bai_giang_th._hoa_sinh_moi_nhat

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TR NG I H C K THU T CNG NGH TP. HCMKHOA MI TR NG & CNG NGH SINH H C

------

BI GING

TH C HNH

HA SINH

TS. Nguy n Hoi H ng

CN. Bi Vn Th Vinh

Dng cho sinh vin ngnh Mi tr ng v Cng ngh Sinh h cNm xu t b n: 2009


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M C L C Trang

Gi i thi u mn h c 4Quy t c lm vi c trong phng th nghi m Ha

sinh

5

1. An ton khi lm vi c v i axit v ki m 5

2. Quy t c lm vi c v i ha ch t th nghi m 6

Bi 1 Cch pha ch cc dung dch dng trong th

nghi m Ha sinh

8

I L thuy t 81. Dung dch 8

2. Dung dch m 14

II Th c hnh 20

III Bi n p 20

Bi 2 nh l ng ng kh b ng ph ng php Acid

dinitro-salicylic (DNS)

21

I L thuy t 211. nh ngha 21

2. Nguyn t c 21

3. X l m u 21

II Th c hnh 22

1. D ng c - ha ch t 22

2. Ti n hnh th nghi m 23

3. Tnh k t qu 24

III Bi n p 24

Bi 3 nh l ng Nit t ng s b ng ph ng php

Kjeldahl

25

I L thuy t 25

1. nh ngha 25


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2. Nguyn t c 26

II Th c hnh 26



3. Tnh k t qu 28

III Bi n p 28

Bi 4 nh l ng protein b ng ph ng php Bradford 30

I L thuy t 30

1. nh ngha 30

2. Nguyn t c 30

II Th c hnh 31

1. D ng c - ha ch t 312. Ti n hnh th nghi m 31

3. Tnh k t qu 32

III Bi n p 32

Bi 5 Ph ng php xc nh ho t tnh enzyme 33

I L thuy t 33

1. Enzyme v n v o ho t tnh enzyme 33

2. Ph ng php xc nh ho t tnh enzyme 343. Enzyme amylase v ph ng php xc nh ho t

tnh

35

II Th c hnh 36



3. Tnh k t qu 38

III Bi n pTi li u tham kh o

3839


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GI I THI U MN H C

Bi gi ng Th c hnh sinh ho dnh cho sinh vin nm th hai khoa Mi

tr ng & Cng ngh Sinh h c, tr ng i h c K Thu t Cng Ngh TP.HCM.

V i th i l ng 30 ti t v tu thu c i u ki n, c s v t ch t c a phng th

nghi m cho php, sinh vin ti n hnh lm 5 bi th c hnh g m cc n i dung chnh c a

h c ph n l thuy t ha sinh c s .

1. Gi i thi u phng th nghi m ha sinh, cch pha ch cc dung dch dung

trong th nghi m ha sinh ;

2. nh l ng ng kh b ng ph ng php acid dinitro-salicylic (DNS);

3. nh l ng nit t ng s b ng ph ng php Kjeldahl;

4. nh l ng protein b ng ph ng php Bradford;

5. Xc nh ho t tnh enzyme amylase.


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QUY T C LM VI C TRONG PHNG TH NGHI M HA

SINH

I. An ton khi lm vi c v i axit v ki m1. An ton khi lm vi c v i axit:

Ph i lm vi c trong t ht b t c khi no un nng axit ho c th c hi n ph n

ng v i cc h i axit t do.

Khi pha long, lun ph i cho axit vo n c tr phi c dng tr c ti p.

Gi axit khng b n vo da ho c m t b ng cch eo kh u trang, gng tay v

knh b o v m t. N u lm vng ln da, l p t c r a ngay b ng m t l ng n c l n.

Lun ph i c k nhn c a chai ng v tnh ch t c a chng.

H2SO4: Lun cho acid vo n c khi pha long, s d ng kh u trang v gng tay

trnh phng khi vng acid

Cc acid d ng h i (HCl) thao tc trong t ht v mang gng tay, knh b o h .

2. An ton khi lm vi c v i ki m

Ki m c th lm chy da, m t gy h i nghim tr ng cho h h h p.

Mang gng tay cao su, kh u trang khi lm vi c v i dung dch ki m m c.

Thao tc trong t ht, mang m t n ch ng c phng ng a b i v h i ki m.

Amoniac: l m t ch t l ng v kh r t n da, mang gng tay cao su, kh u trang,

thi t b b o v h th ng h h p. H i amoniac d chy, ph n ng m nh v i ch t oxy

ho, halogen, axit mnh.

Amoni hydroxyt: ch t l ng n da, t o h n h p n v i nhi u kim lo i n ng: Ag,

Pb, Zn ... v mu i c a chng.

Kim lo i Na, K, Li, Ca: ph n ng c c m nh v i n c, m, CO2, halogen, axit

m nh, d n xu t clo c a hydrocacbon. T o h i n mn khi chy. C n mang d ng c

b o v da m t. Ch s d ng c n kh khi t o dung dch natri alcoholate, cho vo t t .Trnh t o tinh th c ng khi ho tan. T ng t khi ho tan v i n c, ng th i ph i

lm l nh nhanh.

Oxit canxi r t n da, ph n ng c c m nh v i n c, c n b o v da m t, ng

h h p do d nhi m b i oxit.


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Natri v kali hydroxyt: r t n da, ph n ng c c m nh v i n c. Cc bi n php

an ton nh trn, cho t ng vin ho c t b t vo n c ch khng c lm ng c l i.

II. Quy t c lm vi c v i ha ch t th nghi m

1. Ho ch t th nghi m:

Cc ho ch t dng phn tch, lm tiu b n, ti n hnh ph n ng, ... trong phng th

nghi m c g i l ha ch t th nghi m.

Ho ch t c th d ng r n (Na, MgO, NaOH, KCl, (CH3COO)2...; l ng (H2SO4,

aceton, ethanon, chloroform, ...) ho c kh (Cl2, NH3, N2, C2H2 ...) v m c tinh khi t

khc nhau:

- S ch k thu t (P): s ch > 90%

- S ch phn tch (PA): s ch < 99%

- S ch ha h c (PC): s ch > 99%

Ha ch t c tinh khi t khc nhau c s d ng ph h p theo nh ng yu c u khc

nhau v ch nn s d ng ha ch t cn nhn hi u.

2. Nhn hi u ho ch t:

Ha ch t c b o qu n trong chai l th y tinh ho c nh a ng kn c nhn ghi

tn ho ch t, cng th c ha h c, m c s ch, t p ch t, kh i l ng tnh, kh i l ng

phn t , n i s n xu t, i u ki n b o qu n.3. Cch s d ng v b o qu n ho ch t:

Khi lm vi c v i ha ch t, nhn vin phng th nghi m c ng nh sinh vin c n h t s c

c n th n, trnh gy nh ng tai n n ng ti c cho mnh v cho m i ng i. Nh ng i u

c n nh khi s d ng v b o qu n ha ch t c tm t t nh sau:

- Ha ch t ph i c s p x p trong kho hay t theo t ng lo i (h u , v c ,

mu i, acid, baz , kim lo i, ...) hay theo m t th t a, b, c khi c n d tm.

- T t c cc chai l u ph i c nhn ghi, ph i c k nhn hi u ha ch t tr c

khi dng, dng xong ph i tr ng v tr ban u.

- Chai l ha ch t ph i c n p. Tr c khi m chai ha ch t ph i lau s ch n p,

c chai, trnh b i b n l t vo lm h ng ha ch t ng trong chai.

- Cc lo i ha ch t d b thay i ngoi nh sng c n ph i c gi trong chai

l mu vng ho c nu v b o qu n vo ch t i.


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- D ng c dng l y ha ch t ph i th t s ch v dng xong ph i r a ngay,

khng dng l n n p y v d ng c l y ha ch t.

- Khi lm vi c v i ch t d n , d chy khng c g n n i d b t l a. Khi

c n s d ng cc ha ch t d b c h i, c mi,... ph i a vo t ht, ch y kn n p

sau khi l y ha ch t xong.

- Khng ht b ng pipette khi ch cn t ha ch t trong l , khng ng i hay n m

th ha ch t.

- Khi lm vi c v i acid hay base m nh:

Bao gi c ng acid hay base vo n c khi pha long (khng c n c

vo acid hay base);

Khng ht acid hay base b ng mi ng m ph i dng cc d ng c ring nh ng

bp cao su.

Tr ng h p b b ng v i acid hay base r a ngay v i n c l nh r i bi ln v t

b ng NaHCO3 1% (tr ng h p b ng acid) ho c CH3COOH 1% (n u b ng base). N u

b b n vo m t, d i m nh v i n c l nh ho c NaCl 1%.

