DNA Structure, Semi-Conservative Replication & Meselson-Stahl Experiment
Deoxyribonucleic acid (DNA) stores hereditary genetic information across all cellular life via a remarkably stable, antiparallel double helix. During the S-phase of interphase, DNA replicates semi-conservatively through the coordinated action of DNA helicase and DNA polymerase, ensuring each daughter DNA molecule inherits one original parental strand and one newly synthesised strand. Because DNA polymerase has an active site complementary only to the 3' hydroxyl end of a growing strand, synthesis proceeds strictly in the 5' to 3' direction. Examiners frequently assess both the enzymology of replication and the density-gradient centrifugation evidence from Meselson and Stahl (1958) using ¹⁵N and ¹⁴N isotopes.
- DNA & RNA Nucleotide Structure
- Each nucleotide consists of a pentose sugar (deoxyribose in DNA, lacking an oxygen at C2'; ribose in RNA), a phosphate group attached to C5', and a nitrogenous base attached to C1'. DNA bases: Adenine (A) and Guanine (G) are double-ring purines; Cytosine (C) and Thymine (T, replaced by Uracil U in RNA) are single-ring pyrimidines.
- Phosphodiester Backbone & Complementary Base Pairing
- Nucleotides join via condensation reactions between the C3' -OH of one pentose sugar and the C5' phosphate group of the next, forming strong covalent phosphodiester bonds that create a chemically stable sugar-phosphate backbone. Two antiparallel strands (running 5'→3' and 3'→5') are held together by hydrogen bonds: A pairs with T via 2 hydrogen bonds; G pairs with C via 3 hydrogen bonds.
- Semi-conservative replication mechanism
- 1. DNA helicase breaks the hydrogen bonds between complementary bases, unwinding the double helix and exposing both strands as templates. 2. Free activated DNA nucleotides (dNTPs) align opposite exposed bases by complementary base pairing (A-T, G-C). 3. DNA polymerase catalyses condensation reactions to form phosphodiester bonds between adjacent nucleotides.
DNA Structure, Semi-Conservative Replication & Meselson-Stahl Experiment: 3D Molecular, Lattice, DNA & Vector Explorer
Drag directly on the 3D model to rotate it through any angle, or switch tabs to inspect VSEPR bond angles, crystal lattices, DNA base pairing, and 3D space diagonals.
Antiparallel Strands & Complementary Hydrogen Bonds
- Adenine (A) = Thymine (T): Joined by 2 hydrogen bonds (in RNA, Uracil replaces Thymine).
- Guanine (G) ≡ Cytosine (C): Joined by 3 hydrogen bonds, making G-C rich DNA more thermally stable.
- Backbone: Deoxyribose sugar and phosphate groups joined by strong covalent phosphodiester bondsin antiparallel 5'→3' and 3'→5' directions.
- Helicase slider (0%): DNA helicase breaks hydrogen bonds between bases so DNA polymerase can catalyse semi-conservative replication.
Key Concepts & Exam Definitions (6)
- DNA & RNA Nucleotide Structure
- Each nucleotide consists of a pentose sugar (deoxyribose in DNA, lacking an oxygen at C2'; ribose in RNA), a phosphate group attached to C5', and a nitrogenous base attached to C1'. DNA bases: Adenine (A) and Guanine (G) are double-ring purines; Cytosine (C) and Thymine (T, replaced by Uracil U in RNA) are single-ring pyrimidines.
- Phosphodiester Backbone & Complementary Base Pairing
- Nucleotides join via condensation reactions between the C3' -OH of one pentose sugar and the C5' phosphate group of the next, forming strong covalent phosphodiester bonds that create a chemically stable sugar-phosphate backbone. Two antiparallel strands (running 5'→3' and 3'→5') are held together by hydrogen bonds: A pairs with T via 2 hydrogen bonds; G pairs with C via 3 hydrogen bonds.
