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OSU COD 2022- D1 Fall 2018 > Biochem Exam 1 > Flashcards

Biochem Exam 1 Flashcards

1
Q

Name the Amino Acids with Nonpolar, Aliphatic R groups and describe their unique characteristic.

A

1.) glycine 2.) alanine 3.) valine 4.) leucine 5.) isoleucine

R groups consist of CH groups.

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2
Q

Name the Amino Acids with Nonpolar, Aromatic R groups and describe their unique characteristic.

A

1.) Phenylalanine 2.) Tyrosine 3.) Tryptophan

All have aromatic rings.

Tyrosine has OH group

Tryptophan has an NH group

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3
Q

Name the Amino Acids with Uncharged, Polar R groups and describe their unique characteristic.

A

1.) threonine 2.) serine 3.) asparagine 4.) glutamine

All have an OH or C=ONH2 attached to their R groups.

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4
Q

Name the Amino Acids with Acidic R groups and describe their unique characteristic.

A

1.) aspartic acid 2.) glutamic acid

Both have COOH group attached to R group.

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5
Q

Name the Amino Acids with Basic R groups and describe their unique characteristic.

A

1.) histidine 2.) lysine 3.) arginine

All have NH2 or NH3 attached to their R group.

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6
Q

Name the cyclic imino acid and describe some of its unique characteristics.

A

proline

  • Proline does not form a conventional peptide bond
  • Very constrained compared to other aa
  • The sidechain is aliphatic with no functional groups
  • The sidechain covalently bonds to the amide nitrogen eliminating the opportunity for hydrogen bonding at that position
  • Its α-imino group is bonded to its side chain
  • hydrophobic
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7
Q

Name the Amino Acids with sulfur-containing R groups:

A

1.) cysteine 2.) methionine

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8
Q

Amino Acids with aliphatic hydroxyl R groups:

A

1.) serine 2.) threonine

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9
Q

Describe primary protein structure:

A

Primary protein structure is sequence of a chain of amino acids (often thought of as beads on a string).

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10
Q

Describe secondary protein strucutre:

A

2 types of secondary protein structure: 1.) Alpha Helix-hydrogen bonds between the “backbone” amide (NH) and carboxyl (C=O) groups stabilize the alpha helix. There are 3.6 amino acids per turn. 2.) Beta Sheet- the “backbone of the polypeptide chain is extended into a zigzag structure. When the zigzag polypeptide chains are arranged side by side, they form a structure resembling a series of pleats.

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11
Q

Amino Acid with an Amide Derivative R group:

A

1.) asparagine 2.) glutamine

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12
Q

Describe tertiary protein structure:

A

-Tertiary structure refers to the overall 3D structure of the protein. -Amino Acid side chain interactions contribute to protein folding seen in tertiary structure. -4 types of side chain interactions: 1.) disulfide bonds 2.) hydrogen bonds 3.) salt bridges 4.) hydrophobic interactions

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13
Q

Describe quaternary protein structure:

A

Can be composed of multiple polypeptide chains. Quaternary structure refers to the number and arrangement of protein subunits.

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14
Q

Define ionization state:

A

-The ionization state of an amino acid depends on pH. -Amino acids are amphoteric (can act as a base and as an acid). -At low pH, the carboxyl group accepts a proton and becomes uncharged, and the overall charge on the molecule is positive -At high pH, the amino group loses its proton and becomes uncharged, and the overall charge on the molecule is negative +H3N—CH2—COOH ↔ +H3N—CH2—COO- ↔ H2N—CH2–COO-

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15
Q

Define the Henderson Hasselbalch equation:

A
  • Describes the derivation of pH as a measure of acidity in biological and chemical systems
  • useful for estimating the pH of a buffer solution
  • widely used to calculate the isoelectric point of proteins (point at which protein neither accept nor yield proton)
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16
Q

Define zwitterion:

A
  • a molecule or ion that has both positively charged and negatively charged groups
  • in aqueous solution, the amino acid exists as a dipolar molecule
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17
Q

Define pI

A
  • pI is the Isoelectric point.
  • It is the pH at which a protein neither accepts nor yields a proton. (The protein has no net electrical charge.)
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18
Q

Define peptide:

A

-consisting of two or more amino acids linked in a chain, the carboxyl group of each acid being joined to the amino group of the next

