Is there some explanation for $y_t=1$ The Yukawa coupling of the top quark is Dirac-natural in a too excellent way, it is within one sigma experimentally, and within 99.5% in absolute value, of being equal to one. Without some symmetry, it seems too much for a quantity that is supposed to come down from GUT/Planck scale via the renormalization group. Is there some explanation for this?
 A: This is a very naive answer or, in fact, it is not an answer. Among all numbers of order one, is not  $y_t=1$ the most likely value, i.e., the statically expected value? Why do we need an explanation for $y_t=0.995$ and not for, say, $y_t=0.629$?
A: In a new paper, Rodejohann and Zhang write (pages 13 to 14) that in the standard model (with massless neutrinos), the top Yukawa can never RG-evolve to exactly 1, but that this becomes possible once you have massive neutrinos. Then it will grow beyond 1 as you continue to still higher energies. But they also write that attaining the exact value 1 could indicate "the restoration of certain kinds of Yukawa unifications or flavor symmetries". So if you can find a form of symmetry breaking which sometimes occurs when a coupling is exactly unity, and then use it appropriately in a GUT or other model of new physics... then you will have an explanation. 
A: Some lecture notes, e.g. http://arxiv.org/pdf/0906.1271v4.pdf p. 149, are more concrete about the "order one" condition by asking that the $J=0$ channel of the diagram $t\bar t \to ZZ$ that emits two $Z$ must be canceled with the diagram that actually aniquilates $t \bar t$ into $H$ and then decays to two $Z$. The first diagram goes as $\alpha^2 m_t^2 / m_Z^4$, and the second one goes as $Y_t^2 / m_Z^4$. This is, it seems, the unitarity argument of Llewellyn Smith and Christopher Hubert (CERN-TH-1665)
