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Important Diagrams: Molecular Basis of Inheritance | Biology Class 12 - NEET PDF Download

1. Structure of Polynucleotide Chains

(a) Single-Stranded Polynucleotide Chain:

  • Components: Composed of a series of nucleotides, each consisting of a nitrogenous base, a pentose sugar (ribose in RNA, deoxyribose in DNA), and a phosphate group.
  • Nucleoside Formation: A nitrogenous base is linked to the 1'C of the pentose sugar via a N-glycosidic linkage, forming a nucleoside.
  • Nucleotide Formation: The addition of a phosphate group to the 5'C of the nucleoside via a phosphoester linkage results in a nucleotide.
  • Chain Formation: Nucleotides are connected through 3'-5' phosphodiester linkages to form a single polynucleotide chain.
  • Termini: The chain has a free phosphate group at the 5' end of the pentose sugar and a free hydroxyl group at the 3' end.
  • RNA Specificity: In RNA, each ribose sugar has an additional OH group at the 2' position.

Structure of Polynucleotide Chain: Single StrandStructure of Polynucleotide Chain: Single Strand

(b) Double-Stranded Polynucleotide Chain:

  • Backbone: Each strand has a backbone formed by alternating sugar and phosphate groups, with nitrogenous bases extending from the sugar.
  • Anti-Parallel Orientation: The two polynucleotide chains run in opposite directions (anti-parallel), one in a 5'->3' direction and the other in a 3'->5'.
  • Base Pairing: Nitrogenous bases from opposite strands pair via hydrogen bonds; Adenine (A) pairs with Thymine (T) with two hydrogen bonds, and Guanine (G) pairs with Cytosine (C) with three hydrogen bonds.
  • Uniformity and Helix Structure: The pairing of a purine with a pyrimidine maintains a uniform distance between the two strands, forming a right-handed helix.
  • Helical Pitch: The helix has a pitch of 3.4 nm with about 10 base pairs per turn, and the distance between adjacent base pairs is roughly 0.34 nm.
  • Stabilization: The helical structure is stabilized by the stacking of base pairs in addition to the hydrogen bonds.

Structure of Polynucleotide Chain : Double StrandedStructure of Polynucleotide Chain : Double Stranded

DNA Double HelixDNA Double Helix

2. Packaging of DNA Helix

DNA in both prokaryotic and eukaryotic cells is compactly organized despite its large length, which allows it to fit within the microscopic confines of a cell nucleus or nucleoid.

Prokaryotic DNA Packaging:

  • In prokaryotes like E. coli, the DNA is organized into a nucleoid.
  • The nucleoid contains the DNA in large loops held together by proteins, particularly those that are positively charged, which interact with the negatively charged DNA.
  • This organization helps to compact the DNA into a manageable structure without a defined nucleus.

Eukaryotic DNA Packaging:

  • Eukaryotic DNA is more complexly organized around histone proteins.
  • Histones, which are rich in positively charged amino acids like lysine and arginine, form an octamer around which DNA wraps to create a structure called a nucleosome.
  • Each nucleosome consists of about 200 base pairs of DNA.
  • These nucleosomes appear as "beads-on-string" under an electron microscope and form the repeating units of chromatin.
  • Chromatin fibers are further coiled and condensed during the metaphase stage of cell division to form chromosomes.
  • Additional non-histone chromosomal (NHC) proteins help in the further packing of chromatin.
  • Chromatin is differentiated into euchromatin, which is less densely packed and transcriptionally active, and heterochromatin, which is more densely packed and transcriptionally inactive.

NucleosomeNucleosome

EM Picture: Beads on StringEM Picture: Beads on String

3.  Hershey & Chase Experiment

  • The Hershey-Chase experiments in 1952 provided definitive evidence that DNA, not protein, is the genetic material. 
  • They conducted experiments with bacteriophages, viruses that infect bacteria, using two different radioactive markers: phosphorus and sulfur. 
  • Radioactive phosphorus was used to label DNA, as DNA contains phosphorus but not sulfur, which labels proteins. 
  • After allowing the radioactive phages to infect E. coli bacteria, they removed the viral coats through agitation in a blender and separated the virus particles from the bacteria using a centrifuge. 
  • The bacteria infected with phages containing radioactive DNA were themselves radioactive, indicating that DNA had entered the bacteria. 
  • Conversely, bacteria infected with phages that had radioactive proteins were not radioactive, demonstrating that proteins did not enter the bacteria. 
  • This experiment conclusively showed that DNA is the material that transfers genetic information in viruses, settling the debate over whether proteins or DNA were the genetic material.