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19
Q

Describe Homocysteinuria

A
  • disorder of methionine metabolism
  • leading to an abnormal accumulation (excess) of homocysteine and its metabolites in blood and urine
  • clinical presentation: downward dislocation of the lens (ectopia lentis), long limbs, mental retardation, and/or seizures due to low abundance of fibrin protein fibers in lens (disulfide bonds required for lens structure).
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20
Q

Describe Sulfite Oxidase Deficiency

A

-Sulfite oxidase is required to metabolize the sulfur-containing amino acids cysteine and methionine in foods. -Lack of functional sulfite oxidase causes a disease known as sulfite oxidase deficiency. -This rare but fatal disease causes neurological disorders, mental retardation, physical deformities, the degradation of the brain, and death. -Reasons for the lack of functional sulfite oxidase include a genetic defect

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21
Q

Describe Marfan Syndrome

A

-involves a mutation to the gene that makes fibrillin, which results in abnormal connective tissue -specifically, it can result in lens dislocation, where the lens is shifted out of its normal position. This occurs because of weakness in the ciliary zonules, the connective tissue strands which suspend the lens within the eye.

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22
Q

Describe Hyperbilirubinemia

A
  • Heme oxygenase converts heme to biliverdin, which is ultimately converted to bilirubin.
  • If heme oxygenase is overactive you get higher levels of bilirubin resulting in discoloration of teeth
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23
Q

What is the approximate pH of an amino acid that has 2 ionizable groups and is fully protonated?

A

pH<2.4

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24
Q

What is the approximate pH of an amino acid that has 2 ionizable groups and only one group is protonated?

A

between 2.4 and 9.8

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25
Q

What is the approximate pH of an amino acid that has 2 ionizable groups and neither group is protonated?

A

pH>9.8

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26
Q

The isoelectric points for most amino acids is between 5 and about 6.5. There are 5 whose pI fall outside of this range. What are the five and what are their corresponding pI?

A
  1. ) Arginine at 10.76
  2. ) Aspartic Acid at 2.77
  3. ) Glutamic Acid at 3.22
  4. ) Lysine at 9.74
  5. ) Histidine at 7.59
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27
Q

Describe Solubility-Based Purification Methods

A

Salting in- increasing the ionic strength of a solution so that the protein of interest stays in solution. (solubility increases)

Salting out- increasing the ionic strength of a solution so that the protein of interest crashes out of solution. (solubility decreases)

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28
Q

Describe purification of proteins by chromatography

A

Chromatography- purifies and concentrates biomolecules (proteins) based on chemical or physical differences

  1. ) Size- Size Exlusion chromatography
  2. ) Surface Charge-Ion Exchange chromatography
  3. ) Biorecognition (ligand specificity)- Affinity chromatography
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29
Q

Describe Dialysis and Ultrafiltration

A

Proteins of interest is in solution. The membrane of the dialysis tubing has molecular weight cut off. Choose a size of membrane based on what your are trying to accomplish. EX if have 100 kilodalton protein, choose 50 kilodalton membrane. Smaller molecules move in and out of membrane, protein is trapped inside of membrane. Used for buffer exchange and purification

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30
Q

Describe Size Exclusion Chromatography

A

Have resin beads of specific size in column. Pour solution containing protein of interest through the column. Small molecules can slowly travel through beads. Larger molecules (proteins) are too large to move through the beads so they have to move around beads. The result is that larger molecules moves through column faster and elude first. Get a buffer exchange solution.

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31
Q

Describe Ion Exhange Chromatography

A

With a cation exchanger (Resin), has ligand attached that is negatively charged and binds with the positively charged protein

Anion exchanger (Resin), has ligand attached that is positively charged and binds with the negatively charged protein

DEAE (anion exchanger, positively charged). If solution is at pH 7.5, albumin has a negative charge so it will be retained, but immunoglobulin has positive charge at pH 7.5 so it will elude out.

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32
Q

Describe Affinity Chromatography (AC)

A

In AC, a specific interaction of substrate with ligand occurs. An antibody is attached to the resin in the column. When you pour the protein solution through the column, the protein of interest will bind to the resin. Used to purify IgG from blood.

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33
Q

Explain an example of pI protein purification.

