Q.Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Match the following A- Anticodon of methionine B- Anticodon of tryptophan C- Anticodon of tyrosine D- Anticodon of serine (A) A-I, B-IV, C-II, D-III (B) A-IV, B-III, C-I, D-II (C) A-IV, B-III, C-II, D-I (D) A-IV, B-I, C-III, D-II

   ›Reveal solutionSolution

   The anticodon is the tRNA triplet complementary to the mRNA codon, pairing with it antiparallel and conventionally written in the 5′→3′ direction. Matching each amino acid's codon to its anticodon gives: A-IV, B-III, C-II, D-I, so the correct option is (C).

   The key idea is that the anticodon on a tRNA molecule pairs with the mRNA codon during translation. The codon is written 5′→3′; the anticodon binds it antiparallel, so once you write the complementary bases you must reverse their order to express the anticodon in the standard 5′→3′ direction.

   1. Methionine (Met) has a single codon: 5′-AUG-3′. Complementary bases (A↔U, U↔A, G↔C) paired antiparallel give 3′-UAC-5′; reversed to the standard 5′→3′ direction, the anticodon is 5′-CAU-3′. That matches option IV (CAU).

   2. Tryptophan (Trp) has a single codon: 5′-UGG-3′. Complement: 3′-ACC-5′; reversed: 5′-CCA-3′. That matches option III (CCA).

   3. Tyrosine (Tyr) has two codons, 5′-UAU-3′ and 5′-UAC-3′, differing only at the third (wobble) position. A single tRNA can read both via a G at the anticodon's wobble position (G pairs with both C and U). For the codon 5′-UAC-3′: the complement is 3′-AUG-5′, reversed to give 5′-GUA-3′ — the G at the wobble position also lets this same tRNA read the UAU codon. That matches option II (GUA).

   4. Serine (Ser) is six-fold degenerate; for the codon 5′-UCA-3′, the complement is 3′-AGU-5′, reversed to 5′-UGA-3′. That matches option I (UGA).

   Putting the matches together:

   • A (Met) → IV (CAU)
   • B (Trp) → III (CCA)
   • C (Tyr) → II (GUA)
   • D (Ser) → I (UGA)

   That corresponds to option (C).

   Watch out

   A common mistake is to write the anticodon as the direct complement of the codon without reversing the direction. For AUG, the direct complement is UAC, but the actual anticodon is CAU, because the tRNA pairs with the codon antiparallel.

   Tip

   Quick method: write the codon 5′→3′, replace each base with its complement (A↔U, C↔G), then reverse the resulting sequence. That gives the anticodon in the standard 5′→3′ direction.

   ✓Final answer

   The correct option is (C).

2. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Units of marker used in creation of three domains are (A) Amino acids (B) DNA (C) 16s rRNA (D) Monosaccharides

   ›Reveal solutionSolution

   The three-domain system of classification relies on comparing the sequence of a specific ribosomal RNA molecule. The marker used for creating the three domains is 16S rRNA.

   The classification of life into three domains—Archaea, Bacteria, and Eukarya—represents a fundamental shift in our understanding of evolutionary relationships. This system, proposed by Carl Woese and his colleagues in the 1970s, moved beyond traditional morphological comparisons, especially for microorganisms, and instead relied on molecular evidence. The core idea was to find a molecule present in all forms of life that evolves slowly enough to retain historical information, yet fast enough to show differences between distinct lineages.

   Why 16S rRNA is the ideal marker

   Ribosomal RNA (rRNA) molecules are crucial components of ribosomes, the cellular machinery responsible for protein synthesis. Among these, the small subunit ribosomal RNA (SSU rRNA), specifically the 16S rRNA in prokaryotes and 18S rRNA in eukaryotes, proved to be an excellent phylogenetic marker for several key reasons:

   1. Ubiquitous Presence: 16S rRNA is found in all cellular life forms, making it possible to compare organisms across the entire tree of life.
   2. Essential Function: Because rRNA plays a vital, non-redundant role in protein synthesis, its structure and sequence are highly conserved. Significant changes would likely be lethal, meaning it evolves slowly and retains ancient evolutionary signals.
   3. Conserved and Variable Regions: The 16S rRNA molecule contains regions that are highly conserved across all life (allowing for alignment and comparison) as well as regions that are more variable (allowing for differentiation between distinct groups). This combination makes it powerful for both deep evolutionary comparisons and distinguishing closely related species.
   4. Appropriate Size: The 16S rRNA molecule is large enough (approximately 1,500 nucleotides) to contain sufficient phylogenetic information for robust analysis, yet small enough to be sequenced and analyzed efficiently.

