Numerical Problems: Molecular Basis of Inheritance

Problem 1: Calculating Nucleotide Percentages (Chargaff's Rule)

A double-stranded DNA molecule contains 20% Adenine (A). Calculate the percentages of Thymine (T), Guanine (G), and Cytosine (C) in this DNA molecule.

Solution:

According to Chargaff's rules, in a double-stranded DNA molecule:

The amount of Adenine (A) is equal to the amount of Thymine (T). (A = T)
The amount of Guanine (G) is equal to the amount of Cytosine (C). (G = C)
The sum of all bases is 100% (A + T + G + C = 100%).

Given: A = 20%

Calculate T: Since A = T, then T = 20%.
Calculate G + C: A + T + G + C = 100% 20% + 20% + G + C = 100% 40% + G + C = 100% G + C = 100% - 40% = 60%
Calculate G and C: Since G = C, and G + C = 60%: G = 60% / 2 = 30% C = 60% / 2 = 30%

Therefore, the percentages are: Thymine (T) = 20%, Guanine (G) = 30%, and Cytosine (C) = 30%.

Problem 2: Calculating Number of Nucleotides in a DNA Segment

A segment of a DNA molecule is 100 base pairs (bp) long. How many nucleotides are present in this segment?

Solution:

A base pair (bp) consists of two nucleotides (one on each strand) joined by hydrogen bonds.
Therefore, 1 base pair = 2 nucleotides.

Given: Length of DNA segment = 100 base pairs

Number of nucleotides = Length in base pairs * 2 Number of nucleotides = 100 bp * 2 nucleotides/bp = 200 nucleotides

Therefore, there are 200 nucleotides present in a 100 base pair DNA segment.

Question 1: DNA Structure and Chargaff's Rule

A segment of DNA has 1000 base pairs. The number of adenine bases is 200. What is the number of cytosine bases in this segment?

Solution:

Total base pairs: 1000
Total number of bases: 1000 x 2 = 2000
According to Chargaff's rule, the amount of Adenine (A) is equal to the amount of Thymine (T), and the amount of Guanine (G) is equal to the amount of Cytosine (C).
- A = T = 200
Total A + T: 200 + 200 = 400
Remaining bases (G + C): 2000 - 400 = 1600
Since G = C, the number of cytosine bases is:
- C = 1600 / 2 = 800

Therefore, there are 800 cytosine bases in this segment of DNA.

Question 2: DNA Length Calculation

The distance between two consecutive base pairs in a DNA molecule is 0.34 nm. If the length of a DNA molecule is 1.36 mm, calculate the number of base pairs in this DNA.

Solution:

Length of DNA: 1.36 mm = 1.36 x 10⁶ nm (since 1 mm = 10⁶ nm)
Distance between base pairs: 0.34 nm
Number of base pairs = (Total length of DNA) / (Distance between two base pairs)
- Number of base pairs = (1.36 x 10⁶ nm) / (0.34 nm/bp)
- Number of base pairs = 4 x 10⁶ bp

Therefore, there are 4 x 10⁶ base pairs in this DNA molecule.

Question 3: Genetic Code and Polypeptide Chain

A polypeptide chain has 50 amino acids. What is the minimum number of nucleotides in the mRNA that codes for this polypeptide?

Solution:

The genetic code is a triplet code, meaning that 3 nucleotides code for 1 amino acid.
Number of amino acids: 50
Number of nucleotides for amino acids: 50 amino acids x 3 nucleotides/amino acid = 150 nucleotides.
In addition to the codons for the amino acids, the mRNA also has a stop codon (UAA, UAG, or UGA) to terminate translation. A stop codon also consists of 3 nucleotides.
Total number of nucleotides: 150 (for amino acids) + 3 (for stop codon) = 153 nucleotides.

Therefore, the minimum number of nucleotides in the mRNA that codes for this polypeptide is 153.

Question 4: DNA Replication and Isotopes

If a DNA molecule is replicated for two generations in a medium containing ¹⁴N, starting with a hybrid DNA molecule (¹⁵N-¹⁴N), what will be the ratio of hybrid DNA to light DNA in the second generation?

Solution:

Starting DNA: One hybrid molecule (¹⁵N-¹⁴N).
First Generation (in ¹⁴N medium):
- The hybrid DNA will replicate semi-conservatively.
- The two strands (¹⁵N and ¹⁴N) will separate, and each will serve as a template for a new ¹⁴N strand.
- This will result in two DNA molecules:
  - One hybrid molecule (¹⁵N-¹⁴N)
  - One light molecule (¹⁴N-¹⁴N)
Second Generation (in ¹⁴N medium):
- The two molecules from the first generation will replicate.
- The hybrid molecule (¹⁵N-¹⁴N) will produce one hybrid (¹⁵N-¹⁴N) and one light (¹⁴N-¹⁴N) molecule.
- The light molecule (¹⁴N-¹⁴N) will produce two light (¹⁴N-¹⁴N) molecules.
- Total molecules in the second generation: 4
  - 1 hybrid molecule (¹⁵N-¹⁴N)
  - 3 light molecules (¹⁴N-¹⁴N)
Ratio of hybrid to light DNA: 1 : 3

Therefore, the ratio of hybrid DNA to light DNA in the second generation will be 1:3.

Numerical Problems: Molecular Basis of Inheritance

Problem 1: Calculating Nucleotide Percentages (Chargaff's Rule)

A double-stranded DNA molecule contains 20% Adenine (A). Calculate the percentages of Thymine (T), Guanine (G), and Cytosine (C) in this DNA molecule.

