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Nucleic Acids (AHL) - DNA Structure and Replication (HL)

Grade 11IBBiology

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

🔑Concepts

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DNA consists of two polynucleotide strands that are anti-parallel, meaning one strand runs in the 5′→3′5' \rightarrow 3' direction while the other runs 3′→5′3' \rightarrow 5'.

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Nucleosomes consist of DNA wrapped around an octamer of histones (two each of H2AH2A, H2BH2B, H3H3, and H4H4), secured by histone H1H1. This structure helps in supercoiling DNA.

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DNA replication is semi-conservative and occurs in a 5′→3′5' \rightarrow 3' direction because DNA Polymerase III can only add nucleotides to the 3′3' OHOH group.

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Leading strand synthesis is continuous, moving towards the replication fork, while lagging strand synthesis is discontinuous, moving away from the fork and forming Okazaki fragments.

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Key enzymes in HL replication: DNA Gyrase (reduces torsional strain), Helicase (unwinds DNA), Single-Strand Binding (SSB) Proteins (prevent re-annealing), DNA Primase (adds RNA primer), DNA Polymerase III (adds DNA nucleotides), DNA Polymerase I (removes RNA primers), and DNA Ligase (joins Okazaki fragments).

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Non-coding regions of DNA have functions such as acting as promoters, enhancers, silencers, telomeres (protective caps), and coding for tRNAtRNA or rRNArRNA molecules.

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Dideoxynucleotides (ddNTPs) are used in DNA sequencing because they lack the 3′3' OHOH group required for phosphodiester bond formation, thus terminating the DNA chain.

📐Formulae

[A]=[T],[G]=[C][A] = [T], \quad [G] = [C] (Chargaff's Rule)

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

Distance per base pair≈0.34 nm\text{Distance per base pair} \approx 0.34 \text{ nm}

Length of one full turn (10 bp)≈3.4 nm\text{Length of one full turn (10 bp)} \approx 3.4 \text{ nm}

💡Examples

Problem 1:

A double-stranded DNA molecule contains 22%22\% Cytosine. Calculate the percentage of Adenine present in the molecule.

Solution:

If C=22%C = 22\%, then by Chargaff's rule, G=22%G = 22\%. Total G+C=22%+22%=44%G + C = 22\% + 22\% = 44\%. Remaining percentage for A+T=100%−44%=56%A + T = 100\% - 44\% = 56\%. Since A=TA = T, the percentage of A=56%2=28%A = \frac{56\%}{2} = 28\%.

Explanation:

According to Chargaff's rule, base pairing is complementary (G≡CG \equiv C and A=TA = T). Therefore, the sum of purines equals the sum of pyrimidines.

Problem 2:

Identify the sequence of the complementary DNA strand for the following sequence, indicating the directionality: 5′−ATGCCGTA−3′5'-A T G C C G T A-3'.

Solution:

3′−TACGGCAT−5′3'-T A C G G C A T-5' (or 5′−TACGGCAT−3′5'-T A C G G C A T-3' if reversed).

Explanation:

DNA strands are anti-parallel. Adenine (AA) pairs with Thymine (TT), and Guanine (GG) pairs with Cytosine (CC). The 5′5' end of one strand matches the 3′3' end of the other.