DNA STRUCTURE AND REPLICATION

Question 1

Function of DNA

Answer 1

• high capacity for information storage
• chemically stable: less susceptible to chemical/enzyme degradation/separation of double helix
• replicate accurately
• be capable of variation

Question 2

Structure of nucleic acids

Answer 2

1. Exists as polymers known as polynucleotides
2. Each polynucelotide composed of repeating units called nucleotides
3. Each nucleotide consists of a five-carbon pentose + a nitrogenous base + a phosphate group

Question 3

Types of nucleic acids

Answer 3

1. Deoxyribonucleic acid(DNA)
2. Ribonucleic acid(RNA)

Question 4

DNA vs RNA

Answer 4

Type of pentose sugar:
- DNA: deoxyribose, at 2’ carbon, -OH group replaced by H atom
1. partial negative charge of hydroxyl group in ribose repels negative charge of phosphate, preventing RNA chain from couling in as tight as a helix as it does in DNA
2. RNA is more susceptible to chemical and enzyme degradation
- RNA: ribose, at 2’ carbon, H atom replaced by -OH group

Type of nitrogenous base:
- DNA: Adenine, Guanine, Cytosine, Thymine
- RNA: adenine, guanine, cytosine, uracil
1. Thymine VS uracil: presence of methyl substituent at C5 of thymine vs uracil

Type of monomer:
- DNA: deoxyribonucleosides + phosphate group
- RNA: ribonucleosides + phosphate group

Strand:
- DNA: double-stranded
- RNA: single-stranded

Question 5

Structure of + formation of nucleotide in DNA

Answer 5

Structure:
Pentose(deoxyribose) sugar + nitrogenous base + phosphate group
* 4 types of nitrogenous bases, Adenine, Thymine, guanine and cytosine
* 2 classes purines and pyrimidines
* Pure As Gold - A and G are purines
* Purines VS pyrimidines
1. purines: 6 membered ring
2. pyrimidine: 5 membered ring

Formation:
1. 1’ carbon of pentose linked via glycosidic bond to nitrogenous base via condensation reaction to form nuceloside
2. further condensation between nucleoside and phosphte group to form phophoester bond

Question 6

Formation of polynucleotide

Answer 6

1. Formed via condensation between 5’ phosphate group of one nucelotide and 3’ hydroxyl group of the other to form a phosphodiester bond
2. Condensation reaction between nucleotides is repeated several million times to form a polynucleotide
3. Phosphodiester bonds between 5’ phosphate and 3’ hydroxy groups of nucleotdies form a linear, unbranched sugar-phosphate backbone
4. Phosphodiester bonds are strong covalent bonds, confer strength and stability

Question 7

Structure of polynucleotide

Answer 7

1. Linear, unbranched sugar-phosphate backbone
2. Polynucleotide has polarity
   * 5’end iwth a free 5’ carbon carrying a phosphate group
   * 3’ end with a free 3’ carbon carrying a hydroxyl group
   * read in 5’ to 3’ direction

Question 8

Structure of DNA

Answer 8

• consists of 2 polynucleotide strands
• Each strand forms a right-handed helix
• the 2 strands coil around each other to form a double-helix
  1. makes one complete twist every 3.4nm
  2. 10 bases to complete each turn on the helix
  3. diameter of helix is uniformly 2nm, enough space for 1 purine and 1 pyrimidine
• grooves of unequal sizes between sugar-phosphate backbones called major and minor grooves
• run in opposite directions, anti-parallel
• Base composition of DNA of an organism is constant throughout all the somatic cells and is characteristic for a given species
• Consists of 4 types of nucleotides, ATCG
• Always equal propotion of adenine and thymine, equal proportion of guanine and cytosine
  1. Due to specific complementary base pairing between A on one DNA strand and T on other other, and between G on one strand and C on the other
• Has a sugar phosphate backbone
• phosphate groups project outside the double-helix as they are hydrophilic
• nitrogenous bases orientate inwards toward the central axis at almost right angles
• Bases on opposite strands are bonded together by relatively weak hydrogen bonds
  1. 2H bonds present between A and T
  2. 3H bonds present between C and G