- Semi-conservative replication mechanism
- 1. DNA helicase breaks the hydrogen bonds between complementary bases, unwinding the double helix and exposing both strands as templates. 2. Free activated DNA nucleotides (dNTPs) align opposite exposed bases by complementary base pairing (A-T, G-C). 3. DNA polymerase catalyses condensation reactions to form phosphodiester bonds between adjacent nucleotides.
- 5' to 3' Directionality of DNA Polymerase
- The active site of DNA polymerase is only complementary in shape to the 3' (deoxyribose -OH) end of the newly forming DNA strand. Therefore, DNA polymerase moves along the template strand in the 3' to 5' direction while synthesising the new strand in the 5' to 3' direction (continuously on the leading strand and discontinuously in Okazaki fragments joined by DNA ligase on the lagging strand).
- Meselson-Stahl Experiment (1958)
- E. coli bacteria were grown for many generations in heavy nitrogen (¹⁵NH₄Cl) so all DNA was heavy (¹⁵N/¹⁵N), then transferred to light nitrogen (¹⁴NH₄Cl) medium. DNA samples were centrifuged in caesium chloride (CsCl) density gradients: Generation 0 = single lower heavy band (100% ¹⁵N/¹⁵N); Generation 1 = single intermediate hybrid band (100% ¹⁵N/¹⁴N, ruling out conservative replication); Generation 2 = 50% intermediate band (¹⁵N/¹⁴N) and 50% upper light band (¹⁴N/¹⁴N, ruling out dispersive replication).
- Eukaryotic vs Prokaryotic DNA Organization
- Eukaryotic nuclear DNA is long, linear, associated with basic histone proteins to form chromatin/chromosomes, and contains non-coding introns inside genes plus non-coding multiple repeats between genes. Prokaryotic DNA (and mitochondrial/chloroplast DNA) is short, circular, not associated with histones, and lacks introns.
Quick-Fire Flashcards (6)
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How many hydrogen bonds form between Adenine and Thymine, and between Guanine and Cytosine?
What are the specific roles of DNA helicase and DNA polymerase in DNA replication?
Why can DNA polymerase only synthesise a new DNA strand in the 5' to 3' direction?
In Meselson and Stahl's experiment, what bands are observed in the centrifuge tube at Generation 1, 2, and 3 after transferring E. coli from ¹⁵N to ¹⁴N?
If 28% of the bases in a double-stranded DNA molecule are Adenine, what percentage are Cytosine?
Give three structural ways in which an mRNA molecule differs from a DNA molecule.
Exam-Style Questions & Mark Schemes (4)
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- Q1
Explain how five structural features of the DNA molecule make it suited to its function of storing genetic information accurately. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. The sugar-phosphate backbone linked by strong covalent phosphodiester bonds provides structural stability and protects the chemically reactive nitrogenous bases inside the helix [1 mark]. 2. It is a very long/large polymer containing millions of base pairs, allowing it to store a vast amount of genetic information [1 mark]. 3. Complementary base pairing (A with T, G with C) allows exact, accurate semi-conservative replication so genetic information is conserved across cell divisions [1 mark]. 4. Although individual hydrogen bonds are weak (allowing strands to be separated easily by DNA helicase for replication/transcription), the huge number of hydrogen bonds collectively gives the double helix high stability [1 mark]. 5. The double-stranded helix is compact when coiled around histones, fitting inside the nucleus and providing a backup template strand for DNA repair if one strand is damaged [1 mark].
- Q2
Describe the process of semi-conservative DNA replication in a eukaryotic cell. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. The enzyme DNA helicase breaks the hydrogen bonds between the complementary nitrogenous bases of the two polynucleotide strands, causing the double helix to unwind [1 mark]. 2. Both separated parental strands act as templates for the synthesis of new strands [1 mark]. 3. Free DNA nucleotides in the nucleoplasm are attracted to their exposed complementary bases on the template strands (Adenine pairs with Thymine; Guanine pairs with Cytosine) via hydrogen bonding [1 mark]. 4. The enzyme DNA polymerase joins the adjacent nucleotides together via condensation reactions, forming phosphodiester bonds in the 5' to 3' direction [1 mark]. 5. Each of the two resulting DNA molecules contains one original parental strand and one newly synthesised strand [1 mark].