A

Isoelectric Focusing (IEF)

IEF Gels have pH gradient.

A protein that is in a pH region below its pI (protein has a positive charge) will migrate towards the cathode (left to right) until it stops at its pI.

A protein that is in a pH region above its pI (protein has a negative charge) it will migrate towards the anode (right to left) until it stop at its pI.

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34
Q

What are some ways to characterize proteins by separation methods?

A

1.) Once purified for determination for aa composition, the protein is subjected to hydrolysis

  • 6 M HCl
  • 110 ºC
  • 24-48 hours

Chromatogram from an aa analysis by cation exchange chromatography shows relative fluorescence of different amino acids.

  1. ) Edman degradation - N-terminal residue is labeled and cleaved from the peptide sequentially without disrupting the peptide bonds between other amino acid residues
  2. ) Mass Spectrometry- Can do this with mix of proteins or purified. Chop up protein with protease, get peptides, put in liquid chromatography column, separates peptides out. Mass spectrometer will give you intact mass, this will tell you the mass of your peptide which gives clues as to what peptide it might be. Peptide gets fragmented again and get msms spectrum. This is a fragmentation of origin peptide.
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35
Q

What is botulinum toxin and what does it do?

A
  • Botulinum toxin is a proteinaceous toxin produced by the microorganism clostridium botulinum
  • it is internalized and in acid environments the toxic and non-toxic components separate
  • Toxic fragments prevent release of Ach neurotransmitter and cleaves the proteins involved in docking of neurotransmitter vesicles.
  • This causes FLACCID (sagging) PARALYSIS of muscles.
  • Cranial nerves affected first (double vision, difficulty swallowing) then
  • Paralysis descends resulting in respiratory failure
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36
Q

Explain Carbohydrate Chirality

A
  • Numbering of the carbons begins from the end containing the aldehyde or ketone functional group
  • D or L assignment is based on the configuration around the highest numbered asymmetric center (anomeric carbon) (farthest from the aldehyde or keto group)
  • If the hydroxyl (OH) group in the projection formula points to the right, it is defined as a member of the D-family. A left-directed hydroxyl group (the mirror image) represents the L-family
  • NEARLY ALL SUGARS FOUND IN THE BODY HAVE THE D CONFIGURATION
  • AMINO ACIDS HAVE THE L CONFIGURATION
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37
Q

Of the 8 D-aldohexoses, only 3 are found in significant amounts in the body:

A
  1. ) glucos (blood sugar)
  2. ) Mannose
  3. ) Galactose
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38
Q

_______ is the only ketohexose present in siginificant concentration in our diet or in the body

A

fructose

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39
Q

What is the most aboundant carbohydrate in nature?

A

Cellulose

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40
Q

What is the difference between starch and glycogen?

A
  • Starch = amylose + amylopectin
  • It is the most common storage polysaccharide in PLANTS
  • Glucose storage in polymeric form minimizes osmotic effects
  • Starch adopts a helical form in water
  • Glycogen is the animal equivalent to starch.
  • It has more alpha (1-6) brances than starch
  • It promotes rapid glucose release during exercise
  • It is stored in muscle and liver cells as high MW granules
  • It is hydrolyzed by amylases and other enzymes
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41
Q

What is an amino sugar?

A
  • A sugar molecule in which a hydroxyl group has been replaced with an amine group.
  • More than 60 amino sugars are known, with one of the most abundant being N-Acetyl-d-glucosamine, which is the main component of chitin.
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42
Q

Do lipids or carbohydrates yield more energy? Why?

A
  • gram for gram, fats provide more energy than carbohydrates
  • The reason for this is the amount of oxidation that takes place as these compounds are converted to carbon dioxide and water.
  • Carbon for carbon, fats require more oxidation to become CO2 and H2O than do carbohydrates
  • More energy is present in fats - fatty acids are more “reduced” than carbohydrates
  • Fatty acids are nonpolar and not hydrated by water
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43
Q

How many stereoisomers (D and L) can be made from aldoses containing XX carbons with XX asymmetric centers?

A
  • The number of stereoisomers is 2n (n = number of asymmetric centers).
  • 6-C aldoses have 4 asymmetric centers for a total of 16 stereoisomers (8 D-sugars and 8 L-sugars).
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44
Q

Why has glucose been evolutionarily selected as blood sugar?

