Download Selected Solutions to End of Chapter 17 Problems

Selected Solutions to End of Chapter 17 Problems
1. Where is the energy in triacylglycerols? The glycerol or fatty acids? Of course, there is so much
more fatty acids (long carbon chains) than the three carbon glycerol. But is that all there is, NO! All
of glycerol carbons are alcohols, whereas all most all (except for one carbon) all the carbons are
alkanes, with a few alkenes tossed in for good measure. Each of these carbons are more reduced
than the alcohol carbons in glycerol.
2. Fuel reserves in the average, 70 kg, human. Data: the energy in triacylglycerols is 38 kJ/g and
like it or not,15% of the average human is pure fat. The basic energy requirement for the average
human is 2,000 dietary calories/day = 8,400 kJ/day. And, 1 lb = 0.454 kg. A dietary calorie is 103
calories.
a. What is the total energy reserve in the average human?
(7 x 104 g)(0.15)(38 kJ/g) = 4.0 x 105 kJ or in calories = 9.5 x 104 dietary calories.
b. How long could this poor average human being survive on fat? (given water,etc.)
(4.0 x 105 kJ) / (8,400 kJ/day) = 47.6 days
c. What is the weight loss per day under these conditions?
(70 kg)(0.15)
47.6 day
= 0.22 kg/day
or 0.48 lb/day
3. See how the first three reactions in β-oxidation are similar to CAC reactions:
Succinate + FAD  Fumarate + FADH2
ane  ene
Fatty acyl-CoA + FAD  trans-Δ2-enoyl CoA + FADH2
Fumarate + H2O  Malate
ene  hydroxy
trans-Δ2-enoyl CoA  L-β-hydroxyacyl-CoA
Malate + NAD+  oxaloacetate + NADH
hydroxyl  keto
L-β-hydroxyacyl-CoA +NAD+  β-ketoacyl-CoA
4. Each cycle of β-oxidation produces an acetyl-CoA. How many β-oxidation cycles does it take to
breakdown oleic acid, 18:1 (Δ9)?
Answer = 8. (Hint: the last reaction produces 2 acetyl-CoAs)
9. Compartmentalization of β-oxidation (it occurs in the mitochondria). Palmitic acid taken up by a
cell gets converted to palmitoyl-CoA in the cytoplasm. If that Co-A was radioactive ( 14C-CoA ).
Palmitoyl-CoA if isolated from the cytoplasm would be radioactive, but the palmitoyl-CoA isolated
from the mitochondria is not. Why?
To get palmitic acid into the mitochondria, it uses the carnitine shuttle system which transfers
the palmitic acid from the radioactive CoA to acyl-carnitine and then transfers the palmitic acid
to CoA in the mitochondria which is not radioactive.
10. Comparative Biochem in birds…we did problems related to this in earlier chapters. Here’s more
data comparing pigeon to pheasant. Below are the Vmax of various enzymes in these bird’s flight
muscles:
Vmax (µmoles/min/g tissue
Enzyme
Pigeon
Pheasant
Hexokinase
3.0
2.3
Glycogen phosphorylase
18
120
Phosphofructokinase-1
24
143
Citrate synthase
100
15
Triacylglycerol lipase
0.07
0.01
You can see several systems here: glycogen utilization, glycolysis, triacylglycerol utilization, citric acid
cycle (CAC).
a. Compare glycogen and fat metabolism: Pigeon is setup up to utilize fat, the enzyme that
cleaves fatty acids from triacylglycerol is 7X more active than Pheasant. This also goes for
entry to CAC. Pheasant has faster rates for using glycogen and glycolysis.
b. What would you predict the oxygen consumption rates for these two birds would be?
Pigeon would us more oxygen, a full bore CAC!
c. Which is the long distance flyer? Pigeon. Both birds spend time on the ground, but pigeon
can go the distance in air on their fat. Pheasant can have powerful, but short distance flight on
glycogen.
d. Why were these enzymes selected for this problem, what do they have in common? They
are all regulatory enzymes with allosteric sites or regulated by covalent modification. They
need to be “on” during flight, and have much lower activity on the ground.
13. Fatty acids as a source of water. Camels ! store water? No, they store fat in their humps. Fatty
acid oxidation produces water, here’s the example of palmitic acid using β oxidation + CAC + electron
transport:
Palmitoyl-CoA + 23O2 + 108 Pi + 108 ADP  CoA + 16 CO2 + 108 ATP + 23 H2O
So one mole of palmitic acid could produce 23 moles of water!
Problem: How much water can be produced from 1 kg of camel fat, use tripalmitoylglycerol as
the example.
Data: tripalmitoylglycerol has a molecular weight of 885 daltons.
So, 1 kg of fat: 1,000 g fat / 885 g/mole = 1.13 moles of tripalmitoylglycerol which contains 3
palmitic acids. So we can get the moles of water from (we will ignore glycerol for this):
(1.13 mol tripalmitoin)(3 mol of palmitate/mol tripalmitoin)(23 mol water/mole palmitate) = 78 mol H 2O
Now lets get the volume of water:
(78 mol H2O)(18 g/mol) = 1,404 grams
knowing the density of water, this is 1.4 L.
This is an underestimate because we did not include glycerol oxidation of which would produce
a small quantity of water and we did not include longer chain fatty acids. But, it is also an
overestimate because less water would be produced from each point of unsaturation. See
problem 26 below.
16. Fatty acid oxidation in diabetics. Fatty acid oxidation is important for diabetics because they can
not use glucose. Excess acetyl-CoA is converted to ketone bodies: acetoacetate, β-hydroxybutyrate,
and acetone. What problem would arise if acetyl-CoA could not be converted to ketone bodies?
The mitochondrial CoA pool is small, so two acetyl-CoA are converted to acetoacetyl-CoA +
CoA. And in the liver, acetoacetyl-CoA is converted to the ketone bodies: acetoacetate, βhydroxybutyrate, and acetone. Note that acetone is a minor component of the ketone bodies.
Ketone bodies can be metabolized by non-hepatic tissues.
26. Water is produced from the reaction: ADP + Pi  ATP + H2O. But was left out of the equation
we used in the camel problem (13 above). Why did we do that?
Because ATP is recycled and when used as a source of energy consumes the same amount of
water:
ATP + H2O  ADP + Pi
28. Fat loss during hibernation. Data: Bears use about 25 x 10 6 J/day during hibernation, which for
some bears goes on for 7 months! How much weight loss occurs during this hibernation:
We know that fat has about 38 kJ/g which is 3.8 x 104 kJ/kg and,
that the bear needs 2.5 x 104 kJ/day. So:
(2.5 x 104 kJ/day) / (3.8 x 104 kJ/kg) = 0.66 kg/day
For 7 months (210 days):
0.66 kg/day x 210 days = 139 kg weight loss
But what about blood glucose? During hibernation there is enough protein turnover to supply
glucogenic intermediates for gluconeogenesis. What a fun chapter this is: birds, camels and
bears !