Download BCHM 270 Module 4 Test: Carbohydrates (Saccharides) - Detailed Explanation and Exercises and more Exams Biochemistry in PDF only on Docsity!
BCHM 270 Module 4 Test With
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Carbohydrates (saccharides) - ANSWER - the most abundant compound found in nature and
- have many cellular functions: provide energy in the diet provide a storage form of energy form components of cell membrane Classifications of Monosaccharides - ANSWER - simplest carbohydrate and are single sugar molecules Classified according to:
- the name of carbons they contain 3 carbon: triose 4 carbon: tetrose 5 carbon: pentose 6 carbon: hexose 7 carbon: heptose 9 carbon: nonose
- oxidation state of their carbonyl group
aldoses and ketoses aldose - ANSWER - contain an aldehyde which has a hydrogen, R group, and =O attached to the carbon ketoses - ANSWER contain a keto group, which has 2 other carbons attached to the C and =O Cyclization of Monosaccharides - ANSWER - 99% of monosaccharides exists as a ring or cyclic structure
- to form this ring, the aldehyde or keto group react with alcohol on the same molecule closing the molecule into a ring
- the carbon 1 becomes an "anomeric carbon" which is a stereocentre
- the position of the hydroxyl group on the anomeric carbon determines whether the molecule is in the alpha (up) or beta (down) position stereocentre - ANSWER - also known as a chiral centre
- is an atom with 3 or more different attachments where interchanging 2 of these attachments leads to another stereoisomer polysaccharide formation via the glycosidic bond - ANSWER - monosaccharides can be joined to form larger structures by a glycosidic bond
- ex. disaccharides, oligosaccharides, polysaccharide, homopolysaccharide, heteropolysaccharide disaccharide - ANSWER contains 2 monosaccharides lactose (galactose + glucose)
Isomers, Epimers, and Enantiomers - ANSWER Isomers
- compounds that have the same chemical formula but different structures (ex. glucose and galactose are both C6H12O6) Epimers
- compounds the differ in configuration at one specific carbon atom
- epimers and antiheroes are different types of isomers
- ex. glucose and mannose are C2 epimers, meaning the OH is on opposite sides of carbon 2 Enantiomers
- structures that are mirror images of each other. In humans, most sugars are in the D- conformation but can have D- or L- conformations digestion of carbohydrates - ANSWER - our diet contains very little monosaccharide. Instead, the major carbohydrate dietary components are polysaccharides, such as glycogen from animal sources and starch from plants
- enzymes must hydrolyze the glycosidic bonds to break carbohydrates down to their monosaccharides components
- glycosidases are enzymes that cleave the glycosidic bonds with water to break down the polysaccharides. This is an example of a hydrolysis reaction hydrolysis reaction - ANSWER reaction where a molecule of water is added Carbohydrate digestion of the mouth - ANSWER - during chewing, alpha-amylase is released by the salivary glands and begin to break apart the polysaccharide starch by cleaving an alpha-1,4 bond
- humans only produce alpha-1,4 endoglycosidases which cleave bonds 'within' the sugar polymer
- since there are also alpha-1,6 bonds, alpha amylase produces a mis of smaller branched oligosaccharides cellulose - ANSWER - humans cannot digest cellulose because we lack enzymes to break beta-1,4 bonds between glucose molecules in the polysaccharide carbohydrate digestion in the intestine - ANSWER - in the small intestine, pancreatic alpha-amylase continues digestion, picking up where the salivary enzyme left off
- final carbohydrate digestion is completed by enzymes made in the intestinal mucosal cells lining the jejunum (a component of the small intestine). These enzymes are specific to a particular disaccharide (ex. maltose, sucrase, and lactase) and are referred to as bush border enzymes
- intestinal absorption occurs in the duodenum and upper jejunum, where the monosaccharides are passed into the circulatory system to be used in the body digestive enzyme deficiencies - ANSWER - individuals can have a genetic deficiency in the enzymes that catalyze the final cut of dissahcardies in the intestine, resulting in the person being unable to digest the disaccharide (making them intolerant to it)
- dissachardie intolerance can arise from malnutrition, intestinal disease, and drugs that cause damage to the mucosa of the small intestine
- ex. brush border enzymes are lost in people with severe diarrhea, resulting
- Energy Generation Phase
- final 5 reactions
- 2 NADH produced
- 4 ATP produced energy investment phase - ANSWER - breakdown of glucose occurs when the cell is in an energy poor state