Question 9

Chargaff’s rule

Answer 9

amount of A=T and amount of G=C

Question 10

What is CBP

Answer 10

1. specific CBP occurs between A and T to form 2 H bonds and betwen C and G to form 3 H bonds

Question 11

Reason for CBP

Answer 11

1. A-T and C-G pairs are the only ones that can fit the physical dimensions of double helix due to steric restrictions
   * sugar-phosphate backbone of each polynucleotide chain has a regular helical structure
   * double helix has uniform diamter
   * T and C are pyrimidines, single ring
   * A and G are purines, twice as wide
   * pair purine with pyrimidine to fit into diameter of helix
2. In accordance with Chargaff’s rule
3. Hydrogen bond factors
   * each nitrogenous base has chemical side groups with well-defined positions that can form hydrogen bonds with its appropriate patner

Question 12

Significance of CBP

Answer 12

• base sequence in one strand determines base sequence in comlementary strand
  1. use of intact strand as template for repair if base sequence in one of the 2 strands is accidentally altered
• weak hydrogen bonds between CBP make it relatively easy to separate the 2 strands of DNA
  1. A-T pair easier to separate by heating than G-C
  2. A-T pair has 2H, G-C has 3H
• does not restrict base sequence along each DNA strand and linear sequence of 4 bases can be varied in countless ways
  1. 4 to the power of 3 x 10^9 combinations of bases
  2. each gene has a unique base sequence

Question 13

Drawing of DNA

Answer 13

1. Draw 2 polynucleotide strands with uniform diamters
2. Draw nucleotide with deoxyribose + phosphate group + nitrogenous base
3. Laberl polarity of strands (anti-parallel)
4. Size of purine > size of pyrimidines
5. CBP of 2H bonds between A and T and 3H between C and G

Question 14

the 5 levels

Packing of DNA

Answer 14

DNA Double helix -> Chromatin -> Nucleosomes -> chromosomes in extended form -> condensed section of metaphse chromosome -> entire metaphase chromosome

Question 15

Stability of DNA in storing information

Answer 15

• Extensive hydrogen bonds between base pairs
• Hydrophobic interactions between stacked base pairs
• Exposure to outside influences of only the sugar-phosphate backbone
• Nitrogenous bases tucked safely inside double-helix
• DNA is tightly wound around histones to form repeating array of nucleosomes, eventually fold into higher order structures such as chromsomes to prevent physical and thermal damage

Question 16

Store information accurately

Answer 16

• genetic information is redundant in DNA molecules
• CBP means that base sequence in one strand is determined by base sequence in the other strand
  1. use of intact strand as template for repair of strand that is accidentally alterted, damaged strand discarded by cell and integrity of DNA maintained

Question 17

Role of DNA in DNA replication

Answer 17

• Acts as a template
• as the 2 strands are complementary in nature due to specific complementary base pairing of nitrogenous bases in DNA
  1. stores the information necessary to reconstruct the other
• each strand serves as a template for ordering nucleotides into a new, complementary strand

Question 18

Semi-conservative DNA replication

Answer 18

1. 2 DNA strands unwind and separate from each other as hydrogen bonds between complementary base pairs are broken
2. each DNA strand acts as a template for assembly of complementary strand
3. nucleotides line up singly along tempalte DNA strand acc Chargaff’s rule of CBP
4. DNA polymerases join nucleotides together at their sugar-phosphate moieties
5. 2 identical daughter DNA molecules produced from a single parental DNA molecule
6. Each of the 2 daughter DNA molecules consists of one parental DNA strand and one newly-synthesised daughter DNA strand

Question 19

Characteristics of DNA replication

Answer 19

1. complex
2. fast
3. accurate
4. requires large team of enzymes and other proteins, as well as expenditure of ATP

Question 20

6 main stages

Process of DNA replication

Answer 20

Initiation: Formation of Origin of Replication
1. is a specific sequence of nucleotides that are A-T rich(only 2 H bonds, easier to distrupt the bonds as less energy needed to overcome them)
2. initiator proteins recognise and bind to specific oriR sequence
3. DNA double-helix separates(involves other proteins) into 2 strands to form replication ‘bubble’
4. Each end of replication bubble(Y-shaped structure) called a replication fork
5. replication forks move away from oriR as replication proceeds bidirectionally

