2. DNA Replication and Repair

3. Learning Objectives Student will be able to Discuss DNA Replication
Explain DNA Repair
Solve clinical problem

4. DNA Replication
The process of making an identical new DNA copy of a duplex (double-stranded) DNA, using existing DNA as a template.
In humans and other eukaryotes, replication occurs in the cell nucleus.   

5. DNA Replication
The genetic information found in DNA is copied and transmitted to daughter cells through DNA replication.

6. Do you have any Idea about process of DNA Replication ?

7. The flow of information from DNA to RNA to
protein is termed as “central dogma” of
molecular biology

8. Secret behind Central Dogma
The Central Dogma. This states that once ‘information’ has passed into protein it cannot get out again. In more detail, the transfer of information from nucleic acid to nucleic acid, or from nucleic acid to protein may be possible, but transfer from protein to protein, or from protein to nucleic acid is impossible. Information means here the precise determination of sequence, either of bases in the nucleic acid or of amino acid residues in the protein.

10. DNA Replication
DNA replication takes place by separation of the strands of the double helix, and synthesis of two daughter strands complementary to the two parental templates.

11. DNA Replication
DNA replication is called semiconservative because half of the parent structure is retained in each of the daughter duplexes.

12. Separation of the two DNA Strands
For the initiation of Replication Process
The two strands of the parental double stranded DNA dsDNA must first separate (or melt) over a small region.
Because polymerase use only ssDNA as a template.

13. In prokaryotes
DNA replication begins at a single unique nucleotide sequence, a site called the origin of the replication , or ori.
The ori includes short , AT-rich segments that facilitate melting.

14. In eukaryotes
Replication begins at multiple sites along the DNA helix. Having multiple origins of replication
Which provides the mechanism for rapidly replicating the great length of eukaryotic DNA molecule.

16. Proteins Required for DNA strand Separation
Initiation of replication requires the recognition of the origin by a group of the protein that forms a prepriming complex.
These proteins are
DNA A protein
DNA helicases
Single stranded DNA –binding protein

17. DNA A Protein
This protein binds to specific nucleotide sequences (DnaA boxes) within the origin of replication, causing tandamly arranged (one after the other) AT-rich regions in the origin to melt.
Melting is adenosine triphosphate (ATP) dependent and results in strand separation with the formation of localized region of ssDNA.

18. DNA helicases
These enzymes bind to ssDNA near the replication fork and then move into the neighboring double stranded region, forcing the strands apart (in effect , unwinding the double helix).
Helicases require energy provided by ATP.
Unwinding at the replication fork causes supercoiling in other regions of the DNA molecule.

20. Single stranded DNA-binding protein
This protein binds to the ssDNA generated by helicases.
The SSB protein are not enzymes, but rather serve to shift the equilibrium between dsDNA and ssDNA in the direction of a single stranded forms.
These proteins not only keep the strands of DNA separated in the area of the replication origin, but also protects the Dna from nucleases that degrade ssDNA

21. Supercoiling
As the two strands of double helix are separated a problem is encountered namely, the appearance of positive supercoils in the region of DNA ahead of the replication fork as a result of over winding.
And negative supercoils in the region behind the fork.
The accumulating positive supercoils interfere with further unwinding of the double helix.

23. To solve this problem there is a group of enzymes called Dna topoisomerases responsible for removing supercoils in the helix by transiently cleaving one or both of the DNA strand.

24. Type I DNA topoisomerases
These enzyme reversibly cleaves one strand of the double helix.
They have both strand cutting and strand resealing activities.
They do not require ATP but store energy from phosphodiester bond they cleave.
Reuse the energy to reseal the strand

25. Each time a transient “nick” is
created in one DNA strand, the intact DNA strand is passed
through the break before it is resealed, thus relieving (“relaxing”)
accumulated supercoils.

26. Type II DNA topoisomerases
These enzymes bind tightly to the DNA double helix and make transient breaks in both strands.
The enzyme then causes a second stretch of the DNA double helix to pass through the break and, finally, reseals the break (Figure 29.13). As a result, both negative and positive supercoils can be relieved by this ATP-requiring process.

28. Direction of DNA replication
The DNA polymerases responsible for copying the DNA templates are only able to “read” the parental nucleotide sequences in the 3'→5' direction, and they synthesize the new DNA strands only in the 5'→3' (antiparallel) direction.
Therefore, beginning with one parental double helix, the two newly synthesized stretches of nucleotide chains must grow in opposite directions—
one in the 5'→3' direction toward the replication fork and one in the 5'→3‘ direction away from the replication fork.