A
  • Glucose exists largely in nonreactive, inert, cyclic hemiacetal conformations (>99.99% in aqueous solution at pH 7.4 and 37°C)
  • Of all the D-sugars in the world, D-glucose exists to the greatest extent in these cyclic conformations
  • Least oxidizable and least reactive with protein
  • The relative chemical inertness of glucose is the reason for its evolutionary selection as blood sugar.
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45
Q

Explain the relationship between sugars and blood types?

A
  • The blood types are named after antigens that are found on the surface of of the red blood cells, and these antigens are simple chains of sugars.
  • The types of oligosaccharides present on the surface of the red blood cells determine a person’s blood type: if only the 0-type antigen is present, the blood type is 0, if only the antigen A or B is found, the blood is type A or B, respectively, and if both A and B antigens are present, the blood type is AB [1]. The A and B antigens differ only in a sidechain on the terminal sugar.
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46
Q

Describe the complex carbohydrate examples: 1.) homoglycans and 2.) heteroglycans

A
  • Glycans are polysaccharides (carbohydrate with more than 10 sugar molecules)
  • Homoglycans/homopolysaccharides have more than 10 of the same sugar molecule
  • Heteroglycans/heteropolysaccharides have more than 10 of different sugar molecules
47
Q

Examples of homoglycans:

A
  • starch, cellulose, glycogen
  • These are all GLUCOSE homoglycans
48
Q

Examples of heteroglycans:

A
  • gums, mucopolysaccharides, hyaluronate
49
Q

Describe hyaluronate.

A
  • Hyaluronate is a heteroglycan (heteropolysaccharide which has more than 10 sugar different sugar molecules).
  • It is a glucosaminoglycan (Repeating disaccharide of glucuronic acid and N-acetyl-glucosamine).
50
Q

What is the solid storage form of lipids that is found primarily in adipose tissue?

A

Triglycerides (also come in mono or di forms)

51
Q

Explain the relationship between fatty acid structure and physical properties (e.g., melting point, solubility).

A
  • Physical properties (melting point, solubility) are largely determined by the LENGTH and DEGREE OF SATURATION of the hydrocarbon chain.

LENGTH

  • The longer the fatty acid chain length, the poorer the solubility in water
  • Carboxylic acid end is polar, imparting moderate solubility of short-chains (<10 carbons) in water
  • The longer the fatty acid chain, the higher the melting point.

DEGREE OF SATURATION

  • The fewer double bonds in a fatty acid chain, the lower the solubility in water.
  • The more double bonds in fatty acid chain, the lower the melting point.
52
Q

Describe phospholipids.

A

}Major components of cell membranes

  • Polar lipids derived from phosphatidic acid (1,2-diacyl-glycerol-3-phosphate)
  • sn-3 position is occupied by phosphate esterified to amino compound such as:

–Choline

–Serine

–Ethanolamine

–Inositol (Cell signaling)

53
Q

Fatty acids are esterified to what molecule?

A
  • Esterified to glycerol to form triacylglycerol (triglyceride)
  • Advantage to using triacylglycerols as storage fuel rather than polysaccharides: fats yeild more energy than carbohydrates.
54
Q

What are the main complex lipids?

A
  1. ) Sphingolipids
  2. ) Sterols (cholesterol) and hormones (streroid hormones)
  3. ) Terpenes (Mono, tri, tetra)
  4. ) Glycolipids (Lipopolysaccharide (LPS))
  5. ) Prostaglandins and Eicosanoids
  6. ) Leukotrienes
  7. ) Lipoproteins
  8. ) Lipid-Linked Proteins
55
Q

Sphingolipids:

  1. ) building blocks
  2. ) function
A

BUILDING BLOCKS:

  • Sphingosine backbone (not glycerol)
  • fatty acid chain joined via amide linkage (rather than ester linkage with glycerol)

FUNCTION:

  • cell surface recognition (red blood cells)
56
Q

Sterols (cholesterol) and hormones (steroid hormones):

1.) building blocks 2.) function

A

BUILDING BLOCKS

  • four fused rings (Three 6-membered and one 5-membered)

FUNCTION

  • present in membranes of most eukaryotic cells
  • lipid soluble (diffuse freely through the cell membrane into the cytoplasm of target cells)
57
Q

