Download Exam_2005 - The University of Sydney

1812 A
Semester 2, 2005
Page 1 of 46
Faculty of SCIENCE
School of MOLECULAR AND MICROBIAL BIOSCIENCES
BCHM2072 - HUMAN BIOCHEMISTRY
Duration: 3 hours
Reading time: 10 min
SEAT NUMBER: ..................................................................................................................
FULL NAME: ........................................................................................................................
SID: ........................................................................................................................................
INSTRUCTIONS TO CANDIDATES
 Students are permitted to bring in the coloured sheets of ‘stimulus material’
provided by the School of MMB. These sheets may be annotated but MUST
BE HANDED IN WITH THIS PAPER.
 This paper is CONFIDENTIAL. No part of this paper may be removed from
the examination room.
 Only University supplied calculators may be used
 Answer Short Answer Questions in the spaces provided in this booklet
 Answer Multiple Choice Questions on the answer sheet provided.
 All Multiple Choice Questions are graded using the Partial Marking System.
Although each question has been designed to have only ONE option which
carries full marks, each option of each question may carry a partial positive or
negative mark. No negative marks are incurred for questions that are left
blank.
Section A covers the THEORY (lecture and assignment) Recommended time: 2 hours
Part I - SIXTY Multiple Choice Questions at 1.5 marks each.
Part II - SIX short answer questions (at 5 marks each)..
The 120 marks available in Section A will contribute 80-100% of your THEORY
mark. The remaining 0-20% is contributed from the marks that you chose to accept
from the assignments during the semester.
Section B covers the PRACTICAL (lab work) Recommended time, 1 hour
Part I - THIRTY FIVE Multiple Choice Questions at one mark each.
Part II – ONE short answer question (6 marks)
The 41 marks available for this Section will contribute 50% of your PRACTICAL
mark for BCHM2072. The other 50% is contributed from the marks that you
obtained for practical reports and laboratory tasks during the semester.
All marks are considered raw, and may be subjected to scaling, until approved by the
Faculty of Science.
1812 A
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1812 A
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SECTION A - THEORY
PART I – Multiple Choice Questions worth 1.5 marks each
1.
Which of the following is an ANABOLIC reaction which occurs in humans?
A
B
C
D
E
2.
Net fixation of carbon dioxide into carbohydrate
Hydrolysis of DNA into nucleotides
Proteolysis
Glycogenolysis
Lipogenesis
Which statement is most CORRECT?
A
B
C
D
E
3.
1 kg of human tissue, on average, contains somewhere between 0.5 and
5 mg ATP
In a healthy cell, the [ATP] is always much less than the [ADP]
The total adenine nucleotide pool ([ATP] + [ADP] + [AMP]) in cells is
about 5 mM
ATP can be produced in the mitochondria of liver cells and transported in the
blood for use by the muscle
At room temperature, a 5 mM solution of ATP will completely hydrolyse into
ADP and phosphate within 1 minute.
Which statement about fatty acid oxidation is CORRECT?
A
B
C
D
E
1812 A
Carnitine is a protein embedded in the cell membrane that allows fatty acids to
enter from the bloodstream
Fatty acids are covalently attached to Coenzyme A during the FAD/NAD
catalysed oxidation reactions
The oxidation reactions involving FAD/NAD occur only in the cytoplasm
Fatty acids attached to Coenzyme A can move freely across the mitochondrial
membrane
Carnitine is consumed (two carbons at a time) during fatty acid oxidation
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4.
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Which process occurs in the CYTOPLASM?
A
B
C
D
E
5.
Conversion of fatty acyl-CoA to acetyl-CoA
Conversion of acetyl-CoA to malonyl-CoA
Conversion of pyruvate to acetyl-CoA
Conversion of oxaloacetate and acetyl-CoA to citrate
Conversion of acetyl-CoA into ketone bodies
Which description of the operation of the Krebs Cycle is MOST CORRECT?
A
B
C
D
E
6.
The cycle turns acetyl-CoA into ATP
The pathway is located in both the cytoplasm and the mitochondria
The cycle reacts fuel molecules with oxygen to produce carbon dioxide
The cycle generates CoA and NADH
Most of the ATP in the cell is made directly by enzymes of the Krebs Cycle by
substrate level phosphorylation.
Which of the following statements is INCORRECT?
A
B
C
D
E
1812 A
Electrons can move down the electron transport chain even if proton pumping
from the matrix can not occur.
Protons are only pumped from the matrix if electrons are passed down the
electron transport chain.
ATP synthesis at the F1ATPase requires both ADP and phosphate
Protons will only come in through the F0F1ATPase if ATP is simultaneously
being made from ADP
Protons can pass freely across the outer mitochondrial membrane
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7.
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Which of the following statements concerning electron transport and proton
pumping is INCORRECT?
A
B
C
D
E
8.
Proton release occurs when electron carriers receive electrons from hydrogen
carriers
Cytochrome c carries only electrons
Ubiquinone carries hydrogens from Complex I to Complex III
Oxygen is consumed on the matrix side of the inner mitochondrial membrane
The actual protons that move out of the matrix during electron transport come
exclusively from the hydrogens on NADH
Which description of the components in the F0F1ATPase is INCORRECT?
A
B
C
D
E
1812 A
The gamma subunit rotates as protons enter the matrix
Portions of the F0 channel rotate as protons pass through it
The stator portion prevents free rotation of the alpha-and beta-subunits
ATP is made on an active site on the gamma subunit
The F1ATPase portion is located inside the matrix
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1812 A
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The following table is relevant to Questions 9 – 13.
The table lists five situations in which there has been a change in the operation of the
mitochondrial ATP generating system.
Situation
Rate of ATP
Synthesis
Rate of Proton
Pumping
Size of Proton
Gradient
Rate of Oxygen
Consumption
A
STOPS
INCREASES
DISSIPATES
INCREASES
B
STOPS
STOPS
DISSIPATES
STOPS
C
STOPS
STOPS
STAYS HIGH
STOPS
D
CONTINUES
STOPS
STAYS HIGH
INCREASES
E
CONTINUES
CONTINUES
FALLS
SLIGHTLY
STOPS
Which situation (from A-E above) would you expect to result from the following
interventions?
9.
An uncoupler
10.
A lack of oxygen
11.
Exposure to a compound which accepts electrons directly from cytochrome c
12.
An inhibitor of proton movement through the F0 channel
13.
An inhibitor of electron transport at Complex IV
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1812 A
14.
A
B
C
D
E
15.
A
B
C
D
E
16.
A
B
C
D
E
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Which statement regarding Basal Metabolic Rate (BMR) is CORRECT?
Whole body BMR is generally lower in people who weigh more
BMR contributes more to whole body energy expenditure than diet induced
thermogenesis
BMR is assessed by comparing food intake to weight gain
BMR is inversely proportional to lean body mass
Thyroid hormone lowers BMR
Which statement is CORRECT?
