Biochemistry Exam 2 Flashcards

1
Q

What is hemostasis? What are 3 principle components of hemostasis?

A

What is hemostasis? What are 3 principle components of hemostasis?

Hemostasis: Blood Clotting Cascade

Hemostasis has three major steps: 1) vasoconstriction - Vascular Spasm - Damaged blood vessels constrict. Cut smooth muscles, constrict. blood vessels cut. muscles squeezes blood vessel so loos ends come together (don’t tie not!). They stick back together)

2) temporary blockage of a break by a platelet plug (synthesized in the liver!) - Formal elements include megakaryocytes, which can fragment. The cell fragments..

Platelets adhere to damaged endothelium to form platelet plug (primary hemostasis) and then degranulate. This process is regulated through thromboregulation. Platelet Plug Formation: Platelets play one of the biggest factors in the hemostatic process. Being the second step in the sequence they stick together (aggregation) to form a plug that temporarily seals the break in the vessel wall. As platelets adhere to the collagen fibers of a wound they become spiked and much stickier. They then release chemical messengers such as adenosine diphosphate (ADP), serotonin and thromboxane A2. These chemicals are released to cause more platelets to stick to the area and release their contents and enhance vascular spasms. As more chemicals are released more platelets stick and release their chemicals; creating a platelet plug and continuing the process in a positive feedback loop. Platelets alone are responsible for stopping the bleeding of unnoticed wear and tear of our skin on a daily basis.[3]

The second stage of Hemostasis involves platelets that move throughout the blood. When the platelets find an exposed area or an injury, they begin to form what is called a platelet plug. The platelet plug formation is activated by a glycoprotein called the Von Willebrand factor (vWF), which are found in the body’s blood plasma. When the platelets in the blood are activated, they then become very sticky so allowing them to stick to other platelets and adhere to the injured area.[4][5]

There are a dozen proteins that travel along the blood plasma in an inactive state and are known as clotting factors. Once the platelet plug has been formed by the platelets, the clotting factors begin creating the Blood Clot. When this occurs the clotting factors begin to form a collagen fiber called fibrin.Fibrin mesh is then produced all around the platelet plug, which helps hold the fibrin in place. Once this begins, red and white blood cells become caught up in the fibrin mesh which causes the clot to become even stronger.[3]

and

3) blood coagulation, or formation of a fibrin clot. These processes seal the hole until tissues are repaired.

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2
Q
  1. Be able to identify whether atoms are oxidized or are reduced in a simple chemical reaction?
    a. 2K + Cl2 → 2K+ + Cl−
    b. cytochrome c1 (2+) → cytochrome c1 (3+)
    c. pyruvate → lactate
    d. acetaldehyde → ethanol

CH3CH=O → CH3CH2OH

A

a. 2K (becomes oxidized) + Cl2 (becomes reduced) → 2K+ + Cl−
b. cytochrome c1 (2+) → cytochrome c1 (3+) oxidation
c. pyruvate → lactate reduction
d. acetaldehyde → ethanol

CH3CH=O → CH3CH2OH reduction

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3
Q
  1. Calculate the equilibrium constant for the hydrolysis of pyrophosphate to phosphate (Pi) at room temperature.
        PPi  →  2 Pi                 ∆G0/  =  − 8.01 Kcal (Calories).
A
  1. Calculate the equilibrium constant for the hydrolysis of pyrophosphate to phosphate (Pi) at room temperature.
        PPi  →  2 Pi                 ∆G0/  =  − 8.01 Kcal (Calories).

∆G0/ = − 8.01 Kcal = (-2.3)(1.987 x 10-3 kcal/mole oK)(298)log Keq

Log Keq = -8.01/-1.3637 = 5.874; Keq = 7.48 x 10 5

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4
Q
  1. Calculate the ∆G0/ for formation of pyruvate and ATP.

phosphoenolpyruvate + ADP + Pi + H2O ↔ pyruvate + ATP + Pi + H2O

        a. ADP  +  Pi       ↔     ATP  +  H2O   (7.3 kcal/mole)
        b. phosphoenolpyruvate  +  H2O      ↔      pyruvate  +  Pi   (-14.8 kcal/mole)
A
  1. Calculate the ∆G0/ for formation of pyruvate and ATP.

