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Nucleic Acids (AHL) - DNA Structure and Replication

Grade 11IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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DNA structure is a double helix consisting of two anti-parallel strands. Each strand is composed of nucleotides containing a deoxyribose sugar, a phosphate group, and a nitrogenous base (AA, TT, CC, or GG).

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Nucleosomes help to supercoil the DNA and regulate transcription. A nucleosome consists of DNA wrapped around an octamer of eight histone proteins, secured by a ninth histone (H1H1).

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DNA replication is semi-conservative and occurs in a 5′→3′5' \rightarrow 3' direction. DNA Polymerase III can only add nucleotides to the 3′3' end of an existing polynucleotide chain.

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The process involves a leading strand, synthesized continuously towards the replication fork, and a lagging strand, synthesized discontinuously away from the fork in short segments called Okazaki fragments.

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Enzymes involved in replication: Helicase (unwinds the helix), DNA Gyrase (reduces torsional strain/supercoiling), Single-Stranded Binding (SSB) Proteins (prevent re-annealing), DNA Primase (creates RNA primers), DNA Polymerase III (adds DNA nucleotides), DNA Polymerase I (replaces RNA primers with DNA), and DNA Ligase (joins Okazaki fragments).

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Non-coding DNA sequences have specific functions, including promoters, enhancers, silencers, introns, telomeres, and genes for tRNAtRNA and rRNArRNA.

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DNA sequencing uses dideoxynucleotides (ddNTPsddNTPs) which lack the 3′3'-OHOH group required for phosphodiester bond formation, thereby acting as chain terminators.

📐Formulae

Chargaff’s Rule: [A]=[T] and [C]=[G]\text{Chargaff's Rule: } [A] = [T] \text{ and } [C] = [G]

Purines (A, G)=Pyrimidines (T, C)\text{Purines (A, G)} = \text{Pyrimidines (T, C)}

Bonding: A=T (2 Hydrogen bonds), G≡C (3 Hydrogen bonds)\text{Bonding: } A = T \text{ (2 Hydrogen bonds), } G \equiv C \text{ (3 Hydrogen bonds)}

Nucleotide Structure: PO43−+Deoxyribose+Base\text{Nucleotide Structure: } PO_4^{3-} + \text{Deoxyribose} + \text{Base}

💡Examples

Problem 1:

In a sample of double-stranded DNA, 22%22\% of the nitrogenous bases are Cytosine (CC). Calculate the percentage of Thymine (TT) in the sample.

Solution:

If C=22%C = 22\%, then G=22%G = 22\%. Total C+G=44%C + G = 44\%. The remaining 56%56\% must be A+TA + T. Since A=TA = T, the percentage of T=56%2=28%T = \frac{56\%}{2} = 28\%.

Explanation:

This calculation relies on Chargaff's rule, where complementary base pairs exist in equal ratios (C=GC=G and A=TA=T).

Problem 2:

Explain why DNA replication can only occur in a 5′→3′5' \rightarrow 3' direction.

Solution:

DNA Polymerase III requires a free 3′3' hydroxyl (−OH-OH) group on the deoxyribose sugar to catalyze the condensation reaction with the 5′5' phosphate group of the incoming deoxynucleoside triphosphate (dNTPdNTP).

Explanation:

The energy for the bond formation comes from the hydrolysis of the two extra phosphates from the dNTPdNTP. Without the 3′3'-OHOH group, the phosphodiester bond cannot form.

Problem 3:

Identify the role of DNA Gyrase (Topoisomerase) during the initiation of replication.

Solution:

DNA Gyrase moves ahead of the replication fork and releases the torsional strain (tension) created by the unwinding of the double helix by Helicase.

Explanation:

Without Gyrase, the DNA would become over-twisted (supercoiled) ahead of the fork, which would eventually halt the movement of Helicase.

DNA Structure and Replication - Revision Notes & Key Diagrams | IB Grade 11 Biology