Molecular B1 Flashcards

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1
Q

What is a genome,gene,genotype,phenotype,nucleic acid
All life depends on three critical
Molecules name them and their functions

A

Genome: an organism’s genetic material
Gene: a discrete units of hereditary information located on the chromosomes and consisting of DNA.
Genotype: The genetic makeup of an organism
Phenotype: the physical expressed traits of an
organism
Nucleic acid: Biological molecules (RNA and DNA) that allow organisms to reproduce; Nucleic acids are polynucleotides—that is, long chainlike molecules composed of a series of nearly identical building blocks called nucleotides. Each nucleotide consists of a nitrogen-containing aromatic base attached to a pentose (five-carbon) sugar, which is in turn attached to a phosphate group.

All life depends on 3 critical molecules
DNAs:
• Stores and transmit information on how cell works and divide.
RNAs:
• Act to transfer short pieces of information to different parts of cell
• Provide templates to synthesize into protein
Proteins:
• Form enzymes that send signals to other cells and regulate gene activity
• Form body’s major components (e.g. hair, skin, etc.)

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2
Q

What is denaturation
What is the structure of nucleic acids
What is the difference between a nucleoside and a nucleotide

The presence of the 2’OH (the OH is on the second carbon ) confers special chemical and structural properties to RNA compared to DNA true or false(DNA has 2’H (H is on the second carbon) not 2’OH)

While DNA contains deoxyribose, RNA contains ribose, characterised by the presence of the 2′-hydroxyl group on the pentose ring (Figure 5). This hydroxyl group make RNA less stable than DNA because it is more susceptible to hydrolysis true or fakse

A

The process of separating two DNA strands into two single strands is called denaturation.

Nucleic Acids Structure
1.Nucleic acid bases and nucleotides
2-Double Helix Structures of DNA
B-DNA
A-DNA and A-dsRNA
Z-DNA
3-Transition DS(double stranded)   SStranded (single stranded) DNA 
4- Structure of tRNA(transfer RNA) 

Nucleoside: composed of sugar(ribose or deoxyribose) and the base (purine or pyrimidine)

Nucleotide:composed of phosphate 1,2,3 and sugar(ribose or deoxyribose) and the base (purine or pyrimidine)

Or A nucleoside consists of a nitrogenous base covalently attached to a sugar (ribose or deoxyribose) but without the phosphate group. A nucleotide consists of a nitrogenous base, a sugar (ribose or deoxyribose) and one to three phosphate groups

A nucleotide: Pentose+phosphate+base

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3
Q

What is the consequence of Aromaticity of bases (purines and pyrimidines)

Bases are planar true or false
Why are purines and pyrimidines) called bases?
Ring nitrogens of bases are normally not protonated at physiological pH
True or false

A

Pyrimidines:

-Large number of electrons -
in the pi orbital system
Protonation of ring nitrogens:

Purines:

-Delocalization of electrons:
- transient dipole and
attraction between bases

Base stacking
They called bases Because N1 and N3 of pyrimidine, and N1, N3, and N7 of purine can accept protons

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4
Q

State three similarities between DNA and RNA

State seven differences between them

A

Both have adenine, guanine, cytosine • The nucleotides are linked together by
phosphodiester bonds.
• Main function involves protein biosynthesis.

DNA

  1. Bases are A, G, C and T
  2. Pentose sugar is deoxyribose
  3. Present in nucleus and mitochondria but never in cytoplasm
  4. Consist of 2 helical strands
  5. There are A, B, C, D, E and Z forms of DNA

DNA

  1. Large molecules
  2. One strand 3’- 5’ carries genetic information
  3. DNA can form RNA by the process of transcription
  4. Purine and pyrimidine contents are almost equal
  5. Alkali hydrolysis does not give 2’-3’ cyclic diesters

RNA

  1. Bases are A, G, C and U
  2. Pentose sugar is ribose
  3. In addition to nucleus, RNA is found in the cytoplasm
  4. Single stranded
  5. There are tRNA, mRNA, rRNA, snRNA etc

RNA

  1. Only m- and rRNA are large molecules
  2. mRNA transcribed from DNA carries genetic information
  3. RNA does not give rise to DNA except only in the presence of reverse transcriptase
  4. Purine and pyrimidine nucleotides not equal
  5. Alkali hydrolysis gives 2’-3’ cyclic mononucleotides.
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5
Q

State seven differences between mRNA and tRNA

A

mRNA

  1. Large molecular wt
  2. Most heterogenous
  3. Acts as a template for protein synthesis
  4. Carries codons
  5. Shape and size is not constant
  6. The cap structure is found in the 5’ end

RNA

  1. Poly A tail is found on 3’ end 8. Precursor is pre-mRNA
  2. Unusual bases are not found
  3. Stem and loop structure is not found

tRNA

  1. Low molecular wt
  2. Only about 20 different forms, less heterogenous
  3. Acts as a carrier of amino acids
  4. Carries anticodons
  5. Shape and size is constant for all tRNAs
  6. No cap structure.
  7. 3’ end always carries CCA sequence where specific amino acids bind
  8. Pre-tRNA precursor
  9. Unusual bases such as pseudouridine, thymine etc. are found
  10. Stem and loop structure is a consistent feature.
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6
Q

Recombinant DNA technology involves techniques in manipulating DNA such as,
What is a restriction enzyme or restriction endonuclease
What is it’s function

A

Involves techniques in manipulating DNA:-
• Molecular Cloning.
• DNA sequencing.
• Polymerase Chain Reaction (PCR).
• Nucleic acid blotting and hybridization.
(Southern, Northern analysis).
Production of proteins.
Creation of Knock-out, Knock-in and Transgenic mice.
Nucleic acid Microarray (simultaneous monitoring of expression level of each gene in a cell.

restriction enzyme, restriction endonuclease, or restrictase is an enzyme that cleaves DNA into fragments at or near specific recognition sites within molecules known as restriction sites.

A restriction enzyme is a protein isolated from bacteria that cleaves DNA sequences at sequence-specific sites, producing DNA fragments with a known sequence at each end.

The function of restriction endonucleases is mainly protection against foreign genetic material especially against bacteriophage DNA. The other functions attributed to these enzymes are recombination and transposition.

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7
Q

Enzymes are the agents of metabolic function
• Enzymes endow cells with the remarkable capacity to exert kinetic control over thermodynamic potentiality
True or false
What are the properties of enzymes?
Explain catalytic power under enzymes and give an example of an enzyme with catalytic power

Enzyme catalyzed reactions are much faster than the corresponding uncatalyzed reaction
E+S forward arrow and backward arrow ES forward arrow and backward arrow EP forward arrow and backward arrow E+P

A

) High reaction rates

(2) Catalyze reactions at physiological conditions (milder reaction conditions)
(3) Have a high degree of specificity (e.g. only A is converted to B)
(4) Can be regulated (e.g., A is only converted to B under certain conditions)

Catalytic Power
• Enzymes can accelerate reactions as much as 1016 over uncatalyzed rates!
• Urease is a good example: – Catalyzed rate: 3x104/sec
– Uncatalyzed rate: 3x10-10/sec – Ratio is 1x10 to the power 14
Succinyl CoA transferase is 1x10 to the power 13

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8
Q
What is the international classification of enzymes? (State the class of enzyme and the type of reaction they catalyze)
Substrates are converted to products by enzymes and products Can be converted to substrates by enzymes true or false 
What is the function of DNA ligase?
A

Oxidoreductases:transfer of electrons(hydride ions or H atoms)
Transferases:group transfer reactions
Hydrolases: hydrolysis reactions (transfer of functional groups to water )
Lyases:Addition of groups to double bonds or formation of double bonds by the removal of groups
Isomerases: transfer of groups within molecules to yield isomeric forms
Ligases: formation of C-C,C-S,C-O,and C-N bonds by condensation reactions coupled to ATP cleavage

DNA ligase joins pieces of DNA together, mainly joins Okazaki fragments with the main DNA piece.

DNA Ligase enzymes seal the breaks in the backbone of DNA that are caused during DNA replication, DNA damage, or during the DNA repair process.

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9
Q

What’s the difference between cofactors and coenzymes and give an example each

Name four metal inorganic elements that serve as cofactors for enzymes

Name four organic cofactors or coenzymes and an example of each of the chemical group transferred and their dietary precursor in mammals

What’s a prosthetic group
Give an example

A

Cofactors may be metal ions
(such as Zn2+ required for the catalytic activity of carboxypeptidase A)
Coenzymes are organic molecules
(such as the NAD+ in YADH)

Copper is a cofactor for Cytochrome oxidase
Potassium is a cofactor for pyruvate kinase
Magnesium is a cofactor for glucose-6-phosphate,pyruvate kinase
Selenium-glutathione peroxidase
Nickel (II) ion for Urease

Coenzyme
Biocytin
Chemical group transferred- Carbon dioxide
Dietary precursor- biotin

Coenzyme A
Acyl groups
Pantothenic acid

Nicotinamide adenine dinucleotide
Hydride ion
Nicotinic acid or niacin

TetrahydrofolAte
One carbon groups
Folate

Other cofactors
• Known as prosthetic groups
• Permanently attached with their protein • Often by covalent bonds
• Example: Heme in hemoglobin

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10
Q

Explain specificity as a property of enzymes
What controls this property

What is stereospecificity
Why does it arise

The same set of non-covalent interactions that enable a protein to fold are involved in stabilizing
the interaction between the substrate and enzyme. True or false

What is the difference between induced fit mechanism and lock and key mechanism

A

Enzymes selectively recognize proper substrates over other molecules
• Enzymes produce products in very high yields - often much greater than 95%
• Specificity is controlled by structure - the unique fit of substrate with enzyme controls the selectivity for substrate and the product yield

Stereospecificity
Enzymes are highly specific in both in binding chiral substrates & in catalyzing their reactions.
Stereospecificity arises because enzymes virtue of their inherent chirality.
Proteins consists of only L-amino acids.

Induced fit mechanism is most prevalent in enzymes. The enzyme active site adapts its structure to interact with the substrate & transition states.

The specific action of an enzyme with a single substrate can be explained using a Lock and Key analogy first postulated in 1894 by Emil Fischer. In this analogy, the lock is the enzyme and the key is the substrate. Only the correctly sized key (substrate) fits into the key hole (active site) of the lock (enzyme).

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11
Q

How do enzymes accelerate reactions

A
  1. Chemical reactions between the substrate and functional groups on the enzyme can provide alternative, lower-energy reaction pathways.
    (Example:group transfer through an intermediate with the group transiently covalently attached to the enzyme)
  2. Binding energy, DGB, is a major source of free energy used by enzymes to lower the activation energies of reactions.
Enzyme Kinetics
• Enzymes accelerate reactions by lowering the free energy of activation
• Enzymes do this by binding the
transition state of the reaction
better than the substrate

An enzyme helps catalyze a reaction by decreasing the free energy of the transition state. As a result, more product will be made because more molecules will have the energy necessary for the reaction to occur and the reaction will occur at a faster rate

Enzymes perform the critical task of lowering a reaction’s activation energy—that is, the amount of energy that must be put in for the reaction to begin.

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12
Q

Define rate or velocity of a reaction,rate constant,rate law,order of a reaction,molecularity of a reaction

A

Rate:change in concentration of a reactant or product in a given time

Rate is equal to the rate constant x the concentration of reactants

How much the concentration of a reactant decreases over a given period of time or the concentration of a product increases over a given period of time
Appearance of a product and disappearance of a reactant
So rate is equal to Delta(the triangle sign) concentration of the product divided by Delta (triangle sign) time and also equal to negative (delta concentration of the reactant divided by delta time )

It is measured in mol per Litre per unit time(check how it’s actually written)

Rate constant : The rate constant, or the specific rate constant, is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the molar concentrations of the reacting substances.(check the rate constant for first order,second order,third order reactions and zero order reactions)

Rate law: A rate law shows how the rate of a chemical reaction depends on reactant molar concentration. For a reaction such as aA → products, the rate law generally has the form rate = k[A]ⁿ, where k is a proportionality constant called the rate constant and n is the order of the reaction with respect to A. And if it was a product B it’ll be k[B] raised to the power m.

Order of a reaction: The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. The overall order of the reaction is found by adding up the individual orders. For example, if the reaction is first order with respect to both A and B (a = 1 and b = 1), the overall order is 2. We call this an overall second order reaction.

The order of reaction is defined as the dependence of the concentration of all reactants in a chemical reaction on the rate law expression

The molecularity of a reaction is defined as the number of reacting molecules which collide simultaneously to bring about a chemical reaction. In other words, the molecularity of an elementary reaction is defined as the number of reactant molecules taking part in the reaction.

a single-step chemical reaction is said to have a molecularity of 1 if just one molecule transforms into products. We call this a unimolecular reaction. An example is the decomposition of N2 O4. N2 O4 (g) → 2NO2 (g)

is also defined as the number of reactant molecules taking part in a single step of the reaction.

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13
Q

What is the The Michaelis-Menten Equation and what does the theory assume?
Breakdown of enzyme substrate complex to form product sis assumed to be slower than what two things ?

When the S in the equation is low,what happens to the equation rate and when it’s high what happens
What does the Michaelis-Menten equation describe ?

When is Km = [S] ?

What is Michaelis constant?

What is a first order,second order,zero order reaction

What does a Small Km mean?
What does a high Km mean?

A

Equation. v= Vmax[S] divided by Km + [S]

v- velocity of the reaction
Vmax-maximum rate achieved by the system or maximum velocity
[S]-concentration of a substrate S
Km-Michaelis constant (has a unit which is taken according to the unit of the substrate so if unit is mmol/L the. The Km unit is mmol/L)

• Louis Michaelis and Maude Menten’s theory
• It assumes the formation of an enzyme- substrate complex
• ItassumesthattheEScomplexisinrapid equilibrium with free enzyme
• Breakdown of ES to form products is assumed to be slower than
(1) formation of ES and
(2) breakdown of ES to re-form E and S

Combination of zero-order and first-order kinetics
• When [S] is low, the equation for rate is first order in [S]
• When [S] is high, the equation for rate is zero-order in [S]
• The Michaelis-Menten equation describes a rectangular hyperbolic dependence of Vo on [S]

When V = Vmax divided by 2

Km is defined as the concentration of substrate at which enzyme is working at half of maximum velocity. It is also a measure of the affinity that the enzyme has for its substrate.

First order:rate is dependent on the concentration of one reactant
Second order:dependent on conc of two reactants
Zero order: independent on conc of the reactants

Km is a constant
• Km is a constant derived from rate
constants
• Km is, under true Michaelis-Menten conditions, an estimate of the dissociation constant of E from S
• Small Km means tight binding; high Km means weak binding

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14
Q

How can you reach Vmax?
Is it practical or theoretical and why?

What is turnover number
What is the formula

A

The theoretical maximal velocity
• Vmax is a constant
• Vmaxisthetheoreticalmaximalrate of the reaction - but it is NEVER achieved in reality
• To reach Vmax would require that ALL enzyme molecules are tightly bound with substrate
• Vmax is asymptotically approached as substrate is increased

The turnover number
(also known as the molecular activity of the enzyme)
A measure of its maximal catalytic activity
• kcat, the turnover number, is the number of substrate molecules converted to product per enzyme molecule per unit of time, when E is saturated with substrate.

kcat =Vmax divided by [ET]

ET-given enzyme concentration
Vmax- max reaction rate
kcat-turnover number

Unit of kcat is per second so sec raised to the power -1 or 1 divided by sec

• If the M-M model fits, k2 = kcat
kcat = Vmax/Et
K2 is the catalytic rate constant

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15
Q

What is the catalytic efficiency of an enzyme
What is Double-Reciprocal or Lineweaver-Burk Plot(look at a pic of how it looks like)
State the equation issued in this plot

A

Catalytic efficiency of an enzyme
Name for kcat/Km
• An estimate of “how perfect” the enzyme is
• kcat/Km is an apparent second- order rate constant
• It measures how the enzyme performs when S is low
• Catalytic efficiency cannot exceed the diffusion limit - the rate at which E and S diffuse together

The double-reciprocal (also known as the Lineweaver-Burk) plot is created by plotting the inverse initial velocity (1/V0) as a function of the inverse of the substrate concentration (1/[S]). The Vmax can be accurately determined and thus KM can also be determined with accuracy because a straight line is formed. biochemistry, the Lineweaver–Burk plot (or double reciprocal plot) is a graphical representation of the Lineweaver–Burk equation of enzyme kinetics,

1/V =Km/Vmax[S] + 1/Vmax

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16
Q

State the optimum pH of pepsin,catalase,trypsin,Fumarase,ribonuclease,Arginase

State and define the types of enzyme inhibitors
State the classes of inhibition

A
P-1.5
C-7.6
T-7.7
F-7.8
R-7.8
A-9.7

Reversible versus Irreversible
• Reversible inhibitors interact with an enzyme via noncovalent associations
• Irreversible inhibitors interact with an enzyme via covalent associations

Classes of Inhibition
Two real, one hypothetical
• Competitiveinhibition-inhibitor(I) binds only to E, not to ES
Example: Malonate is a strong competitive inhibitor of succinate dehydrogenase
• Uncompetitiveinhibition-inhibitor (I) binds only to ES, not to E. This is a hypothetical case that has never been documented for a real enzyme, but which makes a useful contrast to competitive inhibition
• Noncompetitive(mixed)inhibition - inhibitor (I) binds to E and to ES

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17
Q

What changes in competitive inhibition in the M-M equation and the changes in non competitive or mixed inhibition

How is enzyme activity regulated

Why are regulatory enzymes important?