Brain can use fatty acids
Fatty acids can be converted into glucose
Muscle glycogen is the major carbohydrate reserve for the brain
The total amount of glycogen stored in muscle is less than the total amount
stored in liver
There is no specific protein that represents a store of amino acids
Both indirect calorimetry and the doubly-labeled water (D2O18) methods can
be used to assess whole body energy expenditure. Which statement is
CORRECT?
Only indirect calorimetry can give an indication of which fuels are being burnt
The doubly-labeled water method measures oxygen production
The doubly labeled water method can be done over a shorter time frame than
indirect calorimetry
Only the doubly labeled water method can be used to determine basal
metabolic rate
Indirect calorimetry only works if the subject has consumed carbohydrates,
not fat.
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1812 A
17.
A
B
C
D
E
18.
A
B
C
D
E
19.
A
B
C
D
E
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Which of the following statements regarding leptin is CORRECT?
Leptin is secreted by the β-cells of the pancreas in response to a carbohydrate
meal
Obese subjects have a lower blood leptin concentration than non-obese
subjects
Obese subjects are hyper-responsive to leptin
A lack of leptin will cause ravenous hunger
In rodents, leptin inhibits thermogenesis in brown adipose tissue
Which of the following scenarios would MOST likely lead to flatulence (ie
production of volatile short chain fatty acids and gases in the lower bowel)?
Consumption of sucrose by someone with lactase deficiency
Consumption of amylopectin starch
Consumption of amylose starch
Consumption of dairy products pre-treated with lactase
Consumption of glucose in association with an amylase inhibitor
Which statement is CORRECT?
Two days of continual exposure to a blood glucose concentration of 10 mM
will cause a coma
The reaction between proteins and glucose is not an enzyme catalysed process
Glycosylation does not affect the function of proteins
When blood glucose concentration is 5 mM, the rate of reaction between
proteins and glucose is zero
Hyperglycemia causes problems faster than hypoglycemia
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1812 A
20.
A
B
C
D
E
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Which statement is INCORRECT?
It is not practical to measure the glycemic index of meat
The reference food used in glycemic index determinations is normally glucose
The glycemic index is a relative measure of the peak blood glucose
concentration caused by a food
Legumes have a lower glycemic index than white bread
Sucrose has a lower glycemic index than amylopectin starch
The figures opposite refer to Questions 21 – 23
Each of the graphs shows the response of glucose transporter number to a glucose
load in a particular part of a specific tissue. Assume that the load is completely
cleared from the bloodstream in about 2 hours.
Which graph represents:
21.
The number of GLUT-4 transporters in the vesicles of the Golgi apparatus in
muscle cells
22
The number of GLUT-2 transporters on the cell membrane of liver cells
23
The number of GLUT-1 transporters on the cell membrane of brain cells
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B
Number
Number
A
0
2
Time (h)
0
4
2
Time (h)
D
Number
Number
C
4
0
2
Time (h)
0
4
2
Time (h)
4
Number
E
0
1812 A
2
Time (h)
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4
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24.
A
B
C
D
E
25.
A
B
C
D
E
26.
A
B
C
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Which statement is INCORRECT?
Anomeric forms of monosaccharides result from ring formation
Glyceraldehyde and dihydroxyacteone are both trioses
Interconversion of glucose and galactose occurs spontaneously in solution
Glucose, fructose and galactose are all hexoses
Conversion of glucose 6-phosphate to fructose 6-phosphate involves changing
an aldose into a ketose
Which of the following enzymes is most likely to catalyse a ‘rate limiting
step’ in a pathway
High Vmax enzyme that catalyses irreversible conversion of SP with the
prevailing [S] being about the same as the Km of the enzyme.
High Vmax enzyme that catalyses reversible conversion of SP with the
prevailing [S] being about the same as the Km of the enzyme
High Vmax enzyme that catalyses reversible conversion of SP with the
prevailing [S] being about 2-fold the Km of the enzyme
Moderate Vmax enzyme that catalyses reversible conversion of SP with the
prevailing [S] being about 2-fold the Km of the enzyme
Low Vmax enzyme that catalyses irreversible conversion of SP with the
prevailing [S] being about 20-fold the Km of the enzyme
Which process requires ATP?
D
E
Addition of glucose from UDP-glucose to a growing glycogen chain
Conversion of glucose residues in glycogen to glucose 6-phosphate
ANY of the reactions in the fatty acyl synthase complex (ie, from malonlylCoA to fatty acyl-CoA)
Carboxylation of acetyl-CoA to give malonyl-CoA
Decarboxylation of pyruvate to acetyl-CoA
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1812 A
27.
A
B
C
D
E
28.
A
B
C
D
E
29.
A
B
C
D
E
30.
A
B
C
D
E
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Which statement about the balance between glycogenesis and glycolysis in
muscle after a carbohydrate meal is CORRECT?
Consumption of ATP in glycogenesis stimulates glycolysis
Insulin stimulates dephosphorylation of phosphofructokinase
Glucose 6-phosphate concentration rises about 100-fold and stimulates
glycogen synthase
Production of ATP in glycolysis is the major driving force for glycogenesis
A rise in the level of ATP stimulates glycogen synthase
Which statement describes a genuine feature of glucokinase which is NOT
shared by hexokinase?
Glucokinase irreversibly produces glucose 6-phosphate from glucose
Glucokinase is inhibited by a build up of glucose 6-phosphate
Glucokinase is stimulated by phosphorylation
Glucokinase can work on any hexoses
Glucokinase is not saturated by 10 mM glucose
Which statement about the synthesis of glycogen from glucose is
INCORRECT?
The formation of UDP-glucose is facilitated by the joining of two phosphates
to form pyrophosphate
In UDP-glucose, the glucose residue is joined to UDP at the C-1 end of
glucose.
Glycogen synthesis occurs in the cytoplasm
Activated glucose residues on UDP-glucose form glycosidic bonds with the C4 ends of the growing glycogen chain
UTP is regenerated by the reaction of UDP and ATP
Which statement is CORRECT?
A rise in intracellular glucose concentration is sufficient to increase the rate of
glycogenesis in liver
Liver does not have branching enzyme
The size of a glycogen molecule is unlimited in liver
Glucose transport into liver requires insulin
Liver glycogenesis can occur without a concomitant increase in a catabolic
pathway
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1812 A
31.
A
B
C
D
E
32.
A
B
C
D
E
33.
A
B
C
D
E
34.
A
B
C
D
E
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Which statement is CORRECT?