Hershey and Chase ExperimentHershey and Chase Experiment

4. Semi Conservative DNA Replication ( Watson & Crick Model)

  • James Watson and Francis Crick, upon proposing the double helical structure of DNA in 1953, also introduced the concept of how DNA replicates. They noted that the specific base pairing in DNA suggested a mechanism for copying genetic material. 
  • According to their model, the two strands of the DNA double helix would separate, each serving as a template for the synthesis of a new complementary strand. 
  • This process would result in two DNA molecules, each consisting of one original (parental) strand and one newly synthesized strand. 
  • This method of DNA replication is known as semiconservative replication because each of the new DNA molecules retains one of the original strands, conserving half of the parental DNA molecule in each new DNA molecule formed.

Watson-Crick model for semiconservative DNA replicationWatson-Crick model for semiconservative DNA replication

5. Meselson and Stahl’s Experiment

Meselson and Stahl's experiment in 1958 provided definitive evidence for the semiconservative model of DNA replication. Here's a summary of their method and findings:

  1. Isotopic Labeling: They grew Escherichia coli in a medium containing the heavy nitrogen isotope N15 as the sole nitrogen source over many generations. This allowed N15 to be incorporated into the newly synthesized DNA, making it denser than normal N14 DNA.

  2. Transfer to Normal Nitrogen: The bacteria were then transferred to a medium containing the normal ^14N isotope. DNA samples were taken at specific time intervals as the bacteria continued to reproduce.

  3. Density Gradient Centrifugation: The extracted DNA was centrifuged using a cesium chloride (CsCl) density gradient. This method separates DNA based on its density, allowing differentiation between heavy (N15), light (N14 ), and hybrid (N15/N14 ) DNA.

  4. Results and Analysis:

    • After one generation (20 minutes, since E. coli divides approximately every 20 minutes), the DNA had a hybrid or intermediate density, indicating that each new DNA molecule contained one strand of the original (N15-labeled DNA and one strand of new N14 -labeled DNA.
    • After two generations (40 minutes), the DNA sample contained equal amounts of light and hybrid DNA. This showed that some DNA strands had undergone a second round of replication, incorporating more N14 into new strands.
  5. Conclusion: These findings supported the idea that DNA replication is semiconservative, where each new DNA molecule consists of one old (parental) and one new (daughter) strand. Subsequent experiments in higher organisms confirmed that semiconservative replication is a universal mechanism.

Meselson and Stahl’s ExperimentMeselson and Stahl’s Experiment

6. Replication Fork

During DNA replication in long DNA molecules, the two strands are not fully separated along their entire length due to the high energy requirement for such an unwinding. Instead, replication occurs at specific sections known as replication forks where the DNA helix opens up.

Direction of Polymerization: DNA-dependent DNA polymerases, the enzymes responsible for adding nucleotides to the growing DNA strand, can only synthesize DNA in the 5' to 3' direction. This directional synthesis introduces complexities at the replication fork.

Continuous vs. Discontinuous Synthesis:

  • Leading Strand: On the strand that runs 3' to 5' (the template strand), DNA replication is continuous. This strand is synthesized smoothly in the 5' to 3' direction as the replication fork opens up.
  • Lagging Strand: On the opposite strand that runs 5' to 3', replication is discontinuous. This strand is synthesized in short segments known as Okazaki fragments, each growing in the 5' to 3' direction but initiated at different points as more of the strand is exposed during replication fork progression.

Fragment Joining: The discontinuously synthesized Okazaki fragments on the lagging strand are later connected to form a continuous strand through the action of the enzyme DNA ligase, which links the sugar-phosphate backbones of adjacent fragments.

This method of replication ensures that the DNA is accurately copied despite the directional limitations of the DNA polymerases and the complex structure of the DNA molecule.