Describe Cholesterol

A
  • It is a sterol so it is found in the membrane of eukaryotic animal cells.
  • only found in animal fat
  • synthesized in liver
  • degradation only occurs in the liver
  • Serves as a component of membranes of cells (increases or moderates membrane fluidity)
  • Precursor to steroid hormones and bile acids
58
Q

Describe Steroid Hormones

A
  • Oxidized derivatives of sterols
  • Increased polarity compared to cholesterol
  • Move through the blood stream attached to protein carriers

Five groups based on binding receptors:

◦Glucocorticoids

◦Mineralocorticoids

◦Androgens

◦Estrogens

◦Progestagens

59
Q

Terpenes: 1.) building blocks 2.) function

A

BUILDING BLOCKS

  • Simple lipids that lack fatty acid component
  • Formed by the combination of 2 or more molecules of 2-methyl-1,3-butadiene (isoprene)

Monoterpene = 2 isoprenes

Sesquiterpene = 3 isoprenes

Diterpene = 4 isoprenes

Triterpene = 2 sesquiterpenes or 6 isoprenes

FUNCTION

  • Monoterpenes = flavors and odors
  • Triterpenes = precursors of cholesterol and other steroids
  • Tetraterpenes = colorful compounds (carotenoids, lycopene)
60
Q

Glycolipids: 1.) builiding blocks 2.) funciton

A

BUILDING BLOCKS

  • core oligosaccharide with long O-antigen side chains
  • Lipid A: hydrophobic anchor

FUNCTION

  • core oligosaccharade = bacterial internalization, initiation of immune resistance
  • long O-antigen side chain = resistance to human serum, antibiotics, detergents
  • Lipid A: hydrophobic anchor = initiate inflammatory response
61
Q

Describe Prostoglandins and Eicosanoids

A
  • Local hormones
  • Key mediators of inflammation
  • Vasodilator effects (blood vessel widening/ contracting)
  • Can be used to treat infants with congenital heart defects
62
Q

Describe Leukotrienes

A
  • Synthesized in neutrophils, monocytes, macrophages, mast cells and keratinocytes
  • Also found in lung, spleen, brain and heart
  • Responsible for inflammation in asthma and bronchitis and anaphylaxis
63
Q

Describe Lipoproteins

A

Phospholipid and cholesterol vesicles

Found in plasma

Transport lipids

  • Cholesterol
  • Triglycerides

Contain apolipoproteins

  • Stabilize the complex
  • Help determine lipoprotein’s “fate”
64
Q

Describe Lipid-linked proteins

A
  • Differeint from lipoproteins
  • Proteins with covalently attached lipids
    • peripheral membrane proteins (with lipids covalently attached)
  • 3 common types:
    • prenylated proteins
    • fatty acylated proteins
    • GPI-linked proteins
65
Q

What is a Lipase?

A

an enzyme that hydrolyzes (breaksdown) fats, releasing fatty acids and glycerol

66
Q

What are the major phospholipids?

A

MAJOR PHOSPHOLIPIDS:

  1. ) Glycerophospholipids
  2. ) Sphingolipids
67
Q

How are glycerophospholipids synthesized?

A

Start with Glycerol-3-phosphate OR DHAP

GLYCEROL-3-PHOSPHATE

  1. ) transfer 1 long-chain fatty acid from fatty acyl-CoA to form lysophaosphatidic acid
  2. ) unsaturated fatty acid added to carbon-2 to form phosphatidic acid
  3. ) PA is converted to diacylglycerol (DAG) by phosphatase

DHAP ACYLATION

  1. ) addition of fatty acid to the 1-hydroxyl group
  2. ) reduction
  3. ) acylation to phosphatidic acid
  4. ) PA is converted to diacylglycerol (DAG) by phosphatase
68
Q

Describe the difference between phospholipids and sphingolipids.

A

1.) They have different backbones:

  • glycerophospholipids = glycerol backbone
  • sphingolipids = sphingosine backbone

2.) Head groups

  • glycerophospholipids = polar head groups
  • sphingolipids = sugar head groups

3.) Saponification (glycerides + NaOH = glycerol + fatty acids) (soaps)

  • glycerophospholipids = saponifiable
  • sphingolipids = non-saponifiable
69
Q

What role do phospholipids and sphingolipids play in blood types?

