cach pha hoa chat t

7/22/2019 cach pha hoa chat t

1/39

TRNG I HC K THUT CNG NGH TP. HCMKHOA MI TRNG & CNG NGH SINH HC

------

BI GING

THC HNH

HA SINH

TS. Nguyn Hoi Hng

CN. Bi Vn Th Vinh

Dng cho sinh vin ngnh Mi trng v Cng ngh Sinh hcNm xut bn: 2009


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MC LC Trang

Gii thiu mn hc 4Quy tc lm vic trong phng th nghim Ha

sinh

5

1. An ton khi lm vic vi axit v kim 5

2. Quy tc lm vic vi ha cht th nghim 6

Bi 1 Cch pha ch cc dung dch dng trong th

nghim Ha sinh

8

I L thuyt 81. Dung dch 8

2. Dung dch m 14

II Thc hnh 20

III Bi np 20

Bi 2 nh lng ng kh bng phng php Acid

dinitro-salicylic (DNS)

21

I L thuyt 211. nh ngha 21

2. Nguyn tc 21

3. X l mu 21

II Thc hnh 22

1. Dng c - ha cht 22

2. Tin hnh th nghim 23

3. Tnh kt qu 24

III Bi np 24

Bi 3 nh lng Nit tng s bng phng php

Kjeldahl

25

I L thuyt 25

1. nh ngha 25


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3

2. Nguyn tc 26

II Thc hnh 26



3. Tnh kt qu 28

III Bi np 28

Bi 4 nh lng protein bng phng php Bradford 30

I L thuyt 30

1. nh ngha 30

2. Nguyn tc 30

II Thc hnh 31

1. Dng c - ha cht 312. Tin hnh th nghim 31

3. Tnh kt qu 32

III Bi np 32

Bi 5 Phng php xc nh hot tnh enzyme 33

I L thuyt 33

1. Enzyme v n v o hot tnh enzyme 33

2. Phng php xc nh hot tnh enzyme 343. Enzyme amylase v phng php xc nh hot

tnh

35

II Thc hnh 36



3. Tnh kt qu 38

III Bi npTi liu tham kho

3839


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GII THIU MN HC

Bi ging Thc hnh sinh ho dnh cho sinh vin nm th hai khoa Mi

trng & Cng ngh Sinh hc, trng i hc K Thut Cng Ngh TP.HCM.

Vi thi lng 30 tit v tu thuc iu kin, c s vt cht ca phng th

nghim cho php, sinh vin tin hnh lm 5 bi thc hnh gm cc ni dung chnh ca

hc phn l thuyt ha sinh c s.

1. Gii thiu phng th nghim ha sinh, cch pha ch cc dung dch dung

trong th nghim ha sinh ;

2. nh lng ng kh bng phng php acid dinitro-salicylic (DNS);

3. nh lng nit tng s bng phng php Kjeldahl;

4. nh lng protein bng phng php Bradford;

5. Xc nh hot tnh enzyme amylase.


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QUY TC LM VIC TRONG PHNG TH NGHIM HA

SINH

I. An ton khi lm vic vi axit v kim1. An ton khi lm vic vi axi t:

Phi lm vic trong t ht bt c khi no un nng axit hoc thc hin phn

ng vi cc hi axit t do.

Khi pha long, lun phi cho axit vo nc tr phi c dng trc tip.

Gi axit khng bn vo da hoc mt bng cch eo khu trang, gng tay v

knh bo v mt. Nu lm vng ln da, lp tc ra ngay bng mt lng nc ln.

Lun phi c k nhn ca chai ng v tnh cht ca chng.

H2SO4: Lun cho acid vo nc khi pha long, s dng khu trang v gng tay

trnh phng khi vng acid

Cc acid dng hi (HCl) thao tc trong t ht v mang gng tay, knh bo h.

2. An ton khi lm vic vi k im

Kim c th lm chy da, mt gy hi nghim trng cho h h hp.

Mang gng tay cao su, khu trang khi lm vic vi dung dch kim m c.

Thao tc trong t ht, mang mt n chng c phng nga bi v hi kim.

Amoniac: l mt cht lng v kh rt n da, mang gng tay cao su, khu trang,

thit b bo v h thng h hp. Hi amoniac d chy, phn ng mnh vi cht oxy

ho, halogen, axit mnh.

Amoni hydroxyt: cht lng n da, to hn hp n vi nhiu kim loi nng: Ag,

Pb, Zn ... v mui ca chng.

Kim loi Na, K, Li, Ca: phn ng cc mnh vi nc, m, CO 2, halogen, axit

mnh, dn xut clo ca hydrocacbon. To hi n mn khi chy. Cn mang dng c

bo v da mt. Ch s dng cn kh khi to dung dch natri alcoholate, cho vo t t.Trnh to tinh th cng khi ho tan. Tng t khi ho tan vi nc, ng thi phi

lm lnh nhanh.

Oxit canxi rt n da, phn ng cc mnh vi nc, cn bo v da mt, ng

h hp do d nhim bi oxit.


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Natri v kali hydroxyt: rt n da, phn ng cc mnh vi nc. Cc bin php

an ton nh trn, cho tng vin hoc t bt vo nc ch khng c lm ngc li.

II. Quy tc lm vic vi ha cht th nghim

1. Ho cht thnghim:

Cc ho cht dng phn tch, lm tiu bn, tin hnh phn ng, ... trong phng th

nghim c gi l ha cht th nghim.

Ho cht c th dng rn (Na, MgO, NaOH, KCl, (CH3COO)2...; lng (H2SO4,

aceton, ethanon, chloroform, ...) hoc kh (Cl2, NH3, N2, C2H2 ...) v mc tinh khit

khc nhau:

- Sch k thut (P): sch > 90%

- Sch phn tch (PA): sch < 99%

- Sch ha hc (PC): sch > 99%

Ha cht c tinh khit khc nhau c s dng ph hp theo nhng yu cu khc

nhau v ch nn s dng ha cht cn nhn hiu.