Separation of strands
1. need for continual separation of base pairs so that both DNA strands can act as templates
2. helicases bind to one strand of DNA molecule: use ATP as a source of energy to break the hydrogen bonds holding the 2 strands of DNA together -> unwinds the DNA double-helix -> separates parental DNA strands at region of replication fork
3. allows each of the 2 parental DNA strands to serve as the template for synthesis of a new DNA strand
4. Single-strand DNA-binding proteins (SSB proteins): temporarily stabilises the unwound single-stranded portion of the DNA double-helix by binding to it -> prevets ssDNA from re-annealing to reform the duplex -> keeps 2 parental strands in appropriae ss condition to act as a template + protects ssDNA which is very unstable from being degraded
5. topoisomerases:
* unwinding causes supercoiling ahead of the replication fork, resulting in tension
* cleaves a strand of the helix to create a transient ss nick
* relives strain on DNA molecule by allowing free roation around intact strand
* and then reseals the broken strand

Synthesis of daughter DNA strands via CBP
1. RNA primers with complementary base sequence to template provides a free 3’ OH end that DNA polymerase can extend, priming the synthesis of the daughter DNA strand
2. DNA polymerase with 5’ to 3’ exonuclease activity later replaces RNA nucleotides of primer with DNA versions
3. parental DNA strand separated at replication fork and primed with RNA primer serves as a template for semi-conservative DNA replication
4. DNA polymerase reads the template and assembles the deoxyribonuleoside triphosphates(dNTPS) for newly-synthesised daughter DNA strand based on CBP
5. when incorrect base pair recognised, proofreading carried out where DNA polymerase reverses its direction by one base pair of DNA -> 3’ to 5’ exonuclease activity of enzyme allows incorrect base pair to be excised

Synthesis of strands - via phosphodiester bond formation
1. DNA polymerase catalyses polymerisation of strand
2. all DNA polymerase catalyses phosphodiester bond formation between a free 3’ hydroxyl group of last nucleotide on growing daughter DNA strand and free 5’ phosphate group of an incoming nucleotide/dNTP
3. due to active site specificity of DNA polymerase, synthesis of both strands can only occur in 5’ to 3’ direction
5. incoming dNTP loses a pyrophosphate group when they form phosphoester bond with growing daughter DNA strand
6. energy released from pyrophosphoester bond breakage is coupled to phosphoester bond formation

Leading strand and lagging strand
7. Due to DNA polymerase only being able to add dNTPs to the free 3’ end of a growing DNA strand and not 5’ end, growing strand can only elongate in 5’ to 3’ direction
8. however, the 2 strand of a DNA double-helix are antiparallel and their sugar-phosphate backbone run in opposite directions
9. hence, continuous synthesis of both DNA strands at replication fork is not possible
10. hence, there is the leading strand and lagging strand synthesis
11. leading strand is the complementary DNA strand continously synthesised as a single polymer along template strand, polymerised in mandatory 5’ to 3’ manner towards replication fork
12. lagging strand is the complementary DNA strand that is discontinuously synthesised as a series of short fragments known as okazaki fragments -> each okazaki fragment polymerised in 5’ to 3’ manner against overall direction of replication fork
13. as such, each okazaki fragment requries an RNA primer for strand initiation
14. okazaki fragments ligated
* DNA polymerase I removes RNA primer and replaces it with dNTPs
* DNA ligase catalyses the formation of a phosphoerster bond between 3’ end of each new okazaki fragment and 5’ end of the growing daughter DNA strand