Tr ng h p b ha ch t vo ming hay d dy, n u l acid ph i sc mi ng v

u ng n c l nh c MgO, n u l base ph i sc mi ng v u ng n c l nh c CH3COOH

1%.

L u cc k hi u c nh bo nguy hi m:


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BI 1: CCH PHA CH CC DUNG D CH

DNG TRONG TH NGHI M HA SINH

I. L thuy t1.1. Dung dch

Dung dch l h n h p c a hai hay nhi u ch t tc ng t ng h v i nhau v

mt vt l v ha h c. Trong dung dch g m c ch t ha tan v dung mi. N u ch t

ha tan d ng r n th g i l ch t tan, n u l ch t l ng th g i l dung ch t.

Ty theo tnh ch t c a dung mi m phn thnh dung dch n c v dung dch

khan. Ph n l n cc dung dch acid, base, mu i trong phng th nghi m l dung dch

n c, dng dung mi l n c. M t s ch t khc tan trong dung mi h u c .

Hm l ng ch t ha tan trong dung dch th hi n n ng dung dch. C

nhi u cch bi u th n ng khc nhau. M i cch s ti n d ng trong chu n b, phn

tch v tnh ton khc nhau.

1.1.1. Cc n v n ng dung dch

a) N ng ph n trm, (%)

i) N ng ph n trm kh i l ng - kh i l ng, % (w/w): l s gam ch t tan c

trong 100g dung dch.

V d : dung dch NH4Cl 5% (w/w) l trong 100g dung dch c ch a 5g NH4Cl

ii) N ng % kh i l ng th tch (w/v): l s g ch t tan c trong 100ml dung

dch.

V d : dung dch CuSO4 10% (w/v) l trong 100ml dung dch ch a 10g CuSO4

iii) N ng ph n trm th tch - th tch, % (v/v): l s ml dung ch t c trong

100ml dung dch.

V d : dung dch glycerine 10% (v/v) l trong 100ml dung dch ch a 10ml glycerine.

b) N ng gam-lit, (g/L): l s gam ch t tan c trong 1 lt dung dch.c) N ng phn t gam hay n ng mol, (Mol/L) hay M: l s phn t gam

(hay s mol) ch t tan trong 1 lt dung dch.

V d : Dung dch KH2PO4 M/15 l trong 1000ml dung dch ch a M/15 phn t gam

KH2PO4.


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d) N ng ng l ng (N): l s ng l ng gam (lg) ch t tan c trong 1

lit dung dch.

S ng l ng ch t tan = s mol (n) x h s ng l ng (z)

H s ng l ng (z): ph thu c vo b n ch t c a ch t v ph n ng m

ch t tham gia.

i) N u ph n ng l ph n ng acid, base: z l s ion H+ hay OH- m 1 phn t ,

ion c a ch t tc d ng v a .

V d : Ph n ng gi a HCl v NaOH

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

H2SO4 2 H+ z = 2

NaOH 1 OH- z = 1

ii) N u ph n ng l ph n ng xy ha kh : z l s electron m 1 phn t , ionc a ch t cho hay nh n.

V d : 2 Na2S2O3 + I2 Na2S4O6 + 2 NaI

I + 1e I- z = 1

S2+

- 1e S+ z = 1

e) N ng dung dch bo ha: l n ng dung dch khi t i a ch t ha tan

c m t trong dung dch.

f) n v n ng dngtrong cc php phn tch vi l ng :- N ng mg/mL: s mg ch t tan trong 1mL dung dch

- Miligam ph n trm, mg%: mg ch t ha tan trong 100g dung dch.

- Microgam ph n trm, g%: l s g ch t ha tan trong 100g dung dch.

- Ph n nghn, 0/00: s g ch t ha tan trong 1000g dung dch.

- Ph n tri u, ppm: s mg ch t ha tan trong 1kg hay 1 lt dung dch.

- Ph n t , ppb: s g ch t ha tan c trong 1kg hay 1 lt dung dch.

1.1.2. Cch pha dung dch c n ng xc nh

a) Pha dung dch c n ng ph n trm theo kh i l ng, % (w/w)

i) Ch t tan l ch t r n khan:

V d : Pha 500g dung dch NaOH 40% (w/w)

100g dung dch c n 40g NaOH

500g dung dch c n X g?


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L ng NaOH c n pha dung dch : X= (40*500)/100= 200g

L ng n c c n thi t: 500-200=300g (hay 300ml)

V y, cn 200g NaOH v ong 300 ml n c c t, ha tan ta c 500g dung

dch NaOH 40%

ii) Ch t tan l ch t r n ng m n c (CuSO4.5H2O; Na2HPO4. 12H2O;...)

Khi pha dung dch c n ph i tnh thm l ng n c k t tinh c s n.

V d : Pha 320g dung dch CuSO4 10% (w/w) t CuSO4.5H2O

M(CuSO4) = 160 v M(CuSO4.5H2O) = 250

L ng CuSO4 khan pha dung dch l: X = (10*320)/100 = 32g

L ng CuSO4.5H2O c n dng: Y = (250*32)/160 = 50g

L ng n c c t thm vo: 320 -50 = 270g (hay ml)

V y, cn 50g CuSO4.5H2O, ong 270ml n c c t, ha tan ta c 320g dung

dch CuSO4 10%.

b) Pha dung dch long t m t dung dch m c h n:

V d : Pha 500g dung dch NaOH 5% t dung dch NaOH 10%

L ng NaOH c n pha dung dch 5% l: X = (5*500)/100 = 25g

L ng dung dch NaOH 10% c n dng l: Y = (100*25)/10 = 250g

L ng n c c t thm vo: 500-250 = 250g

V y, ong 250ml n c c t v 250g dd NaOH 10%, ha tan ta c 500gNaOH 5%

c) Pha dung dch bo ho:

L y ch t tan c n pha vo becher, thm m t t n c c t v khu y cho tan. N u

sau khi khu y, ch t tan khng tan h t l ng xu ng th ph n dung dch pha trn l dung

dch bo ha. N u ch t tan tan h t, thm ch t tan v ti p t c khu y, c nh th cho

n khi ch t tan khng cn tan c n a.

d) Pha dung dch c n ng % theo th tch

i) Ch t tan l ch t r n khan

Cn l ng ch t tan c n thi t, chuy n sang bnh nh m c, dng n c c t ha

tan v nh m c n th tch ng.

V d : Pha 1 lt dung dch NaCL 5% (w/v)

L ng NaCl c n pha dung dch: X = (5*1000)/100 = 50g


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V y, cn 50g NaCl, ha tan v nh m c thnh 1 lt b ng n c c t, ta c 1 lt

dung dch NaCl 5%.

ii) Ch t tan l ch t r n ng m n c (CuSO4.5H2O; Na2HPO4. 12H2O;...)

Khi pha dung dch ta c n ph i tnh n l ng n c k t tinh c s n gi ng nh

ph n a.

iii) Ch t tan d ng l ng: M t s ch t tan d ng l ng nh HCl, H2SO4 ...

Vi c cn khng thu n l i, c th a v n v th tch theo cng th c

V = M/d

V: Th tch ch t l ng; M: kh i l ng ch t l ng c n cn; d: t tr ng ch t lng

Ch : Cc ha ch t l ng bn trn th tr ng th ng khng d ng nguyn ch t

m l cc dung dch m c. Gi i h n ha tan t i a c tnh b ng % th tch v

thay i ty theo lo i ha ch t. V d nh H2SO4: 95-98%; HCl: 37%; H3PO4: 65-

85%; NH4OH: 25%. Do khi pha cc dung dch t cc lo i ha ch t ny ta ph i ch

n n ng c a dung dch m c.

V d : Pha 440ml dung dch HCl 1% t dung dch HCl 37% (d=1,19g/ml)

L ng HCl c n pha dung dch 1% l: X= (1*440)/100 = 4,4g

L ng dung dch HCl 37% c n dng l : Y = (100*4,4)/37 = 11,9g

= 11,9/1,19 = 10 ml

L ng n c c t thm vo:440-10 = 430mlV y, dng ng ong l y 10ml dung dch HCl37%, v 430ml n c c t ta c

440 ml HCl 1%

Do vi c s d ng cc lo i bnh nh m c lm cho vi c pha ch dung dch th

nghi m tr nn n gi n v chnh xc v v y ngy nay a s cc dung dch th nghi m

c pha ch theo n ng kh i l ng - th tch (w/v).

e) Pha dung dch n ng phn t gam

i) Ch t tan l ch t r n khan

Mu n pha dung dch n ng 1M c a m t ch t no , ta tnh kh i l ng phn

t ch t (ho c tra b ng) theo n v gam. Cn chnh xc l ng ch t tan, qua ph u

cho vo bnh nh m c c dung tch 1 lt. Cho vo t ng l ng n c c t nh , l c

ha tan hon ton v a n c c t t i m c. Chuy n dung dch sang bnh ch a, l c

tr n u ng nh t.