2. Nhn hiu ho cht:

Ha cht c bo qun trong chai l thy tinh hoc nha ng kn c nhn ghi

tn ho cht, cng thc ha hc, mc sch, tp cht, khi lng tnh, khi lng

phn t, ni sn xut, iu kin bo qun.3. Cch s dng v bo qun ho cht:

Khi lm vic vi ha cht, nhn vin phng th nghim cng nh sinh vin cn ht sc

cn thn, trnh gy nhng tai nn ng tic cho mnh v cho mi ngi. Nhng iu

cn nh khi s dng v bo qun ha cht c tm tt nh sau:

- Ha cht phi c sp xp trong kho hay t theo tng loi (hu , v c,

mui, acid, baz, kim loi, ...) hay theo mt th t a, b, c khi cn d tm.

- Tt c cc chai l u phi c nhn ghi, phi c k nhn hiu ha cht trc

khi dng, dng xong phi tr ng v tr ban u.

- Chai l ha cht phi c np. Trc khi m chai ha cht phi lau sch np,

c chai, trnh bi bn lt vo lm hng ha cht ng trong chai.

- Cc loi ha cht d b thay i ngoi nh sng cn phi c gi trong chai

l mu vng hoc nu v bo qun vo ch ti.


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- Dng c dng ly ha cht phi tht sch v dng xong phi ra ngay,

khng dng ln np y v dng c ly ha cht.

- Khi lm vic vi cht d n, d chy khng c gn ni d bt la. Khi

cn s dng cc ha cht d bc hi, c mi,... phi a vo t ht, ch y kn np

sau khi ly ha cht xong.

- Khng ht bng pipette khi ch cn t ha cht trong l, khng ngi hay nm

th ha cht.

- Khi lm vic vi acid hay base mnh:

Bao gi cng acid hay base vo nc khi pha long (khng c nc

vo acid hay base);

Khng ht acid hay base bng ming m phi dng cc dng c ring nh ng

bp cao su.

Trng hp b bng vi acid hay base ra ngay vi nc lnh ri bi ln vt

bng NaHCO3 1% (trng hp bng acid) hoc CH3COOH 1% (nu bng base). Nu

b bn vo mt, di mnh vi nc lnh hoc NaCl 1%.

Trng hp b ha cht vo ming hay d dy, nu l acid phi sc ming v

ung nc lnh c MgO, nu l base phi sc ming v ung nc lnh c CH3COOH

1%.

Lu cc k hiu cnh bo nguy him:


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BI 1: CCH PHA CH CC DUNG DCH

DNG TRONG TH NGHIM HA SINH

I. L thuyt1.1. Dung dch

Dung dch l hn hp ca hai hay nhiu cht tc ng tng h vi nhau v

mt vt l v ha hc. Trong dung dch gm c cht ha tan v dung mi. Nu cht

ha tan dng rn th gi l cht tan, nu l cht lng th gi l dung cht.

Ty theo tnh cht ca dung mi m phn thnh dung dch nc v dung dch

khan. Phn ln cc dung dch acid, base, mui trong phng th nghim l dung dch

nc, dng dung mi l nc. Mt s cht khc tan trong dung mi hu c.

Hm lng cht ha tan trong dung dch th hin nng dung dch. C

nhiu cch biu th nng khc nhau. Mi cch s tin dng trong chun b, phn

tch v tnh ton khc nhau.

1.1.1. Cc n v nng dung dch

a) Nng phn trm, (%)

i) Nng phn trm khi lng - khi lng, % (w/w): l s gam cht tan c

trong 100g dung dch.

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

ii) Nng % khi lng th tch (w/v): l s g cht tan c trong 100ml dung

dch.

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

iii) Nng phn trm th tch - th tch, % (v/v): l s ml dung cht c trong

100ml dung dch.

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

b) Nng gam-lit, (g/L): l s gam cht tan c trong 1 lt dung dch.c) Nng phn t gam hay nng mol, (Mol/L) hay M: l s phn t gam

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

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

KH2PO4.


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d) Nng ng lng (N): l s ng lng gam (lg) cht tan c trong 1

lit dung dch.

S ng lng cht tan = s mol (n) x h s ng lng (z)

H s ng lng (z): ph thuc vo bn cht ca cht v phn ng m

cht tham gia.

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

ion ca cht tc dng va .

V d: Phn ng gia HCl v NaOH

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

H2SO4 2 H+ z = 2

NaOH 1 OH- z = 1

ii) Nu phn ng l phn ng xy ha kh: z l s electron m 1 phn t, ionca cht cho hay nhn.

V d: 2 Na2S2O3 + I2 Na2S4O6 + 2 NaI

I + 1e I- z = 1

S2+

- 1e S+ z = 1

e) Nng dung dchbo ha: l nng dung dch khi ti a cht ha tan

c mt trong dung dch.

f ) n v nng dng trong cc php phn tch vi lng:- Nng mg/mL: s mg cht tan trong 1mL dung dch

- Miligam phn trm, mg%: mg cht ha tan trong 100g dung dch.

- Microgam phn trm, g%: l s g cht ha tan trong 100g dung dch.

- Phn nghn, 0/00: s g cht ha tan trong 1000g dung dch.

- Phn triu, ppm: s mg cht ha tan trong 1kg hay 1 lt dung dch.

- Phn t, ppb: s g cht ha tan c trong 1kg hay 1 lt dung dch.

1.1.2. Cch pha dung dch c nng xc nh

a) Pha dung dch c nng phn trm theo khi lng, % (w/w)

i) Cht tan l cht rn khan:

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

100g dung dch cn 40g NaOH

500g dung dch cn X g?


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Lng NaOH cn pha dung dch : X= (40*500)/100= 200g

Lng nc cn thit: 500-200=300g (hay 300ml)

Vy, cn 200g NaOH v ong 300 ml nc ct, ha tan ta c 500g dung

dch NaOH 40%

ii) Cht tan l cht rn ngm nc (CuSO4.5H2O; Na2HPO4. 12H2O;...)

