Biochemical
Demonstrations
The General rules of Biochemical Demonstration Course
The practical course occupies three hours’ laboratory work each week and forms an
essential part of the first-year Biochemistry courses. The module examination will include
questions on the practical work. In addition students are required to have
satisfactorily completed their practical course in Biochemistry in order to pass the
course examination.
Marks out of 10 will be awarded for each of the assessed reports, except the LDH
practical, which is marked out of 20. These will be averaged for the whole module and
this average mark will constitute 10% of the final examination marks for the course.
Experiments missed due to absence because of sickness or other good cause will be
omitted when calculating the average but must be supported by a sick note handed in to
the Biochemistry Office. Attendance at and completion of non-assessed practicals is also
compulsory and contributes a further 5% to the overall course mark.
Pro formas are usually completed on the day of the practical. Where a written report
is required, this must be handed in during the following week’s practical. After this, 5%
will be deducted for each working day that the work is handed in late. Late work should
be handed in to the teaching class technicians who will date stamp it. After a peace of
work is two weeks late, a zero mark will be recorded. Again, work handed in late because
of sickness will be excused providing a sick note is handed in to Biochemistry Office.
The School guidelines are:
1. Biochemistry is an experimental science and it is essential that all students acquire
appropriate experimental skill. Attendance at prcaticals is compulsory. Where written
reports are required, that is in the majority of cases, you must hand these in to the
Demonstrator by the due date. Penalties will be made in the event of a late hand-in.
Failure to hand in the report will result in the mark subsequently gained in the module
examination being withheld until this is received.
2. Failure to attend a practical will be condoned if this is supported by a certificated
medical note or other acceptable reason (‘good cause’). In the absence of such a
notified to repeat it (them) at a later date. This may well result in having to attend
such a practical(s) during the next academic year, but note that any timetabling
problems which you may face as a result are entirely your responsibility to resolve. In
any event, the mark for the module examination will be withheld until the practical
record is complete.
3. For each absence from a prescribed teaching session, a School self-certification form
must be completed as soon as you return to your studies. Any absence due to
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ill-health for a period of longer than seven consecutive days must be supported by a
note from Student Health or your medical practitioner. Students who are ill or
absent from exams must in all cases obtain a note from Student Health or a
medical practitioner .
4. The above regulations apply also to other assessed course work. Attendance at the
appropriate teaching session is compulsory and write-ups or reports must be
submitted by the prescribed time.
The practical course will increase your knowledge of Biochemistry and introduce you to
useful practical techniques and scientific methods. Because many of the experiments
require special apparatus, That is, everyone will do the same experiments during the
course, but in different sequences. The timetable sets out this scheme, and you should be
sure to find out which experiment you will be doing, and read it up before coming to the
class. Then you will be able to plan your work efficiently, to fit into the time available. It
is also very important to understand the experiment at the time you are doing it, and not
merely afterwards then writing up. Also take care to read the Safety Motes for your
experiment beforehand. These are located at the end of each experiment in this book.
General rules for laboratory safety and good practice are set out in this manual.
Your Demonstrator is there to help you to gain maximum benefit from the practical course
and to discuss any questions that you may have about Biochemistry in general. Consult
your Demonstrator whenever you are in doubt what to do, or do not fully understand the
experiment.
Your Demonstrator will also mark and discuss with you your written account of each
experiment. Notes and advice on writing up appear in many of the experimental sections
of this book. It is not necessary to describe experimental methods as these are covered in
the book. A brief introduction, results and Results are all that is required. Record all
observations directly as they are made, and explain all calculations fully. Whenever
possible, present the observations and the results of calculations in tables. Graphs will
often be essential and should be hand drawn, not computer generated. Experiments should
be written up on an experiment record book. An example of a good practical write-up is
available in the lab and your Demonstrator will go through it with you.
The success of a large practical class largely depends on a good collaborative attitude by
the students. In the interests of your fellow students (both in your class and in other
classes that use the same laboratories ), please co-operate with the academic and technical
staff in keeping the laboratories tidy and clean and in maintaining pieces of communal
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equipment ( such as spectrophotometers ) in good working order. If you have doubts
about the use of any apparatus, please ask a Demonstrator. Brief notes on the use of
several items of equipment are presented in the following pages.
Please ensure that all mobile phone are switched off during practical sessions.
GOOD LABORATORY PRACTICE AND SAFETY RULES
1. Always wear a laboratory coat, properly fastened. You will not be permitted to
participate in practicals without one
2. Do not smoke, eat , chew, drink or apply cosmetics in the laboratory.
3. Be conscious of hazards. Read the safety notes for each experiment. Wear gloves and
safety glasses when using hazardous materials. Report any accident however slight to
your Demonstrator.
4. Dispose of all broken glass, Pasteur pipettes, chromatography spotting capillaries etc.,
into the specially strong “sharps” containers provided. NEVER put them into the soft
paper waste bins built in under the benches.
5. Most solutions can be safely disposed of down the sinks. Run plenty of water down
after them. But some waste solvents must NOT be put into the sinks: pour these into
the special waste containers provided.
6. keep your area of bench tidy and organized. Mop up spills at once, first consulting a
demonstrator if any hazard in involved. Replace stoppers in bottles: the switching of
stoppers can cross-contaminate solutions. Never put a pipette contaminated with one
solution into another: cross contamination can ruin an experiment for whole group of
students.
7. Label your experimental tubes. Never rely on keeping them in a particular order, or
on remembering which is which. Labelling is the only way to prevent error.
PIPETTING
1. No pipetting by mouth. Learn to use the various types of pipette filler available in
the laboratory.
2. Whichever filler is used, the principles of accurate pipetting are the same.
(a) Choose a suitable size of pipette. If a graduated pipette is used, its total capacity
should not be much greater than the volume to be pipetted.
(b) Examine the graduations carefully. Some pipettes are calibrated to the tip.
(These should be allowed to drain, not blown out ). But most graduated pipettes
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must only be run down to the bottom mark, as the volume below this mark is
unknown.
(c) Fill the pipette above the top mark. Wipe the outside with absorbent paper tissue.
Run down exactly to the mark. Touch the pipette tip against the side of the vessel,
to avoid a hanging drop. Deliver the liquid, running down to the required mark.
Touch down again to remove any hanging drop.
(d) Always hold the pipette vertically and observe the graduations at eye level.
(e) After finishing with a pipette, rinse it out to avoid solid matter forming in the tip
and blocking it. This is particularly important with protein solutions.
AUTOMATIC PIPETTES
For some experiments you will be provided with Gilson automatic pipettes. These come
in three sizes: P20 for samples up to 20 μl (yellow or natural-coloured tips )
P200 for samples up to 0.2 ml ( yellow tips )
P1000 for samples from 0.2 ml to 1.0 ml ( blue tips )
P5000 for samples from 1.0 ml to 5.0 ml (natural-coloured tips )
Always choose a pipette of the appropriate size and not one that is unnecessarily large.
For example, for dispensing 0.1 ml samples, use the P200 and not the P1000.
Volume adjustment is made by turning the knurled ring at the top of the pipette. Care must
be taken with the setting, as the numbers on the scale are sometimes turned out of sight,
especially with the P5000. A chart of volume settings will be provided to help you. When
increasing the setting it is advisable to turn slightly above the desired setting and then turn
back. You will be given more information and a chance to practice setting the pipettes at
the Introductory Practical session.
Method of use
1. Place a disposable tip of the correct colour on the shaft of the pipette. Press on firmly
with a slight twisting motion to ensure a secure seal.
2. Depress the push-button to the first stop.
3. Holding the pipette vertically, immerse the tip a few mm into the sample liquid.
4. Released the push-button SLOWLY to draw up the sample.
5. Wait 1 or 2 seconds and the withdraw the tip from the sample. Wipe any liquid from
the outside of the tip, but take care not to touch the orifice of the tip.
6. To expel the liquid, place the tip end against the inside wall of the receiving vessel
and depress the push-button SLOWLY to the first stop. Wait a couple of seconds and
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then depress the push-button completely to expel any residual liquid.
7. With the push-button still fully depressed, carefully withdraw the pipette, touching
the tip against the wall of the vessel as you do so.
8. Release the push-button.
9. Remove the used tip by depressing the tip ejector button (on P200 and P1000 only )
Precautions
The Gilson pipette gives accurate and reproducible results if used carefully. It is not
accurate if used in slap-sash manner.
In particular, make sure that you:
1. Do not operate a Gilson pipette until its use has been shown to you by a
Demonstrator.
2. Only use the pipette after attachment of a disposable plastic tip.
3. Depress and release the push-button slowly.
4. Hold the pipette vertically.
5. Do not immerse the tip too far into the solution to be dispensed ( but be careful to
immerse it sufficiently far to avoid sucking air into the tip ).
6. Should any liquid accidentally enter the body of the pipette return it immediately to
your Demonstrator to clean. This will prevent corrosion of the pipette.
NOTES ON SPECTROPHOTOMETRY
Many coloured solutions absorb light in proportion to the amount of coloured material
present. The concentration of a coloured substance can therefore be determined by
measuring the amount of light a solution absorbs. As many substances of interest in
Biochemistry are coloured ( or more precisely, absorb light, even though it may be in the
ultraviolet region of the spectrum when they do not appear to have a colour ) this provides
a very important method of biochemical analysis. In principle, one could make up s set of
standard solutions of known concentration and compare them visually against a white
background with the unknown. It is more precise, however, to use an instrument, a
colorimeter or a spectrophotometer, in which a photoelectric cell measures the amount of
light passing through the solution.
In a colorimeter, light from tungsten filament lamp passes through a suitable filter which
transmits light of fairly wide band of wavelengths. In a spectrophotometer, a
monochromator replaces the filter and produces light of a very narrow band of
wavelengths, a few nm at most (1 nm (nanometer) = 10-9m = 10 Å ). A prism or a
5
diffraction grating gives a spectrum, form which a slit selects the chosen wavelength. This
has three main advantages:
1. A spectrophotometer can be used to measure an absorption spectrum. This is a plot of
absorption against wavelength.
2. The simple theory of light absorption applies only to monochromatic light. Therefore
the use of a broad band of wavelengths often involves mere laborious standardization
procedures.
3. Where absorption peaks are sharp, a spectrophotometer can give much greater
sensitivity than a colorimeter, as well as increased selectivity between different
absorbing substances.
Spectrophotometers are widely used, in biochemical research and clinical practice, for
determination of the concentrations of substances in solution. Often a few micrograms of
material can be estimated conveniently and rapidly.
With the simpler instruments only visible light can be used; therefore test compounds
must be coloured. More elaborate spectrophotometers operate also in the ultraviolet and
infra red regions. The UV is particularly useful for the study of proteins and nucleotides.
In all work with colorimeters and spectrophotometer, the relationship between absorbance
and concentration must be carefully studied, using standard solutions of the substance to
be estimated. It is often necessary to measure the absorbance of “blank” solutions
containing all the components except the substance in question, because these other
components may contribute to the measured absorbance.
Theory
The absorbance of the solution (sometimes called optical density or extinction ) is
measured by the spectrophotometer. It is equal to log10 I0/I, where I0is the intensity of the
incident light, and I is the intensity of the light transmitted through the solution. ( Note:
Absorbance is a ratio and has no units ).
Lambert’s Law states that for a given solution the absorbance is proportional to the path
length (I ) of the light through the cell.
Beer’s Law states that the absorbance is proportional to the concentration ( c ) of the
solution.
6
The Beer-Lambert Law combines the two and states that:
Absorbance = εc l
Where c is the concentration, l is the path length and εis a constant. If c is the molar
concentration and I is in cm., then εis called the molar absorption coefficient ( also
called the molar extinction coefficient ). This is the absorbance of a molar solution of the
compound, in a 1 cm cell, at the specified wavelength. Its units are molar-1cm-1since
absorbance has no dimensions.
The scales of colorimeters and spectrophotometers are often calibrated both in absorbance
units ( = optical density = extinction ) and in percentage transmission. The absorbance
scale is used when measuring the concentration of solutions. The Beer-Lambert Law is
normally obeyed, provided the light is monochromatic. Departures usually indicate a
change in molecular constitution with changing concentration.
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NOTES ON pH, ACIDS & BASES
pH: The acidity of a solution is defined by the concentration of [H+] ions it possesses.
Since this concentration is very small, for convenience we use the pH scale, where:
pH = -log10 [H+] = log10
1
[H+]
e.g. In pure water at 25℃, [H+] = 10-7M, pH = -log1010-7 = 7, where [ ] is concentration
in moles l-1
pH Scale
NEUTRAL
[H+]
10-1
10-2
10-3
10-4
10-5
10-6
10-7
pH
1
2
3
4
5
6
7
INCREASING
ALKALINITY
10-8
10-9
10-10
10-11
10-12
10-13
8
9
10
11
12
13
INCREASING
ACIDITY
EMAMPLE
0.1 M HCl or GASTRIC
LEMON JUICE
VINEGAR
ACID SOIL
LYSOSOMES
CYTOPLASM ( MUSCLE CELL )
PURE WATER AT 25℃
FLUID
SEA WATER
ALKALINE SOIL
ALKALINE LAKES
HOUSEHOLD AMMONIA
LIME ( SATURATED )
0.1 M NAOH
The pH Meter
The pH meter measures the potential difference (p.d.) set up between two electrodes
dipping into the solution whose pH is to be determined. One is a reference electrode
( usually a calomel electrode ) and the other, which responds to the pH of the solution, is
usually a glass electrode. The calomel (reference ) electrode makes electrical contact with
the solution by means of saturated potassium chloride solution contained in a glass
sleeve or sintered glass plug.
The glass electrode consists of a thin glass membrane in the form of a bulb. The p.d.
across such a membrane is found to chance linearly with the difference in pH between its
8
inner and outer surfaces. It behaves exactly as though the membrane were permeable only
to hydrogen ions. Inside the glass bulb is 0.1 M HCl and a silver-silver chloride electrode,
which is reversible to chloride ions.
The electron motive force EMF of the cell ( i.e. between the silver and the mercury
electrodes ) is the sum of a number of potential differences, only one of which depends on
the pH of the solution. The pH meter must therefore be standardized before use, with the
electrodes dipping into a buffer solution of accurately known pH. For detailed operating
instructions, see the card placed near the pH meter
GRAPH DRAWING AND PRESENTATION OF RESULTS
The most difficult, time-consuming and expensive part of practical biochemistry is nearly
always the experimental part, i.e. obtaining the results. Nevertheless an equally vital task
is in deciding what inferences may be made and what Resultss drawn from a particular set
of experimental results. In many experiments, the results are presented graphically and the
following notes refer to aspects which often present problems to first year students.
Continuous functions
In nearly all situations met with in biochemistry, and certainly all those you will meet in
this course, graphs represent continuous functions of a variable. This means that the
quantity you are plotting as ordinate varies in a smooth and continuous way with the
quantity you are plotting as abscissa, i.e there are no sharp points or discontinuities. You
must always plot a set of experimental results either as a smooth curve or as a straight
line.
If a smooth curve (or straight line ) cannot be drawn through each of a set of points, this is
due to experimental error which causes “ scatter” of the points. You should then draw
that you regard as the best smooth curve or straight line taking each point into account.
The examples on the next page may help in understanding this.
Note that the best straight line or smooth curve may not actually pass through any of the
experimental points.
Deciding on how to draw the best straight line or smooth curve through s set of points is a
matter of practice and experience, but as a rough rule, the line or curve should be drawn
so that there are approximately equal numbers of points on each side of it.
9
Always consider whether or not the origin of the graph should be taken as a point on the
line.
Example 1 – a straight line
××
×
××
×
××
×
××
×
××
×
××
×
×
×
××
×
××
×
××
×
×
×
×
Experimental points
Incorrectly drawn
×
Correctly drawn
Example 2 – a smooth curve
××
×
×
×
×
××
×
××
×
Experimental points
×
×
×
×
×
××
××
×
Incorrectly drawn
×
×
×
×
××
×
×
Correctly drawn
Graphs should always have titles, and have both axes labeled. Where there is more than
one experimental line on a graph, these should be labeled too. For this module, graphs
should always be hand drawn on graph paper, not by computer.
Your Demonstrator will have an example of what you should be aiming for when writing
your practical report and will go through it with you.
Use the following page to summarize the important features to be included in an
experimental write-up.
10
YOUR NOTES ON THE EXAMPLE PRACTICAL WRITE-UP
TITLE
Introduction
Method
Results
Discussion
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INTRODUCTORY SESSION
PART ONE: BIOCHEMICAL CALCULATIONS
All sciences at University level are numerical subjects, and simple numerical skills are
essential if you are to be able to design and carry out experiments and interpret results.
Although there are specialized areas of biological science where advanced mathematical
skills are required most of you will only need to use skills with which you should already
be familiar from GCSE, and A level chemistry.
However, as many people may be rusty on these and others may lack confidence, the
worksheets in Part 1 of this session are intended as a refresher. It also highlights some of
the conventions and units used in Biochemistry.
In Part 2 you will also have the opportunity to use some basic lab equipment and check
the accuracy of your pipetting.
Please work through this sheet at your own pace, asking for help from your demonstrator
or another student if you need it BUT MAKE SURE YOU UNDERSTAND HOW TO
ARRIVE AT THE CORECT ANSWER.
The aim is to get the answers 100% correct and at the end of the exercise to be confident
in your ability to tackle similar problems at a later stage.
When you have finished ask your Demonstrator to check over the worksheet and give you
a sheet of answers.
PUT YOUR WORK SHEET AND THE ANSWER SHEET IN
YOUR MODULE MANUAL SO THAT YOU CAN REFER
BACK TO IT IN THE FUTURE.
If you need further practice, more questions (an answers ) are available in the 103
Building in which is the office of the staffers of Biochemistry and Molecular Biology
Department.
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SOME HELPFUL INFORMATION
In Biochemistry, units of volume are based on the litre (symbol l, not to be confused with
the number 1).
This is the same quantity as a dm3 with which some of you may be more familiar.
Since we are often dealing with very small quantities, volumes can also be given as
Milliliter (ml ) = 10-3 l
Microliter (μl) = 10-6 l
Similarly, weights are given as
gram (g)
milligram (mg ) = 10-3g
microgram (μg) = 10-6g
CONCENTRATIONS, MOLES AND MOLARITY
1 mole (mol ) of a substance is the relative molecular mass ( Mr ) in grammes
e.g. NaOH
Na = 23, O = 16, H = 1, therefore Mr = 40.
Therefore 1 mole of NaOH = 40.
A solution of 1 mole of a substance dissolved in a total volume
of 1 litre is a 1 molar (1M) solution.
Don’t confuse mol and M. The first is an amount, the second is a concentration.
Concentrations may also be given as mass/volume, e.g. g/l, μg/ml
Handy hint: 1g/l is the same concentration as
1 mg/ml, is the same as
1μg/μl
( convince yourself by doing the calculation!)
To interconvert between volumes, molarities and moles, remember that the molarity (M)
of a solution is the number of moles per liter, i.e.
mol
M=
Vol(in L )
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Examples of calculations:
1. What is the molar concentration of a ribose solution where 15 g of ribose are
dissolved in 100 ml of water? The molecular weight ( Mr ) of ribose is 150.
First, calculate the amount in moles of the ribose dissolved in the solution by dividing
the mass i.e. 15 g by the Mr: 15 ÷ 150 = 0.1 mols.
Next, determine the volume of the solution in liters: 100ml = 0.1 liters
Finally, divide the number of moles by the total volume of the solution to obtain the
concentration in moles per liter: 0.1 mol ÷ 0.1 l = 1 Molar ( or M )
2. How many moles of ribose are there in 200 ml of a 0.5 M solution?
First calculate how many moles are in 1 liter of a 0.5 M solution: 0.5 moles
Next, determine the volume in liters: 200 ml = 0.2 liters
Finally, multiply the number of moles there are in one liter by the volume in liters:
0.5 moles/l × 0.2 l = 0.1 mol i.e. 100 millimol. (mmol)
3. How many g ribose would you need to make 100 ml of 0.25 M solution?
Calculate how many grams you would need to make 1 l 0f 0.25 M ribose by
multiplying the Mr by the required molarity, ie 150 × 0.25 = 37.5 g in 1 l
You require 100 ml, therefore multiply the grams by the volume in liters, i.e. by 0.1,
therefore you need 3.75 g ribose.
Some basic calculations on moles and molarities
Q1. The Mr of NaCl is 58.44
(a) 1 M NaCl
( Show any working out )
=
g/l
(b) 0.5 M NaCl
(c)
(d)
(e)
(f)
(g)
=
g/l
=
mg/ml
=
μg/μl
2.5 mM NaCl
=
mg/l
=
g/l
=
μg/ml
5.844 mg/l NaCl
=
M
=
mM
1 g NaCl in 100ml
=
mM
1μl of a 20 mg/ml solution contains
μg
How many g NaCl in 50 ml of a 2.5 M solution?
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Dilutions
In a biochemistry lab, many frequently used solutions like buffers and reagents are made
up and stored as concentrated stock solutions, and diluted appropriately before use.
In order to carry out any dilution, the dilution factor must be found, ie the factor by which
the original concentration is bigger than the final concentration. It is calculated by
dividing the concentration of stock solution by the concentration of the required solution.
This number is therefore the volume (mls ) into which 1ml of the stock solution must be
diluted in order to obtain the required concentration.
For example
A 4% solution w/v (i.e. 4 g in 100 ml ) copper sulphate solution is required when making
up the reagents for one of the common protein assays. The stock solution may be stored as
a 24% solution.
Using this stock, how would you make up 12 ml at 4% ?
First find the dilution factor:
Divide the stock concentration by the required concentration: 24 ÷ 4 = 6
This tells you that you need to take one part of stock and dilute it to a total of 6 parts i.e. 1
ml stock plus 5 ml water, giving 6.0 ml total.
Since we require 12 ml total, we will need 12 ÷ 6 × 1.0 ml in a total volume of 12 ml,
i.e. 2.0 ml copper sulphate plus 10 ml water, giving 12.0 ml total.
Q2. (a) Given a 0.75 M solution of NaCl, how would you make 100 ml of 0.15 M NaCl?
( This is ‘normal saline’, approximately the concentration of sodium ions in the
blood and is often used in cell culture techniques and in the hospital ‘saline
drip’).
What is the dilution factor?
How many mls 0.75 M saline are required ?
How many mls H2O are required?
15
(b) Given a 1.0 M solution of NaCl, how would you make 500 ml of 10 mM NaCl ?
What is the dilution factor?
How many mls NaCl are required ?
How many mls H2O are required ?
(c) What would the molarity of a solution be if you used 4.0 ml × 2.4 M stock
solution, plus 12.0 ml H2O ?
The previous examples have been worked out using the ‘reasoning ‘ method, a sensible
way to tackle calculations. Some students are attracted to the ‘ formula’ method. This
appears to require less brain power but consequently has greater potential for error. Used
carefully, it should give the same answer!
Try the following calculations using this method:
Q3. ( a ) If the volume (in liters ) of a solution multiplied by its molarity gives the number
of moles of substance in that solution, rearrange the equation
v M = mol
to solve for (i)
v
v=
(ii)
M
M=
( b ) How many moles of NaCl would there be in 375 ml of a 0.5 M solution ? ( be
careful of your units )
( c ) How many moles in 100 μl of a 0.5 M solution ?
(d ) If I have 0.5 moles of NaCl in a 200 ml of solution, what is the molarity ?
16
( e ) I wish to make up 2.5 liters of normal saline (0.15 M ). How many moles of
NaCl will I need ?
( f ) How many ml H2O would need to be added to 5.84 g NaCl to make a 0.25 M
solution ?
( g ) Can you work out why the concentration of water is approximately 55 M ?
( The density of water is 1 kg / l ).
FURTHER QUESTIONS – You may do these at home, but you must bring them to
show your demonstrator next week
A CALCULATION FROM THE HOSPITAL LAB:
Q4. A hospital laboratory technician wants to make some agar plates to test the sensitivity
of bacteria isolated from a patient’s blood sample to some common antibiotics in order to
test which antibiotic to give the patient. Two common antibiotics and their Mr are:
Ampicillin
349.4
Tetracycline
480.9
The technician wants to make three 1 liter batches of agar containing 3 mM, 0.3 mM and
0.03 mM ampicillin. How many milligrammes of ampicilliln needs to be added to each
batch ?
In order to make plates which will set and be reasonably robust the agar concentration
needs to be 2%. How much agar is needed?
Tetracycling is usually used at a working concentration of 25 to 50 μg / ml. What is the
molarity of a 50μg / ml tetracycline solution ?
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PART 2
Introductory Practical Work – A short exercise in pipetting and performing dilutions
This exercise is designed to introduce you to using the pipettes and putting into practice
some of the methods of calculating dilutions. Working INDIVIDUALLY, see how
accurately you are able to use the pipettes. Refer to the instructions on pages 19 and 20.
Your Demonstrator will show you how to use the spectrophotometer.
1. You are provided with a 0.5 M solution of a compound that absorbs at 510 nm. You
must dilute this stock solution appropriately to obtain 5 ml of three solutions, with
concentrations of 0.05 M, 0.01 M and 0.005 M respectively. Fill in the table below,
showing the volumes of water and the 0.5 M solution required to produce each of the
solutions. Also record the absorbance reading you obtain.
2. In addition, dilute your 0.05 M solution by a factor of 10 and record the absorbance in
the table. Compare its absorbance with that of the 0.005 M solution.
Do the absorbance values seem reasonable?
3. Now mix 175 μl of the stock solution with 2.825 ml of water. Record your
absorbance.
Match your accuracy with the correct value, shown on the board !
Required
Concentration
Volume of stock
solution (μl )
Volume of water
175μl
2.825 ml
Absorbance value
0.05 M
0.01 M
0.005 M
0.005 M
(diluted from 0.05 M)
?
SPECTROPHOTOMETRIC DETERMINATION OF CYTOCHROME C
Aim:
18
To determine the concentration of a coloured compound in a solution by comparison of its
light absorption with the light absorption of a standard solution of known concentration.
See theory of light absorption – pages 19-20.
Objectives of this experiment:
1. To learn how to use a simple spectrophotometer to determine the degree of light
absorption by a coloured compound.
2. To plot a SPECRUM of light absorption against wavelength and to decide which
wavelength it is best to use for absorption measurements.
3. To make up a set of standard solutions of known concentration and to use them to
plot a graph of light absorbance against concentration ( calibration graph ).
4. To measure the concentration of an unknown using this curve.
5. To consider what degree of precision can be expected from this method of
determination of concentration.
Biochemists often need to determine the concentration of a compound in solution. If the
compound absorbs light at a given wavelength, this can be measured in a
spectrophotometer and the concentration calculated ( see Notes on Spectrophotometry,
page 19 ).
Cytochrome c is a small protein which has attached to it an organic molecule, haem.
Cytochromes are found in nearly all cells and the one you are using comes from the
mitochondria of animal cells. Cytochromes are important because the iron atom in the
haem group can be oxidized or reduced and the protein operates as an electron carrier in
metabolism. Many other haem-protein complexes are found in living organisms (e.g.
haemoglobin, myoglobin, catalase, cytochrome a ), and all are red-coloured and absorb
light in the region 400~450 nm ( “ Soret band” )
A stock solution of cytochrome c in water, containing 0.1 mg per ml solution, is provided.
Use it to determine the concentration of cytochrome c ( abbreviation: cyt c ) in the
unknown solution ‘X’.
Plot the results of the experiments given below on graph paper and show them to your
demonstrator before you leave. If you need help with the calculations, ask your
Demonstrator during the practical sessions.
METHOD
19
1. Determination of peak wavelength
It is first necessary to find the exact wavelength of maximum absorbance, because the
peak in the absorption spectrum is extremely sharp, and a small change in the
wavelength used will therefore cause a large change in measured absorbance.
You should have two spectrophotometer tubes, suitable for the instrument you will be
using on your bench.
Half fill a spectrophotometer tube with water to act as reference and half fill another
tube with the stock solution of cyt c. Instructions for use of the spectrophotometer will
be found near the instrument. Be sure to keep both the tubes clean on the outside, and
to hold them only near the top, away from the region through which the light beam
passes. Remember to align the mark on the tube with that on the instrument.
Set the spectrophotometer to 380 nm, zero it with the reference (water) tube, and then
measure the absorbance of the cyt c solution. Now change the wavelength to 385 nm,
zero again with the reference tube, and measure the absorbance of the cyt c solution. In
this way, measure the absorbance at 5 nm intervals, from about 380 nm to 420 nm. It is
imperative that you zero the instrument with the water reference at each wavelength.
This is because the light source in the spectrophotometer emits different amounts of
light at different wavelengths. Plot the absorption spectrum that you have measured and
hence find the wavelength of maximum absorbance.
It is important to use the same instrument for the rest of the experiment, as wavelength
settings may vary slightly from instrument to instrument.
2. Calibration graph
Using a P5000 Gilson pipette with great care ( e.g. wiping the outside with tissue
before delivering the solution, bolding vertically ) make accurate dilutions of the stock
cyt c solution in four clean dry boiling tubes as follows and then mix well.
Tube
Stock cyt c soln ( ml )
Deionised water ( ml )
Final concentration of cyt c ( mg / ml )
1
2
8
2
4
6
3
6
4
4
8
2
Set the spectrophotometer to the wavelength of maximum absorbance as already
determined.
Zero the instrument using water, then using the same tube, measure the absorbance of
each solution of cytochrome c and of the undiluted stock. There is no need to wash the
tube between samples, simply drain well and rinse with a little of the new solution.
Tabulate the results.
20
3. Determination of unknown cyt c concentration
Wash the spec tube well and recheck the instrument zero using water. Drain the tube
and fill it with the cyt. C soln X. Measure its absorbance at the same wavelength.
RESULTS
1. The concentration of the stock cyt c solution is 0.10 mg / ml. Calculate the
concentration of each of your dilutions and plot a graph of absorbance against
concentration, including the point for undiluted stock solution and the point at zero
concentration. Note that concentration, which is the independent variable, should be
on the abscissa ( x-axis ) and absorbance, which changes as a result of the chance in
concentration, should be on the ordinate (y-axis ). The graph should be a straight line.
2. Using the graph, find the concentration of the solution X.
DISCUSSION
1. If the spectrophotometer reading is accurate to about ± 0.02 absorbance units,
calculate the approximate % error of the determination of the concentration of X.
( Pipetting and other errors should be much smaller, and therefore negligible. )
2. Take the relative molecular mass of cytochrome c as 13,300 and, assuming that the
cytochrome c is pure, calculate the molarity of the stock solution. The molar
absorption coefficient ( see page 20 ) for cytochrome c is 125,000 at the peak
wavelength and the light path in the spectrophotometer tube is 1 cm. Hence, calculate
the expected absorbance of the stock solution. Compare this with your observed value.
What does this result tell you about the purity of the sample of cytochrome c used ?
SAFETY NOTES FOR EXPERIMENT 1
No hazardous procedures or chemicals are used in this experiment. Read the noted on
Good Laboratory Practice and on Pipetting ( pages 4-5).
