BIOC2000 Lecture Notes - Lecture 8: Dna Polymerase Iii Holoenzyme, Okazaki Fragments, Dna Replication

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21 May 2018

School

Department

Biochemistry

Course

BIOC2000

Professor

Dr Justin Ridge

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Lecture 8: DNA Replication

Enzyme language

Vmax: The maximum rate/’velocity’ at which an enzyme can operate;

maximum amount of substrate > product conversions during a given time

Vmax when:

- temperature is optimal

- pH is optimal

- surplus of substrate (saturating [substrate])

- product may need to be removed; if there is an equilibrium, in a

reversible reaction the product can bind back to the enzyme

- no inhibitors of enzyme in system

T affects the fold of protein

pH affects fold of protein, could also affect protonation/deprotonation of

side chains at the active site – need to ensure fold is optimal and active site

has right charge on it

The cell has a problem – some enzyme kinetics

Cells have a problem of trying to ensure leading and lagging strand go at

the same rate - otherwise one strand will be copied a lot more than the

other – if this continues, a lot of SBB will be required to keep DNA strands

separate, eventually running out

On rate: the rate at which substrate and enzyme associate

3’ end is the substrate for the enzyme

The Pol on the lagging strand has a problem because it has to find the new

strand every time it falls off the other strand

Pol also has a problem with coming off – Off rate: the rate at which product

and enzyme dissociate

The lagging strand is running fast but probably at the same rate as the one

on the leading strand

Solution: Okazaki fragments allow cell to keep up with lagging strand

because there is no wait for entire lagging strand to be unraveled

There is a part of the polymerase that marks where new Okazaki fragments

needs to start (next starting point), so the body of the Pol can find it easily

Body of Pol has higher affinity for the marker than it does for the RNA

primer.

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When DNA Pol jumps off one of these fragments looking for the next

marker, the leading strand can be held back – inhibits its function

So DNA Pol on Okazaki fragments operates at Vmax but one on leading

strand doesn’t.

Ignore all other green dots

DNA polymerases are not simple “blobs”, with one on each

strand of the replicating DNA

DNA polymerase III holoenzyme is the primary enzyme complex involved

in prokaryotic DNA replication.

The primary difference between transcription and DNA replication is that

only a single strand of mRNA is produced during transcription. During

DNA replication both strands need to be copied and DNA polymerase

needs to work on both strands at the same time.

Replication is the process of making two daughter DNA strands from a

single DNA strand. Transcription is the process of synthesizing RNA from

DNA strand/template

Bacteria are adapted to extreme conditions so their proteins do not easily

denature and are easy to work with, unlike human protein.

^ This slide and next is not examinable.

The lagging-strand polymerase is tethered to increase the on-rate and

allow it to keep up with the leading-strand polymerase

DNA replication and transcription are both processes that

involve making nucleic acids.

How does DNA Pol add bases?

When bases are added – it is creating more order, theoretically it shouldn’t

be able to happen according to free energy changes

Only final structure is correct – be able to draw in exam! (without the

bases)

dNMP(n) – nucleotides in a polymer (DNA)

dNTP - triphosphate

dNMP(n+1) – DNA, 1 longer

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Do you need to memorize structures? no

Do you need to be able to recognize? no

- Have an understanding of an enzymatic mechanism behind how bases are

added and why they are

Bases themselves have nothing to do with the enzymatic mechanism for the

addition of bases to the DNA

The part with N in it is the base ‘nitrogenous base’

DNA Polymerase

• DNA replication is an intricate process that involves the interaction

of over twenty different proteins.

DNA (n) + dNTP -> DNA (n+1) + PPi

• So far in the cell, DNA is being replicated and so far there is n

number of nucleotides in DNA polynucleotide chain

• If one more n (nucleotide) is to be added, dNTP needs to be added

on, forming a phosphodiester bond, extending the DNA chain by

one (n+1) – in the process releasing PPi.

• DNA polymerase catalyses the formation of the phosphodiester

bond by adding deoxynucleotide triphosphates (dNTPs) onto the

growing polynucleotide chain. As each deoxribinucleotide is added

onto growing chain in a stepwise fashion, a pyrophosphate molecule

(PPi) is released.