Khi pha dung dch cn phi tnh thm lng nc kt tinh c sn.

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

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

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

Lng CuSO4.5H2O cn dng: Y = (250*32)/160 = 50g

Lng nc ct thm vo: 320 -50 = 270g (hay ml)

Vy, cn 50g CuSO4.5H2O, ong 270ml nc ct, ha tan ta c 320g dungdch CuSO4 10%.

b) Pha dung dch long t mt dung dch m c hn:

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

Lng NaOH cn pha dung dch 5% l: X = (5*500)/100 = 25g

Lng dung dch NaOH 10% cn dng l: Y = (100*25)/10 = 250g

Lng nc ct thm vo: 500-250 = 250g

Vy, ong 250ml nc ct v 250g dd NaOH 10%, ha tan ta c 500gNaOH 5%

c) Pha dung dch bo ho:

Ly cht tan cn pha vo becher, thm mt t nc ct v khuy cho tan. Nu

sau khi khuy, cht tan khng tan ht lng xung th phn dung dch pha trn l dung

dch bo ha. Nu cht tan tan ht, thm cht tan v tip tc khuy, c nh th cho

n khi cht tan khng cn tan c na.

d) Pha dung dch c nng % theo th tch

i) Cht tan l cht rn khan

Cn lng cht tan cn thit, chuyn sang bnh nh mc, dng nc ct ha

tan v nh mc n th tch ng.

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

Lng NaCl cn pha dung dch: X = (5*1000)/100 = 50g


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Vy, cn 50g NaCl, ha tan v nh mc thnh 1 lt bng nc ct, ta c 1 lt

dung dch NaCl 5%.

ii) Cht tan l cht rn ngm nc (CuSO4.5H2O; Na2HPO4. 12H2O;...)

Khi pha dung dch ta cn phi tnh n lng nc kt tinh c sn ging nh

phn a.

iii) Cht tan dng lng: Mt s cht tan dng lng nh HCl, H2SO4 ...

Vic cn khng thun li, c th a v n v th tch theo cng thc

V = M/d

V: Th tch cht lng; M: khi lng cht lng cn cn; d: t trng cht lng

Ch : Cc ha cht lng bn trn th trng thng khng dng nguyn cht

m l cc dung dch m c. Gii hn ha tan ti a c tnh bng % th tch v

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

85%; NH4OH: 25%. Do khi pha cc dung dch t cc loi ha cht ny ta phi ch

n nng ca dung dch m c.

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

Lng HCl cn pha dung dch 1% l: X= (1*440)/100 = 4,4g

Lng dung dch HCl 37% cn dng l : Y = (100*4,4)/37 = 11,9g

= 11,9/1,19 = 10 ml

Lng nc ct thm vo:440-10 = 430mlVy, dng ng ong ly 10ml dung dch HCl37%, v 430ml nc ct ta c

440 ml HCl 1%

Do vic s dng cc loi bnh nh mc lm cho vic pha ch dung dch th

nghim tr nn n gin v chnh xc v vy ngy nay a s cc dung dch th nghim

c pha ch theo nng khi lng - th tch (w/v).

e) Pha dung dch nng phn t gam

i) Cht tan l cht rn khan

Mun pha dung dch nng 1M ca mt cht no , ta tnh khi lng phn

t cht (hoc tra bng) theo n v gam. Cn chnh xc lng cht tan, qua phu

cho vo bnh nh mc c dung tch 1 lt. Cho vo tng lng nc ct nh, lc

ha tan hon ton v a nc ct ti mc. Chuyn dung dch sang bnh cha, lc

trn u ng nht.


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Khi phi un nng dung dch ha tan, hoc qu trnh ha tan c to nhit th

phi ch nhit tr li bnh thng (nhit khng kh) ri mi thm nc ti vch

nh mc.

V d: Pha 1 lt dung dch KOH 1M

Phn t lng ca KOH: MKOH = 39 +16 +1 =56

Lng KOH pha 1 lt dung dch 1M l: 56g

Vy, cn 56g KOH, ha tan trong 1 t nc, cho vo bnh nh mc 1000. y

l phn ng ta nhit, cn lm ngui dung dch trc khi nh mc thnh 1 lt.

Nu mun pha dung dch 2M; 3M hay 0,1M; 0,05M ta cng tin hnh tng t

vi lng cn tng ng.

V d: Pha 500ml dung dch KCl 3M

Phn t lng ca KCl: MKCl = 39 +35,5 = 74,5

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

111,75g

Vy, cn 111,75g KCl, ha tan trong 1 t nc, cho vo bnh nh mc 500,

nh mc n vch.

ii) Cht tan l cht rn ngm nc: khi tnh lng cht tan cn cn phi tnh

lun c khi lng cc phn t nc.

iii) Cht tan dng lng: nu cht tan l dung dch, ta phi tnh ton da vonng dung dch .

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

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

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

Hay 98,65/1,19 = 83ml

Vy, ong 83ml HCl 37% cho vo bnh nh mc 1000 c sn 1 t nc.

nh m c thnh 1 lt. Tin hnh pha trong t Hotte v hi acid bay ln rt c

hi.

f) Pha nng ng l ng(N)

Vic pha dung dch nng ng lng gam (N) cng tng t vi cch pha

dung dch nng phn t gam (M).

1.1.3. Hiu chnh nng dung dch


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Khi pha ha cht, c nhiu nguyn nhn lm cho nng dung dch khng

chnh xc nh vic cn ong khng chnh xc, cc ha cht khng tinh khit hay b

ht m. Thi gian tng tr lu nn cht tan b thng hoa, b oxy ha, dung mi bay

hi, v vy phi kim tra nng thc ca cc dung dch pha sn da vo cc dung

dch c nng chnh xc c gi l dung dch chun (fixanal)

a) Dung dch chun

Dung dch chun l cc dung dch c chun b sn, m bo chnh xc v

c dng nh chun cc dung dch t pha ch khi lm th nghim. Cc cht dng

trong dung dch chun phi kh bn vng sao cho nng ca chng khng thay i

nhanh chng theo thi gian.

