a) n pept.: nHCl = 1:5 ⇔⇔ are 5 grupe -NH2 neimplicate in -CO-NH ; ⇒⇒ cel putin 4 resturi de diaminoaminoacizi ;
b) n pept :nNaOH = 1:3 ⇔⇔ are 3 grupe -COOH ''libere'' ⇒⇒ cel putin 2 resturi de acizi dicarboxilici;
c) n pept.: nCH3-COCl = 1:7 ⇔⇔ are 5 gr. -NH2 libere + 2 grupe -OH ⇒⇒ 2 resturi de serina;
A) A
B) A
C) F
D) F
E) F
