Question

1. O dipeptida simpla contine 12,96% N . Identifica aminoacidul monoaminomonocarboxilic de la care provine .
2. In miosina , glicocolul se afla in proportie de 25 moli / 100 kg proteina . Afla compozitia procentuala a glicocolului in miosina .
3. Keratina din par contine aprocimativ 3 % S provenit din cisteina .
a) Afla continutul procentual al S in cisteina
b) Ce cantitate de cisteina ar fi necesara obtinerii a 500 kg de keratina ?

Answer 1

Aminoacidmonoaminomonocarboxilic: CnH2n+1NO2
2CnH2n+1NO2 -> C2nH4nN2O3 + H2O
Mdipeptida=12(2n)+4n+14*2+16*3=28n+76g/mol
28n+76................100%
28......................12,96%
2800=362,88n + 984,86
362,88n=1815,04 => n=5 => C5H11NO2 - valina
2)
glicocol: NH2-CH2-COOH sau C2H5NO2
Mglicocol=12*2+5+14+16*2=24+5+14+32=75g/mol
n=m/M => m=n*M=25*75=1875g=1,875kg glicocol
100kg..........100%
1,875kg...........x%
x=1,875% glicocol
3)
a)
cisteina: SH-CH2-CH(NH2)-COOH (paranteza reprezinta ramificatie)
Mcisteina=MC3H7NO2S=12*3+7+14+16*2+32=36+7+14+32+32=121g/mol
121g.................100%
32g.......................x%
x=3200/121=26,446281%S
b)
3% din 500kg=3*500/100=15kgS
26,446281% din Y=15000g
26,446281*Y/100=15000g
26,446281Y=1500000g
Y=1500000/26,446281=56718,75g cisteina=56,71875kg cisteina