Pha dung dch chun, ta phi dng ng chun. ng chun l mt ng ampun

thy tinh hay nha ng kn. Bn trong cha mt lng cn chnh xc cht tan hoc

dung dch cht tan. Khi chuyn ht lng cht tan trong ng vo bnh nh mc v pha

thnh 1 lt ta c dung dch chun c nng ghi trn nhn ngoi ng.

Cch pha dung dch chun t ng chun :

- Dng inh thy tinh chc thng ampun, hng ln phu vo bnh nh mc,

dng bnh tia ra sch cht tan c trong ampun vo bnh nh mc 1 lit, va thm

nc ct va lc v a nc ct ti vch mc.

- i vi cc hp cht bn vng, c thnh phn khng thay i nh NaCl,AgNO3, acid oxalic, ... c th pha dung dch chun trc tip bng cch cn chnh xc

cht cn pha, pha long v nh mc ti th tch ng.

- i vi cc cht nh NaOH, HCl, Na2S2O3, ... khng th pha ngay c dung

dch chun, do cc cht ny thng khng bn vng v d thay i thnh phn, v vy

sau khi pha phi hiu chnh li nng .

V d: NaOH thng nhim mt lng Na2CO3 rt d chy nc, HCl d bay

hi, Na2S

2O

3d b mt nc tinh th khi ngoi khng kh.

b) Phng phphiu chnh nng dung dch

i vi cc cht d thay i thnh phn khi dng rn, nu mun pha dung

dch c nng chnh xc, ta pha dung dch c nng gn ng, sau hiu chnh

nng ca dung dch da vo phn ng vi mt dung dch chun thch hp.


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V d: Pha dung dch NaOH 0,1 N t NaOH rn v dng dung dch chun

H2SO4 0,1N chun li.

i vi cc dung dch d thay i trong qu trnh bo qun, mi ln s dng li

phi xc nh li h s hiu chnh nng dung dch

Xt mt phn ng trung ha acid - base, 1 ion gam H+ s phn ng vi 1 ion

gam OH-. Do :

C1V1 = C2V2

Nu gi Cp l nng dung dch nh pha v Ct l nng thc ca dung dch

ta c h s hiu chnh K:

K= Cp/Ct = Vp/Vt

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

Ly 10ml H2SO4 0,1N cho vo erlen, thm ba git ch th mu phenolphtalien,

dng burette chun bng NaOH c 11 ml NaOH

Vy nng thc t NaOH pha l : Ct = (10*0,1)/11 = 0,091 N

H s iu 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 trng lm thay i cu trc khng gian protein v mt s amino acid

c mch nhnh phn ly (COO-

v NH4+

to lin kt ion). Hat ng ti u ca proteinph thuc vo cu trc khng gian nht nh trong mi trng, ngha l ph thuc vo

t l phn ly ca mch nhnh thnh ion, hay ni tm li l ph thuc vo pH mi

trng.

Dung dch m l dung dch c pH khng thay i nhiu lm khi mt lng

nh acid (H+) hoc base (OH-) c thm vo. Nh vy, dung dch m bao gm mt

cp acid base lin hp (acid yu v mui ca acid yu ny hoc base yu v mui ca

base ny) v t l ca chng s quyt nh pH ca dung dch.

Vic pha dung dch m tun theo nguyn tc chn cp acid base c hng s

phn ly pK A/B gn vi pH m mun pha v phi trn chng vi s mol bng nhau.

Th d, c m pH gi tr 4.75, chn cp acid - base CH3COOH

(0,1M)/CH3COONa (0,1M). Nu thm vo dung dch 0.001 mol HCl th pH ca h

vn ch mc 4.748 gn bng 4.75. Ngha l pH thay i rt t.


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

pH ca mu bng cc h m carbonate v phosphate.

V nguyn tc th phi hp cc loi m c cc khong m khc nhau thnh 1

dung dch l khng c vn g. Tuy nhin s dng cc m no cn ty thuc vo

ng dng sao cho thnh phn ca m khng phn ng vi cc cu t trong h phn

ng kho st theo chiu hng c hi. Cc phn ng thng thy l phn ng to kt

ta hay to phc hoc oxy ha kh.

C dung dch cha hn hp 1 acid yu (V d: acid acetic) vi mui ca n (V

d: mui natri acetate), dung dch s cn bng dng nh sau:

CH3COOH + H2O -----> CH3COO-

+ H3O+

C dung dch khc chng hn, cha hn hp base yu (V d: ammoniac) vi

mui ca n (V d: mui amoni clorua),trong dung dch cn bng c dng sau:NH3 + H2O ----------> NH4

++ OH

-

Dung dch m c pH thay i rt t khi thm mt lng acid hoc base.

Khi thm mt lng acid mnh (H3O+), base lin hp kt hp vi n cho acid yu

(cn bng chuyn dch theo chiu nghch) pH ca dung dch gim khng ng k.

Tri li, khi thm mt lng base mnh (OH-), acid in li cho H3O+

(cn bng

chuyn dch theo chiu thun), ion hidroni H3O+

trung ha OH-cho base yu H2O. Kt

qu pH ca dung dch tng khng ng k.Kh nng m ca hn hp m ph thuc vo hm lng acid v base lin

hp c trong hn hp.

1.2.2. Cch pha mt s dung dch m thng dng

a) Dung dch m borat:

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

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

ml

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

dung dch (b) theo bng di y:

A b pH a b pH


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16

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 citr ate (pH = 3,0 6,2)

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

1000 ml.

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

nc n 1000 ml.

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

(a) v s ml dung dch (b) theo bng 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 mc n 1000 ml.

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

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

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

v s ml dung dch (b) nh mc 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 mc n 1000 ml.

