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Unity and diversity - Nucleic acids

Grade 11IBBiology

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

🔑Concepts

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Nucleic acids are polymers called polynucleotides, composed of monomers known as nucleotides. Each nucleotide consists of a pentose sugar, a phosphate group, and a nitrogenous base.

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The pentose sugar is either ribose (C5H10O5C_{5}H_{10}O_{5}) in RNA or deoxyribose (C5H10O4C_{5}H_{10}O_{4}) in DNA. The carbon atoms of the sugar are numbered 1′1' to 5′5'.

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Nucleotides are linked by covalent phosphodiester bonds formed via condensation reactions between the 5′5' phosphate group of one nucleotide and the 3′3' hydroxyl (−OH-OH) group of another.

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DNA consists of two antiparallel strands, 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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Nitrogenous bases are categorized into Purines (Adenine AA and Guanine GG), which have a double-ring structure, and Pyrimidines (Cytosine CC, Thymine TT, and Uracil UU), which have a single-ring structure.

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Complementary base pairing occurs via hydrogen bonds: Adenine pairs with Thymine (A=TA=T) using 22 hydrogen bonds, and Guanine pairs with Cytosine (G≡CG \equiv C) using 33 hydrogen bonds. In RNA, Uracil (UU) replaces Thymine.

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The 'Unity' of nucleic acids refers to the universal nature of the genetic code across all living organisms, while 'Diversity' arises from the unique sequences of nitrogenous bases that encode different proteins.

📐Formulae

%A=%T\%A = \%T

%G=%C\%G = \%C

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

Number of hydrogen bonds=2(nA−T)+3(nG−C)\text{Number of hydrogen bonds} = 2(n_{A-T}) + 3(n_{G-C})

💡Examples

Problem 1:

A double-stranded DNA molecule is analyzed and found to contain 22%22\% Guanine (GG). Calculate the percentage of Adenine (AA) present in this molecule.

Solution:

G=22%  ⟹  C=22%G = 22\% \implies C = 22\% G+C=22%+22%=44%G + C = 22\% + 22\% = 44\% A+T=100%−44%=56%A + T = 100\% - 44\% = 56\% A=56%2=28%A = \frac{56\%}{2} = 28\%

Explanation:

According to Chargaff's rule, the amount of Guanine equals Cytosine, and the amount of Adenine equals Thymine. By subtracting the sum of GG and CC from 100%100\%, we find the total percentage of AA and TT. Dividing by 22 gives the specific percentage for Adenine.

Problem 2:

Identify the complementary DNA sequence for a strand with the sequence: 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'

Explanation:

DNA strands are antiparallel and follow complementary base pairing rules (AA pairs with TT, GG pairs with CC). The sequence must be written in the opposite orientation (3′3' to 5′5').