29. Leading strand: The strand that is being copied in the direction of the advancing replication fork is called the leading strand and is synthesized continuously.
Lagging strand: The strand that is being copied in the direction away from the replication fork is synthesized discontinuously, with small fragments of DNA being copied near the replication fork.
These short stretches of discontinuous DNA, termed Okazaki fragments, are eventually joined (ligated) to become a single, continuous strand. The new strand of DNA produced by this mechanism is termed the lagging strand

31. RNA primer
DNA polymerases cannot initiate synthesis of a complementary strand of DNA on a totally single-stranded template. Rather, they require an RNA primer—that is, a short, double-stranded region consisting of RNA base-paired to the DNA template, with a free hydroxyl group on the 3'-end of the RNA strand (Figure 29.15).
This hydroxyl group serves as the first acceptor of a deoxynucleotide by action of DNA polymerase.

33. Primase
A specific RNA polymerase, called primase (DnaG), synthesizes the short stretches of RNA (approximately ten nucleotides long) that are complementary and antiparallel to the DNA template. In the resulting hybrid duplex, the U in RNA pairs with A in DNA. As shown in Figure, these short RNA sequences are constantly being synthesized at the replication fork on the lagging strand, but only one RNA sequence at the origin of replication is required on the leading strand.

34. The substrates for this process are 5'-ribonucleoside triphosphates, and pyrophosphate is released as each ribonucleoside monophosphate is added through formation of a 3'→5‘ phosphodiester bond.
[Note:The RNA primer is later removed]

35. Primosome
The addition of primase converts the prepriming complex of proteins required for DNA strand separation to a primosome. The primosome makes the RNA primer required for leading strand synthesis, and initiates Okazaki fragment formation in lagging strand synthesis. As with DNA synthesis, the direction of synthesis of the primer is 5'→3'.

36. Chain elongation
Prokaryotic (and eukaryotic) DNA polymerases elongate a new DNA strand by adding deoxyribonucleotides, one at a time, to the 3'- end of the growing chain.
The sequence of nucleotides that are added is dictated by the base sequence of the template strand with which the incoming nucleotides are paired.

37. DNA polymerase III DNA chain elongation is catalyzed by DNA
polymerase III. Using the 3'-hydroxyl group of the RNA primer as the acceptor of the first deoxyribonucleotide, DNA polymerase III begins to add nucleotides along the single-stranded template that specifies the sequence of bases in the newly synthesized chain.

39. DNA polymerase III is a highly processive” enzyme—that is, it remains bound to the template strand as it moves along, and does not diffuse away and then rebind before adding each new nucleotide. The processivity of DNA polymerase III is the result of its β subunit forming a ring that encircles and moves along the
template strand of the DNA, thus serving as a sliding DNA clamp.

40. The new strand grows in the 5 – 3 direction , antiparallel to the parental strand.
All four substrates deoxyadenosine triphosphate, deoxythymidine triphosphate , deoxycytidine triphosphate and deoxyguanosinetriphosphate must be present for DNA elongation to occur. If any one in short supply DNA synthesis will stop.

41. DNA Replication Exonuclease Activities of DNA Polymerases
DNA polymerase I is involved in DNA repair and also removes RNA primers and replaces them with DNA.
Exonucleases degrade nucleic acids by removing 5’ or 3’ terminal nucleotides.

42. The exonuclease activities of DNA polymerase I

43. DNA Replication (18) Initiation of Replication in Eukaryotic Cells
Eukaryotes replicate their genome in small portions (replicons).
Initiation of DNA synthesis in a replicon is regulated.

44. DNA Replication The Eukaryotic Replication Fork
Replication activities are similar in eukaryotes and prokaryotes.
There are several DNA polymerases in eukaryotes.
Eukaryotic DNA polymerases elongate in the 5’-to-3’ direction and require a primer; some have 3’-to-5’ exonuclease activity.

45. Some Proteins Required for Eukaryotic DNA Replication

47. DNA Repair DNA repair is essential for cell survival.
DNA is the cell molecule most susceptible to environmental damage.
Ionizing radiation, common chemicals, UV radiation and thermal energy create spontaneous alteration (lesions) in DNA.
Cells have a number of mechanisms to repair genetic damage.

48. A pyrimidine dimer that has formed within a DNA duplex following UV irradiation

49. Nucleotide excision repair (NER) removes bulky lesions, such as pyrimidine dimers and chemically altered nucleotides.
It consists of two pathways:
A transcription-coupled pathway which is the preferential pathway and selectively repairs genes of greatest importance to the cell.
A global genomic pathway which is less efficient and corrects DNA strands in the remainder of the genome.