Only the liver can dispose of fructose
Fructose is trapped as fructose 6-phosphate in the liver
A fructokinase deficiency would severely deplete ATP levels in liver after a
meal containing fructose
The liver cannot convert fructose into glucose
The aldolase involved in liver fructose metabolism is slow in comparison to
the rate of fructose trapping
In white adipose tissue, which process is NOT stimulated by insulin?
The rate of glucose uptake
Pyruvate dehydrogenase activity
Acetyl-CoA carboxylase acitvity
Expression of fatty acyl synthase
Lipolysis
Hydroxycitrate is an inhibitor of ATP-citrate lyase. What would be the MOST
LIKELY consequence of adding hydroxycitrate to cells undergoing
lipogenesis?
The movement of acetyl-CoA into the cytoplasm would decrease
Production of ATP would stop
Mitochondrial acetyl-CoA levels would rapidly deplete
The rate of fatty acid production would increase
Cytoplasmic acetyl-CoA levels would increase.
Which statement best describes the relationship between the pentose
phosphate pathway (PPP) and lipogenesis?
The PPP produces the glycerol needed for esterification of newly formed fatty
acids
Lipogenesis provides glycerol 3-phosphate for the PPP
Lipogenesis uses NADPH produced by the PPP
The PPP provides ATP to fuel lipogenesis
The PPP is necessary to provide the carbon dioxide needed to produce
malonyl-CoA
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1812 A
35.
A
B
C
D
E
36.
A
B
C
D
E
37.
A
B
C
D
E
38.
A
B
C
D
E
1812 A
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Which statement regarding Fatty Acyl Synthase (FAS) is CORRECT?
FAS is inhibited by insulin
The order of reaction in FAS is (in sequence) oxidation, dehydration, and
oxidation
FAS decides if a fatty acid is to be desaturated during lipogenesis. Once a
saturated fatty acid has been made, it cannot be unsaturated.
FAS incorporates carbon dioxide into the growing fatty acid chain
During the elongation step, the two new carbon atoms are added to the
carboxy- end of the growing fatty acid
Which treatment would LEAST likely affect the uptake of fatty acids into the
epithelial cells of the small intestine after a SINGLE fat meal?
Co-consumption of a drug which prevents emptying of the gall bladder
Co-consumption of a drug to prevent the formation of bile salts in the liver
Substituting 50% of the fat in the meal with Olestra
Co-consumption of a pancreatic lipase inhibitor with the meal
Co-consumption of a compound that prevents the formation of micelles
Which statement BEST DESCRIBES chylomicrons?
Lipoproteins that carry dietary fat to the peripheral tissues
Discs of phosopholipid that mop up loose cholesterol in the blood stream
Milky droplets formed from the churning of a lipid/salt mixture in the small
intestine
Microscopic droplets excreted by tissues that have too much cholesterol
An emulsion of fat and protein in the stomach
Which statement regarding the disposal of dietary fat is CORRECT?
Fat is transported around the bloodstream in micelles made from bile salts
Very Low Density Lipoproteins (VLDL) transport dietary fat from the
intestine to the liver
Unsaturated fat goes to the liver, but saturated fat goes to the peripheral tissues
Lipoproteins are taken up into cells before being acted on by lipoprotein lipase
Peripheral tissues encounter dietary fat before the liver
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1812 A
39.
A
B
C
D
E
40.
A
B
C
D
E
41.
A
B
C
D
E
42.
A
B
C
D
E
1812 A
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What would be a consequence of taking a drug that inhibited Low Density
Lipoprotein (LDL) uptake?
A decrease in the rate of cholesterol synthesis by the peripheral tissues.
An increase in the concentration of LDL in the bloodstream
Decreased intestinal absorption of cholesterol
Decreased intestinal absorption of fat
Prevention of bile salt synthesis
Which statement BEST DESCRIBES the role of High Density Lipoprotein
(HDL)?
HDL is formed by the removal of fat from LDL
HDL is assembled in the peripheral tissues and is secreted into the
bloodstream
HDL is assembled in the peripheral tissues and is secreted into the lymphatic
circulation
A high HDL:LDL ratio is positively correlated with heart disease
HDL is produced by the liver and picks up cholesterol from the periphery
Which statement is CORRECT?
Essential amino acids are made into protein but non-essential amino acids are
used for energy
Consumption of a large amount of non-essential amino acids will not
counteract insufficient consumption of essential amino acids
Essential amino acids are only found in animal products
Aspartate, glutamate and alanine are all essential amino acids
A deficiency in one essential amino acid intake will increase the storage of the
other 19 amino acids
Which statement BEST DESCRIBES the fate of amino groups derived from
the catabolism of amino acids in muscle?
The amino groups are mainly excreted from the muscle as ammonia
The amino groups are mainly excreted from the muscle as urea
The amino groups become linked to pyruvate for transport to the liver
The amino groups are stored on pre-existing proteins by converting glutamate
residues in proteins to glutamine
The amino groups are stored on pre-existing polynucleotides by converting
thymine bases to cytosine
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1812 A
43.
A
B
C
D
E
44.
A
B
C
D
E
45.
A
B
C
D
E
46.
A
B
C
D
E
1812 A
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Which step is NOT INVOLVED the mobilization of liver glycogen?
An increase in the activity of glycogen phosphorylase
The activation of adenylyl cyclase
An increase in intracellular cyclic-AMP (cAMP) concentration
Phosphorylation of protein kinase A
Phosphorylation of glycogen phosphorylase
Why doesn’t muscle glycogen contribute significantly to blood glucose
homeostasis during starvation?
Because muscle does not express glucose 6-phosphatase
Because muscle does not express phosphorylase
Because muscle does not store enough glycogen
Because muscle has non-branched glycogen
Because muscle has a glycogen synthase that is insensitive to G6P
What would be the consequences of inhibition of lipolysis during the first few
days of starvation?
Blood ketone body concentration would rise
Blood glucose concentration would rise
Blood fatty acid concentration would rise
There would be fewer substrates for gluconeogenesis in the liver
Fatty acid oxidation in the muscles would increase
What statement BEST DESCRIBES the regulation of muscle pyruvate
dehydrogenase (PDH) during starvation
During the first three days of starvation, PDH is fully activated but it gets
gradually switched off as starvation progresses
PDH becomes progressively more dephosphorylated during starvation
PDH kinase is activated during the first 48 h and stays active to the end
PDH phosphatase becomes progressively more active as starvation progresses
PDH becomes more active as starvation progresses
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1812 A
47.
A
B
C
D
E
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Which statement about ketone bodies is INCORRECT?
Ketone bodies circulate in the blood stream bound to special carrier proteins
Ketone bodies can be used the peripheral tissues as well as the brain
Ketone bodies can spontaneously decarboxylate to give acetone
Ketone body oxidation requires a source of Coenzyme A
Ketone body oxidation will inhibit glucose oxidation
The following information relates to Questions 48 – 50.