• DNA polymerase:

(A) Require the four types of dNTP molecules

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Document Summary

Vmax: the maximum rate/"velocity" at which an enzyme can operate; maximum amount of substrate > product conversions during a given time. Product may need to be removed; if there is an equilibrium, in a reversible reaction the product can bind back to the enzyme. T affects the fold of protein ph affects fold of protein, could also affect protonation/deprotonation of side chains at the active site need to ensure fold is optimal and active site has right charge on it. The cell has a problem some enzyme kinetics. On rate: the rate at which substrate and enzyme associate. 3" end is the substrate for the enzyme. The pol on the lagging strand has a problem because it has to find the new strand every time it falls off the other strand. Pol also has a problem with coming off off rate: the rate at which product and enzyme dissociate.

Related Questions

An allosteric interaction between a ligand and a protein is one in which: 74. changes in Km are negligible, so Km can be treated as a constant. C. binding of a molecule to a binding site affects binding of additional molecules to the same site. binding of a molecule to a binding site affects binding properties of another site on the protein. binding of the ligand to the protein is covalent. multiple molecules of the same ligand can bind to the same binding site. A. D. VO Vmax. E. varying [S] has no effect on Vo. 78. The steady state assumption, as applied to B. enzyme kinetics, implies: Km = Ks. the enzyme is regulated the ES complex is formed and broken down at equivalent rates. the Km is equivalent to the cellular substrate concentration A. B. C. C. D. D. E. the mximum velocity occurs when two different ligands can bind to the same binding site. E. the enzyme is saturated. The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: 79. 75. One of the enzymes involved in glycolysis aldolase, requires Zn2+ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the: Substrate added vo COmol/min) (mmol/L) A. apoenzyme. B. coenzyme. C. holoenzyme. D. prosthetic group. E. substrate 0.8 217 325 488 647 76. 1,000 The role of an enzyme in an enzyme- catalyzed reaction is to: bind a transition state intermediate, such that it cannot be converted back to substrate. ensure that all of the substrate is converted to product ensure that the product is more stable than the substrate. increase the rate at which substrate is converted into product. make the free-energy change for the reaction more favorable. A. The Km for this enzyme is approximately: B. C. D. E. A. 1 mM B. 1,000 mM. C. 2 mM D. 4 mM E. 6 mM 80. In competitive inhibition, an inhibitor: binds at several different sites on an enzyme. binds covalently to the enzyme. binds only to the ES complex. binds reversibly at the active site. lowers the characteristic Vmax of the enzyme A. B. C. D. E. 77. The benefit of measuring the initial rate of a reaction VO is that at the beginning of a reaction: A. [ES] can be measured accurately. B. changes in [S] are negligible, so [S can be treated as a constant

describe the definitions of substrate, enzyme active site and its general characteristics, and apoand

holo-enzymes.

Â· describe the fact that enzyme catalysis is specific in terms of the type of reaction and the exact

substrate structure.

Â· Explain the difference in the âlock-and-keyâ model and the induced fit model of enzyme

substrate interaction.

Â· describe that reaction rate constant thus reaction rate is dependent on the activation energy.

Â· describe the relationship between the energy of the transition state and the activation energy.

Â· describe the relative binding affinity of a substrate and its transition state to the enzyme.

Â· Explain how enzymes speed up reaction rate, i.e., how do enzymes lower the activation

energy. Note: Factors other than preferential stabilization of transition state also contribute to increased

reaction rate. They include 1) enzymes placing one substrate (reactant) next to the other substrate (reactant)

so that the reaction is no longer dependent on the collision rate of the two substrates; 2) enzymes sequester the

substrate(s), minimizing solvent competition thus speeding up the reaction; 3) in enzyme complexes with each

enzyme catalyzing one step in a multi-step reaction sequence, the product of one reaction is fed to the active site

of another enzyme for the next step without the product diffusing away from the enzyme complex thereby

increasing reaction rate.

Â· describe the Michaelis-Menten equation relating the initial velocity to total substrate

concentration.

Â· describe the assumptions that went into the derivation of the Michaelis-Menten equation. ([E]

is constant, steady state assumption, etc)

Â· describe the relationship between Vmax and kcat and the fact that kcat is a constant for a given

enzyme and substrate at a given temperature and solution condition; whereas Vmax can differ

for a given enzyme and substrate at a given temperature and solution condition, depending

on the enzyme concentration.

Â· describe the definition of Kd (equilibrium dissociation constant for ES into E and S).

Â· describe the relationship between Kd and the affinity of substrate for the enzyme, i.e., the

higher the Kd, the weaker the binding between E and S.

Â· describe the relationship between Kd and Km, i.e., in general, the higher the Kd, the higher theKm.

Â· Use the Michaelis-Menten equation and the relationship between Vmax and kcat to do

calculations.