phosphoenolpyruvate + ADP + Pi + H2O ↔ pyruvate + ATP + Pi + H2O

        a. ADP  +  Pi       ↔     ATP  +  H2O   (7.3 kcal/mole)
        b. phosphoenolpyruvate  +  H2O      ↔      pyruvate  +  Pi   (-14.8 kcal/mole)

∆G0/ = 7.3 + (-14.8) = -7.5 kcal/mole; exergonic

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5
Q
  1. For each of the following reactions, determine whether the reactant has been oxidized or reduced or is unchanged in oxidation state. (Reactions are not necessarily balanced as shown)
A
  1. For each of the following reactions, determine whether the reactant has been oxidized or reduced or is unchanged in oxidation state. (Reactions are not necessarily balanced as shown)
    a. oxidation b. oxidation c. reduction d. reduction e. oxidation
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6
Q
  1. What is the standard free energy change for an electrochemical reaction? Be able to calculate the net reaction potential for a redox couple.
    a. What is the net reaction potential when the [fumarate/succinate] half-reaction is coupled with the [oxygen/H2O] half-reaction.
    b. What is the net reaction potential when the [pyruvate/lactate] half-reaction is coupled with the [acetate/acetaldehyde] half-reaction.
    c. Calculate the standard free energy of reaction ∆G0/ for 5a, 5b.
A
  1. What is the standard free energy change for an electrochemical reaction? Be able to calculate the net reaction potential for a redox couple.
    a. What is the net reaction potential when the [fumarate/succinate] half-reaction is coupled with the [oxygen/H2O] half-reaction.
    b. What is the net reaction potential when the [pyruvate/lactate] half-reaction is coupled with the [acetate/acetaldehyde] half-reaction.
    c. Calculate the standard free energy of reaction ∆G0/ for 5a, 5b.

KNOW: ∆Eo = Eo(oxidant) - Eo(reductant)

∆G^o/ = -nF∆Eo

a. ∆E0/ = 0 .82 volt - 0.03 volt = 0.79 volt; ∆G0/ = (-2 equiv.)(23.061 kcal/volt x equivalents)(.79 volt) = -36.44 kcal
b. ∆E0/ = (-0.19 volt) – (-0.488) = 0.298 volt; ∆G0/ = (-2 equiv.)(23.061 kcal/v x equivalents)(0.298 volt) = - 13.74 kcal

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7
Q

What is a zymogen? What prefix or suffix is used to indicate a zymogen?

A

What is a zymogen? What prefix or suffix is used to indicate a zymogen?

A zymogen (or proenzyme) is an inactive enzyme precursor. Azymogen requires a biochemical change (such as a hydrolysis reaction revealing the active site, or changing the configuration to reveal the active site) for it to become an active enzyme.

Enzymes are activated by proteolytic cleavage of peptide bonds. a - activated form (and zymogenic activation is an irreversible change).

Prefix/suffix for zymogens:

  1. “PRO-“
  2. “-OGEN”
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8
Q

What is the consequence of the blood clotting cascade? What is the unique role of thrombin in the blood clotting cascade?

A

What is the consequence of the blood clotting cascade? What is the unique role of thrombin in the blood clotting cascade?

Eventually, an insoluble meshwork of protein fibers is made over a platelet plug in the damaged blood vessel. There are two pathways that lead to clot formation:

  1. The intrinsic pathway is initiated by exposure of the blood to foreign surfaces or to damaged tissues. Clotting is slow and complex.
  2. Extrinsic pathway is activated by factors released by damaged tissue. Clots within 15 seconds.
  3. The final section of the clotting sequence is a set of reactions common to each pathway.

Thrombin - Is responsbile for inhibition & and activation of clot factors. It activates accelerators 8A and 5A, and is also involved in stopping other accelerators involved in the clot cascade. Thrombin makes fibrin not have a neg. charge, which makes it precipiate out, and this forms a soft clot>>Then we have a hard clot with the activation of 13A.