12
200
8
100
4
0
0
1
Time (h)
Insulin (ng/ml)
Glucose (mM)
The figure below shows the ‘normal’ response of blood glucose (solid line) and
insulin (dashed line) concentrations to a 50 g oral glucose load.
2
Each of the graphs opposite shows the response of a particular individual to a 50 g
oral glucose load. As above, the solid line represents glucose and the dashed line
represents insulin.
48.
Which response is characteristic of a TYPE II DIABETIC (Non-insulin
dependent diabetes)
49.
Which response is characteristic of someone who is glucose tolerant but
insulin resistant
50.
Which response is characteristic of a TYPE I DIABETIC (Insulin dependent
diabetes)
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A
200
100
4
Glucose (mM)
8
0
1
Time (h)
8
100
4
2
0
0
1
Time (h)
200
8
100
0
1
Time (h)
12
200
8
100
4
2
E
12
Glucose (mM)
0
0
1
Time (h)
2
200
8
100
4
Insulin (ng/ml)
4
Glucose (mM)
12
0
0
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2
D
Insulin (ng/ml)
Glucose (mM)
C
0
200
1
Time (h)
2
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Insulin (ng/ml)
0
B
12
Insulin (ng/ml)
Glucose (mM)
12
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Insulin (ng/ml)
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1812 A
51.
A
B
C
D
E
52.
A
B
C
D
E
53.
A
B
C
D
E
1812 A
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Which of the following amino acid sequences would be most likely to form
part of a signal peptide?
arg-glu-asp-asp-lys.
arg-arg-lys-lys.
pro-gly-pro-gly-ser.
val-val-leu-trp-ile.
cys-arg-asp-lys-asp.
A major difference between glycosylphosphatidylinositol (GPI) -anchored
proteins and other lipid-anchored proteins is:
GPI-anchored proteins are normally extra-cellular but other lipid-anchored
proteins are normally intra-cellular.
GPI anchors do not contain sugar residues.
Other lipid anchored proteins are synthesized from GPI-anchored proteins in
the Golgi apparatus.
Glucose residues are trimmed from the GPI anchor of GPI-anchored proteins.
The GPI anchor is linked to the C-terminal of GPI-anchored proteins by a
thiolester bond.
Which of the following statements about protein secretion is correct?
SNARE proteins interact with each other through formation of large β-sheets.
t-SNARE proteins are present on the inner face of cell membranes so they can
directly bind to specific regions of proteins which are destined for secretion.
SNAP proteins are named because they cause t-SNARE and v-SNARE
proteins to snap together.
A key difference between secreted and intra-cellular proteins is the ability of
secreted proteins to activate NSF.
Fusion of vesicles with the cell membrane is associated with a trans to cis
transition of the SNARE proteins
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54.
A
B
C
D
E
55.
A
B
C
D
E
56.
A
B
C
D
E
57.
A
B
C
D
E
1812 A
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Which of the following statements about the JAK-STAT signalling pathway is
correct?
Phosphorylation of a JAK kinase allows the binding of a STAT transcription
factor, mediated by its SH2 domains.
JAK kinases use GTP to phosphorylate small GTP-binding proteins such as
Ras.
Phosphorylation of STAT proteins allows them to dimerise through their SH2
domains.
JAK-dependent phosphorylation of calmodulin leads to inhibition of
Ca2+-dependent protein kinases.
Entry of JAK kinases to the nucleus is followed by phosphorylation of the
2’-OH groups of the deoxyribose residues of chromosomal DNA.
During activation of protein kinase C:
A decrease in the cytoplasm Ca2+ concentration leads to conversion of protein
kinase C into its active form.
Production of IP3 leads to the opening of a ligand-gated ion channel in the
membrane of the endoplasmic reticulum.
Protein kinase C enters the nucleus after a steroid hormone binds to it.
Calmodulin-activated protein kinase binds to the IP3 receptor through its SH2
domain.
Diacyl glycerol is phosphorylated and becomes incorporated into the cell
membrane.
One of the characteristic features of G-protein coupled receptors is that:
Their primary role is signal transduction within cells of the immune system.
GTP binds to G-protein coupled receptors during their activation.
G-protein coupled receptors are heterotrimers of α, β and γ chains.
In most cases, their monomeric forms are inactive.
They contain seven membrane-spanning α-helices.
In general, binding of growth factors to their receptors might be expected to:
Lead to tyrosine kinase-catalysed dephosphorylation of phosphoproteins.
Lead to activation of STAT proteins.
Lead to the activation of protein kinase C because of decreased levels of intracellular Ca2+.
Culminate in the phosphorylation of transcription factors by a MAP kinase.
Have no effect in cells which do not express the Jun transcription factor.
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58.
A
B
C
D
E
59.
A
B
C
D
E
60.
A
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Which of the following statements about targetting of proteins to the nucleus
is correct?
The nuclear localisation signals of proteins are typically rich in hydrophobic
amino acid residues
Proteins which are targetted to the nucleus are usually glycoproteins.
A 50 kDa protein would be expected to enter the nucleus by simple diffusion.
Proteins which are targetted to the nucleus pass through nuclear pores
complexed to transcription factors.
Importin is able to bind to nuclear localisation signals and to nuclear pore
proteins.
The NF-KB transcription factor:
Can not enter the nucleus because it does not contain a nuclear localisation
signal.
Enters the nucleus following phosphorylation of one of its serine residues.
Is activated by proteolytic removal of a signal peptide.
Becomes glycosylated in the Golgi apparatus.
Enters the nucleus after dissociation from a phosphorylated inhibitor protein
In theory, which of the following individuals might you expect to be resistant
to the effects of infection with the cholera bacterium.
D
E
One in which the G proteins found in intestinal cells have a unusually high
ability to hydrolyse GTP.
One in which the levels of NAD+ in intestinal cells are unusually high.
One with mutant Gα proteins, in which their arginine residues replaced by
lysine residues, with little effect on the structure of the proteins.
One in which protein kinase C is unusually active.
One is which biosynthesis of GPI-anchored proteins is defective.
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C
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SECTION A - THEORY
PART II – Short Answer Questions worth 5 marks each
Answer these questions in the spaces provided below.
SAQ1.
What sort of weight change would be observed in an average person if there was 1%
mismatch between energy intake and energy expenditure averaged over the course of
a whole year? Show your working/logic. (2 marks)
List THREE ways of burning fuels without doing work. For each, give a ONE
sentence explanation of how the mechanism works. (3 marks)
1812 A
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1812 A
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SAQ 2.
List FIVE ways (eg, drug or dietary intervention) in which blood cholesterol can be
manipulated. In each case give a ONE sentence description of how the intervention
works.
SAQ 3.