- so the cell must first invest 2 ATP molecules, and there are 2 regulatory enzymes in this process STEP 1: Phosphorylation of Glucose STEP 3: Phosphorylation of Fructose 6-P energy investment phase. Step 1: Phosphorylation of Glucose - ANSWER REGULATORY ENERGY IN
- phosphate from ATP-> ADP+P is added to glucose to form glucose 6-phosphate (G6P)
- the enzyme that catalyzes this reaction is hexokinase in all cells expect the liver and pancreas which are catalyzed by glucokinase
- adding the phosphate helps to trap the glucose within the cell, ensuring its use in glycolysis. This phosphate can be used later on to convert ADP-> ATP
- this reaction is the first step of the 3 regulatory steps within glycolysis
- hexokinase/glucokinase are inhibited by their product of G6P
Hexokinase - ANSWER - expressed in all cells except liver and pancreas
- has a low Km (high affinity) for glucose, so it is efficient
- has a low Vmax, which prevents cells from consuming all their cellular phosphate to phosphorylate sugars
glucokinase - ANSWER - present in liver and pancrea
- has a high km (low affinity) and is most active when glucose concentrations are high in the body (ex. after a carbohydrate rich meal)
- has a high Vmax to allow the liver to quickly remove glucose from circulation after a meal
energy investment phase. Step 3: Phosphorylation of Fructose 6-P - ANSWER REGULATORY ENERGY IN
- the 3rd step of glycolysis is similar to the first, in that a phosphate from ATP is transferred to fructose 6-P to form fructose 1,6-bisphosphate
- enzyme: phosphofructokinase (PFK-1)
- this is the 2nd regulation;atory step in glycolysis, and is the commitment step (after this, the molecule HAS to undergo glycolysis)
- during step 6, both molecules of G3P are oxidized to form 1,3-bisphosphoglycerate
- enzyme: glyceraldehyde 3-P dehydrogenase
- this reaction oxidizes the aldehyde by attaching a phosphate (that doesn't come from ATP)
- in this reaction, 1 NAD+ is reduced to NADH for each glyceraldehyde 3-P, producing 2 NADH total. NADH is used in the ETC to convert to ATP
energy generation phase: Step 7. Dephosphorylation of 1,3-bisphopholycerate - ANSWER ENERGY OUT
- a high energy phosphate is transferred from 1,3-biphosphoglycerate to ADP, forming a 3-phosphoglycerate and ATP
- enzyme: phosphoglycerate kinase
- 1 ATP is produced from each molecule in this step (therefore 2 ATP)
- at this point, the cell has paid off the 2 ATP debt from the investment phase
energy generation phase: step 10. formation of pyruvate - ANSWER REGULATORY ENERGY OUT
- phosphate transferred from phosphoenolpyruvate to ADP.
- This forms 2 molecules of ATP and 2 pyruvate
- 3rd irreversible and regulatory step
- enzyme: pyruvate kinase
- pyruvate kinase is activated by the product of PFK-1 (fructose 1,6-bisphosphate) in a feed-forward mechanism
aerobic vs anaerobic glycolysis - ANSWER - for glycolysis to occur, the cell needs a supply of NAD+ (which is converted to NADH in step 6)
- but the cell only has a bit of NAD+ in the cell so it needs to be converted BACK into NADH for glycolysis to continue
- this can occur through 2 ways:
aerobic conditions
anaerobic conditions
- NADH-dependent conversion of pyruvate to lactate
aerobic glycolysis - ANSWER - aerobic glycolysis occurs when oxygen is
- lactate accumulates in the muscles, lowering the pH, which results in cramps
- eventually, the lactate is diffused to the bloodstream and is converted to glucose by gluconeogensis in the liver
hormonal regulation of glycolysis - ANSWER hormones (mainly insulin and glucagon) influence the amounts of the 3 key enzymes of glycolysis made by cells
hormonal regulation of glycolysis: well-fed state - ANSWER - being in a well-fed state causes insulin levels in blood to side
- insulin causes an increase in the levels of glucokinase, PFK, and PK by increasing their rates of synthesis
- this increases the rate of glycolysis, as the cells state favours the conversion of glucose to pyruvate
hormonal regulation of glycolysis: fasting state - ANSWER - being in the stating state causes glucagon levels in the blood to rise
- glucagon decreases the levels of glucokinase, PFK, and PK by decreasing their synthesis. This decreases the rate of glycolysis as the cell's state does not favour the conversion of glucose to pyruvate
defective pyruvate kinase activity - ANSWER different types of pyruvate
kinase defects:
- changes in Km or Vmas
- alteration in enzyme gene expression
- abnormal respond to the activator fructose 1,6-bisphosphate
PK Defiency leads to hemolytic anemia - ANSWER - since red blood cells lack mitochondria, they are completely dependent on anaerobic glycolysis for ATP synthesis.
