueprint. It is like a photocopy of a single page from the master plan, which a worker can carry to the factory floor. Its functions are:
- Transcription (copying the message): A specific segment of DNA (a gene) is used as a template to make a complementary RNA molecule. This RNA copy is called messenger RNA (mRNA).
- Translation (reading the message to build a protein): The mRNA travels out of the nucleus to the ribosomes (the protein-building machines). Here, another type of RNA called transfer RNA (tRNA) reads the mRNA code and brings the correct amino acids, one by one, to build a protein chain.
- Catalysis (as a biological catalyst): Some RNA molecules, called ribozymes, can act as enzymes and speed up chemical reactions. This is a less well-known but crucial function, especially in the ribosome itself (which is partly made of RNA).
Why This Matters for You
Even if you never touch a test tube, understanding nucleic acid functions helps you grasp:
- Why children resemble their parents: DNA is the hereditary material.
- How vaccines work: Many vaccines use mRNA to instruct your cells to produce a harmless piece of a virus, training your immune system.
- How genetic disorders arise: A mistake in the DNA sequence (a mutation) can lead to a faulty protein, causing diseases like sickle cell anaemia or cystic fibrosis. …
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)
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. …
Part (a)
The scientist is Frederick Griffith (1928), who worked with the bacterium Streptococcus pneumoniae and mice.
- The virulent S-strain (smooth, capsulated) kills mice; the avirulent R-strain (rough, non-capsulated) does not.
- Heat-killed S-strain injected alone does not kill mice.
- But heat-killed S + live R injected together killed the mice, and live S-strain bacteria were recovered from them. …
Part (a): Frederick Griffith's transformation experiment on Streptococcus pneumoniae showed a stable transforming principle (later shown to be DNA) passes from heat-killed S to live R bacteria, converting them to virulent S.
Part (b): selfing the tall-violet plant reveals its genotype — all tall violet ⇒ TTVV; 3:1 tall:dwarf ⇒ TtVV; 3:1 violet:white ⇒ TTVv; and 9:3:3:1 ⇒ TtVv.
Part (a)
Scientist and experiment. The early experiment demonstrating the stability (and transferability) of the genetic material was performed by Frederick Griffith in 1928, using the bacterium Streptococcus pneumoniae (which causes pneumonia) and living mice. He worked with two strains:
- the S-strain — smooth, capsulated and virulent (kills mice);
- the R-strain — rough, non-capsulated and avirulent (does not kill mice).
His observations:
- Mice injected with live S-strain died.
- Mice injected with live R-strain lived.
- Mice injected with heat-killed S-strain lived (the heat killed the bacteria).
- Remarkably, mice injected with a mixture of heat-killed S-strain + live R-strain died, and living S-strain bacteria were recovered from these dead mice. …
Showing the 12 most recent of 23 on this concept.
- CBSE 2026Set 57/1/11 markMCQQ.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.
ImportantThe 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: …
- CBSE 2026Set 57/2/11 markMCQQ.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.
ImportantThe 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 …
- CBSE 2026Set ANNUAL1 markMCQQ.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 …
- CBSE 2025Set 57/6/11 markMCQQ.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. …
- CBSE 2025Set 57/6/11 markMCQQ.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). …
- CBSE 2025Set ANNUAL1 markQ.What kind of charge is present on DNA molecule ?
›Reveal solutionSolution
The phosphate groups in the sugar-phosphate backbone give DNA its negative charge.
Each nucleotide in the DNA backbone contains a phosphate group linked to the 5' and 3' carbons of adjacent deoxyribose sugars via phosphodiester bonds. At physiological pH, these phosphate groups are ionised (each losing a proton), leaving a negatively charged oxygen. Since this repeats along the entire length of both strands, …
- CBSE 2025Set ANNUAL1 markQ.How many types of RNA are there?
›Reveal solutionSolution
RNA occurs in three functional forms — m-RNA, r-RNA and t-RNA — each with a distinct role in protein synthesis.
Unlike DNA, which is essentially one molecule per chromosome, RNA exists in the cell as three chemically similar but functionally different types:
- Messenger RNA (m-RNA) — carries the genetic code copied from DNA (transcription) to the ribosome, specifying the sequence of amino acids to be joined. …
- CBSE 2025Set ANNUAL1 markQ.Who is responsible for heredity?
›Reveal solutionSolution
Heredity is controlled by genes, the functional units of DNA carried on the chromosomes.
Each chromosome in the nucleus of a cell is made of a very long DNA molecule. Specific segments of this DNA, called genes, code for particular traits/proteins. When a cell divides, DNA replicates and an exact copy of these genes is passed on to the daughter cells, and ultimately from parents to offspring during r …
- CBSE 2025Set ANNUAL1 markQ.Define heredity.
›Reveal solutionSolution
Heredity = passing on of hereditary characters from one generation to the next via genes.