Thrombin:

  1. accelerates and amplifies its own creation
  2. converts fibrogen into fibrin to form the clot
  3. activates the tightening of the clot
  4. it ends clot formation by inhibiting its own production
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9
Q

carbohydrates

A

carbohydrates

are poly hydroxylated aldehydes, poly-hydroxylated ketones or compounds that can be hydrolyzed into these compounds. Carbohydrates:

A. are the major source of E in most organisms

B. are structural components of cell walls and membranes

C. serve as metabolic intermediates

D. are a portion of the nucleotides that form RNA, DNA and ATP

E. play a role in cellular communication, lubrication and immunity

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10
Q

aldose, ketose

A

Reactive Group Nomenclature (-ose means we are talking about a sugar)

  1. Aldoses are monosaccharides with an aldehyde reactive group
  2. ketoses are monosaccharides with a ketone reactive group.

>>Can only have an aldehyde at the end of a carbon chain (not in middle). If it’s in middle, it’s a ketone.

EX: Glucose & fructose (Note: You begin counting aldoses at the first Carbon, and you begin counting ketoses at the nearest carbon near the tip). Also, glucose glucos is an aldohexose! =)

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11
Q
  1. What are the major storage forms of polysaccharides? How do they differ from each other?
A
  1. What are the major storage forms of polysaccharides? How do they differ from each other?

Starch, a mixture of amylose and amylopectin, is the storage form of glucose in plants.

A. amylose is a linear, unbranched polymer of alpha D glucose in a repeating sequence of alpha 1,4 glycosidic linkages

B. amylopectin is a branched polymer of alpha D glucose with alpha 1,4 glycosidic linkages: alpha 1,6 brach points are located 25 to 30 sugar residues apart.

We store carbs in glycogen form. Picture: Amylose

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12
Q

α-1,4’ glycoside bond

A

α-1,4’ glycoside bond

glycosidic linkages - complex carbohydrates are formed by glycosidic bonds between sugars (or through links with non saccarides).

Glycosides - when 2 monosaccharides are linked together, the product is a disaccharide (an acetal)

Maltose (one of the most common disaccharides) has an α-1,4’ glycoside bond

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13
Q

D-sugar

A

D-sugar

D/L refer to which way they rotate right!

All monosaccharides contain at least one chiral center and are optically active.

All sugars other than glyceraldehyde are considered to be derived from glyceraldehyde. A D sugar is one that matches the configuration of D glyceraldehyde around the asymmetic carbon that is farthest removed from the aldehyde or ketone group (AKA the OH is on the RIGHT!)

Epimers are stereoisomers (only differ in 3D structure) with configurations that differ at one carbon

D&L isomers are enantiomers, mirror image isomers… EX: Mannose and glucos are epimers, but Mannose and Galactose are not epimers (since 2 carbons are different).

ALSO: The orientation of the Carbon that is furthest from the aldehyde (before CH2OH) makes these three sugars “D sugars”

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14
Q

pentose hexose

A

pentose hexose

monosaccharides are named according to a system that uses the number of carbons as a variable prefix followed by -ose as the suffix.

EX: Glucose is an aldohexose

Also: Ketopentose: 5 C sugar with a ketone reactive group

triose - 3 C

tetrose - 4 C

pentose - 5 C

hexose - 6 C

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15
Q

Haworth projection

Anomeric carbons

A

A Haworth projection is a common way of representing the cyclic structure of monosaccharides with a simple three-dimensional perspective.

Anomeric carbons are the new asymmetric carbons that are created by cyclic hemi-acetal formation in sugars

A. If the newly formed, anomeric carbon is hydroxyl trans to Carbon 6, it is an alpha-anomer

B. If the newly formed, anomeric hydroxyl is cis to carbon 6, it is a beta-anomer (This form is more stable for beta D Glucopyranose)

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16
Q

α-1,6’ glycoside bond

A

α-1,6’ glycoside bond

Amylopectin is a branched polymer of alpha-D-glucose with alpha-1,4-glycosidic linkages: α-1,6 branch points are located 25 to 30 sugar residues apart.

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17
Q
  1. What structural features do carbohydrates have in common? What is a monosaccharide? a disaccharide? a polysaccharide?
A
  1. What structural features do carbohydrates have in common? What is a monosaccharide? a disaccharide? a polysaccharide?

carbohydrates

are poly hydroxylated aldehydes, poly-hydroxylated ketones or compounds that can be hydrolyzed into these compounds. Carbohydrates:

A. are the major source of E in most organisms

B. are structural components of cell walls and membranes

C. serve as metabolic intermediates

D. are a portion of the nucleotides that form RNA, DNA and ATP

E. play a role in cellular communication, lubrication and immunity

Monosaccharides are simple sugars. EX: fructose, glucose

Disaccharides are a combination of two monosaccharides joined by a glycoside linkage. EX: lactose

Oligosaccharides are polymers composed of three to ten monosaccharides

Polysaccharides are polymers of more than 10 monosaccharides

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18
Q
  1. What kind of glycosidic bond is found in trehalose?
A
  1. What kind of glycosidic bond is found in trehalose?

trehalose is an α1,1 glycoside. the anomeric carbon for each unit (each labeled carbon 1) of the disaccharide is involved in an acetal bond so it is not a reducing sugar.