List FIVE ways in which an inhibitor of lipolysis might relieve the symptoms and
metabolic problems associated with Type I diabetes.
1812 A
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1812 A
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SAQ 4.
The table below shows the effect of five different types/periods of exercise on the
relative rate of blood glucose oxidation (glucose uptake through to carbon dioxide)
and energy usage in leg muscles.
Type of exercise
Relative Rate of
Muscle Energy Use
At rest
1
Relative Rate of Blood
Glucose Oxidation by
Muscle
1
A
One minute into a walk
3
3
B
30 minutes into a walk
3
1
C
30 minutes into a light jog
5
1
D
30 minutes into a run
10
5
E
Five seconds into a sprint
15
1
Explain the following. [Hint: focus what is limiting the rate of glucose oxidation in
each circumstance]
The difference between B and A
The difference between C and B
The difference between D and C
The difference between E and D
1812 A
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1812 A
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SAQ 5.
List FIVE treatment strategies for Type II diabetes. In each case, give a single
sentence to describe the basis for the intervention.
SAQ 6.
With the aid of a labelled diagram, explain the signal transduction events which you
would expect to follow binding of a cytokine to its receptor.
1812 A
Semester 2, 2005
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1812 A
Semester 2, 2005
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SECTION B - PRACTICAL
PART I – Multiple Choice Questions worth 1 mark each
Questions on the design of an ENZYME LINKED METABOLITE ASSAY
You wish to measure the concentration of acetyl-CoA in some heart muscle samples.
You find an assay for the enzyme citrate synthase that you think you can manipulate
to become an Enzyme Linked Metabolite Assay (ELMA) for acetyl-CoA. The assay
is described as follows:
“Small frozen pieces of hearts (~10 mg) are homogenized on ice in 10% (final w/v) homogenization
buffer containing (in mM) 20 TRIS, 10 EDTA, pH 7.4. Triton X-100 is then added to 1% (v/v) and,
after mixing, homogenates are then frozen and thawed three times.
The reaction is performed in 1 ml reaction buffer containing (in mM) 20 TRIS, 1 EDTA, 0.1 DTNB,
and 0.1 acetyl-CoA, pH 7.4 at 30°C. The reaction is started by adding 0.05 mM oxaloacetate and
monitored at 412 nm for 3 min.”
You find out that the assay is based on the reaction of CoA with DTNB (5,5’-dithiobis -2-nitrobenzoic acid). As citrate synthase converts acetyl-CoA and oxaloacetate
into citrate, CoA is released and this quickly reacts with DTNB to make the a yellow
product.
Acetyl-CoA + oxaloactate  citrate + CoA (catalysed by citrate synthase)
Then, DTNB (colourless) + CoA  NTB-CoA (yellow) (fast reaction)
The extinction coefficient of the yellow product is 15 mM-1cm-1 at 412 nm.
DTNB does not absorb at this wavelength.
Your assistant resources all the chemicals needed and prepares a table of information,
including their recommendations for stock solutions and other useful comments.
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1812 A
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Page 26 of 46
Chemical
MW
Source
Stock Solution
Comments
Tris [Tris- (hydroxymethyl)-
121.14
Merck
1.0 M
Adjust to pH 8.1 with HCl,
aminomethan]
8382
to obtain Tris-HCl buffer.
Make up using
2.4228 g in 20 ml
water
EDTA
372.2
(ethylenediaminetetraacetic
Sigma
50 mM
E-1644
Chelator for heavy metals, added
to avoid interference with -SH
acid), disodium salt,
Make up using
dihydrate,
186.1 mg in10 ml of
groups.
water
Triton X-100
646.87
Serva
10% (v/v)
37238
Viscous liquid; detergent.
Dissolves membranes.
Add 10 ml to 90 ml
water
Oxaloacetic acid
132.1
Sigma
10 mM
O-4126
Make up using 6.6
mg in 5 ml of water
DTNB [5,5`-dithiobis(2-
396.3
nitrobenzoic acid); 3-
Sigma
1.01 mM
D-8130
TNB-S-S-TNB
(dithionitrobenzoic acid);
carboxy-4- nitrophenyl
Make up using 2 mg
irritant.
disulfide],
in 5 ml of water
Reacts with thiols to give a yellow
Ellman’s reagent,
Acetyl CoA (acetyl
coenzyme A), lithium salt,
compound
816.5
Sigma
12.2 mM
A-2181
Relatively unstable. Needs to be
kept on ice or it will hydrolyse to
Make up using 25
acetate and CoA.
mg in 2.5 ml water.
Citrate synthase, CS
Sigma
Specific activity
Ammonium sulphate fraction from
C-3260
200 U/mg protein
porcine heart. Stored at 4 °C.
but varies with Lot
Supplied as a crystalline
Number.
suspension of 10 mg protein
dissolved in 1 ml 2.2 M
(NH4)2SO4, pH 7.
1812 A
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1812 A
61.
A
B
C
D
E
Semester 2, 2005
Page 27 of 46
What would be the most practical way of making the Tris solution
Weigh out 2.4228 g of Tris, dissolve in 20 ml of water, then adjust the pH to
8.1 by adding concentrated HCl
Dissolve 2.4228 g of Tris in 20 ml water, then adjust the pH to 8.1 by adding
dilute HCl
Dissolve 2.4228 g of Tris in 10 ml water, then adjust the pH to 8.1 with HCl,
then add water to 20 ml
Dissolve about 2.5 g of Tris in about 10 ml water, adjust the pH to 8.1 with
HCl, then add water to give a volume equal to 20 x (weight taken/2.4228)
Accurately measure 20 ml water and adjust its pH to 8.1 with HCl, then add
2.4228 g of Tris
You begin by confirming that you can get the citrate synthase assay system to work.
You want to set up 1 ml cuvettes so that they contain (in order):
20 mM TRIS, 1 mM EDTA, 0.1 mM DTNB, 0.1 mM acetyl-CoA,
0.05 mM oxaloacetate, followed by some citrate synthase enzyme suspension
62.
A
B
C
D
E
1812 A
To achieve the above, how much of the stock acetyl-CoA solution (as
recommended by your assistant) should you add to the cuvette?
120 µl
8.2 µl
1.2 µl
100 µl
2.5 µl
Semester 2, 2005
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1812 A
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To confirm that the assay is ‘working’ you want add enough citrate synthase to get a
fast, but measurable rate of between 0.3 and 1 absorbance units per min.
63.
A
B
C
D
E
64.
A
B
C
D
E
1812 A
If the absorbance change was 0.75 per min, how fast is the AMOUNT of
yellow product increasing in the cuvette?
75 µmol per min
50 nmol per min
5 nmol per min
15 nmol per min
15 µmol per min
You decide that you need 50 mU of citrate synthase (CS) in the cuvette. What
is the MOST PRACTICAL way of doing this?
