VET272 Lab exam Part 2
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SCHOOL OF VETERINARY & LIFE SCIENCES
Lab Test Enzyme assay conditions, enzyme inhibition, blood glucose control, enzyme specificity
NAME: .........................................................................
STUDENT No.: ..............................................................
Time allowed: 50 MINUTES, NO reading time
MULTIPLE CHOICE AND SHORT ANSWER TEST.
USE 1.5 MINUTES PER QUESTION (MCQ) AS A GUIDE FOR TIMING.
ATTEMPT ALL QUESTIONS.
USE BLACK OR BLUE PEN ONLY. TOTAL MARK 40
PASS MARK:
50%
MULTIPLE CHOICE QUESTIONS – 20 MARKS – 5 of each lab
Enzyme assay conditions
1.
What factors would you expect to control the enzyme activity of alkaline
phosphatase? A) Temperature B) pH C)
substrate concentration D) enzyme concentration
E)
All of the above
2.
If an assay is conducted under zero order kinetic conditions, what would be the
effect of doubling the concentration of the enzyme on the amount of product (p-NP)
produced?
A)
When the concentration of enzyme is doubled, the optical density should double and
so the velocity of the reaction will double providing the active site of the enzyme is
not saturated B)
As the enzyme is independent of substrate concentration, doubling the
concentration of the enzyme will have no effect on the amount of product (p-NP)
produced
C)
When the concentration of enzyme is doubled, the optical density should decrease,
hence the velocity will decrease D)
When the concentration of enzyme is doubled, the optical density will only double if
the assay is conducted at the correct pH
E)
A and D are correct
3.
What would you expect to happen to specific activity when enzyme concentration is
doubled if the assay is conducted under zero order kinetic conditions?
A)
Specific activity should be constant under zero order conditions as specific activity
removes the effect of differing amounts of enzyme and makes all activities relative to
a specific constant. B)
Specific activity will increase under zero order conditions as specific activity increases
the effect of differing amounts of enzyme and makes all activities higher. C)
Specific activity will decrease under zero order conditions as specific activity
decreases the effect of differing amounts of enzyme and makes all activities lower.
D)
Specific activity should be constant under zero order conditions as specific activity
removes the effect of differing amounts of substrate and makes all activities relative
to a specific constant.
E)
A and D are correct
4.
In an enzyme catalysed reaction, the velocity of the reaction was 0.54
µmol/hour/tube. Calculate the specific activity (µmol/hour/mg protein
)
in 1ml extract of the
enzyme when the concentration of protein in the enzyme is 0.07 mg per ml and you used
0.75 ml of the enzyme extract for the assay.
A)
8.5 µmol/hour/mg protein
B)
7.9 µmol/hour/mg protein
C)
12.3 µmol/hour/mg protein
D)
5.8 µmol/hour/mg protein
E)
10.3 µmol/hour/mg protein
5
Which of the following statements about Michaelis-Menten kinetics is correct?
A)
Km, the Michaelis constant, is defined as the concentration of substrate required
for the reaction to reach maximum velocity.
B)
Km, the Michaelis constant, is defined as the dissociation constant of the enzyme-substrate complex.
C)
Km, the Michaelis constant, is expressed in terms of the reaction velocity.
D)
Km, the Michaelis constant, is a measure of the affinity the enzyme has for its substrate.
E)
A and D are correct
Enzyme Inhibition
6
Inhibition (or activation) of enzymatic activity by certain small molecules and ions is a
major control mechanism in the animal body. Many drugs and toxins also exert their action by inhibiting particular enzymes. In addition, the effect of inhibitors on enzyme kinetics can provide insight into the nature of enzymes and the mechanism of enzyme action.
