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Introduction
•
The introduction should begin with the relevant biology, physiology and commercial
applications for lactase, and end with its pertinent biochemistry (MW, number of amino acids,
secondary structure, quaternary structure, reaction catalyzed)
Materials and Methods: No Materials and Methods Required
Results
Two sections to the Results for this lab:
•
•
•
Characterization of Lactase catalysis of the formation of o-nitrophenol from ONPG
o Figure: Michaelis-Menten plot
o In the legend, your determined values for Km, Vmax
Characterization of the inhibition by galactose.
o Figure: Michaelis-Menten plots of all uninhibited and inhibited experiments.
o Figure: Lineweaver-Burk plots comparing uninhibited and inhibited by galactose
experiments
o In the legend:
▪ Km and Vmax value for the enzyme inhibited by galactose.
▪ Identify the type of inhibitor (competitive, uncompetitive)
The results section should have professional quality figures and paragraphs summaries of the
uninhibited and inhibited results.
Discussion
1. Look up the structure of lactose and create a figure, labeling glucose, galactose and the
glycosidic bond.
2. Look up the reaction catalyzed by lactase when ONPG is the substrate analogue. Create a figure
showing the structure of ortho-Nitrophenyl-β-galactoside (ONPG) and of the products.
a. In the figure legend, consider the structure of galactose and predict the sort of
interactions would you expect to find between galactose and the enzyme in the active
site?
3. Download and open the PyMol file, Lactase480.pse
4. Press “F1”. The view stored is a surface representation of lactase with lactose shown as sticks.
Galactose is shown with yellow carbons and glucose with pink carbons? Create a figure from this
image and address the questions below in the legend.
a. Based on what you see, why would ortho-Nitrophenyl-β-galactoside (galactose as the
sugar) be a good substrate analogue and ortho-Nitrophenyl-β-glucoside (glucose as the
sugar) be a poor substrate analogue?
b. Similarly, why would we expect galactose to inhibit the reaction with ONPG while
glucose did not?
5. Press F2. The view stored is a close-up of the active site showing interactions between galactose
and the enzyme. Waters are shown as red spheres and a sodium ion as a purple sphere. Create a
figure from this image and address the questions below in the legend.
a. Trp 518 does not have polar contacts with galactose. What might the role of Trp 518 be
in binding?
b. Asn 110 and Glu 393 do not have direct interactions with galactose. Explain how they
indirectly interact with the galactose.
c. Seven amino acids have direct interactions with the galactose. Identify the type of
interaction for each. Some may be ambiguous. If you claim ambiguity, explain why.
d. What two amino acids are near the sodium ion? What is their charge and what might
their role be?
MicroL Substrate
200
250
400
600
100
50
10
Slope inhibited
Slope uninhibited
0.001968
0.003972
0.00154
0.008607
0.003176
0.008411
0.004349
0.006419
0.001188
0.00259
0.0004599
0.005496
0.0001063
0.0005981
Slope v Substrate
0.01
Slope
0.008
0.006
0.004
0.002
0
0
100
200
300
Substrate ul
Slope inhibited
Enzyme concentration – 1.914 mg/ml
Linear (Slope inhibited)
Slope vs Substrate
0.01
0.009
0.008
0.007
0.006
Slope
ope v Substrate
y = 9E-06x + 0.0031
R² = 0.4107
0.005
0.004
0.003
y = 7E-06x + 0.0002
R² = 0.9665
0.002
0.001
400
500
600
Substrate ul
Slope uninhibited
Linear (Slope uninhibited)
700
0
0
100
200
300
400
Substrate Vol (MicroLiters)
s Substrate
Inhibited series
500
600
700
F1
F2
F1
F2
F1
F2
F1
F2
Lactase
UNDERSTANDING THE EXPERIMENT AND ANALYSIS
Prep Stage
DETERMINING APPROPRIATE EXPERIMENTAL CONDITIONS
Prep Stage
The purpose of the prep stage is to determine conditions (enzyme,
substrate, and inhibitor concentrations) that will result in a meaningful
experiment
Experimental design is to vary a single parameter (e.g. enzyme
concentration) to achieve a specific result
Examples:
Find an enzyme concentration that results in ΔAbs = 0.1/min
Determine the approximate Km
Determine the inhibitor concentration that results in ~50% inhibition
Questions to ask:
Why is determining this parameter important for a good experiment?
What would happen if a value far from the desired value were chosen?
Experimental Stage
COLLECT DATA UNDER CONDITIONS THAT SUPPORT ANALYSIS
Experimental Stage
In the Experimental Stage a single parameter is varied (e.g.
substrate concentration) to determine its effect on the measured
value
Example:
Vary substrate concentration and measure its effect on initial velocity
Questions to ask:
How will the values chosen impact downstream analysis (e.g. using only
high concentrations of substrate will result in V = Vmax)
How can the values chosen best support downstream analysis?
Data Processing
Stage
PERFORM MATHEMATICAL OPERATIONS TO TRANSFORM RAW DATA
INTO FORMATS SUITABLE FOR ANALYSIS
Data Processing Stage
Experimental Data often requires manipulation to prepare for
analysis
Examples:
Transform Experimental Units into Units suitable for analysis
Volume of Substrate added to [Substrate]
Δabs/sec to Δ[Product]/sec
Questions to ask:
What is the desired data format for my processed data?
What units should my processed data be in?
What is the relationship between raw data format and the format
desired for analysis?
Analysis Stage
PROCESSED DATA IS ANALYZED AND RESULTS OBTAINED
Analysis Stage
Processed Data is analyzed/modeled and results obtained
Examples:
Determine Km and Vmax
Michaelis/Menten Plot (V vs [S]) and Solver
Lineweaver-Burk Plot (1/v vs 1/[S])
Questions to ask:
What is the mathematical relationship between the data and the
parameters being modeled?
In the case of regression analysis, what is being varied and what is the
criteria to determine model quality?
Conclusion Stage
RESULTS ARE CONSIDERED AND CONCLUSIONS DRAWN
Conclusion Stage
Results are compared and conclusions drawn
Examples:
Internal Comparisons
Uninhibited and inhibited Km and Vmax values are compared to determine
type of inhibition
Uninhibited and inhibited Km and Vmax values are compared to determine α
and α’ values
External Comparison
Results are compared with literature values
Additional Info
INFORMATION COMPILED FROM TEXTBOOK AND LAB DOCUMENTS
Additional Info
kcat
Turnover number
For an enzyme with one active site per molecule, the turnover number is
the number of substrate molecules transformed to product by one
enzyme molecule per unit time.
Mathematically
kcat = Vmax / [Etot]
The enzyme concentration, [Etot],can be determined from the A280 value of
the stock solution (see the “Determine Enzyme” tab) and the extinction
coefficient for tyrosinase). Don’t forget the protein stock was diluted for the
experiment
Additional Info
Relationship between Apparent and Actual Km and Vmax Values for Different types of Inhibitors
Inhibitor type
Apparent Vmax
Apparent Km
None
Vmax
Km
Competitive
Vmax
αKm
Uncompetitive
Vmax/α’
Km/α′
Mixed
Vmax/α’
αKm/α′
Additional Info
Ki
Dependent on inhibitor type
Competitive
Uncompetitive
α = 1 + ([Inhibitor] / Ki)
α’ = 1 + ([Inhibitor] / Ki)
Mixed
α = 1 + ([Inhibitor] / Ki)
α’ = 1 + ([Inhibitor] / Ki)
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