Biochem Week 3

Question 1

Define rate constant of a ligand protein interaction, and correlate rate constants to Ka and Kd

Answer 1

Kd is the dissociation constant given by the equation Kd=k2/k1 . Small k2 compared to k1 favors reaction going forward (enzyme-substrate, or ligand binding formation) and greater affinity of enzyme with substrate.

Question 2

Describe how measurable parameters can be used to determine the initial velocity of an enzyme catalyzed reaction using the Michaelis-Menten equation

Answer 2

Vo=k3[ES] gives initial velocity of the reaction but [ES] cannot be measured so instead use the idea that at a saturating concentration of substrate Vo=Vmax because at this condition [Et]=[ES] therefore Vmax=k3[Et]

Question 3

Define Km, and describe its practical implication in understanding the efficiency of an enzyme-catalyzed reaction

Answer 3

Km=S when Vo=Vmax/2 Km of the enzyme for a substrate is the concentration of substrate required to reach 1/2*Vmax Km is a measure of the apparent affinity of the enzyme for its substrate. Lower Km means substrate has a higher affinity for the enzyme.

Question 4

Deduce the Michaelis-Menten equation from the ligand-binding equation

Answer 4

Ligand-Binding equation: B = (Bmax[H])/(Kd+[H]) Michaelis-Menten equation: Vo = (Vmax[S])/(Km+[S])

Question 5

Distinguish between ligand binding and enzyme catalysis, and describe the relationship between Km and Kd

Answer 5

Ligand binding (ex. Hormone-Receptor) and enzyme catalysis can be described in a similar manner mathematically because both have two reactants favoring a single product. Km and Kd are both ways of showing the favorability of a reaction. Km and Kd are the same when the dissociation of Enzyme-Substrate to Enzyme and Product is ignored.

Question 6

Define Vmax, and explain what is meant by saturation kinetics

Answer 6

The approach to the finite limit of Vmax is called saturation kinetics because velocity cannot increase any further once the enzyme is saturated with substrate.

Question 7

Explain the mechanistic significance of the hyperbolic nature of the Michaelis-Menten curve

Answer 7

According to the curve Vmax will only be reached at an infinite concentration of substrate, which leads to the idea of saturation kinetics.

Question 8

Define isozymes, and explain the physiological significance of different isoforms having different Km’s for substrate

Answer 8

Isozymes are enzymes that catalyze the same reaction but have different primary or quaternary structures, which permits the fine-tuning of metabolism to meet the particular needs of a given tissue or developmental stage. Isozymes generally have different kinetic and regulatory properties, cofactors, and subcellular or tissue distribution

Question 9

Explain hexokinase and glucokinase as they pertain to isozymes

Answer 9

Hexokinase (in the brain) and Glucokinase (in hepatocytes). Hexokinase has low Km and low Vmax for glucose. Glucokinase has higher Km and very high Vmax for glucose. The glucose sparing effect of the liver allows the brain to get the glucose it needs before the liver begins to store glucose (bad things would happen if liver stored glucose before the brain got what it needed).

Question 10

Explain what is meant by reversible inhibition of enzymes, and describe the mechanistic difference between reversible and irreversible inhibitions

Answer 10

Reversible inhibition of enzymes means something other than the substrate binds to the enzyme limiting its effectiveness in producing product. The inhibition is reversible because the inhibitor can leave and the enzyme can return to its effectiveness. Irreversible inhibitors the enzyme will never be able to take part in the reaction to turn substrate into product. This drastically reduces Vmax of the reaction taking place.

Question 11

Explain the Lineweaver-Burke (double reciprocal) transformation of the Michaelis-Menten equation

Answer 11

Enzymes obeying michaelis-menten relationship can be plotted as a straight line by plotting 1/VO vs 1/[S]. The Lineweaver-Burke (double reciprocal) transformation provides: a more accurate determination of Vmax and Km, better distinguishment among certain types of enzymatic reactions, and allows for analysis of enzyme inhibition.

Question 12

Distinguish between competitive, noncompetitive and uncompetitive reversible inhibition

Answer 12

Competitive inhibitor binds to the same active site as the substrate, preventing substrate from binding to enzyme. Noncompetitive inhibitor binds to a different active site than the substrate but its binding essentially lowers the concentration of enzyme available for product formation. According to Marks text, “Uncompetitive inhibition is almost never encountered in medicine and will not be discussed further.” Dr. Dey’s definition of uncompetitive inhibition is when an enzyme has two active binding sites A and B. Substrate A is allowed to bind but Substrate B cannot bind because an uncompetitive inhibitor has bound to site B, thus rendering the enzyme inactive (from the perspective of substrate B this could be classified as competitive inhibition).

