Foot-and-mouth disease virus (FMDV) genome size — Approximately how many nucleotides are present in the single-stranded RNA (ssRNA) molecule of this picornavirus?

Difficulty: Easy

Correct Answer: 8,000

Explanation:

Introduction / Context:Picornaviruses possess positive-sense, single-stranded RNA genomes that function directly as mRNA upon entry. Knowing typical genome sizes helps in designing RT-PCR assays, sequencing strategies, and understanding coding capacity (single large polyprotein strategy). Foot-and-mouth disease virus (FMDV) is a classic member of this family affecting cloven-hoofed animals.

Given Data / Assumptions:

  • The genome is single-stranded, positive-sense RNA.
  • It encodes one large polyprotein that is post-translationally cleaved.
  • Typical picornavirus genomes fall in the ~7,000–9,000 nucleotide range.

Concept / Approach:Recall the canonical size range and select the closest representative value. FMDV genomes are about 8 to 8.5 kilobases in length. Therefore, the best rounded option among those provided is 8,000 nucleotides.

Step-by-Step Solution:

Identify viral family → Picornaviridae (FMDV).Recall genome form → +ssRNA, polyadenylated, covalently linked VPg at 5’ end.Match to size range → approximately 8 kb.

Verification / Alternative check:Reference genomes of various FMDV serotypes consistently report ~8.2–8.5 kb, validating that 8,000 is the closest rounded choice here.

Why Other Options Are Wrong:

1,000 or 5,000: too small to encode the polyprotein and regulatory elements.10,000 or 15,000: exceed typical picornaviral genome length; 15,000 approaches paramyxovirus sizes (−ssRNA), not picornaviruses.

Common Pitfalls:Confusing “kb” (kilobases) with “kDa” (kilodaltons) or mixing up DNA vs RNA viruses with larger genomes.

Final Answer:8,000.

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion