Download Enzymes - Ústav lékařské chemie a biochemie

ÚSTAV LÉKAŘSKÉ BIOCHEMIE 1. LF UK
Enzymes
Practical lesson on medical biochemistry
General Medicine
Jiří Kraml and Jiřina Crkovská
edited by Jan Pláteník
2010/2011
In the following practical lesson the protein nature of enzymes will be demonstrated
on enzyme sucrase (invertase - EC 3.2.1.26) from yeast, and the specificity with respect to the
type of glycosidic bond will be demonstrated on enzymes amylase from saliva and yeast
sucrase.
As to the physical-chemical factors acting on the activity of enzymes, the pH
dependence of pepsin activity will be tested.
On the example of lactate dehydrogenase (EC 1.1.1.27) the dependence of the rate of
enzyme-catalysed reaction on the initial concentration of substrate will be demonstrated (at a
constant enzyme concentration), and also the relationship between Michaelis constant Km and
maximal velocity V will be examined.
Furthermore the activity of potato peroxidase will be demonstrated using benzidine
derivative o-tolidine (dimethylbenzidine) as electron donor, which is oxidized in the presence
of hydrogen peroxide as substrate to a coloured product (blue).
1. Demonstration of protein in sucrase solution
Principle:
The experiment will test the protein nature of sucrase from yeast by biuret reaction and
its precipitation (denaturation) with sulphosalicylic acid.
Yeast sucrase, which will be employed also in the next experiment analysing enzyme
specificity for substrate, splits glycosidic bond between glucose and fructose in sucrose. This
bond is of a mixed type of glycosidic bonds α- (where the 1st carbon of glucopyranose
participates) and β- (with the 2nd carbon of fructofuranose involved). Yeast sucrase (invertase)
is directed to the β-glycosidic bond (it is a β-fructofuranosidase, its systematic name is β-Dfructofuranoside fructohydrolase, EC 3.2.1.26). In this respect it has a specificity different
from mammalian sucrase from the small intestinal mucosa, which belongs to α-glycosidases.
Reagents:
1) Sucrase in solution
2) Sodium hydroxide 80 g/l
3) Fehling´s solution I: copper (II) sulphate CuSO4.5H2O 70 g/l
4) Sulphosalicylic acid 200 g/l
Procedure:
a. Perform biuret reaction with sucrase solution. To 1 ml of sucrase solution add 2 ml of
NaOH solution followed by adding drop by drop a strongly diluted CuSO4 solution
(Fehling I solution diluted about 10-fold).
b. Demonstrate precipitation of sucrase solution by sulphosalicylic acid. Add a few drops
of sulphosalicylic acid solution to approx. 1 ml of sucrase solution.
2
2. Demonstration of enzyme specificity
Principle:
The action of enzymes is highly specific with respect to the substrate as well as to the
type of catalysed reaction. In our experiment, the specificity will be demonstrated on enzyme
amylase from saliva and yeast sucrase. Amylase cleaves α-1,4-glycosidic bonds in starch
giving a mixture of oligosaccharides. Sucrase hydrolyses sucrose to glucose and fructose.
Since reducing saccharides are the resulting products in both cases, Fehling´s test can be
employed to show whether the cleavage of substrate occurred.
Theoretical background to substrate specificity
As to substrate specificity, some enzymes exhibit a very broad substrate specificity,
e.g. esterases splitting both natural and artificial esters; some have a very narrow specificity,
e.g. urease splitting only urea.
Glycosidases belong to the class of hydrolases EC 3 (subsubclass EC 3.2.1 of the
international nomenclature of enzymes). They are specific towards the type of glycosidic
bond (α-glycosidases or β-glycosidases).
Yeast sucrase, as already mentioned, is a β-fructofuranosidase and splits
β-glycosidic bond in sucrose. In this respect it has a specificity different from the
mammalian small intestinal sucrase, sucrose α-D-glucohydrolase (EC 3.2.1.48), which
splits sucrose and maltose and is specific towards the α-glycosidic bond.