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

Dung dch m c pH khc nhau ph thuc 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 mc 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 ly 50 ml dung dch (a) v X ml

dung dch (b) ri nh mc 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 = 8,610,6)

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

ml.

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

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

dung dch (b) ri nh mc 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 mc n

1000 ml.

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

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

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

dung dch (b) v nh mc 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 vn nng 0,1M

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

nc ct n 1000 ml.- Dung dch acid phosphoric 0,1M (b): 6,45 ml H3PO4 m c v nh mc

bng nc ct n 1000 ml.

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

nc ct v nh mc n 1000 ml.


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- Dung dch natri hydroxide 1N (d): ha tan 40 g NaOH trong nc ct v nh

mc n 1000 ml.

Trn ln cc dung dch (a), (b), (c) vi th tch bng nhau nhn c dung dch

m 0,1M c pH = 1,8. iu chnh gi tr pH ca dung dch m trong khong pH =

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

vn nng nng khc c chun b tng t.

II. Thc hnh

Sinh vin pha dung dch m v mt s dung dch s dng trong cc bi th

nghim sau theo hng dn ca ging vin.

III. Bi np

1. Ghi chp phng php pha cc dung dch thc chun b.

2. Lm cc bi tp sau:1. Pha 1 L dung dch NaOH 40% (w/v): cn ..g NaOH kh2. Pha 0,5 L dung dch H2SO4 2M cn bao nhiu ml H2SO4 m c, bit rng H2SO4

l acid 97% c M=98 g/mol, t trng d=1,84 g/ml.Nng mol ca H2SO4 l:Th tch H2SO4 cn s dng l:Lng nc cn b sung l :

3. Pha 250 ml dung dch CuSO 4 1 M, cn .g CuSO4.5H2O4. 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%

Lng NaOH cn pha dung dch 10% l: X =

Lng dung dch NaOH 40% cn dng l: Y =Lng nc ct thm vo:6. Pha 500 ml dung dch HCl 2% t dung dch HCl 37% (d=1,19g/ml)

Lng HCl cn pha dung dch 2% l: X=Lng dung dch HCl 37% cn dng l : Y =Lng nc ct thm vo:


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BI 2. NH LNG NG KH BNG

PHNG PHP ACID DINITRO - SALICYLIC (DNS)

I. L thuyt1. nh ngha

ng kh l cc ng cha nhm aldehyde (-CHO) hoc ketone (-CO) nh

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

khng phi ng kh.

2. Nguyn tc

Phng php ny da trn c s phn ng to mu gia ng kh vi thuc

th acid dinitrosalicylic (DNS). Cng mu ca hn hp phn ng t l thun vi

nng ng kh trong mt phm vi nht nh. So mu tin hnh bc sng

540nm. Da theo th ng chun ca glucose tinh khit vi thuc th DNS s tnh

c hm lng ng kh ca mu nghin cu.

Phng trnh phn ng to mu gia ng kh v DNS acid:

Acid dinitrosalicylic 3-amino, 5- dinitrosalicylic acid

3. X l mu

Ty vo i tng nghin cu (ht, qu,...) m cch chun b dung dch nghin

cu dng nh lng ng c khc nhau i cht, nhng nguyn tc chung nh

sau:

- Trng hp nguyn liu th nghim khng cha qu nhiu tinh bt hoc

inulin, c th chit ng t nguyn liu bng nc. Cn v cho vo ci s 1 g

nguyn liu ht hay cc mu th nghim thc vt kh nh cy, l hoc qu kh,...

c nghin nh (v sy kh n khi lng khng i). Nu l nguyn liu ti (nh

hoa, qu ti) th cn 510 g. Nghin cn thn vi bt thy tinh hay ct sch v 30


22/39

22

ml nc ct nng 70 80oC. Chuyn ton b hn hp vo bnh nh mc dung tch

1000 ml. un cch thy 70 80oC trong 3540 pht, kt ta protein v cc tp cht

bng dung dch ch acetate [Pb(C2H2O2)2.3H2O] hoc ch nitrate [Pb(NO3)2] 10%.

Trnh dng qu d ch acetate (dng 25 ml ch acetate). Sau loi b lng ch

acetate d bng dung dch Na2SO4 bo ha, yn hn hp 10 pht. Tip thm

nc ct ti vch mc v em lc qua giy lc vo cc hay bnh kh. Nc lc dng

lm dung dch th nghim.

- Trng hp nguyn liu cha qu nhiu tinh bt hoc inulin nh khoai lang,

sn, khoai ty,... cn chit ng bng ru 70 80oC, un hn hp cch thy trong

bnh cch thy c lp ng lm lnh khng kh. Trong trng hp ny khng cn kt

ta protein bng ch acetate v lng protein chuyn vo dung dch khng nhiu.

- Trng hp cc nguyn liu cha nhiu acid hu c nh c chua, da, chanh,kh,... cn ch l trong qu tnh un khi chit, ng saccharose c th b thy phn

mt phn. Do cn xc nh ring ng kh v ring saccharose. Trc khi un

cch thy hn hp phi trung ha acid bng dung dch Na2CO3 bo ha ti pH 6,4

7,0.

II. Thc hnh

1. Dng c - ha cht:

a) Dngc,thit b- My so mu hoc quang ph k

- Cuvette d= 1 cm, V = 4 ml

- ng nghim c np v cc dng c thy tinh thng thng khc

- Bp in

b) Ha cht- nguyn liu

- Mu rau qu cha ng (rau ci ngt, da)

- Thuc 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 bng H Cl 0.1N vi chthlaphenolphtalein, neu het 5 -6 ml HCl lac.


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- Dung dch glucose chun: pha sn dung dch glucose chun 10 mg/ ml (bo qun t

lnh vi ngy).

2. Tin hnh

a) Chit xut ng trong thcvt

- 5g rau ci ngt nghin nh cho vo becher 100ml + 10ml cn 900 un cch

thy (si 3 ln) khuy u bng a thy tinh ngui lc (ch gn ly phn

trong).