An assay was set up to determine the effect that two unknown drugs (drug X and drug Y) would have on the activity of alkaline phosphatase. Based on the table below, what type of inhibitors are drugs X and Y. Uninhibited
X
Y
Km (mM)
0.1
2
0.1
Vmax (µmoles/h/mg)
70
70
5
A)
X is a competitive inhibitor and Y is an non-competitive inhibitor
B)
X is a non-competitive inhibitor and Y is a competitive inhibitor
C)
X is an un-competitive inhibitor and Y is a non-competitive inhibitor
D)
X is not an inhibitor and Y is a competitive inhibitor
E)
X is a competitive inhibitor but Y is not an inhibitor
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7
Which of the following statements is correct of a Competitive inhibitor; A)
binds specifically at the active site
B)
Inhibition is irreversible by substrate C)
The Vmax Increases D)
The Km is unchanged
E)
It is structurally different to the substrate
8
A lineweaver-Burk plot is a double reciprocal transformation of the Michaels Menton
plot i.e 1/[S] is plotted against 1/velocity. Based on the following values, what is the Km
and Vmax; X-intercept = -2.51; Y-intercept = 0.34
A)
2.94, 0.39
B)
0.30, 1.94
C)
0.39, 2.94 D)
1.94, 0.30
E)
None of the above
9
Which of the following statements is correct of a non-Competitive inhibitor;
A)
Can bind to enzyme when it is empty and to the enzyme substrate complex,
inhibition cannot be reversed by substrate, Km is unaltered; Vmax is decreased B)
Can bind to the enzyme substrate complex only, inhibition cannot be reversed by
substrate, Km is unaltered; Vmax is decreased
C)
Bind to the catalytic site of the enzyme, inhibition can be reversed by substrate, Km
is unaltered; Vmax is decreased
D)
Can bind to the catalytic site of the enzyme, inhibition can be reversed by substrate,
Km is increased; Vmax is unchanged
E)
None of the above is correct
10
What is the final concentration (mM) of a substrate when the initial concentration of
substrate is 11mM and 1.5ml of this is added to a final volume of 6ml.
A)
0.81
B)
2.75
C)
44
D)
1.22
E)
3.0
Blood glucose control
11. What would you use as a blank for the glucose assay if you work with 400µl
deproteinised blood (taken from mixture containing 2ml blood and 6ml PCA), 0.6ml water
and 5ml GOD (Glucose Oxidase reagent).
A)
5.4ml GOD, 0.3ml 5% PCA (perchloric acid)
B)
5ml GOD, 1.1ml water
C)
0.1ml blood, 5ml GOD, 0.3ml 5% PCA, 0.6ml water
D) 0.4ml blood, 5ml GOD, 0.3ml 5% PCA, 0.6ml water
E)
5ml GOD, 0.3ml PCA, 0.7ml water
0.3 ml PCA (1 in 3 ratio of blood and PCA; in 400ul deprot. blood we will have 0.1ml blood
and 0.3ml PCA)
0.6ml water + 0.1ml to adjust for the blood volume
5 ml GOD
12. After administering insulin, which tissues are primarily involved in taking up glucose? (ie
where is glucose primarily stored after administering insulin).
A)
The liver
B)
The brain and central nervous system. C)
The muscle and adipose tissue
D)
The kidney and disposed through urine.
E)
Insulin stimulates GLUT4 transporters in the liver which causes the liver to absorb all
the glucose. 13. To enable blood sampling a cannula is inserted into the jugular vein of sheep. Following
this the sheep are rested for 30 minutes prior to the collection of blood to determine basal
plasma glucose concentrations. What is the reason for resting the sheep for this 30 minute
period.
Choose the most correct answer.
A)
The sheep may attack you immediately after cannulation because it is unhappy that
it was cannulated.
B)
The process of cannulation, although done carefully, still involves some stress in the
animal which would raise plasma glucose levels giving you a biased reflection of
basal glucose concentration.
C)
The animals need time to acclimate to human contact prior to blood sampling. This
30 minute period provides enough time for acclimation to occur. D)The cannulation process induces hunger in the sheep, which will then eat and
subsequently alter their plasma glucose response.
E) The cannulation process relaxes some animals, subsequently depressing their
response to the adrenalin challenge.
14. The Sheep were administered a single dose of insulin. To ensure the uniformity of this
dose between animals what was this dose? A)
4 units of insulin per sheep
B)
4 units of insulin per hour
C)
4 units of adrenalin per hour
D)
4 units of insulin per kg liveweight.
