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Nucleic Acids

DNA and RNA structure, nucleotides, and genetic information

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🧬 Nucleic Acids

Overview

Nucleic acids store and transmit genetic information.

Two types:

  1. DNA (deoxyribonucleic acid) - stores genetic info
  2. RNA (ribonucleic acid) - transfers genetic info, protein synthesis

Nucleotide Structure

Three components:

  1. Pentose sugar (5-carbon)
    • Deoxyribose (DNA)
    • Ribose (RNA - has extra OH group)
  2. Phosphate group (PO₄³⁻)
  3. Nitrogenous base

Nitrogenous Bases

Purines (double ring):

  • Adenine (A) - DNA & RNA
  • Guanine (G) - DNA & RNA

Pyrimidines (single ring):

  • Cytosine (C) - DNA & RNA
  • Thymine (T) - DNA only
  • Uracil (U) - RNA only (replaces thymine)

DNA Structure

Double helix:

  • Two antiparallel polynucleotide strands
  • Sugar-phosphate backbone (outside)
  • Bases paired in middle
  • Complementary base pairing:
    • A pairs with T (2 hydrogen bonds)
    • G pairs with C (3 hydrogen bonds)

Chargaff's Rules:

  • Amount of A = amount of T
  • Amount of G = amount of C

Directionality:

  • 5' end (phosphate group)
  • 3' end (OH group on sugar)
  • Strands run antiparallel (5'→3' and 3'→5')

RNA Structure

Single-stranded (can fold on itself)

Types of RNA:

  1. mRNA (messenger) - carries genetic code from DNA to ribosomes
  2. tRNA (transfer) - brings amino acids to ribosomes
  3. rRNA (ribosomal) - component of ribosomes

DNA vs RNA

FeatureDNARNA
SugarDeoxyriboseRibose
BasesA, T, G, CA, U, G, C
StrandsDouble (helix)Single
LocationNucleus (eukaryotes)Nucleus & cytoplasm
FunctionStore genetic infoTransfer info, protein synthesis
StabilityVery stableLess stable

Key Concepts

  1. Nucleotides are monomers of nucleic acids
  2. DNA stores genetic information; RNA transfers it
  3. Complementary base pairing: A-T and G-C (DNA); A-U and G-C (RNA)
  4. DNA is double helix; RNA is usually single-stranded
  5. Antiparallel strands in DNA (one 5'→3', other 3'→5')
  6. Chargaff's rules: amount of purines = amount of pyrimidines

📚 Practice Problems

1Problem 1easy

❓ Question:

Compare DNA and RNA in terms of: (a) sugar component, (b) nitrogenous bases, (c) structure, and (d) primary biological functions.

💡 Show Solution

DNA vs RNA Comparison:

(a) Sugar Component:

DNA: Deoxyribose (lacks OH on 2' carbon)

  • Formula: C₅H₁₀O₄
  • 2'-H (hydrogen at position 2)

RNA: Ribose (has OH on 2' carbon)

  • Formula: C₅H₁₀O₅
  • 2'-OH makes RNA more chemically reactive

(b) Nitrogenous Bases:

DNA:

  • Purines: Adenine (A), Guanine (G)
  • Pyrimidines: Cytosine (C), Thymine (T)

RNA:

  • Purines: Adenine (A), Guanine (G)
  • Pyrimidines: Cytosine (C), Uracil (U)

Key difference: Thymine (DNA) vs Uracil (RNA)

  • Thymine = methylated uracil (extra -CH₃ group)

(c) Structure:

DNA:

  • Double-stranded (double helix)
  • Antiparallel strands (5'→3' and 3'→5')
  • Base pairing: A-T (2 H-bonds), G-C (3 H-bonds)
  • Very stable, long-term storage
  • Width: ~2 nm, 10 bp per turn

RNA:

  • Usually single-stranded
  • Can fold into secondary structures (hairpins, loops)
  • Some regions may base pair (A-U, G-C)
  • More flexible, temporary

(d) Primary Biological Functions:

DNA:

  1. Long-term genetic storage

    • Contains hereditary information
    • Passed from parent to offspring
  2. Template for replication

    • Makes identical copies during cell division
  3. Template for transcription

    • Genes transcribed into RNA

RNA:

  1. mRNA (messenger RNA):

    • Carries genetic information from DNA to ribosomes
    • Template for protein synthesis
  2. rRNA (ribosomal RNA):

    • Structural and catalytic component of ribosomes
    • Catalyzes peptide bond formation
  3. tRNA (transfer RNA):