A

Kmchanges while Vmax does not

They increase Km by interfering with the binding of the substrate, but they do not affect Vmax because the inhibitor does not change the catalysis in ES because it cannot bind to ES.

At any given time, only the competitive inhibitor or the substrate can be bound to the enzyme (not both). That is, the inhibitor and substrate compete for the enzyme. Competitive inhibition acts by decreasing the number of enzyme molecules available to bind the substrate.

Km Unchanged Vmax Reduced

In non-competitive inhibition, the inhibitor binds to an allosteric site and prevents the enzyme-substrate complex from performing a chemical reaction. This does not affect the Km (affinity) of the enzyme (for the substrate).

Noncompetitive inhibition, a type of allosteric regulation, is a specific type of enzyme inhibition characterized by an inhibitor binding to an allosteric site resulting in decreased efficacy of the enzyme. An allosteric site is simply a site that differs from the active site- where the substrate binds.

Two ways that this may occur:
1) Control of enzyme availability
Depends on rate of enzyme synthesis & degradation
2) Control of enzyme activity
Enzyme-substrate binding affinity may vary with binding of small molecules called allosteric effectors (ex: BPG for Hb)
Allosteric mechanisms can cause large changes in enzymatic activity

important in controlling flux through metabolic pathways
1. Allosteric enzymes
2. Regulation by covalent modification: Covalent modifications are enzyme-catalysed alterations of synthesised proteins and include the addition or removal of chemical groups. Modifications can target a single type of amino acid or multiple amino acids and will change the chemical properties of the site
3.Regulation by Feedback Inhibition : Feedback inhibition occurs when the end product of a reaction interferes with the enzyme that helped produce it. The inhibitor does this by binding to a second active binding site that’s different from the one attached to the initial reactant. The enzyme then changes its shape and can’t catalyze the reaction anymore. example:
Conversion of L-threonine to L- isoleucine catalyzed by a sequence of five enzymes, E1-E5
L-isoleucine is an inhibitory allosteric modulator of E1

Regulatory enzymes, which facilitate the transferring of phosphate groups to the specific substrates, are called kinases

Regulatory molecules. Enzymes can be regulated by other molecules that either increase or reduce their activity. Molecules that increase the activity of an enzyme are called activators, while molecules that decrease the activity of an enzyme are called inhibitors

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18
Q

What is DNA replication
What is the importance of high fidelity of replication ?
What should germ cells be protected against?
What should somatic cells be protected against?

In all cells, DNA sequences are maintained and replicated with high fidelity.
•The mutation rate, approximately 1 nucleotide change per 109 nucleotides each time the DNA is replicated, is roughly the same for organisms as different as bacteria and humans.
•Because of this remarkable accuracy, the sequence of the human genome (approximately 3x109 nucleotide pairs) is changed by only about 3 nucleotides each time a cell divides.
•This allows most humans to pass accurate genetic instructions from one generation to the next, and also to avoid the changes in somatic cells that lead to cancer.
True or false

A

All organisms must duplicate their DNA with
extraordinary accuracy before each cell division. This process is called DNA replication.
.
• Maintaining order requires the continued surveillance and repair of the genetic information because DNA inside cells is repeatedly damaged by chemicals and radiation from the environment, as well as by thermal accidents and reactive molecules.
• Despite the great efforts that cells make to protect their DNA, occasional changes in DNA sequences do occur. Over time, these changes provide the genetic variation upon which selection pressures act during the evolution of organisms
• Whereas germ cells must be protected against high rates of mutation to maintain the species, the somatic cells of multicellular organisms must be protected from genetic change to safeguard each individual.

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19
Q

Explain DNA replication is semi conservative according to Meselson and Stahl experiment
What are the properties of DNA polymerase

A

DNA replication is Semiconservative
The two old DNA strands serves as a template for the formation of an entire new strand.

properties common to ALL DNA polymerases
1) Catalyze the polymerization of deoxyribonucleotides in the 5’ to 3’ direction.: DNA synthesis catalyzed by DNA polymerase. DNA polymerase catalyzes the stepwise addition of a deoxyribonucleotide to the 3’-OH end of a polynucleotide
chain, the primer strand, that is paired to a second template strand. The newly synthesized DNA strand therefore polymerizes in the 5’-to-3’ direction.
Because each incoming deoxyribonucleoside triphosphate must pair with the template strand to be recognized by the DNA polymerase, this strand determines
which of the four possible deoxyribonucleotides (A, C, G, or T) will be added.
2) Require a template.
3) Require a primer

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20
Q

What are the biological roles of DNA polymerase I

Explain elongation as a step in DNA replication

A
• removes RNA primers
• fills gap with DNA
• DNA repair
• 5’→3’ polymerase fills gap left by repair enzymes which excise regions of DNA containing damaged or mispaired nucleotides (more later)
• Processivity: usually catalyzes ~20 nt additions before falling off template.
Enymatic activities of DNA Pol I
• 5’ -> 3’ Polymerase
• 3’ -> 5’ Exonuclease
• 5’ -> 3’ Exonuclease

DNA polymerase type I cuts out primers in a 5’ to 3’ exonuclease activity to pluck out RNA primers in the leading strand . It reads the DNA strand which the primer has been plucked out from,from 3’ to 5’ and synthesizes from 5’ to 3’. It also proof reads from 3’ to 5’ and if any mistake is found it cuts it out in a 3’ to 5’ exonuclease activity .
Same for the lagging strand.

Elongation :primase makes RNA primers Which enable DNA polymerase type III to make DNA since it needs the 3’ OH of RNA primers to carry out its activity. It reads the first DNA parent strand from 3’ to 5’ and It synthesizes RNA primers in a 5’ to 3’ fashion. DNA polymerase now also reads the DNA strand from 3’ to 5’ and Makes a DNA strand in a 5’ to 3’ fashion. Or opposite fashion of the parent strand .So RNA primers are produced and the DNA polymerase follows and makes the DNA strand so it’s like RNA primer(5’ to 3’) then DNA strand(continues where the RNA primer ended which is at 3’ and continues with 5’ to 3’) . The strand that theDNA polymerase makes and that’s trans it toward the replication fork is called the leading strand. The primase comes to the other parent strand which will be in the 3’ to 5’ and read that strand and synthesize complimentary nucleotides from 5’ to 3’. So now there’s an OH at the 3’ strand and DNA polymerase continues and reads the strand from 3’ to 5 thereby continuing what the primase started and synthesizes the strand from 5’ to 3. This new strand is the lagging strand
But the difference is that there are DNA strands in between the RNA primers meanwhile in the leading strand the RNA primer is only seen before the DNa strand and it’s all DNA. No RNA in between. Okazaki fragments are seen in lagging strand. These fragments are defined as fragments where there are multiple RNA primers and multiple stretches of DNA.
DNA polymerase proof reads to prevent mistakes. It goes back the strands it has read to make sure everything has been paired correctly. It reads from 3’ to 5’. And if there are any mistakes it uses a 3’ to 5’ exonuclease activity to cut it out and put in the correct complimentary nucleotide.
DNA polymerase type I cuts out primers in a 5’ to 3’ exonuclease activity to pluck out RNA primers in the leading strand . It reads the DNA strand which the primer has been plucked out from,from 3’ to 5’ and synthesizes from 5’ to 3’. It also proof reads from 3’ to 5’ and if any mistake is found it cuts it out in a 3’ to 5’ exonuclease activity .
Same for the lagging strand. In the lagging strand though this creates gaps between the DNA strands since that’s where the primers used to be. Ligase enzyme comes on the lagging strand and fuses the DNA ends together. In HIV their T cells are infected. And replication occurs a lot. Drugs target the T cell replication process. Example of such drugs are Nucleoside reverse transcriptase inhibitors(didanosine) . These remove the 3’ OH region so the DNA polymerase III is not able to build on it to make a DNA strand .

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21
Q

Explain the fidelity of DNA replication

A

Fidelity of DNA replication
Low error frequency (≈ 10-9) accounted for by redundant safeguards.
1. Binding pocket of DNA polymerase clamps tightly around the base before catalysis occurs. Wobble pairs (Non Watson/Crick geometry) don’t fit and so catalysis can’t occur.
• But binding pocket is unlikely to be rigid enough to exclude wobble pairs every time.
2. Enol tautomers(Watson/Crick geometry but rare tautomer) are very unstable (keto/enol tautomerization equilbrium constants are in the 10-5 to 10-3 range).
• But this isn’t enough to account for the 10-9 error frequency.
3. DNA polymerases have “editing exonuclease activities” that allow them to erase mistakes and try again
4. Cells contain mismatch repair systems that come along after DNA polymerase to clean up any residual errors

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22
Q

How is replication fidelity increased by mismatched repairs
What is the structural basis for proofreading
What is a replisome
What’s a lagging strand and a leading strand
What is processivity
What is fidelity

Explain mismatch repair mechanism
State the proteins involved

A

DNA methylation
DNA methylation is delayed after replication

Switch between Polymerization and Editing Modes
in DNA Polymerase: Structural Basis for “Proofreading”

Replisome: replicating machinery that moves DNa along at the replication fork
It consists of DNA polymerase III
Primase,sliding clamp,single stranded burning protein,clamp loader,Helicase

A strand that is synthesized discontinuously in the 5’ to 3’ direction away from the replication fork

A strand that is synthesized in the 5’ to 3’ direction towards the replication fork. It is synthesized continuously

Number of catalytic turnovers per binding event ie. the number of nucleotides incorporated before DNa poly erase dissociates from DNA

DNA polymerase I had a processivity number of 10 to 100 nucleotides

DNA poly erase III has a processivity number in the thousands because of the beta sliding clamp(ring shaped protein that encircles the DNA polymerase of the lagging strand and keeps it from falling off when it restarts DNA synthesis at a new Okazaki fragment.

Fidelity is the frequency of errors. Fidelity is high due to the 3’ to 5’ exonuclease activity of DNA polymerase I and DNA polymerase III
They both have a net error of 1 error per 10 raised to the power 6 to 10 raised to the power 8 nucleotide additions.

During DNA replication DNA polymerase may add wrong base during elongation. The wrongly added base results in mismatched nucleotides because of the presence of this mismatch the DNA strands get distorted. Mismatch repair is the mechanism by which mismatched nucleotides are removed .
Mut S,Mut L,Mut H
Mut S (mismatch recognition protein) recognizes mismatched nucleotides in the DNA.
Mut L binds to the Mut S protein
Mut H binds to Mut S and Mut L complex
The parent strand is methylated but the new daughter strand is not methylated
The GATC sequence in the parent strand is methylated
This structure is called hemi methylated DNA meaning one strand is methylated while the other strand is not.

The mismatch proteins can recognize methylated and non methylated strands
Mut H protein is endonuclease and Mut L activates the endonuclease activity
Mut H searches for the nearest GATC sequence with methylated adenine
During the search,the DNA is looped out.
When the sequence is found,the Mut H protein cleaves at the daughter or new strand

Helicase II unwinds the cleaved strand
The unwound strand is removed and cleaved by exonuclease enzyme
DNA polymerase III replicates the DNA and adds correct nucleotides and the gap is sealed by DNA ligase

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23
Q

Explain DNA polymerase proof reading

Explain the exonuclease activity of polymerase I and it’s biological significance

A

DNA polymerase has three domains namely the palm domain (responsible for exonuclease and proofreading activity),fingers domain(polymerization activity), thumb domain(for structural integrity part and holding DNA in the proper shape )
During nucleotide addition when they encounter a wrong nucleotide the polymerase chnages the structure and brings the newly added wrong nucleotide to the palm domain or exonuclease activity domain and this domain cuts the wrongly put nucleotide and the original conformation is gotten back and they start adding the nucleotide sequences again

Biological significance:
- Allows the replacement of damaged or abnormal DNA sequences
by “Nick translation”
- Also allows the removal of RNA sequences embedded in DNA(removal of replication primers)

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24
Q

Explain the two problems posed by the properties of the known DNA polymerases and state the solution
Why is DNA replication semi discontinuous?

Synthesis of DNA on the lagging strand requires continuous synthesis
In bacteria, the Okazaki fragments are about 1000-2000 bp in length true or false

A

Two problems posed by the properties of the known DNA polymerases
1.
The directionality problem. How can DNA polymerase replicate both strands behind each replication fork, when all polymerases operate in the 5’ to 3’ direction?
􏰀 Solution - semidiscontinuous DNA synthesis

2.The priming problem. Since all DNA polymerases require a primer (usually of at least 10 nucleotides in length), where do the primers come from?
􏰀 Solution - primers are made of RNA

Semi discontinuous:
During DNA replication one of the two strands specifically the leading strand is replicated continuously in the 5’ to 3’ direction while the other strand lagging strand is replicated discontinuously or in pieces from the 3’ to 5’ direction
This is important because DNA polymerase (the enzyme that synthesizes a new DNA strand using a template strand) can only add nucleotides to the 3’ end of a polynucleotide strand
Overall DNa replication is semi discontinuous

So when the Helicase unwinds the DNA it exposes the strands as Leading strand template and lagging strand templates
DNA polymerase attaches to the 3’ end of the RNA primers and the polymerase adds dna nucleotides in the 5’ to 3’ direction
DNA ligase eliminates the nick or the gaps formed when the RNA primers are removed and replaced with DNA nucleotides
Primase synthesizes short RNA primers on each template strand from the 5’ to 3’ direction

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25
Q

Explain the replisome of E. coli

Explain the sub units of DNA polymerase III

A

Helicases
Unwind DNA at the replication
fork in a reaction coupled to ATP Hydrolysis
2) Single-stranded DNA
binding proteins (SSB)
Bind and stabilize the DNA in a single stranded conformation
after the melting by helicases 3) The Primosome Synthesizes RNA primers
of the lagging strand Contains Primase
4) DNA Polymerase III : The replicase
5) DNA topoisomerase II Relaxes supercoiled DNA that forms ahead of the replication fork. Decatenates
the final product
6) DNA Polymerase I Replaces RNA primers with DNA by nick translation
7) DNA Ligase Joins the Okazaki
fragments

DNA Polymerase III Core comprises of three sub units namely:
Alpha sub unit:Catalytic site for polymerization
Epsilon sub unit ;3’->5’ editing exonuclease or proofreading activity
Theta sub unit: structural role or stimulate the exonuclease activity

It has two catalytic cores

DNA Polymerase III: is a multisubunit Enzyme

  • gamma Complex:
    4 polypeptides
    ATP-dependent conformational changes facilitates the loading of the β clamp onto DNA or it places the processivity sub unit ie. the beta clamp on the DNA
  • beta Subunit or processivity sub unit:
    A homodimer
    ATP-dependent processivity
    factor = β clamp Pol.III Core is poorly processive by itself. Beta clamp is responsible for holding the catalytic core on their template strand

The clamp controls the association of core enzymes with DNA .
The Helicase creating the replication fork is connected to two DNA polymerase catalytic sub units each of which is held on to DNA by a sliding clamp. The polymerase that synthesizes the leading strand moves continuously. The core on the leading strand is processive because it’s clamp keep it on the DNA . The clamp associated with the core on the lagging strand dissociates at the end of each Okazaki fragment and read so coated with a primer in the single stranded template loop to synthesize the next fragment reassembles for the next fragment
Helicase or DNaB is responsible for interacting with the primase DnaG to imitate each Okazaki fragment

  • tau sub unit Dimerization factor or dimerizing sub unit :Holds two Pol. III catalytic cores together
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26
Q

Polymerase III holoenzyme functions with processivity,catalytic potency and fidelity
Explain these terms with regards to polymerase III

A
  1. Processivity: means the enzyme remain bound to the template for many rounds of nucleotide addition (probably whole synthesis) Pol III core (αεθ) alone only has processivity of 10-15 residues
  2. Catalytic potency: not only processive, but efficient
    Result of 1 & 2: fastest known polymerization reaction in vivo
  3. Fidelity of Pol III replication
    α subunit: 5’→3’ polymerase
    ε subunit: 3’→5’ editing exonuclease
    Why have 3’→5’ exonuclease activity? allows enzyme to proofread newly synthesized DNA.
    If a wrong (unpaired) base is incorporated, polymerase activity is inhibited & 3’→5’ exonuclease excises wrong nucleotide. 3’→5’ exonuclease activated by unpaired 3’- terminal nucleotide with free 3’OH.
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27
Q

Sliding b clamps ensure processivity of DNA polymerase III
True or false

How do we get clamp on and off?
What is nick translation

Summary of DNA replication paradigms
• Semiconservative
• Bidirectional
• One origin per bacterial chromosome • Semidiscontinuous
• RNA primed
True kr false
A

True

Topoisomerases relax positive supercoils ahead of the replication fork and decatenate the final products
1.Topoisomerase required to relax positive supercoils ahead of replication fork
2.Not enough room between the converging replication forks for topoisomerase
If topo II is absent at this stage,
the daughter chromosomes can’t
3.Type II topoisomerase required to decatenate the final products

Pol I excises the RNA primers by “nick translation”(Nick translation is the name given to a reaction that is used to replace cold nucleoside triphosphates in a double-stranded DNA molecule with radioactive ones (1,2). Free 3’-hydroxyl groups are created within the unlabeled DNA (nicks) by deoxyribonuclease 1 (DNAse 1). Or nick translation is tagging DNa with fluorescence labels or labeled dNTPs . The DNA to be nick translated is treated with DNAase 1 enzyme which produces small single stranded nicks or cuts in the DNA. The nicked DNa is then treated with DNA polymerase I enzyme. DNa pol I Carrie’s out two simultaneous reactions: additions of nucleotides in 5’ to 3’ direction and removal of nucleotides in 5’ to 3’ direction. Hence the sample DNA is labeled. Once the nick translation is completed the nicks are sealed with DNA ligase enzyme

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28
Q

The eukaryotic replisome is homologous in many respects to the bacterial replisome!!!
True or false

A

Pol 􏰃 - the eukaryotic replicase
Pol alpha /primase - contains both primase and DNA polymerase activities
PCNA - trimeric sliding clamp Replication Factor C (RFC) - the
clamp loader

Each chromosome has multiple origins. Why?
• Time for DNA replication is limited - e.g., mammalian S phase lasts 6-8 hours.
• Eukaryotic forks move at only about 1/10th the rate of bacterial forks
• Chromosomes can be in excess of 108 bp
􏰅 Completion of replication in the allotted time necessitates multiple
origins.
Origin Recognition Complex (ORC) - a complex of 6 ATPases
- the functional equivalent of DnaA

MCMs - a heterohexameric helicase
Replication Protein A (RPA) = SSB
RNase H - nuclease that is specific for RNA in RNA/DNA hybrids - excises primers

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29
Q

What is telomerase

What’s the clinical relevance of telomerase

A
Telomerase: a reverse
5’
transcriptase, with a built- etc.
in RNA template
Humans: TTAGGG 
Tetrahymena: TTGGGG 10-1000 repeats.