- Cho tip 10ml cn 800 vo cc ng b, khuy u, un si 2 ln trn ni un cch

thy ngui lc (lp li 2 ln)

- a phn b ln lc v ra bng ru nng (800)

- Ru qua lc c cho bay hi trong phng hoc trn ni cch thy un nh. Sau

khi cho bay hi ru, mu c th lu trong bnh ht m.- Cn kh trong cc c pha long thnh th tch V = 50ml vi nc ct ta c

dung dch ng gc. Pha lang dung dch ng gc vi mt h s pha lang n thch

hp.

b) Dng ngchun v xc nh hm lng ng

- Cho vo 6 ng nghim sch vi cc cht c th tch nh bng sau:

Ho cht 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

Nc ct (ml) 10 9.8 9.6 9.4 9.2 9.0

Nng

(mg/ml)

OD540nm

(Sinh vin in nng glucose chun vo bng trn)

- T cc ng nghim trn ly 1ml mi ng vo 6 ng nghim khc v thm vo mi

ng 3ml thuc th DNS.

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


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- Lm lnh n nhit phng.

- o mt quang bc sng 540 nm vi mu trng pha t ng i chng.

V ng chun glucose vi trc tung l mt quang (OD540nm), trc honh l nng

glucose. Tm phng trnh biu din ng chun dng y = ax + b vi y =

OD540nm; x = [glucose] (mg/ml) v h s tng quan R2 nh phn mm Excel.

- Tng t ht 1 ml dung dch ng pha lang X cho vo ng nghim, thm 3 ml

DNS, un si 5 pht. Sau khi ngui o mt quang (OD), s dng mu trng

trn.

Ch : tin hnh cc mu dng ng chun v mu th nghim ng thi.

3. Tnh kt qu:

- T phng trnh th ng cong chun tnh c X mg/ml ng kh trong dung

dch ng pha lang.

- Nhn vi h s pha lang n c lng ng trong 1 ml dung dch gc.

- Chn h s pha lang n sao cho OD nm trong gii hn ng chun.

- Tnh hm lng ng trong nguyn liu (mg/g) =

V = th tch dch ng gc (ml)

m = khi lng mu cn (g)

III. Bi np:

1. V ng chun glucose2. Vit phng trnh ng chun y = ax + b

3. Tnh R2

=

4. Bin lun cc h s pha lang.


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BI 3. NH LNG NIT TNG S

BNG PHNG PHP KJELDAHL

I. L thuyt

1. nh ngha

Tt c cc dng nit c trong c th hay trong cc m c gi l nit tng s.

Nit c trong thnh phn amino acid ca protein l nit protein. Nit khng c trong

thnh phn protein nh ca cc mui v c, acid nitric, cc amino acid t do, cc

peptide, urea v cc dn xut ca urea, cc alkaloid, cc base purine v pyrimidine,...

l nit phi protein.

Nit tng s = Nit protein + Nit phi protein

m tng s hay protein tng s l nit tng s nhn vi h s chuyn i. H

s ny ph thuc vo hm lng nit trong protein. Thng thng nit chim 16%

protein nn h s chuyn i thng c s dng l 100/16 = 6.25.

m tng s = Nit tng s x h s chuyn i

Bng di y biu din h s chuyn i c th cho nhiu i tng mu khc

nhau.

Mu H s chuyn i

Ngun gc ng vt 6.25

Ht bng 5.30

u phng 5.46

u nnh 5.71

Ht hng dng 5.3

Cm da 5.3Ht m 5.3

Bp 6.25

Go 5.95

La m 5.83


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2. Nguyn tc

a) V c ha mu

- Trc tin mu c v c ha bng H2SO4 c nhit cao v c cht xc

tc. Cc phn ng ca qu trnh v c ha xy ra nh sau:

2H2SO4 2H2O +2SO2+ O2

- Oxi to thnh trong phn ng li oxi ha cc nguyn t khc. Cc phn t

cha nit di tc dng ca H2SO4 to thnh NH3. V d cc protein b thy phn

thnh acid amine; carbon v hidro ca acid amine to thnh CO2 v H2O; cn nit

c gii phng di dng NH3 kt hp vi H2SO4 d to thnh (NH4)2SO4 tan trong

dung dch

2NH3 + H2SO4 (NH4)2SO4

Cc nguyn t P, K, Ca, Mg,... chuyn thnh dng oxide: P2O5, K2O, CaO,

MgO,...

b) Chng ct m:

- ui NH3 ra khi dung dch bng NaOH

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

- NH3 bay ra c lm lnh bin i thnh NH4OH ri vo bnh hng, bnh

hng cha H2SO4 0.1N

2NH4OH + H2SO4 (NH4)2SO4 + 2H2O + H2SO4 dc) Chun H2SO4 d

- Chun H2SO4 d bng NaOH 0.1N

H2SO4 d + NaOH Na2SO4 + H2O

II. Thc hnh

1. Dng c - ha cht

a) Dng c

- Bnh ph mu

- Bp un

- B chng ct Kjeldahl nh lng Nit gm:

Bnh ct m (bnh Kjeldahl)

ng dn kh

ng sinh hn


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Bnh hng (becher 250ml)

Bi thy tinh

- Pipet 20ml; pipet 10ml

- Erlen 250ml

B chng ctKjeldahl

b) Ha cht + nguyn liu

- Bt u nnh (0.1g), nc tng 1-2 ml

- H2SO4 m c

- NaOH 40%

- H2SO4 0.1N chun- NaOH 0.1N chun

- Thuc th Tashiro

- Cht xc tc: hn hp K2SO4 : CuSO4 (3:1)

2. Tin hnh

a) V c ha mu

- Cn 100mg bt u nnh sy kh tuyt i cho vo bnh Kjeldahl, cho tip

5ml H2SO4 m c s thy xut hin mu nu en (do nguyn liu b oxi ha).Cho thm vo 200mg cht xc tc, lc nh, y kn khong 3 pht. t bnh

Kjeldahl ln bp un, y ming bnh bng mt phu thy tinh.

