1 Chapter 16 DNA Replication, Repair, and Recombination: The Efficiency of Cellular Chemistry

2 Different Proteins Evolve At Different Rates (the number in parentheses indicates how many million years it has taken, on average, for one acceptable amino change to appear for every 100 amino acids that the protein contains.) Both E. coli and humans have similar mutation rates for one round of DNA replication, about 1 nucleotide change per 10 9 nucleotides each time that DNA is replicated.

3 Semi-conservative Replication DNA Polymerase: first DNA polymerizing enzyme (1957)

4

5 Movie: Adding of bases in 5’ to 3’ direction.

6 E. Coli DNA polymerase looks like a “right hand.”

7 Semiconservative Replication

8 Bacterial Chromosome The replication fork and its associated DNA polymerase is really a multi-enzyme complex of which DNA polymerase is a part.

9 An Incorrect Model: new bases are not just added “continuously” on both strands.

10 The Structure of the Replication Fork Movie: Replication of Leading and Lagging Strands

11

12

13 Sites for Polymerizing (P) and Editing (E): so DNA polymerase functions as a “self-correcting” enzyme.

14

15 It’s an RNA primer made by DNA primase!!!

16 Synthesis of DNA fragments on the lagging strand. Why an erasable RNA primer and not just a DNA primer??

17 The action of DNA ligase

18 DNA helicase has a prying apart rate of up to 1000 nucleotide pairs per second. DNA helicases NOT DNA helicase!!

19 The 6 subunits of the helicase hydrolyze ATP and thereby propels itself along a DNA single strand.

20 Hairpin Helices need to be “straightened.”

21 SSB protein of humans: DNA is in red

22 DNA polymerase would actually fall off the DNA after synthesizing a short string of nucleotides. So Okazaki fragments are fine; but the leading strand would have some trouble.

23

24 Bacterial DNA replication fork (a similar complex is present in eukaryotes)

25

26 Strand-Directed Mismatch Repair In E. Coli The proofreading exonuclease will miss some mismatched bases. This produces distortions in the DNA helix. So how does the repair system know which base to remove? The detection system could only be right 50% of the time. The proofreading mechanism must be able to distinguish between the mismatched nucleotide only on the newly synthesized strand where the error actually occurred. Methyl groups (CH 3 -) are added (methylation) onto A residues some time after the new A has been added. Proteins recognize the unmethylated GATC sequences near the replication fork. So recognition is complete. It is then excised and resynthesis occurs.

27 Strand-Directed Mismatch Repair In humans and hereditary nonpolyposis colon cancer HNPCC is caused by one defective copy of a mismatch repair gene. Mismatch recognition occurs NOT BY METHYLATION but the new strands ARE PREFERENTIALLY NICKED. So a mismatch repair protein triggers the breakdown of the DNA strand as far back as the nick

28

29 Take two intertwined ropes and pull one set of ends apart. What happens farther down the rope or at the other end? To unravel the DNA would require lots of energy. Types of Topoisomerases (I told you enzymes were important!!) Generally, topoisomerases break the phosphodiester bond and reseal it.

30 Topoisomerase I Single strand nick One strand can then rotate around the other to relieve the tension. Resealing is quick.

31

32 Topoisomerase II Acts where the two DNA strands cross over (note the positions of the red and gold strands) ATP causes two parts of the enzyme (two domains) to join (dimerize). A “double-strand” break is made in the helix.

33

34

35

36 Mammalian Replication Fork

37 Summary: DNA replication takes place at a ___-shaped structure called a ________________ fork. A self-correcting DNA polymerase enzyme catalyzes nucleotide polymerization in a ___________ direction, copying a DNA __________ strand with remarkable fidelity. Since the two strands of a DNA double helix are antiparallel, this 5’-to-3’ DNA synthesis can take place _______________ on only one of the strands at a replication fork (the _____________ strand). On the ___________ strand, short DNA fragments must be made by a “backstitching” process. Because the self- correcting DNA polymerase cannot start a new chain, these lagging- strand DNA fragments are primed by short _____ primer molecules that are subsequently erased and replaced with DNA. DNA replication requires the cooperation of many proteins. These include (1) DNA _____________ and DNA primase to catalyze nucleoside triphosphate polymerization; (2) _____ helicases and _________ - _________ DNA ______________ proteins to help in opening up the DNA helix so that it can be copied; (3) DNA _________ and an enzyme that degrades RNA primers to seal together the discontinuously synthesized lagging-strand DNA fragments; and (4) DNA __________________________ to help to relieve helical winding and DNA tangling problems. Many of these proteins associate with each other at a replication fork to form a highly efficient “replication machine,” through which the activities and spatial movements of the individual components are coordinated.

38 The Concept of the Replication Bubble Replication Origin Initiator proteins Each strand, having served as a template, separates with the new copy.

39 40 minutes for E. Coli to duplicate its DNA. 4.6 x 10 6 nucleotide pairs are duplicated at a rate of about nucleotides/sec. Bacterial genomes have a single origin or replication.