Telomerase–aging, cancer, and disease
􏰄 Most somatic cells have low or undetectable level of telomerase activity.
􏰄 Telomere length is correlated with cellular aging.
􏰄 Telomerase is active in some germline, epithelial, and
stem cells (haemopoeitic cells), and in >90% of cancer cell lines.
􏰄 Mutations in the RNA component of human telomerase have been linked to autosomal dominant dyskeratosis congenita and some forms of aplastic anemia.
􏰄 Telomerase is required for telomere maintenance.
􏰄 Telomerase is responsible for the immortal phenotype of
cancer cells.
After a few generations, telomerase-mutant mice exhibit reduced fertility, signs of premature aging, and shortened
life-span.

Psychological stress leads to reduced telomerase activity and increased telomere shortening! This presumably results in premature aging!!
Telomere length
Study conducted on mothers under stress due to need to care for a chronically ill child.
• Telomere shortening in high stress subjects is the equivalent of one decade of additional aging!

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30
Q

What are the differences between replication in eukaryotes and prokaryotes

A

Prokaryotes
1.One specific initiation
point (Ori)
2. Three types of DNA polymerases, I, II and III
3. Diverse functional variety specially that of DNA Pol I
4. Not applicable

• Prok
5. DNA Pol I is main repair
enzyme
6. Few replication forks
7. Theta structure observed
8. Accessory proteins few with limited function
9. Only unwinding takes place
10. No histones, few replication bubbles
11. RNA as primer
12. Primase makes primer

• Eukaryotes

  1. Multiple specific initiation points but different from prokaryotes
  2. Five types of DNA polymerases, α,ε,γ,β and δ
  3. Functional variety of DNA Pol is specific.
  4. DNA Pol. γ replicates mitochondrial DNA

• Euk
5. DNA Pol β is the main repair
enzyme
6. Many replication forks
7. Theta structure not observed
8. Many accessory proteins with diverse functions
9. Histone separation from DNA as well as unwinding takes place
10. Many replication bubbles 11. RNA/DNA as primer
12. Pol α/ primase makes primer

Summary of DNA Replication
• Identification of the initiation site of replication (OriC)
• Unwinding of parental DNA (DS -> SSDNA)
• Formation of replication fork
• Synthesis of RNA primer complementary to the DNA template by primase
• Leading strand is synthesized in the 5’-3’ direction by DNA Pol.
• Lagging strand is synthesized as Okazaki fragments
• RNA primers removed when polymerization is done
• The gaps are filled by dNTPs and the pieces are joined by DNA ligase which requires energy from ATP.

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31
Q

State five things that cause DNA damage
State various DNA damages that need to be repaired
What is nitrosonium ion

A

DNA damage
A gallery of horrors
– 1. UV damage
– 2. Environmental Chemicals
e.g. alkylating agents –
3. Normal Physiological agents H20 (Hydrolytic deamination)
depurination 02 Oxidation)
nitrites (Oxidative deamination) Alkylation
- 4. Replication errors (wrong base) – Et cetera

various DNA Damages that need to be repaired:
-Pyrimidine dimers
(UV light)

UV-induced formation of pyrimidine dimers in DNA is a major deleterious event in both eukaryotic and prokaryotic cells. Accumulation of cyclobutane pyrimidine dimers and pyrimidine (6-4) pyrimidone photoproducts can lead to cell death or be at the origin of mutations.

  • Alkylation of bases
    Methylation of guanine N6 G-> O6meG
  • Hydrolysis of glycosidic bond
    (Depurination)
  • Deamination of bases

Another common type of DNA damage that occurs under physiological conditions is the hydrolytic deamination of cytosine to form uracil (5).
•Spontaneous: Spontaneous deamination is the hydrolysis reaction of cytosine into uracil, releasing ammonia in the process. This can occur in vitro through the use of bisulfite, which deaminates cytosine, but not 5-methylcytosine.
•Chemically induced
•C->U
•5 meC -> T
•A->HX: Deamination of adenine results in the formation of hypoxanthine. Hypoxanthine, in a manner analogous to the imine tautomer of adenine, selectively base pairs with cytosine instead of thymine. This results in a post-replicative transition mutation, where the original A-T base pair transforms into a G-C base pair.

  • Oxidative damages
    •G -> 8 oxoguanine
    •Strand Break

Nitrosonium ion
•Electron hungry
•Formed from: nitrates and nitrites (common food preservatives, but also naturally occurring in foods such as spinach)
•Also formed from nitrosamines (byproducts of rubber production)

The nitrosonium ion is a critical step in the creation of carcinogenic nitrosamines, as well as the deamination of DNA or amine-containing amino acid residues in proteins.

So processed foods are dangerous.
But so is water!

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32
Q

Why was uracil not selected as a natural base for DNA?
Explain oxidative damage in DNA
What is the consequence of oxidative damage in dna
How is oxidative damage reduced

A

as a natural base in DNA:
If Uracil were a natural
base, the DNA repair machinery would not know whether these Uracil are “normal” uracil that
come from incorporation by polymerase, or “non-natural”
uracil coming from deamination
of cytosines.
Since there is no way to discri- minate between these, the
genetic systems did not select uracil as a natural base (exceptions)
in DNA
’TGUA Due to incorporation of dUTP by polymerase

Source of oxidative agents

Oxidative DNA damage provides direct routes to mutations. While guanine usually pairs with cytosine, 8-oxo-7,8-dihydroguanine (8-oxoG), the most frequent type of oxidative base damage, may cause mispairing with adenine through a conformational change. This is one route to oxidative DNA damage induced mutations.

8-oxo-guanine generates replication block or G-C -> T:A transversions after DNA Replication
G-C
DNA Polymerases tend to incorporate A opposite to 8-oxoG because of the tendency of 8-oxoG to switch in the syn conformation; A is the only nt that can form a base pair
8xoG•C
H2O with syn8-oxoG whose geometry vaguely 2 resembles that of a Watson-Crick base pair

Oxidative damage, produced by intracellular ROS, results in DNA base modifications, single- and double-strand breaks, and the formation of apurinic/apyrimidinic lesions, many of which are toxic and/or mutagenic

• Respiratory Chain:

Accordingly, cells have evolved a balanced system to neutralize the extra ROS, namely antioxidant systems that consist of enzymatic antioxidants such as superoxide dismutase (SOD)(breaks down oxygen and hydrogen into hydrogen peroxide and oxygen) , catalase (CAT) (converts hydrogen,oxygen and hydroxide to water and oxygen)and glutathione peroxidases (GPxs), thioredoxin (Trx) as well as the non-enzymatic antioxidants which collectively reduce

No cellular Neutralization !!! -> Main source of Oxidative agent

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33
Q

Explain alkylation in dna damage

Consequences to dna ,examples of alkyl agents

A

Any nucleophilic atom on the DNA (e.g., N7, N6, N3, O2, O6, etc.) can be alkylated sometimes leading to changes in basepairing specificity. For example, alkylation of guanine O6 can lead to GC to AT transition mutations.

Alkylation damage of DNA is one of the major types of insults which cells must repair to remain viable. One way alkylation damaged ring nitrogens are repaired is via the Base Excision Repair (BER) pathway.

Alkylation of guanine or other bases results in abnormal base pairing as well as the excision of these bases, which in turn leads to strand breakage.

Alkylation damage repair involves multiple partially redundant pathways, which include direct reversal by O6-methylguanine DNA methyltransferase (MGMT), the ALKB family of demethylases, and base excision repair (BER).

Alkylating agents exert cytotoxic effects by transferring alkyl groups to DNA, thereby damaging the DNA and interfering with DNA transcription and cell division.

Some favorite alkylating agents
Ethylmethane sulfonate (EMS) - a favorite among geneticists MNNG - another tool of geneticists - a nitrosamine
Cl-
Nitrogen mustard
Alkyl groups transferred are shown in red
Mustard gas

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34
Q

State five ways to minimize damage to the dna
State five dna repair strategies
Explain three

Three excision repair pathways exist to repair single stranded DNA damage: Nucleotide excision repair (NER), base excision repair (BER), and DNA mismatch repair (MMR). True or false

A

How to minimize damage to your DNA
• Avoid chemical warfare • Avoid processed foods • Avoid the highways
• Avoid sunlight
• Avoid aerobic activities
• Avoid water
But if all these precautions fail, our cells have multiple DNA repair pathways to undo the damage!!!

Some DNA repair strategies
• Direct reversal of the damage
• Base excision repair
• Nucleotide excision repair
• Methyl directed mismatch repair
• SOS repair
• Double-strand Break Repair
• Recombination repair

-Bypass of lesions: avoids DNA replication stalls
• bypass of 8 oxoG by DNA polymerase 􏰆
-> not really a “repair”, more like a quick fix..

  • Direct Reversal of Damage
    • Photolyase reversion of Y dimers
    • Dealkylation of 1mA and 3mC by AlkB

Direct reversal of the damage
• DNA Photolyase - uses energy from sunlight to split pyrimidine dimers
– Absent in placental mammals
– In one study (Current Biology 15, 105-15), transgenic mice carrying a bacterial photolyase gene were found to be resistant to sunlight induced skin cancer
– The solution to the thinning ozone layer?!! • Dealkylating enzymes
– O6-methyl guanine methyl transferase (MGMT) - a suicidal “enzyme” that dealkylates G O6
– AlkB - promotes dealkyation of A N1 and C N3

  • Base excision repair
    • Uracil-N glycosylase
    • 8-oxoG glycosylase
  • Nucleotide excision repair
    • Bacteria: UvrA, UvrB, UvrC, Helicase II
    (UvrD)
    • DNA pol. I, DNA ligase
    • Eukaryotes : Xeroderma pigmentosum proteins,
    TFIIH
  • Methyl directed Mismatch
    repair
    • Dealkylation of guanines by suicidal MGMTase
    Dealkylation of guanines by Methyl Guanine Methyl Transferase: Mutations of Human Homologues of O6MGMT (methyl guanine methyl transferase) linked to cancer : another proof that maintaining DNA information is required for
    tumor suppression
    • The “inactivated” enzyme serves as a transcription factor to induce expression of DNA repair genes ->”biosensor” for DNA repair
    • MutS, MutL, MutH
    For more information regarding the mechanisms of DNA repair enzymes, http://www.scripps.edu/~cliff/class/class.html

Oxidative Demethylation
by AlkB; Oxidation of 1-methyladenine and 3-methylcytosine by AlkB requires O2, - ketoglutarate (KG) and Fe(II), and generates CO2 and succinate. Oxidized methyl groups are released as formaldehyde resulting in direct reversal of the lesions to the unmodified base residues.

Base excision repair pathway
X = damaged base (e.g.,uracil,hypoxanthine,8-oxoguanine,etc)
There are many dozen such
enzymes each one recognizing
a different X (e.g., uracil-N-
glycosylase)
In bacteria, the exonuclease and polymerase functions are both provided by Pol I through the process of nick translation

Base excision repair (BER) corrects small base lesions that do not significantly distort the DNA helix structure. It is initiated by a DNA glycosylase that recognizes and removes the damaged base, leaving an abasic site which is further processed by short-patch repair or long-patch repair. BER takes place by short-patch repair or long-patch repair that largely use different proteins downstream of the base excision. The repair process takes place in five core steps: (1) excision of the base, (2) incision, (3) end processing, and (4) repair synthesis, including gap filling and ligation.

Nucleotide excision repair:
In humans, this step catalyzed by XP-A, XP-C, XP-F, XP-G
TFIIH (contains XP-B and XP-D)
DNA polymerase b
 XP = xeroderma
pigmentosum

The main difference between base excision repair and nucleotide excision repair is that the base excision repair pathway corrects only the damaged bases, which are non-bulky lesions, whereas the nucleotide excision repair pathway corrects bulky DNA adducts through the removal of a short-single stranded DNA segment . Nucleotide excision repair (NER) is the main pathway used by mammals to remove bulky DNA lesions such as those formed by UV light, environmental mutagens, and some cancer chemotherapeutic adducts from DNA. nucleotide excision repair (NER), damaged bases are cut out within a string of nucleotides, and replaced with DNA as directed by the undamaged template strand. This repair system is used to remove pyrimidine dimers formed by UV radiation as well as nucleotides modified by bulky chemical adducts.

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35
Q

What is xeroxerma pigmentosum

Methyl-directed mismatch repair in prokaryotes
Uvr genes = genes that promote UV Resistance
Mut genes = When these genes ar mutated, bacteria show increased
rates of mutations true or false

A

A defect that is due to a defect in nucleotide excision repair(XP group A to 6). It’s a rare skin disorder marked by extreme sensitivity to sunlight
NER is the sole pathway for repairing thymine dimers in humans
NER is also involved in repairing oxidative damage
When NER is defective,translesion bypass mechanism is used
NER is coupled to transcription and the NER pathway in humans is the same in E. coli but involves different proteins in genetic groups

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36
Q

Explain mismatch repair

A

Mismatch repair
Eukaryotes
• Similar mismatch repair system contains MutL and MutS homologues.
• In humans, mutations in these genes are responsible for Hereditary nonpolyposis colorectal cancer ( HNPCC), an inherited predisposition to colon cancer.
• Eukaryotes lack both Dam methylase and mutH
• It’s not known how the mismatch repair machinery differentiates between the parental and daughter strands.

37
Q

State the differences between A DNa ,B DNa and Z dna

H20 is essential in the transition A <–> B DNa
A water spine is present in the minor
groove of B-DNA
True or false

Z- DNA
• Occurs in DNA sequences with stretches of consecutive G-C base pairs
• Left-handedhelix
• Jaggedbackbone
• Requires high salt
• G nucleotides switch from C2’ endo to C3’ endo and no change in C nucleotide sugar pucker.
True or false

A

Sugar pucker
B-DNA :C2’-endo
A-DNA: C3’-endo

 Rise/residue Bdna-  3.4Å  Adna- 2.6Å

Residues/turn
B- 10.5
A-11

Helical twist
B- 34Þ
A-33Þ

Diameter
B-20Å
A- 26Å

Tilt
B- 6Þ
A-20Þ

Propellor twist B-   12Þ  A-15Þ

Base pairs are more tilted in A-DNA.

Major differences between A and B-DNA

1) A-DNA is shorter due to different sugar pucker
2) Bases shifted away from helical axis in A-DNA:
a) Results in cavernous major groove and shallow minor groove
b) Results in 6 Å hole
3) Base pairs dramatically tilted in A-DNA

Type of helix for the types of DNA
A dna- right handed
B DNa- right handed
Z dna-left handed

Helical diameter:
A- 2.55nanometer (nm)
B- 2.37 nm
Z - 1.84

Rise per base pair(nm)
A-0.29
B-0.34
Z-0.37 means Z dna contains more bases per turn

Distance per complete turn(pitch) value is (nm)
A-3.2
B-3.4
Z-4.5 means a complete turn for Z DNa is longer than the others

Number of base pairs per complete turn:
A-11
B-10
Z-12 this is seen cuz the rise is also higher because per turn they carry more base pairs that’s why they have a higher pitch

Topology of major groove:
A- narrow but deep
B- wide and deep
Z-flat

Topology of minor groove
A- broad but shallow
B-narrow but shallow
Z-narrow but deep

38
Q

Explain the Relative stability of A- vs. B-form helices

A

DNA
1. In aqueous solution, B-DNA is favored over A-DNA, apparently due to B-DNA’s “spine of hydration”.
2. Reduced water activity in solutions containing high concentrations of organic solvents (or in partially dried out DNA fibers favors A-DNA.
RNA - Steric crowding with 2’-OH group in the C2’ endo sugar pucker forces RNA double helices into A-conformation

39
Q

Helix/Coil Transition in DNA
What influences the equilibrium?