21
Protein separation methods
Aim
To illustrate the principles of protein separation:
(i) on the basis of size difference; (ii) using difference in net charge.
Objectives of this experiment
1. To understand the practical and theoretical basis of molecular exclusion
chromatography.
2. To understand the practical and theoretical basis of ion exchange
chromatography.
3. To gain experience in operation columns and collecting fractions of
eluate.
4. To understand the terms ‘ bed volume’ and ‘void volume’ and how they
are measured.
5. To be able to apply simple tests to identify the elute proteins.
6. To appreciate some applications of the techniques.
Proteins may be separated on the basis of a number of properties, such as size, charge of
affinity for other molecules. Some methods allow the proteins to be recovered and used
after separation----preparative methods----whilst others give good separation of very
small quantifies for identification purposes----analytical methods.
This practical will demonstrate two preparative methods which rely on different
properties to effect the separation.
1. GEL FILTRATION ( MOLECULAR EXCLUSION ) CHROMATOGRAPHY
WITH SEPHADEX: Separation by size
How gel filtration works
Gel filtration is a method of separation substances according to their molecular sizes. A
column is filled with very small beads of an insoluble but highly hydrated carbohydrate
polymer, suspended in buffer. Each bead consists of a network of cross-linked fibres
resembling a microscopic ball of gauze. By altering the degree of cross-linking, beads can
be produced with different sized pores. Within each grade of beads, the pore size is not
22
uniform but covers a specific range. When a mixture of molecules of various sizes in
applied to the top of the column, followed by buffer solution, the molecules move
downwards at different rates, depending on their sizes, and can be collected as they
emerge ( are eluted ) from the bottom of the column. Small molecules are able to enter the
beads of gel, passing freely between the fibres, but molecules which are larger than the
largest pores in the beads cannot enter the fibre network and are said to be excluded. They
remain in the liquid outside the beads.
The total bed volume (Vt) of a gel filtration column consists of the volume of the hydrated
beads ( Vi) and the volume of the liquid between the beads, which is known as V0, the
void volume
Vt = V i + V0
The buffer solution passes down the column through the spaces between the stationary
beads. Large molecules that are excluded from the gel travel with the moving liquid in the
void volume. They are eluted when the void volume has just been replaced by new liquid
run in from the top, i.e. when the volume of eluate collected up to this time is V0.
In contrast, small molecules which can penetrate all the pores of the gel beads distribute
themselves between the liquid inside and outside the beads. Molecular diffusion carries
them rapidly to and fro, so that their concentrations in the water inside and outside the
beads remain equal. While they are outside they move downwards with the flow of liquid,
but while inside the stationary beads they are temporarily retarded. Their net rate of
movement down the column therefore depends on the proportion of time spent inside and
outside the gel beads, which in turn is determined only by the relative volumes of water
inside and outside. Small molecules have a larger volume of liquid available to them and
therefore take longer to pass down the column. They are eluted when a volume of liquid
equal to the whole bed volume ( Vt ) has passed through the column.
If molecules of intermediate size are present, they will be able to pass through only the
larger of the pores of the gel but not the smaller. ( The size of the pores in the gel beads in
not uniform but covers a considerable range, being the result of random chemical
cross-liking of the fibres ). Intermediate sized molecules will be able to enter only part of
the gel volume and will therefore spend an intermediate proportion of their time within
the stationary beads. They will be eluted after the large molecules that are completely
excluded from the gel and before the small molecules that penetrate the whole gel.
Each size of molecules effectively occupies a particular volume of liquid ( ranging from
23
V0 to Vt ) and reaches the bottom of the column when this volume of eluate has ren
through. Thus substances emerge in order of decreasing molecular size.
In this experiment the gel is Sephadex, a cross-linked polysaccharide. It is available in
many different network sized, depending on the degree of cross-linkage. The size must be
chosen according to the molecular weights of the substances to be separated.
Using Sephadex to remove small molecules from a protein solution
Gel filtration is a frequently used technique in the biochemistry lab. Here is an example:
Often during the preparation or study of a purified protein it is necessary to remove a salt
from a protein solution. For example, during the purification of an enzyme, a 50%
saturation with ammonium sulphate may have been used to precipitate unwanted protein.
In order to assay the enzyme, now in the supernatant liquid, it is necessary to remove the
ammonium sulphate. One way in which this could be done easily and quickly is by means
to a Sephadex column on which the large molecules ( enzyme ) are separated from the
small molecules ( ammonium sulphate ). This is known as desalting.
In the above example, both the enzyme and the ammonium sulphate may be colourless.
We will be doing an experiment with coloured compounds so that you can see what
happens on the Sephadex column.
We will separate haemoglobin (large molecule, reddish brown ) from a dye, methylene
blue (small molecule, blue ). The mixture is labeled Hb / dye
Experimental method
Packing the column
1. Tighten the screw clip a the bottom of the glass column provided and fill the column
with water to ensure that the clip is capable of stopping the solvent flow. Empty out
the water.
2. Wearing gloves, roll a piece of wet glass wool between finger and thumb to produce a
small ball just sufficient to block the outlet of the column. Drop it into the bottom of
the column and pack it down with a glass rod to make a flat surface. Now set up the
column in a burette stand, making sure that the column is vertical and that the screw
clip is shut.
You are provided with a slurry of dextran gel ( Sephadex 50, medium )
24
3. Invert the bottle of Sephaex several times so that the beads are thoroughly suspended
in the buffer ( 0.05 M phosphate, pH 6.5 ), and two-thirds fill a 50 ml beaker with the
slurry, Swirl the contents of the beaker and carefully pour the slurry into the column
until it is full. Please pour any unused slurry back into the bottle ( Sephadex is very
expensive ).
4. In a few minutes, the gel ‘bed’ in the column will have started to settle, leaving 1 to 2
cm of clear solution above it. Open the screw clip slightly and allow the buffer to drip
slowly from the column into a waste beaker.
5. The level of the Sephadex gel will fall until the bed is fully packed and the volume of
the clear solution above the bed will start to decrease. When approximately 2 cm of
the clear solution remains above the top of the Sephadex bed, tighten the screw clip
again. On no account allow the top of the gel to run dry.
6. If the Sephadex bed fills less than half the column, you should pour in more slurry
and allow it to pack as before. Aim at a packed bed filling about tow-thirds of the
tube. The column must now be equilibrated with buffer.
Washing the column
1. Using a Pasteur pipette, very slowly pipette 2 ml of phosphate buffer down the walls
of the column on to the Sephadex, ensuring the top of the bed remains even. Continue
until the column is full. ( If the Sephadex surface is uneven, it may be improved by
stirring up the top 2 cm of the gel with a fine glass rod and allowing it to settle
again. )
2. Open the clip and allow the buffer meniscus to sink just to the top of the bed. Tighten
the clip immediately.
Loading the column
1. Place a clean 10 ml cylinder under the outlet tube. Using a 1 ml graduated pipette, run
0.3 ml of the haemoglobin / dye sample very slowly on to the top of the column,
taking not less than 30 seconds.
2. Open the clip slightly until all the sample has just entered the gel, then tighten the
screw immediately.
3. Slowly wash down the walls of the column with 0.3 ml buffer. Allow this to run in
until the meniscus just sinks into the gel bed. Tighten the screw clip again.
Eluting the column
1. Fill up the column with buffer, adding it very slowly, especially at first, and being
extremely careful not to disturb the bed surface.
2. Now adjust the screw clip so that the buffer drips form the column at a rate of one
25
3.
4.
5.
6.
7.
8.
drop every 3-4 seconds. ( If the flow rate is faster than this, equilibration between the
liquid in the gel particles and the interstitial liquid will not occur and the bands will
spread out ).
As the buffer flows through the column, two band should become visible and move
down the column. Do not, at any time, allow the column to run dry. Always keep
it topped up with buffer. ( There is less disturbance of the bed surface if more buffer
is added before the level has sunk too low ). Do the two bands look alike?
Collect all the eluate in the 10 ml cylinder until the first (lower ) coloured hand is
about to emerge. Quickly remove the cylinder and replace it with a second, clean 10
ml cylinder.
Collect the first coloured band in the second 10 ml cylinder and then change to a third
10 ml cylinder.
Collect the eluate in this 10 ml cylinder until the second band is about to emerge.
Finally, collect the second coloured band in a clean 25 ml cylinder.
Note the volume collected in all four cylinders. Measure the height of the packed
Sephadex.
IMPORTANT: The Sephadex used in these experiments is recycled. When you have
finished, please empty your column into the beakers provided. Use a little buffer to wash
out the Sephadex if necessary ( no glass wool ! ). The lab staff will wash it well for reuse.
Identification of the protein in the eluted fractions
The color of the two bands should tell you which contains the protein and which the
smaller dye molecules. Test this identification by means of a precipitation experiment.
Precipitation with tricholoracetic acid
Strong acids which yield large multivalent anions ( such as trichloroacetic acid ) denature
and precipitate proteins. Because the solution is acid, proteins are positively charged and
will be bound together in a precipitate by the large multivalent negative ions. N.B.
Wearing gloves and safety glasses, to a sampled of each eluted fraction add an equal
volume of 10 % ( 0.6 M ) trichloroacetic acid.
A precipitate or cloudiness ( within 5 minutes indicates the presence of protein. )
Results
Record the volumes collected in the cylinders, the height of the gel and the results of the
precipitation tests o the proforma. Complete the calculations and answer the follow-up
questions. Check with your Demonstrator of you have any queries about this section then
move on to the second separation method.
26
SAFETY NOTES FOR GEL FILTRATION CHROMATOGRAPHY
Sephadex gel
Phosphate buffer
Haemoglobin with methylene blue
All contain 0.02 % sodium azide as a
preservative. This is toxic by ingestion.
10 % trichloroacetic acid solution
Causes burns to eyes and skin and is toxic by
ingestion and by skin contact.
Glass wool
Irritation to eyes and skin by contact.
2.
SEPARATION
CHROMATOGRAPHY:
Separation by charge
OF
PROTEINS
USING
ION-EXCHANGE
How ion exchange chromatography works
Proteins can be separated on the basis of their net charge by ion-exchange
chromatography. A protein with a net positive charge (a cationic protein ) will normally
bind to a solid phase containing negatively charged carboxylate groups ( a cation
exchanger ). Alternatively, if the solid phase contains positively charged tertiary amino
groups ( an anion exchanger ), negatively charged ( anionic ) proteins will bind but
cationic proteins will not. Chemically codified celluloses are often used as solid phases
for protein separations.
Cellulose
Or
Agarose
Cellulose
Or
Agarose
O
CARBOXY-METHYL (CM ) GROUP
(CATION EXCHANGER )
-CH2-C
O
C2H5
-CH2-CH2-N H
C2H5
+
diethyl amino-ethyl ( DEAE )
(ANION EXCHANGER )
The ion exchange celluloses are most effectively used by packing them into glass or
plastic columns. Protein mixtures are adjusted to a chosen pH then applied to the top of
the packed column. Once bound, they may be eluted by one of two methods.
27
(a) changing the pH of the buffer to reverse the net charge on the bound proteins. At
higher pH, proteins become negatively charged, whilst at lower pH they become
positively charged, Each protein has a characteristic isoionic pH (the pI ) at which it
has a zero net charge, this depending on the number of basic and acidic side groups
present in the protein. If the pI values of two proteins differ appreciably, a buffer of a
particular pH can be chosen for elution, such that one protein is negatively charged
and the other positively charged. Separation by ion exchange then becomes
possible.
(b) An alternative method is to use a buffer containing a high concentration of a neutral
salt, such as NaCl. The high concentration of cations ( Na+) will compete with
cationic proteins for the negatively charged binding sites on the cation exchange
column; this will cause the cationic proteins to be eluted, although they remain
positively charged. Similarly, high concentrations of anions ( Cl-) will compete with
anionic proteins and cause their elution from an anion exchange column. By
increasing the concentration of salt in the buffers (producing a salt gradient ) it is
possible to elute weakly bound proteins first ( wit a low density of net charge )
followed by strongly bound proteins ( with a high density of net charge ), thus
achieving separation.
In this experiment we will be separating two proteins from a mixture using a cation
exchange column. You will also have to suggest and explain a protocol for separating
these proteins on an anion exchange column.
A practical example of the use of ion exchange chromatography in medical diagnosis and
monitoring is also given on the proforma and you are asked to choose and explain a
suitable system for effecting the required separation.
Experimental method
You are provided with the following materials:
1. Suspension of carboxyethyl cellulose CM23 in 0.01 M acetate buffer, pH 5.0.
2. Protein mixture containing 10 mg /ml cytochrome c and 30 mg /ml haemoglobin, in
0.01 M acetate buffer, pH 5.0
3. Buffer: 0.01 M acetate, pH 5.0
0.01 M Tris, pH 8.0
0.1 M Tris, pH 8.0, containing 1.5 M NaCl.
Data
Cytochrome c pI = 10.7
Haemoglobin pI = 6.9
28
Packing the column
1. Insert a glass fibre disc into the barrel of a 5 ml syringe ( we are using a syringe as the
column ). Using a glass rod gently push the disc down the barrel until it is lying flat
on the base.
2. Mount the syringe on the clamp stand. Make sure the clip on the syringe outlet is
closed.
3. Invert the suspension of CM-cellulose gently several times, to resuspend, and quickly
pour some into the syringe barrel, filling it the top.
4. Place a beaker under the column, open the clip, and allow the buffer to flow through
as fast as possible. With this type of column, you don’t have to worry about the
column running “dry”
5. Allow the buffer to drain through until none remains above the surface of the packed
cellulose. The packed column volume should be about 4-5 ml. If it is much less
then top up with additional CM-cellulose.
6. Close the clip and insert a second glass fibre disc into the syringe, pushing it down
gently to rest on the top of the packed cellulose. This will help prevent the surface
being disturbed.
Washing the column
1. Using a Pasteur pipette, gently fill up the syringe barrel with 0.01 M acetate buffer,
pH 5.0.
2. Open the clip and allow this buffer to flow through until none remains above the
surface.
Loading the column
1. Very slowly apply 0.3 ml of the protein mixture to the top of the glass fibre disc.
2. Open the clip and allow the protein mixture to drain into the column.
3. Close the clip, slowly add 0.5 ml 0.01 M acetate buffer, pH 5.0, then open the clip
and allow this buffer to drain through, The glass fibre disc should now appear white.
Having loaded the protein mixture onto the column, you are now ready to effect
separation. Decide which buffer you are going to use; if you are at all unsure, check
with your Demonstrator.
Eluting the column
1. Gently fill up the barrel with your chosen buffer.
2. Open the clip slightly and adjust the elution rate to about 1 drop every 4 or 5 seconds.
Collect the colourless eluate in a waste beaker; this can be discarded. Add more of
29
the same buffer as necessary to keep the column running.
3. When a coloured band is about to elute from the column, collect it in one or more
tubes. Try to collect about 1 ml from the center of the band, ie a concentrated sample
of the protein.
4. When you have collected a concentrated sample, switch back to the waste beaker and
allow all the buffer to drain into the column. Close the clip.
5. Now you need to decide which buffer you will use to elute the remaining
coloured protein band from the column. Repeat steps 1-4 with your chosen
buffer.
Identifying the eluted proteins
A simple colour test ( Based on differences in structure ) will help confirm the identity of
the two proteins fractions. On treatment with solid sodium dithionite ( a reducing agent ),
cytochrome c will be reduced to the Fe2+ form, which is bright pink, whilst haemoglobin
will become deoxygenated and hence a dull purple colour.
Wearing gloves and safety glasses, take your concentrated protein sample and add a small
quantity of sodium dithionite using a microspatula. Cover the tube with Parafilm, and
invert gently until the solid dithionite has dissolved ( do not shake or all the dithionite
may become inactivated by reacting with molecular oxygen ).