Â· describe the other names for the catalytic rate constant, kcat (i.e., the turn over number; also

called krelease) and why it is so called (kcat reports on per unit time per enzyme molecule the

number of product molecules formed/released, thus the name turnover number or krelease).

Â· Determine Vmax and Km using both the Michaelis-Menten plot (initial velocity versus [S]) and

the Lineweaver-Burk Plot (the double reciprocal plot), given the data of initial velocity and

[S].

Â· Draw the Michaelis-Menten plot AND the double reciprocal plot (Lineweaver Burk), given

the Vmax and Km values.

Â· describe the fact that the term kcat/Km is the best parameter for measuring the catalytic

capability (efficiency) of an enzyme.

Â· describe the existence of irreversible enzyme inhibition and the general mechanisms of

irreversible inhibition.

Â· describe the different types of reversible inhibitions, namely the competitive, uncompetitive,

and mixed inhibitions (noncompetitive inhibition is a special case in which the inhibitor

binds E and ES with equal affinity therefore Î± is equal to Î±â â Km is not affected).

Â· Describe each type of inhibition in terms of inhibitor binding site (active site or other than

the active site), how the inhibitor affects vmax and Km and the quantitative relationship of

each parameter with and without the inhibitor present and their relationship with [I] and KI

or KI

Â· Show the characteristics of each type of inhibition on a Michaelis-Menten plot (initial

velocity versus total substrate concentration), i.e. draw the curves with and without the

competitive inhibitor.

Â· Show the characteristics of each type of inhibition on a double reciprocal plot (LineweaverBurk

Plot) by drawing the lines with and without the competitive inhibitor.

Â· Given the Michaelis-Menten or double reciprocal plot with and without the inhibitor,

determine the type of inhibitor.

Â· describe the fact that transition state analogs are potent inhibitors.

Â· Explain why transition state analogs are potent inhibitors.

Â· describe that not all enzymes obey Michaelis-Menten kinetics.

Â· describe situations when an enzyme will not obey Michaelis-Menten kinetics (when the enzyme

catalyzes reactions with multiple substrates or when the enzyme is allosteric, i.e. binding of

the substrate at one active site affects the other active site(s) for the same substrate).

Â· Name the four types of catalytic mechanisms. (electrostatic catalysis, general/specific acid

base catalysis, metal ion catalysis, covalent catalysis)

Â· Contrast acid/base catalysis to general acid/base catalysis, i.e., comment on their similarity

and difference.

Â· Name the eight common general acid/base pairs in proteins and write the formula of the

acid and base form for each pair (hint: consider the eight amino acids that have ionizable

side chainsâ¦)

Â· describe the meaning of covalent catalysis.

Â· describe the reaction catalyzed by chymotrypsin and the three catalytic residues in

chymotrypsin.

Â· Draw the interactions among the three catalytic residues (catalytic triad).

Â· State the catalytic role of each of the catalytic residues and how they accomplish that.

Â· describe the steps in the formation of the acyl-enzyme intermediate, and the substrate and

products in this first part of the chymotrypsin mechanism.

Â· Draw the transition state structure for the formation of the acyl-enzyme intermediate and

indicate by drawing how it is preferentially stabilized.

Â· describe the steps in the regeneration of chymotrypsin (deacylation), the substrates and

product of this second part of the chymotrypsin mechanism.

Â· Draw the transition state for deacylation and explain how it is preferentially bound to or

stabilized by the enzyme.

Â· Explain what general acid and base catalysis refers to and what groups act as the general acid

and base in the chymotrypsin mechanism.

Â· Explain what covalent catalysis refers to in the chymotrypsin mechanism.

Â· Explain the basis for the substrate specificity of chymotrypsin (consider the composition of

the active site â chymotrypsin specifically cleaves peptide bonds adjacent to aromatic

residues)

Â· Explain the term âserine proteaseâ. Answer: these are proteases (enzymes that

hydrolyze peptide bonds) that have serine as a catalytic residue (that acts as a

nucleophile, which can attack the carbonyl carbon of the peptide bond).

Â· Key features of Hexokinase, enolase and lysozyme mechanisms that were discussed in class

(specific methods of catalysis utilized by these enzymes; general description of overall

catalytic transformation)

Â· describe the four regulatory strategies of enzyme catalysis

Â· describe the definition of feedback inhibition (mechanism of cellular regulation that regards an

enzyme that catalyzes the production of a given substance that then inhibits the cell when

concentration of that substance has reached an appropriate level â this balances/regulates

amount of product provided with amount needed)

Â· describe the most common type of covalent modification and the types of enzymes

responsible for the addition and removal of the chemical group in this covalent

modification.