Amylase, both of plant and animal origin, splits 1,4-α-glycosidic bonds of starch,
glycogen and polysaccharides. It does not split 1,4-β-glycosidic bonds, as in cellulose.
α-Amylase is an enzyme present in saliva and pancreatic juice of animals. It splits by
endohydrolysis 1,4-α-glycosidic bonds in starch, glycogen and polysaccharides containing 3
or more glucose units bound as 1,4-α-glycosides. Thus it is named oficially 1,4-α-D-glucan
glucanohydrolase (EC 3.2.1.1). Amylose is split hydrolytically to a mixture of maltose and
glucose with hemiacetal hydroxyles released as α-anomers. Amylopectin and glycogen are
split randomly on 1,4-α-glycosidic bonds, whereas 1,6-α-bonds remain intact, thus the
resulting products are branched and unbranched oligosaccharides. Enzymatic hydrolysis of
starch passes through various stages, which can be demonstrated by the reaction with iodine.
Starch gives with iodine a dark-blue colour, polysaccharide split products – dextrins are
stained violet (amylodextrin), purple or red (erythrodextrin), the shorter ones do not give a
reaction with iodine at all (achrodextrin). At the same time the amount of reducing sugars in
the hydrolytic mixture increases.
β-Amylase is an enzyme of plant origin, contained also in malt, and splits
1,4-α-glycosidic bonds from the non-reducing end of polysaccharide chain. Thus amylose
gives practically quantitatively maltose, released as anomer β- , leading to optical inversion
in polarimetry. The enzyme is systematically designated as 1,4-α-D-glucan maltohydrolase
(EC 3.2.1.2). Amylopectin (and glycogen) are split from the non-reducing ends as well, up to
the 1,6-α- branching, the 1,6-α-bonds remaining intact. Dextrin originated from enzymatic
hydrolysis of amylopectin by β-amylase is called limit-dextrin. Neither animal α-amylase
nor plant β-amylase can split 1,6-α- or 1,4-β-glycosidic bonds.
γ-Amylase (glucoamylase) is a membrane bound enzyme in brush borders on the
surface of enterocytes of the small intestinal mucosa. It is relatively thermostable
(thermostable maltase) and splits polysaccharides, and still better maltose from the nonreducing end on the 1,4-α-bonds, releasing β-glucose. (Its official name is 1,4-α-D-glucoside
glucohydrolase, EC 3.2.1.20). It appears that γ-amylase has a slight specificity also to 1,6-αbonds in isomaltose, amylopectin and glycogen. In oligosaccharides released in the small
intestine by digestion of starch and glycogen by α-amylase these bonds are split by
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isomaltase subunit of sucrase-isomaltase complex of the brush-border, named oligosaccharide α-glucohydrolase, EC 3.2.1.10 (oligo-1,6-glucosidase). Sucrase-isomaltase
complex of the brush border is originally synthesized as a single polypeptide chain, the
subunits originate secondarily, by proteolytic action of other proteases secreted into the
intestinal lumen, and have a high degree of homology in their primary structures. The gene of
the precursor pro-sucrase-isomaltase apparently evolved by duplication of a single ancestral
gene. Both subunits split maltose (representing about 80% of maltase activity of smallintestinal mucosa), moreover sucrase splits sucrose and the isomaltase activity is specific for
the bonds 1,6-α- (isomaltose, dextrins).
