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Nucleic Acids (AHL) - DNA structure and replication

Grade 12IBBiology

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 strands of nucleotides running antiparallel to each other in the 5′→3′5' \rightarrow 3' and 3′→5′3' \rightarrow 5' directions.

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Nucleosomes consist of DNA wrapped twice around an octamer of eight histone proteins (two each of H2AH2A, H2BH2B, H3H3, and H4H4), stabilized by the H1H1 histone protein. This facilitates the supercoiling of DNA.

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DNA replication is semi-conservative and occurs only in a 5′→3′5' \rightarrow 3' direction because DNA Polymerase III can only add nucleotides to the 3′3' hydroxyl (−OH-OH) group of the deoxyribose sugar.

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The leading strand is synthesized continuously towards the replication fork, whereas the lagging strand is synthesized discontinuously away from the fork in short segments called Okazaki fragments.

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Enzymatic roles in replication: DNA Helicase unwinds the helix; DNA Gyrase (topoisomerase) reduces torsional strain; Single-Strand Binding (SSB) proteins prevent re-annealing; DNA Primase adds RNA primers; DNA Polymerase III synthesizes the new strand; DNA Polymerase I removes RNA primers and replaces them with DNA; DNA Ligase joins Okazaki fragments.

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Non-coding regions of DNA include promoters (binding sites for RNA polymerase), enhancers, silencers, introns, telomeres (protective caps), and genes for tRNAtRNA and rRNArRNA.

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The Hershey-Chase experiment used radioactive isotopes 32P^{32}P (to label DNA) and 35S^{35}S (to label protein) to prove that DNA is the genetic material of the T2T2 bacteriophage.

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Rosalind Franklin and Maurice Wilkins used X-ray diffraction to determine the helical structure, the distance between base pairs (0.34 nm0.34\text{ nm}), and the repeat distance (3.4 nm3.4\text{ nm}) of the DNA molecule.

📐Formulae

[A]=[T][A] = [T]

[G]=[C][G] = [C]

[A]+[G]=[T]+[C]=50%[A] + [G] = [T] + [C] = 50\%

Nucleotide=Phosphate Group+Deoxyribose Sugar+Nitrogenous Base\text{Nucleotide} = \text{Phosphate Group} + \text{Deoxyribose Sugar} + \text{Nitrogenous Base}

dNTP→DNAn+PPi\text{dNTP} \rightarrow \text{DNA}_n + PP_i

💡Examples

Problem 1:

A double-stranded DNA molecule is analyzed and found to contain 18%18\% Guanine (GG). Calculate the percentage of Thymine (TT) in this DNA molecule.

Solution:

The percentage of Thymine is 32%32\%.

Explanation:

According to Chargaff's rule, [G]=[C][G] = [C], so [C]=18%[C] = 18\%. This means G+C=36%G + C = 36\%. The remaining 100%−36%=64%100\% - 36\% = 64\% must be composed of Adenine (AA) and Thymine (TT). Since [A]=[T][A] = [T], the percentage of TT is 64%2=32%\frac{64\%}{2} = 32\%.

Problem 2:

Explain why DNA replication is described as occurring in a 5′→3′5' \rightarrow 3' direction.

Solution:

New nucleotides are added to the 3′3' carbon of the growing strand.

Explanation:

DNA Polymerase III requires a free −OH-OH group on the 3′3' carbon of the pentose sugar to catalyze the formation of a phosphodiester bond. Therefore, the phosphate group (5′5') of the incoming deoxynucleoside triphosphate (dNTPdNTP) is attached to the 3′3' end of the existing chain.

Problem 3:

Identify the significance of the 32P^{32}P isotope in the Hershey-Chase experiment.

Solution:

It specifically labeled the DNA of the bacteriophage.

Explanation:

Phosphorus is a component of the DNA backbone but is not found in amino acids (proteins). By using 32P^{32}P, Hershey and Chase could track whether the viral DNA entered the bacterial cell during infection, confirming DNA as the hereditary material when the radioactive signal was found in the pellet (bacteria).

DNA structure and replication - Revision Notes & Key Diagrams | IB Grade 12 Biology