   By comparing the sequences of 16S rRNA, Woese discovered that prokaryotes were not a single, unified group but comprised two distinct lineages: Bacteria and Archaea, which are as different from each other as they are from Eukaryotes.

   Here's a step-by-step breakdown of why 16S rRNA is the correct answer and why other options are not:

   1. Understanding the "Three Domains" Concept: Before the three-domain system, life was often classified into five kingdoms (Monera, Protista, Fungi, Plantae, Animalia). However, this system struggled with the diversity and relationships among microorganisms, particularly prokaryotes. Carl Woese's work revealed that the "Monera" kingdom was actually composed of two fundamentally different groups, leading to the establishment of the three domains: Bacteria, Archaea, and Eukarya.

   2. The Need for a Molecular Marker: Traditional classification based on observable traits (morphology, metabolism) was insufficient for understanding the deep evolutionary relationships, especially among microorganisms that often look similar but are genetically very different. A molecular "clock" was needed—a molecule whose sequence could be compared across diverse organisms to infer their evolutionary history.

   3. Carl Woese's Discovery: In the 1970s, Carl Woese pioneered the use of ribosomal RNA sequences for phylogenetic analysis. He chose rRNA because of its universal presence and critical function, which ensures its sequence evolves at a relatively constant rate.

   4. Why 16S rRNA is the Marker: Woese specifically focused on the small subunit ribosomal RNA (16S rRNA in prokaryotes, 18S rRNA in eukaryotes). By comparing the sequences of this molecule from various organisms, he was able to construct a phylogenetic tree that revealed the three distinct domains of life. The sequence differences in 16S rRNA provided the molecular evidence to separate Archaea from Bacteria and Eukarya.

   5. Evaluating Other Options:

      • (A) Amino acids: Amino acids are the building blocks of proteins. While protein sequences can be used for phylogenetic analysis, specific proteins are not as universally conserved or as consistently used for the initial establishment of the three domains as 16S rRNA. Also, the question asks for "units of marker," and amino acids are monomers, not the primary molecular marker itself in this context.
      • (B) DNA: DNA is the genetic material, and the 16S rRNA sequence is encoded by DNA. However, the marker itself, the molecule whose sequence is directly compared for this classification, is the RNA molecule. Comparing entire genomes was not feasible when the three-domain system was established, and even now, specific gene sequences (like the 16S rRNA gene) are often used for phylogenetic studies rather than whole genomes for broad classification. The question refers to the "units of marker," implying the molecule being analyzed.
      • (D) Monosaccharides: Monosaccharides are simple sugars, the building blocks of carbohydrates. Carbohydrate structures are highly variable and do not serve as a reliable universal phylogenetic marker for deep evolutionary relationships.

   ✓Final answer

   The units of marker used in the creation of the three domains are (C) 16S rRNA.

3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Which of the following is correct answer regarding the structure of a section of cilia / flagella? (A) Peripheral microtubules (doublets): $9+0$ \quad Central microtubules (singlets): $2$ \quad Radial spokes: $8$ \quad Central sheath: $1$ (B) Peripheral microtubules (doublets): $9+2$ \quad Central microtubules (singlets): $9+0$ \quad Radial spokes: $9$ \quad Central sheath: $2$ (C) Peripheral microtubules (doublets): $9$ \quad Central microtubules (singlets): $2$ \quad Radial spokes: $9$ \quad Central sheath: $1$ (D) Peripheral microtubules (doublets): $3$ \quad Central microtubules (singlets): $6$ \quad Radial spokes: $9$ \quad Central sheath: $1$

   ›Reveal solutionSolution

   The classic “9+2” arrangement of cilia/flagella refers to 9 peripheral doublet microtubules and 2 central singlet microtubules, with 9 radial spokes and 1 central sheath. The correct option is (C).

   The question tests your recall of the standard ultrastructure of cilia and flagella — a topic that appears in cell biology (eukaryotic cells). The “9+2” pattern is one of the most iconic motifs in biology, but the numbers in the options go beyond just the microtubules: they also ask about radial spokes and the central sheath. You need to know each component’s count precisely.

   Let’s break it down piece by piece.

   1. Peripheral microtubules (doublets)

      In a cross-section of a cilium or flagellum, you see a ring of 9 pairs of microtubules around the outside. Each pair is a doublet (one complete A-tubule and one partial B-tubule). So the number of peripheral doublets is 9, not 9+0 or 9+2 — those notations describe the overall pattern. The “9” here is simply the count of doublets.