Hyperchromic Effect: SS DNA > native DNA
True or false

A

In favor of single-stranded DNA
1. Electrostaticrepulsion
2. Conformationaland translational entropy
In favor of double- stranded DNA
1. H-bonds (minor component)
2. Basestacking(induced dipole interaction

40
Q

What are the major types of RNA ?

Red elements (Cap, polyA tail) are not encoded with The genes: they are added after transcription for a mature eukaryotic mRNA after processing

A
Major types of RNA in Humans
• mRNA 
• rRNA 
• tRNA
• The primary structure of RNA is defined as the number and sequence of ribonucleotides in the chain.

tRNAs are adaptor molecules
tRNA structure (cont.)
Seconday structure
• The clover leaf
Tertiary structure
• Two A-form helices arranged in an L- shape
• Tertiary contacts between conserved residues in the D and TC loops, which meet at the bend in the L, stabilize the structure

41
Q

• Requirements for protein synthesis; ribosome, mRNA, soluble protein factors and aminoacyl-tRNA.
• The genetic code and the role of tRNA in protein synthesis.
• Charging of amino acids to their cognate tRNAs, the role of aminoacyl-tRNA synthetases.
• Ribosomes as the engine for protein synthesis and the peptidyl transferase reaction.
• Initiation, elongation and termination of protein synthesis in prokaryotes.
• Comparing protein synthesis in prokaryotes and eukaryotes and how the differences are explored by the use of antibiotics.
True or false
Explain what translation is

A

The template-strand of DNA is copied into mRNA
“coding” strand “template” strand Transcript
Coding strand: 5’- ATGCCAGTAGGCCACTTGTCA-3’
Template strand:
3’- TACGGTCATCCGGTGAACAGT-5’

Both the coding and template strand are DNA
Transcript:
5’- AUGCCAGUAGGCCACUUGUCA-3’ mRNA

RNA polymerases bind to the “template- strand” (by definition)
-they read the template strand in a 3’ to 5’ direction while
synthesizing mRNA in the 5’ to 3’ direction.

Translation provides the framework for the second statement of the Central Dogma (RNA specifies Protein).

The template strand is also called the “antisense” strand. This is a useful terminology since “antisense RNA probes/drugs” are complementary and therefore bind, to RNA.

42
Q

Why do we need translation

What is the function of GTP in translation

A

What do we need for protein synthesis?

  1. mRNA, properly processed and spliced.
  2. Ribosomes, which acts as the engines for protein synthesis.
  3. Amino-acyl tRNAs, which acts as the adaptors to bring amino acids to the ribosome in a high energy state.
  4. Soluble protein factors

GTPases that couple conformational changes to GTP hydrolysis are ubiquitous in biology and
particularly in signal transduction (e.g., the Ras oncoprotein is another such enzyme).
in translation, they act as GTP fueled motors to drive the process of protein synthesis forward.

Role of GTP hydrolysis
Can get translation in absence of GTP, but slowly. GTP hydrolysis is not associated with covalent bond formation;
rather, it is coupled to conformational changes of translational machinery.
⇒ important for making reactions fast and irreversible (conformational change to inactive state when GTP → GDP)
We’ll see this for all stages of translation.
⇒important for fidelity (accuracy) of protein synthesis (proofreading after GTP hydrolysis on EF-Tu).
We won’t have time to talk about this

43
Q

What is mRNA

Eukaryotic mRNAs are usually what?

A

mRNA – the carrier of genetic information from DNA to the site of translation (the cytoplasm of eukaryotic cells)

Eukaryotic mRNA’s are:
- Usually monocistronic, encoding one polypeptide -Usually spliced ( eukaryotic DNA sequence is
not “co-linear” with mature RNA).
- Capped at the 5’ end and polyadenylated at the 3’ end.
- Structured such that information specifying the sequence
of protein is presented in an “ open-reading-frame” (ORF).
-Structured such that the ORF is specified by a translation initiation signal which usually begins at the AUG sequence closest to the 5’ end and terminates in one of three sequences or stop codons (UAA(you are away),UGA(you go away), UAG(you are gone))

44
Q

What is the features of the genetic code
Explain how How will the translational machinery know when to start and stop
State six important lessons to note in the genetic code

A

The “Genetic Code”:
- General Features:
- It is a triplet code (specified by DNA but read from RNA)
- Each triplet is called a “codon”.
- 43 = 64 possible triplet combination of 4 bases (A, G, C, U).
-61 of these are coding (or sense) codons
(coding triplets specify one of the 20 common AA’s)
-3 are non-coding (or “nonsense”) codons (these specify “stop”).
- The code is degenerate (more than one codon codes for each AA).

Examples of
reading the
 code:
1st 2nd 3rd positions
    C(1st) A(2nd) U(3rd) = HIS 
UGG = TRP
Learn how to read the code

The Code is Degenerate!
AUG forms part of the initiating signal

The first (5’) AUG in mRNA is “usually” selected as the START codon by ribosomes.
-The “ORF” specifies the amino acid sequence of the primary
“protein transcript” (from Start (AUG) to Stop (UGA, UAG, UAA).
-Once the initiating MET is specified (as AUG), the rest of the protei is translated in the same frame until a stop codon is encountered.
-At this point, the primary transcript is complete and the protein -will fall off (disengage) the ribosome.

Important Lessons of the Genetic
Code:
• The Genetic Code:
• 1. is read as a sequence of ribonucleotide “triplets”.
• 2. is “degenerate”.
- most amino acids are specified by >1 codons.
- this feature reduces errors in translation AND reduces the influence of mutations.
3. is “universal”.
- “Basically”, one nucleotide sequence will produce the same protein product in a human cell or bacterium (yeast, mouse, etc).
- this feature speeds up characterization of gene products and the development of genetic therapies and drugs.
- (exceptions: mitochondria have a few differences in codon meaning, and different species have distinct preferences in codon usage)






• • • •
Important Lessons of the Genetic Code (cont’d) The Genetic Code (also):
- provides “START” and “STOP” signals which define an “open reading frame” (ORF).
- within an ORF, mRNA sequence directly equates with protein sequence.
- changes in the 1st and 2nd base usually result in an altered amino acid.
- changes in the 3rd base have much less effect.
- e.g. CAU and CAC both encodes HIS (etc)
- the third position is called the “Wobble” base.
- “wobble” is a feature permitted by tRNAs.
-it allows one tRNA to read more than one codon.

45
Q

What is mutation
State and explain the types of mutation
Explain Cricks adaptor hypothesis

A
The Basis for Mutation:
-A mutation is a change in DNA sequence within a gene. (Most terminology reflects
a protein-centric “world view”. But terminology can be used to describe mutations
that affect RNA processing and expression. Types of mutations:
- point mutation, changes in a single base pair of DNA.
- silent mutation, changes in DNA which do not affect protein expression
or sequence (some polymorphism).
-mis-sense mutations, changes in DNA which lead to a change in an amino acid (some polymorphism).
-nonsense mutations, changes in DNA which “create” a ‘termination codon’ and thus stop translation.
-read-through mutations, changes in DNA which “eliminate” a ‘termination codon’ to add new sequence.
-Insertion or deletion mutations, changes in DNA which introduce or remove bases and thus “change the frame” of codon usage. -others, mutations in regulatory or splicing regions, chromosomal
rearrangements, etc.

Crick, in his adaptor hypothesis, proposed that small RNA molecules would be the adaptors that could be charged with amino acids by specific enzymes and that could identify the codons (triplets of nucleotides) of the mRNA by base-pairing.

Primary structure of tRNA
•60 to 95 nt
•20% of the bases are covalently modified Two examples of covalent modifications

46
Q

State the steps of chargin in tRNA and the enzyme that catalyses the reactions in charging
What are the two classes of tRNA synthetase
The anticodon always recognizes the corresponding codon regardless of what the tRNa is charged with
True or false
State the functional comparison between DNA Polymerase
and tRNA synthetases

A

The enzyme is amino-acyl tRNA synthetase (RS)
It’s a two step reaction ,ester formation

  1. Activation
  2. covalent attachment to tRNA

Class II:attachment at 3’ OH
Class I: attachment at 2’ OH
But equilibriums between these two positions
The 3’ position is usually used in ribosomes
Each class of the enzyme must recognize:
1.correct aa(amino acyl)
2.correct tRNA

Recognition of aa depends on what things:
The size(if it’s too big it’s excluded)
Specific H bonds
Van der waals 
Ionic interactions 

By itself these factors don’t lead to sufficient discrimination to accojnt for low error rate of aa charging;
Additional editing step is kinetic proofreading

  1. Recognition of tRNA by tRNA synthetase(RS)
    All tRNAs have similar 3 structure ,differ in 1 sequence,variable loop,anticodon
    Recognition features vary among synthetases
    Class I RS tend to read the codon directly
    Class II RS often recognize elements of the acceptor stem

DNA polymerase has a polymerase domain and an exonuclease domain but tRNA synthetase has a synthetic domain and an editing domain

Two extreme views on how it works
1) RNA molecules just serve as a scaffold for assembly of proteins. Proteins do all the work.
2) RNA molecules are the catalysts. Proteins just help them fold up in the proper way.
Recent findings suggest that view #2 may be closer to the truth.
That is, the ribosome is a ribozyme!!

47
Q

What is the peptidyl transferase reaction

A

Peptidyl transferase is the activity responsible for peptide bond formation during protein synthesis. This enzyme activity catalyzes the reaction between the amino group of the aminoacyl-tRNA in the A site and the carboxyl carbon of the peptidyl-tRNA in the P site, forming a peptide bond from an ester bond.

The ribosomal peptidyl transferase center (PTC) resides in the large ribosomal subunit and catalyzes the two principal chemical reactions of protein synthesis: peptide bond formation and peptide release.

The peptidyl transferase activity of the ribosome catalyzes peptide bond formation between the adjacent amino acids. Once fMet is bound to the second amino acid, it no longer binds to its tRNA. The ribosome translocates (facilitated by elongation factors) towards the 3′ end of the mRNA by one codon.

The enzyme that catalyzes peptide bond formation is known as peptidyl transferase. The catalytic center for the peptidyl transferase sits in the large subunit of the ribosome.

So I’m this reaction the amino acid from the P site is moved tk the A site
Making A site have two amino acids and p site have tRNA with no amino acid
Example in the A site is amino acyl-tRNA and the P site is peptidyl tRNA
So I’m the reaction,the peptidyl is moved to the A site
Leaving an uncharged tRNa in the P site and in the A site is now peptidyl tRNA plus amino acyl tRNA

48
Q

What evidence suggest ribosome is a ribozyme
What is initiator tRNA
What is initiation signals in eukaryotes and in prokaryotes

A

Evidence suggesting that the ribosome is a ribozyme:
􏰀 Known that RNA can be a catalyst
􏰀 rRNA sequence is more conserved than
ribosomal proteins
􏰀Large subunit still catalytic after extensive
• proteolysis
􏰀Large subunit not catalytic after treatment with RNase T1

Initiator tRNA
Bacteria: fMet 􏰃 tRNA
Eukaryotes: Met 􏰃 tRNAi

These tRNAs above are
Different from the tRNAs used to read an internal Met codon (recognized by different soluble translation factors)
􏰁
Note: prokaryotic proteins are
post translationally modified to remove f or fmet.
Recognition of start codon fmet-tRNAfMet recognizes AUG (met) (& sometimes GUG (val) in prokaryotes only)

Chain Initiation - the initiation signal
Eukaryotes
• Translation begins at the first AUG at the beginning of an
open reading frame.
• In vertebrates, the initiation efficiency is enhanced if the AUG is in a preferred sequence context: the Kozak consensus = RCCAUGG
• Because of the nature of the initiation signal, eukaryotic mRNAs are almost invariably monocistronic

Prokaryotic translation initiation
Approx. 30-35 non-coding mRNA nucleotides bind to ribosomal RNA
The non-coding region contains a run of 3-10 purines centered ~10 nt from start (AUG) codon consensus: AGGAG “Shine-Dalgarno” sequence
This RNA/RNA interaction is an important determinant of initiation frequency
Translation of a polycistronic mRNA in
prokaryotes
Translation starts
Ribosome Binds to Shine-Dalgarno sequence
Translation stops
Ribosome Dissociates

49
Q

What is translation initiation in prokaryotes

Explain the steps of elongation

A

Translation Initiation
in Prokaryotes
Initiation factors:
IF-1 Binds A site assists IF-3 binding may prevent tRNA from binding
during initiation
IF-2 Binds initiator tRNA & GTP
IF-3 Releases 30S subunit from inactive ribosome and aids mRNA binding

Elongation 3 steps (overview)
• • •

• •

a) aminoacyl tRNA binding
b) peptide bond formation: transpeptidation c) translocation: shifts register by one codon
Elongation factors (proteins)
EF-Tu Binds aminoacyl tRNA & GTP [GTPase] EF-Ts Displaces GDP from EF-Tu
EF-G Promotes translocation by binding with GTP to ribosome [GTPase]
EF-Tu: most abundant E. coli protein. ~10,000/cell, ~same as # of tRNAs,
∴ most tRNA in cells is bound by EF-Tu binds tightly to all aatRNAaa except fmet•tRNAfmet

Peptide Bond Formation:
The Peptidyl Transferase Reaction

50
Q

Explain protein synthesis in eukaryotes

What is a nonsense mutation

A

Protein synthesis in Eukaryotes
Initiation
Different from prokaryotes due to mRNA processing and transport to cytosol
Requires more initiation factors (at least 11)
Initiation signal: first AUG at beginning of an ORF; no S-D
Initiator tRNA: met-tRNAiMet
Important for translation regulation
Elongation very similar to procaryotes E. coli
eEF1α ≅ EF-Tu -Docks aatRNA at A site.
eEF1B ≅ EF-Ts -Helps recycle eEF-1α
eEF2 ≅ EF-G-Translocation

Termination:
One release factor, eRF1, recognizes all 3 codons

Eukaryotic translation initiation

1) Binding of the mRNA by the 40S subunit+initiation factors
2) Scanning of the mRNA to search for AUG
3) Binding of the 60S subunit

Eukaryotic initiation:

40S ribosome + Met-tRNAiMet
 Binding to CAP mediated by eIF-4E
Scanning
•AUG encountered
 •60S subunit binds
 •Translation begins

Translation termination:

3 termination codons
UAG recognized by RF-1 (36 kD) UAA
UGA recognized by RF-2 (38 kD)
RF-3 releases bound RF-1 or -2; (46 kD) requires GTP
RRF (ribosome recycling factor); helps induce disso of L and S subunits
Termination codons are not recognized by any tRNA

A nonsense mutation is a sequence that is mutated to involve a stop codon

51
Q

State some inhibitors of protein synthesis

State some inhibitors of protein biosynthesis(inhibitor,process affected,site of action)

A

Inhibitors of Protein Synthesis
>experimentally useful in dissecting steps in protein synthesis >medically useful if specifically inhibit proc. vs. euk. protein synthesis
Examples of antibiotic inhibitors
Aminoglycosides:inhibit prokaryotic ribosomes in a variety of ways
streptomycin: causes misreading; inhibits initiation in proc. paromomycin: binds A-site RNA; causes misreading (increases error rate)
erythromycin: binds 50S subunit; inhibits translocation in prokaryotes
tetracycline: prevents binding of aatRNA in prokaryotes and eukaryotes in vitro; prokaryotes only in vivo since doesn’t get across plasma membrane; binds 30S
chloramphenicol: inhibits peptidyl transferase in prokaryotes;
binds 50S
puromycin: aatRNA analog;inhibits elongation in prokaryotes and euks cycloheximide: inhibits peptidyl transferase in eukaryotes;
binds large subunit

Inhibitor:streptomycin
Process affected:initiation,elongation
Site of action:prokaryotes:30s subunits

Neomycin
Translation
Prokaryotes,multiple sites

Tetracyclines
aminoacyl-tRNA binding
30s or 40s sub units

Puromycin
Peptide transfer
70s or 80s ribosomes

Erythromycin
Translation
Prokaryotes 50s subunits

Fusidic acid
Translocation
Prokaryotes:EF-G

Cycloheximide
Elongation
Eukaryotes: 80S ribosomes

Ricin
Multiple
Eukaryotes: 60S subunit

52
Q

Explain the wobble hypothesis

What is the function of the 50s large sub unit of ribosomes and the 30s small sub unit of ribosomes

A

The Wobble Hypothesis
The A-site imposes Watson-Crick geometry upon the first two basepairs in the codon/anticodon duplex, but more flexibility is allowed in the third position.
These relaxed basepairing rules make it possible to recognize the 61 sense codons with as few as 31 tRNA species (although organisms generally have more than 31).
Example: many bacteria contain only a single tRNAIle:
Anticodon: 3’ UAI 5’ Reads all 3
Codon: 5’ AUU 3’
5’ AUC 3’
5’ AUA 3’

But E. coli has two tRNAIle (anticodons UAG and UAL).