- Q3
Meselson and Stahl grew E. coli in a medium containing heavy nitrogen (¹⁵N) and then transferred them to a medium containing light nitrogen (¹⁴N). Explain how the results after one and two generations disproved the conservative and dispersive models of replication. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. After one generation in ¹⁴N, centrifugation in CsCl produced a single DNA band at an intermediate density (hybrid ¹⁵N/¹⁴N) [1 mark]. 2. This disproved the conservative model, which predicted two separate bands at Generation 1: one 100% heavy (¹⁵N/¹⁵N) band and one 100% light (¹⁴N/¹⁴N) band [1 mark]. 3. After two generations in ¹⁴N, two bands of equal thickness appeared: one at the intermediate density (¹⁵N/¹⁴N) and one higher up at the light density (¹⁴N/¹⁴N) [1 mark]. 4. This disproved the dispersive model, which predicted a single band at Generation 2 that would merely shift slightly higher (3/4 ¹⁴N, 1/4 ¹⁵N) rather than separating into distinct hybrid and light bands [1 mark].
- Q4
A sample of double-stranded viral DNA was analysed and found to have a Guanine content of 34% and a higher melting temperature than a second DNA sample of equal length with a Guanine content of 18%. Calculate the percentage of Thymine in the first sample and explain why it has a higher melting temperature. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Since Guanine (G) pairs with Cytosine (C), C = 34%, so G + C = 68% of total bases [1 mark]. 2. The remaining 32% consists equally of Adenine (A) and Thymine (T), so Thymine = 32 ÷ 2 = 16% [1 mark]. 3. Guanine and Cytosine pairs are joined by 3 hydrogen bonds, whereas Adenine and Thymine pairs are joined by only 2 hydrogen bonds [1 mark]. 4. Therefore, the DNA sample with 34% Guanine (68% G-C pairs) has significantly more hydrogen bonds in total than the 18% Guanine sample, requiring more thermal energy to break the bonds and separate the strands [1 mark].
Common Exam Questions (4)
Why did Meselson and Stahl use nitrogen isotopes (¹⁵N and ¹⁴N) to label DNA?
Nitrogen is a key component of the purine and pyrimidine nitrogenous bases in every nucleotide (Adenine, Thymine, Guanine, Cytosine), whereas carbohydrates and lipids contain no nitrogen. Incorporating ¹⁵N vs ¹⁴N alters the mass density of the DNA without changing its chemical properties or replication ability.
What is the difference between leading strand and lagging strand replication?
Because DNA strands are antiparallel and DNA polymerase can only synthesise in the 5' to 3' direction, the leading strand (3' to 5' template) is synthesised continuously toward the replication fork. The lagging strand (5' to 3' template) must be synthesised discontinuously away from the fork in short Okazaki fragments, which are later covalently sealed together by DNA ligase.
How do prokaryotic chromosomes differ from eukaryotic chromosomes?
Prokaryotic DNA is a single, circular molecule located free in the cytoplasm (nucleoid region), is not wrapped around histone proteins, and contains no non-coding introns. Eukaryotic nuclear DNA consists of multiple linear molecules tightly wound around histone proteins inside a membrane-bound nucleus and contains both introns and exons.
Why is RNA much less stable than DNA inside cells?
RNA is single-stranded (leaving its bases exposed to enzymatic attack) and contains ribose sugar, which has a reactive 2'-hydroxyl (-OH) group that makes the phosphodiester backbone susceptible to alkaline hydrolysis. This rapid degradation is useful for mRNA so that cells can quickly stop producing a protein when it is no longer needed.
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