Replication ForkReplication Fork

7. Transcription Unit

A transcription unit in DNA comprises three main elements:

  1. Promoter: Located upstream of the gene, it provides a binding site for RNA polymerase and dictates the start of transcription.
  2. Structural Gene: The segment of DNA transcribed into RNA, containing the genetic information for protein production.
  3. Terminator: Situated downstream of the gene on the coding strand, it signals RNA polymerase to stop transcription.

The template strand (3'→5') acts as the mold for RNA synthesis, while the coding strand (5'→3') mirrors the RNA sequence (except for thymine in place of uracil) and serves as a reference for gene regulation. The arrangement of promoter and terminator defines which strand serves as the template, and additional regulatory sequences may influence gene expression further.

Schematic Structure of Transcription UnitSchematic Structure of Transcription Unit

8. Transcription in Prokaryotes ( Bacteria)

  • In bacteria, transcription of all RNA types—mRNA, tRNA, and rRNA—necessary for protein synthesis is performed by a single DNA-dependent RNA polymerase. 
  • This enzyme initiates transcription at promoters, elongates by adding nucleotides based on template complementarity, and terminates at specific terminator sequences. 
  • It interacts with specific factors: initiation-factor (σ) for starting transcription and termination-factor (ρ) for ending it, enhancing its functional specificity. 
  • Due to the absence of a nuclear membrane in bacteria, transcription and translation processes are closely coupled, allowing translation to start even before the transcription of mRNA is fully completed.

Process of Transcription in BacteriaProcess of Transcription in Bacteria

9. Transcription in Eukaryotes

  • In eukaryotes, transcription is more complex than in prokaryotes due to the presence of three distinct RNA polymerases in the nucleus, each specialized for different types of RNA: Polymerase I for rRNA, Polymerase II for precursor mRNA (hnRNA), and Polymerase III for tRNA and other small RNAs. 
  • Additionally, primary transcripts undergo significant processing, including splicing to remove introns and join exons, capping the 5'-end, and adding a poly-A tail at the 3'-end, transforming hnRNA into functional mRNA. This intricate regulation and processing hint at the evolutionary significance of split genes and the ancient dominance of RNA in cellular mechanisms.

Process of Transcription in EukaryotesProcess of Transcription in Eukaryotes

10. Genetic Code

The salient features of genetic code are as follows:

 (i) The codon is triplet. 61 codons code for amino acids and 3 codons do not code for any amino acids, hence they function as stop codons. 

(ii) Some amino acids are coded by more than one codon, hence the code is degenerate. 

(iii) The codon is read in mRNA in a contiguous fashion. There are no punctuations. 

(iv) The code is nearly universal: for example, from bacteria to human UUU would code for Phenylalanine (phe). Some exceptions to this rule have been found in mitochondrial codons, and in some protozoans. 

(v) AUG has dual functions. It codes for Methionine (met) , and it also act as initiator codon. (vi) UAA, UAG, UGA are stop terminator codons.

The Codons for the Various Amino AcidsThe Codons for the Various Amino Acids

11. tRNA– the Adapter Molecule

  • Francis Crick proposed that an adapter molecule was necessary to translate genetic code into proteins by linking specific amino acids. This role is fulfilled by tRNA (formerly known as sRNA or soluble RNA), which reads the nucleotide code through its anticodon loop and binds amino acids at its acceptor end. 
  • Each tRNA is specific to a particular amino acid. While there is a specialized initiator tRNA for starting protein synthesis, there are no tRNAs for stop codons. Structurally, tRNA is often depicted as a clover leaf in diagrams but is more compact, resembling an inverted L in its actual three-dimensional form.

tRNA– the Adapter MoleculetRNA– the Adapter Molecule

12.  Translation

  • Translation is the process where amino acids are polymerized to form a polypeptide, dictated by mRNA base sequences. Initially, amino acids are activated with ATP and attached to their specific tRNA, a process known as tRNA charging. 
  • During translation, ribosomes, which consist of RNA and proteins and split into two subunits, catalyze peptide bond formation between amino acids at specific sites on the large subunit. 
  • Translation begins when the small ribosomal subunit binds to mRNA at a start codon (AUG), recognized by initiator tRNA. It progresses through elongation, where tRNA-amino acid complexes match with corresponding mRNA codons, adding amino acids sequentially to the growing polypeptide chain. 
  • Translation ends when a release factor binds to a stop codon, causing the ribosome to release the completed polypeptide.

