Chem*3560
1
Question set 8
Mar 1-5, 2004
How do the regulatory properties of glucokinase and hexokinase determine the
distribution and fate of glucose a) in a physically active person b) in a person with
sedentary lifestyle?
Hexokinase I is distributed in most tissues. It has KM = 0.4 mM for glucose, so is nearly
saturated at typical blood glucose concentration of 5 mM. However, it is subject to product
inhibition. This is a consequence of the enzyme having relatively high affinity for its product,
glucose-6-phosphate so that it does not leave the enzyme quickly, and competes with the
substrate for occupancy of the catalytic site.
Product inhibition strictly involves a product of the enzyme reaction, and is distinct from end
product inhibition, in which a substance that is not a product of the enzyme, but is a product of
the pathway as a whole acts as a negative allosteric effector. CTP is an end product inhibitor
of ATCase, because ATCase is at the start of the pyrimidine synthesis pathway, and CTP is a
pyrimidine, but CTP itself is not a direct product of the ATCase reaction.
As a result of product inhibition by glucose-6-phospate, hexokinase I can only continue to act
on glucose if the glucose-6-phosphate product is being used in subsequent processes
such as glycogen storage or glycolysis, and this will be the case in an active person whose
glycogen reserves have recently been depleted. In the sedentary person, the glycolysis pathway
is inactive, and glycogen reserves are already fully built up, so there will be no need for
glucose-6-phosphate, so glucose utilization ceases in muscles and peripheral tissues.
The liver contains hexokinase D, also known as hexokinase IV or glucokinase, with a high KM
for glucose. The enzyme is relatively inactive at normal blood levels of glucose, which gives the
peripheral tissues "first look" at the blood glucose. After a starch or glucose rich meal, blood
glucose rises sharply, and this increases the rate at which glucokinase makes
glucose-6-phosphate in liver. Glucokinase is not subject to product inhibition, so will continue
to phosphorylate glucose even if there is no immediate use.
If blood glucose is high, insulin is secreted by the pancreas, and will stimulate production of
fructose-2,6-bisphosphate in the liver. This results in active glycolysis even though the energy is
not immediately needed, and the pyuvate produced enters the mitochondrion. In the active
person, glucose is cleared from the blood by peripheral tissues, but in the sedentary person
citrate accumulates because low ADP levels fail to activate isocitrate dehydrogenase. Excess
citrate is exported to the cytoplasm, and activates acetyl-CoA carboxylase. These are the
conditions that favour fatty acid biosynthesis.
Paté de foie gras (fatty goose liver) is a French delicacy made by force-feeding geese with
starch-rich grains. The geese's livers overdevelop and become loaded with fat deposits.
Chem*3560
2.
Question set 8
Mar 1-5, 2004
Arachidonate is 20:4 (∆5, 8, 11, 14) in the shorthand notation for unsaturated fatty acids.
What does this mean? Draw the structure.
Arachidonate is 20:4 (∆5,8,11,14) which means a straight chain of 20 C atoms including and count
from the carboxylate. There are 4 cis double bonds, positioned at bond numbers 5, 8, 11, and
14. Bond 5 lies between C atoms 5 and 6, etc.
Arachidonate, 20:4 (∆5, 8, 11, 14), is made from linoleate, 18:2 (∆9, 12). How do the double
bonds move to the positions indicated?
Arachidonate is made in animals by a sequence involving 5/6 desaturase to make γ -linolenate
18:3 (∆
∆ 66 ,, 9, 12). Desaturase places the double bond at C-6 because that makes the double
bond allylic, 3 positions away from the neighbouring double bond.
Elongase then extends the chain by two more C atoms (donated from malonyl-CoA), and these
are added at the carboxylate end. The chain is renumbered counting from the newly added
carboxylate, and this adds 2 to all atoms positions inherited from γ -linolenate. The
double bonds remain associated with exactly the same C atoms as before, and are not moved.
When the eicosatrienoate encounters 5/6 desaturase again, the new double bond is placed at
C-5, making it allylic with respect to the double bond at C-8 (∆8 ).