Results
Use the proforma to record your methods and results and answer the follow-up questions.
The completed proforma ( Sephadex plus Ion Exchage ) must be handed in at the next
session.
Safety notes for ion exchange chromatography
Haemoglobin solution ( 30 mg /ml )
Tris buffer, 0.1 M, pH 8.0
Sodium dithionite
Irritating to eyes and skin
May be irritating to eyes and skin
Harmful by ingestion or of inhaled as dust.
Irritating to eyes and skin
30
Protein Separation Methods
Student Name…………………
Demonstrator……………………………..
1. Gel filtration ( Molecular exclusion ) chromatography with sephadex
Eluate
Vol
Colour
Reaction with TCA
1
2
3
4
Height of Sephadex gel:
From the data obtained above:
(i)
What is the evidence for separation of the haem protein and free dye?
(ii)
Which eluate contained the free dye and which one the protein?
(iii)
In your own words and referring to the theory behind this separation method,
explain why the two substances came through in this order.
(iv)
From the height of the sephadex gel and its diameter ( 1.2 cm ), calculate the
volume of the sephadex bed (Vt)
(v)
When gel exclusion chromatography is used to separate unspecified
mixtures of proteins, it is often necessary to know the void volume of the
31
column ( i.e. the volume of liquid between the beads, see Introduction
section ). This can be calculated from the data you have.
While moving down a column, all bands will spread by diffusion. This is
likely to occur equally upwards and downwards, giving a symmetrical peak
of concentration. The center of the band represents the ‘true’ position where
all molecules would be found if diffusion had not occurred, and hence half
the volume of this coloured band is really a part of the void volume.
Assuming that the protein is totally excluded from the gel, and taking
diffusion into account, find V0, the void volume of the column.
(vi)
What % of the bed volume ( Vt) is the void volume?
(vii)
The pore size of the Sephadex used in this experiment totally excludes all
haem protein molecules, but Sephadex can be used to separate a number of
proteins in a mixture, using a single column. To do this, a different grade of
Sephadex ( with a different range of pore sizes ) would have to be used.
Would the pores have to be larger or smaller than in the class gel? Explain
your reasoning and indicate the size order in which the proteins will elute.
(viii)
Use this information to explain the following se=aration profile obtained for
human serum.
The void volume of the gel filtration column = 7.0 ml.
32
Protein
concentration
E
D
C
A
5
B
10
15
20
25
Elution volume (ml)


2.
In which peaks would you expect to find:
Protein
Mr
g/l serum
Albumin
Immunoglobulin M (Ig M )
Ig A
Ig G
44.0
0.9
2.1
12.0
68,000
900,000
160,000
150,000
Peak
Explain your reasoning
SEPARATIONOF
PROTEINS
USING
ION-EXCHANGE
CHROMATOGRAPHY
On loading the protein mixture onto the column, in pH 5.0 acetate buffer:
(i)
Was there any evidence for separation of the two proteins?
33
(ii)
What is the state of the two proteins ( bound or free ). Give your reasoning.
FIRST ELUTION
(iii)
BUFFER =
Explain your rationale for your choice of first buffer. Did it achieve a
separation?
SECOND ELUTION
BUFFER =
(iv)
Explain your rationale for the choice of the second eluting buffer. Did it
elute the remaining protein?
(v)
Now summarize your results in the table below:
Nature and pH
of buffer
Charge
on
eluted protein
Result
of
dithionite test;
colour
of
protein solution
Identity
of
protein. Cyt c
or Hb ?
1st
eluting
buffer
2nd
eluting
buffer
(vi ) Instead of using a CM 23 cation exchange column, now devise a protocol for
34
the separation of haemoglobin and cytochrome c using a DEAE anion exchange
column. List the stages, state your reasoning behind your choice of buffer and give
your expected observations.
Stage
Reasoning
Expected observation
35
3. APPLICATION OF ION EXCHANGE CHROMATOGRAPHY IN DIAGNOSIS
AND MONITORING OF DIABETES
Haemoglobin of normal adult consists of about 90% of one major species, haemoglobin A
(Hb A ), composed of two pairs of different polypeptides (a tetramer of 2α plus 2β
globin chains ), each with a bound haem group. Among a number of quantitatively less
important haemoglobins, which are also found in blood, are the glycosylated forms of
HbA, the most common ( HbA1c ) having a glucose moiety covalently attached to the
amino group of te N-terminal valine residue of eachβ-globin chain. Glycosylation of
HbA to HbA1c takes place during the lifespan of the erythrocyte, in a non-enzymic
reaction with blood glucose. Normal subjects have about 5-7% of their hemoglobin in the
form of HbA1c ; however, in diabetics, this may be increased 2 to 3-fold. Although the
exact clinical significance of glycosylated haemoglobins remains unclear, the level of
HbA1c is not particularly affected by dietary changes, exercise or recently administered
anti-biabetic drugs.
(i)
At physiological pH, what would be the charge on the N-terminal valine
residue?
(ii)
What difference will the covalent bonding of glucose to the N-terminal
valine residue make to the overall charge density on the haemoglobin
molecule?
(iii)
Will the pI of HbA1c be greater or less than HbA?
36
Choose one type of ion exchange resin and state how you could separate HbA from
HbA1c. You do not need to give precise pH or [ionic], but discuss it in relation to pI,
density of charge and binding to the resin.
(iv)
Type of resin
(v)
At what pH (approximately ) would you load the sample ? Explain why.
(vi)
What buffer(s) would you use to elute the proteins?
(vii)
In which order would they elute?
(viii) Explain the theoretical basis for your choice of buffer (s) and the order of
elution.
37
EXPERIMENT 1
ASSAY OF PROTEIN CONCENTRATION
------COOMASSIE BRILLIANT BLUE G-250
Aims
To learn how to assay protein concentration with Coomassie Brilliant blue G-250 dyeing
method
Principle
In acid environment, Coomassie Brilliant G-250 can combine protein, then its colour can
change from yellow to blue. The extent of the blue colour is proportional to the protein
concentration.
Protocol
1. Make a standard curve
(mL)
reagent
1
2
3
4
5
6
Protein standard solution
(0.1mg/ml)
NaCl2(9g/L)
Protein reaction solution
corresponding
protein
concentration (mg/ml)
A595
0.1
0.3
0.5
0.7
0.9
0
0.9
5.0
0.01
0.7
5.0
0.03
0.5
5.0
0.05
0.3
5.0
0.07
0.1
5.0
0.09
1.0
5.0
0
Draw the standard curve by using corresponding protein concentration as x-coordinate
and A595 as y-coordinate.
2. Assay of the blood serum specimen
(1) Dilute 200 times for the serum
(2) According to method of making standard curve to manipulate .
(3) Read the protein concentration from the standard curve according to A595.
Calculation
Blood serum protein concentration(g/l)= data located from the standard curve×200/0.1
200-----dilution times
0.1------ dosage of dilution serum specimen
38
Reagents
1. Protein standard solution ( 0.1mg/ml )
2. Protein reaction solution (Coomassie Brilliant Blue G-250, 0.1g/L)
3. 9g/L NaCl ( Sodium chloride solution, 9g/L)
Results
Discussion
39
EXPERIMENT 2
ELECTROPHORESIS OF SERUM
PROTEIN (CAME)
Aims
1. To understand and master the principle of CAME ( Cellulose Acetate Membrane
Electrophoresis ).
2. To learn how to analyze relative concentration of various protein in serum
Principle
When placed in an electric field, molecules with a net charge such as proteins, will move
towards one electrode or the other, a phenomenon know as electrophoresis.If a protein has
much net charge it will run faster ,otherwise it will run slower. According to this
phenomenon, the proteins in serum can be separated into 5 compositions, which
are albumin (A)、α1 、α2 、β、and γ-globulin.
Protocol
1. Preparation before Electrophoresis
(1) Dip the piece of CAM in the buffer(PH=8.6) for ten minutes
(2) Put some serum (2~3) on the CAM piece by pipette.
(3) Take the piece of CAM on the Electrophoresis instrument (caution: the side with
serum must be put on (-)electrode.
2. Electrophoresis
voltage: about 100~120v
time: 40~60min
3. Dyeing and rinsing ( washing )
When Electrophoresis finish, take out the piece and put it into the protein dyeing for
10~15min then rinsing with the washing solution 1 till its background is cleaned.
4. Determination of the amount of proteins
To cut each protein strip and blank strip, put them into 5ml the washing
solution 2 respectively, shake up several times, read A500 after half an hour.
T=A+α1+α2+β+ γ
A%=A/T*100
α1%=α1/T*100
α2%=α2/T*100
β%=β/T*100
γ%=γ/T*100
40
Reagents
1. Barbital buffer (pH8.6)
2. Protein dyeing solution
3. Washing solution1 (C2H5OH 45ml + CH3COOH 5ml + H2O 50ml)
4. Washing solution 2 (0.4mol/L NaOH)
Caution
1. The end of piece of CAM with serum should be put on the negative polar on the
electric field.
2. Take care about the voltage, don’t leave it too high which will damage the electric
field because of overheating.
Results
Discussion
41
EXPERIMENT 3
ASSAY OF PROTEIN PI ( ISOELECTRIC POINT
—— PRECIPITATION METHOD
Aims
1. To strengthen the comprehension of the protein ionization and pI
2. To learn how to assay protein pI
Principle
When a protein is at its pI, its net charge is zero and hence its solubility become minimum.
According to this, we can place casein in a series of solutions whose pH are different. By
observation of the extent of the precipitation, the protein pI will be obtained.
Protocol
(The unit: ml)
Tube
1
H2O
1.00M HAc
0.10M HAc
0.01M HAc
casein
Final pH
Precipitation
2.4
1.6
2
4
3.0
4.0
1.0
3.5
3
1.0
4.1
5
3.5
1.0
1.0
4.7
2.5
1.0
5.3
0.5
1.0
5.9
Observation precipitation, Using “+”、 “++” “+++” “++++” to show the extent of the
Precipitation
Results
Judge the protein pI according to the amount of precipitation (“+”most),
Reagents
1. 1mol/L CH3COOH ( Acetic acid solution )
2. 0.1 mol/L CH3COOH ( Acetic acid solution )
3.0.01 mol/L CH3COOH ( Acetic acid solution )
4. Casein- CH3COONa solution ( Casein-Sodium acetate )
42
EXPERIMENT 4
ASSAY OF ACTIVITY OF ALKALINE PHOSPHATE
Aims
1. To master the basic method of the assay of the AKP (Alkaline phosphatase ). To
strengthen the comprehension of enzyme activity and its inhibition.
2. To be familiar in the meaning of the activity unit, total activity , ratio activity and
how to do the calculation of AKP activity.
Principle
Alkaline phosphatase (AKP ) can catalyze p-nitrophenol phosphate decomposition in
basic conditions and produce p-nitrophenol whose clour is yellow .This yellow production
has the most absorbance atλ405. According to this we can calculate AKP enzyme unit.
An enzyme unit is that amount of enzyme which will catalyze the transformation of 1μ
mol of substrate per minute at 25oC under optimal conditions for that enzyme.
Total activity refers to the total unit of enzyme units of enzyme in the sample, whereas
the specific activity is the number of enzyme units per milligram of protein(unit/mg).
Mg2+
Na2PO4
EDTA
+
+H2O
OH
－
AKP
+ Na2HPO4
OH
K3Fe(CN)6
4-aminoantipyrene +
Red
compound
This red compound has the most absorbance atλ510nm. According to this we can
calculate AKP enzyme unit.
An enzyme unit is that amount of enzyme which will catalyze the transformation of 1μ
mol of substrate per minute at 37oC under optimal conditions for that enzyme.
Total activity refers to the total unit of enzyme units of enzyme in the sample, whereas the
specific activity is the number of enzyme units per milligram of protein(unit/mg).
Protocol
43
1. Make the standard curve
( Unit: ml )
Reagents
blank
1
2
3
4
5
6
H2CO3-NaHCO3 buffer (pH10)
Phenol standard solution
H2O
1.0
0
1.0
1.0
0.05
0.95
1.0
0.1
0.9
1.0
0.2
0.8
1.0
0.3
0.7
1.0
0.4
0.6
1.0
0.5
0.5
1.0
1.0
2.0
1.0
1.0
2.0
1.0
1.0
2.0
1.0
1.0
2.0
37oC, 25min
NaOH
4-aminoantipyrene
K3Fe(CN)6
1.0
1.0
2.0
1.0
1.0
2.0
1.0
1.0
2.0
Mixing, ten minutes wait，read A510
A510
Phenol (μmol)
0
0.05
0.1
0.2
0.3
0.4
0.5
Correspond enzyme unit
0
2
4
8
12
16
20
Draw the standard curve by using A510 as y-coordinate and correspond enzyme unit as
x-coordinate
2. Assay of AKP activity
Reagent
Blank
Assay1
Assay2
(with Mg2+)
Assay3
(with EDTA)
Substrate (0.02mol/L) (no
Mg2+ and EDTA)
Mg2+(0.012mol/L)
EDTA(0.012mol/L)
H2O
Enzyme (0.5mg/ml)
H2CO3 buffer (pH10)
0
0.3
0.3
0.3
0
0
1
0.1
0.9
0
0.1
0
0
0
0.1
0.7
0.6
0.6
0.1
0.1
0.1
0.9
0.9
0.9
37oC, 25min
NaOH
1.0
1.0
1.0
1.0
4-aminoantipyrene
1.0
1.0
1.0
1.0
K3Fe(CN)6
2.0
2.0
2.0
2.0
Mixing, let the test tubes stand on the table for ten minutes，then measure A510 each tube
A510
Total activity(mU/tube)
0
mU/ml
0
mU/mg
0
44
Reagents
1.
2.
3.
4.
5.
6.
7.
0.1mol/L pH10 H2CO3-Na2CO3 buffer ( Carbonate buffer )
0.02mol/L C6H5Na2PO4 ( Disodium phenyl phosphate )
1μmol/ml C6H5OH ( Phenol standard solution )
0.012mol/L MgCl2 ( Magnesium chloride )
0.012mol/L EDTA ( Disodium ethylene diamine tetraacetate )
0.2mol/L NaOH (Sodium hydroxide )
0.5mg/ml snake venom
8. 0.3% 4-aminoantipyrene
9. 0.5% K3Fe(CN)6 ( Potassium ferricyanide )
Caution
Results
Discussion
45
EXPERIMENT 5
COMPETITIVE INHIBITION OF ENZYME
Aims
To strengthen the comprehension of competitive inhibition of enzyme.
Principle
CH2COOH
CH2COOH
FAD
Methylene white
SDH
HOOC-CH
HC-COOH
(-)
FAD·2H
Methylene blue
malonate
A competitive inhibitor has close structural similarities to the normal substrate for
the enzyme. Thus it competes with substrate molecules to bind to the active site, so
enzyme activity is inhibited.
A good example of competitive inhibition is provided by succinate
dehydrogenase. This enzyme uses succinate as its substrate and is
competitively inhibited by .but at high substrate concentrations , the action of
a competitive inhibitor is overcome because a sufficiently high substrate
concentration will successfully compete out the inhibitor molecule in binding
to the active site.
Protocol
1. Prepare for the homogenate of rabbit’s heart cells by homogenization.
2. Manipulate according to the following table .
(Unit: drops)
Tube
Homogenate
Succinate
(15g/L)
Malonate
(10g/L)
H2O
Methylene
blue
1
2
3
4
10
10
0
10
10
10
10
20
0
10
10
10
20
10
20
5
5
5
5
Result
3. Shake up and add 10 drops of mineral oil ( paraffin oil ) to the tubes.
46
4. Observe the results and analyze.
Reagents
1. Succinate (15g/L)
2. Malonate (10g/L)
3. Methylene (0.2g/L)
4. PBS (pH7.4, 0.1mol/L)
5. Homogenate of rabbit’s cardiac muscle.
Caution
1. Don’t shake up the tubes again after the addition of paraffin oil. Why? Thinking about
this please.
2. The homogenization should be enough.
Results
Discussion
47
EXPERIMENT 6
PREPARATION OF TOTAL RNA FROM CULTURE CELLS
Aims
1. To learn how to extract total RNA from culture cells.
2. To have an appreciation of the keys of the process for high-quality RNA isolation.
Principle
The extraction of high-quality RNA for Northern blot and enzymatic reactions is an
important prerequisite for studies of gene expression processes in organisms.
Ribonuclease present in the environment and within your sample can rapidly degrade
RNA resulting in low yield and poor quality. To avoid RNA degradation, the activity of
ribonuclese should be strictly controlled during the process of RNA isolation. The
structure of cells are disrupted quickly by using high concentration of
isothiocyanocarbamidine resulting in the release of RNA from cells. Meantime,
isothiocyanocarbamidine and β-mercaptoethanol can denature the activities of various
ribonucleases and protect RNA from degradation. N-lauryl sarcosine, carbamidine
hydrogen chloride can make ribosome proteins separation from RNA. There are RNA,
DNA, proteins and cell debris in the lysis buffer and RNA can be separated from other
cell ingredients through the handling, centrifugation, isopropyl alcohol precipitation
resulting in pure total RNA. DEPC ( diethyl pyrocarbonate ) is a kind of strong inhibitor
but not completely, which can react with the imidazole cycle in the active group of
histidine of ribonuclease and inhibit the activities of ribonuclease.
Protocol
1. Collect the cells 1×107 ~ 1×107, centrifugate for 5 minutes at 1 500 r/min.
2. Add 2 ml of lysis buffer and gently shake up, making the cells disrupted completely.
3. Add 0.2 ml of 2 mol/L sodium acetic acid (pH 4.0 ), then transfer the solution into a 5
ml centrifuge tube.
4. Add 2.2 ml of phenol: chloroform: isopentyl alcohol =25: 24:1, shaking up strongly for
10 seconds, then put it into ice bath for 10 minutes.
5. Centrifugate at 4℃for 20 minutes at 12 000 r/min
6. Transfer carefully the supernatants from step 5 to a new 5 ml centrifugate tube.
7. Add the same volume of isopropoyl alcohol as the solution from step 5
8. Centrifugate at 4℃ for 15 minutes at 12 000 r/min
9. Discard the supernatant, resuspend the RNA pellet in 2 ml of cell lysis buffer and make
the pellet completely dissolved.
48
10. Add the same volume of isopropoyl alcohol as step 9 and set for 30 mins at -20℃
11. Centrifugate at 4℃for 15 min at 12 000 r/min
12. Discard the supernatant, add 4 ml of 70% ethanol alcohol to wash the RNA pellet. If
the pellet become resuspend, centrifugate at 4℃ for 10 min at 10 000 r/min.
13. Discard the supernatant, drying the RNA pellet at vacuum for 10 ~ 20 min. Note: not
dry to more, otherwise the RNA is difficult
14. Add 100 μl of ddistilled water without ribonuclease to resuspend the RNA pellet,
storage at miner 70℃.
Reagents
1. Cell lysis buffer
2. CSB buffer
3. 2 mol/L Sodium acetic acid ( pH 4.0 )
4. isopropyl alcohol
5. Rnase-free water
6. Ethanol alcohol
7. 70% ethanol alcohol
8. phenol: chloroform: isopentyl = 25: 24 :1
9. 0.1 % DEPC (0.1% diethyl pyrocarbonate )
Results
Discussion
49
EXPERIMENT 7
ASSAY OF Km VALUE OF ALKALINE PHOSPHATASE
Aims
1. To understand the effect of the changes of substrate concentrations on the velocity of
enzyme-catalyzed reaction.
2. To learn the principle and methods of determining Km value and its role at enzyme
kinetics.
Principle
An initial reaction velocity (V0) of enzyme-catalyzed reaction at low substrate
concentrations is directly proportional to [S] under the constants of temperature, pH and
enzyme concentrations, while at high substrate concentrations the velocity t Michaelis
ends towards a maximum value (Vmax), which is independent of [S]. The relationship
between substrate concentration [S] and initial reaction velocity (V0) is described by
Michaelis-Menten equation.
V0 =
Vmax·[S]
Km+[S]
In this equation, Km is Michaelis constant, Vmax is maximum reaction velocity, [S] is
substrate concentration. When V= Vmax/2, Km=[S]. So, the units of Km are the same
as substrate concentration’s. Km is character constant of enzyme. The Km values of most
enzymes are between 0.01~100mmol/L. Km can be calculated by Lineweaver-Burk
double-reciprocal plot in which 1/V0 against 1/[S] is made, and then a straight line is gave
with the intercept on the X-axis equal to –1/km. So, Km can be calculated.
1
=
V0
Vmax
1
Km
1
+
Vmax
·
[S]
In this experiment the Km of alkaline phosphatase (AKP) was determined by
Lineweaver-Burk double-reciprocal plot. Disodium phenyl-phosphate as a substrate was
catalyzed to produce phenol and Disodium hydrogen phosphate , phenol reacted with 4aminoantipyrene in alkaline solution to form a red derivative of quinone with the
oxidation of potassium ferricyanide, and then activity of AKP was assayed according to
the absorption of the color at 510nm.
50
AKP, OH
phenol
Disodium phenyl-phosphate +H2O
Phenol + 4-aminoantipyrene
K3Fe (CN)6·OH-
derivative of quinone（red）
Protocol
1. Add reagents in each of 8 tubes according to the following table.
(Unit: ml)
No
1
2
3
4
5
6
7
0.02mol/L
substrate
0.05 0.10
0.20
0.40
0.60
0.80
1.0
pH10 carbonate
0.90 0.90
0.90
0.90
0.90
0.90
0.90
8
0.00
0.90
buffer
H2O
0.95
0.90
0.80
0.60
0.40
0.20
0.00
1.00
Mixing , incubate in 37℃ water bath for 5 mins.
0.1
Enzyme solution
0.1
0.1
0.1
0.1
0.1
0.1
0.1
(0.5mg/L)
Mixing immediately, incubate in 37℃ water bath for 15mins exactly.
Alkaline
1.0
solution
4-aminoantipyrene 1.0
K3Fe(CN)6
2.0
1.0
1.0
1.0
1.0
1.0
2.0
1.0
2.0
1.0
2.0
1.0
2.0
1.0
1.0
2.0
1.0
1.0
2.0
1.0
1.0
2.0
Mixing, place at room temperature for 10 mins, and record A510 with No 8 tube as
blank
A510
No
1
2
3
4
5
6
7
8
8.0
12.5
10
10
0
0
Calculation of data
[S]mmol/L
1/[S]×10-2
0.5
200
1.0
100
2.0
50
4.0
25
6.0
16.7
1/A510
Note:
making plot with 1/[S] as abscissa, 1/A510 as ordinate
Calculation
According –1/Km, calculate the value of Km.
51
Reagents
1. The enzyme solution: alkaline phosphatase (AKP) 5mg in 100ml buffer (or
GuangXi cobra venom 0.5mg/ml).
2. 0.1mol/L pH10 H2CO3-Na2CO3 buffer ( Carbonate buffer )
3. 0.02mol/L C6H5Na2PO4 ( Disodium phenyl phosphate )
4. 1μmol/ml C6H5OH ( Phenol standard solution )
5. 0.2mol/L NaOH (Sodium hydroxide )
6. 0.5mg/ml snake venom
7. 0.3% 4-aminoantipyrene
8. 0.5% K3Fe(CN)6 ( Potassium ferricyanide )
Caution
Results
Discussion
52
EXPERIMENT 8
ASSAY OF GLUCOSE CONCENTRATION IN SERUM
(THE METHOD OF GLUCOSE OXIDASE-PEROXIDASE)
Aims
1. To grasp the principle and methods related to the assay of glucose concentration in
serum with glucose oxidase-peroxidase.
2. To learn about normal value of glucose concentration in serum and its clinical
significances.
Principle
Glucose is oxidized by glucose oxidase to form gluconic acid and peroxide of hydrogen
(H2O2), and then H2O2 react with phenol and 4-aminoantipyene to form a red complex.
The glucose concentration can be calculated by measuring absorbance of complex at
505nm. The whole reactions are showed following:
Glucose+O2+H2O
H2O2 + phenol
+ 4-aminoantipyrine
Glucose Oxidase
Gluconic acid + H2O2
peroxidase
Red complex + H2O
Protocol
Take out 3 tubes, label and add reagents as following table:
(Unit: ml)
Reagent
Serum
Standard of glucose (5.55mmol/L)
Mixture of enzyme and phenol
Sample
Standard
0.02
__
3.0
__
0.02
3.0
Blank
__
__
3.0
A505
Mixing, place at 37℃ water bath for 20 min, take the tubes out, cool down to room
temperature, and measure A505.
Calculation
A505 of sample
Glucose concentration (mmol/L)=
A505 of standard
×5.55
53
Normal reference value
3.89~6.11mmol/L
Reagents
1. The commodity kit contains following reagent:
(1) Enzyme reagent, having some suitable stable reagent.
(2)1% phenol solution.
2. Glucose standard solution (5.55mmol/L).
3. Mixing of enzyme and phenol
(1) 1% phenol solution will be diluted to 0.1% with distilled water.
(2) 0.1% phenol solution is mixed with enzyme reagent at same volume that is called
mixing of enzyme and phenol.
Caution
Results
Discussion
54
EXPERIMENT 9
DETERMINATION OF GLUCOSE IN URINE (THE
METHOD OF COPPER ION REDUCTION)
Aims
1. To understand the clinic significance of the assay about glucose in urine.
2. To have an appreciation of the principle of the reaction related to the experiment.
Principle
Glucose has a characteristics of reduction, it reduces Cu2+ to Cu2O and produces brick red
precipitation.
Protocol
1. Add glycosuria reagent 1ml into tube, heat to boil, not precipitation would be seen.
2. Add clear urine 3~4 drops, heat to boil, observe the result after nature cooling.
3. Judge guideline of result
(-) No color change (No glucose in urine or infinitesimal).
(Tittle) The color is green but no precipitation. (The content of glucose is about 1g/L).
(+) Have a little green precipitation. (The content of glucose is about 1~5g/L).
(++) Appear a little green-yellow precipitation within boiling 1 minute. (The content of
glucose is about 5~14g/L).
(+++) Appear a mud yellow precipitation after boiling 10~15 seconds. (The content of
glucose is about 15~20g/L).
(++++) Immediately appear yellow and change into brick red gradually after boiling. (The
content of glucose is more than 20g/L).
Normal reference value
The glucose in urine for health people is very tittle (0.1~0.3g every day), result is (-).
Clinic significance
Abnormal glucosuria can be seen in the patients with diabetes, and furthermore, the
patients with hyperparathyroidism, some nephropathy, pressure of skull increasing,
cerebral concussion et al also can be seen.
Caution
1. A error positive would be seen if the patients took a lot of reductive medicines such as
55
streptomycin, vitamin C. The patients would be stopped taking these medicines several
days before examined.
2. If there were proteins in urine the reaction would be interfered with. The urine would
be examined again after adding acid into urine and boiling and filtering it.
3. Lots of uric acid sodium also interferes with reaction (uric acid has a weak reduction.).
Placed urine at icebox till uric acid sodium separated out, and then took supernate to
examine again.
Reagents
1. Glycosuria reagent
A solution: citrate 173g, anhydrous sodium carbonate 100g, distill water 700ml, mixed
in a large beaker, heated to speed dissolution. If the solution is turbid or precipitating
the liquor will not be used.
B solution: bluestone (CuSO4·5H2O) 17.3g, distill water 100ml, heated to speed
dissolution. If the solution is turbid or precipitating the liquor will not be used.
When A and B solution are cooling down, put B solution into A solution slowly, mix at
the same time, add water to 1000ml.If the liquor is not clear, filter and use filterable
liquor.
Caution
Results
Discussion
56
Influence of adrenalin and insulin on Glucose concentration in serum
Principle
Hormone can regulate blood sugar,such as insulin for decreasing blood sugar
levels,epinephrine(adrenalin) for increasing blood sugar levels.In this
experiment,we observe the change level of blood sugar after insulin or
adrenalin was injected to the rabbit,So as to know the effect of hormone on
blood sugar levels.
Protocol
1.Blood sample preparation
Rabbit 1
Rabbit 2
weight
weight
Taking about 2 ml blood from vein
Taking about 2 ml blood from vein
Injecting insulin(1.3U/kg)(40U/ml)
injecting adrenalin (0.2mg/kg)
Taking 2 ml blood again after 35 min
Taking 2 ml blood again after 30 min
2.
Serum sample preparation
Centrifuging the four blood samples and get the Serum samples
3. Determination of blood sugar concentration in serum
4. Calculation:
Change rate of blood sugar concentration(%) =
BS
BS(before injection)
57
EXPERIMENT 10
ASSAY OF TOTAL CHOLESTEROL CONCENTRATION
IN SERUM (THE METHOD OF CHOLESTEROL
OXIDASE- PEROXIDASE)
Aims
1. To grasp the principle and methods of Assay of cholesterol concentration in serum
using cholesterol oxidase-peroxidase.
2. To learn about normal value of cholesterol concentration in serum and its clinical
significance.
Principle
58
Cholesterol ester in serum is hydrolyzed to form cholesterol by Cholesterol esterase, and
cholesterol oxidase catalyze cholesterol to produce H2O2, and then H2O2 react with
hydroxybenzene and 4-aminoantipyene in the action of peroxidase to form a red complex,
quinonimine. The cholesterol concentration can be calculated by measuring absorbence of
complex at 500nm wavelength.
Cholesterol
Cholesterol oxidase
H2O2
H2O2+hydroxybenzene peroxidase
+4-aminoantipyene
Red complex + H2O
Protocol
Take out 3 tubes, label and do as follow table:
(Unit: ml)
Reagent
standard
sample
blank
0.02
0
0
1.5
0
1.5
Serum
0
Standard Solution of
Cholesterol (5.18mmol/L) 0.02
Work Solution of Enzyme
1.5
A500
Mix, place at 37℃ water bath for 15 min, read absorbence of each tube at 500nm
wavelength with regulating absorbance of blank as zero.
Calculation
A500of sample
Total cholesterol concentration (mmol/L)=
A500of standard
×5.18
Caution
1. It is right to phlebotomize with empty stomach in the morning and does not allow high
fat diet before night. Separating serum should avoid hemolysis. It is not affected with a
little Hemolysis.
2. The enzyme reagent is stored at dark 2℃~8℃ that can be used for half a year.
3. The work solution of enzyme can be used for 7 days if only it is stored at 2℃~8℃.
Reagents
59
1.The components of commodity reagent box: enzyme reagent 20 ml, dilute reagent 80 ml,
standard solution of cholesterol 5.18 mmol/L (2.0g/L).
2. The work solution of enzyme: enzyme reagent: dilute reagent = 1:4, dilute enzyme
reagent before use.
Clinic significance
1.There are two cholesterol forms in the blood, one is free another is ester form. The
young people’s total cholesterol in our country is seldom over 5 mmol/L, but after
middle age, total cholesterol in a few people is up to 6.25 mmol/L.
2. Hypercholesterinemia: often be seen in the patients with atherosclerosis, family
hypercholesterinemia, diabetes nephrotic syndrome, hypothyroidism, obstructive
jaundice and obesity.