    • Brings amino acids to ribosome during translation
    • Has anticodon that pairs with mRNA codon
  4. Other RNAs:

    • miRNA, siRNA (gene regulation)
    • snRNA (splicing)
    • Ribozymes (catalytic RNA)

Summary Table:

FeatureDNARNA
SugarDeoxyriboseRibose
BasesA, T, G, CA, U, G, C
StrandsDoubleUsually single
StabilityVery stableLess stable
FunctionStorageProtein synthesis, regulation

DNA: stable storage (T, deoxyribose); RNA: functional (U, ribose)\boxed{\text{DNA: stable storage (T, deoxyribose); RNA: functional (U, ribose)}}

2Problem 2medium

❓ Question:

A segment of DNA has the sequence 5'-ATGCGATACG-3' on one strand. (a) Write the complementary strand with proper directionality, (b) explain Chargaff's rules and verify they apply to this double-stranded segment, and (c) calculate the percentage of G-C base pairs.

💡 Show Solution

Given: 5'-ATGCGATACG-3'

(a) Complementary strand:

Rules:

  • Strands are antiparallel
  • A pairs with T (2 H-bonds)
  • G pairs with C (3 H-bonds)

Original: 5'-ATGCGATACG-3' Complement: 3'-TACGCTATGC-5'

Or written in conventional 5' to 3' direction:

Complement: 5’-CGTATCGCAT-3′\boxed{\text{Complement: } 5\text{'-CGTATCGCAT-}3'}

(b) Chargaff's Rules:

Statement: In double-stranded DNA:

  1. Amount of adenine (A) = Amount of thymine (T)
  2. Amount of guanine (G) = Amount of cytosine (C)
  3. Amount of purines (A+G) = Amount of pyrimidines (T+C)
  4. The ratio (A+T)/(G+C) varies by species but is constant within species

Verification for this segment:

Count bases in both strands:

Original strand: A=3, T=2, G=3, C=2 Complement: A=2, T=3, G=2, C=3

Total (double-stranded):

  • A = 3 + 2 = 5
  • T = 2 + 3 = 5 ✓ (A = T)
  • G = 3 + 2 = 5
  • C = 2 + 3 = 5 ✓ (G = C)
  • Purines (A+G) = 5 + 5 = 10
  • Pyrimidines (T+C) = 5 + 5 = 10 ✓

Chargaff’s rules verified: A=T=5, G=C=5\boxed{\text{Chargaff's rules verified: A=T=5, G=C=5}}

(c) Percentage of G-C base pairs:

Total base pairs = 10 bp (double-stranded segment)

G-C base pairs = 5

%GC=G-C pairsTotal pairs×100%\%GC = \frac{\text{G-C pairs}}{\text{Total pairs}} \times 100\%

%GC=510×100%\%GC = \frac{5}{10} \times 100\%

%GC=50%\boxed{\%GC = 50\%}

Biological Significance:

  • G-C content affects DNA stability
  • 3 H-bonds (G-C) vs 2 H-bonds (A-T)
  • Higher GC% → higher melting temperature (T_m)
  • This segment: 50% GC = moderate stability

Calculation for melting temperature: Tm≈81.5+0.41(%GC)T_m \approx 81.5 + 0.41(\%GC) Tm≈81.5+0.41(50)=102°CT_m \approx 81.5 + 0.41(50) = 102°C

(For longer DNA; short oligos use different formula)

Explain using:

📋 AP Biology — Exam Format Guide

⏱ 3 hours📝 66 questions📊 3 sections
SectionFormatQuestionsTimeWeightCalculator
Multiple ChoiceMCQ6090 min50%🚫
Free Response (Long)FRQ250 min30%🚫
Free Response (Short)FRQ440 min20%🚫

📊 Scoring: 1-5

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3
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2
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1
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💡 Key Test-Day Tips

  • ✓Focus on experimental design
  • ✓Know data analysis
  • ✓Practice graph interpretation

⚠️ Common Mistakes: Nucleic Acids

Avoid these 3 frequent errors

🌍 Real-World Applications: Nucleic Acids

See how this math is used in the real world

📌 Related Topics in Chemistry of Life

❓ Frequently Asked Questions

What is Nucleic Acids?▾
DNA and RNA structure, nucleotides, and genetic information
How can I study Nucleic Acids effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 2 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Nucleic Acids study guide free?▾
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What course covers Nucleic Acids?▾
Nucleic Acids is part of the AP Biology course on Study Mondo, specifically in the Chemistry of Life section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Nucleic Acids?▾
Yes, this page includes 2 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.