Adding 0.025 µl of the stock CS directly to the cuvette
Adding 25 µl of the stock CS directly to the cuvette
Mixing 1 µl of the stock CS with 999 ul Tris buffer and then adding 25 ul of
this diluted CS to the cuvette
Mixing 10 µl of stock CS with 990 ul Tris buffer and then adding 2.5 ul of this
diluted CS to the cuvette
Adding 2.5 µl of the stock CS directly to the cuvette
Semester 2, 2005
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1812 A
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Page 29 of 46
On adding all the reagents together in the order, you notice that the solution in the
cuvette shows a tinge of yellow on adding the acetyl-CoA. But, after adding the CS,
it rapidly turns dark yellow, even before you have a chance to put in the oxaloacetate.
This is shown in the diagram below.
CS
Dark yellow
EDTA
Light yellow
Tris
DTNB
Acetyl-CoA
Colourless
0
5
10
15
Time (min)
65.
A
B
C
D
E
1812 A
What is NOT a plausible interpretation of these observations?
Something in the citrate synthase stock has a lot of free thiol groups
The acetyl CoA solution that you made up has partially degraded to give CoA
You have added a lot more citrate synthase than you thought
The DTNB spontaneously degrades at the pH of the Tris
Your acetyl CoA solution contains some free thiol groups
Semester 2, 2005
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1812 A
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Page 30 of 46
After some trouble-shooting, you have convinced yourself that you can now make the
citrate synthase assay work.
The scheme that you have come up with is to set up a cuvette containing
20 mM TRIS, 1 mM EDTA, 0.1 mM DTNB, 0.1 mM acetyl-CoA
and 50 mU of citrate synthase (total volume of 1 ml). Then you mix the cuvette
contents, put it in the spectrophotometer and zero its absorbance. Then you add
0.05 mM oxaloaceate and record the absorbance change with time.
Over the first 30 seconds, you find that the absorbance rises by about 0.35.
66.
A
B
C
D
E
Which statement is the MOST LIKELY prediction of what will happen next?
The rate of increase in absorbance will remain constant for at least 10 minutes
The absorbance will plateau when the acetyl-CoA runs out
The rate will fall after about 5 minutes as the citrate synthase becomes
degraded
The absorbance will fall as the all the DTNB gets used up
In less than five minutes all the oxaloacetate will run out
To make this system measure acetyl-CoA, you replace the stock acetyl-CoA with a
volume of an ‘unkonwn’ sample (eg, a tissue extract)
67.
A
B
C
What else do you have to do to make this system useful as an assay for
acetyl-CoA?
D
E
Ensure that oxaloacetate is never in excess
Ensure that DTNB is in excess
Ensure that citrate synthase does not drive the reaction to completion in less
than 5 minutes
Ensure that the absorbance of the system plateaus above 2
Ensure that the unknown acetyl-CoA is added in excess
1812 A
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1812 A
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Page 31 of 46
Now you try your assay system on a sample of heart extract. This extract has been
prepared as described in the original paper ie,
“Small frozen pieces of hearts (~10 mg) are homogenized on ice in 10% (final w/v) homogenization
buffer containing (in mM) 20 TRIS, 10 EDTA, pH 7.4. Triton X-100 is then added to 1% (v/v) and,
after mixing, homogenates are then frozen and thawed three times.”
68.
A
B
C
D
E
69.
A
B
C
D
E
70.
A
B
C
D
E
1812 A
How does the Triton X-100 and freezing/thawing help you get the most
accurate results from the tissue samples?
It activates citrate synthase
It increases the release of citrate synthase from the mitochondrial matrix
It causes the hydrolysis of acetyl-CoA
It increases the release of metabolites from the cell
It increases the turbidity of the extract
When you use the real tissue extract, you anticipate problems will occur.
Which is NOT likely to be a cause of such problems?.
Tissue extracts can contain pigments that absorb in the visible range
Tissue extracts may contain CoA
Tissue extracts contain citrate synthase
The acetyl CoA concentration in the tissue extracts may be too low
Tissue extracts can be cloudy
If the acetyl-CoA concentration of a sample of heart homogenate is found to
be 50 µM, what is the concentration of acetyl-CoA in the tissue?
1000 nmol g-1
50 nmol g-1
10 nmol g-1
5000 nmol g-1
100 nmol g-1
Semester 2, 2005
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1812 A
71.
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Page 32 of 46
What would give you the most confidence that your acetyl-CoA assay was
working well?
A
B
C
D
Each tissue extract gives a different final result for [acetyl-CoA]
Different volumes of the same tissue extract plateau at the same absorbance
The final result for [acetyl-CoA] for each tissue extract is identical
For a particular tissue extract, the plateau absorbance is proportional to the
amount of extract added
The absorbance of each assay plateaus at the same value
E
THE SPACE BELOW HAS DELIBERATELY BEEN LEFT BLANK
1812 A
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1812 A
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Questions on the design of an COLORIMETRIC assay
You want to implement a colorimetric method for measuring tyrosine. It involves the
reaction of tyrosine with 1-nitroso-2-napthol (NN) in the presence of a mixture of
nitric acid and sodium nitrite. The reaction produces a coloured compound that
absorbs maximally at 450 nm. The instructions are as follows:
1) Prepare the following solutions
Solution A:
0.1% (w/v)1-nitroso-2-naphthol solution
Solution B:
2.6 M nitric acid containing 0.05% (w/v) sodium nitrite.
Solution C:
1 mM tyrosine
2) For each sample (‘unknown’ and ‘standard’) create a reaction mixture of
1 ml of ‘unknown or standard’
0.5 ml Solution A
0.5 ml Solution B
3) Mix well and heat at 55°C for 30 minutes.
4) Cool and then measure the absorbance of all tubes at 450 nm
To determine the effective range of the standard curve, you set up the following tubes
Tube #
1
2
3
4
5
6
Solution A (ml)
0.5
0.5
0.5
0.5
0.5
0.5
Solution B (ml)
0.5
0.5
0.5
0.5
0.5
0.5
0
100
200
300
400
500
1000
900
800
700
600
500
0
0.21
0.39
0.60
0.81
0.99
1 mM tyrosine (µl)
Water (µl)
Absorbance 450 nm
1812 A
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1812 A
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Page 34 of 46
Using this data, your assistant plots the following standard curve
Tyrosine Standard Curve
1.2
y = 0.1989x - 0.196
R2 = 0.9995
Absorbance 450 nm
1
0.8
0.6
0.4
0.2
0
1
2
4
3
5
6
Tube
72.
A
B
C
D
E
73.
A
B
C
D
E
1812 A
What is the amount of tyrosine in Tube #6
500 nmol
0.05 mmol
0.029 mmol
50 nmol
0.29 mmol
What is the approximate E450nm (mM-1 cm-1) for the coloured compound
formed in the assay
0.199
250
4
2
0.196
Semester 2, 2005
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1812 A
74.