Process of TranslationProcess of Translation

13. The Lac Operon

  • The lac operon is a transcriptionally regulated system discovered by Francois Jacob and Jacque Monod, common in bacteria, and controls the metabolism of lactose. 
  • It consists of a regulatory gene (i gene) and three structural genes (z, y, a) that code for beta-galactosidase, permease, and transacetylase, respectively.
  •  Beta-galactosidase breaks down lactose into galactose and glucose, while permease facilitates the entry of lactose into cells. 
  • The operon is usually off but can be activated by lactose, which serves as an inducer. In the presence of lactose, it binds to the repressor protein (produced by the i gene), inactivating it and allowing RNA polymerase to initiate transcription of the operon.
  •  This system exemplifies negative regulation, where the repressor inhibits gene expression until deactivated by an inducer.

The Lac OperonThe Lac Operon

14.  Human Genome Project

The Human Genome Project (HGP), launched in 1990 and completed in 2003, was a large-scale initiative aimed at mapping all human genes and sequencing the 3 billion base pairs in human DNA.

 Key achievements include identifying 20,000-25,000 genes, revealing that less than 2% of the genome codes for proteins, and highlighting extensive repeated sequences that contribute to genomic structure and evolution. 

Techniques like automated DNA sequencing and specialized computer programs were essential for managing the massive data involved. The project's insights are foundational for advancing personalized medicine, understanding genetic disorders, and exploring human evolution.

 Salient Features of Human Genome

(i) The human genome contains 3164.7 million bp.

(ii) The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.

(iii) The total number of genes is estimated at 30,000–much lower than previous estimates of 80,000 to 1,40,000 genes. Almost all

(99.9 per cent) nucleotide bases are exactly the same in all people.

(iv) The functions are unknown for over 50 per cent of the discovered genes.

(v) Less than 2 per cent of the genome codes for proteins.

(vi) Repeated sequences make up very large portion of the human genome.

(vii) Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.

(viii) Chromosome 1 has most genes (2968), and the Y has the fewest (231).

(ix) Scientists have identified about 1.4 million locations where singlebase DNA differences (SNPs – single nucleotide polymorphism, pronounced as ‘snips’) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease-associated sequences and tracing human history.

A representative diagram of human genome projectA representative diagram of human genome project

15. DNA Fingerprinting

  • DNA fingerprinting is a technique that identifies genetic differences by analyzing specific regions of repetitive DNA in an individual's genome. 
  • These regions, often non-coding, display high polymorphism, making them ideal for unique identification in forensic and paternity testing. 
  • The process involves isolating DNA, digesting it with restriction enzymes, separating fragments via electrophoresis, and detecting variations with labeled probes like VNTRs (Variable Number of Tandem Repeats). Initially developed by Alec Jeffreys, DNA fingerprinting has evolved with enhancements like PCR, allowing analysis from just a single cell and extending its applications beyond forensics to include population and genetic diversity studies.
DNA FingerprintingDNA Fingerprinting

Diagram Based Questions NEET


Q1: A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and down stream end;     (NEET 2024)
(a) Repressor, Operator gene, Structural gene
(b) Structural gene, Transposons, Operator gene
(c) Inducer, Repressor, Structural gene
(d) Promotor, Structural gene, Terminator

Ans: (d)

Important Diagrams: Molecular Basis of Inheritance | Biology Class 12 - NEETA transcription unit in DNA is critical for the process of transcription, wherein a particular segment of DNA is copied into RNA (especially mRNA) by the enzyme RNA polymerase. This unit is composed of sequences that include both coding regions, which are directly transcribed into RNA, and regulatory regions, which ensure that transcription is initiated and terminated at the correct locations on the DNA.The correct answer is: Option D: Promotor, Structural gene, Terminator
Here's a detailed explanation of each component:
Promoter: The promoter is a sequence in DNA that signals the RNA polymerase to start transcription. It is located at the upstream end (5' end) of the gene. Promoters are essential for transcription initiation and are typically found just before the genes they regulate.

