Why does the inhibition of translation initiation cause the accumulation of 80S ribosomal monosomes?

Question

As I have read in (1), inhibition of translation initiation will increase the number of 80S ribosomes while decreasing the fraction of polysomes due to the polysome-runoff. The net effect is to decrease the polysome/monosome (P/M) ratio and we observe it for example by inducing ER stress by tunicamycin in (2).

I understand the reduction in polysomes - since we don't interefere with elongation, polysomes work as normal, fallling of eventually, but new monosomes cannot proceed to polysomal stage due to inhibited initation.

However, why do the 80S ribosomes accumulate? I have always thought that 80S ribosome is always bound to mRNA, otherwise it dissociates as separate 40S and 60S subunits. Does this effect therefore mean that the accumulated 80S fraction is mRNA-free and therefore "vacant 80S ribosomes", as decribed in (1), can exist?

Thanks a lot!

1. https://books.google.co.uk/books?id=6sEq1xLu_hcC&pg=PA130&lpg=PA130&dq=polysome+translation+initiation&source=bl&ots=Un6JjImBJX&sig=ln9mn0KoJDJaVOGK8LbrEmMz9XE&hl=cs&sa=X&ei=x2JiVci8N4jX7Ab02YLgBQ&ved=0CGQQ6AEwCw#v=onepage&q=polysome%20translation%20initiation&f=false

2. http://jcs.biologists.org/content/115/11/2443/F4.large.jpg

Answer 1

It will depend on the drug you are using. There is a pathway that forms the translational initiation complex that starts with binding the mRNA's cap, the small subunit scans to find the AUG, then the large subunit binds, and so on. The 80 S peaks on the density gradients are not vacant, they are poised on mRNAs waiting to initiate (if you removed the inhibitor).