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Khi ph i un nng dung dch ha tan, ho c qu trnh ha tan c to nhi t th

ph i ch nhi t tr l i bnh th ng (nhi t khng kh) r i m i thm n c t i v ch

nh m c.

V d : Pha 1 lt dung dch KOH 1M

Phn t l ng c a KOH: MKOH = 39 +16 +1 =56

L ng KOH pha 1 lt dung dch 1M l: 56g

V y, cn 56g KOH, ha tan trong 1 t n c, cho vo bnh nh m c 1000. y

l ph n ng t a nhi t, c n lm ngu i dung dch tr c khi nh m c thnh 1 lt.

N u mu n pha dung dch 2M; 3M hay 0,1M; 0,05M ta c ng ti n hnh t ng t

v i l ng cn t ng ng.

V d : Pha 500ml dung dch KCl 3M

Phn t l ng c a KCl: MKCl = 39 +35,5 = 74,5

L ng KCl pha 500ml dung dch 3M l: X= (74,5 *3*500)/1000 =

111,75g

V y, cn 111,75g KCl, ha tan trong 1 t n c, cho vo bnh nh m c 500,

nh m c n v ch.

ii) Ch t tan l ch t r n ng m n c: khi tnh l ng ch t tan c n cn ph i tnh

lun c kh i l ng cc phn t n c.

iii) Ch t tan d ng l ng: n u ch t tan l dung dch, ta ph i tnh ton d a von ng dung dch .

V d : 8) Pha 1 lt dung dch HCl 1M t HCl 37%

Phn t l ng HCl: MHCl = 1+35,5 = 36,5

L ng HCl 37% pha dung dch 1M l: X = (36,5*100)/37 = 98,65g

Hay 98,65/1,19 = 83ml

V y, ong 83ml HCl 37% cho vo bnh nh m c 1000 c s n 1 t n c.

nh m c thnh 1 lt. Ti n hnh pha trong t Hotte v h i acid bay ln r t c

h i.

f) Pha n ng ng l ng (N)

Vi c pha dung dch n ng ng l ng gam (N) c ng t ng t v i cch pha

dung dch n ng phn t gam (M).

1.1.3. Hi u chnh n ng dung dch


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Khi pha ha ch t, c nhi u nguyn nhn lm cho n ng dung dch khng

chnh xc nh vi c cn ong khng chnh xc, cc ha ch t khng tinh khi t hay b

ht m. Th i gian tng tr lu nn ch t tan b thng hoa, b oxy ha, dung mi bay

h i, v v y ph i ki m tra n ng th c c a cc dung dch pha s n d a vo cc dung

dch c n ng chnh xc c g i l dung dch chu n (fixanal)

a) Dung dch chu n

Dung dch chu n l cc dung dch c chu n b s n, m b o chnh xc v

c dng nh chu n cc dung dch t pha ch khi lm th nghi m. Cc ch t dng

trong dung dch chu n ph i kh b n v ng sao cho n ng c a chng khng thay i

nhanh chng theo th i gian.

Pha dung dch chu n, ta ph i dng ng chu n. ng chu n l m t ng ampun

th y tinh hay nh a ng kn. Bn trong ch a m t l ng cn chnh xc ch t tan ho c

dung dch ch t tan. Khi chuy n h t l ng ch t tan trong ng vo bnh nh m c v pha

thnh 1 lt ta c dung dch chu n c n ng ghi trn nhn ngoi ng.

Cch pha dung dch chu n t ng chu n :

- Dng inh th y tinh ch c th ng ampun, h ng ln ph u vo bnh nh m c,

dng bnh tia r a s ch ch t tan c trong ampun vo bnh nh m c 1 lit, v a thm

n c c t v a l c v a n c c t t i v ch m c.

- i v i cc h p ch t b n v ng, c thnh ph n khng thay i nh NaCl,

AgNO3, acid oxalic, ... c th pha dung dch chu n tr c ti p b ng cch cn chnh xc

ch t c n pha, pha long v nh m c t i th tch ng.

- i v i cc ch t nh NaOH, HCl, Na2S2O3, ... khng th pha ngay c dung

dch chu n, do cc ch t ny th ng khng b n v ng v d thay i thnh ph n, v v y

sau khi pha ph i hi u chnh l i n ng .

V d : NaOH th ng nhi m m t l ng Na2CO3 r t d ch y n c, HCl d bay

h i, Na2S

2O

3d b m t n c tinh th khi ngoi khng kh.

b)Ph ng phphi u chnh n ng dung dch

i v i cc ch t d thay i thnh ph n khi d ng r n, n u mu n pha dung

dch c n ng chnh xc, ta pha dung dch c n ng g n ng, sau hi u chnh

n ng c a dung dch d a vo ph n ng v i m t dung dch chu n thch h p.


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V d : Pha dung dch NaOH 0,1 N t NaOH r n v dng dung dch chu n

H2SO4 0,1N chu n l i.

i v i cc dung dch d thay i trong qu trnh b o qu n, m i l n s d ng l i

ph i xc nh l i h s hi u chnh n ng dung dch

Xt m t ph n ng trung ha acid - base, 1 ion gam H+ s ph n ng v i 1 ion

gam OH-. Do :

C1V1 = C2V2

N u g i Cp l n ng dung dch nh pha v Ct l n ng th c c a dung dch

ta c h s hi u chnh K:

K= Cp/Ct = Vp/Vt

V d : Dung dch chu n l H2SO4 0,1N, dung dch nh pha l NaOH 0,1N.

L y 10ml H2SO4 0,1N cho vo erlen, thm ba gi t ch th mu phenolphtalien,

dng burette chu n b ng NaOH c 11 ml NaOH

V y n ng th c t NaOH pha l : Ct = (10*0,1)/11 = 0,091 N

H s i u chnh K: K= 0,091/0,1 = 10/11 = 0,91

1.2. Dung dch m:

1.2.1. nh ngha dung dch m

pH mi tr ng lm thay i c u trc khng gian protein v m t s amino acid

c m ch nhnh phn ly (COO

-

v NH4+

t o lin k t ion). H at ng t i u c a proteinph thu c vo c u trc khng gian nh t nh trong mi tr ng, ngha l ph thu c vo

t l phn ly c a m ch nhnh thnh ion, hay ni tm l i l ph thu c vo pH mi

tr ng.

Dung dch m l dung dch c pH khng thay i nhi u l m khi m t l ng

nh acid (H+) ho c base (OH

-) c thm vo. Nh v y, dung dch m bao g m m t

c p acid base lin h p (acid y u v mu i c a acid y u ny ho c base y u v mu i c a

base ny) v t l c a chng s quy t nh pH c a dung dch.

Vi c pha dung dch m tun theo nguyn t c ch n c p acid base c h ng s

phn ly pK A/B g n v i pH m mu n pha v ph i tr n chng v i s mol b ng nhau.

Th d , c m pH gi tr 4.75, ch n c p acid - base CH3COOH

(0,1M)/CH3COONa (0,1M). N u thm vo dung dch 0.001 mol HCl th pH c a h

v n ch m c 4.748 g n b ng 4.75. Ngha l pH thay i r t t.


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Trong c th s ng, dung dch m ng vai tr quan tr ng, v d nh n nh

pH c a mu b ng cc h m carbonate v phosphate.

V nguyn t c th ph i h p cc lo i m c cc kho ng m khc nhau thnh 1

dung dch l khng c v n g. Tuy nhin s d ng cc m no cn ty thu c vo

ng d ng sao cho thnh ph n c a m khng ph n ng v i cc c u t trong h ph n

ng kh o st theo chi u h ng c h i. Cc ph n ng th ng th y l ph n ng t o k t

t a hay t o ph c ho c oxy ha kh .

C dung dch ch a h n h p 1 acid y u (V d : acid acetic) v i mu i c a n (V

d : mu i natri acetate), dung dch s cn b ng d ng nh sau:

CH3COOH + H2O -----> CH3COO-

+ H3O+

C dung dch khc ch ng h n, ch a h n h p base y u (V d : ammoniac) v i

mu i c a n (V d : mu i amoni clorua),trong dung dch cn b ng c d ng sau:

NH3 + H2O ----------> NH4+

+ OH-

Dung dch m c pH thay i r t t khi thm m t l ng acid ho c base.

