6.08 End of Chapter Questions

B) Sugar-phosphate chain
Both DNA and mRNA contain a sugar-phosphate backbone, which provides structural support to the molecules.

C) Thymine can pair with either adenine or uracil.
This statement is incorrect because thymine pairs only with adenine in DNA, while uracil pairs with adenine in RNA.

D) 4³
With four different bases and three positions in a triplet, the number of possible triplets is calculated as 4×4×4=4³=64

D) Nucleotides
Nucleic acids, such as DNA and RNA, are made up of chains of nucleotides, which consist of a sugar, a phosphate group, and a nitrogenous base.

• Unwinding of the DNA double helix:
  • The enzyme helicase unwinds and separates the two strands of the DNA molecule by breaking the hydrogen bonds between the complementary nitrogenous bases, creating two single-stranded templates.
• Binding of RNA primer:
  • The enzyme primase synthesizes a short RNA primer on each template strand, providing a starting point for DNA synthesis.
• Elongation of the new strand:
  • DNA polymerase adds nucleotides to the 3′ end of the RNA primer, following the base-pairing rules (A with T, and C with G).
  • On the leading strand, DNA polymerase synthesizes the strand continuously in the 5′ to 3′ direction.
  • On the lagging strand, DNA polymerase synthesizes the strand in short segments called Okazaki fragments, which are later joined by DNA ligase.
• Proofreading and finalising:
  • DNA polymerase also has a proofreading function to correct any errors.
  • After the entire DNA molecule is replicated, the RNA primers are replaced with DNA, and the strands are sealed by DNA ligase to form two identical DNA molecules.

Explanation:

The tRNA anticodon binds to the complementary mRNA codon using base pairing rules (A pairs with U, T pairs with A, C pairs with G, and G pairs with C).

The DNA triplet is the sequence that codes for the corresponding mRNA codon using the complementary base pairing rules (A pairs with T, C pairs with G, and G pairs with C).

a) Missense mutation
This is a missense mutation because it results in a change in the amino acid coded for by the triplet (from aspartic acid to histidine).

This type of mutation is silent in some cases because of the degeneracy of the genetic code. Many amino acids are coded for by more than one codon.

The mutation may change the DNA triplet, but if the new triplet codes for the same amino acid due to the redundancy in the codon system, there is no effect on the protein. This is called a silent mutation.

The mutation could be harmful because it leads to a change in the amino acid sequence of the protein.

If the new amino acid has different chemical properties, it can affect the protein’s structure and function, potentially disrupting its biological role.

For example, if the protein is an enzyme or structural protein, the change could impair its ability to function properly, leading to health problems or disease.

The mutation is more likely to be harmful because it occurred in a gamete-producing cell (germline cell), meaning it can be passed on to offspring.

If the mutation is inherited, it could affect multiple generations, leading to a potential inherited disorder or disease that affects the organism throughout its life.

The addition or deletion of three nucleotides often has less effect because it causes the insertion or deletion of an entire amino acid in the polypeptide chain, but does not shift the reading frame of the codons. This is known as a frame-preserving mutation.

However, the addition or deletion of a single nucleotide results in a frameshift mutation, which shifts the entire reading frame, causing changes in every amino acid downstream from the mutation. This often results in a nonfunctional protein or a protein with an altered function, making it potentially more harmful.

Complete the table below to distinguish between the processes of transcription and translation.

Translation: The process where mRNA is used to assemble amino acids into a polypeptide chain. It occurs in the cytoplasm on ribosomes, where tRNA molecules bring the appropriate amino acids to the mRNA. The polypeptide is made up of amino acids, which are the building blocks for proteins.

Transcription: The process of copying a gene’s DNA sequence to make an RNA molecule (mRNA). It takes place in the nucleus, with RNA polymerase catalysing the reaction. The template for mRNA synthesis is DNA, and the end product is mRNA made up of nucleotides.

U: mRNA (messenger RNA) – The mRNA strand serves as the template for translation.

V: Ribosome – The ribosome is responsible for translating the mRNA into a polypeptide.

W: Amino acid – This is the amino acid being carried by the tRNA at the A-site of the ribosome.

X: tRNA (transfer RNA) – The tRNA molecule brings the amino acid to the ribosome to match with the codon on the mRNA.

Y: Polypeptide bond – The chemical bond between two amino acids in the growing polypeptide chain, which forms at the ribosome.

The bond formed is a peptide bond. This bond connects two adjacent amino acids in the polypeptide chain.

As the ribosome moves, the A-site and P-site change positions, and the tRNA in the A-site shifts to the P-site, enabling the continuation of polypeptide elongation.

The position of V (the ribosome) has moved because the ribosome is moving along the mRNA strand during translation.

As the ribosome reads the mRNA codons, it shifts along the mRNA in a 5′ to 3′ direction, allowing for the addition of amino acids to the growing polypeptide chain.

The relative numbers of bases in the ox spleen and ox thymus are very similar, indicating that the DNA in these tissues has a consistent base composition. This is expected because both tissues are from the same organism, and DNA base composition does not vary significantly across different tissues in an individual.

The relative numbers of bases in yeast differ from those in ox spleen and thymus because yeast is a different organism with a distinct genetic makeup. Different species often have variations in the proportions of the four nucleotide bases in their DNA due to differences in their genome structure and sequence.

The ratio of A to T and C to G is equal to 1 in ox and yeast because A pairs with T and C pairs with G in the DNA double helix. In double-stranded DNA, these base pairings must occur in equal amounts, so the quantities of A and T, and C and G, should be approximately equal. This is consistent with Chargaff’s rules.

In the virus with single-stranded DNA, the ratio of A to T and C to G is not equal to 1 because single-stranded DNA does not follow Chargaff’s rule. In single-stranded DNA, the base composition is not constrained by complementary base pairing, leading to unequal proportions of bases.

STUDY NOTES:

Chargaff’s rule may or may not apply to viral DNA, depending on the type of viral genome. Here’s a detailed breakdown:

1. Double-Stranded DNA (dsDNA) Viruses:

• Chargaff’s rule applies because base pairing occurs in the same way as in cellular DNA:
  • Adenine (A) pairs with Thymine (T).
  • Cytosine (C) pairs with Guanine (G).
• This results in equal amounts of A = T and C = G in the genome.

Examples:

• Herpesviruses (e.g., HSV-1, Varicella-Zoster virus)
• Adenoviruses
• Bacteriophages with dsDNA (e.g., T4 phage)

2. Single-Stranded DNA (ssDNA) Viruses:

• Chargaff’s rule does not apply to ssDNA because the DNA is single-stranded and does not form complementary base pairs within the genome.
• The proportions of A, T, C, and G can vary and are not constrained by complementary base pairing.

Examples:

• Parvoviruses

3. RNA Viruses:

• Chargaff’s rule does not apply to RNA viruses (both single-stranded and double-stranded) because they use uracil (U) instead of thymine (T), and base pairing, if present, may not follow the same constraints as in DNA.