3. Hypocholesterinemia: often be seen in the patients with severe anaemia,
hyperthyroidism and malnutrition.
Results
Discussion
EXPERIMENT 11
DETERMINATION OF KETONE BODIES IN URINE (THE
METHOD OF THE SODIUM NITROFERRICYANIDE)
Aims
1. To grasp the method of the determination of ketone bodies in urine.
2. To understand the clinic significances related to the assay.
Principle
The acetoacetate and acetone in ketone bodies can react with sodium nitroferricyanide to
produce aubergine substance on alkaline condition. Acetoacetate is more sensitive for this
60
reaction about 15 to 20 times.
Protocol
Take the powder reagent of acetone bodies a little (about mung bean size) to place on 2cm
×2cm filter paper and smear it as a rotundity with 0.5cm. of its diameter.
1. Add sample of urine a drop on the powder reagent.
2. Wait 1~2 minute, and observe the result.
3. Judge guideline of result
① No aubergine appearing: (Negative) the acetone bodies of urine is less than 100mg/L.
② Aubergine appearing: (Positive) the acetone bodies of urine is more than 100mg/L.
③ According to the shade color of aubergine, use (+), (++), (+++), (++++) to represent
the content of acetone bodies of urine respectively.
Normal reference value
The acetone bodies excreted by health people every day are less than 100mg/L,
mainly is acetone. So, the reaction is negative.
Clinic significance
The positive reaction of acetone bodies in the urine often be seen in the
patients with diabetes acetoacidosis, acute rheumatic fever, tuberculosis, blood
poisoning, acute gastroenteritis with severe dehydrate, after anaesthesia and
toxic shock.
Reagents
The powder reagent for the determination of ketone bodies: sodium nitroferricyanide 1g,
dry ammonium sulphate 20g and anhydrous sodium carbonate 20g, were baked at 110℃
for 1 hour respectively, and then would be ground respectively, mixed together. Stored it
at dry place. It should be used up in two months. If the reagent was wet or yellow that
should be prepared again.
Caution
Results
61
Discussion
EXPERIMENT 12
ASSAY OF ALANINE TRANSAMINASE (ALT) ACTIVITY
IN SERUM
Aims
1. To know about the method of the assay of alanine transaminase activity in
serum.
2. To grasp the clinic significances of the assay of alanine transaminase
activity in serum.
Principle
62
Alanine and -ketoglutarate as substrate are catalyzed and converted into pyruvate and
glutamate by alanine transaminase. Pyruvate may combine with 2,4-dinitrophenyl
hydrazine to produce pyruvate- dinitrophenylhydrazone in which pyruvatedinitrophenylhydrazone appear brown in alkali solution. So, that color can be used for
colorimetry.
CH3
CHNH2
COOH
+
GPT
CH3
C=N—NH—
OH—
—NO2
H2N-N—
COOH
(CH2)2
CHNH2
COOH
+
pyruvate
NO2
+
CH3
C=O
COOH
-ketoglutarate
Alanine
CH3
C=O
COOH
COOH
(CH2)2
C=O
COOH
glutamate
NO2
—NO2
+ H2O
COOH
Pyruvate
2,4-dinitrophenyl hydrazine
pyruvate-dinitrophenylhydrazone
-ketoglutarate may also combine with 2,4-dinitrophenylhydrazine to produce
-ketoglutarate - dinitrophenylhydrazone and interfere with the colorimetric result, but
the absorbance of latter in the alkali solution is differ from pyruvatedinitrophenylhydrazone, and is lower than pyruvate- dinitrophenylhydrazone at 520nm
wavelength. After reaction, -ketoglutarate will be reduced and pyruvate will be
increased in which the interference will be weak. So, the enhancing degree of absorbance
at 520nm is directly proportional to content of pyruvate in the reaction system.
Protocol
1. Draw a standard curve: take out 6 tubes, label and do as following
table.
(Unit: ml)
Reagent
Pyruvate Standard
Solution (2mmol/L)
Substrate Solution
0.1mol/L PBS(pH7.4)
control
1
2
3
4
5
0
0.05
0.10
0.15
0.20
0.25
0.50
0.1
0.45
0.1
0.40
0.1
0.35
0.1
0.30
0.1
0.25
0.1
Mix, place in water bath at 37℃ for 5 min, take out and add below reagent
63
2,4-dinitrophenylhydrazine 0.5
0.5
0.5
0.5
0.5
0.5
Mix, place in water bath at 37℃ for 20min again, take out and add below reagent
5.0
Corresponding to ALT Unit 0
0.4mol/L NaOH
5.0
28
5.0
57
5.0
97
5.0
150
5.0
200
A520
Mix, let stand for 10 minute at room temperature, read the absorbance of each tube at
520nm wavelength with regulating absorbance of control tube as zero, and then, draw a
standard curve using absorbance as Y-axis, ALT unit of each tube as X-axis.
2. Assay of ALT activity:
Take out 2 tubes, label and do as following table.
(Unit: ml)
Reagent
Serum
0.1mol/L PBS (pH7.4)
Substrate Solution
blank
sample
0
0.1
0.5
0.1
0
0.5
Mix, put in water bath at 37℃ for 30 min, take out and add reagent below
2,4-dinitrophenylhydrazine
0.5
0.5
Mix, place at 37℃ water bath for 20min again, take out and add below reagent
0.4mol/L NaOH
5.0
5.0
Read the absorbance of sample tube at 520nm wavelength with regulating absorbance
of blank tube as zero. Check up the unit of ALT activity from standard curve.
Normal reference value
Below 40u (<40u/L)
Reagents
1. 0.1mol/L phosphate buffer (PBS), pH7.4
2. Substrate Solution: Weigh out DL-alanine 1.79g and -ketoglutarate 29.2g exactly,
dissolve them in about 50ml 0.1mol/L PBS, and then adjust pH to 7.4 using 0.1mol/L
NaOH, add 0.1mol/L PBS to 100ml. Mix the liquor fully and divide it into small bottles
and store at refrigeratory.
3. Store solution of 2,4-dinitrophenylhydrazine: Weigh out 2,4-dinitrophenylhydrazine
396mg exactly and dissolve it in 10mol/L sulphuric acid (H2SO4) 100ml, and then store
it in brown bottle can be preserved for a long time.
64
4. Work solution of 2,4-dinitrophenylhydrazine: Take out store solution of
2,4-dinitrophenylhydrazine 10ml and dilute it to 200ml using distill water.
4. 0.4mol/L NaOH.
Clinic significances
1. ALT in serum is high remarkably in the patients with acute hepatitis and hepatotoxicity
necrosis.
2. ALT in serum is middle high in the patients with chronic hepatitis, liver cancer,
hepatocirrhosis and myocardial infarction.
3. ALT in serum is middle high in the patients with obstructive jaundice and cholangitis.
Note
Establishing of ALT activity unit in standard curve:
Karman used NADH to make pyruvate reduce to lactate, at the same time, NADH
dehydrogenates and oxidized to form NAD+. Because NADH has a absorbance at 340nm
wavelength but NAD+ has not a absorbance at this wavelength, so the absorbance at
340nm wavelength decreases. The decrease degree of A340 is inverse proportional to
content of pyruvate in the reaction of transamination and is not affected by
-ketoglutarate. Under the condition of D=1, 25℃, substrate and NADH were sufficient,
total reaction volume was 3 ml, 1 ml serum would make A340 decrease 0.001 in 1minute
which enzyme activity was known as a Karman unit. Karman unit could reflect ALT
actual activity objectively but it needed expensive apparatus and reagent (NADH) as well
as the experiment was complex. So it is difficult for the clinical examination, only used
for revising the value of pyruvate in colorimetric reaction and worked out “pyruvate
relatively coefficient”, and then use this coefficient to make sure ALT activity unit with
correspond to the concentration of pyruvate in standard curve.
According to above, in standard curve, the concentration of pyruvate was regular change
but corresponding ALT activity was not regular change.
-ketoglutarate
alanine
NADH + H+
ALT
glutamate
pyruvate
NAD+
A340
lactate
Results
65
Discussion
EXPERIMENT 13
ASSAY OF UREA CONCENTRATION IN SERUM
(Diacetyl monohydroxime Method)
Aims
1. To have an appreciate understand of the assay method of urea in serum and its clinic
significances.
2. To review the mechanism of the producing of urea in liver and its significances.
Principle
66
By the catalytic of Fe2+ in acid environment, the urea can react with diacetyl
NOH
monohydroxime ( CH3—C—CO—CH3 ) to produce a kind of carmine compound when
amino thiocarbamide exists in the solution. Then to measure the absorbance of the
carmine compound at 520 nm and to calculate and determine the concentration of urea in
serum or in plasma. It is unnecessary to get rid of proteins in plasma before the
measurement.
NOH
H3C
C
C
H3C
NH2
C O
O
Diacetyl
monohydroxime
NH3
H3C
Fe3+
2H2O
C
C
H3C
Urea
N
N
H3C
C
NH
C O
C
H3C
N
N
N
C
OH
carmine compound
Protocol
1. Use three middle test tubes, and mark on the tubes, respectively.
2. Add the reagents into each tube according to the following table:
（volume unit：ml）
Reagents
blank
standard
unknown
Serum to be measured
0
0
0.1
standard urea application solution
0
0.1
0
6.1
6.0
6.0
Urea color reaction reagent
A520
0
3. Mix the solution in each tube gently, and then put the tubes into boiling water bath for
10 minutes.
4. Cool down the temperature of the tubes in flowing water to room temperature.
5. Zero the spectrophotometer 721 with the blank tube and measure the absorbance of
other tubes at 520 nm. Record the results in the blanks of the table.
Caution
The measurement should be finished in half of hour otherwise the absorbance would
decrease if the assay were done in two hours later, which the veracity of the result would
be affected.
Calculation
Au
As
67
Urea in serum（mmol/L）=
×5.0
Reference normal value of urea in serum
3.2mmol/L～7.0mmol/L（19mg%～42mg%）
Clinic significances
Increase: Kidney function failure; Acute upalimentary tract bleeding; Standing
hyperpyrexia etc.
Decrease: Liver functions have been severely damaged.
Why? Review the mechanism of producing of urea in liver
You should answer the questions after the experiment in your report. Write down the
answer on your lab book and hand it to your demonstrator.
Reagents
1. Urea color reagent
(i) Solution 1:
Gently and slowly pour 50 ml of H2SO4 (AR ) and 50 ml of H3PO4 into 800 ml of
water, and add 0.05 ml of FeCl3·6H2O (100g/L) and mix it. Then transfer the solution
into a vessel of 1000ml and add water exactly to the scale.
(ii) Solution 2:
0.4g of amino thiocarbamide + 2.0g of diacetyl monohydroxime, dilute to100 ml
with water.
Make the equal volume of solution 1 and solution 2 to be mixture well, store the
mixture of solution 1 and solution 2 into refrigeratory.
2. Urea standard stock solution
Weigh up 3.0 g of the dry urea, and dissolve it into water, then add water to exactly
the scale of 1000 ml.
3. Urea standard application solution
10 ml of standard urea stock solution + water to 100 ml. The concentration of urea
in the standard application solution should be 5.0 mmol/L.
Evaluation on the method
1. Although there are several assay methods about BUN ( Blood Urea Nitrogen ), the
method in this experiment and Urease method are often used. The latter is more
specific and precise, but it is more laborious and taking time as well as hard to be
prevalent. The former is simpler and more exact.
68
2. The significances of the assay of BUN is nearly as same as NPN, but much simpler.
So, the assay of NPN has been replaced progressively by the assay of BUN in clinic.
3. The urea concentration in serum can be transformed to BUN according to the
following equation:
BUN (mmol/L ) = urea conc.(mmol/L) ×
28
60
Results
Discussion
EXPERIMENT 14
DETERMINATION OF PROTEINS IN URINE
Aims
1. To get more sight into the physicochemical properties of proteins.
2. To practice the assay of protein in urine.
Principle
It can make the proteins denaturation and to be clot by heating. In acidic solution, the pH
is more close to the pI of proteins in urine, and the precipitation of proteins will be more
completely. Meanwhile, in acidic solution, it can be avoided the effect of the precipitation
of basic salts which can bring about the solution turbid, and it’s useful to observe the
69
result of the reaction.
Protocol
1. Take a big test tube, add 5 ml of clear urine ( to occupy half of the test tube volume )
2. Clamp the up part of the test tube with a test tube clamp, heating the urine to boiling
with an alcohol burner.
3. Drop a few of drops of 5% acetic acid to the surface layer of the urine, gently mix the
top of the urine, continue to heating to boiling.
4. Observe the reaction and record the results ( Take the bottom of the test tube without
heating as control part )
5. The outcome will be estimated according to the following guidelines:
(-) negative: the urine appearance is still clear and transparent;
(with a little of protein ): turbid a little;
(+): visibly turbid, but there are no granules;
(++):visibly turbid and granules appearing, but no floccule;
(+++): turbid more visibly and there are much floccule and some of them
precipitated.
(++++): turbid severely and a lot of floccule precipitation
Reference normal value
The reaction of normal people urine should be negative because there are no protein or
only a little in the normal urine.
Caution
1. To be careful with the alcohol burner.
2. The urine to be assayed should be clear, sometimes it can be centrifuged first and to
take the top layer solution out carefully.
3. Sometimes there are false negative or positive grade decreased results because of
dietetic with sub-salt chronically. Add 1-2 drops of saturated NaOH solution to the
urine and mix it before the assay.
Clinic significances
Pathological albuminuria have been seen in nephritis, nephrosis, pyelitis, the whole body
illnesses ( including hyperpyrexia, sapremia etc. )
Reagents
5% acetic acid solution:
5 ml of iced acetic acid + water to be 100 ml, mix it.
70
Results
Discussion
EXPERIMENT 15
LDH AND ITS COENZYME
Aims
1. To understand the actions of lactate dehydrogenase ( LDH ) and its coenzyme.
2. To have an appreciation of the extract method about the raw lactate dehydrogenase
apoenzyme and its coenzyme.
Principle
Lactate dehydrogenase is an enzyme which is present in many different kinds of living
cells. In mammals it is particularly active in liver and muscle. It is a kind of conjugated
protein with NAD+ as its coenzyme. It catalyses the following reversible reaction:
71
CH3CH(OH)COO- + NAD+
Lactate
CH3COCOO- + NADH + H+
pyruvate
NAD+ represents nicotinamide adenine dinucleotide, the coenzyme for the reaction. This
coenzyme acts as the hydrogen acceptor, being reduced to NADH when lactate is
oxidized to pyruvate.
In organisms, the hydrogen atoms carried by NADH will be delivered through the
oxidative respiration chains to oxygen to produce water and energy.
In vitro, the artificial oxidation system can be used in which methylene blue replaced the
site of CoQ in respiration chain, the hydrogen atoms are delivered to methylene blue from
flavoprotein. After accepting hydrogen, methylene blue is reduced to methylene white.
Add paraffin oil into test tubes to isolate the solution from air to avoid oxidation of
methylene white again.
The reactions are following:
1. In organism:
lactate
2H
NAD+
2H
2CytFe2+
CoQ
FMNH2
LDH
pyruvate
(Fe-S)
NADH
2H
+
H+
(Fe-S)
2e
1/2O2
+
bc1caa3
CoQH2
FMN
2e
2CytFe3+
O2-
+
2H
+
H2O
2. In vitro:
Sodium 2H
latctate
(Fe-S)
NAD+
FMNH2
2H
Methylene blue
LDH
Sodium
pyruvate
NADH+H+
Fe-S
2H
FMN
Methylene white
The degree of depigmentation of methylene blue is the indicator of redox reaction
process.