Khi thm m t l ng acid m nh (H3O+), base lin h p k t h p v i n cho acid y u

(cn b ng chuy n dch theo chi u nghch) pH c a dung dch gi m khng ng k .

Tri l i, khi thm m t l ng base m nh (OH-), acid i n li cho H3O

+(cn b ng

chuy n dch theo chi u thu n), ion hidroni H3O+

trung ha OH-cho base y u H2O. K t

qu pH c a dung dch tng khng ng k .Kh nng m c a h n h p m ph thu c vo hm l ng acid v base lin

h p c trong h n h p.

1.2.2. Cch pha m t s dung dch m th ng dng

a) Dung dch m borat:

- Dung dch acid boric (a): 12,404 g H3BO3 ha tan v nh m c n 1000 ml.

- Dung dch borat (b): 19,108 g Na2B4O7.10H2O ha tan v nh m c n 1000

ml

Dung dch m borat c pH khc nhau ph thu c vo s ml dung dch (a) v

dung dch (b) theo b ng d i y:

A b pH a b pH


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9,80 0,2 6,60 6,50 3,50 8,20

9,70 0,6 6,77 6,00 4,00 8,31

9,40 0,6 7,09 5,50 4,50 8,41

9,00 1,0 7,36 5,00 5,00 8,51

8,75 1,25 7,50 4,50 5,50 8,60

8,50 1,50 7,60 4,00 6,00 8,69

8,00 2,0 7,78 3,00 7,00 8,84

7,70 2,30 7,88 2,00 8,00 8,98

7,50 2,50 7,94 1,00 9,00 9,11

7,00 3,00 8,08 0,00 10,0 9,24

b) Dung dch m citrate (pH = 3,0 6,2)

- Dung dch acid citric 0,1M (a): 21,01 g C6H8O7.H2O ha tan v nh m c n

1000 ml.

- Dung dch trinatri citrate 0,1M (b): 29,41 g C6H5O7Na3.2H2O ha tan v d n

n c n 1000 ml.

Dung dch m citrate c gi tr pH khc nhau ph thu c vo s ml dung dch

(a) v s ml dung dch (b) theo b ng sau:

A b pH A b pH

46,5 3,50 3,0 23,0 27,0 4,8

43,8 6,20 3,2 20,5 29,5 5,0

40,0 10,0 3,4 18,0 32,0 5,2

37,0 13,0 3,6 16,0 34,0 5,4

35,0 15,0 3,8 13,7 36,3 5,6

33,0 17,0 4,0 11,8 38,2 5,8

31,0 19,0 4,2 9,5 40,5 6,0

28,0 22,0 4,4 7,2 42,8 6,2

25,5 24,5 4,6 - - -


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c) Dung dch m phosphate (pH = 5,7 8,0)

- Dung dch mononatri orthophosphate 0,2M (a): 27,8 g NaH2PO4 ha tan v

nh m c n 1000 ml.

- Dung dch dinatri hydrophosphate 0,2M (b): 53,05 g Na2HPO4.7H2O ho c

71,7 g Na2HPO4.12H2O ha tan v nh m c n 1000 ml.

Dung dch m phosphate c pH khc nhau ph thu c vo s ml dung dch (a)

v s ml dung dch (b) nh m c n 200 ml.

a b pH A b pH

93,5 6,50 5,6 45,0 55,0 6,9

92,0 8,00 5,8 39,0 61,0 7,0

90,0 10,0 5,9 33,0 67,0 7,1

87,7 12,3 6,0 28,0 72,0 7,2

85,0 15,0 6,1 23,0 77,0 7,3

81,5 18,5 6,2 19,0 81,0 7,4

77,5 22,5 6,3 16,0 84,0 7,5

73,5 26,5 6,4 13,0 87,0 7,6

68,5 31,5 6,5 10,5 89,5 7,7

62,5 37,5 6,6 8,50 91,5 7,8

56,5 53,5 6,7 7,00 93,0 7,9

51,0 49,0 6,8 5,30 94,7 8,0

d) Dung dch m Na2HPO4 KH2PO4 (pH = 5,0 8,0)

- Dung dch dinatri hydrophosphate 1/15M (a): 23,9 g Na2HPO4.12H2O ha tan

v nh m c n 1000 ml.

- Dung dch kali dihydrophosphate 1/15M (b): 9,07 g KH2PO4 ha tan v nhm c n 1000 ml.

Dung dch m c pH khc nhau ph thu c vo s ml dung dch (a) v s ml

dung dch (b).


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a b pH a b pH

10 990 5,0 372 628 6,6

18 982 5,2 492 508 6,8

30 970 5,4 612 388 7,0

49 951 5,6 726 274 7,2

79 921 5,8 818 182 7,4

121 879 6,0 885 115 7,6

184 816 6,2 936 64 7,8

264 736 6,4 969 31 8,0

e) Dung dch m Glycine HCl (pH = 2,2 3,6)

- Dung dch glycine 0,2 M (a): 15,01g glycine ha tan v nh m c n 1000

ml.

- Dung dch HCl 0,2 M (b): 16,8 ml HCl c (37%) pha thnh 1000 ml.

Dung dch m Glycine c pH khc nhau khi l y 50 ml dung dch (a) v X ml

dung dch (b) r i nh m c n 200 ml.

X pH X pH

5,0 3,6 16,8 2,8

6,4 3,4 24,2 2,6

8,2 3,2 32,4 2,4

11,4 3,0 44,0 2,2

f) Dung dch m Glycine NaOH (pH = 8,6 10,6)

- Dung dch glycine 0,2 M (a): 15,01g glycine ha tan v nh m c n 1000

ml.

- Dung dch NaOH 0,2 M (b): 8 g NaOH ha tan v nh m c n 1000 ml.

Dung dch m Glycine c pH khc nhau khi l y 50 ml dung dch (a) v X ml

dung dch (b) r i nh m c n 200 ml.


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X pH X pH

4,0 8,6 22,4 9,6

6,0 8,8 72,2 9,8

8,8 9,0 32,0 10,0

12,0 9,2 38,6 10,4

16,8 9,4 45,5 10,6

g) Dung dch m acetate (pH = 3,6 5,6)

- Dung dch acid acetic 0,2M (a): 11,55 ml CH3COOH c v nh m c n

1000 ml.

- Dung dch natri acetate 0,2M (b): 16,4 g CH3COONa ho c 27,2 g

CH3COONa.3H2O c ha tan v nh m c n 1000 ml.

Dung dch m c pH khc nhau ph thu c vo X ml dung dch (a) v Y ml

dung dch (b) v nh m c n 100 ml.

a b pH A b pH

46,3 3,7 3,6 20,0 30,0 4,8

44,0 6,0 3,8 14,8 35,2 5,0

41,0 9,0 4,0 10,5 39,5 5,2

36,8 13,2 4,2 8,8 41,2 5,4

30,5 19,5 4,4 4,8 45,2 5,6

25,5 24,5 4,6

h)Dung dch m v n nng 0,1M

- Dung dch acid acetic 0,1M (a): 5,7 ml CH3COOH c v nh m c b ng

n c c t n 1000 ml.

- Dung dch acid phosphoric 0,1M (b): 6,45 ml H3PO4 m c v nh m c

b ng n c c t n 1000 ml.

- Dung dch acid ortho-boric 0,1M (c): ha tan 6,18 g acid ortho-boric trong

n c c t v nh m c n 1000 ml.


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- Dung dch natri hydroxide 1N (d): ha tan 40 g NaOH trong n c c t v nh

m c n 1000 ml.

Tr n l n cc dung dch (a), (b), (c) v i th tch b ng nhau nh n c dung dch

m 0,1M c pH = 1,8. i u chnh gi tr pH c a dung dch m trong kho ng pH =

1,8 n pH = 12,0 b ng cch b sung dung dch NaOH 1N (d). Cc dung dch m

v n nng n ng khc c chu n b t ng t .

II. Th c hnh

Sinh vin pha dung dch m v m t s dung dch s d ng trong cc bi th

nghi m sau theo h ng d n c a gi ng vin.

III. Bi n p

1. Ghi chp ph ng php pha cc dung dch th c chu n b.

2. Lm cc bi t p sau:

1. Pha 1 L dung dch NaOH 40% (w/v): cn ..g NaOH kh

2. Pha 0,5 L dung dch H2SO4 2M c n bao nhiu ml H2SO4 m c, bi t r ng H2SO4

l acid 97% c M=98 g/mol, t tr ng d=1,84 g/ml.