E)
4 units of insulin per day 15. Which of the following is not correct.
A) Following administration of the adrenalin challenge, plasma glucose
concentrations rise due to the impact of adrenalin causing glycogenolysis in
the liver.
B)
Following the administration of the adrenalin challenge, muscle glycogen is
not
mobilized because muscle contains no receptors for adrenalin.
C)
Following administration of the adrenalin challenge, plasma glucose
concentrations rise due to the release of glucose stored as glycogen from the
liver.
D)
Following the administration of the adrenalin challenge, the muscle glycogen
depot is mobilized but this does not appear in the blood as glucose because
muscle does not express glucose 6 phosphatase.
E)
Following administration of the adrenalin challenge, plasma glucose
concentrations initially rise but after about 1 hour they start to fall again due
to the endogenous release of insulin.
Enzyme specificity
16. Choose the most correct statement.
A)
On a per-gram of tissue basis the gluconeogenic capacity of the liver is similar
to the kidney, however on a per organ basis the gluconeogenic capacity of
the liver is greater than the kidney.
B)
On a per-gram of tissue basis the gluconeogenic capacity of the kidney is
similar to the liver, however on a per organ basis the gluconeogenic capacity
of the kidney is greater than the liver.
C)
On a per-gram of tissue basis the gluconeogenic capacity of the liver is far
greater than the kidney, however on a per organ basis the gluconeogenic
capacity of the kidney is greater than the liver.
D)
On a per-gram of tissue basis the gluconeogenic capacity of the liver is far
greater than the kidney, however on a per organ basis the gluconeogenic
capacity of the kidney and liver is similar.
E)
None of the above statements are correct.
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17. Choose the most correct answer:
A)
Arginase activity was relatively consistent across all tissues sampled.
B)
Arginase activity is only found in the liver as this is where the urea cycle is
expressed.
C)
Arginase activity was high in the liver, kidney and spleen as these are the only
tissues where the urea cycle is expressed.
D)
Arginase activity was not present in the muscle because the urea cycle is not
expressed in muscle.
E)
Both B) and D) are correct. 18. Acid phosphatase is strongly associated with which intracellular organelle
A) nucleus
B)
mitochondria
C)
sarcoplasmic reticulum
D)
Golgi apparatus.
E)
Lysosomes
19. Within both the alanine transaminase and acid phosphatase assays the homogenates
are incubated for a specified amount of time, and then NaOH is added to the tubes. Why is
this done?
A) NaOH enables a colour reaction to occur which can be read in the
spectrophotometer
B)
NaOH is an essential buffer holding the pH at neutral which is the optimal
conditions for the assay.
C)
NaOH is a preservative that inhibits microbial growth enabling the samples to
be processed without contamination
D)
NaOH stops the reaction at the pre-determined time enabling “enzyme
activity per minute” to be calculated.
E)
The assay must be run in fairly basic conditions, which NaOH provides.
20. Glucose-6-phosphatase is an essential enzyme for which of the following pathways?
A) Gluconeogenesis
B)
Urea cycle
C)
Glycolysis
D) Lipogenesis.
E)
Lipolysis
SHORT ANSWER QUESTIONS – 20 MARKS – 5marks per lab
Enzyme assay conditions
1.
Briefly describe the reasons why the rate of product formation changes over time
(list at least 3) (3 marks)
2.
In an assay measuring enzyme activity, what are the conditions under which the
assay should be run? (2 marks). In the table below, fill in the appropriate
corresponding condition.
Type of kinetics
Substrate concentration
Enzyme concentration
Incubation time
Enzyme Inhibition
1.
There are 2 broad classes of enzyme inhibitors. Give one example of an irreversible
enzyme inhibitor and briefly outline the reactions affected and the enzyme that is
inhibited (3 marks).
2.
What is K
m
? (1 mark)
3.
What is V
max
? (1 mark)
Blood glucose control
The tables below contain the analysis of blood glucose control results of sheep administered
with insulin. The approach of the analysis is to determine whether glucose concentration
(mmol/L) changes over time in response to the hormone challenge. Based on the results below.
