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Why viruses "die" outside the host after sufficient time? I want to know the biological process behind it. What role being exposed to air plays in virus destruction? What role sunlight plays in the same?

Now there might be many biological processes behind this phenomenon, please mention the same and explain about any specific virus.

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  • $\begingroup$ Answers on that post doesn't explain the biological process involved. $\endgroup$ – fghjkl May 23 at 14:04
  • $\begingroup$ But the question covers your point. Sorry, the message is autogenerated to be polite, but in doing so misses the point that SE Biology doesn't do duplicate questions as the idea (at least in principle) is to amass a definitive collection of questions. $\endgroup$ – David May 23 at 16:37
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    $\begingroup$ @fghjkl: Virus "death" isn't really a biological process, it's a chemical one. Think of viruses as complex chemicals, and those chemicals degrading in the presence of UV light & oxygen, or just drying up because they aren't in a liquid environment. $\endgroup$ – jamesqf Jun 2 at 15:40
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Viruses are metabolically inactive outside the host cell. They bind to the specific receptor to be activated. They cannot replicate without a host. For example, in case of HIV, the protein envelope of the virus binds to CD4 receptor fo T helper cells and thus supressing the immune system gradually.These viruses are only active when it binds to CD4 receptor and not any other receptor.

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