Biological Peptide Formation is NOT a condensation reaction
Example of condensation: glycine plus glycine gives diglycine

Small Subunit
mRNA binding
Error correction

Large Subunit
Peptidyl transferase

Translation immediately follows transcription in Prokaryotes

53
Q

What is a nuclear envelope made up of?
Give a three line description of transcription
What are lamins? Mutation in lamin A causes what ?
What’s the function of nuclear pores and give an example of molecules that go out and go in the cell
What is the function of the nucleolus?
What produces ribosomes
What’s the function of chromatin and what’s it made up of?
What are the types of chromatin and their functions
Histones have what amino acids?
What is epigenetics
What are CPG islands and what’s their relevance to DNA ?
What things promote transcription and which reduce it

A

Nuclear envelope:
Outer layer of nucleus-ribosomes are on the outer layer

DNA to mRNA and mRNA moves out via the nuclear pores and binds to the ribosomes and then moves to the rough ER
The ribosome bound to the mRNA undergoes translation (taking the RNA and making proteins )

Inner membrane-lamins are proteins that line the inner membrane and control the structure of the nuclear envelope,interact w the chromatin
Mutation in Lamin A causes individuals to age quickly than normal

Nuclear pores-for transport between the nucleus to cytoplasm and cytoplasm to nucleus . Nucleotide going in the nucleus and mRNA going out the nucleus

Nucleolus-site of rRNA synthesis,ribosome sub units which combine to form ribosomes
rRNA combined w other or small proteins produces ribosomes

Chromatin(in the nucleolus)- made up of DNA and proteins specifically histones
Types of chromatin:
Euchromatin -a loose chromatin . There’s a weak attraction between the histones and the DNA. Due to this there’s a space where the RNA polymerase can go in there to make RNA so transcription occurs easily
Heterochromatin -is closer towards the inner membrane of the nuclear envelope and is highly condensed so it’s hard for transcription to occur.

Chromatin condenses into chromosomes during replication to easily pass the genetic material on to daughter cells

H2A,H2B,H3,H4 -double of these four makes an Octomer of histone proteins

Histones have lysine and arginine which are positively charged amino acids
DNA is negatively charged

Epigenetics is when you regulate the expression of genes through out the lifetime from parental to daughter cells by modifying the interaction between DNA and histone proteins .
Interaction between the lysine and arginine on histone proteins and phosphate group on DNA makes the DNa condensed

Plenty Cytosine and Guanine are located in CPG islands
Enzymes add methyl group o wherever these CGp islands are found and inhibits that area of DNA from being expressed thus preventing it from transcribing,making proteins
This modifies the DNA thus making you able to
control which gene you want to be expressed in particular cells (epigenetics)

Modifying histones will promote transcription by addition of an acetyl group which relaxes the histone proteins from the DNA but addition of methyl group prevents transcription which will make the histone proteins and DNA interaction tighter

H1 linker protein which is a type of histone links the DNa nucleosomoes between one another and is the most positively charged Histones

54
Q

What is replication and transcription
Nucleotide is made up of?
What are the types of nitrogenous bases?
Which carbon has the nitrogenous base and which has the phosphate group and which carbon has the OH attached to it
What is a nucleoside and a nucleotide,
Explain complimentarity of DNA
Explain anti parallel arrangement in DNA
Where do enzymes bind to cause replication
What’s the difference between rough ER and smooth ER
What’s the function of peroxisomes
What is rRNA,tRNA,mRNa

A

DNA-DNA: replication
DNA-RNA:transcription

Nucleotide is made up of the nitrogenous base and the sugar phosphate backbone
Bunch of nucleotides together make a nucleic acid

Nitrogenous bases are of two types namely -two rings or a heterocyclic ring called purines(adenine and guanine) and a single ring structure called pyrimidines (cytosine,uracil(only in RNA),thymine) (CUT Pi)

Pentose sugars(first 1 carbon connects to the nitrogenous base in all and the phosphate group connects at the 5 carbon or fifth carbon in all and OH at the third carbon in all whether DNA or RNA) oxyribose or ribose(there’s an OH on the second carbon) and deoxyribose(no OH on the second carbon but will be H)

Phosphate group is negatively charged and binds to the fifth carbon on the pentose sugar

Nucleoside :pentose sugar and a nitrogenous base

Nucleotide:pentose sugar,nitrogenous base,phosphate group

Complementarity:
Hydrogen bonds are weak bonds
A interacts w T using 2 hydrogen bonds (easier to break the bonds between A T or Thymine than between G C)
G interacts w C using 3 hydrogen bonds

Strong bonds example- bond between the phosphate and hydroxyl group (phosphodiester bond . A covalent bond)
Formed between the 5’ end of one nucleotide and 3’ end of another nucleotide
The phosphate group is on the 5’end and the OH group is on the 3’end

Anti parallel arrangement in DNA:
If Left side is arranged from top to bottom jn 5’-3’ then right side should be arranged from top to bottom in 3’-5’

Enzymes bind to the minor groove of the helix structure of DNA to cause replication

Presence of ribosomes on the outer membrane in the rough ER differentiates it from smooth ER

Rough ER-site of protein synthesis
(Proteins such as proteins in lysosomes,proteins in other organelles,excreted proteins)
Helps with folding process of proteins or protein folding
Glycosylation- N type

Peroxisome-spherical organelles that contain enzymes catalase and oxidase and other metabolic enzymes
These enzymes are important for free radicals
The free radicals that is usually accumulated in the peroxisome hydrogen peroxide
Break down the hydrogen peroxide into water and oxygen
Catalase breaks fatty acids down or fatty acid oxidation into acetyl coA and the coA is used to make lipids(example plasmalogen is a lipid that is important for the myelin in the white matter in the brain ) and cholesterol( alpha and beta fatty acid oxidation) and breaks ethanol

RNA is single stranded so it can easily leave the nucleus but DNa can’t easily leave cuz it’s double stranded and big
RNA -It leaves the nucleus as mRNA to go to the ribosomes in the cytoplasm.
(Messenger RNA) mRNA- copy of a gene sequence from one strand of a DNA molecule. When compelled it leaves the nucleus and goes to the ribosome in the cytoplasm

Ribosomal RNA(rRNA)- structural molecule that is part of a ribosomes structures and assist in the production of proteins during the translation process

tRNA(transfer RNA)- assists in production of Proteins by transferring amino acids(building blocks of proteins) to the ribsosomes when needed

55
Q

Which phase of the cell cycle does the DNA replication primarily occur
Explain The features of DNA replication
Explain the steps in DNA replication
What are telomeres

A

Replication or cell cycle
G1phase-Sphase-G2phase:mitosis to make two cells
DNA replication primarily occurs in the S phase

  • DNA replication occursim a sem conservative model:replication occurs in a. Complimentary pattern . New strands which are complimentary to the old strand s
  • It occurs in a specific direction.pccurs from the five prime end (5’end) to the three prime end (3’end)
  • DNA replication is bidirectional : replication fork is created (its Y shaped ) on each side when you pull apart the DNA strands before you replicate them. Helicase unwinds the DNA on both sides moving it in opposite directions from the origin of replication

DNA polymerase follows helicase and synthesize new DNA off the parental strands in a bi directional fashion.

DNA replication steps:
Initiation of DNA replication: the area where you decide to separate in order to create two separate parent strands and make templates of new strands is the origin of replication.
Adenine and thymine rich areas are chosen for origins of replications because the bonds between A and T are easier to break than in G and C thus less energy will be required
In eukaryotic cells there are multiples origins of replication
Pre -replication protein complex binds to origin of replication and separates the A and T nitrogenous based (DNA ).
Replication bubble is formed.
Single stranded binding protein binds to each of the single parental strands that are separated and prevents them from coming back together or reconnecting to one another. The proteins also protect the strands from exo or endonucleases that try to eat or break the phosphodiester bonds. Helicase enzymes works at both of the replication fork areas which are at the ends of the replication bubble to unwind the DNA in front of them. Helicase enzyme requires A lot of ATP. Super coils are created as Helicase unwinds the DNA
These make it difficult for the Helicase to unwind the DNA in front of them
Nuclease domain on the Topoisomerases (we have type I,II,IV. Type I and II are usually in eukaryotic cells and II and IV are in prokaryotic cells. Type I doesn’t need ATP to unwind the super coils but Type II and IV need ATP . Drugs target topoisomerases in cancer cells to prevent them from causing replication. Example of such drugs is Irinotecon and topotecon and they inhibit type I in eukaryotic cells and etoposide inhibit type II in eukaryotic cells . Drugs such as frluoroquinalones example ciproflaxicin inhibit type II in prokaryotic cells if there’s a bacterial infection that’s replicating anyhow by increasing the activity of the nuclease domain and inhibit the ligase domain) has two arms. One arm cuts the phosphodiester bonds in the DNA strands that have coiled up so it can easily uncoil and the other arm called the ligase domain restitches the area that has been broken by the nuclease domain
Elongation :primase makes RNA primers Which enable DNA polymerase type III to make DNA since it needs the 3’ OH of RNA primers to carry out its activity. It reads the first DNA parent strand from 3’ to 5’ and It synthesizes RNA primers in a 5’ to 3’ fashion. DNA polymerase now also reads the DNA strand from 3’ to 5’ and Makes a DNA strand in a 5’ to 3’ fashion. Or opposite fashion of the parent strand .So RNA primers are produced and the DNA polymerase follows and makes the DNA strand so it’s like RNA primer(5’ to 3’) then DNA strand(continues where the RNA primer ended which is at 3’ and continues with 5’ to 3’) . The strand that theDNA polymerase makes and that’s trans it toward the replication fork is called the leading strand. The primase comes to the other parent strand which will be in the 3’ to 5’ and read that strand and synthesize complimentary nucleotides from 5’ to 3’. So now there’s an OH at the 3’ strand and DNA polymerase continues and reads the strand from 3’ to 5 thereby continuing what the primase started and synthesizes the strand from 5’ to 3. This new strand is the lagging strand
But the difference is that there are DNA strands in between the RNA primers meanwhile in the leading strand the RNA primer is only seen before the DNa strand and it’s all DNA. No RNA in between. Okazaki fragments are seen in lagging strand. These fragments are defined as fragments where there are multiple RNA primers and multiple stretches of DNA.
DNA polymerase proof reads to prevent mistakes. It goes back the strands it has read to make sure everything has been paired correctly. It reads from 3’ to 5’. And if there are any mistakes it uses a 3’ to 5’ exonuclease activity to cut it out and put in the correct complimentary nucleotide.
DNA polymerase type I cuts out primers in a 5’ to 3’ exonuclease activity to pluck out RNA primers in the leading strand . It reads the DNA strand which the primer has been plucked out from,from 3’ to 5’ and synthesizes from 5’ to 3’. It also proof reads from 3’ to 5’ and if any mistake is found it cuts it out in a 3’ to 5’ exonuclease activity .
Same for the lagging strand. In the lagging strand though this creates gaps between the DNA strands since that’s where the primers used to be. Ligase enzyme comes on the lagging strand and fuses the DNA ends together. In HIV their T cells are infected. And replication occurs a lot. Drugs target the T cell replication process. Example of such drugs are Nucleoside reverse transcriptase inhibitors(didanosine) . These remove the 3’ OH region so the DNA polymerase III is not able to build on it to make a DNA strand .
Termination:so when the origins of replication hit each other, the DNA replication at that point stops. When DNA polymerase move towards each other at a replication fork and they just stop at that point. They hop off and don’t perform their function again.
ends of chromosomes (telomeres).

56
Q

What are centromeres?
What’s the function of telomeres clinically
What is Hayflick limit
How do telomeres operate
What sequence of nucleotides is always seen on telomeres
What is done to prevent the shortening of telomeres

A

Telomeres shorten overtime
Centromeres (center if the chromosome) and the telomeres. Over time as the cell replicates,the telomeres get shorter. Telomeres don’t code for any RNA or you can’t take the DNA from the telomere to make RNA or proteins. Telomeres prevent gene loss. Hayflick limit: max amount of times DNA can replicate before it starts to involve genes.
Telomerase has two arms. It takes one arm and expresses nucleotides specifically complimentary nucleotides that are always seen on telomeres (TTAGGG. Tell them all genes got to go. These are the repeat nucleotides seen on the telomeres constantly). So the arm expresses the RNA compliment of TTAGGG which is AAUCCC
The other arm uses the RNA strand made which is AAUCCC to make DNA that’s complimentary to it . So it’ll be TTAGGG. This elongates the 3’ end thereby preventing the shortening of telomeres
This is called reverse transcription since RNA is used to make DNA and DNA to RNA is transcription.
Highly replicating cells need high amount or activity of telomerase enzyme. Example of such cells is stem cells,hematopoietic cells,cancer cells.
Cancer cells up regulate the activity of their telomerase enzymes this elongating the ends of the DNA on the chromosome which allows them to continue replicating without shortening the telomeres enough that it starts to involve genes within that cells.

57
Q

What is translation
What are codons
How many types of codon are there?
How many read for a specific type of amino acid
AUG codon codes for which amino acid?
How many codons do not code for any amino acid?
What then do they code for?
What are they called?
Explain the structure of tRNA
What are the characteristics of the genetic code
What is wobble effect
Explain tRNA charging
What are the two sub units of ribosomes in prokaryote and eukaryotes

A

Translation(protein synthesis):
Taking mRNA from DNA to make proteins.
mRNA,tRNA and rRNA are important for making proteins
mRNA has specific sequence of three nucleotides so three nucleotides (codons) each along the mRNA molecule .

Codons: triplets of nucleotides
There are 64 different types of codons.
61 of those codons read or
These codons code for a specific type of amino acids
Example: AUG codon codes for Methionine which is an amino acid
The remaining three codons left do not code for any amino acid. They code for terminating the translation process
They are called stop codons.
The remaining three codons are UGA(you go away),UAA(You are away),UAG(you are gone).

tRNA contains anti-codons (they are triplets of nucleotides that are complimentary to the codons in mRNA ).
Example codon is AUG so anticodon will be UAC.
There’s a specific amino acid domain in this tRNA that Carries the amino acid specific to the codon
Structure of tRNA: bottom loop contains anticodons
Upper portion is the 5’ end
The y shape contains the 3’ end. The Y shape is the amino acid holding or binding domain and it contains specific sequence of nucleotides . (CCA).
The CCA in the Holding domain binds to the amino acid.

Characteristics of the genetic code :
mRNA is read from the 5’ end to 3’ end.
Commeless: codon is read and amino acids are made. there can be nucleotides between the codons you want to read and it’s not normal. It’s only seen in viruses.

Non overlapping: occurs in viruses not in normal cells

Redundant or degenerate:
Example- Isoleucine has three different types of codons that can code for it
AUA,AUC,AUU.
Exceptions :methionine and tryptophan each have only one codon that can code for them while all the other types of amino acids have three different types or multiple types of codons that can code for them
Wobble effect or wobble phenomenon:
Ianosine is complimentary to A,U,C. This decreases the risk of mutations
All the codon differ in the third position and first position on the tRNA in the anti codon.
So redundancy works because of the wobble effect

tRNA charging:
Portion near the 3’ end is the T arm (it tethers the tRNA to the ribosome
The portion near the 5’ end is the D arm(it allows for the identification of the tRNA by the enzyme called the amino acyl tRNA or tRNA synthetase enzyme Variable domain.
Charging; how to get the amino acid to bind to the 3’ end.
Add ATP molecule to the amino acid. When ATP is added ,two phosphate groups are broken off the ATP thus giving a pyrophosphate.so the only thing hanging on to the amino acid is AMP. (Amino acyl AMP molecule-amino acid + AMP)
An enzyme with two pockets
One holding the tRNA and the other pocket holding the amino acyl AMP molecule. Then it checks if the anti codon is appropriate to the mRNA codon needed
If yes,it clicks the amino acid w the AMP to the 3’ CCA region. The amino acid is put onthe 3’ end and AMP is released during this process. So a charged tRNA is produced
This enzyme is called the amino acyl tRNA synthetase

Ribosomes : ribosomes have two sub units. Large ribosomal sub unit and small ribosomal sub unit
For eukaryotes for large units they’re called 60 S and small units are 40 s
Together called 80 s ribosomes in eukaryotes

For prokaryotes; large units are 50 s and the small is 30s
Together it gives 70 s
Ribosomes contain rRNA and proteins
Clinical relevance: antibiotics such as tetracyclines and aminoglycosides target and inhibit the 30 s ribosomal sub unit in prokaryotes thus inhibiting protein synthesis

58
Q

Explain the phases of translation.
In prokaryotes and eukaryotes
Explain how the translation process can occur in the rough ER instead of free ribosomes
What kind of proteins from free ribosomes are produced?
What kind of proteins from rough ER are produced ?

A

Phases of translation:
Initiation: difference between prokaryotic initiation and translation and eukaryotic initiation and translation

5’ to 3’ mRNA for prokaryotes. Start codons example AUG code for N-formal methionine or f met not methionine.
But eukaryotic cells AUG code for methionine
There are some nucleotide bases towards the 5’ end from the start codon called the Shine Delgano sequence. Small ribosomal sub units will bind to the Shine area on the mRNA and uses the initiation factors. These initiating factors identify the shine Dalgarno (SD) sequence and These factors bind to the Shine Dalgarno sequence and moves towards the start codon. So both small ribosomal sub units and initiation factors bind to the shine delgano sequence and move till the hit the start codon.
2.There’s a molecule called tRNA which has anti codons specific to the start codon and this anti codon carries an amino acid with it. The anti codon interacts with the start codons . Initiation factor brings the initiator tRNA which contains the amino acid to where the start codon is and binds to tRNA
3. GTP molecule is bound to the initiation factor . The factor breaks the molecule into GDP and inorganic phosphate which creates a lot of energy. At this same time,the large ribosomal sub unit comes over and binds to this. So at the end of this process you’ll get the large and small ribosome bound to the mRNA with the tRNA sitting in the ribosome. Sites in the ribosome( A site,P site,E site). The tRNA sits in the P site.