A
B
C
D
E
Semester 2, 2005
Page 35 of 46
What is the minimum concentration of tyrosine that you could reliably
measure in an unknown solution? [Hint: for reliable measurements, the
absorbance of a tube must be between 0.1 and 1)
~0.02 µM
~1 µM.
~5 µM
~50 µM
~200 µM
You have access to a new type of spectrophotometer that
uses optical fibres to measure the absorbance of drops as
small as 20 µl in volume.
Cut optical fibre
20 l drop
In the spectrophotometer, drops of solution are placed
between two cut ends of optical fibre which are the
stretched out to give a path length of 1 cm.
Stretch to 1 cm
You measure the absorbance of Tubes 1 -6 on this new
machine
75.
A
B
C
D
E
1812 A
Which statement is CORRECT?
Using the new machine, the absorbance of the contents of Tube 6 will be
greatly increased
Using the new machine, the extinction coefficient of the coloured compound
will be decreased
This new machine will be able to give absorbance readings to at least FIVE
decimal places with great accuracy
The new machine will be able to determine if tyrosine itself absorbs at 460 nm
The new machine will enable us to create a workable standard curve with
100-fold smaller volumes
Semester 2, 2005
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1812 A
76.
A
B
C
D
E
Semester 2, 2005
Page 36 of 46
Your assistant repeats your set up but, although their absorbance readings
increase linearly from Tube 1 to Tube 6, the slope of their standard curve is
80% of yours (ie, their highest absorbance is 0.8). In your diagnosis of their
mistake, what can you RULE OUT (ie, what would NOT have caused this)?
They measured the samples at the wrong wavelength
They used a pipette for the tyrosine solution that consistently delivered 20%
less than it was meant to
They used a pipette for the tyrosine solution that consistently delivered 100 µl
too little
They incubated their tubes for too short a time
They incubated their tubes for too long
One of your colleagues has made up a stock tyrosine solution but cannot remember its
concentration. You set up a series of tubes as follows:
Tube #
1
2
3
4
5
6
Solution A (ml)
0.5
0.5
0.5
0.5
0.5
0.5
Solution B (ml)
0.5
0.5
0.5
0.5
0.5
0.5
Unknown tyrosine sol (µl)
0
10
50
250
500
1000
Water (µl)
1000
990
950
750
500
0
Absorbance 450 nm
0
0.41
1.85
2.05
2.1
2.09
77.
A
B
C
D
E
1812 A
Which statement is CORRECT?
This data shows that the real relationship between absorbance and
concentration is complex and cannot be solved with a simple linear regression
The concentration of your colleague’s solution is too high for you to determine
using your assay
Your colleague’s solution is too dilute for you to determine using your assay
Your colleague’s solution contains something that interferes with your assay
These results enable you to give your colleague a reasonable estimate of the
concentration of their solution
Semester 2, 2005
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1812 A
Semester 2, 2005
Page 37 of 46
You now use your assay to measure the concentration of tyrosine in various plasma
samples. Plasma is blood that has been centrifuged to remove red blood cells.
Plasma should be clear and colourless, but burst red blood cells can make it pink, and
excessive fat/lipoproteins can make it milky.
78.
A
B
C
D
E
How could you control for the extra problems associated with using plasma?
The complete absorption spectrum of each plasma sample needs to be
determined before the assay can be done
There is no need to introduce any extra controls because the pink colour or
milkyness will not affect the absorbance at 450 nm
If samples look pink or milky we should just use smaller volumes of these
plasma samples in the assay
Blanks containing plasma but no Solution A or B need to be set up
All the standards need to be spiked with plasma samples before assay
Plasma tyrosine concentration is normally about 80 µM. Your colleague has a theory
that plasma tyrosine concentrations are about 60% higher in people with certain
fatigue syndromes. They have collected about 3 mls of plasma from a several control
and fatigued patients and they would like you to assay these samples for tyrosine.
79.
A
B
C
D
E
1812 A
Their samples of plasma are precious and they do not want you to use any
more than you need. What do you tell them?
“Easy. I only need to use a total of 10 µl of each sample to get accurate
enough measurements to see the difference”
“My current assay is sensitive enough to get accurate results with just 10 µl of
each sample, but the differences you’re expecting to see are just too small”
“I’ll need to use 0.5 to 1 ml of plasma in each assay tube, so by the time I’ve
done replicates and controls, I may have used several mls of your plasma”
“It’s impossible. Even if you buy me one of those new fibre-optic machines”
“The changes that you anticipate cannot be detected by any colorimetric
method”
Semester 2, 2005
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1812 A
Semester 2, 2005
Page 38 of 46
Questions on the use of RADIOACTIVITY to measure metabolic fluxes
Useful Information
1 nCi = 2,200 dpm = 37 Bq
1 µCi = 2.2 million dpm = 37 kBq
1 ALI (Annual limit of Intake) for 14C is 40 MBq
Measuring Glucose uptake using the radio-labelled 2-deoxyglucose method
You wish to measure the rate of glucose uptake into muscle cells in response to
insulin. You have read about a method that uses an analog of glucose called
2-deoxyglucose (2DG). 2DG differs from glucose in that it has an hydrogen instead
of a hydoxyl at the C-2 position. Cells take up 2DG and convert it into
2-deoxyglucose 6-phosphate (2DGP). The 2DGP cannot proceed down glycolysis
and is ‘trapped’ in the cells after uptake. This is because 2DGP cannot be made into
fructose 6-phosphate. It can, however, be made into glycogen.
In the method, cells are incubated in the presence of normal glucose containing a
TRACER amount of radioactively labeled 2DG. After a particular time, the
incubation is stopped so as to determine the amount of radioactivity that has become
trapped in the cells.
1812 A
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1812 A
80.
A
B
C
D
E
Semester 2, 2005
Page 39 of 46
For the 2DG method to work properly, which of the following must be TRUE?
It is important that hexokinase works faster on 2DG that it does on glucose
The presence of trapped 2DGP should not affect the normal metabolism of
glucose
Muscle glucose transporters should behave differently towards 2DG than they
do towards glucose
2DGP should be able to desphosphorylate back to 2DG
2DGP should be able to move freely in and out of the muscle cells
You purchase 50 µCi of [U14-C] 2-deoxyglucose. It arrives as a aqueous solution
contained in a 3 mm thick glass vial. The label on the vial is as shown in the picture.
[U-14C] 2deoxyglucose
50 Ci
0.2 mCi/ml
200 Ci/mol
81.
A
B
C
D
E
1812 A
Which statement is CORRECT?
