Nucleic Acids

DNA and RNA structure, nucleotides, and genetic information

🧬 Nucleic Acids

Overview

Nucleic acids store and transmit genetic information.

Two types:

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

Nucleotide Structure

Three components:

Pentose sugar (5-carbon)
- Deoxyribose (DNA)
- Ribose (RNA - has extra OH group)
Phosphate group (PO₄³⁻)
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:

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

DNA vs RNA

| Feature | DNA | RNA | |---------|-----|-----| | Sugar | Deoxyribose | Ribose | | Bases | A, T, G, C | A, U, G, C | | Strands | Double (helix) | Single | | Location | Nucleus (eukaryotes) | Nucleus & cytoplasm | | Function | Store genetic info | Transfer info, protein synthesis | | Stability | Very stable | Less stable |

Key Concepts

Nucleotides are monomers of nucleic acids
DNA stores genetic information; RNA transfers it
Complementary base pairing: A-T and G-C (DNA); A-U and G-C (RNA)
DNA is double helix; RNA is usually single-stranded
Antiparallel strands in DNA (one 5'→3', other 3'→5')
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:

Long-term genetic storage
- Contains hereditary information
- Passed from parent to offspring
Template for replication
- Makes identical copies during cell division
Template for transcription
- Genes transcribed into RNA

RNA:

mRNA (messenger RNA):
- Carries genetic information from DNA to ribosomes
- Template for protein synthesis
rRNA (ribosomal RNA):
- Structural and catalytic component of ribosomes
- Catalyzes peptide bond formation
tRNA (transfer RNA):
- Brings amino acids to ribosome during translation
- Has anticodon that pairs with mRNA codon
Other RNAs:
- miRNA, siRNA (gene regulation)
- snRNA (splicing)
- Ribozymes (catalytic RNA)

Summary Table:

| Feature | DNA | RNA | |---------|-----|-----| | Sugar | Deoxyribose | Ribose | | Bases | A, T, G, C | A, U, G, C | | Strands | Double | Usually single | | Stability | Very stable | Less stable | | Function | Storage | Protein synthesis, regulation |

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

2Problem 2easy

❓ 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:

Long-term genetic storage
- Contains hereditary information
- Passed from parent to offspring
Template for replication
- Makes identical copies during cell division
Template for transcription
- Genes transcribed into RNA

RNA:

mRNA (messenger RNA):
- Carries genetic information from DNA to ribosomes
- Template for protein synthesis
rRNA (ribosomal RNA):
- Structural and catalytic component of ribosomes
- Catalyzes peptide bond formation
tRNA (transfer RNA):
- Brings amino acids to ribosome during translation
- Has anticodon that pairs with mRNA codon
Other RNAs:
- miRNA, siRNA (gene regulation)
- snRNA (splicing)
- Ribozymes (catalytic RNA)

Summary Table:

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

3Problem 3easy

❓ 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:

Long-term genetic storage
- Contains hereditary information
- Passed from parent to offspring
Template for replication
- Makes identical copies during cell division
Template for transcription
- Genes transcribed into RNA

RNA:

mRNA (messenger RNA):
- Carries genetic information from DNA to ribosomes
- Template for protein synthesis
rRNA (ribosomal RNA):
- Structural and catalytic component of ribosomes
- Catalyzes peptide bond formation
tRNA (transfer RNA):
- Brings amino acids to ribosome during translation
- Has anticodon that pairs with mRNA codon
Other RNAs:
- miRNA, siRNA (gene regulation)
- snRNA (splicing)
- Ribozymes (catalytic RNA)

Summary Table:

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

4Problem 4easy

❓ 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:

Long-term genetic storage
- Contains hereditary information
- Passed from parent to offspring
Template for replication
- Makes identical copies during cell division
Template for transcription
- Genes transcribed into RNA

RNA:

mRNA (messenger RNA):
- Carries genetic information from DNA to ribosomes
- Template for protein synthesis
rRNA (ribosomal RNA):
- Structural and catalytic component of ribosomes
- Catalyzes peptide bond formation
tRNA (transfer RNA):
- Brings amino acids to ribosome during translation
- Has anticodon that pairs with mRNA codon
Other RNAs:
- miRNA, siRNA (gene regulation)
- snRNA (splicing)
- Ribozymes (catalytic RNA)

Summary Table:

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

5Problem 5medium

❓ 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:

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

(b) Chargaff's Rules:

Statement: In double-stranded DNA:

Amount of adenine (A) = Amount of thymine (T)
Amount of guanine (G) = Amount of cytosine (C)
Amount of purines (A+G) = Amount of pyrimidines (T+C)
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 ✓

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

(c) Percentage of G-C base pairs:

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

G-C base pairs = 5

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

$\%GC = \frac{5}{10} \times 100\%$

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

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: $T_m \approx 81.5 + 0.41(\%GC)$ $T_m \approx 81.5 + 0.41(50) = 102°C$

(For longer DNA; short oligos use different formula)

6Problem 6medium

❓ Question:

💡 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:

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

(b) Chargaff's Rules:

Statement: In double-stranded DNA:

Amount of adenine (A) = Amount of thymine (T)
Amount of guanine (G) = Amount of cytosine (C)
Amount of purines (A+G) = Amount of pyrimidines (T+C)
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 ✓

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

(c) Percentage of G-C base pairs:

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

G-C base pairs = 5

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

$\%GC = \frac{5}{10} \times 100\%$

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

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: $T_m \approx 81.5 + 0.41(\%GC)$ $T_m \approx 81.5 + 0.41(50) = 102°C$

(For longer DNA; short oligos use different formula)

7Problem 7medium

❓ Question:

💡 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:

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

(b) Chargaff's Rules:

Statement: In double-stranded DNA:

Amount of adenine (A) = Amount of thymine (T)
Amount of guanine (G) = Amount of cytosine (C)
Amount of purines (A+G) = Amount of pyrimidines (T+C)
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 ✓

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

(c) Percentage of G-C base pairs:

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

G-C base pairs = 5

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

$\%GC = \frac{5}{10} \times 100\%$

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

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: $T_m \approx 81.5 + 0.41(\%GC)$ $T_m \approx 81.5 + 0.41(50) = 102°C$

(For longer DNA; short oligos use different formula)

8Problem 8medium

❓ Question:

💡 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:

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

(b) Chargaff's Rules:

Statement: In double-stranded DNA:

Amount of adenine (A) = Amount of thymine (T)
Amount of guanine (G) = Amount of cytosine (C)
Amount of purines (A+G) = Amount of pyrimidines (T+C)
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 ✓

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

(c) Percentage of G-C base pairs:

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

G-C base pairs = 5

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

$\%GC = \frac{5}{10} \times 100\%$

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

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: $T_m \approx 81.5 + 0.41(\%GC)$ $T_m \approx 81.5 + 0.41(50) = 102°C$

(For longer DNA; short oligos use different formula)

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