N ng mol c a H2SO4 l:

Th tch H2SO4 c n s d ng l:

L ng n c c n b sung l :

3. Pha 250 ml dung dch CuSO 4 1 M, cn .g CuSO4.5H2O

4. Pha 1 L dung dch H2SO4 0.1N t dung dch H2SO4 2M:

5. Pha 500 g dung dch NaOH 10% t dung dch NaOH 40%

L ng NaOH c n pha dung dch 10% l: X =

L ng dung dch NaOH 40% c n dng l: Y =L ng n c c t thm vo:

6. Pha 500 ml dung dch HCl 2% t dung dch HCl 37% (d=1,19g/ml)

L ng HCl c n pha dung dch 2% l: X=

L ng dung dch HCl 37% c n dng l : Y =

L ng n c c t thm vo:


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BI 2. NH L NG NG KH B NG

PH NG PHP ACID DINITRO - SALICYLIC (DNS)

I. L thuy t1. nh ngha

ng kh l cc ng ch a nhm aldehyde (-CHO) ho c ketone (-CO) nh

glucose, fructose, arabinose, maltose, lactose; trong khi saccharose, trehalose

khng ph i ng kh .

2. Nguyn t c

Ph ng php ny d a trn c s ph n ng t o mu gi a ng kh v i thu c

th acid dinitrosalicylic (DNS). C ng mu c a h n h p ph n ng t l thu n v i

n ng ng kh trong m t ph m vi nh t nh. So mu ti n hnh b c sng

540nm. D a theo th ng chu n c a glucose tinh khi t v i thu c th DNS s tnh

c hm l ng ng kh c a m u nghin c u.

Ph ng trnh ph n ng t o mu gi a ng kh v DNS acid:

Acid dinitrosalicylic 3-amino, 5- dinitrosalicylic acid

3. X l m u

Ty vo i t ng nghin c u (h t, qu ,...) m cch chu n b dung dch nghin

c u dng nh l ng ng c khc nhau i cht, nh ng nguyn t c chung nh

sau:

- Tr ng h p nguyn li u th nghi m khng ch a qu nhi u tinh b t ho c

inulin, c th chi t ng t nguyn li u b ng n c. Cn v cho vo c i s 1 g

nguyn li u h t hay cc m u th nghi m th c v t kh nh cy, l ho c qu kh,...

c nghi n nh (v s y kh n kh i l ng khng i). N u l nguyn li u t i (nh

hoa, qu t i) th cn 5 10 g. Nghi n c n th n v i b t th y tinh hay ct s ch v 30


22/39

22

ml n c c t nng 70 80oC. Chuy n ton b h n h p vo bnh nh m c dung tch

1000 ml. un cch th y 70 80oC trong 35 40 pht, k t t a protein v cc t p ch t

b ng dung dch ch acetate [Pb(C2H2O2)2.3H2O] ho c ch nitrate [Pb(NO3)2] 10%.

Trnh dng qu d ch acetate (dng 2 5 ml ch acetate). Sau lo i b l ng ch

acetate d b ng dung dch Na2SO4 bo ha, yn h n h p 10 pht. Ti p thm

n c c t t i v ch m c v em l c qua gi y l c vo c c hay bnh kh. N c l c dng

lm dung dch th nghi m.

- Tr ng h p nguyn li u ch a qu nhi u tinh b t ho c inulin nh khoai lang,

s n, khoai ty,... c n chi t ng b ng r u 70 80oC, un h n h p cch th y trong

bnh cch th y c l p ng lm l nh khng kh. Trong tr ng h p ny khng c n k t

t a protein b ng ch acetate v l ng protein chuy n vo dung dch khng nhi u.

- Tr ng h p cc nguyn li u ch a nhi u acid h u c nh c chua, d a, chanh,

kh ,... c n ch l trong qu tnh un khi chi t, ng saccharose c th b th y phn

m t ph n. Do c n xc nh ring ng kh v ring saccharose. Tr c khi un

cch th y h n h p ph i trung ha acid b ng dung dch Na2CO3 bo ha t i pH 6,4

7,0.

II. Th c hnh

1. D ng c - ha ch t:

a) D ng c ,thi t b- My so mu ho c quang ph k

- Cuvette d= 1 cm, V = 4 ml

- ng nghi m c n p v cc d ng c th y tinh thng th ng khc

- B p i n

b) Ha ch t - nguyn li u

- M u rau qu ch a ng (rau c i ng t, d a)

- Thu c th DNS

Can 5g DNS va300 ml nc cat vao coc, hoa tan 500C, sau ocho them 5ml

dung dch NaOH 4M . Cuoi cung cho them 150g muoi tartrat kep hoa tan roi cho

vao bnh nh mc v athem nc cat u500ml, ng trong lo thuy tinh sam mau.

Chuan 3 ml thuoc thDNS b ng H Cl 0.1N vi chthlaphenolphtalein, neu het 5 -

6 ml HCl lac.


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- Dung dch glucose chu n: pha s n dung dch glucose chu n 10 mg/ ml (b o qu n t

l nh vi ngy).

2. Ti n hnh

a) Chi t xu t ng trong th c v t

- 5g rau c i ng t nghi n nh cho vo becher 100ml + 10ml c n 900 un cch

th y (si 3 l n) khu y u b ng a th y tinh ngu i l c (ch g n l y ph n

trong).

- Cho ti p 10ml c n 800

vo c c ng b, khu y u, un si 2 l n trn n i un cch

th y ngu i l c (l p l i 2 l n)

- a ph n b ln l c v r a b ng r u nng (800)

- R u qua l c c cho bay h i trong phng ho c trn n i cch th y un nh . Sau

khi cho bay h i r u, m u c th lu trong bnh ht m.

- C n kh trong c c c pha long thnh th tch V = 50ml v i n c c t ta c

dung dch ng g c. Pha lang dung dch ng g c v i m t h s pha lang n thch

h p.

b) D ng ngchu n v xcnh hm l ng ng

- Cho vo 6 ng nghi m s ch v i cc ch t c th tch nh b ng sau:

Ho ch t ngC

ng 1 ng 2 ng 3 ng 4 ng 5

Glucose 10

mg/ml (ml)

0 0.2 0.4 0.6 0.8 1.0

N c c t (ml) 10 9.8 9.6 9.4 9.2 9.0

N ng

(mg/ml)

OD540nm

(Sinh vin i n n ng glucose chu n vo b ng trn)

- T cc ng nghi m trn l y 1ml m i ng vo 6 ng nghi m khc v thm vo m i

ng 3ml thu c th DNS.

- un si ng 5 pht (c y nt).


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- Lm l nh n nhi t phng.

- o m t quang b c sng 540 nm v i m u tr ng pha t ng i ch ng.

V ng chu n glucose v i tr c tung l m t quang (OD540nm), tr c honh l n ng

glucose. Tm ph ng trnh bi u di n ng chu n d ng y = ax + b v i y =

OD540nm; x = [glucose] (mg/ml) v h s t ng quan R2 nh ph n m m Excel.

- T ng t ht 1 ml dung dch ng pha lang X cho vo ng nghi m, thm 3 ml

DNS, un si 5 pht. Sau khi ngu i o m t quang (OD), s d ng m u tr ng

trn.

Ch : ti n hnh cc m u d ng ng chu n v m u th nghi m ng th i.

3. Tnh k t qu :

- T ph ng trnh th ng cong chu n tnh c X mg/ml ng kh trong dung

dch ng pha lang.

- Nhn v i h s pha lang n c l ng ng trong 1 ml dung dch g c.

- Ch n h s pha lang n sao cho OD n m trong gi i h n ng chu n.

- Tnh hm l ng ng trong nguyn li u (mg/g) =

V = th tch dch ng g c (ml)

m = kh i l ng m u cn (g)

III. Bi n p:

1. V ng chu n glucose2. Vi t ph ng trnh ng chu n y = ax + b

3. Tnh R2

=

4. Bi n lu n cc h s pha lang.


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BI 3. NH L NG NIT T NG S

B NG PH NG PHP KJELDAHL

I. L thuy t

1. nh ngha

T t c cc d ng nit c trong c th hay trong cc m c g i l nit t ng s .

Nit c trong thnh ph n amino acid c a protein l nit protein. Nit khng c trong

thnh ph n protein nh c a cc mu i v c , acid nitric, cc amino acid t do, cc

peptide, urea v cc d n xu t c a urea, cc alkaloid, cc base purine v pyrimidine,...

l nit phi protein.

Nit t ng s = Nit protein + Nit phi protein

m t ng s hay protein t ng s l nit t ng s nhn v i h s chuy n i. H

s ny ph thu c vo hm l ng nit trong protein. Thng th ng nit chi m 16%

protein nn h s chuy n i th ng c s d ng l 100/16 = 6.25.

m t ng s = Nit t ng s x h s chuy n i

B ng d i y bi u di n h s chuy n i c th cho nhi u i t ng m u khc

nhau.