   2. Central microtubules (singlets)

      At the very center of the ring, there are two single microtubules (not doublets), arranged side by side. These are the central pair. So the number of central singlet microtubules is 2.

   3. Radial spokes

      Each peripheral doublet is connected to the central sheath by a radial spoke. Since there are 9 doublets, there are 9 radial spokes — one per doublet. (Each spoke has a head that contacts the central sheath.)

   4. Central sheath

      The two central microtubules are surrounded by a single, continuous central sheath (sometimes called the inner sheath). It is 1 structure, not two.

   Now match these numbers to the options:

   • Option (A): Peripheral = 9+0 (wrong — that’s the pattern for centrioles, not cilia), Central = 2 (correct), Radial spokes = 8 (wrong), Central sheath = 1 (correct).
   • Option (B): Peripheral = 9+2 (misleading — that’s the overall pattern, not the count of doublets), Central = 9+0 (wrong), Radial spokes = 9 (correct), Central sheath = 2 (wrong).
   • Option (C): Peripheral = 9 (correct), Central = 2 (correct), Radial spokes = 9 (correct), Central sheath = 1 (correct).
   • Option (D): Peripheral = 3 (wrong), Central = 6 (wrong), Radial spokes = 9 (correct), Central sheath = 1 (correct).

   Only option (C) gives all four numbers correctly.

   Watch out

   A common mistake is to confuse the “9+2” notation with the count of peripheral doublets. The “9” in “9+2” is the number of peripheral doublets, and the “2” is the number of central singlets — so the correct counts are simply 9 and 2, not “9+2” as a single entry.

   Tip

   To remember: think of a wagon wheel — 9 spokes (radial spokes) connecting the rim (peripheral doublets) to the hub (central sheath around the 2 central microtubules). The hub has one sheath, not two.

   ✓Final answer

   The correct option is (C).

4. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Chemical nature of primers of DNA / RNA (A) Oligosaccharides (B) Oligonucleotides (C) Polypeptides (D) Disaccharides

   ›Reveal solutionSolution

   Primers are short nucleic-acid sequences that provide a free 3'-OH end for DNA polymerase to extend — they are oligonucleotides, not sugars or proteins.

   The question tests a basic but crucial fact about molecular biology: what are primers made of? In DNA replication, the enzyme DNA polymerase cannot start synthesis from scratch — it needs an existing strand with a free 3'-hydroxyl group to add nucleotides to. That starter strand is the primer.

   Primers are short sequences of nucleotides, typically around 10–30 bases long. In living cells, the primer is made of RNA (synthesized by primase), but in laboratory techniques like PCR, we use synthetic DNA primers. Either way, the chemical building blocks are nucleotides — so the primer is a short nucleic acid chain.

   1. Oligosaccharides are short chains of sugars (e.g., maltose, raffinose). They have nothing to do with genetic material — eliminate (A) and (D).

   2. Polypeptides are chains of amino acids — that's proteins. Primers are not proteins; they are nucleic acids. Eliminate (C).

   3. Oligonucleotides are short chains of nucleotides (the monomers of DNA and RNA). A primer, whether RNA or DNA, is exactly that: a short oligonucleotide. This matches (B).

   Watch out

   A common mistake is to think primers are "made of DNA" and therefore not oligonucleotides — but "oligonucleotide" simply means a short nucleic acid chain, which includes both DNA and RNA. The term is not limited to one type.

   ✓Final answer

   The correct option is (B) Oligonucleotides.

5. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Match the following Disorder Hormone A. Acromegaly I. Vasopressin B. Cretinism II. Somatotropin C. Diabetes insipidus III. Insulin D. Cushing's syndrome IV. Thyroxine V. Cortisol (A) A – II, B – IV, C – III, D – V (B) A – II, B – IV, C – I, D – V (C) A – V, B – I, C – IV, D – II (D) A – III, B – II, C – I, D – IV

   ›Reveal solutionSolution

   Each disorder is caused by a specific hormonal imbalance — matching them correctly requires knowing which hormone is deficient or in excess. The correct pairing is: Acromegaly → Somatotropin, Cretinism → Thyroxine, Diabetes insipidus → Vasopressin, Cushing’s syndrome → Cortisol.

   The question tests your grasp of endocrine disorders and their root hormonal causes. Instead of memorising lists, think about what each disorder actually does to the body — that tells you which hormone is involved.

   Acromegaly is the overgrowth of bones and tissues in adults, caused by too much growth hormone. Growth hormone is somatotropin. So A goes with II.

   Cretinism is a condition of stunted physical and mental development in children, due to severe deficiency of thyroid hormone. That hormone is thyroxine. So B goes with IV.

   Diabetes insipidus is not about blood sugar — it’s about excessive thirst and dilute urine. The problem is a lack of antidiuretic hormone (ADH), which is vasopressin. So C goes with I.

   Cushing’s syndrome results from prolonged exposure to high levels of cortisol, often from a pituitary tumour or steroid medication. So D goes with V.

   Now check the options. Only one matches A–II, B–IV, C–I, D–V: that’s option (B).

   Watch out

   A common mistake is confusing diabetes insipidus with diabetes mellitus. The latter involves insulin (III), but diabetes insipidus is about vasopressin (ADH). Don’t match C with III.

   ✓Final answer

   The correct option is (B).

6. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.Match the following List-1 (Codon) List-2 (Amino acid) A. AUG I. Tryptophan B. UAA II. Phenylalanine C. UUU III. End codon D. UGG IV. Methionine (A) A – IV, B – III, C – II, D – I (B) A – IV, B – III, C – I, D – II (C) A – III, B – IV, C – II, D – I (D) A – II, B – I, C – III, D – IV

   ›Reveal solutionSolution

   This question tests your knowledge of the genetic code, specifically identifying the amino acids or functions associated with given mRNA codons. The correct match is A-IV, B-III, C-II, D-I.

   The genetic code is a set of rules by which information encoded in genetic material (DNA or RNA sequences) is translated into proteins (amino acid sequences) by living cells. A codon is a sequence of three nucleotides that forms a unit of the genetic code in DNA or RNA. Each codon specifies a particular amino acid or signals the termination of protein synthesis. Understanding these specific assignments is fundamental to molecular biology.

   Here's how we match each codon to its corresponding amino acid or function:

   1. Analyze Codon A: AUG

      • AUG is universally known as the start codon. It initiates protein synthesis.
      • It codes for the amino acid Methionine (Met) in eukaryotes and formylmethionine in prokaryotes.
      • Therefore, A (AUG) matches with IV (Methionine).

   2. Analyze Codon B: UAA

      • UAA is one of the three stop codons (the others being UAG and UGA).
      • Stop codons do not code for any amino acid; instead, they signal the termination of translation (protein synthesis). They are also referred to as "nonsense codons" or "end codons".
      • Therefore, B (UAA) matches with III (End codon).

   3. Analyze Codon C: UUU

      • UUU is a codon that specifies the amino acid Phenylalanine (Phe).
      • Another codon, UUC, also codes for Phenylalanine. This is an example of the degeneracy of the genetic code, where multiple codons can code for the same amino acid.
      • Therefore, C (UUU) matches with II (Phenylalanine).

   4. Analyze Codon D: UGG

      • UGG is a codon that specifies the amino acid Tryptophan (Trp).
      • Tryptophan is unique in that it is coded by only a single codon (UGG), unlike most other amino acids which have multiple codons.
      • Therefore, D (UGG) matches with I (Tryptophan).

   5. Consolidate the matches and select the correct option:

      • A $\rightarrow$ IV
      • B $\rightarrow$ III
      • C $\rightarrow$ II
      • D $\rightarrow$ I

      Comparing this with the given options:

      (A) A – IV, B – III, C – II, D – I

      (B) A – IV, B – III, C – I, D – II

      (C) A – III, B – IV, C – II, D – I

      (D) A – II, B – I, C – III, D – IV

      The correct option is (A).

   ✓Final answer

   The correct option is (A).

7. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Chemical bonds cut by restriction endonucleases (A) Hydrogen bonds (B) Glycosidic bonds (C) Phosphodiester bonds (D) Peptide bonds

   ›Reveal solutionSolution

   Restriction endonucleases cut DNA at specific sequences by breaking the phosphodiester bonds between adjacent nucleotides in the sugar-phosphate backbone.

   Restriction endonucleases are enzymes that act like molecular scissors. Their job is to recognize a specific short sequence of DNA (usually 4–8 base pairs long) and make a cut within or near that sequence. To understand which bond they break, you need to picture the structure of a DNA strand.