The sodium pyruvate produced by dehydrogenase of sodium lactate should be fixed by
potassium cyanide otherwise the reaction can only reach balance, not to complete.
72
LDH
CH3CHOHCOONa + NAD+
CH3COCOONa + NADH + H+
OK
CH3COCOONa + KCN
C
CH3
COONa
CN
（sodium pyruvate potassium cyanhydrin）
There are enough flavoproteins which could react with methylene blue in the extraction of
apoenzyme from lactate dehydrogenase.
Protocol
1. The preparation of muscle homogenate:
Peel the muscle off from the rear legs of 6 rats, and put the muscle fragments into a
big beaker, then the muscle fragments are homogenized for 3 minutes with 100 ml of
0.1 mol/L PBS (pH 7.4 ). After that, make up the slurry to 200 ml with 0.1 mol/L
PBS for the use of whole lab.
2. The raw extract of NAD+:
Pour 100 ml of slurry of rat muscle into a big beaker, boiling the slurry for 5 minutes.
Then the slurry is separated to 4 batches holding in the centrifugal tubes and to
centrifuge for 10 minutes at 2 500 runs/min. Take the supernatant out carefully and
remain it to be used.
3. The preparing of apoenzyme of lactate dehydrogenase:
Centrifuge 100 ml of another batch of slurry of rat muscle in big plastic centrifugal
tubes at 2 500 runs/min for 10 minutes, transfer the supernatant into another beaker
carefully and add 1.0 g of activated carbon following mixing gently. After staying up
30 minutes, filtrate the solution and remain the filtrate for further experiment
4. Observe the actions of apoenzyme of lactate dehydrogenase and its coenzyme:
(i) Mark 4 small test tubes and add the reagents according to the following table:
(The unit: drop)
Tube
No.
Extract of
LDH
apoenzyme
Extract
+
of NAD
5.0g/L
KCN
50g/L
sodium
lactate
H 2O
0.2g/L
outcome
methylene
blue
1
10
0
5
5
10
5
73
2
10
10
5
5
0
5
3
10
10
5
0
5
5
4
0
10
5
5
10
5
(ii) After mixing each tube, add 10 drops of paraffin oil along the wall of tubes to
isolate the solution in tubes from air. Never shake the tubes again.
(iii) Pay attention to the phenomena of the reaction in each tube, record the time,
order and degree of depigmentation of each tube .
(iv) Explain the results with your knowledge about the bio-oxidation and lactate
dehydrogenase.
Caution
1. Because KCN ( potassium cyanide ) is virulent, you must wash your hands after
using the reagent.
2. The waste solution must be poured into an appointed vessel specially after the
experiment finish to avoid the KCN flow into the drainer.
3. If the speed of depigmentation is too slow, the test tubes can be incubated into 40℃
bath to accelerate the reactions until the colour of solution fades.
Reagents
1. 0.1 mol/L phosphate buffer solution ( PBS, pH 7.4, see appendix 1 ).
2. Activated carbon
3. 5.0 g/L KCN ( Virulent. Ask your demonstrator if the solution has to be used. Hand in
KCN solution to your demonstrator when you finish the pipetting. )
4. 50 g/L sodium lactate ( or 50 g/L lithium lactate )
5. 0.2 g/L methylene blue
6. Liquid paraffin oil
Results
Discussion
1. Which tube should depigment first and which one should be last? Can you list the
74
order of depigmentation? Explain the phenomena of the experiment.
2. What’s the actions of KCN in the experiment?
EXPERIMENT 16
FUNCTION AND INHIBITION OF CYTOCHROME
Aims
To understand the actions of cytochrome system and the inhibition of cyanide on
cytochrome oxidase.
Principle
Cytochrome system includes several kinds of cytochrome and cytochrome oxidase, a
class of apoprotein which contain ferrum in their chemical structures. Iron porphyrins are
the prosthetic of cytochromes in which the iron can go through reversible redox reactions.
Cytochromes belong to the important electron transfer system in cells and bridge between
75
flavoprotein and oxygen which distribute universally in organism.
It contains abundant cytochromes and cytochrome oxidase in cardiac muscle which
cytochrome oxidase can make cytochrome oxidized and cyanide can combine an empty
site of iron atom in the prosthetic and result in the inhibition of electron transfer chain
which is the mechanism of cyanide poisoning.
In vitro and under aerobic conditions, p-diamino benzene can be oxidized to be brown
p-diimino benzene by cytochrome and cytochrome oxidase. Therefore the color
exchange of solution can be used to observe the actions of cytochrome and cytochrome
oxidase. The reaction is following:
2H+
NH2 2e
H2N
2Cyt-Fe3+
2Cytaa3Fe2+
2Cyt-Fe2+ 2e
2Cytaa3Fe3+
2e
½ O2
p-diamino benzene
HN
NH
p-diimino benzene
O2-
H2O
（-）
CNProtocol
1. Use a fresh heart of rat, cut the heart and rinse blood off with physiological saline.
2. Cut the heart to be small fragments and put the heart fragments into a mortar. Then
add 0.1 mol/L PBS (pH 7.4 ) 9 ml, pestle the fragments adequately to be slurry
( about 10% slurry )
3. Mark 4 small test tubes and add the reagents according to the following table:
( The unit: drop )
Test tube No.
Heart muscle
slurry
1
10
2
50g/L
KCN
H2O
10g/L
p-diamino benzene
0
10
10
10
0
20
0
3
10
10
0
10
4
0
0
20
10
Phenomena
4. Mix the solution in each tube, staying up 10 minutes at room temperature. Observe the
phenomena and record the outcome into the above table. ( With incubating the test tubes
into a 37℃ bath, it could accelerate the reactions. )
Caution
76
1. KCN is virulent, so it must pay more attention to the application of the reagent (see
the front reaction ).
2. When pestling the heart muscle with a little of fine sand, it is necessary to add the
PBS solution several times, 2-3 ml once respectively.
Reagents
1. 0.1 mol/L phosphate buffer solution (PBS, pH 7.4 ).
2. 5.0 g/L KCN ( Potassium cyanide ).
3. 10 g/L p-diamino benzene ( It should be prepared freshly )
Results
Discussion
1. What’s the KCN action in the experiment? Is it the same as in the front experiment?
2. Why does the heart muscle slurry be used into the experiment?
EXPERIMENT 17
DESIGN OF EXPERIMENT ABOUT THE BEST
CONDITIONS FOR AMYLASE ACTIVITY
Aims
1. You should get the thread how to design a project about experiment research by the
design of experiment and should be able to put it in practice.
2. You would get deeper insight into some methods of enzyme kinetic research, for
example, how does the change of pH affect the reactions catalyzed by enzyme
Procedure ( in your free time before the experiment )
77
1. Review the contents about the main factors influencing the reactions catalyzed by
enzyme.
2. Consult the research state related to amylase at present, including the functions,
secretion, activity determination of amylase and its relationship with some clinic
illnesses. You can use the library or/and internet.
3. Two students as a group, to discuss the project of experiment design about amylase.
You have to decide the concentration of the enzyme or substrate as well as reaction
time about the experiment. You should find the best pH buffer for the activity of
saliva amylase by putting your experiment project in practice.
4. Prepare for some experiment conditions, such as reagents, instruments,
spectrophotometer 721. Write down the steps how to get the saliva amylase, which
diluted multiple of saliva amylase should be used.
5. Hand in your project to your demonstrator before the experiment. Get some advise
about the procedures from your demonstrator.
Protocol ( two students as a group )
1. Read your project carefully again in order to understand clearly the significance and
the aims of the experiment design.
2. Get saliva amylase by yourself and carefully considered dilute it.
3. Make up the diverse pH for the experiment.
4. Measure the activity of saliva amylase in different pH solution. Draw the graph
according to the experiment data in order to find the best pH for the saliva amylase.
5. Amend the experiment conditions such as amylase concentration, substrate
concentration and their amount in the reaction until getting appropriate data to
determine the best pH of amylase.
Caution
1. You have to get some useful information related to amylase before the experiment
design.
2. You have to prepare for some reagents, especially diverse pH buffer solutions before
manipulating the normal experiment.
Reagents ( supplied by the lab )
1.
2.
3.
4.
2.5 g/L starch ( stock solution )
Iodine application solution
Acetic acid, sodium acetic acid, glycine, phosphate, sodium hydroxide
Distilled water, mineral water
5. spectrophotometer 721, test tubes, test tube rack, small beakers, pipettes, label paper,
water bath etc.
78
Results
Record the data you get from the experiment into the following table
which only used as a reference. ( You can change the contents in the table )
pH
5.2
5.4
5.6
5.8
6.0
6.2
6.4
6.6
6.8
pH buffer
0.8ml
0.8ml
0.8ml
0.8ml
0.8ml
0.8ml
0.8ml
0.8ml
0.8ml
Substrate (starch)
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
1:5
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
0.2ml
Enzyme
Put the test tubes into water bath at 37℃ for 7.5 minutes, then add the reagents as following
Iodine application
solution
1.0ml
1.0ml
1.0ml
1.0ml
1.0ml
1.0ml
1.0ml
1.0ml
1.0ml
Distilled water
6.0ml
6.0ml
6.0ml
6.0ml
6.0ml
6.0ml
6.0ml
6.0ml
6.0ml
Mix and then measure the absorbance at 660 nm with 721 meter
A660
Table filled by yourself
pH
pH buffer
Substrate (starch)
Enzyme
Put the test tubes into water bath at 37℃ for 7.5 minutes, then add the reagents as following
Iodine application
solution
Distilled water
Mix and then measure the absorbance at 660 nm with 721 meter
79
A660
Discussion
1. Which pH buffer is the best for the activity of saliva amylase? Why?
2. Discuss the outcome of the experiment each other, write down a short summary about
the experiment design.
EXPERIMENT 18
PREPARATION OF DNA FROM BLOOD WHITE CELLS
Aims
To Learn the technique about DNA extract from white blood cell (WBC).
Principle
First, the WBC is separated with gelatin from whole blood. Second, the WBC is disrupted
by SDS, then the DNA in the WBC can be released into the TES. The third, DNA can be
separated from other ingredients by phenol, chloroform and ethanol (alcohol).
Protocol
80
1. Isolation of WBC
(1) Mix with 2 ml blood and 2 ml 3% gelatin in a glass tube with a dropper, setting the
mixed solution at 37℃ for 5 to 10 minutes.
(2) Put the upper aqueous phase (containing WBC) to another tube and centrifugate at 3
000 rpm for 5 minutes.
(3) The pellets can be used after upper aqueous phase be discarded .
2. Breaking of WBC
Dissolve the pellets in 2 ml TES. Put in 10 drops 10% SDS to the tube and mix the
solution with the dropper until the WBC broken.
3. Separation of DNA
(1) Put in 2 ml saturated phenol (pH7.8) to the tube contained WBC dissolved solution
and completely mixing with dropper, centrifuged the mixing solution at 3000 rpm for
5 minutes.
(2) Remove the upper aqueous phase to another tube, Add the equal volume solution of
chloroform/isoamyl alcohol (24:1, v/v) into the aqueous phase solution tube. Mix the
solution completely with the dropper and centrifuge at 3 000 rpm for 5 minutes
(3) Remove the upper aqueous phase to another tube. Add 2.5-fold volume of ethanol
into the tube containing water solution. It should be careful to mix the water solution
with the ethanol. It must draw the upper ethanol to rinse the lower water solution,
then the DNA can be precipitated and isolated.
(4) Remove the DNA pellets to another tube with dropper and wash 2 times with 70%
ethanol. Let stand the DNA pellets at 4℃ for a period of time so that the remained
ethanol can be volatilized.
4. DNA storage
Above DNA pellets can dissolved in 200 μ l TE (or distilled water) for next
experiment or can be set at 4℃ for storage.
Reagents
1. 30g/L gelatin: 3g gelatin and 0.5g EDTA-Na2 are compounded with NS solution
to whole volume 100ml.
2. TES: 0.5mol/L EDTA 15ml + 1mol/L Tris-HCl 7.5ml + 3mol/L NaCl 2.5ml add
H2O to 500ml.
3. 100g/L SDS: 10g SDS are dissolved in distilled water to 100ml.
4. TAE buffer(50 × TAE): Tris 12.1g + ice-acetic acid 2.9ml + 0.5ml/L
EDTA(pH8.0)5.0ml are added distilled water to 50ml. When the buffer is need to
use, 10ml buffer is compounded with distilled water to 500ml.
5. Loading buffer: 0.2% bromophenol blue and 50% sucrose or glycerol.
81
Equipment
centrifuge; water bath
Results
Discussion
EXPERIMENT 19
ELECTROPHORESIS OF DNA ON AGAROSE GEL
Aims
To learn the base principle and the experiment method of DNA Electrophoresis.
Principle
In the solution with pH higher than 8.0, DNA molecules become electricly negative
charge. So DNA molecules would move to the positive polar in an electric field when
DNA molecules are putted into agarose gel. On the other hand, the smaller the size of
DNA, the faster the DNA would migrate on the agarose gel. That means the speed of the
82
migration of DNA mainly dependents on the size and the amount of electric charges with
DNA molecules in the electric field. Therefore, if there are different molecular weight and
shape, DNA molecules can be separated from each other by electrophoresis on agarose
gel. The density of agarose gel can affect the speed of migration of DNA in electric field
too so that different density of agarose gel should be used for the different size and shape
of DNA. It is showed in the following table:
Conc. of Agarose Gel (%)
The ranges of DNA molecular (kb)
0.7
1.0
1.5
2.0
0.8~10
0.4~6
0.2~4
0.1~3
Protocol
1. Select a proper electrophoretic set and place it on the level. Check up the electrical
source and the circuitry.
2. Select a proper comb and put it vertically on the negative polar end of electrophoretic
set, making the space between the bottom side of the comb and the electrophoretic set
on the level is about 1.0mm.
3. Prepare for the agarose gel. For a 100 ml of 0.8 % gel melt 0.8g of agarose in about
90 ml 1×TBE. Put in microwave oven or hot water bath until agarose melting. Add 1
×TBE to be 100 ml.
Seal the four sides of the electrophoretic set with sticky tape and pour gently the gel
after ethidium bromide has been added ( the final conc. of EB is 0.5 μg/ ml ) and
mixed when the temperature drop to about 60℃. Leave the gel to set for 30 minutes.
4. Add 1×TBE to the electrophoretic tank after the gel set, then take out the comb gently
and keep the wells well.
5. Mix the DNA sample with 1/5 volume of loading buffer ( 0.2 % bromophenol blue
and 50 % sucrose or glycerol ) and load samples into the wells on the gel. Record the
order and the amount of samples.
6. Run the gel in 1×TBE for 1~2 hour ( the time dependent on the concentration, quality,
size of DNA sample ) at 5 voltage / cm.
83
7. Stop the electrophoresis and take out the electrophoretic set. Put it on ultraviolet light
facility to record the phenomena or take pictures.
Reagents
1. 0.8%Agarose gel: prepared with electrophoresis buffer 1×TBE.
2. electrophoresis buffer 10×TBE ( 0.89mol/L Tris, 0.89mol/L Boric acid, 0.02 mmol/L
EDTA.Na2 pH 8.0 ), and dilute to 1×TBE before using.
3. Ethidium bromide ( EB ) 10 mg/ml, the final concentration should be 0.5μg/ ml when
it is used in a gel.
4. Loading buffer: 0.2% Bromophenol blue, 50% sucrose or glycerol.
Results
Discussion
A PPENDIX
APPENDIX 1 THE PREPARATION OF BUFFER
The concentration of the hydrogen ion of the solution which consist of certain materials
84
nearly keep unchanged or only change a little when certain acid or basic solution have
been added. This kind of solution is termed buffer solution, which the action is called
buffer action, and the materials in the solution is called buffer.