What would happen if it was elongated to the 22:4 fatty acid?
Assuming that the second elongation happens after the second desaturation, then the product
would have all double bonds shifted up by 2: 22:4 (∆
∆ 7, 10, 13, 16).
Chem*3560
3.
Question set 8
What product whould you get from linolenate 18:3 (∆9, 12, 15) if you took it through the
same steps that convert linoleate to arachidonate?
5/6 desaturase makes 18:4(∆
∆
Elongase makes 20:4 (∆
∆
66 ,, 9, 12, 15
88 ,, 11, 14, 17
5/6 desaturate makes 20:5 (∆
∆
4.
Mar 1-5, 2004
)
)
55 ,, 8, 11, 14,17
), eicosapentaenoate.
Reports in the media keep talking about the health benefits of omega-3 fatty acids (ω-3).
The omega number is the bond position counting back from the terminal -CH3 group
instead of forward from the carboxylate group. Which fatty acid out palmitoleate,
oleate, linoleate, α-linolenate, γ-linolenate, eicosatrienoate and arachidonate (Lehninger
Fig. 21-13, p.781) qualifies as ω-3?
The ω bond in C18 fatty acids lies between C-17 and C-18
The ω -2 bond in 18-C fatty acids lies between C-16 and C-17
The ω -3 bond in 18-C fatty acids lies between C-15 and C-16
Therefore α -linolenate, 18:3(∆
∆ 9,12,15) is an ω -3 unsaturated fatty acid
The ω -3 bond in C20 fatty acids lies between C-17 and C-18
None of those listed fit this description, but eicosapentaenoate 20:5 (∆
∆
55 ,, 8, 11, 14,17
) would.
Omega, ω or Ω, is the last letter in the Greek alphabet. Therefore the ω atom is the highest
conventionally numbered atom in a molecular chain. This is convenient to describe a group
which is positioned relative to the high numbered end of the chain.
5.
What precursors are needed to make a) triacylglycerol? b) phosphatidic acid c)
phosphatidyl choline?
a) Triacylglycerol is made from scratch starting with dihydroxyacetone phosphate and three fatty
acids. Triacylglyceral may also be recycled from existing 1,2-diacylglycerol or phosphatidic
acid by adding the third fatty acid.
dihydroxyacetone
reduced with NADH ↓
phosphate glycerol phosphate
add acyl-CoA
↓
1-acylglycerol-3 phosphate
add acyl-CoA
↓
phosphatidic acid
remove Pi
↓
diacylglycerol
add acyl-CoA
↓
triacylglycerol
Chem*3560
5.
Question set 8
Mar 1-5, 2004
b) Phosphatidic acid is made by the sequence described for triacylglycerol, minus the last two
steps.
c) See #7
6.
What is the difference between phosphocholine and
phosphatidyl choline?
Phosphocholine is the ester of choline and phosphate:
Phosphatidyl choline is the complete phospholipid
with choline as headgroup; the phosphatidyl
radical includes the complete
1,2-diacylglycerol-3-phosphate:
7.
What are the two ways to make phosphatidyl choline? Why is one known as the salvage
pathway and why is it taken in preference to the alternative route if at all possible.
Phosphatidyl choline:
salvage synthesis in animals.
choline
uses ATP
↓
phosphocholine
add CTP
↓
CDP-choline
add diacylglycerol ↓
phosphatidyl choline
de novo synthesis in bacteria
phosphatidic acid
add CTP
↓
CDP-diacylglycerol
add serine
↓
phospahtidyl serine
remove CO2
↓
phosphatidyl ethanolamine
add SAM × 3
↓
phosphatidyl choline
new choline is made in animals
ethanolamine
use ATP
↓
phosphoethanolamine
add CTP
↓
CDP-ethanolamine
add diacylglycerol ↓
phosphatidyl ethanolamine
add SAM × 3
↓
phosphatidyl choline
Each S-adenosyl methionine (SAM)
used requires one molecule of
methionine, which animals can't
synthesize for themselves.
