67
YECHISH: 180g-----2
4500 ----x
X=4500*2 =9000: 180= 50 mol
180
JAVOB: 4500 g C6 H12 O6 parchalanishidan 50 mol sut kislota hosil bo`ladi.
3) Muskullarda 7 mol glyukoza parchalandi. Shundan 3
mol glyukoza kislorod
ishtirokida, 4 mol glyukoza kislorod ishtirokisiz parchalandi. Qancha CO
2
, H
2
O,
sut kislota hosil bo`ladi.
BERILGAN: 7mol glukoza
3 mol C6 H12 O6 aerob parchalangan
4 mol C6 H12 O6 anaerob parchalangan
CO
2
=? H
2
O=? Sut kislota =?
YECHISH: C6 H12 O6 + 6O
2
→ 6CO
2
+ 6H
2
O
3 C6 H12 O6 + 18O
2
→ 18CO
2
+ 18H
2
O
C6 H12 O6
→ 2C
3
H
6
O
3
4 C6 H12 O6
→ 8C
3
H
6
O
3
JAVOB: 3 mol glyukoza kislorodli
muhitda parchalansa, 18 mol CO
2
va 18
mol H
2
O
ajraladi. 4 mol glyukoza kislorodsiz muhitda parchalansa, CO
2
va H
2
O hosil bo`lmaydi. Faqat 8 mol sut kislota (C
3
H
6
O
3
) hosil bo`ladi.
4 ) Anaerob nafas olish jarayonida sitoplazmada 14 molekula sut kislota hosil
bo’ldi. Parchalangan glyukozaning miqdorini aniqlang .
BERILGAN: O
2
li muhutda 14 molekula sut kislota
Parchalangan C
6
H
12
O
6
=?
YECHISH: 1mol C6 H12 O6
→ 2mol C
3
H
6
O
3
x C6 H12 O6
→ 14 mol C
3
H
6
O
3
x= 14*1 =14:2=7
2
JAVOB: 7 molekula glyukoza parchalangan.
68
5) Dissimilatsiya jarayonida 7 mol glukoza parchalangan. Agar 2 mol glukoza
to`liq parchalangan bo`lsa, qancha (mol) ATF sintezlangan?
BERILGAN: 7 mol C6 H12 O6 parchalangan
2 mol C6 H12 O6 to`liq parchalansa,
ATF=?
YECHISH:
1mol C
6
H
12
O
6
→38mol ATF 1mol C
6
H
12
O
6
→2mol
ATF
2 mol C
6
H
12
O
6
→ xmol ATF 5 mol C
6
H
12
O
6
→ x mol
ATF
X= 38*2= 76:1= 76 mol X= 5*2=10:1=10 mol
ATF
7 mol C
6
H
12
O
6
2 mol to`liq parchalangan
76+10=86 ATF
5 mol to`liqsiz parchalangan
JAVOB: 2 mol glyukoza to`liq parchalansa, 86 mol ATF hosil bo`ladi.
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