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Biochemistry 2 > lec2 glycolysis > Flashcards

lec2 glycolysis Flashcards

1
Q

the only metabolite that begins its digestion from the mouth is…….

A

CHO

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2
Q

the CHO is digested in the mouth via………. enzyme

A

a-amylase enzyme

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3
Q

the a-amylase enzyme breaks …………. resulting in a mixture of ……….. and ………

A

a(1>4)bonds of starch

short branched and unbranched oligosaccharides (dextrin)

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4
Q

Humans can not digest cellulose. Why?

A

Salivary β-amylase that breaks β glycosidic
bonds are not found in humans, therefore humans
cannot digest cellulose.

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5
Q

there is No digestion of carbohydrates In the stomach why?

A

because high acidity inactivates salivary amylase.

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6
Q

……..it is an enzyme continues the digestion of CHO

A

Pancreatic α-amylase

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7
Q

The final digestive processes occur at………….

A

jejunum

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8
Q

write a short note about lactose intolerance.

A

 due to genetic lactase deficiency.
Any deficiency in disaccharides activity of intestinal mucosa causes passage of undigested CHO into large intestine causing osmotic diarrhea→ bacterial fermentation of CHO to 2 or 3-carbon compounds + CO2 + H2 gas→ abdominal cramps and flatulence.
 Treatment: remove lactose and give lactase pills, Eating yoghurts, cheeses and green vegetables to ensure adequate calcium intake.

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9
Q

Glycolysis means…………….

A

oxidation of glucose to give pyruvate (in the presence of oxygen) or lactate (in the absence of oxygen)

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10
Q

the glycolysis occurs in …….

A

cytosol

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11
Q

the end product of the 1st step in glycolysis is……

A

glucose 6-phosphate

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12
Q

the 1st step in glycolysis is a reversible state

T or F?

A

F

irreversible

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13
Q

the initial substrate in 1st step is…….

A

glucose

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14
Q

the enzyme used in 1st step is ………

A

kinase enzyme (hexokinase or glucokinase)

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15
Q

write a short note about 1st step

A

Phosphorylation of glucose into glucose 6-phosphate, using an ATP and a kinase enzyme (glucokinase or hexokinase),
(Irreversible step).

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16
Q

the 2nd step is the irreversible step

T or F?

A

F reversible

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17
Q

the initial substrate in the 2nd step is…….

A

glucose 6-phosphate

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18
Q

the enzyme used in the 2nd step is………

A

phosphohexose isomerase

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19
Q

the product in the 2nd step is……

A

fructose 6-phosphate

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20
Q

write a short note about the 2nd step

A

Isomerization of glucose 6-phosphate into fructose 6-

phosphate by phosphohexose isomerase (reversible step).

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21
Q

the 3rd step is the irreversible step

T or F?

A

T

22
Q

the initial substrate in the 3rd step is…….

A

fructose 6-phosphate

23
Q

the product in the 3rd step is……

A

fructose1,6 -bisphosphate

24
Q

the enzyme used in the 3rd step is………

A

phosphofructokinase-1

25
Q

write a short note about the 3rd step

A

Phosphorylation of fructose 6-phosphate into fructose 1,6-bisphosphate with phosphofructokinase-1 (PFK-1), (Irreversible step, rate limiting step).

26
Q

what is the name of the process that occurs in step 4?

A

cleavage

27
Q

the initial substrate in the 4th step is…….

A

fructose1,6-bisphosphate

28
Q

the product in the 4th step is……

A

dihydroxyacetone phosphate + glyceraldehyde-3 -phosphate

29
Q

the 4th step is the irreversible step

T or F?

A

F reversible

30
Q

the enzyme used in the 4th step is………

A

aldolase

31
Q

………….it is an enzyme used in the cleavage

A

aldolase

32
Q

write a short note about the 4th step

A

Cleavage of fructose 1,6-bisphosphate by aldolase to
dihydroxyacetone phosphate and glyceraldehyde-3-phosphate,
(reversible step).

33
Q

the enzyme used in the 5th step is………

A

triosephosphate isomerase

34
Q

the product in the 5th step is……

A

2 molecules of glyceraldehyde 3-phosphate

35
Q

the initial substrate in the 5th step is…….

A

dihydroxyacetone phosphate

36
Q

write a short note about the 5th step

A

Isomerization of dihydroxyacetone phosphate into another glyceraldehyde 3-phosphate by triose phosphate isomerase. So, the result is 2 molecules of glyceraldehyde-3-phosphate.

37
Q

the enzyme used in the 6th step is………

A

glyceraldehyde 3-phosphate dehydrogenase

38
Q

write a short note about the 6th step

A

Oxidation of glyceraldehyde-3-phosphate to 1,3 bisphosphoglycerate (1,3 BPG) by glyceraldehyde-3-phosphate dehydrogenase.

39
Q

the product in the 6th step is……

A

2 MOLECULES OF 1,3-bisphosphoglycerate

40
Q

the initial substrate in the 7th step is…….

A

2 MOLECULES OF 1,3 bisphposphoglycerate +ADP

41
Q

the product in the 7th step is……

A

2 MOLECULES OF 3-phosphoglycerat + 2ATP

42
Q

write a short note about the 7th step

A

Dephosphorylation; formation of 3-phosphoglycerate and ATP from 1,3 BPG and ADP. • 1,3 BPG is a high energy phosphate compound used in the synthesis of ATP in a reaction catalyzed by phosphoglycerate kinase. • This reaction gives 2 ATP and 2 compounds of 3-PG. (substrate-level phosphorylation). Unlike most other kinases, the reaction is reversible.

43
Q

the 4th step is the irreversible step

T or F?

A

F

UNLIKE most other kinases, the reaction in step 7 is reversible.

44
Q

write a short note about the 8th step

A

The shift of the phosphate group from C-3 to C-2 to form 2-PG by Phosphoglycerate mutase, (reversible step).

45
Q

the enzyme used in the 8th step is………

A

Phosphoglycerate mutase

46
Q

what dose the enolase enzyme responsible for?

A

dehydration of 2PG to form phosphoenolpyruvate

47
Q

the enzyme used in the 9th step is………

A

enolase

48
Q

………. is a high energy phosphate compound

A

phosphoenolpyruvate

49
Q

the enzyme used in the 10th step is………

A

Pyruvate kinase

50
Q

the 10th step is the rreversible step

T or F?

A

F irrreversible

51
Q

the product in the 10th step is……

A

2 molecules of pyruvate

and 2 ATP

Biochemistry 2 (7 decks)

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