Unless I’m missing something, this is a simple matter of counting. The heterozygote makes two distinguishable chains, call them alpha and beta. If all types of dimers are equally likely, there are three types of dimers possible: pure alpha, pure beta and alpha-beta heterodimers. If each are equally likely, then 33.3% of the mixture should be represented by each type of dimer. I guess you can argue there are five types of proteins, the two monomers and the three dimers, but there is no way to know how many monomers are present without knowing more information (like turnover rates and association constants, etc).

Sounds as though you just did somebody’s homework for them there blcr11…

congrats blcr11 🙂
And be careful, ADH is the accepted abreviation of antidiurhetic hormone, which is only one polipeptide chain with 9 aminoacids.

Well, if they copied my answer verbatim without thinking it through for themselves, they probably got it wrong, because I think I did it wrong. I was acting as if the chains leapt into existence as pre-made dimers, which is, obviously, not the case. They are made as individual molecules that later associate to form dimers. If you look at it as a pseudo-Punnet square, then 25% of the dimers are of each homodimer and 50% are heterodimers (alpha associates with beta or beta associates with alpha). There are four possible associations, two of them are the same heterodimer, so there are three distinguishable proteins, ignoring residual monomers. Sorry to mislead anyone, but this is a caveat emptor board.