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Molecular Biology - Structure of DNA and RNA

Grade 11IBBiology

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

🔑Concepts

Nucleic acids, including DNADNA and RNARNA, are polymers made of repeating units called nucleotides. Each nucleotide consists of a pentose sugar, a phosphate group (PO43PO_4^{3-}), and a nitrogenous base.

The two types of pentose sugars are deoxyribose (found in DNADNA) and ribose (found in RNARNA). Ribose has a hydroxyl group (OH-OH) on the 22' carbon, whereas deoxyribose has a hydrogen atom (H-H).

Nitrogenous bases are categorized into purines (AdenineAdenine and GuanineGuanine, which have a double-ring structure) and pyrimidines (CytosineCytosine, ThymineThymine, and UracilUracil, which have a single-ring structure).

Nucleotides are linked by covalent bonds called phosphodiester bonds, which form between the phosphate group at the 55' carbon of one sugar and the hydroxyl group at the 33' carbon of the adjacent sugar.

DNADNA is a double-stranded molecule forming a double helix. The two strands are antiparallel, meaning one runs in the 535' \rightarrow 3' direction while the other runs 353' \rightarrow 5'.

Complementary base pairing occurs via hydrogen bonding: AdenineAdenine (AA) pairs with ThymineThymine (TT) via 22 hydrogen bonds, and GuanineGuanine (GG) pairs with CytosineCytosine (CC) via 33 hydrogen bonds. In RNARNA, UracilUracil (UU) replaces ThymineThymine.

Crick and Watson's model of DNADNA structure was supported by Rosalind Franklin’s X-ray diffraction data, which suggested a helical shape with a consistent width.

📐Formulae

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

%A+%T+%C+%G=100%\%A + \%T + \%C + \%G = 100\%

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

Number of Phosphodiester Bonds=Total NucleotidesNumber of Strands\text{Number of Phosphodiester Bonds} = \text{Total Nucleotides} - \text{Number of Strands}

💡Examples

Problem 1:

A double-stranded DNADNA molecule is found to contain 22%22\% AdenineAdenine. Calculate the percentage of CytosineCytosine present in this molecule.

Solution:

If [A]=22%[A] = 22\%, then by Chargaff's rule, [T]=22%[T] = 22\%. Together, A+T=22%+22%=44%A + T = 22\% + 22\% = 44\%. The remaining percentage for G+CG + C is 100%44%=56%100\% - 44\% = 56\%. Since [G]=[C][G] = [C], the percentage of CytosineCytosine is 56%2=28%\frac{56\%}{2} = 28\%.

Explanation:

In double-stranded DNADNA, AdenineAdenine always pairs with ThymineThymine and GuanineGuanine always pairs with CytosineCytosine. Therefore, their molar ratios are equal.

Problem 2:

Given the DNADNA template strand sequence 5ATGCCG35'-ATGCCG-3', determine the sequence of the complementary DNADNA strand.

Solution:

The complementary strand must be antiparallel and follow base-pairing rules (AT,GCA-T, G-C). The sequence is 3TACGGC53'-TACGGC-5', which is conventionally written as 5CGGCAT35'-CGGCAT-3'.

Explanation:

The strands run in opposite directions. We match AA with TT and GG with CC, then reverse the orientation to write it in the standard 55' to 33' notation.

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