Advanced Numerical Problems: Molecular Basis of Inheritance

Mastering the molecular basis of inheritance requires a deep understanding of DNA/RNA structure, replication, and the genetic code. These advanced numerical problems are designed to test your conceptual clarity, analytical skills, and ability to navigate common traps.

Type 1: Base Composition and Chargaff’s Rules

Problem 1.1: The Double-Stranded vs. Single-Stranded Trap

You are given two nucleic acid samples:

Sample A is isolated from an E. coli cell. It has 22% Adenine.
Sample B is isolated from a $\phi$ X174 bacteriophage. It has 24% Adenine.

Question: Calculate the percentage of Cytosine in both Sample A and Sample B.

Solution:

Step 1: Analyze Sample A (E. coli)

E. coli has double-stranded DNA (dsDNA). Therefore, Chargaff's rules apply.
Rule: $A = T$ and $G = C$ . Total = $100\%$ .
Given $A = 22\%$ , so $T = 22\%$ .
$A + T = 44\%$ .
Remaining bases ( $G + C$ ) $= 100\% - 44\% = 56\%$ .
Since $G = C$ , $C = 56\% / 2 = 28\%$ .
Sample A Cytosine = 28%

Step 2: Analyze Sample B ( $\phi$ X174 bacteriophage)

$\phi$ X174 is a well-known virus with a single-stranded DNA (ssDNA) genome.
Crucial Concept: Chargaff's rules ( $A=T$ , $G=C$ ) do not apply to single-stranded nucleic acids because base pairing is not mandatory.
Therefore, the percentage of Cytosine cannot be determined solely from the percentage of Adenine.
Sample B Cytosine = Cannot be determined.

[!CAUTION] Common Pitfall Students frequently blindly apply Chargaff's formulas ( $A=T, G=C$ ) to any given percentage. Always verify if the genetic material is double-stranded. Single-stranded DNA (like in $\phi$ X174 or Parvoviruses) or single-stranded RNA (most plant viruses, HIV) do not follow Chargaff's rule.

Problem 1.2: Hydrogen Bonds and Helix Geometry

A linear double-stranded DNA molecule is $340\text{ nm}$ long and has a $30\%$ Adenine content. Calculate:

The total number of nucleotides in this DNA molecule.
The total number of complete helical turns.
The total number of hydrogen bonds holding the two strands together.

Solution:

Step 1: Calculate Total Base Pairs (bp) and Nucleotides

In a B-DNA double helix, the distance between two consecutive base pairs is $0.34\text{ nm}$ ( $3.4\text{ \AA}$ ).
Total Length = $340\text{ nm}$ .
Total base pairs = $\frac{\text{Total Length}}{\text{Distance per bp}} = \frac{340\text{ nm}}{0.34\text{ nm/bp}} = 1000\text{ bp}$ .
Since each base pair consists of 2 nucleotides (one on each strand), the total number of nucleotides is $1000 \times 2 = 2000$ .

Step 2: Calculate Helical Turns

In B-DNA, there are exactly 10 base pairs per complete helical turn (pitch = $3.4\text{ nm}$ ).
Number of turns = $\frac{\text{Total bp}}{10} = \frac{1000}{10} = 100\text{ turns}$ .

Step 3: Calculate Total Hydrogen Bonds

Given $A = 30\%$ , then $T = 30\%$ . Total $A+T = 60\%$ .
Therefore, $G+C = 40\%$ . Since $G=C$ , $G = 20\%$ and $C = 20\%$ .
Total number of $A-T$ pairs = $30\%$ of $1000\text{ bp} = 300\text{ pairs}$ .
Total number of $G-C$ pairs = $20\%$ of $1000\text{ bp} = 200\text{ pairs}$ .
$A-T$ pairs have 2 hydrogen bonds.
$G-C$ pairs have 3 hydrogen bonds.
Total H-bonds = $(300 \times 2) + (200 \times 3) = 600 + 600 = 1200\text{ hydrogen bonds}$ .