Question 13

Demonstrate how the double reciprocal plot can be used to discriminate one mechanism (competitive and non-competitive inhibition) from the other

Answer 13

Competitive inhibitor rotates slope of double reciprocal plot line along the 1/V axis (slope changes, 1/V intercept doesn’t change). - Lines meet on Y axis (Vmax doesn’t change, Km is increased) Noncompetitive inhibitor rotates slope of double reciprocal plot line along the 1/[S] (slope changes, 1/[S] intercept doesn’t change). - lines meet on X axis (Km (affinity) doesn’t change, Vmax is decreased) Uncompetitive inhibitor changes the 1/V and 1/[S] slope intercepts but not the slope itself.

Question 14

Define/describe allosteric regulation of enzymes:

Answer 14

Allosteric regulation of enzymes: • Regulation by reversible noncovalent binding of regulatory compounds • The regulatory compounds are called allosteric modulators • Modulators induce conformational changes switching between more active and less active forms • Modulators could be inhibitory or stimulatory

Question 15

What is the T and R states of an allosteric enzyme?

Answer 15

Structurally distinct enzyme forms that occur after conformational change between a low-activity, low-affinity “tense” or T state and a high-activity, high-affinity “relaxed” or R state. Also can thought of as the first conformation = T state; the second = R state. • Higher concentrations of substrate favor the conversion of the T state to the R state.

Question 16

Distinguish between homotropic and heterotropic allosteric regulation

Answer 16

Homotropic: A homotropic allosteric modulator is a substrate for its target enzyme, as well as a regulatory molecule of the enzyme’s activity. It is typically an activator of the enzyme. • Example: O2 is a homotropic allosteric modulator of hemoglobin. Heterotropic: A heterotropic allosteric modulator is a regulatory molecule that is not also the enzyme’s substrate. It may be either an activator or an inhibitor of the enzyme. • Example: H+, CO2, and2,3-bisphosphoglycerate are heterotropic allosteric modulators of hemoglobin.

Question 17

Describe cooperativity in substrate binding to multisubunit protein with reference to hemoglobin

Answer 17

Cooperativity is a phenomenon displayed by systems involving identical or near-identical elements, which act non-independently of each other, relative to a hypothetical standard non-interacting system in which the individual elements are acting independently. One manifestation of this are enzymes or receptors that have multiple binding sites where the affinity of the binding sites for a ligand is apparently increased, positive cooperativity, or decreased, negative cooperativity, upon the binding of a ligand to a binding site. • Example: The affinity of hemoglobin’s four binding sites for oxygen is increased above that of the unbound hemoglobin when the first oxygen molecule binds.

Question 18

Distinguish the action of allosteric activators and inhibitors by inspection of a V0/ Vmax versus [S] plot

Answer 18

Allosteric inhibition: occurs when the binding of one ligand decreases the affinity for substrate at other active sites. For example, when 2,3-BPG binds to an allosteric site on hemoglobin, the affinity for oxygen of all subunits decreases. This is when a regulator is absent from the binding site. Allosteric activation: occurs when the binding of one ligand enhances the attraction between substrate molecules and other binding sites. An example is the binding of oxygen molecules to hemoglobin, where oxygen is effectively both the substrate and the effector. The allosteric, or “other”, site is the active site of an adjoining protein subunit. The binding of oxygen to one subunit induces a conformational change in that subunit that interacts with the remaining active sites to enhance their oxygen affinity.

Question 19

How do covalent modifications regulate enzymes?

Answer 19

Enzyme activity may increase or decrease after the covalent addition of a chemical group. 1. Phosphorylation affects many enzymes. a. Pyruvate dehydrogenase and glycogen synthase are inhibited by phosphorylation. b. Glycogen phosphorylase is activated by phosphorylation. 2. Phosphatases that remove the phosphate groups alter the activities of these enzymes. 3. Phosphorylation introduces negative charges to the protein, which may alter the secondary and tertiary structures.

Question 20

How does the protein-protein interaction with calmodulin regulate an enzyme?