1.Now in eukaryotes there’s a 5’ end and a 3’ end. But there’s no shine delgano sequence. They just have the start codon.
Eukaryotic initiation factor type IV identifies the mRNA and binds to the 5’ end.
2.Small ribosomal sub units interact with the mRNA and bind to it with the help of other initiating factors .
3. Eukaryotic initiating factor type II will bind the tRNA, containing the anticodon with the amino acid (example methionine not f met as in prokaryotes) ,to the portion on the start codon
4. GTP molecule is bound the eukaryotic initiating factor type II. Will be broken down into GDP and an inorganic phosphate. The large ribosomal sub unit binds to the mRNA . GDP and inorganic phosphate is released and the eukaryotic initiating factors are also released
Elongation(second phase in translation):
Primarily used in eukaryotic cells
A site is the amino acyl site(entry site) ,P site is the peptidle site (synthesis site) and E is the exit site (EPA)
In order to bring the tRNa into the A site we need a eukaryotic elongation factor type I to help bring it to that area.
Eukaryotic elongation factor type I carries a GTP molecule so GDP and inorganic phosphate is released as well as the Eukaryotic elongation factor type I.
There’s a nitrogen in the amino Acid in the A site and attacks the carbon end of the amino acid on the P site. Thus bringing the amino acid from the P site into the A site.so there are two amino acids in the A site . For this process to occur,the ribosome has an enzyme called a peptidile transferase. So the elongation step is catalyzed by the peptidile transferase .
Translocation: this process in elongation needs energy. The eukaryotic elongation factor type II contains GTP and brings it j to the translocation process thus releasing GDP and inorganic phosphate and shifting what was in the P site into the E site and what was in the A site into the E site . So now the E site will have tRNa with no amino acid ,P site will have tRNa which contains the two amino acids and nothing will be in the A site. Then the E site will spit out the tRNA with no amino acid and will be charged for a new amino acid to be added to it and it will enter again through the A site. And the process starts all over again

This causes elongation of the peptide. Termination: Eventually you hit a stop codon. So if you bind to a stop codon in the A site ,there won’t be any tRNa entering the A site instead a release factor will be added and it will interact with the stop codon and prevent the ribosome from continuing to move along the mRNA and continuing to translate it. The peptide is cleaves from the tRNa in the P site.
The release factor separates the tRNa from the peptide.

The translation process can occur on free ribosomes or it can occur on the rough ER(membrane bound ribosomes)
The process occurs on the rough ER for proteins that are going to be 1.secreted from the cell,2.incorporated into the cell membrane and 3.then proteins that will be part of the lysosomes.

There is a specific sequence of amino acids on the peptide that become an identifier on the peptide. This sequence is called the signal sequence. This sequence is recognized by a particular protein. This protein is called signal recognition particle or protein. This binds to the signal sequence and has a high affinity for the receptors located on the rough ER thus dragging the signal sequence to the rough ER membrane.
So the whole translation process occurred on a free ribosome. Now we’re moving it towards the rough ER membrane
There are two proteins on the rough ER membrane namely 1.signal recognition protein particle receptor and this binds to the signal recognition particle which is bound to the signal sequence which is on the peptide and the growing peptide which is from the free ribosome
The mRNA is sandwiched between the sub units of the ribosomes.
2.The trans locon protein is closed at this stage. The signal recognition particle and the signal recognition receptor contain GTP molecules bound to them.
This molecule is broken down into GDP and inorganic phosphate so two GTPs will be broken down . This opens the trans locon protein. The ribosome lines up perfectly with the trans locon and the growing peptide is pushed through the trans locon into the lumen of the Rough ER. Signal recognition particle is released.
Signal peptidase cuts off the signal sequence in the lumen of the rough ER. And the signal sequence is degraded
This occurs till the translation process hits a stop codon. And this will make the trans locon close ,the peptide is released into the rough ER lumen and the ribosomes and mRNA will disassociate from the rough ER and will be degraded .
These proteins need to be sent to the rough ER cuz they’ll need to go to the golgi to be packaged by the golgi apparatus into vesicles which will go to the cell membrane to be incorporated(membrane bound protein),go to the cell membrane to be secreted ,or they can become a lysosome.

Free ribosomes that don’t bind to the rough ER produce 1.proteins for the cytosol(the metabolic processes in the cytosol),2.proteins incorporated into the nucleus,3.proteins involved in the mitochondria or mitochondrial enzymes,3.enzymes involved in peroxisomes.

Proteins afterward must be modified :
Types of modifications-
if a sugar molecule is added to the proteins(glycosylation) example for antigens(proteins with sugar residues on them),for transporters,for voltage gated channels
and a lipid is added too(lipidation) example for proteins to be incorporated into the cell membrane and organelle membranes such as the Golgi apparatus
and
a phosphate group is added too(phosphorylation) example protein kinase A or cyclin dependent kinase
and a hydroxyl group is added too(hydroxylation) example for making collagen
,cutting some of the amino acids off(trimming ) example cutting trypsinogen to turn it into trypsin
,adding a methyl group(methylation) example methylation of a histone protein to decrease transcription ,
adding an acetyl group(acetylation) example acetylation of a histone protein to increase transcription

59
Q

What is a turn,rise,tilt

A

Structure of DNA:
Has Major groove and minor groove

From One minor groove to another minor groove or one major groove to another major groove is called a complete turn

So you’re comparing the grooves on the same line not the grooves adjacent
So on the same line the smaller groove is the minor groove and the larger is the major groove

B DNa contains 10 nucleotides per turn
B DNa structure is the most common type of DNA structure
Distance between two adjacent base pairs is called rise

In A DNA the minor groove is deeper than the minor groove of B DNA

Tilt is the angle with which two consecutive base pairs are placed with one another

So for A dna the major groove is smaller than the minor groove.

If you compress B DNA it becomes A
If you stretch B dna it becomes Z

B dna is the most common

60
Q

Transcription:

Transcription in bacteria – RNA polymerase
– Promoters
– The transcription cycle
– Mechanisms of activation and repression
• Transcription in eukaryotes
– Eukaryotic RNA polymerases
– General transcription factors
– The mediator of transcription
– Epigenetic control of transcription • Eukaryotic RNA processing:
-5’ capping, 3’ poly A tailing and splicing

Who is the gatekeeper of gene expression?
Why is regulation of gene expression important?

Transcription and translation in eukaryotic cells are separated in space and time.
Extensive processing of primary RNA transcripts in eukaryotic cells.

A

Transcription.
Protein is the target of most drugs interaction and basis of disease

Why regulation of gene expression is important?
Target of most drugs interaction and basis of diseases
􏰀Cellular function is largely dictated by the set of macromolecules inside the cell 􏰀Different macromolecules accumulate to different levels under different growth conditions and in different cell types.
􏰀Diseases can be caused by aberrant control of gene expression: too much or too little of a protein; wrong time and wrong place for a protein.

61
Q

Transcription and translation in eukaryotic cells are separated in space and time.
Extensive processing of primary RNA transcripts in eukaryotic cells.
True or false
State and define the function of the types of RNAs
What is a promoter
What are the four common properties of RNA polymerases
What is the function of the bacterial RNA polymerase core and rna polymerase holoenzyme

A

The RNAs carrying the code for protein synthesis are called “coding RNAs” or “messenger RNAs (mRNAs)

Messenger RNA (mRNA) carries the genetic code from DNA in a form that can be recognized to make proteins. The coding sequence of the mRNA determines the amino acid sequence in the protein produced. Once transcribed from DNA, eukaryotic mRNA briefly exists in a form called “precursor mRNA (pre-mRNA)” before it is fully processed into mature mRNA.

This processing step, which is called “RNA splicing”, removes the introns—non-coding sections of the pre-mRNA.

Non-coding RNA (ncRNA)
Ribosomal RNA (rRNA):
Ribosomal RNA is the catalytic component of ribosomes. In the cytoplasm, rRNAs and protein components combine to form a nucleoprotein complex called the ribosome which binds mRNA and synthesizes proteins (also called translation).

Transfer RNA (tRNA):
Transfer RNA is a small RNA chain of about 80 nucleotides. During translation, tRNA transfers specific amino acids that correspond to the mRNA sequence into the growing polypeptide chain at the ribosome.

Small nuclear RNAs (snRNA; 150 nt):
Small nuclear RNAs are always associated with a group of specific proteins to form the complexes referred to as “small nuclear ribonucleoproteins (snRNP)” in the nucleus. Their primary function is to process the precursor mRNA (pre-mRNA). Splicing of pre-mRNA

Small nucleolar RNAs (snoRNA; 60-300 nt):
Small nucleolar RNAs are components of small nucleolar ribonucleoproteins (snoRNPs), which are complexes that are responsible for sequence-specific nucleotide modification. Or process and chemically modify rRNAs

Non-Coding = Template = Antisense Strand Strand Strand
Coding= Non Template = Sense Strand Strand Strand
Promoter = The region near the 5’ end of the transcribed region to which the RNA polymerase binds prior to the initiation of transcription
UTR = Untranslated region

RNA polymerases
4 properties common to most RNA polymerases
1. Synthesize RNA in the 5’ to 3’ direction
2. Require a DNA template
3. Can synthesize RNA strands de novo (no primer required). Low fidelity compared to DNA Pols
4. They are processive but slower than DNA Pols
The elongation cycle

 •
•
•
Bacterial RNA Polymerase
RNA Polymerase Core - 􏰁2􏰂􏰂’
Core only capable of initiating transcription
randomly (e.g., from nicks and gaps)
RNA Polymerase Holoenzyme - 􏰁2􏰂􏰂’􏰃
– 􏰃 subunit enables polymerase to recognize and
initiate transcription at promoters
– Multiple 􏰃’s. The “housekeeping” 
􏰃 in E. coli is 􏰃70 (70 kDa)
62
Q

How is the transcription complex formed in bacteria?
Explain a promoter
State the stages of transcription

A

Formation of the transcription complex
• Inbacteria:
• Sigma factor recognizes the promoter sequences
• RNA polymerase attaches to the promoter region
• RNA polymerase σ factor melts the helical structure and separates 2 strands of DNA locally
• RNA polymerase initiates RNA synthesis. The site at which the first nucleotide is incorporated is called the Start site or start point.

Promoter
• A promoter, is a special sequence of nucleotides indicating the starting point for RNA synthesis
• General transcription factors for the RNA polymerase recognizes this DNA sequence by making specific contacts with the portions of the bases that are exposed on the outside of the helix
• In bacteria, RNA polymerase is recognized and recruited onto the promoter by the σ(sigma) factors

Stages of Transcription
• Formation of transcription complex (of DNA and RNA polymerase
• Initiation
• Elongation
 • Termination
63
Q

Explain the transcription cycle(a 5 step porgram)

What step is essentially irreversible? G

A
Step 1: closed complex formation
promoter +
  RNAP holoenzyme
Initial binding is non-specific
    1-D search for promoter
Closed complex not stable (t1/2 < 1 sec)

Transcription cycle (cont.)
Step 2: open complex formation
Close complex undergoes Local melting forming an Open complex
(kinetically stable, t1/2 = hours)

RNAP is functioning like a helicase
• 􏰃54 holoenzyme requires ATP at this point
• 􏰃70 holoenzyme does not require ATP
This step is essentially irreversible

Step 1: closed complex formation
promoter
  RNAP holoenzyme
Initial binding is non-specific
    1-D search for promoter
Closed complex not stable (t1/2 < 1 sec)
 Transcription cycle (cont.)
Step 2: open complex formation
Close complex
Local melting
  Open complex
(kinetically stable, t1/2 = hours)
~-10   ~+3
RNAP is functioning like a helicase
• 􏰀54 holoenzyme requires ATP at this point
• 􏰀70 holoenzyme does not require ATP
This step is essentially irreversible
  1. Initiation
    • Core enzyme starts transcription at the separated DNA strands of an initiation complex.
    • It is observed that the subunit of RNA polymerase has two binding sites for the binding of NTPs
    • Formation of H-bonds is always as per the base pairing rules. The binding is with 3’ end of the NTP leaving the 5’ end to be free.
    • The second NTP binds to the elongation site on the polymerase.
    • First base is then dissociated from initiation site and that marks the completion of initiation.

Transcription cycle (cont.)
Step 3: initial transcribing complex (ITC) formation
Open comple x
ITC
• Association of the transcript is weak, often leading to abortive initiation
• There can be hundreds of cycles of abortive initiation before proceeding to the next step.
• RNAP maintains contact with promoter throughout the period of abortive cycling.
NTP PPi
Abortive transcript
Abortive initiation
Bubble has moved a few notches to the right, and 2-10 nt long transcript has been
synthesized.

  1. Elongation
    • The core enzyme polymerase moves in 3’-5’ direction of the non-coding strand and it adds successive NTPs at the 3’-OH.
    • The incoming NTP forms a phosphodiester bond with 3’-OH end of the preceding NTP.
    • The DNA helix rewinds after RNA polymerase transcribes through it and growing RNA chain dissociates from the DNA

Transcription cycle (cont.)
Step 4: Ternary elongation complex (TEC) formation (promoter clearance)
ITC
Occurs when the transcript
􏰀 achieves a length of ~ 10 bp.
TEC
Core polymerase Nascent transcript
• Ejection of 􏰀 from active site results in a dramatic increase in the stability of the ternary complex. The enzyme is now highly processive, rarely releasing the transcript before reaching a termination signal.
• This transition is associated with the release of RNAP from the promoter.

Step 5 Termination
• Specific sequences on the DNA molecule function as the signal for termination of the transcription process.
• The signal could be two inverted GC rich regions separated by intervening region followed by AT rich sequences
• A sequence of Adenine that codes for 6 to 8 Uracil residues
• Rho protein can also bring about termination (Rho dependent
termination).
• Rho factor dissociates RNA and RNA polymerase from the DNA.

64
Q

State four inhibitors of transcription
Explain how rifamycin inhibits transcription
State some antibiotics that inhibit RNA polymerase and explain how they do so

A

Inhibitors of Transcription
• Rifamycin: Rifampicin and streptovaricin bind with β subunit of the polymerase to block the initiation of transcription.
• Actinomycin D: It forms a complex with double stranded DNA and prevents the movement of core enzyme and as a result inhibits the process of chain elongation.
• Streptoglydigin: It binds with the β subunit of prokaryotic RNA pol and thus inhibits the elongation.
• Heparin: It is a polyanion that binds to the β’ subunit and inhibits transcription in vitro. The α subunit has no known role in the process

Antibiotics that inhibit RNA polymerase:
Rifampicin
It is a semisynthetic compound derived from Streptomyces mediterranei
• Important antibiotic in the treatment of tuberculosis (causative agent = Mycobacterium tuberculosis)
• Rif resistance is a serious problem in TB treatment
• Information about the mechanism of rif resistance might be useful in optimizing therapies
Microcin J25
• An antibiotic produced by some strains of E. coli - 21 amino acid peptide lariat.
• It kills other E. coli giving the microcin J25- producing strain a competitive advantage.
Amide linkage between
Gly1 and Glu8 makes lariat

 Mechanism of Rifampicin
 inhibition
Actinomycin D and Acridine inhibits transcription both in prok. and Euk. by intercalating into the ds DNA.
This prevents the movement
of the RNA pol along the template
Rifampicin also binds to the β subunit
of RNAP and prevents promoter clearance
65
Q

State the differences in transcriptional machinery between bacteria and eukaryotes
How is the Regulation of bacterial Transcription done

A

Regulatory machinery:eukaryotes:activators and repressors and co activators and co repressors
Bacteria: activators and repressors

Basal machinery: eukaryotes:core RNa polymerase and general factors(General factors are analogous to bacterial 􏰂 factors in that they direct core polymerases to promoters. But while a single polypeptide suffices for this job in bacteria, multiple polypeptides are required in eukaryotes.)
Bacteria: core RNa polymerase and sigma factors

Regulation of bacterial Transcription:The lac operon
LacZ = 􏰃-galactosidase LacY = lactose permease Lac A= transacetylase
I = Inducer
P = Promoter O = operator
Physiological inducer
􏰃-gal
lactose allolactose
Gratuitous inducer
IPTG
66
Q

What is a ribozyme

Give examples of natural ribozymes ,state the name ,the reaction and genomic location

A

Ribozymes = RNA molecule capable of catalyzing a (bio)chemical reaction
Examples of natural ribozymes
Name:Group I Introns-(the reaction)self-splicing-(genomic location)Nuclear rRNA Tetrahymena organelles fungi and plants bacteriophages, bacteria

Group II introns-self-splicing ,splicing pathway is identical to nuclear pre-mRNA introns-organelles fungi and plants bacteriophages,
bacteria
RNase P -5’ end cleavage of pre-tRNAs-Eukaryotes ,Prokaryotes

Ribosome-Peptidyl Transferase-Eukaryotes ,Prokaryotes

67
Q

What is the importance of a cap structure ,a poly(A) tail and what’s their function

A

Why a cap structure ?

  • Protects mRNA against degradation by 5’->3’ exonucleases
  • Enhances translation initiation by binding of eIF4E

Why a poly(A) tail ?
-Protects mRNA against degradation by 3’->5’ exonucleases
(AAA)n
PABI
- Facilitates mRNA export out of the nucleus
- Facilitates Translation : PAB is a component of the translation machinery

Why a poly(A) tail ?
-Enhances translation initiation by binding of PAB

68
Q

What are restriction endonucleases?
Explain the statement The recognition sequences are usually palindromes?
What are restriction-recognition sites ?
Why are restriction enzymes used to digest fragments of DNA?