Reagents:
1) Starch solution 10 g/l (1 g of starch is shaken or heated with water; pour the resulting
suspension slowly to 100 ml of boiling water, keep boiling 2-3 minutes)
2) Sucrose 10 g/l (daily fresh)
3) Physiological solution (NaCl 9 g/l)
4) Yeast sucrase
5) Fehling′s solution I: copper sulphate 70 g/l
6) Fehling′s solution II: sodium hydroxide 250 g/l
sodium potassium tartrate 350 g/l
7) Lugol solution: iodine 13 g/l and potassium iodide 25 g/l
Procedure:
a. Collect about 1 ml of your own saliva to a test tube, and shake with 9 ml of physiological
solution. This will serve as a preparation of amylase.
b. Recall that Fehling′s reagent is prepared by combining Fehling I (CuSO4) with Fehling II
[sodium potassium tartrate (Seignett’s salt) with NaOH]. A sample is mixed with
Fehling′s reagent and heated to boiling. Red colour of Cu2O demonstrates presence of
reducing compounds.
b. Perform Fehling′s test with supplied solutions of starch and sucrose to check that these
substrates do not contain reducing substances.
c. Pipette following volumes into four tubes:
Tube No
1
2
3
4
Starch
5 ml
-
5 ml
-
Sucrose
-
5 ml
-
5 ml
Amylase
1 ml
1 ml
-
-
Sucrase
-
-
1 ml
1 ml
d. Incubate the tubes for 30 minutes at 37 °C (thermo block).
e. Transfer about a half of the volume of the test tubes into other four tubes and perform
Fehling′s test to know in which tube the substrate has been hydrolysed.
f. With tubes 1 and 3 (where starch was used) also test for starch using Lugol solution (prediluted with water: 4 drops/10 ml of water).
Evaluation: Indicate to the table in your lab report where the cleavage of the substrate
occurred using symbols + (substrate cleaved) or − (substrate not cleaved).
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3. Demonstration of dependence of enzyme activity on pH
Principle:
The influence of pH will be demonstrated on the example of pepsin. There are
different forms of active pepsin which differ depending on species and are also named
differently. All of them belong to the subsubclass of aspartate endopeptidases (EC 3.4.23),
i.e. they possess in their active sites two aspartate residues mutually oriented towards the
opposite partner, which participate in the acid base catalysis, and have an acidic pH optimum.
Due to their similarities they constitute so called peptidase family A1 (EC 3.4.23.1, EC
3.4.23.2, EC 3.4.23.3). The originally described properties of human pepsin correspond to the
porcine enzyme named pepsin A (EC 3.4.23.1), which is formed in acidic medium of gastric
lumen by limited proteolysis from the inactive precursor – pepsinogen A, product of
zymogenic cells of gastric mucosa located mainly in fundus. The proteolysis proceeds at pH
lower than 5 as an intramolecular reaction (autoactivation), or due to active pepsin itself
(autocatalysis). At pH higher than 2 the released peptides remain bound to pepsin and act as
inhibitors of pepsin activity. This inhibition is removed at pH below 2.
Human pepsin A has 5 molecular forms. It represents the main pepsin activity in the
gastric juice of vertebrates and its specificity is directed to hydrophobic (Leu), preferentially
aromatic (Phe) amino acid residues before and after the split bond (endohydrolysis). Pepsin B
(EC 3.4.23.2) is formed by a similar mechanism as pepsin A, but from pepsinogen B, and
compared to pepsin A it has a low proteolytic activity to hemoglobin as substrate, however,
splits gelatin very efficiently. It is produced mainly in pylorus. Pepsin C (EC 3.4.23.2) is
named in humans as gastricsin and is formed from the precursor progastricsin, which is
secreted in gastric fundus, antrum and in proximal duodenum. It splits by endohydrolysis
preferentially peptide bonds after the amino acid residue Tyr, is less specific than pepsin A
and has a high activity against hemoglobin as substrate.
For pepsin A the pH optimum depends on substrates: for egg albumin it is 1.5, for
casein it is 1.8 and for hemoglobin the value is 2.3. Gastricsin has pH optimum for
hemoglobin as substrate 3.2.
In the present exercise the proteolytic action of pepsin will be demonstrated on eggwhite proteins as substrate.