End replication problem
1. DNA polymerase is incapable to completely replicating all the way to the ends of lagging strand, leading to shortening of telomeres
2. when final RNA primer at end of lagging strand is removed, there is no upstream strand with 3’ OH group to which DNA polymerase can add nucleosides to fill resulting gap
3. a small section of extreme 3’ end of parental strand does not undergo DNA replication
4. daughter DNA strand resulting from lagging strand synthesis shortened with each round of synthesis
5. telomerase, a ribonuceloprotein consisting of RNA sequence template + protein component
* RNA sequence template has sequence 3’AAUCCC 5’, complementary to telomere repeat sequence 5’ TTAGGG 3’ and binds to 3’ overhang of parental strand
* RNA template acts as template for insertion of dNTP onto exisiting 3’ overhang strand of the telomeres, lengthening overhang
* telomerase extends 3’ overhang of parental DNA strand in 5’ to 3’ direction by adding sequence repeats of 5’TTAGGG 3’ via CBP
* protein component, Telomere Reverse Transcriptase(TERT) is a reverse transcriptase enzyme that provides catalytic action to synthesis DNA from a RNA template
* hence, synthesis of shorter DNA strand extended during next round of replication, resulting in longer telomere
* helps maintain the number of repeats at the telomeres

Question 21

oriR

Answer 21

Initiation: Formation of Origin of Replication
1. is a specific sequence of nucleotides that are A-T rich(only 2 H bonds, easier to distrupt the bonds as less energy needed to overcome them)
2. initiator proteins recognise and bind to specific oriR sequence
3. DNA double-helix separates(involves other proteins) into 2 strands to form replication ‘bubble’
4. Each end of replciation bubble(Y-shaped structure) called a replication fork
5. replication forks move away from oriR as replication proceeds bidirectionally

Question 22

Separation of DNA strands

Answer 22

1. need for continual separation of base pairs so that both DNA strands can act as templates
2. helicases bind to one strand of DNA molecule: use ATP as a source of energy to break the hydrogen bonds holding the 2 strands of DNA together -> unwinds the DNA double-helix -> separates parental DNA strands at region of replication fork
3. allows each of the 2 parental DNA strands to seve as the template for synthesis of a new DNA strand
4. Single-strand DNA-binding proteins (SSB proteins): temporarily stabilises the unwound single-stranded portion of the DNA double-helix by binding to it -> prevets ssDNA from re-annealing to reform the duplex -> keeps 2 parental strands in appropriae ss condition to act as a template + protects ssDNA which is very unstable from being degraded
5. topoisomerases:
   * unwinding causes supercoiling ahead of the replication fork, resulting in tension
   * cleaves a strand of the helix to creae a transient ss nick
   * relives strain on DNA molecule by allowing free roation around intact strand. and then reseals the broken strand

Question 23

Synthesis of daughter DNA strands via CBP

Answer 23

1. RNA primers with complementary base sequence to template provides a free 3’ OH end that DNA polymerase can extend, priming the synthesis of the daughter DNA strand
2. DNA polymerase with 5’ to 3’ exonuclease activity later replaces RNA nucleotides of primer with DNA versions
3. parental DNA strand separated at replication fork and primed with RNA primer serves as a template for semi-conservative DNA replication
4. DNA polymerase reads the template and assembles the deoxyribonuleoside triphosphates(dNTPS) for newly-synthesised daughter DNA strand based on CBP
5. when incorrect base pair recognised, proofreading carried out where DNA polymerase reverses its direction by one base pair of DNA -> 3’ to 5’ exonuclease activity of enzyme allows incorrect base pair to be excised

Question 24

Synthesis of strands - via phosphodiester bond formation

Answer 24

1. DNA polymerase catalyses polymerisation of strand
2. all DNA polymerase catalyses phosphodiester bond formation between a growing daughter DNA strand and an incoming nucleotide/dNTP
3. due to active site specificity of DNA polymerase, synthesis of both strands can only occur in 5’ to 3’ direction
4. formation of phosphoester bond between the 3’ hydroxyl group of the last nucleotide in the growing strand and the free 5’ phosphate group of the incoming dNTP
5. incoming dNTP loses a pyrophosphate group when they form phosphoester bond with growing daughter DNA strand
6. energy released from pyrophosphoester bond breakage is coupled to phosphoester bond formation

