Foot-and-mouth disease virus (FMDV) genome size — Approximately how many nucleotides are present in the single-stranded RNA (ssRNA) molecule of this picornavirus?
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A1,000
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B5,000
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C8,000
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D10,000
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E15,000
Answer
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.