   DNA is a polymer of nucleotides. Each nucleotide has three parts: a phosphate group, a deoxyribose sugar, and a nitrogenous base. The backbone of each strand is built from alternating sugar and phosphate molecules, linked together by phosphodiester bonds. These are covalent bonds formed between the 3' carbon of one sugar and the 5' carbon of the next sugar's phosphate group. The two strands of the double helix are held together by hydrogen bonds between complementary bases.

   Now, what does a restriction enzyme actually do? It cuts the DNA backbone. It does not unzip the strands or break the base pairing. It severs the covalent linkage that holds the sugar-phosphate chain together. That linkage is the phosphodiester bond.

   Let’s go through the options:

   1. Hydrogen bonds – These hold the two strands together (A–T and G–C pairs). Restriction enzymes do not break these; they cut each strand individually. Hydrogen bonds are weak and can be broken by heat or changes in pH, not by restriction enzymes.

   2. Glycosidic bonds – These connect the nitrogenous base to the deoxyribose sugar (forming a nucleoside). Breaking this would remove the base from the sugar, which is not what restriction enzymes do. They leave the bases intact.

   3. Phosphodiester bonds – This is the bond between the phosphate group of one nucleotide and the sugar of the next. It forms the continuous backbone. Restriction enzymes hydrolyze this specific covalent bond, creating a break in the DNA strand. This is the correct answer.

   4. Peptide bonds – These link amino acids together in proteins. DNA has no peptide bonds, so this is irrelevant.

   Watch out

   A common mistake is to think restriction enzymes cut hydrogen bonds because they "cut DNA." But cutting DNA means breaking the backbone, not separating the strands. Hydrogen bonds are broken by helicases during replication, not by restriction enzymes.

   Tip

   To remember: Restriction enzymes cut the sugar-phosphate backbone — that's the phosphodiester bond. The bases and their hydrogen bonds remain untouched until the DNA is denatured.

   ✓Final answer

   The correct option is (C) Phosphodiester bonds.

8. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.Dual functions of codon AUG (A) Initiator codon and code for phenylalanine (B) Universal codon and code for phenylalanine (C) End codon and code for methionine (D) Initiator codon and code for methionine

   ›Reveal solutionSolution

   The codon AUG has two distinct roles: it serves as the start (initiator) codon that signals the beginning of translation, and it codes for the amino acid methionine. The correct option is (D).

   The genetic code is a set of rules that maps each three-nucleotide sequence (codon) in mRNA to a specific amino acid or a stop signal. Most codons have a single, unambiguous meaning — they code for one amino acid and nothing else. AUG is a special exception: it is the only codon that carries out a dual function.

   Why does this duality exist? In translation, the ribosome needs a clear signal to know where to start reading the mRNA. That signal is the start codon, almost always AUG. At the same time, the cell must actually insert an amino acid at that position to begin the polypeptide chain. Nature solved this by making AUG code for methionine, which becomes the first amino acid of every newly synthesized protein. So AUG is both a signal (initiator) and a code (for methionine).

   Let’s walk through the options to see why only one fits.

   1. Option (A) — Initiator codon and code for phenylalanine

      Phenylalanine is coded by UUU and UUC, not by AUG. This is factually wrong. The initiator role is correct, but the amino acid assignment is not.

   2. Option (B) — Universal codon and code for phenylalanine

      "Universal codon" is a vague term — the genetic code is nearly universal across organisms, but that applies to all codons, not just AUG. More importantly, AUG does not code for phenylalanine. So this option fails on both counts.

   3. Option (C) — End codon and code for methionine

      End (stop) codons are UAA, UAG, and UGA — they signal termination and do not code for any amino acid. AUG is a start codon, not a stop codon. This is completely incorrect.

   4. Option (D) — Initiator codon and code for methionine

      This is exactly right. AUG is the primary start codon in both prokaryotes and eukaryotes, and it codes for methionine (or formylmethionine in bacteria, which is a modified methionine). No other codon has this dual role.

   Watch out

   A common mistake is to confuse the initiator codon with the idea that it codes for a "special" amino acid. It does not — it codes for methionine, the same amino acid that internal AUG codons also code for. The special part is its position at the start of the reading frame, not the amino acid itself.

   Tip

   In eukaryotes, the first methionine is often removed later by a processing enzyme, so the final protein may not start with methionine. But during translation, the first amino acid inserted is always methionine (coded by AUG).

   ✓Final answer

   The correct option is (D): AUG functions as an initiator codon and codes for methionine.

9. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.Match the following A. Bacteriophage $\emptyset \times 174$ I. $3.3 \times 10^9$ base pairs B. Bacteriophage Lambda II. $4.6 \times 10^6$ base pairs C. Escherichia coli III. 48502 base pairs D. Human DNA haploid IV. 5386 Nucleotides The correct answer is (A) A – IV, B – III, C – II, D – I (B) A – IV, B – III, C – I, D – II (C) A – III, B – II, C – IV, D – I (D) A – II, B – IV, C – III, D – I

   ›Reveal solutionSolution

   Genome sizes climb steeply with complexity: $\phi$X174 = 5386 nucleotides, phage $\lambda$ = 48502 bp, E. coli = $4.6\times10^6$ bp, human haploid = $3.3\times10^9$ bp. That gives A–IV, B–III, C–II, D–I — option (A).

   The concept first: the ladder of genome sizes

   These four numbers are among the most quotable in all of molecular biology, and they are memorable precisely because they form a clean staircase, each step roughly a thousand-fold above the last:

   $\underbrace{5.4 \times 10^3}_{\phi\text{X174}} \ \ll \ \underbrace{4.9 \times 10^4}_{\lambda} \ \ll \ \underbrace{4.6 \times 10^6}_{E.\ coli} \ \ll \ \underbrace{3.3 \times 10^9}_{\text{human haploid}}$

   The biological logic is simple: the more an organism has to do, the more genes it needs, and the more DNA it must carry. A virus outsources almost all its machinery to the host, so it needs only a handful of genes. A free-living bacterium must build its own ribosomes, metabolic enzymes and cell wall, so it needs a few thousand genes. A human must additionally build a nervous system, an immune system and a developmental programme — and carries vast amounts of non-coding DNA besides.

   So before you even look at the options, you can be confident that the ordering of the match must be:

   $\text{smallest} \to \phi\text{X174} \ < \ \lambda \ < \ E.\ coli \ < \ \text{human} \to \text{largest}$

   Half the battle is already won. Now attach the numbers.

   Step-by-step matching

   Step 1 — A. Bacteriophage $\phi$X174 $\rightarrow$ IV (5386 nucleotides).

   This phage has a special place in history: in 1978 Frederick Sanger sequenced its genome, the first genome ever fully sequenced. Two details lock the match in:

   • Its genome is single-stranded DNA, which is exactly why item IV is quoted in "nucleotides" and not in "base pairs" — with one strand there are no pairs. That unit is a deliberate and honest clue planted by the examiner.
   • The figure 5386 is small enough that it can belong to nothing else in the list.

   Step 2 — B. Bacteriophage Lambda $\rightarrow$ III (48502 base pairs).

   Phage $\lambda$ is the celebrated double-stranded DNA phage of E. coli, the workhorse of the lysogeny/lysis story and a classic cloning vector. Its genome, 48502 bp, is about nine times larger than $\phi$X174's — still a virus, but a far more elaborate one. Note the unit is now correctly "base pairs", consistent with double-stranded DNA.

   Step 3 — C. Escherichia coli $\rightarrow$ II ($4.6 \times 10^6$ base pairs).

   E. coli is a free-living bacterium. Its single circular chromosome contains about $4.6 \times 10^6$ bp ($\approx 4.6$ million), coding for roughly 4,000 genes. This is about a hundred times larger than phage $\lambda$ — the jump from "parasite of a cell" to "a whole cell".

   Step 4 — D. Human DNA (haploid) $\rightarrow$ I ($3.3 \times 10^9$ base pairs).

   The haploid human genome — the DNA in a single set of 23 chromosomes, i.e. what one gamete carries — is about $3.3 \times 10^9$ bp (3.3 billion, often quoted as "about 3 billion"). A diploid somatic cell contains twice this, $\approx 6.6 \times 10^9$ bp, which is why the question is careful to specify haploid.

   Step 5 — Assemble the answer.

   $\text{A} - \text{IV}, \quad \text{B} - \text{III}, \quad \text{C} - \text{II}, \quad \text{D} - \text{I}$

   Notice the pleasing structure: the match is a perfect reversal — A pairs with the last item, D with the first. That is the examiner's small joke, and it is also a useful self-check.

   Step 6 — Locate this in the printed options.

   • (A) A – IV, B – III, C – II, D – I ✓ — exactly our result.
   • (B) A–IV, B–III, C–I, D–II — swaps E. coli and human, i.e. claims a bacterium has 3.3 billion bp and a human only 4.6 million. Biologically absurd.
   • (C) A–III, B–II, C–IV, D–I — gives $\phi$X174 the $\lambda$ figure and gives E. coli only 5386 nucleotides. Wrong.
   • (D) A–II, B–IV, C–III, D–I — gives the tiny phage $\phi$X174 a bacterial-sized genome. Wrong.