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19
Q
  1. Be able to distinguish D and L isomers on both Fischer and Haworth projections.
A
  1. Be able to distinguish D and L isomers on both Fischer and Haworth projections.

Picture: Fischer & Haworth projections

D: OH is on the right side @ the assymetric C that is furthest removed from the adlehyde or ketone group

L: On the left side

For Haworth projections: ?

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20
Q

mutarotation

A

mutarotation is the process by which anomers slowly interconvert in solution. Cyclic forms of monosaccharides occur much more frequently in solution and in solids than do open chair sugar structures.

21
Q
  1. What are the elements of carbohydrate digestion. (enzymes, their locations in GI tract, substrate types and products)
A
  1. What are the elements of carbohydrate digestion. (enzymes, their locations in GI tract, substrate types and products)

Note: Endoglycosidase –> Cuts bonds!

22
Q
  1. The following structure is that of gulose. Is it an a or b anomer? Is it a D sugar or an L sugar?
A
  1. The following structure is that of gulose. Is it an a or b anomer? Is it a D sugar or an L sugar?

The gulose isomer shown is a β–anomer (the anomeric hydroxyl is on the same side of the ring as the methoxyl group, i.e., carbon 6.

It is a D-sugar because the methoxyl group (carbon 6) is in the equatorial position. The group would have been axial or down if this had been L-gulose

23
Q
  1. Draw D-ribulose in its five-membered cyclic b hemiacetal form (see open chain structure in Q9 above).
A
  1. Draw D-ribulose in its five-membered cyclic b hemiacetal form (see open chain structure in Q9 above).

ribulose is a pentose; the carbon 5 alcohol group is used to form the ring

24
Q
  1. Name the three most common disaccharides.
A
  1. Name the three most common disaccharides.

Sucrose, lactose, and maltose are the most common disaccharides

25
Q

Be able to convert tetrahedral into a fischer projection

A

Be able to convert tetrahedral into a fischer projection

26
Q
  1. Draw b-D-galactose (anomeric form).
A
  1. Draw b-D-galactose (anomeric form).

locate the anomeric carbon as the ring member attached to 2 oxygen atoms. The hydroxyl on this carbon is in the beta position/above the ring in the Haworth projection

27
Q
  1. What is a “reducing sugar”? Why is D-galactose a reducing sugar? Is trehalose a reducing sugar? (see question 15 below for structure)
A
  1. What is a “reducing sugar”? Why is D-galactose a reducing sugar? Is trehalose a reducing sugar? (see question 15 below for structure)

D-galactose is a reducing sugar because the open chain form contains an aldehyde functional group,

a requirement for a sugar to be oxidized to an acid.

A reducing sugar is any sugar that either has an aldehyde group or is capable of forming one in solution through isomerism. The aldehyde functional group allows the sugar to act as a reducing agent, for example in the Tollens’ test or Benedict’s reagent, or the Maillard reaction, important in the browning of many foods.

28
Q

Regulation of Enzyme Activity

>>>>Allosteric Regulation and control of metabolic pathways

What are other types of regulation of enzyme activity?

A

Regulation of Enzyme Activity

>>>>Allosteric Regulation and control of metabolic pathways

What are other types of regulation of enzyme activity?