Reagents:
1) Pepsin solution 10 g/l
2) Hydrochloric acid 0.1 mol/l
3) Egg-white suspension (boiled egg-white is homogenized in physiological solution – final
protein concentration about 300 g/l)
5
Procedure:
Measure the following volumes:
Tube No
1
2
3
4
Final pH
1.2
1.5
2.5
control
Pepsin
ml
2.0
2.0
2.0
-
HCl
ml
4.0
2.0
0.2
2.0
Distilled water
ml
-
2.0
3.8
4.0
Egg-white
ml
1.0
1.0
1.0
1.0
Incubate the tubes for 5-10 minutes at 37 °C (thermo block). Read the results after 5 minutes
of incubation, and again 5 minutes later, after 10 minutes of incubation.
Interpretation:
The egg-white suspension is cloudy and digestion with pepsin makes the solution
clearer. Evaluate visually the appearance of solution in the tubes and write the results to the
table in your lab report. Use the symbols + for complete clarification, ± for partial
clarification and − for persistent turbidity.
The optimal pH is in the test tube with the clearest solution.
3. Effect of substrate concentration on the rate of enzyme catalysed
reaction: the Michaelis constant
Principle:
The rate of an enzyme-catalysed reaction is defined as the amount of product formed
per unit of time:
∆ [P ]
∆t
The rate depends on the availability of substrate, i.e. on substrate concentration.
Unless the substrate is in great excess, the reaction velocity decreases with time as the
substrate is consumed in the reaction and its concentration lowers. In further consideration,
we will work with initial velocity - v0, i.e. the reaction rate in first minutes of the reaction
when the substrate has not been significantly consumed yet and the effect of time is
negligible.
The relationship of the initial velocity on substrate concentration follows MichaelisMenten equation:
reaction rate
Vmax × S
Km + S
maximal velocity (in unlimited availability of substrate)
substrate concentration
Michaelis constant
v
Vmax
S
Km
=
=
6
Mathematically the relationship has a form of rectangular hyperbola (graph I on p. 10).
The Michaelis constant is a very important term introduced by this equation. It is
defined as the concentration of substrate required to achieve half the maximum velocity
of the enzyme reaction. By definition it is expressed in units of concentration. It does not
depend on concentration of the enzyme in reaction mixture and remains constant for a given
couple enzyme-substrate. Therefore, the Michaelis constant is actually a measure of the
affinity of the enzyme to the substrate. The lower the value the higher is the affinity.
In the present experiment the preparation of lactate dehydrogenase (L-lactate: NAD+oxidoreductase, EC 1.1.1.27) will be used. Lactate will serve as a substrate, which is
converted in the presence in NAD+ as the coenzyme of lactate dehydrogenase (LD) to
pyruvate.
CH3-CHOH-COOH + NAD+
← →
CH3-CO-COOH + NADH + H+
This reaction represents the final stage of glycolysis. Lactate dehydrogenase has 5 isoenzymes
(LD1 – LD5) and their distribution in the body reflects the metabolic specialisation of
individual organs, e.g. LD1 prevails in the heart, whereas LD5 is dominant in skeletal muscle.
The latter form prevails also in the liver. Individual isoenzymes may be separated e.g. by
electrophoresis, which enables an organ specific diagnosis. In our experiments the preparation
of lactate dehydrogenase will serve for demonstration of quantitative relationship between the
enzyme and the substrate (L-lactate). The task will be to follow the influence of various
concentrations of substrate (S) on the initial velocity (v0). Therefore the substrate
concentration will range between 12.5 – 200 mmol/l, whereas the dilution of the enzyme
solution will remain constant.
The estimation of initial velocity v0 is based on the photometry of 2,4-dinitrophenylhydrazone of pyruvic acid, originating by the reaction of the oxo- group of pyruvate with the
free amino group of dinitrophenylhydrazine in acidic medium (at the same time low pH stops
the enzymatic reaction). From this reaction 2,4-dinitrophenylhydrazone results, which after
alkalisation gives a brown-orange colour suitable for photometric estimation at 505 nm.
