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Nutritional physiology > 8. Carbohydrate metabolism 2 > Flashcards

8. Carbohydrate metabolism 2 Flashcards

1
Q

TCA cycle needs/does not need oxygen to work

A

does not

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2
Q

Glucose -> pyruvate: last step is irreversible (PEP -> pyruvate). What if you want to convert pyruvate into glucose? How is this process called?

A

Pyruvate can be converted into oxaloacetate into PEP with the help of the enzymes pyruvate carboxylase and PEP carboxykinase. = Gluconeogenesis

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3
Q

OAA occurs both in the cytoplasm and mitochondrion, but cannot pass te mitochondrial membrane. What needs to be done? ATP cost?

A

OAA is converted to malate (-2.5 ATP), transported outside of the membrane and then converted into OAA again (+1.5 ATP)

Cost = 1 ATP

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4
Q

How many TCA cycles before citrate is fully oxidated?

A

Two, because in one cycle two carbons are released and citrate is a C4 molecule.

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5
Q

How do you fully oxidize citrate?

A

Citrate 6C -> TCA cycle until malate 4C > outside membrane OAA 4C > 3C PEP > 3C pyruvate> 2C acetyl-CoA > TCA cycle

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6
Q

What is the ATP yield when you fully oxidize citrate?

A

21.5 ATP

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7
Q

Why is citrate ATP yield less efficient (21.5) than glucose (30 ATP)?

A

 Citric acid is related to glucose (C6 both). Nevertheless it only yields 21.5 ATP instead of 30, because citrate does not enter glycolysis and misses all ATP production there

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8
Q

What is the difference in ATP yield between oxaloacetate (OAA) and citrate upon complete oxidation?

A

21.5 - 11.5 = 10 ATP

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9
Q

What is propionic acid?

A

volatile fatty acid (VFA). One of the SCFA’s. Fermentation in large intestine.
> produced by bacteria as waste product. Our bodies can use it.

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10
Q

Conversion of propionic acid to succinyl-CoA via two steps costs .. ATP

A

3 ATP
(first ATP -> AMP)
(then ATP -> ADP)

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11
Q

Where does propionic acid enter the cycle?

A

Succinyl-CoA

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12
Q

What is the ATP yield of propionic acid/propionate: complete oxidation?

A

13.5 ATP

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13
Q

Storage of glucose/other nutrients: Tell me about these organs

  1. Brain
  2. Muscle
  3. Adipose tissue
  4. Blood
A

 Brain only uses glucose, or ketone bodies in fasting conditions. It has only a small amount of glucose stored.
 Muscle has a lot of glucose and protein stored and some triacylglycerols
 Adipose tissue: lot of triacylglycerols, some glucose and proteins
 Blood: very constant amount of 250 kj glucose

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14
Q

What can only the liver do and why?

A

 Liver can only convert glucose-6-ph into glucose because it has the glucose-6-phosphatase enzyme

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15
Q

Glucose homeostasis in blood plasma:

A

Constant, always 3-5 mM (1 gr/l)

With 2.5 liter plasma, only 2.5 gr circulates

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16
Q

How is glucose homeostasis maintained in fed/fasted state?

A

 Fed state: good regulation by condensation to liver and muscle glycogen
 Fasted state: release from liver, not muscle, into circulation

17
Q

Changes in blood levels only reflect difference in … but is a poor indicator of …

A

Changes in blood levels only reflect difference between input – output.
Blood level is a poor indicator of total flux (throughput)

18
Q

How does the glucose tolerance test work?

A

Fasting condition, give drink, check blood glucose levels.

19
Q

Glucose from the diet can be stored as glycogen. First, the gluose molecule has to be …. , which takes …. ATP

A

2 ATP

(glucose -> gl-6-p = 1 ATP)
(gl-6-p -> gl-1-p = 0 ATP)
(gl-1-p -> UDP tag = 1 UDP = 1 ATP)

By UDP tagging (attaching to the gl-1-p)

20
Q

Then, the molecule can be added to the linear glucose molecule by ..

A

release of UDP. STILL, 2 ATP used

21
Q

Direct glycolysis = ?
Indirect glycolysis = ?

A

Indirect = from diet you obtain glucose > g-6-p -> g-1-p -> glycogen 2 ATP cost
Direct = glucose -> g-6-p = 1 ATP cost

22
Q

 Energy costs for storage and mobilisation of blood glucose via hepatic (=lever) glycogen = x ATP

A

2 ATP (indirect)
(glucose -> glycogen -> glucose)

23
Q

 ATP yield of (blood) glucose in leg muscle after direct oxidation = ?

A

30 ATP

24
Q

 ATP yield of (blood) glucose in leg muscle after indirect oxidation (via leg muscle glycogen) = ?

A

29 ATP (invest one more than direct)

25
Q

 ATP yield of (blood) glucose in leg muscle after indirect oxidation (via arm muscle glycogen):

A

0 (arm cannot give glucose to leg)

26
Q

 ATP yield of (blood) glucose in leg muscle after indirect oxidation (via liver):

A

28 ATP. Liver glucose > G-6-P > glycogen > glu = 2 ATP. It is released as glucose.

Then, in the leg muscle, glucose -> oxidation , so glucose -> g-6-p = 1 ATP ‘extra’

27
Q

 ATP yield of (blood) glucose in leg muscle after indirect oxidation (after 5 glycogenolysis/glycogen synthesis cycles in the leg muscle)?

A

glycogen > gl-6-p > glycogen (1 ATP)
x 5 = 5 ATP
30 - 5 = 25 ATP

28
Q

Regulation of storage or glycolysis of gl-6-p is done by..

A

PFK phosphofructokinase

29
Q

What does PFK usually help with in glycolysis?

A

fr-6-p to fr-1,6-p
It decides the faith of gl-6-p

30
Q

PFK is inhibited by citrate. But what kind?

A

Not the citrate of the mitochondria.

From the cytosol, which is also coming from fatty acid synthesis. Lot of citrate in cytosol = lot of energy.

31
Q

What is the pasteur effect?

A

Switch from slow to fast glucose use

32
Q

How was the pasteur effect discovered?

A

Discovered by Pasteur, that yeast fermented at a lower speed in the presence of oxygen.

33
Q

Oxidative phosphorylation is …. in low oxygen conditions

A

Blocked

34
Q

What happens in the glycolysis when there is low oxygen?

A

Normally, glucose > pyruvate, ATP + NADH produced. NADH needs to be oxidated back to NAD+ so the glycolysis can continue. Happens in the mitochondria (shuttle).
> However, with low oxygen, no regeneration of NAD+.

35
Q

What is a solution for low oxygen which inhibits NAD regeneration? Does this cost energy?

A

Regeneration of NADH to NAD+ by cytoplasmic conversion of pyruvate to lactate (No ATP!)
Pyruvate + NADH > lactate + NAD+

36
Q

Normally, under aerobic conditions, glucose is oxidized which yields 30 ATP, 6 CO2 and 6 H2O. What is yielded with anaerobic conditions?

A

Glucose in anaerobic conditions yields 2 ATP and 2 lactate

37
Q

Pasteur effect increases/decreases glucose utilisation by …

A

increases, 15x

38
Q

Energy efficiency of glucose aerobic oxidation was 35%. For lactate, it is …

A

75%
GE glucose = 2816 kj/mol
GE lactate = 1364 kj/mol
ATP = 33 kj/mol

2 * 1264 = 2728 kj
2816 - 2728 = 88 kj lost
trapped 2 ATP = 66 kj
66/88 = 75%

Nutritional physiology (26 decks)

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