Type 2: DNA Replication & Isotope Tracing (Meselson-Stahl)

Problem 2.1: Multi-Generation Isotope Shift

A culture of E. coli cells grown continuously in a heavy nitrogen medium ( $^{15}N$ ) is transferred to a normal light nitrogen medium ( $^{14}N$ ) for 60 minutes. The cells are then immediately transferred back to the heavy nitrogen medium ( $^{15}N$ ) for another 20 minutes. Assuming a generation time of 20 minutes for E. coli, calculate the ratio of Light : Hybrid : Heavy DNA molecules at the end of the experiment.

Solution:

Step 1: Identify Generations

Generation time = 20 minutes.
Time in $^{14}N$ = 60 minutes $\rightarrow$ 3 generations.
Time in $^{15}N$ = 20 minutes $\rightarrow$ 1 generation.
Total generations = 4.

Step 2: Track the DNA Strands (Semi-conservative Replication)

Let Heavy strand = H ( $^{15}N$ ), Light strand = L ( $^{14}N$ ).
Initial state: $100\%$ Heavy DNA (HH). Let's start with 1 molecule for simplicity.
Gen 1 (in L medium, 20 mins): The HH molecule splits, each H strand pairs with a newly synthesized L strand. Result: 2 HL (Hybrid) molecules.
Gen 2 (in L medium, 40 mins): The 2 HL molecules split into 2 H strands and 2 L strands. Each pairs with new L strands.
- 2 H strands $\rightarrow$ 2 HL (Hybrid)
- 2 L strands $\rightarrow$ 2 LL (Light)
Gen 3 (in L medium, 60 mins): The 4 molecules (2 HL, 2 LL) split into 2 H strands and 6 L strands. Each pairs with new L strands.
- 2 H strands $\rightarrow$ 2 HL (Hybrid)
- 6 L strands $\rightarrow$ 6 LL (Light)
- Total: 8 molecules.
Gen 4 (Transferred to H medium, 80 mins total): The 8 molecules (2 HL, 6 LL) split into 2 H strands and 14 L strands. Since the medium is now H ( $^{15}N$ $^{15} N$ ), all newly synthesized strands are H.
- 2 H strands + new H $\rightarrow$ 2 HH (Heavy)
- 14 L strands + new H $\rightarrow$ 14 HL (Hybrid)

Step 3: Calculate the Final Ratio

Light (LL) : Hybrid (HL) : Heavy (HH)
0 LL : 14 HL : 2 HH
Simplifying the ratio: 0 : 7 : 1

DNA Generation Tracking:

Initial: 1 HH (Heavy)
$\xrightarrow{\text{Gen 1: in }^{14}\text{N}}$ 2 HL (Hybrid)
$\xrightarrow{\text{Gen 2: in }^{14}\text{N}}$ 2 HL + 2 LL (Hybrid + Light)
$\xrightarrow{\text{Gen 3: in }^{14}\text{N}}$ 2 HL + 6 LL (Hybrid + Light)
$\xrightarrow{\text{Gen 4: in }^{15}\text{N}}$ 2 HH + 14 HL (Heavy + Hybrid)

[!TIP] The "Conserved Strand" Shortcut In any purely semi-conservative replication experiment where cells are moved to a new medium, the original strands are perfectly conserved. In Gen 3 (8 molecules total), the 2 original H strands form 2 Hybrid molecules. The rest ( $8-2=6$ ) are Light. When moved back to H medium in Gen 4 (16 molecules total), the 14 L strands will form 14 Hybrids, and the 2 H strands will form 2 Heavys.

Type 3: Transcription and The Genetic Code

Problem 3.1: Phosphodiester Bonds and Reading Frames

A functional mRNA is synthesized to produce a continuous polypeptide chain containing 120 amino acids. Calculate:

The minimum number of nucleotides required in the coding region (open reading frame) of the mRNA.
The number of phosphodiester bonds present strictly within this coding region of the mRNA.