（i）Standard buffer solution
1．Preparation of standard buffer solution
（1）pH4.01 standard buffer solution：
Weigh up 10.21 g of potassium hydrogen phthalic acid (GR ) ( after baked at 115±5℃
for 2-3 hours, cooling in a drier for 45 minutes ),and dissolve in 1000 ml di-distilled
water( The resistance of dd-water is larger than 1mΩ)
（2）pH6.86 standard buffer solution：
Weigh up 3.39 g of high grade potassium dihydrogen phosphate and 3.53 g of high grade
disodium hydrogen phosphate ( The two reagents must be baked at 115℃ for 2-3 hours,
cooling for 45 minutes. It should be hurry when you weigh up the latter ). Dissolve them
into 1000 ml of dd-water and mix it.
（3）pH9.18 standard buffer solution：
Weigh up 3.80 g of high grade pure borax ( Borax couldn’t be baked, so it should be hurry
when weighing up borax ), dissolve the borax into 1000 ml of dd-water without carbon
dioxide.
2．Comparison of the relationship between pH and temperature
Temperatur
pH
e（℃）
phthalic acid
standard solution
Phosphate standard
Borax standard
solution
solution
0
4.01
6.98
9.46
85
5
4.01
6.95
9.39
10
4.00
6.92
9.33
15
4.00
6.90
9.27
20
4.01
6.88
9.22
25
4.01
6.86
9.18
30
4.01
6.85
9.14
35
4.02
6.84
9.10
40
4.03
6.84
9.07
45
4.04
6.83
9.04
50
4.06
6.83
9.01
55
4.07
6.84
8.99
60
4.09
6.84
8.96
（ii）The buffers in common use
1．Potassium chloride- hydrochloric acid buffer（pH 1.0～2.2）
0.2mol/L potassium chloride Xml+0.2mol/L HCl Yml，add distilled water to
100 ml.
86
pH（25℃）
X
Y
1.0
25.0
67.0
1.1
25.0
58.8
1.2
25.0
42.5
1.3
25.0
33.6
1.4
25.0
26.6
1.5
25.0
20.7
1.6
25.0
16.2
1.7
25.0
13.0
1.8
25.0
10.2
1.9
25.0
8.1
2.0
25.0
6.5
2.1
25.0
5.1
2.2
25.0
3.9
Potassium chloride（KCl），MR＝74.55；0.2mol/L solution
≌
14.91g/L
2．Glycine-hydrochloric acid buffer（pH2.2～3.6）
0.2mol/L glycine Xml＋0.2mol/L HCl Yml，add distilled water to 100 ml.
pH（25℃）
X
Y
2.2
25.0
22.0
2.4
25.0
16.2
2.6
25.0
12.1
2.8
25.0
8.4
3.0
25.0
5.7
3.2
25.0
4.1
3.4
25.0
3.2
3.6
25.0
2.5
glycine（C2H5NO2），Mr＝75.07；0.2mol/L solution
≌
15.01g/L.
3．Citric acid 一 disodium hydrogen phosphate buffer （pH2.6～7.6）
pH
0.1mol/L(citric acid)
0.2mol/L Na2HPO4
2.6
89.10
10.90
2.8
84.15
15.85
3.0
79.45
20.55
87
3.2
75.30
24.70
3.4
71.50
28.50
3.6
67.80
32.20
3.8
64.50
35.50
4.0
61.45
38.55
4.2
58.60
41.40
4.4
55.90
44.10
4.6
53.25
46.75
4.8
50.70
49.30
5.0
48.50
51.50
5.2
46.40
53.60
5.4
44.25
55.75
5.6
42.00
58.00
5.8
39.55
60.45
6.0
36.85
63.15
6.2
33.90
66.10
6.4
30.75
69.25
6.6
27.25
72.75
6.8
22.75
77.25
7.0
17.65
82.35
7.2
13.05
86.95
7.4
9.15
90.85
7.6
6.35
93.65
Citric acid（C6H8O7．H2O），Mr＝210.04；0.1mol/L solution ≌ 21.01g/L.
NaHPO4,
Mr=141.98；0.2mol/L solution ≌ 28.40g/L.
NaHPO4．2H2O，Mr＝178.05；0.2mol/L solution ≌ 35.61g/L.
4．Citric acid--sodium citrate buffer（pH3.0～6.2）
pH
0.1mol/L citric acid (ml)
0.1mol/L sodium citrate（ml）
3.0
82.0
18.0
3.2
77.5
22.5
88
3.4
73.0
27.0
3.6
68.5
31.5
3.8
63.5
36.5
4.0
59.0
41.0
4.2
54.0
46.0
4.4
49.5
50.5
4.6
44.5
55.5
4.8
40.0
60.0
5.0
35.0
65.0
5.2
30.5
69.5
5.4
25.5
74.5
5.6
21.0
79.0
5.8
16.0
84.0
6.0
11.5
88.5
6.2
8.0
92.0
Citric acid（C6H8O7．H2O）， Mr＝210.14；0.1mol/L solution ≌ 21.01g/L.
Sodium citrate（Na3C6H5O7 ．2H2O）, Mr=294.12; 0.1mol/L solution ≌
29.41g/L.
5．Acetic acid----sodium acetate buffer (pH3.7～5.8）
pH
0.2mol/L NaAc
0.2mol/L HAc
3.7
10.0
90.0
3.8
12.0
88.0
89
4.0
18.0
82.0
4.2
26.5
73.5
4.4
37.0
63.0
4.6
49.0
51.0
4.8
59.0
41.0
5.0
70.0
30.0
5.2
79.0
21.0
5.4
86.0
14.0
5.6
91.0
9.0
5.8
94.0
6.0
Sodium acetate（CH3COONa．3H2O），Mr=136.09；0.2mol/L solution ≌
27.22g／L.
6．Disodium hydrogen phosphate----sodium dihydrogen phosphate buffer
（pH5.8－8.0）
0.2mol/L Na2HPO4Xml + 0.2mol/L NaH2PO4Yml，add distilled water to 100 ml.
90
pH
0.2mol/L Na2HPO4
0.2mol/L NaH2PO4
5.8
4.00
46.00
6.0
6.15
43.85
6.2
9.25
40.75
6.4
13.25
36.75
6.6
18.75
31.25
6.8
24.50
25.50
7.0
30.50
19.50
7.2
36.00
14.00
7.4
40.50
9.50
7.6
43.50
6.50
7.8
45.75
4.25
8.0
47.35
2.65
Na2HPO4 . 2H2O，Mr＝178.05；0.2mol/L ≌ 35.61g/L
Na2HPO4 . 12H2O，Mr＝385.22；0.2mol/L ≌ 71.64g/L
NaH2PO4 . H2O，Mr＝138.01；0.2mol/L ≌ 27.6g/L
NaH2PO4 . 2H2O，Mr＝156.03；0.2mol/L ≌ 31.21g/L
7 ． Disodium hydrogen phosphate----potassium dihydrogen phosphate
（pH5.8～8.2）
pH
1/15mol/L Na2HPO4 (ml)
1/15mol/L KH2PO4 (ml)
91
5.8
8.0
92.0
6.0
12.2
87.8
6.2
18.6
81.4
6.4
26.7
73.3
6.6
37.5
62.5
6.8
49.6
50.4
7.0
61.1
38.9
7.2
72.0
28.0
7.4
80.8
19.2
7.6
87.0
13.0
7.8
91.5
8.5
8.0
94.7
5.3
8.2
97.0
3.0
Disodium hydrogen phosphate（Na2HPO4 . 2H2O），Mr＝178.05；1/15mol/L
≌ 11.876g/L.
Potassium dihydrogen phosphate（KH2PO4 ），Mr＝136.09；1/15mol/L ≌
9.078g/L.
8．Sodium barbital----hydrochloric acid buffer（pH6.8－9.6）
pH(18℃)
0.1mol/L sodium barbital（ml）
0.2mol/L HCl (ml)
6.8
52.2
47.8
92
7.0
53.6
46.4
7.2
55.4
44.6
7.4
58.1
41.9
7.6
61.5
38.5
7.8
66.2
33.8
8.0
71.6
28.4
8.2
76.9
23.1
8.4
82.3
17.7
8.6
87.1
12.9
8.8
90.8
9.2
9.0
93.6
6.4
9.2
95.2
4.8
9.4
97.4
2.6
9.6
98.5
1.5
Sodium barbital （C8H11N2NaO3），Mr=206.18；0.1mol/L ≌ 20.618g/L。
9．Boracic acid ----borax buffer（pH7.4－9.0）
pH
0.05mol/L borax（ml）
0.1mol/L boracic acid (ml)
7.4
1.0
9.0
7.6
1.5
8.5
7.8
2.0
8.0
8.0
3.0
7.0
8.2
3.5
6.5
8.4
4.5
5.5
8.7
6.0
4.0
9.0
8.0
2.0
Borax（Na2B4O7 .10H2O）,Mr=381.43；0.05mol/L ≌ 19.07g/L.
Boracic acid（H3BO3）
，Mr=61.84 ；0.2mol/L ≌ 12.37g/L.
Borax is ease to lose crystal water, so it must be kept into a special bottle with
a tight stopper.
10．Borax----hydrochloric acid buffer（pH8.1～9.0）
0.025mol/L borax Xml+0.1 mol/L HCl Yml，add distilled water to 100ml.
pH (25℃)
Borax X（ml）
HCl Y(ml)
93
8.1
50.0
19.7
8.2
50.0
18.8
8.3
50.0
17.7
8.4
50.0
16.6
8.5
50.0
15.2
8.6
50.0
13.5
8.7
50.0
11.6
8.8
50.0
9.4
8.9
50.0
7.1
9.0
50.0
4.6
Borax（Na2B4O7 .10H2O）, Mr=381.43；0.025mol/L ≌ 9.536g/L。
11．Barbital----sodium barbital buffer（pH8.6）
pH
Ion intensity（I） Barbital （g）
Sodium barbital
Add
distilled
（g）
water to （ml）
8.6
0.1
3.68
20.60
1000
8.6
0.075
2.76
15.45
1000
8.6
0.06
2.21
12.36
1000
8.6
0.05
1.84
10.30
1000
12．Glycine－Sodium hydroxide buffer（pH8.6～10.6）
0.2mol/L glycine Xml＋0.2mol/L sodium hydroxide Yml，add distilled water
94
to 100ml.
pH（25℃）
Glycine X（ml）
Sodium hydroxide Y（ml）
8.6
25.0
2.0
8.8
25.0
3.0
9.0
25.0
4.4
9.2
25.0
6.0
9.4
25.0
8.4
9.6
25.0
11.2
9.8
25.0
13.6
10.0
25.0
16.0
10.4
25.0
19.3
10.6
25.0
22.8
Glycine（C2H5NO2），Mr＝75.07，0.2mol/L ≌ 15.01g/L。
13．Sodium carbonate－sodium bicarbonate buffer（pH9.2－10.8）
（It couldn’t be used when there are Ca2+ or Mg2+ in solution.）
pH
0.1mol/L Na2CO3
0.1mol/L Na2HCO2
8.8
10
90
9.4
9.1
20
80
9.5
9.4
30
70
9.8
9.5
40
60
9.9
9.7
50
50
10.1
9.9
60
40
10.3
10.1
70
30
10.5
10.3
80
20
10.8
10.6
90
10
20℃
37℃
9.2
Na2CO3·10H2O，Mr＝286.2；0.1mol/L ≌ 28.62g/L; Na2CO3，Mr＝105.99；
0.1mol/L ≌ 10.6g/L; NaHCO3，Mr＝84；0.1mol/L ≌ 8.40g/L.
14 ． Tris (hydroxymethyl)aminomethane----hydrochloric acid buffer
（pH7.2－8.8）
95
0.1mol/L Tris (hydroxymethyl)aminomethane Xml + 0.1mol/L HCl Yml，add
distilled water to 100ml.
pH（25℃）
0.1 mol/L TrisX（ml）
0.1 mol/L HCl Y（ml）
7.2
50.0
44.7
7.4
50.0
42.0
7.6
50.0
38.5
7.8
50.0
34.5
8.0
50.0
29.2
8.2
50.0
22.9
8.4
50.0
17.2
8.6
50.0
12.7
8.8
50.0
8.5
Tris (hydroxymethyl)aminomethane,（abbreviated as Tris ,C4H11NO3）
Mr = 121.14 ; 0.1mol/L ≌ 12.114g/L. Tris solution can absorb carbon
dioxide from air, so the bottle should be stoppered
HOCH2
C
HOCH2
tightly when it is nonuse.
CH2OH
H2
NH2
APPENDIX 2
Molecular weight ( Mr ) and isoelectric point (pI ) of some proteins in
96
common use
Proteins
Mr
pI
胰岛素
(Insulin)(cattle)
5,733
5.35
细胞色素 C
(CytochromeC)(cattle heart)
13,000
10.6
糜蛋白酶
(Chymotrypsin)
13,000
8.1
溶菌酶
(Lysozyme)
13,900
11.2
核糖核酸酶
(Ribonuclease)(cattle pancreas)
14,000
7.8
血红蛋白
(Hemoglobin)
15,500
7.07
肌红蛋白
(Myoglobin)
17,000
7.0
β－乳球蛋白
(β-Lactoglobulin)
18,400
5.2
生长激素
(Somatotropin)(人)
21,500
6.9
胰蛋白酶
(Trypsin)
23,3000
糜蛋白酶原
(Chymotrypsinogen)
25,000
碳酸酐酶
(Carbonic anhydrase)
30,000
羧基肽酶
(Carboxypeptidase)
134,000
6.0
胃蛋白酶
(Pepsin)
35,500
1.0 左右
乳酸脱氢酶
(Lactic dehydrogenase)
36,000
卵白蛋白
(Ovaldumin)
40,000
4.6
血清白蛋白
(Serum allumin)
68,000
4.9
醇脱氢酶
(Alcohol dehydrogenase)
80,000
烯醇酶
(Enolae)
90,000
已糖激酶
(Hexokinase)
96,000
组氨酸脱羧酶
(Histidine decarboxylase)
109,000
3－磷酸甘油醛脱
Glylceraldehyde 3-phos-
145,000
氢酶
phate dehydrogenase
血浆铜兰蛋白
(Ceruloplasmin)
151,000
Y－球蛋白
(r-globulin)
156,000
延胡索酸酶
(Fumarase)
194,000
Proteins
Mr
6－磷酸葡萄糖脱
(Glucose-6-phosphate
氢酶
Dehydrogenase)
9.5
6.6
pI
240,000
97
过氧化氢酶
(Catalase)
250,000
纤维蛋白原
(Fibrinogen)
330,000
脲
(Urease)
480,000
谷氨酰胺合成酶
(Glutamine synthetase)
592,000
甲状腺球蛋白
(Thyrogobulin)
660,000
谷氨酸脱氢酶
(Glutamate dehydrogenase)
1,000,000
β-脂蛋白
(β1－Lipoprotein)
1,300,000
乙酰 CoA 羧化酶
(AcetylCoA carbxylase)
2,000,000
血蓝蛋白
(Hemocyanin octopus)
2,800,000
a1－粘蛋白
(α-Mucoprotein)
1.8～2.7
胎球蛋白
(Fetuin)
3.4～3.5
胶原蛋白
(Collagen)
6.6～6.8
鲑精蛋白
(Salmine)
12.1
酶
5.6
5.1
5.4
2002/8/28
Wu Yaosheng, Zhou Sufang, Lai Xiangjin, Ling Ming
Department of Biochemistry and Molecular Biology
School of Pre Clinic Science
Guangxi Medical University
98