Methionine must be obtained in
the diet, and is one of the less
abundant amino acids.
Chem*3560
8.
Question set 8
Mar 1-5, 2004
What is the energy cost of making phosphatidyl choline by the salvage pathway, starting
from glycerol phosphate, two molecules of fatty acid and one of choline?
a)
the two molecules of fatty acid must be activated
to acyl-CoA, at a cost of 2 ATP each
4 ATP
glycerol-3-phosphate + 2 acyl-CoA → phosphatidic acid
b)
c)
choline must be converted to choline phosphate (uses ATP)
1 ATP
and then to CDP-choline using CTP (equivalent to 1 ATP)
1 ATP
phosphatidic acid + H2O → diacylglycerol + Pi
diacylglycerol + CDP-choline → phosphatidyl choline + CMP
overall cost is 6 ATP
9.
What is wrong with the phospholipid structure in Lehninger Fig 11-16, p. 378?
This shows a phospholipid, headgroup unspecified, with arachidonate as the 2-acyl group.
The natural arachidonate, 20:4 (∆ 5,8,11,14) should have all cis- double bonds, but the figure
shows all trans- in the phospholipid. The free arachidonic acid should underneath has the
proper cis- double bonds.
10.
Why do phospholipases A2 , C and D have adverse effects on membrane structure?
Neutral phospholipid forms bilayer spontaneously because the cross-sectional area occupied by
headgroup is well balanced with the two acyl tails. The molecules pack together like cylinders,
so that planar bilayer structures can form.
Phospholipid molecules become wedge-like if one acyl chain is removed by phospholipase A2 .
Wedge shaped molecules tend to pack together as a micelle.
Phospholipase C removes the headgroup-phosphate ester, leaving diacylglycerol. The single
-OH group in diacylglycerol is not sufficiently polar to balance against the two acyl chains, so
diacylglycerol tends to separate as an oil-like phase rather than forming either bilayer or micelle.
Phospholipase D removes only the headgroup, leaving phosphatidic acid. The double negative
charge of the phosphates repel each other, spreading the headgroup portions into wedge shape.
Phosphatidic acid tends to break up into micelles.
Chem*3560
11.
Question set 8
Mar 1-5, 2004
What do you find in the middle of a micelle? In the middle of a vesicle or liposome?
The core of the micelle consists of the
packed hydrocarbon chains of the fatty
acid tails.
A vesicle or liposome is a complete
bilayer that encloses a small spherical
volume. The contents will be the same
as the surrounding solution at the time
that the vesicle closed off.
12.
Why is the membrane described as a fluid?
A fluid is a state of matter in which molecules are free to move in space. If the molecules are
distant and do not cohere, the fluid is a gas. If the molecules cohere and are packed together
close enough for van der Waals forces to act, then the fluid is a liquid.
A solid consists of molecules which are in a fixed spatial relationship with their neighbours.
The membrane bilayer needs to contain phospholipid molecules which are free to move in the
plane of their leaflet of the bilayer, but cannot escape from their own leaflet because the
headgroup can't enter the core of the bilayer, and the hydrocarbon tails of the fatty acids can't
enter the surrounding aqueous phase. In other words the phospholipid molecules should have
freedom to move in two dimensions, but not the third dimension. This fluidity allows membrane
proteins the necessary freedom to rearrange themselves as required in a living cell.
How does this relate to degree of unsaturation of membrane lipids?
The fatty acid tails in a bilayer have a transition temperature similar to the melting point of a
solid. Below this temperature, the hydrocarbon chains pack together like a solid, and lack
freedom of movement, and this can impair the function of membrane proteins.
The cis- double bonds of unsaturated fatty acids kink the fatty acid chains, increasing the level
of disorder. Organisms can adjust the degree of unsaturation of newly synthesized fatty acids.
Fatty acids can also be exchanged on and off the phospholipid molecules by the action of
phospholipase A2 (removes 2-acyl) and acyl transferase (replaces 2-acyl) to lower the
transition temperature and maintain the fluid state.