Solution:

Step 1: Calculate Minimum Nucleotides

The genetic code is a triplet code (1 codon = 3 nucleotides = 1 amino acid).
Nucleotides for 120 amino acids = $120 \times 3 = 360\text{ nucleotides}$ .
Trap: Do not forget the STOP codon! A stop codon (UAA, UAG, or UGA) is required to terminate translation but does not code for an amino acid.
Stop codon = 3 nucleotides.
Minimum total nucleotides in the coding region = $360 + 3 = 363\text{ nucleotides}$ .

Step 2: Calculate Phosphodiester Bonds

A polynucleotide chain of $N$ nucleotides is held together by $N - 1$ phosphodiester bonds.
Number of nucleotides ( $N$ ) = $363$ .
Number of phosphodiester bonds = $363 - 1 = 362$ .

[!CAUTION] The "N-1" Rule Pitfall A common error is assuming $N$ nucleotides have $N$ bonds. Linear nucleic acids always have $N-1$ phosphodiester bonds. However, if a problem specifies a circular DNA (like a bacterial plasmid), the number of bonds equals the number of nucleotides ( $N$ ).

Problem 3.2: Introns, Exons, and Mature mRNA

A eukaryotic structural gene consists of 4 exons and 3 introns. The lengths of the exons are 150 bp, 200 bp, 100 bp, and 250 bp respectively. The lengths of the introns are 300 bp, 400 bp, and 150 bp.

What is the length of the primary transcript (hnRNA)?
What is the length of the fully mature mRNA, assuming a 5' cap of 1 nucleotide and a 3' poly-A tail of 200 nucleotides are added?
What is the maximum number of amino acids in the polypeptide translated from this mature mRNA? (Assume the entire exon sequence is coding, ignoring UTRs for maximum theoretical yield, and one stop codon is present).

Solution:

Step 1: Length of Primary Transcript (hnRNA)

The primary transcript contains both exons (coding) and introns (non-coding).
Total length = Sum of all exons + Sum of all introns
Exons = $150 + 200 + 100 + 250 = 700\text{ nucleotides}$
Introns = $300 + 400 + 150 = 850\text{ nucleotides}$
hnRNA Length = $700 + 850 = 1550\text{ nucleotides}$ .

Step 2: Length of Mature mRNA

During splicing, introns are removed, and exons are joined together.
Length of spliced RNA = Sum of exons = $700\text{ nucleotides}$ .
Post-transcriptional modifications include adding the 5' cap and 3' Poly-A tail.
Mature mRNA Length = Spliced RNA + 5' Cap + Poly-A Tail
Mature mRNA Length = $700 + 1 + 200 = 901\text{ nucleotides}$ .

Step 3: Maximum Polypeptide Length

Translation only occurs on the spliced exon sequences. The 5' cap and poly-A tail are not translated.
Total coding nucleotides = $700$ .
Since the last 3 nucleotides must be a stop codon, coding nucleotides for amino acids = $700 - 3 = 697$ .
Number of amino acids = $\lfloor 697 / 3 \rfloor$ (we take the floor because partial codons don't code for anything).
$697 / 3 = 232$ with 1 base left over.
Therefore, the maximum number of amino acids is 232.

Eukaryotic mRNA Processing Flow:

DNA / hnRNA Structure: Exon 1 (150bp) - Intron 1 (300bp) - Exon 2 (200bp) - Intron 2 (400bp) - Exon 3 (100bp) - Intron 3 (150bp) - Exon 4 (250bp)

Mature mRNA Structure: 5' Cap (1nt) - Exon 1 - Exon 2 - Exon 3 - Exon 4 - Poly-A Tail (200nt)

[!NOTE] The Gantt chart conceptually aligns the relative lengths of the elements. In biological reality, the 5' cap and Poly-A tail are added to the ends of the ligated exons.

Advanced Numerical Problems: Molecular Basis of Inheritance

Type 1: Base Composition and Chargaff’s Rules

Problem 1.1: The Double-Stranded vs. Single-Stranded Trap

You are given two nucleic acid samples:

Sample A is isolated from an E. coli cell. It has 22% Adenine.
Sample B is isolated from a $\phi$ X174 bacteriophage. It has 24% Adenine.