NO 2
COOH
C
Condensation
O + H 2N NH
CH3
NO 2
COOH
C
N NH
NO2
NO2 + H2O
CH3
2,4-dinitrophenylhydrazine
Pyruvic acid
2,4-dinitrophenylhydrazone of pyruvic acid
Reagents:
1)
2)
3)
4)
Rat liver homogenate (contains lactate dehydrogenase).
Sodium lactate solution 0.6 mol/l
TRIS-HCI 0.05 mol/l buffer, pH 8.5 containing 1 mg of bovine serum albumin in 1 ml.
NAD+ solution 3 mmol/l in TRIS-HCl buffer (same as above). Prepare daily fresh!
5) 2,4-dinitrophenylhydrazine (DNPH) solution in HCl 1 mol/l.
6) NaOH 0.4 mol/l
7
Prepare daily fresh!
Procedure:
a. Prepare working solutions of sodium lactate by serial diluting the 0.6 mol/l stock as
follows (solution from the preceding tube is used for each next dilution):
Tube No
Lactate
mol/l
ml
1
0.6000 (from stock)
2
-
0.6000
2
0.6000 (from tube 1)
1.0
1.0
0.3000
3
0.3000 (from tube 2)
1.0
1.0
0.1500
4
0.1500 (from tube 3)
1.0
1.0
0.0750
5
0.0750 (from tube 4)
1.0
1.0
0.0375
Buffer ml
Final lactate
concentration mol/l
b. Prepare a set of 6 marked test tubes as follows:
Note: Numbers of test tubes correspond to the numbers of working concentrations of
lactate in the table above. The tube No 6 serves as a blank, use the lowest lactate
concentration for this addition. Keep 20 sec intervals between the individual additions of
the enzyme and DNPH.
Time
Tube No
1 2 3 4 5 6
Notes
(min)
Prepare
Substrate 1 - 5 (tab 1)
(ml)
0.2 0.2 0.2 0.2 0.2 0.2 Time
outside
independent
thermo
(ml)
0.3 0.3 0.3 0.3 0.3 0.3
NAD+
block
Preincubate at 37° C - 5 min
0
In
thermo
0.1 0.1 0.1 0.1 0.1 5
Enzyme (liver homogenate) (ml)
block
Mix + incubate - exactly 15 min
20
DNPH
(ml)
Enzyme (liver homogenate)
(ml)
0.5 0.5 0.5 0.5 0.5 0.5
-
-
-
-
-
0.1
Allow to stand at room temperature
NaOH
(ml)
20
21
35
5.0 5.0 5.0 5.0 5.0 5.0
Do
outside
thermo
block
36
Measure the absorbances of the mixtures from each of the tubes
1 – 5 at wavelength 505 nm against the blank (tube 6)
Results:
Write the results to the table in your lab report form.
Evaluation - estimation of the Michaelis constant:
Two graphical methods and one numerical are offered for the estimation of Michaelis
constant. Perform all three of them using your experimental results to obtain the value of
Michaelis constant for lactate dehydrogenase and lactate.
8
A. Double reciprocal plot (Lineweaver-Burke)
When 1/v is plotted against 1/S, a linear function is obtained (see graph II on p. 10).
The intercept of the function with x-axis represents -1/Km, while intercept with y-axis gives
1/Vmax.
1
v
=
Km
Vmax
×
1
S
+
1
Vmax
Plot 1/v (1/A505) against1/S as in graph II, fit a straight line and obtain Km
and Vmax from the intercepts.
The reaction rate v can be plotted directly as A505, because this value is directly related to
the amount of product formed in nanomoles. It is also possible to read the amount of the
product in nanomoles from a calibration graph and convert to the rate v expressed in
nanokatals (nanomoles per second). One student from each pair can plot the graph as the
relationship between 1/A505 and 1/S. The other student can plot the values on the y axis as 1/v
in nanomoles of product formed per 1 second (nkatals). Comparing the graphs, both students
can verify that the value of Km (expressed as molar concentration of lactate) obtained in both
ways of plotting is identical and does not depend on the units chosen for the reaction rate.