40 Proteins initiate bacterial genome replication.

41

42 Eukaryotic Replication Average linear human chromosome has about 150 million nucleotide pairs. At a rate of 50 bases/sec it would take about 800 hours to replicate the chromosome but we know the S phase normally takes about 8 hours. So...

43 Replication origins are activated in clusters called replication units. New replication units are activated at different times during the S phase. The structure of the chromatin determines the order of activation. Heterochromatin (highly condensed DNA) is replicated late “Housekeeping Genes” are genes active in all cells and these replicate very early in the S phase in all cells. Within these units, replication origins are spaced out at intervals of 30,000 – 300,000 nucleotide pairs. Replication forks form, form bubbles and move in opposite directions as they do in bacteria.

44

45 In search of Replication Origins. Each of the yeast replication origins contains a binding site for intiator proteins called the origin recognition complex (ORC) as well as a binding site for proteins that attract the ORC. How is eukaryotic replication triggered? We don’t know. How is it ensured that the replication origin is used only once in S phase? We don’t know.

46 Telomerase Replicates the Ends of Chromosomes When replicating a linear chromosome the lagging strand will have no place to put its RNA primer to begin the last Okazaki fragment Eukaryotes have a special sequence at the end of their DNA called telomeres. In humans the repeated sequence is GGGTTA and the telomere extends for about 10,000 nucleotides. So how do you get those ends replicated? Telomerase

47

48

49 Summary: The proteins the initiate DNA replication bind to DNA sequences at a replication ________ to catalyze the formation of a replication _________ with two outward-moving replication forks. The process begins when an initiator protein- DNA complex is formed that subsequently loads a DNA helicase onto the DNA template. Other proteins are then added to form the multienzyme “replication machine” that catalyzes DNA synthesis at each replication fork. In bacteria and some simple eukaryotes, origins are specified by specific DNA sequences that are only several hundred nucleotide pairs long. In other eukaryotes, such as humans, the sequences needed to specify an origin of DNA replication seem to be less well defined, and the origin can span several thousand nucleotide pairs. Bacteria typically have a _____ origin of replication in a ________ chromosome. With fork speeds up to 1000 nucleotides per second, they can replicate their genome in less than an hour. Eukaryotic DNA replication takes place in only one part of the cell cycle, the ___ phase. The replication fork in eukaryotes moves about 10 times more slowly than the bacterial replication fork, and the much longer eukaryotic chromosomes each require many replication origins to complete their replication in a typical 8-hour S phase. The different __________ origins in these eukaryotic chromosome are activated in a sequence, determined in part by the structure of the chromatin, with the mot condensed regions of chromatin beginning their replication ______. Eukaryotes solve the problem of replicating the ______ of their ______ chromosomes by a specialized end structure, the ____________, which requires a special enzyme, _________________. Telomerase extends the telomere DNA by using an _______ template that is an integral part of the enzyme itself, producing a highly repeated DNA sequence that typically extends for 10,000 nucleotide pairs or more at each chromosome end.

50 Cells invest a lot into DNA repair enzymes Several % of bacterial and yeast genome coding regions are involved in DNA repair functions. Increased rate of mutation follows the inactivation of a DNA repair gene. Variety of human diseases are linked to decreased DNA repair.

51

52 DNA undergoes major changes as a result of thermal fluctuations: 5000 purine bases (A and G) are lost every day due to depurination (loss of a purine) Spontaneous deaminations: C’s get deaminated and become uracils at a rate of 100 bases per cell per day. Environmental chemicals: can also damage DNA UV Radiation: will form thymine dimers (link up two thymines on one strand)

53

54 Thymine Dimers

55 How Chemical Modifications of Nucleotides Produce Mutations

56

57 “There’s More Than One Way to Skin a Cat” or DNA Damage Can be Removed by More Than One Pathway.

58

59 The Deamination of DNA Nucleotides.

60 Cytosines are commonly “methylated” because this is a means by which gene expression (transcription) is controlled. If a methylated cytosine is deaminated, a thymine results. This thymine is now H bonded to a guanine forming a mismatched base pair.

61 Double Strand Breaks

62 Summary: Genetic information can be stored stably in DNA sequences only because a large set of DNA _______ enzymes continuously scan the DNA and replace any damaged nucleotides. Most types of DNA repair depend on the presence of a separate copy of the genetic information in each of the two stands of the DNA double helix. An accidental lesion on one strand can therefore be cut out by a _______ enzyme and a corrected strand resynthesized by reference to the information in the ________ strand. Most of the damage to DNA bases is excised by one of two major DNA repair pathways. In _____ excision repair, the altered base is removed by a DNA glycosylase enzyme, followed by _________ of the resulting sugar phosphate. In _____________ excision repair, a small section of the DNA strand surrounding the damage is removed from the DNA double helix as an oligonucleotide. In both cases, a gap left in the DNA helix is filled in by the sequential action of DNA ____________ and DNA _____, using the undamaged DNA strand as the __________. Other critical repair systems-based on either nonhomologous or homologous end-joining mechanisms-reseal the accidental double-strand breaks that occur in the DNA helix. In most cells, an elevated level of DNA damage causes both an increased synthesis of repair enzymes and a delay in the cell cycle. Both factors help to ensure that DNA damage is repaired before a cell divides.