Question: Calculate the percentage of Cytosine in both Sample A and Sample B.

Solution:

Step 1: Analyze Sample A (E. coli)

E. coli has double-stranded DNA (dsDNA). Therefore, Chargaff's rules apply.
Rule: $A = T$ and $G = C$ . Total = $100\%$ .
Given $A = 22\%$ , so $T = 22\%$ .
$A + T = 44\%$ .
Remaining bases ( $G + C$ ) $= 100\% - 44\% = 56\%$ .
Since $G = C$ , $C = 56\% / 2 = 28\%$ .
Sample A Cytosine = 28%

Step 2: Analyze Sample B ( $\phi$ X174 bacteriophage)

$\phi$ X174 is a well-known virus with a single-stranded DNA (ssDNA) genome.
Crucial Concept: Chargaff's rules ( $A=T$ , $G=C$ ) do not apply to single-stranded nucleic acids because base pairing is not mandatory.
Therefore, the percentage of Cytosine cannot be determined solely from the percentage of Adenine.
Sample B Cytosine = Cannot be determined.

[!CAUTION] Common Pitfall Students frequently blindly apply Chargaff's formulas ( $A=T, G=C$ ) to any given percentage. Always verify if the genetic material is double-stranded. Single-stranded DNA (like in $\phi$ X174 or Parvoviruses) or single-stranded RNA (most plant viruses, HIV) do not follow Chargaff's rule.

Problem 1.2: Hydrogen Bonds and Helix Geometry

A linear double-stranded DNA molecule is $340\text{ nm}$ long and has a $30\%$ Adenine content. Calculate:

The total number of nucleotides in this DNA molecule.
The total number of complete helical turns.
The total number of hydrogen bonds holding the two strands together.

Solution:

Step 1: Calculate Total Base Pairs (bp) and Nucleotides

In a B-DNA double helix, the distance between two consecutive base pairs is $0.34\text{ nm}$ ( $3.4\text{ \AA}$ ).
Total Length = $340\text{ nm}$ .
Total base pairs = $\frac{\text{Total Length}}{\text{Distance per bp}} = \frac{340\text{ nm}}{0.34\text{ nm/bp}} = 1000\text{ bp}$ .
Since each base pair consists of 2 nucleotides (one on each strand), the total number of nucleotides is $1000 \times 2 = 2000$ .

Step 2: Calculate Helical Turns

In B-DNA, there are exactly 10 base pairs per complete helical turn (pitch = $3.4\text{ nm}$ ).
Number of turns = $\frac{\text{Total bp}}{10} = \frac{1000}{10} = 100\text{ turns}$ .

Step 3: Calculate Total Hydrogen Bonds

Given $A = 30\%$ , then $T = 30\%$ . Total $A+T = 60\%$ .
Therefore, $G+C = 40\%$ . Since $G=C$ , $G = 20\%$ and $C = 20\%$ .
Total number of $A-T$ pairs = $30\%$ of $1000\text{ bp} = 300\text{ pairs}$ .
Total number of $G-C$ pairs = $20\%$ of $1000\text{ bp} = 200\text{ pairs}$ .
$A-T$ pairs have 2 hydrogen bonds.
$G-C$ pairs have 3 hydrogen bonds.
Total H-bonds = $(300 \times 2) + (200 \times 3) = 600 + 600 = 1200\text{ hydrogen bonds}$ .

Type 2: DNA Replication & Isotope Tracing (Meselson-Stahl)

Problem 2.1: Multi-Generation Isotope Shift

Solution:

Step 1: Identify Generations

Generation time = 20 minutes.
Time in $^{14}N$ = 60 minutes $\rightarrow$ 3 generations.
Time in $^{15}N$ = 20 minutes $\rightarrow$ 1 generation.
Total generations = 4.