To avoid exposure to beta-particles, the glass vial should be encased in a lead
container from now on
Beta-particles will not penetrate the glass and gloved hands
Drinking the entire contents will exceed your Annual Limit of Intake (ALI) for
radiation
Spilling the contents of this vial on your hands will require hospitalization
You should maintain a distance of at least one metre away from this vial
Semester 2, 2005
Page 39 of 46
1812 A
82.
A
B
C
D
E
83.
A
B
C
D
E
84.
A
B
C
D
E
1812 A
Semester 2, 2005
Page 40 of 46
What is the volume of 2DG solution in the vial?
100 µl
50 µl
200 µl
250 µl
25 µl
How many dpm are contained in 1 µl of the 2DG solution?
50,000
200,000
2.2 million
440,000
5 million
What is the concentration of 2DG in the vial?
200 mM
1 mM
200 µM
50 µM
200 nM
Semester 2, 2005
Page 40 of 46
1812 A
Semester 2, 2005
Page 41 of 46
You have a suspension of muscle cells in a suitable buffer (Buffer X) at a
concentration of 20,000 cells/ml. You set up some incubations in Eppendorf tubes as
shown in the table below. Each incubation contains a final volume of 1 ml and a final
concentration of 5 mM glucose. The glucose is associated with a tracer amount of
[U-14C]-2DG to give a (planned) final specific activity of about 500 dpm per nmol.
Some of the incubations contain insulin.
Tube
1
2
3
4
5
Cell suspension (ml)
0.5
0.5
0.5
0.5
0.5
Buffer X (ml)
0.5
0.4
0.4
0
0
200 ng/ml Insulin solution (ml)
0
0
0
0.4
0.4
2DG/glucose mixture (ml)
0
0.1
0.1
0.1
0.1
The tubes were set up was as follows:

Firstly, the cells, buffer and/or insulin were added to the tubes..

The tubes were then placed in a water bath at 30°C.

At time zero, the reaction in Tube #2 was started by adding 0.1 ml of the
2DG/glucose mixture.

At 1 min intervals, Tubes 3, 4 and 5 were started in a similar manner (ie, at a
clock time of 1, 2 and 3 min respectively).

At 5 min on the clock, Tube #2 was centrifuged for 30 seconds to form a pellet
of the cells. The supernatant was immediately aspirated (removed) and
discarded. The remaining pellet was counted for 14C for 10 minutes.

Tubes 4, 3 and 5 were treated in the same way when the clock read 7, 11 and
13 minutes respectively.
1812 A
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Page 41 of 46
1812 A
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Page 42 of 46
You decide that you will need 1 ml of the radioactive 2DG/glucose mixture.
You dissolve 9 mg of glucose in 1 ml water to make a 50 mM solution.
85.
A
B
C
D
E
86.
A
B
C
D
E
87.
A
B
C
D
E
1812 A
How many dpm SHOULD be taken from the 2DG vial to make this glucose
solution the desired specific activity?
500
500,000
25 million
50,000
1 million
Which of the following BEST DESCRIBRES the APPROXIMATE ratio of
2DG molecules to glucose molecules in the above 2DG/glucose mixture.
One 2DG to every thousand glucoses
A thousand 2DGs to every glucose
One 2DG to every glucose
One 2DG to every five glucoses
One 2DG to every million glucoses
After YOUR ASSISTANT makes up the 2DG/glucose mixture, you count 10
and 20 µl aliquots and find that these contain 25,123 and 52,715 dpm
respectively. What is the ACTUAL specific activity of the mixture?
288 dpm/nmol
52 dpm/nmol
25 dpm/nmol
500 dpm/nmol
144 dpm/nmol
Semester 2, 2005
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1812 A
Semester 2, 2005
Page 43 of 46
The results of the experiment are shown in the table below.
Tube
88.
A
B
C
D
E
89.
A
B
C
D
E
90.
A
B
C
D
E
1812 A
Incubation
dpm in cell
time (min)
pellet
1
0
30
2
5
4151
3
10
7330
4
5
22030
5
10
42732
APPROXIMATELY how much 2DG/glucose is trapped in the cells in
Tube #4?
110,000 nmol
8.8 nmol
22 nmol
44 nmol
425 nmol
What is the approximate rate of glucose uptake in Tube #4
(in nmol/min/10,000 cells)?
85
4.4
425
8.8
43
In Tube #4, what percentage of the total dpm put into the tube has been taken
up by the cells?
<1%
5 - 15%
20 - 30 %
50 – 80 %
>90%
Semester 2, 2005
Page 43 of 46
1812 A
Semester 2, 2005
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The following information relates to Questions 91 - 95
Consider TUBE #3 and the following table of options.
Property
A
B
C
D
E
dpm in cell pellet
INCREASE
INCREASE
INCREASE
INCREASE
SAME
nmol 2DG/glucose in
INCREASE
INCREASE
INCREASE
SAME
SAME
INCREASE
INCREASE
SAME
SAME
SAME
INCREASE
SAME
SAME
SAME
SAME
cell pellet
rate of 2DG/glucose
uptake into pellet
(nmol/min)
rate of 2DG/glucose
uptake into pellet
(nmol/min/10,000 cells)
Which option BEST represents the pattern of changes that would result if Tube #3
was altered in the following ways:
91.
Performing the centrifugation step when the clock showed 21 minutes
92.
Doubling the specific activity of the 2DG/glucose mixture
93.
Counting the cell pellet in the scintillation counter for twice as long
94.
Doubling the volume of all additions (ie, 2 ml incubation volume, 1 ml cells,
0.2 ml 2DG/glucose mixture and 0.8 ml buffer)
95.
Adding insulin insead of buffer X (ie, making it identical to Tube #5)
1812 A
Semester 2, 2005
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1812 A
Semester 2, 2005
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SECTION B - PRACTICAL
PART II – Short answer question worth 6 marks
A kibitzing* colleague suggests that your results are not valid because a significant
amount of ‘free’ 2DG/glucose gets transferred into the scintillation vial. In other
words, they think that 2DG/glucose associated with the outside of the cells or solution
simply left with the cell pellet is interfering with your results. They also point out that
2DG/glucose uptake can still occur during the centrifugation and transfer steps.
*
Kibitz; to look on and offer intrusive comments and unwanted, usually meddlesome, advice to others.
.
They suggest the following improvements. Comment on each.
i)
“You should re-suspend the pellet of cells in 1 ml fresh buffer and centrifuge
them again before you count them.”
ii)
“You should formally ‘stop’ the reactions by adding something that will break
open the cells on demand (eg, strong acid).
iii)
“You should include incubations with dead cells as a control”
1812 A
Semester 2, 2005
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1812 A
Semester 2, 2005
Page 46 of 46
THIS IS THE END OF ALL THE QUESTIONS
1812 A
Semester 2, 2005
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