7.05 Recitation Schedule

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1 7.05 Spring 004 February 13, Recitation Schedule Contact Information TA: Victor Sai Recitation: Friday, 3-4pm, ffice ours: Friday, 4-5pm, -13 Spring 004 Calendar Sun Monday Tuesday Wednesday Thursday Friday Sat Feb Rec * Rec0 March 1 Rec Review 1 5 * 8 Exam Rec Rec05 3 Spring 4 Break 5 Week 6 April 9 30 Rec06 31 Review 1 Exam * Rec Rec08 19 Review Exam 3 3 Rec Rec10 May Rec11 * ormal Friday recitation canceled Recitation/Exam Date Lectures Covered Recitation #1 Friday, February 13 1,, 3 Recitation # Friday, February 7 4, 5, 6 Recitation #3 Tuesday, March, 5pm, -13 6, 7, 8 Exam 1 Review Thursday, March 4, 7pm, Exam 1 Monday, March 8, 9:30-11am,Walker 1-8 Recitation #4 Friday, March 1 9, 10 Recitation #5 Friday, March 19 11, 1, 13, 14 Recitation #6 Tuesday, March 30, 5pm, , 16, 17, 18 Exam Review Wednesday, March 31, 7pm, Exam Friday, April, 9:30-11am, Walker 9-18 Recitation #7 Friday, April 9 19, 0, 1, Recitation #8 Friday, April 16 3, 4, 5, 6 Exam 3 Review Monday, April 19, 7pm, Exam 3 Wednesday, April 1, 9:30-11am, Walker 19-6 Recitation #9 Friday, April 3 7, 8 Recitation #10 Friday, April 30 9, 30, 31 Recitation #11 Friday, May 7 3, 33, 34, 35, 36 Exam 4 Review TBA 7-36 Exam 4 TBA 7-36 Recitation Schedule age 1 of 1

2 7.05 Spring 004 February 13, 004 Recitation #1 Contact Information TA: Victor Sai Recitation: Friday, 3-4pm, ffice ours: Friday, 4-5pm, -13 Unit 1 Schedule Recitation/Exam Date Lectures covered Recitation #1 Friday, February 13 1,, 3 Recitation # Friday, February 7 4, 5 Recitation #3 Tuesday, March, 5pm, -13 6, 7, 8 Exam 1 Review Thursday, March 4, 7pm, Exam 1 Monday, March 8, 9:30-11am,Walker 1-8 Recitation verview Topic roblems 1. Structure and Constituents of DA and RA 1, 4. DA Replication, 3, 4 3. Transcription 4 roblems 1. (00 Exam 1 Question, 15 points) (a) Draw the chemical structure of -deoxyguanosine 3 -monophosphate. (b) n the drawing in part (a), circle the atoms that form hydrogen bonds when guanosine is part of a Watson-Crick helix. (c) n the drawing in part (a), box the atoms of guanosine that could form hydrogen bonds with residues of a transcription regulatory protein that binds the major groove of DA. Recitation 1 age 1 of 3

3 7.05 Spring 004 February 13, 004. (1996 Exam 1 Question 6, 0 points) rotein engineering refers to an area of structural biology in which scientists try to alter the properties of proteins. Suppose you were able to engineer a DA polymerase that contained a modified 3 5 exonuclease activity with the following property: the exonuclease activity required that the 3 -nucleotide was part of a double helix. What would you expect the properties of this engineered enzyme to be, in terms of fidelity and overall speed of replication, as compared to the otherwise identical DA polymerase that completely lacks 3 5 exonuclease activity? Explain. 3. (00 Exam 1 Question 6, 15 points) Consider an E. coli cell that has just acquired a mutation that abolishes the 3 -exonuclease activity of DA pol I. (a) (5 pts) When this cell divides, the two daughter cells will acquire additional mutations. Explain why. (b) (10 pts) ne of the daughter cells will have mutations in one half of its chromosome, the other daughter cell will have mutations in the opposite half of its chromosome. Explain this. Include diagrams of the daughter chromosomes in your explanation. Recitation 1 age of 3

4 7.05 Spring 004 February 13, (001 Exam 1 Question 3, 0 points) 3 -deoxyadenosine 5 -triphosphate (3 -dat), is the triphosphate form of the antibiotic nucleoside, cordycepin. It is a potent inhibitor of RA synthesis. (a) (4 pts) Draw the chemical structure of 3 -dat (b) (4 pts) n the drawing in part (a), circle the atoms that would hydrogen bond with the template strand when 3 -dat is used as a substrate during RA synthesis. (c) (4 pts) Explain why the incorporation of cordycepin (3 -deoxyadenosine) inhibits RA synthesis. (d) (4 pts) 3 -dat does not inhibit DA synthesis by DA ol I or DA ol III. Explain why these enzymes would not be expected to use 3 -dat, whereas E. coli RA polymerase does use the inhibitor. (e) (4 pts) Although 3 -dat does not inhibit DA synthesis by DA ol I or DA ol III, this nucleotide is expected to directly inhibit the process of DA replication in E. coli. Explain why. Recitation 1 age 3 of 3

5 Structure and Constituents of DA and RA Victor Sai 7.05 Spring 004 Adenine (A) Guanine (G) BASES urines 1 * * denotes -bond donor * denotes -bond acceptor Cytosine (C) Attachment site for sugar Thymine (T) Uracil (U) (In DA nly) (In RA nly) yrimidines * 3 * * C Attachment site for sugar Ribose (In RA only) Deoxyribose (In DA only) SUGARS In nucleotides: binds to phosphate groups In RA: forms phosphodiester bonds In nucleotides: binds to phosphate groups In DA: forms phosphodiester bonds 5' 5' C C 4' 1' 4' 1' 3' ' Attachment site for base 3' ' Attachment site for base Forms phosphodiester bond age 1 of 4