Answer 20

Calmodulin (CaM) (an abbreviation for CALcium-MODULated proteIN) is a calcium-binding messenger protein expressed in all eukaryotic cells. • Calmodulin induces substantial conformational change in the protein • The calcium-induced conformational change exposes hydrophobic surfaces in calmodulin to mediate interaction with other proteins and subsequently modulate their function

Question 21

What are the major differences in Hb and Mb

Answer 21

Differences in Hb and Mb - Mb is a storage protein – binds O2 avidly, dissociates slowly - Mb is not cooperative - Mb is 1 polypeptide vs Hb is multiple subunits = tertiary structure

Question 22

Compare and contrast myoglobin and hemoglobin structure:

Answer 22

II. MYOGLOBIN: an oxygen carrier in muscle cells A. Physico-chemical properties i. 153 amino acids – single polypeptide chain ii. Very compact: globular structure →little empty space for solution to get in iii. Tertiary structure: 8 alpha helices (A-H), 4 helices terminated by proline residues iv. About 75% is in alpha helical structure v. Polar side chains on outside of protein →interact with solution vi. Myoglobin = storage protein →mainly in skeletal muscle vii. High O2 affinity – does not change with concentration viii. Monomer →no cooperativity Myoglobin globin fold A single polypeptide chain. Built primarily of a-helices and turns Presence of heme (prosthetic group) allows reversible binding of oxygen Heme consists a protoporphyrin and a central iron atom Interior consists almost entirely of hydrophobic residues The only polar residues inside are the two Histidines that play critical role in binding of heme oxygen Interaction of O2 with heme is often accompanied by conformational changes Protein structure affects how ligands bind Bound O2 is hydrogen bonded to a Histidine (His E7) residue Myoglobin and b-Subunit of Hemoglobin Have Structurally Related Globin Folds III. HEMOGLOBIN A. Structure i. Primary, secondary and tertiary structures are same as Mb ii. More than 1 subunit →quaternary structure: A tetrameric Protein with four heme prosthetic groups Adult hemoglobin contains two a- and two b- chains Strong interaction between unlike subunits Hydrophobic interaction predominates Hemoglobin is an oxygen carrier in blood Binding of oxygen induces a conformational change in hemoglobin which is transmitted to other subunits iv. Geometric explanation 1. Hemes are dome-shaped 2. O2 binds →pulls Fe down →dome becomes flat →pulls helix

Question 23

Interpret the saturation curve of oxygen binding by hemoglobin and myoglobin at increasing partial pressure of O2, and explain the physiological significance of cooperativity in O2 binding to hemoglobin

Answer 23

As demonstrated by the graph a single subunit has very low affinity even as pO2 increases and only reaches about 50% saturation. Contrast this to the rapid rise of hemoglobin, once activated from the T to the R state it quickly is able to bind O2 easily.

Question 24

Describe the role of heme in oxygen binding by hemoglobin and illustrate why free heme cannot substitute for hemoglobin in transporting oxygen

Answer 24

Heme improves affinity of oxygen in the hemoglobin molecule, example of cooperativity. When oxygen binds to the iron complex, it causes the iron atom to move back toward the center of the plane of the porphyrin ring. At the same time, the imidazole side-chain of the histidine residue interacting at the other pole of the iron is pulled toward the porphyrin ring. This interaction forces the plane of the ring sideways toward the outside of the tetramer, and also induces a strain in the protein helix containing the histidine as it moves nearer to the iron atom. This strain is transmitted to the remaining three monomers in the tetramer, where it induces a similar conformational change in the other heme sites such that binding of oxygen to these sites becomes easier.

Question 25

Explain the molecular basis of sickle cell anemia, and the mechanistic basis of the existing modalities of treatment

Answer 25

A mutation substituting a Glutamate residue with a Valine residue at position 6 of the b-subunit of hemoglobin causes the mutated protein to aggregate with each other The aggregation of the hemoglobin molecules leads to deformation and premature destruction of the red blood cells (RBC) causing an anemic state known as Sickle Cell Anemia The Valine side chain creates a hydrophobic patch on the surface of HbS molecules which leads to intermolecular aggregation

Question 26

Explain why enzyme catalyzed reactions are pH- and temperature-sensitive

Answer 26

pH: The activity of enzymes is dependent on pH. Each enzyme has an optimal pH at which it is maximally active. * due to ionization properties of the active-site amino acid side chains Temperature: In general, as temperature increases, an enzyme’s activity also increases (the heat increases the kinetic energy in the system). * Above a certain temperature, enzymatic activity rapidly decreases. This is due to protein denaturation.