A
  • Blood groups are determined in part by the type of sugars located on the head groups of phospholipids (phospholipid bilayer)
  • N-acetylgalactosamine group = A antigen
  • Galactose group = B antigen
  • N-acetylgalactosamine + Galactose group = AB antigen
  • No NAG or Gal groups = O antigen
70
Q

What are the building blocks of cholesterol?

A

-Isoprene units

  • Three molecules of acetyl-CoA are converted into 6-carbon mevalonate
  • Two condensation steps
  1. Acetoacetyl-CoA thiolase
  2. HMG-CoA synthase
    * Third rate-limiting reaction
  3. Irreversible formation of mevalonate catalyzed by HMG-CoA reductase
    * –Regulated by insulin, glucagon and cortisol
71
Q

Explain the how cholesterol is regulated.

A

Regulation of intracellular cholesterol concentration:

  1. Reduce activity and expression of HMG-CoA reductase, which decreases cholesterol synthesis
  2. Downregulation of LDL receptors, which limits further cellular entry of cholesterol
  3. Increase cholesterol and phospholipid efflux from cell to apoproteins
  4. Increase the rate of conversion of cholesterol to bile acids so it is excreted.
72
Q

What are bile acids?

A

The liver removes cholesterol as bile acids. There are four primary bile acids and two secondary bile acids.

73
Q

Describe the function and control of bile acids.

A

FUNCTION

  • Bile acids are found in bile.
  • BA assist in digestion of dietary fat
  • BA act as detergents (emulsify ingested lipids)

CONTROL

  • Up to 30g of bile acids pass into intestine each day
  • Total BA pool is only 3g
  • 2% (~0.5g) lost in feces
  • Majority of BA are deconjugated and reabsorbed in enterohepatic circulation 5-10 times per day.
74
Q

What are the differences between primary bile acids and secondary bile acids?

A

PRIMARY bile acids are synthesized in the liver then transported to intestine.

SECONDARY bile acids form in the intestines through the action of anaerobic bacteria.

75
Q

What are the 3 main steroid hormones and how are they secreted?

A

Three groups:

  1. ) Corticosteroids
  2. ) Androgens
  3. ) Estrogens

Secretion occurs in only three organs:

  1. ) Adrenal cortex -corticosteroids
  2. ) Testes in men – androgens
  3. ) Ovaries in women - estrogens

All three organs capable of secreting small amounts of steroids belonging to other groups

76
Q

What are the main lipoprotein classes and what are the differences between them?

A
  1. ) Chylomicrons: main component is triacylglycerides; carries cholesterol and triacylglycerols from liver to adipose tissue; also take in cholesterol from diet
  2. ) VLDL (Very low density lipoproteins): main component is triacylglycerides; chylomicrons are repacked in the liver to form VLDLs
  3. ) IDL (intermediate): triacylglycerides and cholesterol “remnants”
  4. ) LDL (low density lipoproteins): main component is cholesterol; carry lipids around the body
  5. ) HDL (High density lipoprotein): main component is protein; carries phospholipids, cholesterol, triglycerides, etc. back to liver where they are transferred to bile acids and excreted (“Good” lipoproteins)
77
Q

Describe lipoprotein receptors:

A
  • LDL (apoB/E) receptor specifically mediates cellular uptake of intact LDL particles
  • Expression of LDL receptors is regulated by intracellular cholesterol concentration
78
Q

What are the “fuel transport pathway” and the “overflow pathway?”

A
  1. ) VLDLs/chylomicrons transport fuel (dietary triacylglycerols and cholesterol from the intestines and the liver to peripheral tissues
  2. ) The remnants return to the liver.
  3. ) Part of VLDL/chylomicron remnants and IDL are further converted into LDL, which enter the overflow pathway.
  4. ) The overflow pathway: LDL travel from the liver through the peripheral tissues and back to the liver.
79
Q

What is the most critical function of an enzyme?

A

catalyze biological reactions

80
Q

Common features of enzymes:

A
  1. Catalyze only thermodynamically possible reactions
  2. Are not used or changed during the reaction.
  3. Don’t change the position of equilibrium and direction of the reaction
  4. Usually act by forming a transient complex with the reactant, thus stabilizing the transition state
81
Q

Describe the influence of temperature and pH on an enzyme.