A

Restriction Endonucleases
• Their discovery revolutionized recombinant DNA technology because they can cut DNA double helix at specific sites defined by their local nucleotide sequence.
• They are purified from bacteria.
• These enzymes recognize short sequences of DNA usually 4-8bp long and cut
both strands.
• These sequences, where they occur in the genome of the bacterium itself, are protected from cleavage by methylation at an A or a C residue; the sequences in foreign DNA are generally not methylated and so are cleaved by the restriction enzymes.
• The recognition sequences are usually palindromes-that is, the sequence is the same if read from one direction on one strand and the opposite direction on the other strand.
• Some restriction enzymes produce staggered cuts ( leaving what are known as “sticky ends” or “cohesive”) and others produce blunt cuts (leaving “blunt ends”).
• The fragments of DNA produced by cleavage with a restriction enzyme are known as restriction fragments.
• In the case of an enzyme that produces sticky ends, the ends can transiently base-pair. The base-paired ends can be covalently joined by the action of DNA ligase.

Restriction-recognition sites are short DNA sequences recognized and cleaved by various restriction endonucleases.

Restriction Enzymes are used to digest (cut) fragments of DNA for the purposes of:
• fractionating genomic DNA by size (Southern blot).
• cloning a specific sequence (generate library).
• mapping a region of DNA (restriction site map).
• isolating a specific fragment of DNA (to make a labeled probe).

69
Q

What do restriction fragments do?
How often do they cut?
How is the desired DNA fragment isolated?
Explain four reasons why medical researchers clone genes and how they contribute to medicine

A

RESTRICTION FRAGMENT
Cut the DNA with RE
• How often do they cut?
• → If the DNA sequence is completely random (it’s not), the probability of a sequence occuring is:
6 bp recognition sequence : e.g. EcoRI (1/4)6 = 1/4096 ~ 1 cut every 4 kb
8bp““(1/4)8 ~1““65kb
4bp““(1/4)4 ~1““250bases
Have isolated >3000 Type II RE with >200 sequence specificities.
• IsolatethedesiredDNAfragment.
• Usually done by gel electrophoresis on agarose gels (non-denaturing).

Why do Medical Researchers Clone Genes and how does that contribute to the field of Medicine?
• A. To define inherited genetic mutations that cause/predispose to disease making it possible to:
• i.developdiagnosticreagents.
• ii. Develop pre-natal testing and counseling.
• iii. Define the precise biochemical defect.
• iv.Developtherapiesthattreatthedisease instead of the symptoms; a target therapy repairs/overcomes the inherited defect.

B. To isolate functional (“normal”) genes that can be harnessed to produce therapeutic molecules
• a. Proteins produced by recombinant DNA are made outside the body and not contaminated with infectious blood products such as Hepatitis, HIV, etc. For example the clotting factor Factor VIII needed by hemophiliacs.
• b. Proteins may be used not merely to replace a low level or nonfunctional protein, but also generate a desired effect in persons with a functional gene. For example, certain blood cell growth factors are given to allow harvest of bone marrow progenitor cells from peripheral blood.
• c. Genes can be isolated from infectious organism to allow vaccine development. Eliminates the possibility of reversion to a wild-type from an attenuated organism.

C. Isolate genes which contain somatic mutations (not inherited but acquired in a specific cell population)

a. Develop a diagnostic for certain rearrangements or mutations in tumors that leads to the selection of a specific type of therapy.
b. Use as a marker to measure response to therapy or disease progression.
c. Target specific therapies to recognize the altered gene.
i. antibodies against surface proteins ii. Tyrosine kinase inhibitors
iii. Gene therapy
iv. Stimulate the immune system.

D. To obtain an enhanced understanding of biochemical pathways in order to expand our knowledge base (signal transduction, immunity, development, the cell cycle)
• A. Integrate the discovery of a new gene into the existing pathway to make predictions about its function.
• B. “knowledge is power” basic research-Clinical research-New modalities of patient care.

70
Q

How do you clone from protein to gene and from gene to protein?

A

Protein to gene:
Isolate the protein on the basis of its molecular function example enzymatic or hormonal activity then you determine the partial amino acid sequence of the protein then you synthesize the oligonucleotides that correspond to portions of the amino acid sequence then you use the oligonucleotides as probes to select the cDNA or genomic clone encoding the protein from the library then you sequence the isolated protein

Gene to protein:
Isolate the genomic clone corresponding to an altered trait in mutants example nutritional auxotrophy,inherited diseases,developments defects then you use the genomic DNA to isolate a cDNA for the mRNA encoded by the gene then you sequence the cDNA to deduce the amino acid sequence of the encoded protein then you compare the deduced amino acid sequence with that of the known proteins to gain insight into the function of the protein then you use expression vector to produce the encoded protein

71
Q

What is molecular cloning
State four important features to consider when choosing a vector
What is the approximate max length of DNA that can be cloned in a vector such as plasmid, bacterial artificial chromosome (BAC)

A

Molecular Cloning.
• To study and manipulate a particular DNA sequence or gene, it is usually necessary to amplify it.
• The amplification process is called cloning (more precisely-molecular cloning) since a single recombinant DNA molecule is replicated into million of identical copies.
• This is different from cloning organisms which is the process of taking a single cell from an organism and using it to regenerate an identical copy of the organism.
• Molecular cloning involves joining a specific DNA fragment to a DNA molecule that can replicate in a host cell-usually the bacterium E. coli.
• The replicating DNA molecule is called a vector.
• Since the vector can replicate in the bacterial cells, any heterologous (foreign) DNA covalently joined to the vector will also be replicated.
• Common vectors are Plasmids and bacterial viruses. In eukaryotic cells, vectors can also be either plasmids or viruses (retroviruses).

The important features to consider when choosing a vector are:
• should be small and easy to handle.
• able to generate large amounts of DNA through replication in an appropriate organism such as bacteria.
• ifnecessarycandirectexpressionofa recombinant protein in appropriate host cell.
• can accommodate DNA of the required size.
• has an appropriate selection marker (antibiotic resistance)

Plasmid-20kb
BAC-300kb

72
Q

What is the General procedure for cloning a DNA fragment in a plasmid vector

A

. A
The plasmid fragment to be cloned is enzymatically inserted into the plasmid vector . A recombinant plasmid is formed . Mix E. Coli cells with plasmid in the presence of CaCl2 . Culture on nutrient agar plates containing ampicillin. Cells that do not take up the plasmid die on ampicillin plates and transformed E. coli cells with the plasmid survive . The surviving cells undergo independent plasmid replication ,there is cell multiplication.a colony of cells is Produced each containing copies of the same recombinant plasmid
The plasmid DNA can then be isolated
from cells in the colony and characterized or otherwise manipulated.

B
Isolation of DNA fragments from a mixture by cloning in a plasmid vector.

73
Q

State some uses of cloning
What are DNA libraries?
What are genomic libraries?
What are cDNA libraries

A

Some uses of cloning:
• 1. Isolate and characterize a specific gene or mRNA from a complex mixture (the genome or the total RNA in a cell).
• 2. Generate probes to detect homologous sequences-useful in disease diagnosis.
• 3. Express large amounts of protein-useful in producing therapeutic proteins.
• 4. Generate mutations to develop animal models of particular diseases.

DNA Libraries
• Libraries are a complex collection of DNA fragments (for example, the DNA fragments produced by cutting human DNA with a restriction fragment) individually inserted into a vector.

  • Genomic Libraries
  • Genomic libraries consist of DNA derived from the genome of an organism.
  • Genomic libraries therefore contain all elements of the genome-coding regions, intron, regulatory control regions, regions between gene, etc.

cDNA Libraries
• cDNA libraries contain double-stranded copies of DNA copied from mRNA.
• cDNA library only contains the sequences that are present in the mRNAs.
• It also contains only those sequences that are expressed-therefore two cDNA libraries made from two cell types in an organism will have overlapping but distinct sets of sequences.
• cDNA is synthesized using reverse-transcriptase-an enzyme that synthesizes a single strand DNA copy of an mRNA. The single stranded DNA can then be converted into a double-stranded molecule using a DNA polymerase and then cloned into a vector.

74
Q

How is Construction of a human Genomic DNA library done

Why should you construct and screen a genomic library?

A

Human double stranded DNA cleaves with restriction nuclease . Millions of genomic DNA fragments are inserted into plasmids . Recombinant DNa molecule is formed. Introduction of plasmids into bacteria . Genomic DNa library is formed

Genomic library contains all of the genetic information found in the genome of an organism.
• In the case of human DNA introns, exons and noncoding regions would all be included.
• Reasons to construct and screen a genomic library include:
• 1. to clone the whole gene including introns and 5’ regulatory sequences (promoter), for example to study gene expression.
• 2. the gene will be cloned in the context of surrounding DNA making it possible to move along a chromosome and clone a new gene whose approximate chromosomal location is known.
• 3. to make knock-out animal model by replacing the functional gene with a nonfunctional copy.

75
Q

How is cDNA synthesized?

When thinking about constructing or using a cDNA library what should you remember?

The differences between cDNA clones and genomic DNA clones derived from the same region of DNA true or false

A

When thinking about constructing or using a cDNA library remember:
• There are no upstream regulatory sequences controlling gene expression because they are not transcribed into the mRNA.
• ThecellularsourceofmRNAiscritical!Ifthecell does not express the gene , no mRNA, no cDNA and it won’t be in the library.
• Since there are no introns, cDNA clones can be “translated” to find open reading frames and predict protein sequence.
• Itisnotpossibletodeterminechromosomal “context” or identify relationship to genes that map nearby.

76
Q

What is nucleic acid hybridization

What is it used for

A

Nucleic Acid Hybridization
• A common method to detect a specific nucleotide sequence in a complex mixture such as a library is hybridization.
• Hybridizationisbasedontheabilityofasingle- stranded nucleic acid (probe) to specifically anneal by base pairing to a complementary sequence present among many copies of non- complementary sequences.
• To detect the hybridization product, the probe is modified so that it can be detected-for example with radioactive or fluorescent nucleotides.

Membrane hybridization assay for detecting nucleic acids.
This assay can be used to detect both DNA and RNA, and the radiolabeled complementary probe can be either DNA or RNA.

77
Q

What is southern and northern blotting
How are they performed?
What is southern blot used for?
What happens in northern blotting?

A

Southern and Northern Blotting
• These are procedures in which mixtures of nucleic acids are separated by size and then probed by hybridization.
• To perform a southern blot, the genomic DNA is first digested to completion with one or more restriction endonucleases.
• The size separation is carried out by gel electrophoresis in which DNA or RNA is applied to a porous gel and then subjected to an electric field. The negatively charged DNA or RNA migrates towards the positive electrode. The smallest fragments migrate the fastest and therefore at the bottom of the gel.
• The molecules are then transferred to a solid support such as nitrocellulose filter.
• Before the transfer, the DNA gel is treated with an alkaline solution to denature the DNA, this separates the two complementary strands making them available for hybridization with the probe.
• The blot is then hybridized to a labeled DNA probe (can be either radioactive or non radioactive system) to detect the gene or RNA of interest.

Southern blot technique for detecting the presence of specific DNA sequences
In Northern blotting, the total cellular RNA is denatured by treatment with an agent (e.g formaldehyde) that prevents H-bonding between base pairs, ensuring that all the RNA molecules have unfolded, linear conformation.

78
Q

What is southern blot used to detect?
What does southern blot not detect?
What can northern blot detect?

A

Southern Blot can be used to Detect:
• large insertion or deletion (50 to 100bp) of chromosomal DNA into the gene; the restriction fragments which hybridize to the probe will be either larger or smaller than expected.
• gross gene amplification, this means that rather than the normal number of 2 copies per diploid cell there may be 5 to 20 or more copies per cell. (such as the HER-2/neu gene which plays a causative role in the development of a highly aggressive breast cancer).
• genomic rearrangements caused by chromosomal translocation. This occurs when two chromosomes actually break and re-join with the wrong chromosome (an example is the BCR-Abl translocation that plays a causative role in the development of chronic myeloid leukemia (CML)).
• mutation of a single nucleotide can only be seen on a Southern blot if it creates or destroys a restriction enzyme site (Hemophilia A and sickle cell diseases are examples).

A Southern Blot does not detect:
• whether or not a gene is expressed
• point mutations (other than restriction fragment length polymorphism)
• small deletions.

A Northern Blot can detect:
• expression of a specific gene which likely changes from tissue to tissue (except for “housekeeping” genes).
• the size of the RNA transcript. An aberrant sized mRNA could arise as a result of deletion or insertion in the gene or a chromosomal translocation.
• the relative levels of expression in different samples. Increased or decreased levels of mRNA are associated with many diseases states
• northern can detect deletions or insertions in genes as well as splicing defects.

79
Q

What is polymerase chain reaction

A

Polymerase Chain Reaction
• Procedure that can amplify a specific sequence without cloning into a vector.
• It is very useful in clinical and forensic settings where limited amounts of DNA are available.
• Use short, chemically-synthesized primers that flank region to be amplified.
• Use enzyme from bacteria that lives at high temperature such as Thermus aquaticus.
• Template is melted at high temperature and annealed to excess amounts of primers at lower temperatures.
• Copies are made by the polymerase using dNTPs.
• The cycle is repeated about 20 to 40 cycles.
• Since each round results in a replication cycle, if carry out 20 cycles will get 220= 1,048,576 copies.

PCR continued.
• Because of the amplification properties of PCR it is a very powerful technique to detect very low levels of DNA or RNA. For example it is used as a very sensitive test for HIV infection since PCR can be used to amplify sequences from the RNA genome of the virus.
• It is also sensitive enough to detect specific sequences in small amounts of material available from crime scenes. It is so sensitive that a fingerprint or saliva left on the back of a stamp contains enough DNA for analysis by PCR.
• PCR is very fast compared to Southern or Northern blotting so it has replaced these techniques in many areas.

80
Q

What are the key steps in PCR reaction and state ten uses of PCR

A

The Key Steps in the PCR reaction are:
• 1. denature the template DNA by heating to 94o for 5 minutes.
• 2. anneal the specific primer by lowering the temperature to around 50o.
• 3. extend the DNA strand using DNA polymerase (Taq) for 5 minutes (72-75o).
• 4. heat 20 seconds to separate the newly synthesized short strands.
• 5. go to step 2 and repeat 25 to 40 cycles.

Uses of PCR
• To look for mutations either inherited or acquired, by doing a PCR reaction followed by DNA sequencing.
• To detect Minimal Residual Disease in leukemia patients using primers made to recognize DNA sequence on either side of the chromosomal breakpoint. The presence of a small number of leukemic cells in the midst of a proportionally larger number of normal cells is detected by the presence of a positive signal.
• PCR may be used to determine the gender of embryonic cells generated by in vitro fertilization in families carrying X-linked mutations.
• It can also be used for in vitro mutagenesis to test the effect of a specific mutation on the function of the gene product.

Uses of PCR Cont’d
• Gene cloning.
• Carrier screening.
• Clinical diagnosis/confirmation.
• Newborn screening.
• Presymptomatic diagnosis/predisposition screening.
• Transplant engraftment-distinguishing donor from recipient cells after bone marrow or organ transplant.
• Parentage/paternity testing-excluding or not excluding an alleged father (or mother).
• Forensic identification-matching a suspect’s DNA to that found in some crime evidence in order to solve cases of murder, rape, mail fraud, etc.; identifying victims of mass disasters (bombings, airline crashes, wars, genocides).






• • • •
PCR uses continued
Twin zygosity-distinguishing mono-from dizygotic twins (since only monozygotic twins should match at all polymorphic loci tested).
Early detection of HIV, the virus that causes AIDS.
Surgical specimen mix-ups-to match a mislabeled or unlabeled biopsy specimen to the patient from which it came.
DNA fingerprinting
Prenatal diagnosis-several sample types are possible, depending on the clinical situation:
Amniocentesis
Chorionic villus samples (CVS)
Embryonic blastomeres for preimplantation diagnosis Fetal cells circulating in maternal blood.

81
Q

How is PCR used in forensic science?(Three suspects A,B and C, producing
6 DNA bands for each person after PAGE. The band pattern therefore save as a “fingerprint” to identify an individual
nearly uniquely.)
How is it used. Obtain a genome or cDNA clone?

A
82
Q

Explain the types of DNA polymorphisms

A

Types of DNA Polymorphism
• RFLP=Restriction Fragment Length Polymorphism-restriction endonuclease recognition sites that are present in some people but not in others (can be detected by Southern blotting or PCR).
• VNTR=Variable Number of Tandem Repeats-strings of nucleotide sequences (usually of 16 or more nucleotides each), of different repeat number in different people, existing between fixed restriction endonuclease sites; also called minisatellites.
• STR= Short Tandem Repeats-string of oligonucleotide sequences that are shorter that VNTR (usually of 2,3,4 or 8 nucleotides each), of different repeat number in different people, not necessarily lying between restriction endonuclease sites; also called microsatellites or Simple Sequence Repeats (SSR). The more highly related two individuals are, the more likely it is that a given SSR will be present in the same numbers. SSR variations can be detected by PCR, using primers that lie outside of the repeat region followed by accurate gel electrophoresis of the products. The size of the PCR product then reflects the number of repeats.