Insulin Results example
Type 3 test of fixed effects
Effect
Num DF
Den DF
F Value
Pr > F
Time
6
15
50.15
<.0001
Means data
Effect
Time
Estimated
Mean
Standard
Error
DF
t Value
Pr > |t|
Time
0
4.86
0.3795
15
12.79
<.0001
Time
15
3.45
0.3795
15
9.09
<.0001
Time
30
2.18
0.3795
15
5.75
<.0001
Time
60
1.55
0.3795
15
4.08
0.001
Time
90
1.36
0.3795
15
3.58
0.0028
Time
120
1.42
0.3795
15
3.73
0.002
Differences of least square means
Effect
Time
v Time
Difference
Estimate
Standard
Error
DF
t Value
Pr > |t|
Time
0
120
3.44
0.2814
15
12.22
<.0001
Time
0
15
1.405
0.2814
15
4.99
0.0002
Time
0
30
2.6725
0.2814
15
9.5
<.0001
Time
0
60
3.305
0.2814
15
11.75
<.0001
Time
0
90
3.4975
0.2814
15
12.43
<.0001
Time
15
30
1.2675
0.2814
15
4.5
0.0004
Time
15
60
1.9
0.2814
15
6.75
<.0001
Time
15
90
2.0925
0.2814
15
7.44
<.0001
Time
30
60
0.6325
0.2814
15
2.25
0.0401
Time
30
90
0.825
0.2814
15
2.93
0.0103
Time
60
90
0.1925
0.2814
15
0.68
0.5044
Time
120
15
-2.035
0.2814
15
-7.23
<.0001
Time
120
30
-0.7675
0.2814
15
-2.73
0.0156
Time
120
60
-0.135
0.2814
15
-0.48
0.6383
Time
120
90
0.0575
0.2814
15
0.2
0.8408
1.
What was the mean blood glucose at 90 minutes (include unit)? (1mark)
2.
Did the blood glucose level remain significantly different after 60 minutes. Explain
briefly using p-value. (1mark)
3.
At what time point did the blood glucose level return to the basal level. Explain
briefly
using p-value. (1mark)
4.
Did insulin have a statistically significant effect on the blood glucose level over time?
Briefly justify your answer
using p-value.
(1mark)
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5.
How long after the administration of insulin did it take for blood glucose levels to
drop? Indicate this with numbers from the table. (1mark)
Enzyme specificity
For Arginase, the following method has been carried out, starting with 1 g of tissue homogenised in 9 ml of buffer (1/10 dilution). (5marks)
1.
Standard curve:
Varying concentrations ornithine in a volume of 1 ml were prepared and
2.0 ml of ninhydrin reagent was then added and mixed thoroughly. The tubes were then covered with a piece of aluminium foil and place in a boiling water bath for 40 minutes. After cooling the tubes in water 4.0 ml of glacial acetic acid was added and the optical density recorded at 517nm. This standard curve was linear such that 1OD = 0.26 µmole ornithine.
Method:
-Dilute the tissue homogenates 1/50 with 14 mM MnSO4 solution. -Add 2.2 ml of 0.1 M arginine substrate to each tube and pre-incubate for 5 minutes at 37°C.
-Add 0.3 ml of respective diluted tissue homogenate to the test tubes. Mix and incubate for exactly 10 minutes at 37°C. -Stop the enzyme reaction by adding 2.0 ml of 10% TCA to all tubes.
-Mix all tubes and centrifuge at 3000 rpm for 5 minutes -Carefully pipette 0.2 ml aliquot of the supernatant from each tube into a fresh set of
tubes. Add 0.8 ml of deionized water.
-To all tubes add 2.0 ml of ninhydrin reagent. Mix thoroughly, cover each tube with a
piece of aluminium foil and place in a boiling water bath for 40 minutes.
-Cool the tubes in water and carefully add 4.0 ml of glacial acetic acid.
-Zero the spectrophotometer at 517 nm with distilled water. Mix the samples/blanks
carefully but thoroughly and read the OD at 517 nm.
Calculate the enzyme activity (step by step) as µmole/h per g tissue if 1OD = 0.26 µmole ornithine, and the net OD is 0.8.