- Genetic defects in PK leads to hemolytic anemia (lowered ability for blood to carry oxygen because the red blood cells are removed before they were ready)
production of acetyl-CoA from pyruvate - ANSWER REGULATED
- once produced by glycolysis in the cytosol, pyruvate is transported to the mitochondrial matrix by a specific transporter molecule
- enzyme: pyruvate dehydrogenase (PDH) complex catalyzes the oxidative decarboxylation (removes a CO2) of pyruvate, and transfers the acetyl group to coenzyme A
- the PDH complex is highly regulated
- the irreversible conversion of pyruvate into acetyl-CoA is the link between glycolysis and the TCA cycle. The reaction Is critical in metabolism because it commits the carbons from glycolysis to oxidation by the TCA cycle or synthesis of lipids
Active abd Inactive forms of PDH - ANSWER - the next level of PDH complex regulation is covalent modification
Active
- active when unphosphorylated
- this is favoured by the energy poor state and the phosphate group removed by PDH phosphate
Inactive
- inactive when phosphorylated
- this is favoured by the energy rich state and the phosphate group added by PDH Kinase
PDH Kinase - ANSWER - inactivates the PDH by phosphorylation
- PDH kinase activity is increased by ATP, acetyl-CoA, and NADH
- PDH kinase activity is allosterically decreased by pyruvate, ADP, and NAD+
PDH Phosphatase - ANSWER - PDH phosphatase is activated by the PDH complex by dephosphorylating it
- PDH phosphatase is active by CA2+, which is produced during muscle
contraction
glucose homeostasis in the body of the cell - ANSWER - a constant blood glucose supply is essential for human life
- glucose is a preferred source of energy for the brain, in cells lack mitochondria (i.e red blood cells), and in excersizing muscle where it is the substrate for anaerobic glycolysis
blood glucose is obtained from 3 main sources - ANSWER diet
- not always reliable gluconeogenesis
- slow in responding to falling glucose levels glucagon breaks down
- glucose is stored as glycogen
depletion of glycogen stores results in glucose synthesis by gluconeogenesis using amino acids from body's protein as substrates
glucose metabolism - ANSWER - glucose metabolism is the net production of glucose (gluconeogenesis) and breakdown of glucose (glycolysis)
The Cori Cycle - ANSWER - the cori cycle describes the metabolic pathway where lactate produced by anaerobic glycolysis in excersizing muscle cells is transported to the liver to be converted back to glucose
STEPS
- glucose from the blood is taken up by excersizing skeletal muscle
- anaerobic glycolysis converts glucose to lactate
- lactate diffuses into the blood and is taken up by the liver
- liver synthesizes glucose from lactate via gluconeogensis
- glucose is released back into the blood
reactions unique to gluconeogensis - ANSWER - gluconeogensis is made up of 11 reactions. 7 are the reversible reactions of glycolysis. 4 are reactions unique to gluconeogensis (to bypass the 3 irreversible steps of glycolysis
- conversion of pyruvate to oxaloacetate
- formation of phosphoenolpyruvate
- dephosphorylation of fructose 1,6-bisphosphate
- dephosphorylation of glucose 6-phosphate
unique reaction 1: carboxylation of pyruvate - ANSWER - pyruvate is carbonylated into oxaloacetate in the mitochondria of liver and kidney cells
- enzyme: pyruvate carboxylase
- the coenzyme is biotin
- step 1: ATP-> ADP powers the formation of the enzyme-biotin-CO intermediate
- step 2: the CO2 is transferred to oxaloacetate and releases the enzyme-biotin complex
regulation of pyruvate carboxylase - ANSWER - pyruvate carboxylase is activated by acetyl-CoA, because large amounts of acetyl-CoA are produced by the degradation of fat (a process called beta oxidation)
- fats are the primary fuel source for the liver when glucose levels are low, generating an increased amount of acetyl-CoA, and stimulating gluconeogenesis through this enzyme (pyruvate carboxylase)
- when acetyl-CoA levels are low, the pyruvate carboxylase is less active, favouring glycolysis and the TCA cycle
- pyruvate carboxylase is inhibited by ADP, ensuring that when ATP levels are low, resources are used to produce ATP.
transport of oxalocetate to cytosol - ANSWER - the oxaloacetate produced by pyruvate crboxulase in the mitochondria cannot pass the mitochondrial membrane; it is this reduced to malate, and NADH is oxidized to NAD+