6 SATES Attaches to 5' carbon of ribose or deoxyribose - - R - R R In ucleotides Monophosphate Diphosphate Triphosphate R In DA and RA Attaches to 3' carbon of ribose or deoxyribose - Attaches to 5' carbon of ribose or deoxyribose R UCLETIDES hosphates Base utting them all together... Sugar Example: dat deoxyribose triphosphate ' 5' 1' adenosine 3' ' Example: CM ribose by default monophosphate - - 4' 5' 1' cytidine 3' ' age of 4

7 DA Structure Example: 5'-A-C-G-T-3' 3'-T-G-C-A-5' 3' end * right-handed * has major and minor grooves C C adenine - C cytosine - C guanine - C C 3 thymine 5' end C thymine C guanine - C cytosine adenine - - C - 5' end - 3' end age 3 of 4

8 RA Structure Example: 5'-A-C-G-U-3' Differences from DA Structure: 1) ' (ribose sugar) ) Uracil in place of Thymine 3) Single-stranded (no base pairing) - 5' end - C adenine - C cytosine - C guanine - C uracil 3' end age 4 of 4

9 DA Replication Victor Sai 7.05 Spring Enzymes needed elicase: unwinds double helix DA gyrase (DA topoisomerase II): introduces negative supercoils to compensate for the positive supercoiling caused by helicase unwinding the double helix SSB (single-stranded binding proteins): stabilize single stranded regions rimase (RA olymerase): synthesizes short RA primers that allow DA olymerase to begin replication DA olymerase III holoenzyme: synthesizes DA DA olymerase I: removes RA primers and fills in gaps DA ligase: connects the ends of DA chains; catalyzes formation of a phosphodiester bond between 3 - group at one end and 5 -phosphate group at other end DA topoisomerase I: removes negative supercoils. rder of Events 1) DA strands are unwound by helicase and replication forks form. ) As DA is unwound, DA gyrase introduces negative supercoils to balance the positive supercoils generated from unwinding 3) SSB binds to single-stranded regions and stabilizes them. 4) Replication begins at the origin site, where primase synthesizes a short RA primer and DA olymerase III then synthesizes the DA. nly one primer is needed on the leading strand, but many primers are continuously synthesized on the lagging strand. 5) Later DA olymerase I erases the RA primers and fills in gaps with DA. 6) DA ligase then seals the disconnected DA chains. DA olymerases General roperties: Requires a primer with a free 3 - Requires a template (DA-dependent DA polymerase) Requires dgt, dat, dtt, dct, and Mg + ewly added nucleotide must base pair to template Synthesizes in 5 3 direction. Reaction: (dm) n + dt (dm) n+1 + i Comparison of prokaryotic DA olymerases: olymerase Type Exonucleases o. of subunits rocessivity Function I Low to moderate rimer removal, gap fill-in II Moderate DA repair III a. Core b. oloenzyme 3 5 a. 3 b. 10 a. moderate b. high Chromosome replication DA olymerase I: a single polypeptide of 103 kd, which can be cleaved into two regions by proteolytic treatment. The larger cleavage product (68 kd) is called the Klenow fragment, which contains the polymerase and the 3 5 exonuclease activities. The smaller fragment (35 kd) possesses a 5 3 exonuclease activity and excises small groups of nucleotides at ~10 bases at a time. Activities: 1) 5 3 polymerase: prefers 5 -nucleotide IS complementary to template ) 3 5 exonuclease: prefers 3 -nucleotide IS T complementary to template 3) 5 3 exonuclease: prefers 5 -nucleotide IS paired age 1 of

10 Transcription 1. Initiation RA olymerase recognizes a promoter as the start location. Sigma factors (σ factors) are proteins in the holoenzyme that specify which types of promoters RA polymerase should look for. RA polymerase must have a sigma factor to identify a promoter. RA olymerase will randomly bind to DA and slide along it until it finds a promoter. At this point it will separate the two strands to form a transcription bubble and begin synthesizing RA.\. Elongation Ts for mra production are selected by their complementarity to the template strand. RA olymerase is extremely processive and will generally fall off if a specific signal is given. owever, no proofreading takes place (error rate: ). 3. Termination Two ways to terminate transcription in E. coli: 1) Factor Independent: Transcription terminates when it reaches the termination signal: a sequence rich in Gs and Cs that can form a hairpin loop by complementary base-pairing. The stable hairpin is followed by a sequence of four or more U residues. Unstable U-A base pairs cause the nascent RA to dissociate from the DA template and then from RA polymerase. ) Factor Dependent: Transcription terminates with the help of Rho (ρ): a protein that recognizes a certain sequence on the newly made RA. Its ATase activity acquires the energy to move towards the transcription site and eventually reaches and pulls the RA from the DA template. RA olymerases General roperties: Does T require a primer Requires a template (DA-dependent RA polymerase) Requires GT, AT, UT, CT, and Mg + ewly added nucleotide must base pair to template Does not have an exonuclease fragment, therefore no proofreading occurs Synthesizes in 5 3 direction. Reaction: (M) n + T (M) n+1 + i E. coli RA olymerase: Core enzyme: 4 subunits (α ββ ) oloenzyme: Core + σ factor (specificity factor) Comparison of eukaryotic RA olymerases: RA olymerase Type % total RA Type of RA synthesized I 80% Large ribosomal RAs II 1-5% mra III 15-0% Small structural RAs (transfer RA) Comparison of Eukaryotic and Bacterial RA: The 5 end of eukaryotic RA is capped and the 3 end is poly(a) tailed after the eukaryotic gene is transcribed. Eukaryotic RA also contains introns that are later removed. age of