Question

How many all possible optically active tripeptides can be formed using only glycine, L-alanine and L-Tyrosine?

Answer 1

The possible combination of amino acids would be:
$H-\left(gly/ala/tyr\right)-\left(gly/ala/tyr\right)-\left(gly/ala/tyr\right)-OH$
Thus no of tripeptides formed $=3\times 3\times 3=27$
Since $H-gly-gly-gly-OH$ will be optically inactive due to the absence of a chiral centre. Hence only 26 will be optically active.
Note: only L-amino acids are given so D-amino cannot be used.