Lu giai on ny phi thc hin trong t Hotte, t bnh hi nghing trn

bp, trnh trng hp khi si mnh ha cht bn ra ngoi, khi si gi nhit bp

un va phi trnh ha cht tro ra ngoi v khng b bay mt ammoniac.


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- Trong khi un, theo di s mt mu en ca dung dch trong bnh un, khi

thy dung dch gn nh trong sut th c th lc nh bnh ko ht cc phn t trn

thnh bnh cn cha b oxi ho vo trong dung dch. Tip tc un cho n khi dung

dch trong hon ton. ngui bnh ri chuyn ton b dung dch sang bnh nh mc

100 ml, dng nc ct v m trng li bnh Kjeldahl v nh mc n vch.

b) Ct m

- Chuyn 50ml dung dch trong bnh nh mc trn vo bnh ct m c sn

50ml nc ct v 3 git thuc th Tashiro lc ny trong bnh c mu tm hng. Tip

tc cho vo bnh ct 20ml NaOH 40% cho n khi ton b dung dch chuyn sang

mu xanh l m (thm 5ml NaOH 40% nu dung dch trong bnh cha chuyn ht

sang mu xanh l m).

- Tin hnh lp h thng ct m, cho vo bnh hng 20ml H2SO4 0.1N v 3

git thuc th Tashiro (dung dch c mu tm hng). t bnh hng sao cho ngp u

ng sinh hn. Bt cng tc ct m. Kim tra xem hi ra t u ng sinh hn c lm

xanh giy qu khng, nu khng thm NaOH 40%.

- Sau khi ct m 20-25 pht kim tra xem NH4OH cn c to ra khng,

dng giy qy th u ng sinh hn. Nu giy qy khng i sang mu xanh l

c. Ngng ct m, i h thng ngui mi tho h thng em i ra.

c) Chun :- Chun H2SO4 d trong bnh hng bng NaOH 0.1N cho n khi mt mu

tm hng v chuyn sang mu xanh l m. Ghi nhn th tch NaOH 0.1 N s dng.

3. Tnh kt qu:

Hm lng % nit tng s c tnh theo cng thc

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

V1: s ml H2SO4 cho vo bnh hng

V2: s ml NaOH 0.1N chun

a s miligam nguyn liu

1,42: h s, c 1 ml H2SO4 dng trung ha NH4OH th tng ng vi

1,42 mg Nit

III. Bi np

1. Gii thch ngha cc bc trong th nghim.


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2. Tho lun cc yu t dn n sai s.

3. Nu thay H2SO4 trong bnh hng bng acid boric th kt qu c g thay i khng ?


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30

BI 4. NH LNG PROTEIN BNG PHNG PHP

BRADFORD

I. L thuyt1. nh ngha

Protein c nh lng bng phng php Bradford l protein tan trong dung

dch. V vy, mun xc nh hm lng protein mt mu ta phi trch ly protein. Hm

lng protein xc nh s ph thuc vo hiu qu trch ly (kh nng ha tan ca

protein trong dung mi trch ly).

2. Nguyn tc

Phng php ny da trn s thay i bc sng hp thu cc i ca thuc

nhum Coomassie Brilliant Blue khi to phc hp vi protein. Trong dung dch mang

tnh acid, khi cha kt ni vi protein th thuc nhum c bc sng hp thu cc i

465 nm; khi kt hp vi protein th thuc nhum hp thu cc i bc sng 595 nm.

hp th bc sng 595 nm c lin h mt cch trc tip ti nng protein.

xc nh protein trong mu, u tin ta xy dng mt ng chun vi

dung dch protein chun bit trc nng . Dung dch protein chun thng l

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

nhum, mu s xut hin sau 2 pht v bn ti 1 gi. Tin hnh o mt quang hn

hp dung dch bng my quang ph k.

Cng thc phn t ca Coomasie Brilliant Blue G-250


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II. Thc hnh

1. Dng c - ha cht

a) Dng c

- My o quang ph

- ng nghim (14)

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

- ng h bm giy

- Giy thm

- Gi ng nghim

- Bnh tia ng cn

b) Ha cht

- Cn 900

- Dung dch protein chun: cn 10 mg albumin bng cn phn tch, pha trong 1

ml nc ct, lc u cho tan. Gi -200C. Khi dng pha long ra 100 ln, c dung

dch c nng 0,1 mg/ml.

- Dung dch thuc th Bradford: dung dch thuc th c thnh phn trong 1 l

nh sau:

Coomassie Brilliant Blue: 0,05g

Methanol: 50 mlPhosphoric acid 85%: 100 ml

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

lang vi nc ct thnh 500 ml dc dung dch (1)ng trong chai mu ti c

np, 100 ml H3PO4 85% pha lang thnh 500 ml c dung dch (2). Khi cn s dng

ha 1 V (1) v 1V (2) ri lc ly dch trong.

2. Tin hnh

- Thu dch trch ly protein nh trong bi enzyme (10 g malt trch ly bng nc

thu c V = 100 ml). Pha lang mu vi h s pha lang n ln c mu phn

tch (mu X) (n=1, 2 i vi malt).

- Lp mt lot 6 ng theo s th t 0, 1, 2, 3, 4, 5 v 1 ng nghim cha mu

cn phn tch X


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- Dng ng h bm giy, canh thi gian 0 pht cho 5ml thuc th vo ng

nghim 0, lc u yn. thi im 1 pht cho 5ml thuc th vo ng nghim 1,

lc u yn,... c tip tc cho n ht.

ng nghim 0 1 2 3 4 5 X

Nc ct, ml 1 0.9 0.9 0.7 0.6 0.5 -

Albumin chun (0,1mg/ml),

ml0 0.1 0.2 0.3 0.4 0.5 -

TT Bradford, ml 5 5 5 5 5 5 5

Nng Albumin (g/ml) ?