Structural gene: This region of the transcription unit is actually expressed or translated into protein (or functional RNA), depending on the kind of gene. These genes contain the functional sequences that are copied during the transcription process.

Terminator: The terminator is found at the downstream end (3' end) of the transcription unit and includes sequences that signal the RNA polymerase enzyme to stop transcription. This ensures that the newly synthesized RNA contains only the necessary genetic message.

The other options contain components that do not accurately define the typical structure of a transcription unit:

Option A mixes regulatory proteins and DNA regions, which does not accurately represent the structural components of a transcription unit.
Option B includes "transposons" which are genetic elements that can move around within the genomes but are not typically part of the transcription unit.
Option C again refers to regulatory proteins (inducer and repressor) along with structural genes, confusing the functions of proteins and DNA regions.
Therefore, Option D correctly represents the standard components of a transcription unit in the context of gene transcription in DNA.

Q2: Match List - I with List - II. (NEET 2022 Phase 2)

Important Diagrams: Molecular Basis of Inheritance | Biology Class 12 - NEET

Choose the correct answer from the options given below
(a) (a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)
(b) (a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
(c) (a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)
(d) (a) - (iv), (b) - (i), (c) - (iii), (d) - (ii)
Ans: 
(c)

Important Diagrams: Molecular Basis of Inheritance | Biology Class 12 - NEETIn Iac operon,

  • The i gene codes for repressor protein.
  • The z gene codes for β-galactosidase.
  • The y gene codes for permease and the a gene codes for transacetylase.
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FAQs on Important Diagrams: Molecular Basis of Inheritance - Biology Class 12 - NEET

1. What is the structure of polynucleotide chains in DNA?
Ans. Polynucleotide chains in DNA consist of long sequences of nucleotides, which are the building blocks of DNA. Each nucleotide is made up of three components: a phosphate group, a deoxyribose sugar, and a nitrogenous base (adenine, thymine, cytosine, or guanine). The nucleotides are linked together by phosphodiester bonds, forming a sugar-phosphate backbone, with the nitrogenous bases extending from the sugar. The two strands of DNA run in opposite directions (antiparallel) and twist to form a double helix.
2. How does DNA get packaged into a compact structure?
Ans. DNA is packaged into a compact structure through a series of hierarchical levels of organization. Initially, DNA wraps around histone proteins to form nucleosomes, resembling "beads on a string." These nucleosomes further coil and fold into a more compact structure called chromatin. During cell division, chromatin condenses to form visible chromosomes, allowing for efficient segregation of genetic material.
3. What was the significance of the Hershey and Chase experiment?
Ans. The Hershey and Chase experiment, conducted in 1952, provided crucial evidence that DNA is the genetic material in organisms. By using radioactive labeling, they tracked the DNA and protein components of bacteriophages (viruses that infect bacteria). Their results showed that only the viral DNA entered the bacterial cells and directed the production of new viruses, while the protein coat remained outside. This experiment confirmed that DNA carries genetic information.
4. What is semi-conservative DNA replication according to the Watson and Crick model?
Ans. Semi-conservative DNA replication is a process where each of the two strands of the original DNA molecule serves as a template for the formation of new complementary strands. According to the Watson and Crick model, when DNA replicates, the double helix unwinds, and each strand is used to synthesize a new strand. As a result, each daughter DNA molecule consists of one old (parental) strand and one newly synthesized strand, preserving half of the original DNA.
5. What are the key differences in transcription between prokaryotes and eukaryotes?
Ans. Transcription in prokaryotes occurs in the cytoplasm and involves a single RNA polymerase that synthesizes mRNA directly from the DNA template. In contrast, eukaryotic transcription takes place in the nucleus, where the primary mRNA undergoes processing (capping, polyadenylation, and splicing) before it is exported to the cytoplasm. Additionally, eukaryotic transcription involves multiple RNA polymerases and various transcription factors, whereas prokaryotes rely on simpler mechanisms.
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