M u H s chuy n i

Ngu n g c ng v t 6.25

H t bng 5.30

u ph ng 5.46

u nnh 5.71

H t h ng d ng 5.3

C m d a 5.3H t m 5.3

B p 6.25

G o 5.95

La m 5.83


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26

2. Nguyn t c

a) V c ha m u

- Tr c tin m u c v c ha b ng H2SO4 c nhi t cao v c ch t xc

tc. Cc ph n ng c a qu trnh v c ha x y ra nh sau:

2H2SO4 2H2O +2SO2 + O2

- Oxi t o thnh trong ph n ng l i oxi ha cc nguyn t khc. Cc phn t

ch a nit d i tc d ng c a H2SO4 t o thnh NH3. V d cc protein b th y phn

thnh acid amine; carbon v hidro c a acid amine t o thnh CO2 v H2O; cn nit

c gi i phng d i d ng NH3 k t h p v i H2SO4 d t o thnh (NH4)2SO4 tan trong

dung dch

2NH3 + H2SO4 (NH4)2SO4

Cc nguyn t P, K, Ca, Mg,... chuy n thnh d ng oxide: P2O5, K2O, CaO,

MgO,...

b) Ch ng c t m:

- u i NH3 ra kh i dung dch b ng NaOH

(NH4)2SO4 + 2 NaOH = Na2SO4 +H2O + 2NH3

- NH3 bay ra c lm l nh bi n i thnh NH4OH r i vo bnh h ng, bnh

h ng ch a H2SO4 0.1N

2NH4OH + H2SO4 (NH4)2SO4 + 2H2O + H2SO4 dc) Chu n H 2SO4 d

- Chu n H2SO4 d b ng NaOH 0.1N

H2SO4 d + NaOH Na2SO4 + H2O

II. Th c hnh

1. D ng c - ha ch t

a) D ng c

- Bnh ph m u

- B p un

- B ch ng c t Kjeldahl nh l ng Nit g m:

Bnh c t m (bnh Kjeldahl)

ng d n kh

ng sinh hn


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Bnh h ng (becher 250ml)

Bi th y tinh

- Pipet 20ml; pipet 10ml

- Erlen 250ml

B ch ng c t Kjeldahl

b) Ha ch t + nguyn li u

- B t u nnh (0.1g), n c t ng 1-2 ml

- H2SO4 m c

- NaOH 40%

- H2SO4 0.1N chu n

- NaOH 0.1N chu n

- Thu c th Tashiro

- Ch t xc tc: h n h p K2SO4 : CuSO4 (3:1)

2. Ti n hnh

a) V c ha m u

- Cn 100mg b t u nnh s y kh tuy t i cho vo bnh Kjeldahl, cho ti p

5ml H2SO4 m c s th y xu t hi n mu nu en (do nguyn li u b oxi ha).Cho thm vo 200mg ch t xc tc, l c nh , y kn kho ng 3 pht. t bnh

Kjeldahl ln b p un, y mi ng bnh b ng m t ph u th y tinh.

L u giai o n ny ph i th c hi n trong t Hotte, t bnh h i nghing trn

b p, trnh tr ng h p khi si m nh ha ch t b n ra ngoi, khi si gi nhi t b p

un v a ph i trnh ha ch t tro ra ngoi v khng b bay m t ammoniac.


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- Trong khi un, theo di s m t mu en c a dung dch trong bnh un, khi

th y dung dch g n nh trong su t th c th l c nh bnh ko h t cc phn t trn

thnh bnh cn ch a b oxi ho vo trong dung dch. Ti p t c un cho n khi dung

dch trong hon ton. ngu i bnh r i chuy n ton b dung dch sang bnh nh m c

100 ml, dng n c c t v m trng l i bnh Kjeldahl v nh m c n v ch.

b) C t m

- Chuy n 50ml dung dch trong bnh nh m c trn vo bnh c t m c s n

50ml n c c t v 3 gi t thu c th Tashiro lc ny trong bnh c mu tm h ng. Ti p

t c cho vo bnh c t 20ml NaOH 40% cho n khi ton b dung dch chuy n sang

mu xanh l m (thm 5ml NaOH 40% n u dung dch trong bnh ch a chuy n h t

sang mu xanh l m ).

- Ti n hnh l p h th ng c t m, cho vo bnh h ng 20ml H2SO4 0.1N v 3

gi t thu c th Tashiro (dung dch c mu tm h ng). t bnh h ng sao cho ng p u

ng sinh hn. B t cng t c c t m. Ki m tra xem h i ra t u ng sinh hn c lm

xanh gi y qu khng, n u khng thm NaOH 40%.

- Sau khi c t m 20-25 pht ki m tra xem NH4OH cn c t o ra khng,

dng gi y qy th u ng sinh hn. N u gi y qy khng i sang mu xanh l

c. Ng ng c t m, i h th ng ngu i m i tho h th ng em i r a.

c) Chu n :- Chu n H2SO4 d trong bnh h ng b ng NaOH 0.1N cho n khi m t mu

tm h ng v chuy n sang mu xanh l m . Ghi nh n th tch NaOH 0.1 N s d ng.

3. Tnh k t qu :

Hm l ng % nit t ng s c tnh theo cng th c

N (mg%) = {1,42 * (V1-V2)*100/a}*2

V1: s ml H2SO4 cho vo bnh h ng

V2: s ml NaOH 0.1N chu n

a s miligam nguyn li u

1,42: h s , c 1 ml H2SO4 dng trung ha NH4OH th t ng ng v i

1,42 mg Nit

III. Bi n p

1. Gi i thch ngha cc b c trong th nghi m.


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2. Th o lu n cc y u t d n n sai s .

3. N u thay H2SO4 trong bnh h ng b ng acid boric th k t qu c g thay i khng ?


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BI 4. NH L NG PROTEIN B NG PH NG PHP

BRADFORD

I. L thuy t1. nh ngha

Protein c nh l ng b ng ph ng php Bradford l protein tan trong dung

dch. V v y, mu n xc nh hm l ng protein m t m u ta ph i trch ly protein. Hm

l ng protein xc nh s ph thu c vo hi u qu trch ly (kh nng ha tan c a

protein trong dung mi trch ly).

2. Nguyn t c

Ph ng php ny d a trn s thay i b c sng h p thu c c i c a thu c

nhu m Coomassie Brilliant Blue khi t o ph c h p v i protein. Trong dung dch mang

tnh acid, khi ch a k t n i v i protein th thu c nhu m c b c sng h p thu c c i

465 nm; khi k t h p v i protein th thu c nhu m h p thu c c i b c sng 595 nm.

h p th b c sng 595 nm c lin h m t cch tr c ti p t i n ng protein.

xc nh protein trong m u, u tin ta xy d ng m t ng chu n v i

dung dch protein chu n bi t tr c n ng . Dung dch protein chu n th ng l

bovine serum albumin (BSA). Sau khi cho dung dch protein vo dung dch thu c

nhu m, mu s xu t hi n sau 2 pht v b n t i 1 gi . Ti n hnh o m t quang h n

h p dung dch b ng my quang ph k .

Cng th c phn t c a Coomasie Brilliant Blue G-250


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31

II. Th c hnh

1. D ng c - ha ch t

a) D ng c

- My o quang ph

- ng nghi m (14)

- Pipette 5ml, 1ml (c th thay b ng pipetteman)

- ng h b m giy

- Gi y th m

- Gi ng nghi m

- Bnh tia ng c n

b) Ha ch t

- C n 900

- Dung dch protein chu n: cn 10 mg albumin b ng cn phn tch, pha trong 1

ml n c c t, l c u cho tan. Gi -200C. Khi dng pha long ra 100 l n, c dung

dch c n ng 0,1 mg/ml.

- Dung dch thu c th Bradford: dung dch thu c th c thnh ph n trong 1 l

nh sau:

Coomassie Brilliant Blue: 0,05g

Methanol: 50 mlPhosphoric acid 85%: 100 ml

Ph m mu Coomasie Brilliant Blue 0,1 g c lm tan trong 50 ml methanol, pha

lang v i n c c t thnh 500 ml d c dung dch (1) ng trong chai mu t i c

n p, 100 ml H3PO4 85% pha lang thnh 500 ml c dung dch (2). Khi c n s d ng

ha 1 V (1) v 1V (2) r i l c l y dch trong.