Step 2: Track the DNA Strands (Semi-conservative Replication)

Let Heavy strand = H ( $^{15}N$ ), Light strand = L ( $^{14}N$ ).
Initial state: $100\%$ Heavy DNA (HH). Let's start with 1 molecule for simplicity.
Gen 1 (in L medium, 20 mins): The HH molecule splits, each H strand pairs with a newly synthesized L strand. Result: 2 HL (Hybrid) molecules.
Gen 2 (in L medium, 40 mins): The 2 HL molecules split into 2 H strands and 2 L strands. Each pairs with new L strands.
- 2 H strands $\rightarrow$ 2 HL (Hybrid)
- 2 L strands $\rightarrow$ 2 LL (Light)
Gen 3 (in L medium, 60 mins): The 4 molecules (2 HL, 2 LL) split into 2 H strands and 6 L strands. Each pairs with new L strands.
- 2 H strands $\rightarrow$ 2 HL (Hybrid)
- 6 L strands $\rightarrow$ 6 LL (Light)
- Total: 8 molecules.
Gen 4 (Transferred to H medium, 80 mins total): The 8 molecules (2 HL, 6 LL) split into 2 H strands and 14 L strands. Since the medium is now H ( $^{15}N$ $^{15} N$ ), all newly synthesized strands are H.
- 2 H strands + new H $\rightarrow$ 2 HH (Heavy)
- 14 L strands + new H $\rightarrow$ 14 HL (Hybrid)

Step 3: Calculate the Final Ratio

Light (LL) : Hybrid (HL) : Heavy (HH)
0 LL : 14 HL : 2 HH
Simplifying the ratio: 0 : 7 : 1

DNA Generation Tracking:

Initial: 1 HH (Heavy)
$\xrightarrow{\text{Gen 1: in }^{14}\text{N}}$ 2 HL (Hybrid)
$\xrightarrow{\text{Gen 2: in }^{14}\text{N}}$ 2 HL + 2 LL (Hybrid + Light)
$\xrightarrow{\text{Gen 3: in }^{14}\text{N}}$ 2 HL + 6 LL (Hybrid + Light)
$\xrightarrow{\text{Gen 4: in }^{15}\text{N}}$ 2 HH + 14 HL (Heavy + Hybrid)

[!TIP] The "Conserved Strand" Shortcut In any purely semi-conservative replication experiment where cells are moved to a new medium, the original strands are perfectly conserved. In Gen 3 (8 molecules total), the 2 original H strands form 2 Hybrid molecules. The rest ( $8-2=6$ ) are Light. When moved back to H medium in Gen 4 (16 molecules total), the 14 L strands will form 14 Hybrids, and the 2 H strands will form 2 Heavys.

Type 3: Transcription and The Genetic Code

Problem 3.1: Phosphodiester Bonds and Reading Frames

A functional mRNA is synthesized to produce a continuous polypeptide chain containing 120 amino acids. Calculate:

The minimum number of nucleotides required in the coding region (open reading frame) of the mRNA.
The number of phosphodiester bonds present strictly within this coding region of the mRNA.

Solution:

Step 1: Calculate Minimum Nucleotides

The genetic code is a triplet code (1 codon = 3 nucleotides = 1 amino acid).
Nucleotides for 120 amino acids = $120 \times 3 = 360\text{ nucleotides}$ .
Trap: Do not forget the STOP codon! A stop codon (UAA, UAG, or UGA) is required to terminate translation but does not code for an amino acid.
Stop codon = 3 nucleotides.
Minimum total nucleotides in the coding region = $360 + 3 = 363\text{ nucleotides}$ .

Step 2: Calculate Phosphodiester Bonds

A polynucleotide chain of $N$ nucleotides is held together by $N - 1$ phosphodiester bonds.
Number of nucleotides ( $N$ ) = $363$ .
Number of phosphodiester bonds = $363 - 1 = 362$ .

[!CAUTION] The "N-1" Rule Pitfall A common error is assuming $N$ nucleotides have $N$ bonds. Linear nucleic acids always have $N-1$ phosphodiester bonds. However, if a problem specifies a circular DNA (like a bacterial plasmid), the number of bonds equals the number of nucleotides ( $N$ ).