Heredity is defined as the biological process by which physical and other characteristics (traits) of parents are transmitted to their offspring. This transmission takes place through genes, which are specific segments of the DNA molecule present on chromosomes, …
- CBSE 2024Set 57/1/11 markMCQQ.The type of bond represented by the dotted line '– – – – –' in a schematic polynucleotide chain is: [Schematic polynucleotide chain shown with P (phosphate), S (sugar) and B (base); dotted lines connect S to B.] (A) Hydrogen bond (B) Peptide bond (C) N-glycosidic linkage (D) Phosphodiester bond
›Reveal solutionSolution
The dotted line connecting sugar (S) to base (B) in a polynucleotide represents the N-glycosidic linkage, the covalent bond that attaches nitrogenous bases to the pentose sugar. The answer is (C).
A polynucleotide chain has three repeating components: a phosphate group, a pentose sugar, and a nitrogenous base. Understanding how these connect is fundamental to DNA and RNA structure.
The sugar-phosphate backbone forms the structural spine of nucleic acids, with phosphodiester bonds linking one sugar's 5′ carbon to the next sugar's 3′ carbon through a phosphate group. But the bases — the information-carrying units — must attach to this backbone somehow. That attachment is what we're identifying here.
The N-glycosidic linkage is a covalent bond between the anomeric carbon (C1′) of the pentose sugar and a nitrogen atom of the nitrogenous base. In purines (adenine, guanine), this nitrogen is N9; in pyrimidines (cytosine, thymine, uracil), it's N1. This bond is called "glycosidic" because it resembles the linkage in carbohydrates, and "N-" specifies that it involves a nitrogen atom rather than oxygen.
Let's eliminate the other options systematically:
-
Hydrogen bonds (A) are weak, non-covalent interactions that hold complementary base pairs together across the two strands of a DNA double helix (A–T, G–C). They do NOT attach bases to the sugar within a single strand. The schematic shows a single polynucleotide chain, not inter-strand pairing.
-
Peptide bonds (B) link amino acids in proteins through a C–N bond between a carboxyl group and an amino group. Nucleic acids contain no peptide bonds — this is a protein-specific linkage. …
-
- CBSE 2024Set 57/1/11 markMCQQ.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.
ImportantThe 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. …
- CBSE 2024Set 57/2/11 markMCQQ.In the double helical structure of DNA molecule, the strands are : (A) identical and complementary (B) identical and non-complementary (C) anti-parallel and complementary (D) anti-parallel and non-complementary
›Reveal solutionSolution
DNA's double helix features two strands running in opposite directions (anti-parallel) with bases pairing by Watson-Crick rules (complementary). The answer is (C).
Why DNA strands must be both anti-parallel and complementary
The architecture of DNA isn't arbitrary—it's dictated by the chemistry of how nucleotides bond and how bases recognize each other. Understanding these two properties separately, then seeing why they must coexist, reveals the elegance of the double helix.
The complementarity principle
When Watson and Crick solved DNA's structure in 1953, the breakthrough was recognizing that bases pair in a specific way: adenine (A) always pairs with thymine (T) through two hydrogen bonds, while guanine (G) pairs with cytosine (C) through three hydrogen bonds. This isn't random preference—it's geometric necessity.
The purine bases (A and G, with their double-ring structure) are larger than the pyrimidine bases (T and C, single-ring). If two purines tried to pair, they'd be too bulky and distort the helix. If two pyrimidines paired, they'd be too small to bridge the gap. Only purine-pyrimidine pairs maintain the uniform diameter of the helix (about 2 nm).
Beyond size, the hydrogen-bonding patterns are specific:
- A and T have exactly the right donor and acceptor groups to form two stable H-bonds
- G and C form three H-bonds in perfect alignment
- Other combinations either can't form enough bonds or have steric clashes
This means if one strand reads 5'-ATGC-3', the other must read 3'-TACG-5' to satisfy base-pairing rules. The strands are complementary, not identical.
The anti-parallel orientation
Now consider the sugar-phosphate backbone. Each nucleotide has a deoxyribose sugar with a phosphate group attached to its 5' carbon and the next nucleotide's sugar attached via its 3' carbon. This creates directionality: one end of a strand has a free 5' phosphate, the other a free 3' hydroxyl.
In the double helix, the two strands run in opposite directions—one goes 5' → 3' while its partner goes 3' → 5'. This anti-parallel arrangement is required because:
- The geometry of base pairing only works when the glycosidic bonds (connecting base to sugar) are positioned correctly relative to each other
- The major and minor grooves of the helix form only when strands are anti-parallel
- The hydrogen bonds between bases align properly only in this configuration
If the strands were parallel (both 5' → 3'), the bases couldn't pair—the geometry would be all wrong, with the glycosidic bonds pointing in incompatible directions. …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.