2. Ti n hnh

- Thu dch trch ly protein nh trong bi enzyme (10 g malt trch ly b ng n c

thu c V = 100 ml). Pha lang m u v i h s pha lang n l n c m u phn

tch (m u X) (n=1, 2 i v i malt).

- L p m t lo t 6 ng theo s th t 0, 1, 2, 3, 4, 5 v 1 ng nghi m ch a m u

c n phn tch X


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32

- Dng ng h b m giy, canh th i gian 0 pht cho 5ml thu c th vo ng

nghi m 0, l c u yn. th i i m 1 pht cho 5ml thu c th vo ng nghi m 1,

l c u yn,... c ti p t c cho n h t.

ng nghi m 0 1 2 3 4 5 X

N c c t, ml 1 0.9 0.9 0.7 0.6 0.5 -

Albumin chu n (0,1mg/ml),

ml0 0.1 0.2 0.3 0.4 0.5 -

TT Bradford, ml 5 5 5 5 5 5 5

N ng Albumin (g/ml) ?

OD595nm

(Sinh vin i n n ng albumin (g/ml) vo b ng trn)

- Tnh th i gian ng 0 c 20 pht, ti n hnh o h p thu c a dung dch

b c sng 595 nm. Ta c gi tr OD0, th i i m 21 pht o ng 1 (OD1),... t ng t

c cch m t pht o cho h t cc ng. Ghi l i gi tr OD.

3. Tnh k t qu :

- V ng tuy n tnh gi a n ng protein (g/ml) v i m t quang OD595

- D a vo ph ng trnh ng tuy n tnh gi a n ng protein (g/ml) v i m t

quang OD595 trn ta tnh c hm l ng protein P trong m u phn tch X.

P = X*n

tnh tan hm l ng protein trong 1g m u cn:

Protein (mg/g) = P x V/m

V = th tch dung dch trch ly, ml

m = kh i l ng m u, g.

III. Bi n p

1. V ng chu n protein theo Bradford

2. Tm ph ng trnh ng chu n y = ax +b

3. Tnh R2

=

4. Tnh tan k t qu th nghi m, ch n h s pha lang no?


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33

BI 5. PH NG PHP XC NH HO T TNH ENZYME

I. L thuy t

1.1. Enzyme v n v o ho t tnh enzyme

a)nh ngha

Enzyme l ch t xc tc sinh h c c b n ch t l protein v c tnh c hi u cao.

M i enzyme c kh nng xc tc cho m t ho c m t s ph n ng nh t nh. Ho t ng

hay ho t tnh c a enzyme cng m nh th l ng c ch t c chuy n ha ho c l ng

s n ph m t o thnh trn m t n v th i gian cng l n. V v y c th nh gi ho t

tnh xc tc c a enzyme b ng cch xc nh t c chuy n ha c ch t ho c t c

tch l y s n ph m ph n ng.

V nguyn t c c th hai nhm ph ng php chnh sau:

- o l ng c ch t b m t i hay l ng s n ph m c t o thnh trong m t th i

gian nh t nh ng v i m t n ng enzyme xc nh.

- o th i gian c n thi t thu c m t l ng bi n thin nh t nh c a c ch t

hay s n ph m t ng ng v i m t n ng enzyme xc nh.

th c hi n c m c ch trn, nhi u ph ng php phn tch khc nhau c

s d ng: ph ng php o quang ph , o phn c c, p su t, nh t, ph ng php

s c k v ph ng php ha h c.

b) n v ho t tnh enzyme

M c ch xc nh ho t tnh enzyme l xc nh s n v ho t tnh. M t n v

h at tnh enzyme c nh ngha theo nhi u ph ng php.

i) n v n v qu c t (Enzyme Unit, vi t t t U) do Hi p h i Ha sinh Qu c

t (International Union of Biochemistry IUB) nh ngha:

M t n v chu n c a enzyme (1 U) l l ng enzyme xc tc chuy n ha c

1 mol c ch t sau 1 pht i u ki n tiu chu n.

1 U = 1 mol s n ph m = 1 mol c ch t (10-6

mol)/pht

ii) Katal:

Nm 1979, H i ng Danh php c a IUB khuy n co nn s d ng Katal lm

n v c b n c a ho t tnh enzyme.


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34

M t Katal l l ng enzyme xc tc chuy n ha c 1 mol c ch t sau 1 giy

i u ki n tiu chu n.

1 Kat = 1 mol c ch t/ giy = 60 mol/ pht = 60 x 106mol/ pht = 6 x 10

7U

1 U = 1/60 x 10-6

Kat = 16,67 nKat (nanokatal)

n v Katal c khuy n co v n m trong h n v o l ng Qu c t (SI).

iii) n v t t: n v ho t tnh d a vo s thay i c tnh h n h p ph n

ng, v d s thay i c, nh t ...trong m t n v th i gian. Tr ng h p c

ch t v s n ph m l m t h n h p ph c t p th p d ng n v ho t tnh ny.

N h v y , c n h i u n v h o t t n h e n z y m e . i u q u a n tr n g n h t l c n n h

n g h a r r n g n v h o t t n h .

c) Ho t tnh ring (specific activity) c a enzyme

Ho t tnh ring c a m t ch ph m enzyme c tr ng cho tinh s ch c a ch

ph m enzyme. Ho t tnh ring c bi u th b ng s n v enzyme/mg protein

protein (U/mg protein) ho c Kat/kg protein, trong hm l ng protein c xc

nh theo ph ng php Lowry ho c Bradford.

1.2. Ph ng php xc nh h at tnh enzyme

Khi ti n hnh th nghi m o h at tnh enzyme c n ch nh ng i m sau:

- C n t r n h n h n g y u t c t h b i n t n h p r o te i n e n z y m e .

- C c th n g s n h i t , p H , n n g i o n v t h n h p h n d u n g d c h m n h h n gl n h o t t n h e n z y m e . T h h o t t n h e n z y m e p h i c t i n h n h t r o n g i u k i n t h c h

h p n h i u k i n s i n h l , i u k i n t n t r t h c p h m h o c i u k i n m h o t l c c

t h t t i u .

- V i n h n g e n z y m e c n c c h t h o t h a h o c c h t n n h t h p h i c h o c c c h t

n y v o e n z y m e t r c k h i c h o c c h t v o h n h p p h n n g .

- N n g c c h t t r o n g p h n n g e n z y m e p h i t r o n g g i i h n t h c h h p ,

t h a b o e n z y m e , n h n g k h n g q u c a o k m h m e n z y m e . S a u k h i d n g p h n

n g l n g c c h t c c h u y n h a 2 0 - 3 0 % .

- T h i g i a n x c n h h o t t n h th n g 5 - 3 0 p h t . T t n h t l x c n h t c b a n

u c a p h n n g (3 0 - 6 0 g i y ) , v g i a i o n n y t c p h n n g l n n h t , s a u b t

u g i m ( H n h 1 ) .


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35

- K h i x c n h h o t t n h p h i l m m u i c h n g s o n g s o n g v i m u t h n g h i m .

T r o n g m u i c h n g e n z y m e p h i b b t h o t t r c k h i t i p x c v i c c h t .

H n h 1 . ng h c ph n n g en zy m e

1 . 3 . E n z y m e a m y la s e v p h n g p h p x c n h h o t t n h

a ) K h a i q u t v e n zy m e a m y l a se

A m y l a s e l c c e n z y m e x c t c c h o c c p h n n g t h y p h n t i n h b t , g l y c o g e n ,

v c c p o ly s a c c h a r i d e t n g t . A m y l a s e c h i a l m 3 l o i c h n h :

- - a m y l a s e (Endo -1,4-glucanase; E C 3 . 2 . 1 . 1 ) c t r o n g n c b t , h t h a t h o

n y m m , t y t n g , n m m c , v i k h u n . E n z y m e n y p h n g i i l i n k t 1 , 4 -

g l y c o s i d e g i a c h u i p o l y s a c c h a r id e ( h o t t n h e n d o a m y l a s e ) t o t h n h

m a l to s e , d e x t r i n p h n t t h p . D i t c d n g c a e n z y m e n y , d u n g d c h t i n h

b t n h a n h c h n g b m t k h n n g t o m u v i d u n g d c h i o d v b g i m n h t .

- a m y l a s e b n v i n h i t , n h n g k m b n v i a c id .

- - a m y l a s e (Exo -1,4-glucanase; EC 3.2.1.2) c nhi u th c v t (h t, c ), xc

tc cho ph n ng th y phn lin k t 1,4-glycoside t u khng kh t o thnh

maltose v dextrin phn t l n. Enzyme ny m t ho t tnh nhi t trn 70oC,

nh ng b n v i acid h n - a m y l a s e .