Polymorphism cont’d
• SNP=single Nucleotide Polymorphism-single nucleotide differences between individuals that are not associated with restriction endonuclease sites. Best studied by PCR followed by accurate gel electrophoresis of the products.
• HLA=the histocompatibility family of genes located on chromosome 6p; although not “junk” DNA, these genes are naturally so highly polymorphic that they can be used for identity studies at either the DNA level or using antibodies to distinguish varieties in the encoded proteins.

83
Q

Explain the Sanger sequencing method

A

Sanger (Dideoxy) Sequencing Method.
• Also called dideoxy sequencing because it involves use of 2’,3’-dideoxynucleoside triphosphates (ddNTPs), which lacks a 3’-hydroxy group.
• In this method, the single-stranded DNA to be sequenced serves as the template strand for in vitro DNA synthesis, a synthetic 5’-end labeled oligodeoxynucleotide is used as the primer.
• Four separate polymerization reactions are performed, each with a low (100 lower) concentration of one of the 4 ddNTPs in addition to higher concentrations of the normal deoxynucleotide triphosphates (dNTPs).
•.

Sanger’s Method cont’d
• In each reaction, the ddNTP is randomly incorporated at the positions of the corresponding dNTP; such addition of a ddNTP terminates polymerization because the absence of a 3’ hydroxy prevents addition of the next nucleotide.
• The mixture of terminated fragments from each of the four reactions is subjected to gel electrophoresis in parallel or complementary sequence;the separated fragments then are detected by autoradiography.
• The sequence of the original DNA template strand can be read directly from the resulting autoradiogram.

Ithis method is Used more frequently than
the Maxam-Gilbert method.
can be used to sequence longer fragments of DNA
than the Maxam-Gilbert method

84
Q

What is the enzymatic method of sequencing DNA
What is the strategy for automating DNA sequencing reactions.

Explain Fluorescent Automated DNA sequencing

A

The enzymatic—or dideoxy— method of sequencing DNA.
This reaction mixture will eventually produce a set of DNAs of different lengths complementary to the template DNA that is being sequenced and terminating at each of the different A’s. The exact lengths of the DNA synthesis products can then be used to determine the position of each A in the growing chain.
In each lane, the bands represent fragments that have terminated at
a given nucleotide (e.g., A in the
leftmost lane) but
at different positions
in the DNA

Strategy for automating DNA sequencing reactions.
Each dideoxynucleotide used in the Sanger method can be linked to
a fluorescent molecule that gives all the fragments terminating in that
nucleotide a particular color.
All four labeled ddNTPs are added to a single tube.
The resulting colored DNA fragments are then separated
by size in a single electrophoretic gel contained in a capillary tube (a
refinement of gel electrophoresis that allows for faster separations).
All fragments of a given length migrate through the capillary gel in a single peak, and the color associated with each peak is detected using a laser beam.
The DNA sequence is read by determining the sequence of colors in the peaks as they pass the detector.
This information is fed directly to a computer, which determines the sequence.

85
Q

Nnnn

A

Production of High levels of Proteins from Cloned cDNA
• Once the desired cDNA is cloned, large amounts of the encoded protein can be synthesized in engineered E. coli cells.
• Example, granulocyte colony-stimulating factor can be produced at high levels in expression vectors designed to produce full length proteins at high levels.

A simple E. coli expression vector utilizing the lac operon
The lacZ gene can be replaced with the G-CSF. When the resulting plasmid is transformed into E. coli cells, addition of IPTG and subsequent transcription from the lac promoter produces G-CSF mRNA, which is translated into G-CSF protein.

Isolation of mouse embryonic stem (ES) cells with a gene targeted disruption by positive and negative selection
When exogenous DNA is introduced into ES cells, random insertion via nonhomologous recombination occurs more frequently than gene-targeted insertion via homologous recombination.
Recombinant cells in which one copy of the gene X is disrupted can be obtained by using a recombinant vector that carries gene X disrupted with neor, a neomycin- resistance gene, and outside the region of homology, tkHSV, the thymidine kinase gene from herpes simplex virus
Nonhomologous insertions includes the tkHSV gene, whereas homologous insertions doesn’t; therefore, only cells with nonhomologous insertion are sensitive to ganciclovir.

General procedure for producing gene- targeted knockout mice
Embryonic stem (ES) cells heterozygous for a knockout mutation in a gene of interest (X) and homozygous for a marker gene (here, black color) are transplanted into the blastocoel cavity of a 4.5-day embryos that are homozygous for an alternative marker (here, white color).
The early embryos then are implanted into a peudopregnant female. Some of the resulting progeny are chimeras, indicated by their black and white coats.
Chimeric mice then are backcrossed to white mice; black progeny from this mating have ES-derived cells in their germ line.
by isolating DNA from a small amount of tail tissue, it is possible to identify black mice heterozygous for the black allele. Intercrossing
of these black mice produces individuals homozygous for the disrupted allele, that is, knockout mice.

Creation of a Knockout animal

General procedure for producing transgenic mice
Frequency of random integration of exogenous DNA into the mice genome at nonhomologous sites is very high.
Production of transgenic mice is a highly efficient and straightforward process.

86
Q

Give some examples of recombinant DNA products in medicine

A

Product- anticoagulant
Example or uses-tissue plasminogen activator. This activated plasmin ,an enzyme involved in dissolving clots ,effective in treating heart attack patients

Product-vaccines
Example or uses:proteins derived from viral coats are as effective in priming an immune system as is the killed virus more traditionally used for vaccine

87
Q

Explain how recombinant DNa works

A

DNA cloning is a molecular biology technique that makes many identical copies of a piece of DNA, such as a gene.
In a typical cloning experiment, a target gene is inserted into a circular piece of DNA called a plasmid.
The plasmid is introduced into bacteria via a process called transformation, and bacteria carrying the plasmid are selected using antibiotics.
Bacteria with the correct plasmid are used to make more plasmid DNA or, in some cases, induced to express the gene and make protein.
Introduction

When you hear the word “cloning,” you may think of the cloning of whole organisms, such as Dolly the sheep. However, all it means to clone something is to make a genetically exact copy of it. In a molecular biology lab, what’s most often cloned is a gene or other small piece of DNA.
If your friend the molecular biologist says that her “cloning” isn’t working, she’s almost certainly talking about copying bits of DNA, not making the next Dolly!
Overview of DNA cloning

DNA cloning is the process of making multiple, identical copies of a particular piece of DNA. In a typical DNA cloning procedure, the gene or other DNA fragment of interest (perhaps a gene for a medically important human protein) is first inserted into a circular piece of DNA called a plasmid. The insertion is done using enzymes that “cut and paste” DNA, and it produces a molecule of recombinant DNA, or DNA assembled out of fragments from multiple sources.
Diagram showing the construction of a recombinant DNA molecule. A circular piece of plasmid DNA has overhangs on its ends that match those of a gene fragment. The plasmid and gene fragment are joined together to produce a gene-containing plasmid. This gene-containing plasmid is an example of recombinant DNA, or a DNA molecule assembled from DNA from multiple sources.
Diagram showing the construction of a recombinant DNA molecule. A circular piece of plasmid DNA has overhangs on its ends that match those of a gene fragment. The plasmid and gene fragment are joined together to produce a gene-containing plasmid. This gene-containing plasmid is an example of recombinant DNA, or a DNA molecule assembled from DNA from multiple sources.
Next, the recombinant plasmid is introduced into bacteria. Bacteria carrying the plasmid are selected and grown up. As they reproduce, they replicate the plasmid and pass it on to their offspring, making copies of the DNA it contains.
What is the point of making many copies of a DNA sequence in a plasmid? In some cases, we need lots of DNA copies to conduct experiments or build new plasmids. In other cases, the piece of DNA encodes a useful protein, and the bacteria are used as “factories” to make the protein. For instance, the human insulin gene is expressed in E. coli bacteria to make insulin used by diabetics.
More about insulin and diabetes
Steps of DNA cloning

DNA cloning is used for many purposes. As an example, let’s see how DNA cloning can be used to synthesize a protein (such as human insulin) in bacteria. The basic steps are:
Cut open the plasmid and “paste” in the gene. This process relies on restriction enzymes (which cut DNA) and DNA ligase (which joins DNA).
Insert the plasmid into bacteria. Use antibiotic selection to identify the bacteria that took up the plasmid.
Grow up lots of plasmid-carrying bacteria and use them as “factories” to make the protein. Harvest the protein from the bacteria and purify it.
Let’s take a closer look at each step.
1. Cutting and pasting DNA

How can pieces of DNA from different sources be joined together? A common method uses two types of enzymes: restriction enzymes and DNA ligase.
A restriction enzyme is a DNA-cutting enzyme that recognizes a specific target sequence and cuts DNA into two pieces at or near that site. Many restriction enzymes produce cut ends with short, single-stranded overhangs. If two molecules have matching overhangs, they can base-pair and stick together. However, they won’t combine to form an unbroken DNA molecule until they are joined by DNA ligase, which seals gaps in the DNA backbone.
See a diagram of restriction enzymes and DNA ligase
Our goal in cloning is to insert a target gene (e.g., for human insulin) into a plasmid. Using a carefully chosen restriction enzyme, we digest:
The plasmid, which has a single cut site
The target gene fragment, which has a cut site near each end
Then, we combine the fragments with DNA ligase, which links them to make a recombinant plasmid containing the gene.
Diagram depicting restriction digestion and ligation in a simplified schematic.

We start with a circular bacterial plasmid and a target gene. On the two ends of the target gene are restriction sites, or DNA sequences recognized by a particular restriction enzyme. In the plasmid, there is also a restriction site recognized by that same enzyme, right after a promoter that will drive expression in bacteria.

Both the plasmid and the target gene are (separately) digested with the restriction enzyme. The fragments are purified and combined. They have matching “sticky ends,” or single-stranded DNA overhangs, so they can stick together.

The enzyme DNA ligase joins the fragments with matching ends together to form a single, unbroken molecule of DNA. This produces a recombinant plasmid that contains the target gene.
Diagram depicting restriction digestion and ligation in a simplified schematic.
We start with a circular bacterial plasmid and a target gene. On the two ends of the target gene are restriction sites, or DNA sequences recognized by a particular restriction enzyme. In the plasmid, there is also a restriction site recognized by that same enzyme, right after a promoter that will drive expression in bacteria.
Both the plasmid and the target gene are (separately) digested with the restriction enzyme. The fragments are purified and combined. They have matching “sticky ends,” or single-stranded DNA overhangs, so they can stick together.
The enzyme DNA ligase joins the fragments with matching ends together to form a single, unbroken molecule of DNA. This produces a recombinant plasmid that contains the target gene.
2. Bacterial transformation and selection

Plasmids and other DNA can be introduced into bacteria, such as the harmless E. coli used in labs, in a process called transformation. During transformation, specially prepared bacterial cells are given a shock (such as high temperature) that encourages them to take up foreign DNA.

Why does a heat shock make bacteria take up DNA?
A plasmid typically contains an antibiotic resistance gene, which allows bacteria to survive in the presence of a specific antibiotic. Thus, bacteria that took up the plasmid can be selected on nutrient plates containing the antibiotic. Bacteria without a plasmid will die, while bacteria carrying a plasmid can live and reproduce. Each surviving bacterium will give rise to a small, dot-like group, or colony, of identical bacteria that all carry the same plasmid.
Left panel: Diagram of plasmid, showing that it contains an antibiotic resistance gene.

Right panel: all the bacteria from the transformation are placed on an antibiotic plate. Bacteria without a plasmid will die due to the antibiotic. Each bacterium with a plasmid makes a colony, or a group of clonal bacteria that all contain the same plasmid. A typical colony looks like a small, whitish dot the size of a pinhead.
Left panel: Diagram of plasmid, showing that it contains an antibiotic resistance gene.
Right panel: all the bacteria from the transformation are placed on an antibiotic plate. Bacteria without a plasmid will die due to the antibiotic. Each bacterium with a plasmid makes a colony, or a group of clonal bacteria that all contain the same plasmid. A typical colony looks like a small, whitish dot the size of a pinhead.
Not all colonies will necessarily contain the right plasmid. That’s because, during a ligation, DNA fragments don’t always get “pasted” in exactly the way we intend. Instead, we must collect DNA from several colonies and see whether each one contain the right plasmid. Methods like restriction enzyme digestion and PCR are commonly used to check the plasmids.
3. Protein production

Once we have found a bacterial colony with the right plasmid, we can grow a large culture of plasmid-bearing bacteria. Then, we give the bacteria a chemical signal that instructs them to make the target protein.
The bacteria serve as miniature “factories,” churning out large amounts of protein. For instance, if our plasmid contained the human insulin gene, the bacteria would start transcribing the gene and translating the mRNA to produce many molecules of human insulin protein. More about expressing human genes in bacteria
A selected colony is grown up in a large culture (e.g., a 1-liter flask). The bacteria in the large culture are induced to express the gene contained in the plasmid, causing the gene to be transcribed into mRNA, and the mRNA to be translated into protein. The protein encoded by the gene accumulates inside of the bacteria.
A selected colony is grown up in a large culture (e.g., a 1-liter flask). The bacteria in the large culture are induced to express the gene contained in the plasmid, causing the gene to be transcribed into mRNA, and the mRNA to be translated into protein. The protein encoded by the gene accumulates inside of the bacteria.
Once the protein has been produced, the bacterial cells can be split open to release it. There are many other proteins and macromolecules floating around in bacteria besides the target protein (e.g., insulin). Because of this, the target protein must be purified, or separated from the other contents of the cells by biochemical techniques. The purified protein can be used for experiments or, in the case of insulin, administered to patients. Is it really that simple to make insulin?

The DNA produced by ligation (which may be a mix of desired plasmids, side-product plasmids, and linear DNA pieces) is added to bacteria. The bacteria are given a heat shock, which makes them more apt to take up DNA by transformation. However, only a tiny minority of the bacteria will successfully take up a plasmid.

88
Q

Explain Recombinant DNA technology (rDNa )

A

Is a technique use sim genetic engineering that involves the identification,isolation,insertion of the gene of interest into a vector such as a plasmid or a bacteriophage to form a recombinant DNA molecule and the production of large quantities of that gene fragment or product encoded by that gene

An Example of a product synthesized using rDNA technology
Humulin, is insulin developed using rDNA technology which is used to treat diabetes. Here insulin is synthesized inside bacterium where we introduced human insulin gene. Thus bacterial system just works as biofactories for the synthesis of insulin.
Watch video here: How Humulin is synthesized using rDNA technology?
How this is achieved? We will be discussing the basic steps involved in rDNA technology, gene cloning or genetic engineering.
Steps in Recombinant technology
Step1: Identification and isolation of gene of interest
From where we get the desired gene?
From
Genomic library
cDNA library
Chemical synthesis of gene if we know the sequence
If the number of copies of the desired gene is not enough for gene cloning we can opt for gene amplification techniques like PCR
Step II: joining of this gene into a suitable vector (construction of recombinant DNA)
What is a Gene Cloning Vector?
A vector is any DNA molecule which is capable of multiplying inside the host to which our gene of interest is integrated for cloning. The selection of vector depends upon the size of the fragments to be cloned.
Common vectors include plasmids (Eg: pBR 322) and phage vectors.
In the process, restriction enzymes functions as scissors for cutting DNA molecules. Ligase enzyme is the joining enzyme that joins the vector DNA with gene of interest. The resulting DNA is called the recombinant DNA, chimera or recombinant vector. Step III: Introduction of this vector into a suitable organism
Introduction of recombinant vector into host cell is achieved by different gene transfer methods
a. Physical gene transfer methods:
Electroporation
Microinjection
Liposome mediated gene transfer
Silicon Carbide fibre mediated gene transfer
Ultrasound mediated gene transfer
DNA transfer via pollen
b. Chemical gene transfer methods:
Poly Ethylene Glycol mediated (PEG mediated),
Calcium Chloride mediated
DEAE dextran mediated gene transfer
c. DNA imbibitions by cells, tissues or organs: Transformation
d. Virus mediated gene transfer: Transduction

Step VI: Selection of transformed recombinant cells with gene of interest
The number of cells with recombinant vector will be very less. So the next step is to select the transformed recombinant cells with our gene of interest from the sea of non transformed cells. Several methods are employed for selection of transformed cells: Antibiotic resistance Watch video: How selectable markers helps in selection?
Visible characters,
Assay for biological activity,
Colony hybridization,
Blotting test.
The selected cells are cultured in large scale.
Step V: Multiplication or expression of the gene of interest
The objective of gene cloning is either to make numerous copies of the desired gene or to produce the protein coded by the desires gene. The inserted gene along with the vector will replicate inside the host so that many copies of the desired gene is synthesized.
For expression of the desired gene, expression vector is used (vector with control elements like promoter, operator etc). The product is synthesized in mass cultures in large quantities. This is how insulin is produced in large quantities in cell cultures.

89
Q

Why is heat resistant DNA polymerase needed for PCR to occur?
What is cDNA ?
Is the nucleotide sequence of cDNA identical to the homologous DNA sequence in the genome?

A

Because PCR relies on cycles of denaturation ,annealing and elongation ,a heat resistant DNA polymerase is necessary to survive the denaturation step

Complimentary DNa produced from an mRNA template

No because it doesn’t include introns