Question 27

Summarize various ways enzymes can be regulate:

Answer 27

pH: The activity of enzymes is dependent on pH. Each enzyme has an optimal pH at which it is maximally active. Temperature: In general, as temperature increases, an enzyme’s activity also increases (the heat increases the kinetic energy in the system). * Above a certain temperature, enzymatic activity rapidly decreases. This is due to protein denaturation. Concentration (of enzyme) Covalent Modification An enzyme’s activity can be altered by the attachment or removal of other molecules. Such additions or subtractions may change the enzyme’s structure or other properties, resulting in a change in enzyme activity. Phosphorylation and Dephosphorylation Each of these processes can increase or decrease enzymatic activity, depending on the particular enzyme. Phosphorylation occurs at serine, threonine, and tyrosine residues. Kinases are enzymes that catalyze phosphorylation. Phosphatases are enzymes that catalyze dephosphorylation. Acetylation and dephosphorylation (eg, COX-2) γ-Decarboxylation (eg, thrombin) ADP-ribosylation (eg, RNA polymerase) Zymogens (Proenzymes): Inactive precursors to enzymes that must be cleaved in some way to achieve their active form. Allosteric Regulation: An enzyme’s activity can be modified by the binding of a ligand to an allosteric site (a site distinct from the active site).

Question 28

Define: Facilitated diffusion

Answer 28

Also known as carrier-mediated diffusion: a substance transported in this manner diffuses through the membrane using a specific protein to help; saturable protein binding with high specificity (e.g. glucose transport)

Question 29

Define: Simple diffusion

Answer 29

spontaneous Brownian (random) movement (oxygen, selected nutrients and other molecules enter and leave capillaries by diffusion; 
a substance moves "down" its concentration gradient

Question 30

Describe the mechanism of **facilitated diffusion **

Answer 30

• Reversible binding and conformational change are the critical factors.*
• High concentration on the outside leads to binding to the transporter based on the dissociation constant (Kd).
• Binding leads to conformational change.
• Low concentration on the inside leads to dissociation.
• Dissociation leads to reversal of the conformational change.

Question 31

Define: Osmosis

Answer 31

The simple diffusion of water, which is controlled by osmotic pressure. Cell can counteract the effects of increases or decreases in cell volume by multiple mechanisms.
Osmolarity is determined by the total concentrations of all solutes present.

Question 32

Describe: Phagocytosis

Answer 32

Binding of “extracellular” particles; accomplished by specialized cells such as macrophages

Question 33

Describe: Primary Active Transport

Answer 33

Direct use of metabolic energy in the form of ATP to carry out transport.

Examples:

The most abundant primary active transport mechanism in higher organisms is sodium-potassium pump or Na+/K+ -ATPase.

Other such pumps/ATPases include 1) H+/K+ - ATPases, 2) Calcium pumps 3) Proton pumps or H+-ATPases and 4) ATP-binding cassette (ABC) transporters.

Question 34

Describe: Secondary Active Transport

Answer 34

Transport energy derived secondarily from energy that has been stored in the form of ionic concentration differences of secondary molecular or ionic substances between the two sides of the cell membrane (difference created by primary active transport)

Two Types:

Co- and Counter-transport

Question 35

Describe: Co-Transport

Answer 35

type of Secondary Active Transport

Differences in sodium concentrations can “pull” other substances along with the sodium through the cell membrane (e.g. glucose and amino acids); move in the same direction

Question 36

Describe: Counter-Transport

Answer 36

**type of Secondary Active Transport **

Sodium ions again attempt to diffuse to the interior of the cell, but the “co-substance” to be transported moves to exterior to the cell as sodium moves in (Calcium and hydrogen ions); subtance moves in the opposite direction

Question 37

Define: Endocytosis

Answer 37

involves bulk fluid transport as well as receptor-mediated processes (e.g. LDL receptor)

Question 38

Define: Exocytosis

Answer 38

Vesicle fuses with plasma membrane and contents are release extracellularly

Constitutive: Goblet cells in the liver

Regulated: Pancreatic enzymes, peptide hormones, neurotransmitters