A. Product inhibition and negative feedback regulation

B. Allosteric Regulation and control of metabolic pathways

C. Covalent Modification

Allosteric Regulation and control of metabolic pathways

  1. means “other site”/other than substrate binding site
  2. allosteric enzymes catalyze critical step(s) in a metabolic pathway
  3. Allosteric enzymes display sigmoidal kinetics
  4. binding of the substrate stabilizes the active (R) confirmation so that additional substrate binds more readily
  5. activators and inhibitors bind to “other sites”
  6. activator binds R state leading to more hyperbolic kinetics, lower Km than normal
  7. inhibitor stabilizes T state (inactive) with more sigmoidal kinetics, higher Km than normal

T - doesn’t bind

R - will bind W/ inhibitor, 1/2 Vmax shifts to the right (binding)

Activator is more myoglobin like, and more hyperbolic - rapid binding even @ low level

29
Q

Regulation of Enzyme Activity:

Covalent modification

A

Regulation of Enzyme Activity:

Covalent modification

30
Q

common pathway

  1. step 1
  2. step 2
  3. step 3
  4. common why?
A

common pathway

  1. step 1
  2. step 2
  3. step 3
  4. common why?
  5. prothrombin activated into thrombin
  6. thrombin activates fibrin
  7. XIIIa (activated by thrombin) tightens the fibrin clot
  8. both intrinsic and extrinsic pathways lead to this; it completes the clot cascade
31
Q

How are the extrinsic and intrinsic clotting pathways activated?

A

How are the extrinsic and intrinsic clotting pathways activated?

Extrinsic Pathway

  1. trauma; deep damage (activated by factors released by damaged tissues)
  2. much quicker and less complex (10-15 seconds)
  3. because factors needed to initiate it are from outside the blood
  4. seems to be more important than intrisnic pathway

Intrinsic pathway

  1. contact activation: it is initiated by exposure of the blood to foreign surfaces or to damaged internal tissues (probably due to inflamation)
  2. clotting is slow and complex (60 seconds or more)
32
Q

What is the importance of factors VIIIa & Va?

A

What is the importance of factors VIIIa & Va?

they are both important accelerators of the zymogenic conversion of prothrombin (II) into thrombin (IIa):

  1. VIIIa binds with protease IXa in intrinsic pathway to accelerate activation of protease Xa
  2. Va works with protease Xa to accelerate conversion of prothrmobin into thrombin
33
Q

What is the role of Vitamin K in the blood clotting process?

A

What is the role of Vitamin K in the blood clotting process?

it facilitates the activation of various clotting factors in the liver.

(in order to be bioactive, glutamate residues on Factors II, VII, IX, and X as well as protein C and protein S must be carboxylated at the gamma carbon.

Carboxylation requires vitamin K as a cofactor

34
Q

Where does carboxylation of the clotting factors occur? Why is carboxylation important for the clotting process?

A

Where does carboxylation of the clotting factors occur? Why is carboxylation important for the clotting process?

Carboxylation of the clotting factors occurs in the liver.

Carboxylation of the clotting factors allows the clotting factors to “chelate,” or bind, Ca++, which allows them to anchor onto the surfaces of platelets, which is where they can be activated

(due to the resulting terminal positive charge, this allows a “calcium bridge” to form between the clotting factor and the negative surface of platelet membranes)

35
Q

What is fibrinogen? Why is the removal of fibrinopeptides central to clot formation?

A

What is fibrinogen? Why is the removal of fibrinopeptides central to clot formation?

  • Fibrinogen is a protein composed of two triple helices joined at their N-termini
  • central region contains negatively charged fibrinopeptides

Thrombin hydrolizes the fibrinopeptides (peptide cleavage) at the central region of the fibrinogen, releasing the fibrin monomer.

The removal of fibrinopeptides is central to clot formation because their removal exposes the central fibrin regions, which are complementary to other fibrin monomers. this allows them to assemble in a staggered array to form a soft fibrin clot layered over the platelet plug in the vasculature

the clot is then stabilized by cross links catalyzed by factor XIIIa

36
Q

How is clotting terminated?

A

How is clotting terminated?

2 ways clotting is terminated:

  1. activated clotting factors don’t last long
  2. inhibitors stop it directly

>>>1. alpha-antithrombin forms irreversible complexes thereby inhibiting several clotting factors

  1. heparin is a negatively charged polysaccharide found in mast cells near the walls of blood vessels and endothelial surfaces. It is necessary for antithrombin activation.
  2. Thrombomodulin binds thrombin, preventing clot encroachment on areas with an intact endothelial lining.
  3. Activated protein C inhibits formation of thrombin by blocking formation of VIIIa and Va
37
Q

How are clots dissolved?

A

How are clots dissolved?