- Glucoamylase (Exo -1,4-g l u c a n a s e ; EC 3.2.1.3) xc tc cho ph n ng th y

phn lin k t 1,4- v 1,6-glycoside t u khng kh c a chu i polysaccharide.

S n ph m ch y u l glucose v dextrin. Enzyme ny m t ho t tnh nhi t

trn 70oC. Nhi u glucoamylase ho t ng m nh pH 3,5-5,5.

b ) C c p h n g p h p o h o t t n h e n z y m e - a m y l a s e


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36

N g u y n t c c h u n g d a t r n c s t h y p h n t i n h b t b i e n z y m e t r o n g d u n g d c h

e n z y m e n g h i n c u t h n h c c d e x t r i n c p h n t l n g k h c n h a u . o c n g m u

t o t h n h g i a t i n h b t v c c s n p h m t h y p h n c a n v i i o d b n g m y s o m u s

t n h c h o t t n h e n z y m e .

n v a m y l a s e ( th e o S m i t h v R o e ) l l n g e n z y m e c n t h i t t h y p h n

h o n t o n 1 0 m g t i n h b t s a u t h i g i a n p h n n g 3 0 p h t t r o n g i u k i n t h n g h i m .

I I . T h c h n h

1 . H a c h t D n g c

a ) H a c h t

- D u n g d c h m

- D u n g d c h m a c e ta t e p H = 4 , 7 ( x c n h h o t t n h e n z y m e n m m c ) : t r n 1 t h

t c h C H 3 C O O H 1 N v i 1 t h t c h C H 3 C O O N a 1 N . K i m t r a p H .

- D u n g d c h m p h o s p h a te p H = 4 , 9 ( x c n h h o t t n h e n z y m e c a m a l t ) : t r n 1 0

m l d u n g d c h N a 2 H P O 4 1 / 1 5 M v i 9 9 0 m l d u n g d c h K H 2 P O 4 1 / 1 5 M n h n c 1 l

d u n g d c h m p h o s p h a te p H = 4 , 9 4 . K i m t r a p H .

- D u n g d c h m g l y c i n e N a O H 0 , 1 M p H = 1 0 ( d n g x c n h h o t t n h -

a m y l a s e k i m .

- D u n g d c h H C l 0 , 1 N .

- D u n g d c h i o d : h a t a n 3 0 m g K I v 3 m g I 2 v i m t l n g n h n c c t . L c n h h n h p h a t a n h o n t o n , s a u c h u y n d u n g d c h s a n g b n h n h m c 1 0 0 0 m l ,

b s u n g n c c t n v c h m c . B o q u n d u n g d c h tr o n g b n h m u n u c h t i .

- D u n g d c h t i n h b t 1 % : h a t a n 1 g t in h b t ( t h e o c h t k h t u y t i ) v i 5 0 m l

n c c t t r o n g b n h n h m c 1 0 0 m l , l c u . t v o b p c c h t h y a n g s i , l c l i n

t c c h o n k h i t i n h b t t a n h o n t o n . S a u l m n g u i v b s u n g 1 0 m l m a c e ta t e

p H = 4 , 7 ( h o c 1 0 m l d u n g d c h m p h o s p h a te p H = 4 , 9 ) b s u n g n c c t n v c h

m c , l c u . D u n g d c h c c h u n b t r o n g n g y s d n g .

- D u n g d c h e n z y m e g c

2 . T i n h n h :

a ) T r c h l y e n z y m e : e n z y m e c t r c h l y t c h p h m h a y t b o , m n g th c

v t b n g d u n g d c h m c p H t h c h h p .

i ) T r c h ly e n zy me t v i kh u n : C n m = 0 ,1 g c h p h m n g h i n c u , d n g a


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37

t h y t i n h c h c n t h n c h p h m t r o n g m t c c th y t i n h d u n g t c h 5 0 m l v i m t l n g

n c n h . S a u c h u y n t o n b h n h p v o b n h n h m c d u n g t c h 1 0 0 m l , b

s u n g n c c t t i v c h m c . L c v t h u h i d c h t r c h l y e n z y m e . B o q u n d u n g d c h

e n z y m e g c 2 4o

C t ro n g t h i g i a n 1 n g y .

i i) T r c h l y e n z yme n m m c : C n m = 5 g c h p h m e n z y m e t h n g h i n s

b , d n g a t h y t i n h c h c n t h n c h p h m t r o n g m t c c d u n g t c h 5 0 m l . S a u

c h u y n t o n b c h p h m v o b n h t a m g i c d u n g t c h 2 5 0 m l , b s u n g 5 0 m l n c c t

v 1 0 m l d c h m a c e ta te p H = 4 , 7 . G i h n h p n h i t 3 0o

C t ro n g t h i g i a n 1 g i

c k h u y o n h k . L c , r a t h u h i d c h t r c h l y e n z y m e s a o c h o V = 1 0 0 m l . B o

q u n d u n g d c h e n z y m e g c 2 4o

C t r o n g t h i g i a n 1 n g y .

i ii ) T r c h l y e n zy me t m alt : C n m = 5 - 1 0 g m a l t n g h i n n h , c h o v o b n h

n n d u n g t c h 2 5 0 m l , b s u n g 5 0 m l n c c t v 1 0 m l d u n g d c h m p h o s p h a te p H =

4 , 9 . G i h n h p n h i t 3 0o

C t r o n g t h i g i a n 1 g i c k h u y o n h k . L c , r a

t h u h i d c h t r c h l y e n z y m e s a o c h o V = 1 0 0 m l . B o q u n d u n g d c h e n z y m e g c 2

4o

C t ro n g t h i g i a n 1 n g y .

b ) P h a l a n g e n z y m e

P h a l o n g d u n g d c h e n z y m e g c ( h s p h a lo n g n = 25 , 50 , 100 ) b n g d u n g

d c h m s a o c h o t r o n g 1 m l d u n g d c h e n z y m e p h n t c h c c h a m t l n g e n z y m e

t h y p h n 2 0 % - 3 0 % t i n h b t t r o n g d u n g d c h i u k i n x c n h . N h v y , p h a l o n g p h t h u c v o h o t t n h c h p h m e n z y m e c n p h n t c h .

c ) X c n h h a t t n h

C h u n b m u t h n g h i m , m u i c h n g v m u t r n g t h e o b n g s a u :

M u t h n g h i m M u i c h n g M u t r n g

D u n g d c h t i n h b t 1 % 2 m l 2 m l 0

N c c t 0 0 2 m l

m 1 m l 1 m l 1 m l

D u n g d c h N a C l 3 % 0 , 5 m l 0 , 5 m l 0 , 5 m l

4 0o

C , 1 0 p h t t n h i t

D u n g d c h e n z y m e 1 m l 0 0

4 0o

C , 3 0 p h t x y r a p h n n g


38/39

38

H C l 1 N 1 m l 1 m l 1 m l

E n z y m e 0 1 m l 1 m l

C h u y n s a n g b n h n h m c 1 0 0 m l

N c c t n v c h n v c h n v c h

I o d e 0 , 5 m l 0 , 5 m 1 0 , 5 m 1

- D u n g d c h i c h n g c m u x a n h

- D u n g d c h t h n g h i m c m u t m v i c n g k h c n h a u t y v o l n g t i n h

b t c h a b th y p h n .

- o m t q u a n g ( O D ) c a c c d u n g d c h b c s n g = 6 2 0 n m s o v i m u

t r n g .

3 . T n h k t q u :

S m g t i n h b t b t h y p h n S = x 2 0

Ho t tnh tan ph n ( n v h at tnh/g)=m

VnS

10

H at tnh ring ( n v h at tnh/mg protein) = h at tnh ton ph n/mg protein

(Bradford).

trong :

n = h s pha long

V = th tch dch trch ly, ml

m = kh i l ng enzyme, g

III. Bi n p

1. n v h at tnh enzyme amylase trong bi l g ?

2. So snh cc h s pha lang, ch n h s pha lang no ?

3. C th s d ng ph ng php o ng kh b ng DNS acid o h at tnh enzyme

amylase khng ? T i sao ?


39/39

TI LI U THAM KH O

1. Nguy n Vn Mi. Ha sinh h c. Nh xu t b n i h c Qu c gia H n i, 2001

2. Holtzhauer M. Basic Methods for the Biochemical Lab. Springer, 2006.

3. Ronald E. Current Protocols in Food Analytical Chemistry. John Wiley &

Sons, Inc., 2003

4. Smith B, Roe J. A photometric method for the determination of -amylase in

blood and urine, with the use of the starch-iodine color.J. Biol. Chem. 179, 53

(1949).

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