Question 25

Leading strand and lagging strand

Answer 25

7. Due to DNA polymerase only beingn able to add dNTPs to the free 3' end of a growing DNA strand and not 5' end, growing strand can only elongate in 5' to 3' direction 8. however, the 2 strand of a DNA double-helix are antiparallel and their sugar-phosphate backbone run in opposite directions 9. hence, continuous synthesis of both DNA strands at replication fork is not possible 10. hence, there is the leading strand and lagging strand synthesis 11. leading strand is the complementary DNA strand continously synthesised as a single polymer along template strand, polymerised in mandatory 5' to 3' manner towards replication fork 12. lagging strand is the complementary DNA strand that is discontinuously synthesised as a series of short fragments known as okazaki fragments -> each okazaki fragment polymerised in 5' to 3' manner against overall direction of replication fork 13. as such, each okazaki fragment requries an RNA primer for strand initiation 14. okazaki fragments ligated * DNA polymerase I removes RNA primer and replaces it with dNTPs * DNA ligase catalyses the formation of a phosphoerster bond between 3' end of each new okazaki fragment and 5' end of the growing daughter DNA strand

Question 26

Features and mechanism of oriR in prokaryotes vs in eukarotyes

Answer 26

Number of sites: * prokaryotes: 1 * eukaryotes: multiple origins of replication 1. multiple regions of chromosome undergo replication at the same time 2. multiple replication bubbles form and fuse, speeding up the copying of very long DNA molecule Nature of direction of replication: both bidirectional and bubbles expand laterally End of replication: * prokaryotes: ends at termination site * eukaryotes: replication bubbles fuse

Question 27

End replication problem

Answer 27

1. DNA polymerase is incapable to completely replicating all the way to the ends of lagging strand, leading to shortening of telomeres 2. when final RNA primer at end of lagging strand, the terminal okazaki fragment, is removed, there is no upstream strand with 3' OH group to which DNA polymerase can add nucleosides to fill resulting gap 3. a small section of extreme 3' end of parental strand does not undergo DNA replication 4. daughter DNA strand resulting from lagging strand syntheis shortened with each round of synthesis 5. telomerase, a ribonuceloprotein consisting of RNA sequence template + protein component * RNA sequence template has seuqence 3'AAUCCC 5', complementary to telomere repeat sequence 5' TTAGGG 3' and binds to 3' overhang of parental strand * RNA template acts as template for insertion of dNTP onto exisiting 3' overhang strand of the telomeres, lengthening overhang * telomerase extends 3' overhang of parental DNA strand in 5' to 3' direction by adding sequence repeats of 5'TTAGGG 3' via CBP * protein component, Telomere Reverse Transcriptase(TERT) is a reverse transcriptase enzyme that provides catalytic action to synthesis DNA from a RNA template * hence, synthesis of shorter DNA strand extended during next round of replication, resulting in longer telomere * helps maintain the number of repeats at the telomeres

Question 28

Synthesis + function of RNA primers

Answer 28

1. portion of parental DNA strand serves as template for making RNA primer with complementary base sequence 2. primase joins ribonucleotides to make the primer using energy from hydrolysis of ATP 3. RNA primer provides free 3' OH end that DNA polymerase can extend 4. DNA polymerase with 5' to 3' exonuclease activity replicates the RNA nucleotides of the primers with DNA versions

Question 29

# there's 2 Limitations of DNA polymerase

Answer 29

1. none of the DNA polymerases can initiate the synthesis of a DNA strand on its own, hence synthesis of a DNA strand requires presence of an RNA primer 2. due to activte site specificity of DNA polymerase, synthesis of both daughter DNA strands can only occur in one direction, 5'->3' direction

Question 30

Prokaryotic VS Eukaryotic genome

Answer 30

Organisation of genes: Eukaryotes: functionally related genes in human chromosomes not clustered together Prokaryotes: functional related genes clustered together as operons Type of genes: Eukaryotes: contains introns, telomeres, centromeres Prokaryotes: absent Organisation of DNA: Eukaryotes: wound around histone proteins to form nucleosomes Prokaryotes: wound around histone-like proteins Shape: Eukaryotes: linear Prokaryotes: circular