Clot dissolution

A. When a clot forms, circulating plasminogen, produced by the kidneys, is trapped in the clot. About 2 days after a clot forms, injured tissues and vascular endothelium produce tissue plasminogen activator (t-PA, which is an enzyme that converts plasminogen into plasmin

B. Plasmin hydrolyzes peptide bonds in the triple helical coils of fibrin clots. Plasmin also destroys fibrinogen. The meshwork and its precursor protein are converted into soluble components.

>>PLasmin is a protesase that breaks down fibrin into pieces. They are now soluble, and can move out of the blood.

38
Q

Explain how heparin and citrate work as anticoagulants

A

Explain how heparin and citrate work as anticoagulants

Heparin: it binds to antithrombin and induces a conformational change that activates antithrombin, making it highly reactive.

this drastically increases (up to 1000x) the rate of inactivation of thrombin and Xa

Citrate: it is a calcium chelator (it binds to calcium in the blood, reducing the amount available to the extent that calcium-bridges can not be formed between clot factors and platelet membranes. thus, it prevents the activation of clot factors)

39
Q

What factors are defective in hemophilia A and hemophilia B? What is the clinical consequence of these defects?

A

What factors are defective in hemophilia A and hemophilia B? What is the clinical consequence of these defects?

  1. Hemophilia A: Clotting factor VIII
  2. Hemophilia B: Clotting factor IX
  3. Clinical consequence: blood cannot clot properly to control bleeding (therefore, great difficulty stopping bleeding once it stops; clots do not form)
40
Q

Vitamin K Antagonists

A

Vitamin K Antagonists

  • warfarin
  • dicoumarol
  • many herbs
41
Q

The overall plan of carbohydrate digestion

A

The overall plan of carbohydrate digestion

>>3 segments of small intestines: 1. duodenum, 2. Jejunum (longer), 3. Ileum

II. Carbohydrate digestion: enzymes in the digestive tract convert complex carbohydrates into monosaccharides.

A. Oral carbohydrate digestion

  1. Alpha amylase initiatees hydrolysis of starch and glycogen
  2. salivary mucin acts as a lubricating agent
  3. gastric acid terminates activity of salivary alpha-amylase

B. Intestinal Digestion of carbohydrates

C. Intestinal and cellular absorption of monosaccharides

42
Q

Carbohydrate digestion: enzymes in the digestive tract convert complex carbohydrates into monosaccharides

A

Carbohydrate digestion: enzymes in the digestive tract convert complex carbohydrates into monosaccharides

B. Intestinal digestion of carbohydrates

  1. Pancreatic alpha amylase continues carbohydrate digestion in the duodenum
  2. intestinal brush border enzymes complete digestion of carbohydrates

(endoglycosidase cuts bonds…)

A. sucrase-isomerase is an endoglycosidase that cleaves alpha 1,4 and alpha 1,6 glycoside bonds as well as alpha 1,2 bonds in sucros. Activity highest in jejunum

B. Glucoamylase breaks alpha 1,4 glycoside bonds. highest activity in ileum

C. Lactase is a beta glycosidase; highest level in jejunum>>>responsible for hypolactasia (lactos intolerance)

d. Trehalase is an alpha 1,1 glycosidase.

43
Q

Carbohydrate digestion:

C. Intestinal and cellular absorption of monosaccharides

1

A

Carbohydrate digestion:

C. Intestinal and cellular absorption of monosaccharides

  1. Sodium-dependent absorption occurs via secondary active transport
  2. Facilitated diffusion occurs via GLUT (glucose) transporters.

>>Microvilli increase SA of GI tract.

GLUT = glucose transporter, and they differ by Km. This tells us that neurons in the center are very effective at capturing glucose (more than liver because have a higher affinity)

44
Q

Carbohydrate digestion:

C. Intestinal and cellular absorption of monosaccharides

II

A

Carbohydrate digestion:

C. Intestinal and cellular absorption of monosaccharides

  1. Sodium-dependent absorption occurs via secondary active transport
  2. Facilitated diffusion occurs via GLUT (glucose) transporters.

>>Microvilli increase SA of GI tract.

Picture: Na dependent transporter - way of getting glucose/galactose in. This transporter is an active mech because it’s going to move Na downhill and glucose uphill. Na will go down its concentration gradient. When 2 materials go in a transporter in opposite directions, this is cotransport. Depends on Na gradient from outside vs. inside. High outside, and low inside the cell. This gradient is maintained by the Na/K ATPase.