   The takeaway

   Commit these four numbers as a ladder, not as four isolated facts — 5386 nt $\to$ 48502 bp $\to$ 4.6 Mb $\to$ 3.3 Gb — and any permutation of this match-the-following becomes instantly answerable. And remember the unit tell-tale: "nucleotides" signals a single-stranded genome ($\phi$X174), while "base pairs" signals double-stranded DNA.

   ✓Final answer

   Genome sizes rise from $\phi$X174 (5386 nucleotides) to phage $\lambda$ (48502 bp) to E. coli ($4.6 \times 10^6$ bp) to the haploid human genome ($3.3 \times 10^9$ bp), giving the match A–IV, B–III, C–II, D–I.

   ANSWER: A

10. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
    Q.Select the viruses in the same order based on the morphology given below. • Spikes • Roughly spherical envelop • Rigid long rods (A) Rabies virus, Polio virus, Measles virus (B) Polio virus, Herpes simplex virus, Rabies virus (C) Measles virus, Influenza virus, Rabies virus (D) Rabies virus, Tobacco mosaic virus, Polio virus

    ›Reveal solutionSolution

    Matching the three morphological descriptions in order — spikes = measles virus, roughly spherical envelope = influenza virus, rigid long rods = rabies virus — gives option (C).

    The concept first: what actually determines a virus's shape

    A virion is a nucleic acid wrapped in a protein capsid, sometimes with a lipid envelope stolen from the host membrane. Two structural facts decide what it looks like:

    1. Capsid symmetry.
       • Helical capsids — the protein subunits spiral around the nucleic acid, producing rods (rigid, like tobacco mosaic virus) or flexible filaments.
       • Icosahedral / polyhedral capsids — subunits close into a near-sphere, producing the polyhedral, roughly spherical particles (poliovirus is the standard example, and it is naked, i.e. has no envelope).
    2. Envelope and its projections. If the virus buds through a host membrane it acquires an envelope studded with virus-coded glycoprotein spikes — these are what the description "spikes" refers to.

    So "spikes" $\Rightarrow$ enveloped virus with prominent projections; "roughly spherical envelope" $\Rightarrow$ enveloped, approximately spherical particle; "rigid long rods" $\Rightarrow$ elongated, rod-like virion.

    Step-by-step matching

    1. Spikes $\rightarrow$ Measles virus. Measles is an enveloped paramyxovirus whose envelope bears prominent glycoprotein projections (the haemagglutinin and fusion proteins). It is described morphologically by its spikes.
    2. Roughly spherical envelope $\rightarrow$ Influenza virus. Influenza virions are described as roughly spherical enveloped particles.
    3. Rigid long rods $\rightarrow$ Rabies virus. Rabies belongs to the family Rhabdoviridae, and the family is named for its rod shape (Greek rhabdos, a rod) — an elongated, rigid, rod/bullet-shaped virion. It is the only rod-like virus among the ones on offer in this option.
    4. Why the other options fail. Poliovirus is a small naked, polyhedral virus — it has no envelope and no spikes, so any option assigning "spikes" or "roughly spherical envelope" to polio is structurally wrong. Likewise, assigning "rigid long rods" to polio or to measles contradicts their icosahedral/pleomorphic-enveloped structure.
    5. In the given order (spikes, roughly spherical envelope, rigid long rods) we get Measles virus, Influenza virus, Rabies virus.

    ✓Final answer

    The three morphologies correspond, in order, to measles virus, influenza virus and rabies virus. The correct option is (C).

    ANSWER: C

11. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
    Q.Chromosomal DNA with 40% Adenine and 60% Guanine replicated during S phase of interphase. Then ratio of Thymine and Cytosine in that DNA after replication respectively (A) 1:1 (B) 3:2 (C) 2:3 (D) 2:4

    ›Reveal solutionSolution

    Chargaff’s rule tells us that in double-stranded DNA, A = T and G = C. Given 40% A and 60% G, the original strand has T = 40% and C = 60%. After replication, each daughter DNA is identical to the parent, so the T:C ratio remains 40:60 = 2:3. The correct option is (C).

    The key here is Chargaff’s rule — a cornerstone of DNA structure. In any double-stranded DNA, the number of adenine (A) bases equals the number of thymine (T) bases, and guanine (G) equals cytosine (C). This is because A pairs with T via two hydrogen bonds, and G pairs with C via three. So the percentages of A and T are always equal, and those of G and C are always equal.