B. Direct linear plot (Eisenthal, Cornish-Bowden)
Plot v (A505) against substrate concentration S as in graph I (page 10) and fit a
rectangular hyperbola.
Then plot S value onto the x-axis (horizontal) and the corresponding v onto the y-axis
(vertical). Connect these two points by a straight line. Repeat the process for each pair of
points. Hence n lines are obtained for n pairs of points.
If data fit equation of rectangular hyperbola exactly, all the lines intersect in one point,
its distance from x-axis is Vmax and that one from y-axis is -Km.
In practice when data contain measurement errors, a number of different intersections
are obtained. The coordinates of each intersection provide estimates of Km and Vmax; the
medians of these estimates are the best-fit values of Km and Vmax.
Note: If the data are ordered according to their numerical values, the median can be found as
the middle one of an odd-numbered data set or the mean of the middle pair in an evennumbered data set.
C. Numerical method
The Michaelis constant can be calculated directly from the results using mathematical
description of hyperbolic function. Your experimental data represent a set of values of v
(A505) and corresponding substrate concentrations S:
v1
S1
v2
S2
v3
S3
v4
S4
v5
S5
9
Taking any two pairs of corresponding values, e.g. v1 S1 and v2 S2, the Michaelis constant
can be calculated from the equation:
Km = S1
v2 − v1
× S2 
(v1 × S2) − (v2 × S1)
Calculate Km for all possible pairs of corresponding values v and S and then find a median
value (see above) from these results.
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4. Demonstration of peroxidase reaction
Principle:
Hydrogen peroxide H2O2 as a by-product of the oxidative processes in the cell is
subsequently removed by specialised enzymes called peroxidases and catalases. They belong
to the group of hemin enzymes (hemoproteins). The reaction mechanisms of the two groups
of enzymes are different. Peroxidases catalyse the oxidation of a substrate by hydrogen
peroxide releasing water molecules as a by-product:
Substred + H2O2 → Substox + H2O
Peroxidases (EC 1.11.1.7) are found in plant and animal organisms.
Catalase (EC 1.11.1.6) decomposes hydrogen peroxide to water and molecular
oxygen:
2 H2O2 → 2 H2O + O2
However, catalase may act also as a peroxidase. It is present in greater or smaller amounts in
all tissues (in so called peroxisomes) and body fluids. Its role is mainly to decompose H2O2,
which is harmful for the cells and may lead to oxidative stress.
In the present experiment the peroxidase reaction will be demonstrated qualitatively by
the oxidation of a derivative of benzidine (4,4´-diaminobiphenyl) to a blue product. Because
benzidine is cancerogenic, tolidine, a derivative of benzidine (3,3´-dimethylbenzidine) will be
used instead.
H2N
NH2
H3C
+
POD
H2O2
HN
NH
H3C
CH3
CH3
Reagents:
1) Extract from potatoes (with peroxidase activity)
2) Solution of o-tolidine (3,3´-dimethylbenzidine) 30 g/l in glacial acetic acid
3) Hydrogen peroxide 30 g/l
Procedure:
a. Take about 2 ml of the potato extract to a separate tube, place the aliquot to water bath
and heat to boiling. This denatured extract will serve as one of the negative controls.
b. Prepare the following reaction mixtures to 4 test tubes:
1
2
3
4
1.0
1.0 (boiled)
-
1.0
o-tolidine in acetic acid
4 drops
4 drops
4 drops
4 drops
H2O2
4 drops
4 drops
4 drops
-
-
-
1.0
-
Potato extract
Deionised H2O
(ml)
(ml)
Read the results immediately and write them to the table in your lab report form.
Evaluation: Compare the result in test tube 1 with the appearance of mixtures in tubes 2, 3
and 4. Explain the differences.
11