Problem 3.2: Introns, Exons, and Mature mRNA

What is the length of the primary transcript (hnRNA)?
What is the length of the fully mature mRNA, assuming a 5' cap of 1 nucleotide and a 3' poly-A tail of 200 nucleotides are added?
What is the maximum number of amino acids in the polypeptide translated from this mature mRNA? (Assume the entire exon sequence is coding, ignoring UTRs for maximum theoretical yield, and one stop codon is present).

Solution:

Step 1: Length of Primary Transcript (hnRNA)

The primary transcript contains both exons (coding) and introns (non-coding).
Total length = Sum of all exons + Sum of all introns
Exons = $150 + 200 + 100 + 250 = 700\text{ nucleotides}$
Introns = $300 + 400 + 150 = 850\text{ nucleotides}$
hnRNA Length = $700 + 850 = 1550\text{ nucleotides}$ .

Step 2: Length of Mature mRNA

During splicing, introns are removed, and exons are joined together.
Length of spliced RNA = Sum of exons = $700\text{ nucleotides}$ .
Post-transcriptional modifications include adding the 5' cap and 3' Poly-A tail.
Mature mRNA Length = Spliced RNA + 5' Cap + Poly-A Tail
Mature mRNA Length = $700 + 1 + 200 = 901\text{ nucleotides}$ .

Step 3: Maximum Polypeptide Length

Translation only occurs on the spliced exon sequences. The 5' cap and poly-A tail are not translated.
Total coding nucleotides = $700$ .
Since the last 3 nucleotides must be a stop codon, coding nucleotides for amino acids = $700 - 3 = 697$ .
Number of amino acids = $\lfloor 697 / 3 \rfloor$ (we take the floor because partial codons don't code for anything).
$697 / 3 = 232$ with 1 base left over.
Therefore, the maximum number of amino acids is 232.

Eukaryotic mRNA Processing Flow:

DNA / hnRNA Structure: Exon 1 (150bp) - Intron 1 (300bp) - Exon 2 (200bp) - Intron 2 (400bp) - Exon 3 (100bp) - Intron 3 (150bp) - Exon 4 (250bp)

Mature mRNA Structure: 5' Cap (1nt) - Exon 1 - Exon 2 - Exon 3 - Exon 4 - Poly-A Tail (200nt)

[!NOTE] The Gantt chart conceptually aligns the relative lengths of the elements. In biological reality, the 5' cap and Poly-A tail are added to the ends of the ligated exons.

Numerical Problems - Molecular Basis of Inheritance

Advanced Numerical Problems: Molecular Basis of Inheritance

Type 1: Base Composition and Chargaff’s Rules

Problem 1.1: The Double-Stranded vs. Single-Stranded Trap

Solution:

Problem 1.2: Hydrogen Bonds and Helix Geometry

Solution:

Type 2: DNA Replication & Isotope Tracing (Meselson-Stahl)

Problem 2.1: Multi-Generation Isotope Shift

Solution:

Type 3: Transcription and The Genetic Code

Problem 3.1: Phosphodiester Bonds and Reading Frames

Solution:

Problem 3.2: Introns, Exons, and Mature mRNA

Solution:

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Numerical Problems - Molecular Basis of Inheritance

Advanced Numerical Problems: Molecular Basis of Inheritance

Type 1: Base Composition and Chargaff’s Rules

Problem 1.1: The Double-Stranded vs. Single-Stranded Trap

Solution:

Problem 1.2: Hydrogen Bonds and Helix Geometry

Solution:

Type 2: DNA Replication & Isotope Tracing (Meselson-Stahl)

Problem 2.1: Multi-Generation Isotope Shift

Solution:

Type 3: Transcription and The Genetic Code

Problem 3.1: Phosphodiester Bonds and Reading Frames

Solution:

Problem 3.2: Introns, Exons, and Mature mRNA

Solution:

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