A

TEMPERATURE:

  1. ) Enzymes have an optimum temperature at which they function most efficiently (most enzymes have an optimum temp of 37°C)
  2. ) Like all proteins, enzymes denature at high temperature and lose activity (most enzymes will denature above 45-50°C)

pH:

  1. ) Each enzyme has an optimum pH at which they function most efficiently
  2. ) For most enzymes the optimum pH is ~7 (there are exceptions)
  3. ) Every enzyme has an optimum pH because ionizable amino acids, such as histidine, glutamate, and cysteine, participate in the catalytic reactions.
82
Q

Describe the structure of complex enzymes.

A
  1. ) Complex enzymes (holoenzymes) have a protein part and a non protein part.
  2. ) The protein part is the apoenzyme.
  3. ) The nonprotein part is the cofactor.
  4. ) Cofactors can be prosthetic groups (inorganic molecule or atom) or they can be coenzymes (large organic molecules)
83
Q

Function of oxidoreductases

A

catalyzes oxidation/reduction reactions

Ared + Box→Aox + Bred

84
Q

Function of transferases

A

catalyze transfer of funcitonal group

A-B + C→A + B-C

85
Q

Function of hydrolases

A

catalyze reactions by using water to cleave chemical bonds and water is the acceptor of the transferred group

A-B + H2O→A-H + B-OH

86
Q

Function of lyases

A

catalyze lysis of a substrate and often form new chemical bonds (double bond or ring) without using water

X-A-B-Y→A = B + XY

87
Q

Function of isomerases

A

catalyze the formation of a substrate’s isomer.

A ⇌ isoA

88
Q

Function of ligases

A

catalyze the ligation of two substrates (often require chemical energy ATP)

catalyze repair of irregularities or breaks in the backbone of double-stranded DNA molecules

A + B + ATP→A-B + ADP + Pi

89
Q
A
90
Q

Define Active site

A

specific region in the enzyme to which substrate molecule is bound

91
Q

Explain absolute and relative specificities of enzymes.

A
  • The substrate interacts with a small 3D region of the enzyme called the active site. Substrate interacts with only three to five amino acid residues. Residues can be far apart in sequence.
  • Enzyme/active site binds substrates through multiple weak interactions (noncovalent bonds)
  • There are contact and catalytic regions in the active site
  1. Absolute – one enzyme acts only on one substrate (example: urease decomposes only urea; arginase splits only arginine)
  2. Relative – one enzyme acts on different substrates which have the same bond type (example: pepsin splits different proteins)
  3. Stereospecificity – some enzymes can catalyze the transformation only substrates which are in certain geometrical configuration, cis- or trans-
92
Q

Define enzyme activity

A

One international unit (IU) of enzyme catalyzes conversion of 1 µmol of substrate to product per minute

The specific activity of an enzyme is a measure of the number of IU/mg protein

93
Q

What is Km and Vmax?

A

Km is the substrate concentration [S] at which the reaction rate is half of V max (max velocity).

Vo = Vmax[S]/Km + [S]

94
Q

What is the Michaelis Menten equation?

A

Vo = Vmax[S]/Km + [S]

Km is the [S] at which reaction rate is half Vmax

95
Q

What is reversible inhibition?

A

Reversible inhibitors – after combining with enzyme (EI complex is formed) it can rapidly dissociate

Enzyme is inactive only when bound to inhibitor

EI complex is held together by weak, noncovalent interaction.

There are three basic types of reversible inhibition: Competitive, Uncompetitive, Noncompetitive

96
Q

Describe competitive inhibition:

A
  • Inhibitor has a structure similar to the substrate thus can bind to the same active site
  • The enzyme cannot differentiate between the two compounds
  • When inhibitor binds, prevents the substrate from binding
  • Inhibitor can be released by increasing substrate concentration
97
Q

Describe noncompetitive inhibition

A
  • Binds to an enzyme site different from the active site
  • Inhibitor and substrate can bind enzyme at the same time
  • Cannot be overcome by increasing the substrate concentration
98
Q

Describe Uncompetitive Inhibition

A
  • Uncompetitive inhibitors bind to ES complex not to free E
  • This type of inhibition usually only occurs in multisubstrate reactions
99
Q

What is irreversible inhibition?