OD595nm

(Sinh vin in nng albumin (g/ml) vo bng trn)

- Tnh thi gian ng 0 c 20 pht, tin hnh o hp thu ca dung dch

bc sng 595 nm. Ta c gi tr OD0, thi im 21 pht o ng 1 (OD1),... tng t

c cch mt pht o cho ht cc ng. Ghi li gi tr OD.

3. Tnh kt qu:

- V ng tuyn tnh gia nng protein (g/ml) vi mt quang OD595

- Da vo phng trnh ng tuyn tnh gia nng protein (g/ml) vi mt

quang OD595 trn ta tnh c hm lng protein P trong mu phn tch X.

P = X*n

tnh tan hm lng protein trong 1g mu cn:

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

V = th tch dung dch trch ly, ml

m = khi lng mu, g.

III. Bi np

1. V ng chun protein theo Bradford

2. Tm phng trnh ng chun y = ax +b

3. Tnh R2

=

4. Tnh tan kt qu th nghim, chn h s pha lang no?


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BI 5. PHNG PHP XC NH HOT TNH ENZYME

I. L thuyt

1.1. Enzyme v n v o hot tnh enzyme

a) nh ngha

Enzyme l cht xc tc sinh hc c bn cht l protein v c tnh c hiu cao.

Mi enzyme c kh nng xc tc cho mt hoc mt s phn ng nht nh. Hot ng

hay hot tnh ca enzyme cng mnh th lng c cht c chuyn ha hoc lng

sn phm to thnh trn mt n v thi gian cng ln. V vy c th nh gi hot

tnh xc tc ca enzyme bng cch xc nh tc chuyn ha c cht hoc tc

tch ly sn phm phn ng.

V nguyn tc c th hai nhm phng php chnh sau:

- o lng c cht b mt i hay lng sn phm c to thnh trong mt thi

gian nht nh ng vi mt nng enzyme xc nh.

- o thi gian cn thit thu c mt lng bin thin nht nh ca c cht

hay sn phm tng ng vi mt nng enzyme xc nh.

thc hin c mc ch trn, nhiu phng php phn tch khc nhau c

s dng: phng php o quang ph, o phn cc, p sut, nht, phng php

sc k v phng php ha hc.b) n v hot tnh enzyme

Mc ch xc nh hot tnh enzyme l xc nh s n v hot tnh. Mt n v

hat tnh enzyme c nh ngha theo nhiu phng php.

i)n v n v quc t (Enzyme Unit, vit tt U) do Hip hi Ha sinh Quc

t (International Union of Biochemistry IUB) nh ngha:

Mt n v chun ca enzyme (1 U) l lng enzyme xc tc chuyn ha c

1 mol c cht sau 1 pht iu kin tiu chun.1 U = 1 mol sn phm = 1 mol c cht (10-6 mol)/pht

ii) Katal:

Nm 1979, Hi ng Danh php ca IUB khuyn co nn s dng Katal lm

n v c bn ca hot tnh enzyme.


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Mt Katal l lng enzyme xc tc chuyn ha c 1 mol c cht sau 1 giy

iu kin tiu chun.

1 Kat = 1 mol c cht/ giy = 60 mol/ pht = 60 x 106 mol/ pht = 6 x 107 U

1 U = 1/60 x 10-6

Kat = 16,67 nKat (nanokatal)

n v Katal c khuyn co v nm trong h n v o lng Quc t (SI).

iii)n v t t: n v hot tnh da vo s thay i c tnh hn hp phn

ng, v d s thay i c, nht ...trong mt n v thi gian. Trng hp c

cht v sn phm l mt hn hp phc tp th p dng n v hot 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 t r 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) Hot tnh r ing (specific activity) ca enzyme

Hot tnh ring ca mt ch phm enzyme c trng cho tinh sch ca chphm enzyme. Hot tnh ring c biu th bng s n v enzyme/mg protein

protein (U/mg protein) hoc Kat/kg protein, trong hm lng protein c xc

nh theo phng php Lowry hoc Bradford.

1.2. Phng php xc nh hat tnh enzyme

Khi tin hnh th nghim o hat tnh enzyme cn ch nhng im 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 t e 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 t h 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 z y m e a m y la s e

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 l y 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 i d 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 i d .

- - a m y l a s e (Exo -1,4-glucanase; EC 3.2.1.2) c nhiu thc vt (ht, c), xc

tc cho phn ng thy phn lin kt 1,4-glycoside t u khng kh to thnh

maltose v dextrin phn t ln. Enzyme ny mt hot tnh nhit trn 70oC,

nhng bn vi acid hn - 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 phn ng thy

phn lin kt 1,4- v 1,6-glycoside t u khng kh ca chui polysaccharide.

Sn phm ch yu l glucose v dextrin. Enzyme ny mt hot tnh nhit

trn 70oC. Nhiu glucoamylase hot ng mnh pH 3,5-5,5.

b ) C c p h n g p h p o h o t t n h en 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 io 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 t a 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 t e 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 t e 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 t r 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 ro 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 t a 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 t e 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 t h 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 khun: 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 t h y t in 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 4 o 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 in 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 t a t e p H = 4 , 7 . G i h n h p n h i t 3 0 o 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 4 o 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 a l t : 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 0 o 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

4 o 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 l o 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 ) Xc 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 in 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 0 o 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 0 o C , 3 0 p h t x y r a p h n n g


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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 t h 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

Hot tnh tan phn (n v hat tnh/g)=m

VnS

10

Hat tnh ring (n v hat tnh/mg protein) = hat tnh ton phn/mg protein

(Bradford).

trong :

n = h s pha long

V = th tch dch trch ly, ml

m = khi lng enzyme, g

III. Bi np

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

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

3. C th s dng phng php o ng kh bng DNS acid o hat tnh enzymeamylase khng ? Ti sao ?


39/39

TI LIU THAM KHO

1. Nguyn Vn Mi. Ha sinh hc. Nh xut bn i hc Quc gia H ni, 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).

cach pha hoa chat t

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