Â· Explain the general mechanism of phosphorylation in regulating enzyme activity.

Â· describe the general naming scheme for active and inactive proteases.

Â· describe that protease inhibitors exist to inhibit protease.

Â· describe that enzymes are regulated through covalent and non-covalent processes.

Â· Noncovalent regulation

o Describe noncovalent modification in general terms. Elaborate on the relation

among enzyme activity, allostery, effector molecules (small molecules) and control

proteins.

o Concerning allosteric regulation, some enzymes have multiple active sites for the

same substrate and the binding of the substrate increases the activity of the enzyme

by stabilizing the R (more active) state over the T (relatively inactive) state. (Lecture

13: âNoncovalent Modification: Allosteric Regulatorsâ)

Ã Draw a curve for the initial reaction velocity versus substrate concentration.

Ã describe and explain the specific shape of the v0 vs. [S] plot.

Ã Draw on the same graph above a v0 vs. [S] curve for the enzyme in the

absence of allostery. and how th i.e., in the absence of conformational change

upon substrate binding. Comment on the shape of this curve.

Ã Comment on the benefit of allosteric activation.

Ã When an inhibitor is present for such an allosteric enzyme, the v0 versus [S]

plot will shift. Draw a v0 vs. [S] curve with and without the inhibitor.

Ã When an activator is present for such an allosteric enzyme, the v0 versus [S]

plot will shift. Draw a v0 vs. [S] curve with and without the inhibitor.

Ã When an activator is present for such an allosteric enzyme, the v0 versus.

Ã describe the relationship between feedback inhibition and allosteric regulation

and the reason behind the term âfeedback inhibitionâ.

Â· Covalent regulation

o describe the most common covalent modification, the enzymes that catalyze the

modification and the removal of that modification, respectively and the amino acid

residues that tend to be the site of this modification.

o Write the structure of the added group bonded to those residues and the reaction for

this covalent modification.

o Rationalize the effectiveness of phosphorylation on altering the activity of an

enzyme.

o describe the mechanism of chymotrypsinogen activation to chymotrypsin (cleavage of

zymogen by trypsin).

Enzyme Activity

1. Briefly explain the principle of spectrophotometry.

2. Which of the following statements is true about enzymes?(check all correct answers)

		__ enzymes act on (are specific for) one or just a fewsubstrates
		__ enzymes can be denatured (loose their tertiary/quaternarystructure and activity) by extreme conditions such as heat, high orlow pH,or presence of organic solvents
		__most enzymes are DNA
		__ enzymes are protein
		__enzymes perform best at specific conditions of temperature, pHand salt concentration 5. Which enzymes break the following macromolecules? Match the following with their correponding answers: ___ Nucleic acids A. proteases or proteinases such as pepsin ___ Cellulose B. none ___ Triglycerides C. lipases ___ Proteins D. amylases ___ Starch E. nucleases 6.Match the following enzymes totheir products. Answers ___ pepsin, trypsin and other digestive proteases A.Nucleotides ___ salivary and pancreatic amylases and other enzymes B. AminoAcids ___ lipases C. Monosaccharides ___ nucleases D. fatty acids and glycerol 7. Based on experiments, what is the effect of extremeconditions of pH and temperature (boiling tube) on enzymeactivity? 8. ATP stores energy in the form of___ ___ Chemical energy ___ Kinetic energy ___ Thermal energy ___ Activation energy 9. An allosteric site in anenzyme is____ __ the site where reactions occur __ the site that binds aregulator __ the site that binds thesubstrate __ the site that binds the product 10. Enzymes increase the rate of reactions by making thetransition state less stable. __ True __ False 11. Enzymes increase the rate of reactions by placing reactingparts of the molecule in favorable positions. __ True __ False 12. Which of the followingis a mechanism for regulation of enzymes? __ covalent modification __ feedback inhibition __ allosteric regulation __ all of the above 13. Which of the following represents a correct order of eventsin an enzyme catalytic cycle? 1.enzyme changes shape 2.substrate binds to active site 3.substrate is converted to products 4. products are released from activesite __ 1>2>3>4 __ 2>1>3>4 __ 3>2>1>4 __ 2>4>3>1

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BIOC2000 Lecture Notes - Lecture 8: Dna Polymerase Iii Holoenzyme, Okazaki Fragments, Dna Replication

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