A
  • After the inhibitor binds there is very slow dissociation of EI complex
  • Tightly bound through covalent or noncovalent interactions

3 types:

  1. ) group-specific reagents
  2. ) substrate analogs
  3. ) suicide inhibitors
100
Q

Describe group specific reagents

A
  • React with specific R groups of amino acids
  • It comes in and forms covalent bond and makes it permanently inactive
101
Q

Describe substrate analogs

A
  • structurally similar to the substrate for the enzyme
  • covalently modify active site residues
102
Q

Describe suicide inhibitors

A
  • Inhibitor binds as a substrate and is initially processed by the normal catalytic mechanism
  • It then generates a chemically reactive intermediate that inactivates the enzyme through covalent modification
  • Suicide because enzyme participates in its own irreversible inhibition
103
Q

List the methods of regulation of enzyme activity

A
  • Allosteric control
  • Reversible covalent modification
  • Isozymes (isoenzymes)
  • Proteolytic activation
104
Q

What is Allosteric regulation?

A
  • Allosteric enzymes have a second regulatory site (allosteric site) distinct from the active site
  • Allosteric enzymes contain more than one polypeptide chain (have quaternary structure).
  • Allosteric modulators bind noncovalently to allosteric site and regulate enzyme activity via conformational changes
  • modulator can increase or decrease activity
105
Q

What is covalent modification regulation?

A

Covalent attachment of a molecule to an amino acid side chain of a protein can modify activity of enzyme

106
Q

Describe regulation with isoenzymes

A

Some metabolic processes are regulated by enzymes that exist in different molecular forms - isoenzymes

Isoenzymes - multiple forms of an enzyme which differ in amino acid sequence but catalyze the same reaction

Isoenzymes can differ in:

§ kinetics,

§ regulatory properties,

§ the form of coenzyme they prefer and

§ distribution in cell and tissues

107
Q

What is regulation by proteolytic cleavage?

A
  • Many enzymes are synthesized as inactive precursors (zymogens) that are activated by proteolytic cleavage
  • Proteolytic activation only occurs once in the life of an enzyme molecule

EXAMPLE

•Digestive enzymes

–Synthesized as zymogens in stomach and pancreas

108
Q

What is metabolic channeling?

A
  • Metabolite channeling - “channeling” of reactants between active sites
  • Occurs when the product of one reaction is transferred directly to the next active site without entering the bulk solvent
  • Can greatly increase rate of a reactions
  • Channeling is possible in multienzyme complexes and multifunctional enzymes
109
Q

Glycolysis starts from ______ and ends at ______. ____ ATP are generated.

A
  • Glycolysis starts from glucose and ends at lactate
  • going from 6 carbon to 3 carbon
  • GENERATE 2 ATP
110
Q

What is the investment part of Glycoloysis?

A
  1. ) Hexokinase catches glucose by adding phosphate to make Glucose-6-phosphate (uses ATP) and brings it into the cell.
  2. ) Isomerase changes Glc-6-phosphate to fructose
  3. ) Glucose can be trapped in a cell in the form of Glc-6-P
  4. ) Phosphofructokinase1 (PFK 1) (this is rate limiting enzyme of glycolysis)-adds another phosphate to fructose (uses ATP) to make fructose-1,6-bisphosphate
111
Q

What is the yield part of Glycoloysis?

A
  • GA3PDH is the enzyme that allows you to put phosphate on glyceraldehyde-3-phosphate to generate 1,3-bisphosphate glycerate
  • pyruvate kinase generates ATP
  • YEILDS 4 ATP, BUT 2 WERE INVESTED IN INVESTMENT PHASE SO NET 2 ATP
112
Q

______ and __________ ________ catalyze the yield of ATP in glycolysis

A
  • PGK and pyruvate kinase catalyze the yield of ATP in glycolysis
113
Q

Pyruvate can be transformed to ______ in order to regenerate ______.

A

Pyruvate can be transformed to lactate in order to regenerate NDA+

114
Q

Hexokinase can be inhibited by ________

A
  • Hexokinase can be inhibited by Glc-6-P (if too much accumulates in cell, hk won’t take in anymore glucose)

OSU COD 2022- D1 Fall 2018 (15 decks)

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