    The problem gives you the percentages of A and G in the chromosomal DNA before replication. That DNA is double-stranded, so we can immediately find T and C using Chargaff’s rule. Then replication during S phase produces two identical daughter DNA molecules — each with the same base composition as the parent. So the ratio of T to C in the replicated DNA is the same as in the parent.

    Let’s work it out step by step.

    1. Find T and C in the parent DNA.

       Given: A = 40%, G = 60%.

       By Chargaff’s rule:

       • T = A = 40%
       • C = G = 60% So the parent DNA has T = 40% and C = 60%.

    2. Understand what replication does.

       During S phase, the parent DNA unwinds and each strand serves as a template for a new complementary strand. The result is two daughter DNA molecules, each identical in sequence (and therefore in base composition) to the parent. So each daughter DNA also has A = 40%, T = 40%, G = 60%, C = 60%.

    3. Find the required ratio.

       The question asks for the ratio of Thymine to Cytosine after replication. That’s T : C = 40% : 60% = 40 : 60 = 2 : 3.

    Watch out

    A common mistake is to think that replication changes the base percentages — for example, that the new strands have different compositions. But replication is semi-conservative: each daughter DNA gets one old strand and one new strand, and the new strand is complementary to the old one. So the overall percentages in each daughter DNA are exactly the same as in the parent.

    Tip

    You don’t need to know the total number of bases — percentages are enough. Chargaff’s rule works with percentages directly because the total is 100%.

    ✓Final answer

    The ratio of Thymine to Cytosine after replication is 2:3, which corresponds to option (C).

12. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
    Q.Match the following List I A) Serine B) Methionine C) Lysine D) Glycine List II I) AUG II) GGG III) UCU IV) AAA (A) A-II, B-I, C-IV, D-III (B) A-II, B-IV, C-I, D-III (C) A-III, B-I, C-IV, D-II (D) A-III, B-IV, C-I, D-II

    ›Reveal solutionSolution

    This question requires matching specific amino acids to their corresponding mRNA codons based on the genetic code. The correct match is Serine-UCU, Methionine-AUG, Lysine-AAA, and Glycine-GGG, leading to option (C).

    The central concept here is the genetic code, which dictates how sequences of nucleotides in mRNA are translated into sequences of amino acids to form proteins. This code is read in groups of three nucleotides, called codons. Each codon specifies a particular amino acid (or signals termination of translation).

    Understanding the genetic code is fundamental to molecular biology. While there are 64 possible codons ($4^3$), only 20 standard amino acids are commonly found in proteins. This means that most amino acids are specified by more than one codon, a property known as degeneracy of the genetic code. However, each codon is specific, meaning it codes for only one particular amino acid.

    To solve this problem, we need to recall the standard mRNA codons for each amino acid listed.

    1. Identify the codon for Serine (Ser):

       Serine is an amino acid that is coded by several codons. Among the options provided in List II, UCU is a known codon for Serine. Other codons for Serine include UCC, UCA, UCG, AGU, and AGC.

       Therefore, A matches with III.

    2. Identify the codon for Methionine (Met):

       Methionine is unique because it is coded by only one specific codon, AUG. This codon also serves as the start codon, signaling the beginning of protein synthesis.

       Therefore, B matches with I.

       Important

       The codon AUG has a dual function: it codes for Methionine and also acts as the start codon for translation.

    3. Identify the codon for Lysine (Lys):

       Lysine is an amino acid coded by two codons: AAA and AAG. From List II, AAA is available.

       Therefore, C matches with IV.

    4. Identify the codon for Glycine (Gly):

       Glycine is an amino acid coded by four codons: GGU, GGC, GGA, and GGG. From List II, GGG is available.

       Therefore, D matches with II.

    Combining these matches, we get:

    • A - III (Serine - UCU)
    • B - I (Methionine - AUG)
    • C - IV (Lysine - AAA)
    • D - II (Glycine - GGG)

    Now, let's compare this with the given options:

    (A) A-II, B-I, C-IV, D-III

    (B) A-II, B-IV, C-I, D-III

    (C) A-III, B-I, C-IV, D-II

    (D) A-III, B-IV, C-I, D-II

    Our derived mapping matches option (C).

    ✓Final answer

    The correct match between the amino acids and their mRNA codons is A-III, B-I, C-IV, D-II, which corresponds to option (C).