50. DNA Repair Nucleotide excision repair (continued)
TFIIH is a key component of the repair machinery and is also involved in the initiation for transcription. It links transcription and DNA repair.
A pair of endonucleases cut on both sides of the lesion, and the damaged strand is removed by helicase.
The gap is filled by a DNA polymerase and sealed by DNA ligase.

51. DNA Repair Base Excision Repair
Base excision repair (BER) removes altered nucleotides that produce distortions of the double helix.
DNA glycosylase recognizes the alteration and cleaves the base form the sugar.
DNA glycosylases are specific for a particular type of altered base.

52. Base excision repair DNA glycosylase removes the altered bases.
Once the altered base is removed, an endonuclease cleaves the DNA backbone and a polymerase fills the gap by inserting a nucleotide complementary to the undamaged strand.
The strand is sealed by DNA ligase

53. DNA Repair
Mismatch repair (MMR) is the correction of mistakes that escape the DNA polymerase proofreading activity.
Repair enzymes recognize distortions caused by mismatched bases.
In bacteria, the parental strands are recognized from daughter strands by the presence of methylated bases.
Several MMR pathways have been identified in eukaryotes.

54. DNA Repair Double-Strand Breakage Repair
Ionizing radiation (X-rays, gamma rays) along with some chemicals cause double-strand breaks (DSBs).
DSBs can be repaired by a pathway in mammalian cells called nonhomologous end joining (NHEJ) in which proteins bind to the broken ends and catalyze reaction to rejoin the broken ends.

55. Repairing DSBs by NHEJ

56. DNA Repair Double-strand breakage repair (continued)
Cells that lack one of the proteins required for NHEJ are very sensitive to ionizing radiation.
Another DSB repair pathway is homologous recombination, and requires a homologous chromosome to serve as a template for repair of the broken strand.
Defects in both repair pathways have been linked to increased cancer susceptibility.

57. The Human Perspective: The Consequences of DNA Repair Deficiencies (1)
Xeroderma pigmentosum (XP) patients cannot repair sun-damaged DNA.
Some help for XP patients may become available in the form of skin creams that contain DNA repair enzymes.

58. Xeroderma pigmentosum

59. The Human Perspective: The Consequences of DNA Repair Deficiencies (2)
Skin cells with optimal levels of repair enzymes are subject to lesions that fail to be excised and repaired.
Skin cancer is not the only disease promoted by deficiency or overworked DNA repair systems.
Some colon cancer cases are due to mutations in mismatch repair genes.

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64. END

65. PRACTICE QUESTIONS

66. Q1. In replication, the template is read in the direction of:- a
Q1. In replication, the template is read in the direction of:- a. 3’→5’ b. 5’→3’ c. 2’→3’ d. 3’→2’

67. Key: A
The DNA template is read in 3′ to 5′ direction whereas a new strand is synthesized in the 5′ to 3′ direction—this is often confused.

68. Q2. In a newly synthesized DNA strand, the Primer is removed by:- a
Q2. In a newly synthesized DNA strand, the Primer is removed by:- a. DNA polymerase I b. DNA polymerase II c. DNA polymerase III d. DNA ligase

69. Key: A a. DNA polymerase I: it removes the primer
b. DNA polymerase II: It catalyzes the transcription of DNA to synthesize precursors of mRNA.
c. DNA polymerase III: it is the main polymerizing enzyme in prokaryotes.
d. DNA ligase is an enzyme that repairs irregularities or breaks in the backbone of double-stranded DNA molecules

70. Q3.The section of DNA that codes for a protein is: a. Exon b. Intron c. Operon d. Regulatory sequence

71. Key: A
a. Exon is a coding region. b. Intron is a non-coding region c. Operon is a functioning unit of DNA containing a cluster of genes under the control of a single promoter. d. Regulatory sequence is a is capable of increasing or decreasing the expression of specific genes

72. Q4 . The following enzyme unwinds the DNA preceding the replication fork:- a. Helicase b. DNA Polymerase 1 c. Gyrase d. Ligase

73. Key: A a. Helicase: unwinds the DNA
b. DNA Polymerase 1: it removes the primer
c. Gyrase: The enzyme causes negative supercoiling of the DNA or relaxes positive supercoils.
d. DNA ligase is an enzyme that repairs irregularities or breaks in the backbone of double-stranded DNA molecules

74. 5. Which of the following type of bonds lead to base pairing in DNA. a
5. Which of the following type of bonds lead to base pairing in DNA? a. Hydrogen bonds between A = G & T ≡ C b. Hydrogen bonds between A ≡ T & G = C c. Covalent bonds between A = T & G ≡ C d. Hydrogen bonds between A = T & G ≡ C

75. Key: D
Adenine is always form bond with thymine through two hydrogen bond and guinine always form bond with cytosine with three hydrogen bond.