45
Q
  1. What is the oxidation step of glycolysis?
A
  1. What is the oxidation step of glycolysis?

The conversion of NADH to NAD+ is an oxidation step. This occurs in anaerobic glycolysis & the conversion of glyceraldehyde 3 phosphate to 1,3 bisphosphoglycerate (w/ the help of glyceraldehyde 3 phosphate dehydrogenase). This last step is also a phosphorylation!

46
Q
  1. Identify the metabolites that connect glycolysis to the Krebs cycle, to glycogen metabolism, to the pentose pathway, to the metabolism of fatty acids and to alanine metabolism?
A
  1. Identify the metabolites that connect glycolysis to the Krebs cycle, to glycogen metabolism, to the pentose pathway, to the metabolism of fatty acids and to alanine metabolism?

Pyruvate - connects glycolysis to anaerobic glycolysis, the Kreb’s cycle, gluconeogenesis, Amino Acid Synthesis

>>>Gluconeogenesis: Gluconeogenesis (GNG) is a metabolic pathway that results in the generation of glucose from non-carbohydrate carbon substrates such as pyruvate,lactate, glycerol, and glucogenic amino acids. While primarily odd-chain fatty acids can be converted into glucose, it is possible for at least some even-chain fatty acids.

Glucose 6 Phosphate (G6P!) connects glycolysis to the pentose pathway, glycogen synthesis, and other pathways.

47
Q
  1. What advantage does the ability to carry out anaerobic glycolysis have for cells? How does anaerobic respiration differ from anaerobic glycolysis?
A
  1. What advantage does the ability to carry out anaerobic glycolysis have for cells? How does anaerobic respiration differ from anaerobic glycolysis?

Anaerobic respiration as compared to fermentation[edit]

Cellular respiration (both aerobic and anaerobic) utilizes highly reduced species such as NADH and FADH2 (for example produced during glycolysis and the citric acid cycle) to establish an electrochemical gradient (often a proton gradient) across a membrane, resulting in an electrical potential or ionconcentration difference across the membrane. The reduced species are oxidized by a series of respiratory integral membrane proteins with sequentially increasing reduction potentials with the final electron acceptor being oxygen (in aerobic respiration) or another species (in anaerobic respiration). The membrane in question is the inner mitochondrial membrane in eukaryotes and the cell membrane in prokaryotes. A proton motive force or pmf drivesprotons down the gradient (across the membrane) through the proton channel of ATP synthase. The resulting current drives ATP synthesis from ADP and inorganic phosphate.

Fermentation in contrast, does not utilize an electrochemical gradient. Fermentation instead only uses substrate-level phosphorylation to produce ATP. The electron acceptor NAD+ is regenerated from NADH formed in oxidative steps of the fermentation pathway by the reduction of oxidized compounds. These oxidized compounds are often formed during the fermentation pathway itself, but may also be external. For example, in homofermentative lactic acid bacteria, NADH formed during the oxidation of glyceraldehyde-3-phosphate is oxidized back to NAD+ by the reduction of pyruvate to lactic acid at a later stage in the pathway. In yeast, acetaldehyde is reduced

48
Q
  1. What hormones and small molecule metabolites regulate glycolysis?
A
  1. What hormones and small molecule metabolites regulate glycolysis?

Magnesium is used in phosphoglycerate mutase

Glucose 6 phosphate inhibits hexokinase

For phosphofractokinase 1: + >> AMP, F2,6BP, F6P

  • >>ATP, cAMP, citrate

Pyruvate kinase: +>>Fructose 1,6 Bisphosphate (feed forward regulator!)

->>ATP, cAMP

49
Q
  1. What are the 3 steps of glycolysis that yield large negative standard free energy changes?
A
  1. What are the 3 steps of glycolysis that yield large negative standard free energy changes?

Three of the steps — the ones with large negative free energy changes — are not in equilibrium and are referred to as irreversible; such steps are often subject to regulation.

D-glucose –(hexokinase/glucokinase)—>glucose 6 phosphate

Fructose 6 phosphate —(phosphofructokinase 1) —>fructose 1,6bisphosphate

phosphoenol pyruvate —-(pyruvate kinase)—>pyruvate