Solution:

According to Chargaff's rules, in a double-stranded DNA molecule:

The amount of Adenine (A) is equal to the amount of Thymine (T). (A = T)
The amount of Guanine (G) is equal to the amount of Cytosine (C). (G = C)
The sum of all bases is 100% (A + T + G + C = 100%).

Given: A = 20%

Calculate T: Since A = T, then T = 20%.
Calculate G + C: A + T + G + C = 100% 20% + 20% + G + C = 100% 40% + G + C = 100% G + C = 100% - 40% = 60%
Calculate G and C: Since G = C, and G + C = 60%: G = 60% / 2 = 30% C = 60% / 2 = 30%

Therefore, the percentages are: Thymine (T) = 20%, Guanine (G) = 30%, and Cytosine (C) = 30%.

Problem 2: Calculating Number of Nucleotides in a DNA Segment

A segment of a DNA molecule is 100 base pairs (bp) long. How many nucleotides are present in this segment?

Solution:

A base pair (bp) consists of two nucleotides (one on each strand) joined by hydrogen bonds.
Therefore, 1 base pair = 2 nucleotides.

Given: Length of DNA segment = 100 base pairs

Number of nucleotides = Length in base pairs * 2 Number of nucleotides = 100 bp * 2 nucleotides/bp = 200 nucleotides

Therefore, there are 200 nucleotides present in a 100 base pair DNA segment.

Question 1: DNA Structure and Chargaff's Rule

A segment of DNA has 1000 base pairs. The number of adenine bases is 200. What is the number of cytosine bases in this segment?

Solution:

Total base pairs: 1000
Total number of bases: 1000 x 2 = 2000
According to Chargaff's rule, the amount of Adenine (A) is equal to the amount of Thymine (T), and the amount of Guanine (G) is equal to the amount of Cytosine (C).
- A = T = 200
Total A + T: 200 + 200 = 400
Remaining bases (G + C): 2000 - 400 = 1600
Since G = C, the number of cytosine bases is:
- C = 1600 / 2 = 800

Therefore, there are 800 cytosine bases in this segment of DNA.

Question 2: DNA Length Calculation

The distance between two consecutive base pairs in a DNA molecule is 0.34 nm. If the length of a DNA molecule is 1.36 mm, calculate the number of base pairs in this DNA.

Solution:

Length of DNA: 1.36 mm = 1.36 x 10⁶ nm (since 1 mm = 10⁶ nm)
Distance between base pairs: 0.34 nm
Number of base pairs = (Total length of DNA) / (Distance between two base pairs)
- Number of base pairs = (1.36 x 10⁶ nm) / (0.34 nm/bp)
- Number of base pairs = 4 x 10⁶ bp

Therefore, there are 4 x 10⁶ base pairs in this DNA molecule.

Question 3: Genetic Code and Polypeptide Chain

A polypeptide chain has 50 amino acids. What is the minimum number of nucleotides in the mRNA that codes for this polypeptide?

Solution:

The genetic code is a triplet code, meaning that 3 nucleotides code for 1 amino acid.
Number of amino acids: 50
Number of nucleotides for amino acids: 50 amino acids x 3 nucleotides/amino acid = 150 nucleotides.
In addition to the codons for the amino acids, the mRNA also has a stop codon (UAA, UAG, or UGA) to terminate translation. A stop codon also consists of 3 nucleotides.
Total number of nucleotides: 150 (for amino acids) + 3 (for stop codon) = 153 nucleotides.

Therefore, the minimum number of nucleotides in the mRNA that codes for this polypeptide is 153.

Question 4: DNA Replication and Isotopes

Solution:

Starting DNA: One hybrid molecule (¹⁵N-¹⁴N).
First Generation (in ¹⁴N medium):
- The hybrid DNA will replicate semi-conservatively.
- The two strands (¹⁵N and ¹⁴N) will separate, and each will serve as a template for a new ¹⁴N strand.
- This will result in two DNA molecules:
  - One hybrid molecule (¹⁵N-¹⁴N)
  - One light molecule (¹⁴N-¹⁴N)
Second Generation (in ¹⁴N medium):
- The two molecules from the first generation will replicate.
- The hybrid molecule (¹⁵N-¹⁴N) will produce one hybrid (¹⁵N-¹⁴N) and one light (¹⁴N-¹⁴N) molecule.
- The light molecule (¹⁴N-¹⁴N) will produce two light (¹⁴N-¹⁴N) molecules.
- Total molecules in the second generation: 4
  - 1 hybrid molecule (¹⁵N-¹⁴N)
  - 3 light molecules (¹⁴N-¹⁴N)
Ratio of hybrid to light DNA: 1 : 3

Therefore, the ratio of hybrid DNA to light DNA in the second generation will be 1:3.

Numerical Problems - Molecular Basis of Inheritance

Numerical Problems: Molecular Basis of Inheritance

Problem 1: Calculating Nucleotide Percentages (Chargaff's Rule)

Problem 2: Calculating Number of Nucleotides in a DNA Segment

Question 1: DNA Structure and Chargaff's Rule

Question 2: DNA Length Calculation

Question 3: Genetic Code and Polypeptide Chain

Question 4: DNA Replication and Isotopes

On this page

Numerical Problems - Molecular Basis of Inheritance

Numerical Problems: Molecular Basis of Inheritance

Problem 1: Calculating Nucleotide Percentages (Chargaff's Rule)

Problem 2: Calculating Number of Nucleotides in a DNA Segment

Question 1: DNA Structure and Chargaff's Rule

Question 2: DNA Length Calculation

Question 3: Genetic Code and Polypeptide Chain

Question 4: DNA Replication and Isotopes

On this page