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Lecture11_NABPty2012 (1)

Lecture11_NABPty2012 (1) - Introduction to DNA and RNA...

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Unformatted text preview: Introduction to DNA and RNA structure Introduction Please view the “The DNA Story” on the streaming AV media reserves. 1. Go to http://tinyurl.com/uweres. (You will need to enter your UW Net ID.) 2. Search for and select BIOC 441 (Winter 2011 - Young). 3. Read the copyright statement. If you agree to the statement click “Accept”. 4. Select “Streaming AV Media Reserves”. 5. You will be prompted for a password. Use “DNA”. 6. Select and watch the video. O C HN H2N-C 5’ end C C 5’ HO-CH2 N N 3’ O OPO N NH2 C 1’ N 2’ O CH2 O 2’-deoxyribose sugars Phosphodiester linkages Directional chain (5’ to 3’) 4 Bases purines: adenine & guanine pyrimidines: cytosine & thymine Chargaff’s rule: [A] = [T], [G] = [C] DNA is a polymer of 2’-deoxyribonucleotides CH O 4’ G CH C CH N C GCTAp O O C HN O OPO O CH2 O C N CH3 C CH NH2 O C N O OPO HC C N CH2 O C O 3’ 3’ end T O-PO32 N A CH N O C HN H2N-C 5’ end C C 5’ HO-CH2 N N 3’ O OPO N NH2 C 1’ 2’ OH N O CH2 O ribose sugars Phosphodiester linkages RNA is a polymer of ribonucleotides CH O 4’ G CH C CH N C O O C HN O OPO OH CH C CH O CH2 O purines: adenine & guanine pyrimidines: cytosine & uracil N C N O U NH2 O Directional chain (5’ to 3’) 4 Bases GCUAp C N HC C N OH N OPO CH2 O CH O 3’ 3’ end A O-PO32 OH RNA is easily hydrolyzed under alkaline conditions . . . . . . . . O O . . . . O P O-CH2 O O O P O-CH2 O O O OPO O O O O OH OH CH 2 O RNA O P O-CH2 N O N P O N O OPO O O OH O P OH O O N O H2O H+ OH ... HOCH2 N O OPO O mixture of 2’- and 3’- monophosphate derivatives O OH ... shortened RNA The reaction proceeds through a 2’,3’-cyclic monophosphate intermediate. The Enzymatic hydrolysis of RNA by RNase proceeds through a similar intermediate. Because DNA lacks the 2’-OH group, it is stable under alkaline conditions. Why does DNA contain T rather than U? Cytosine deaminates non-enzymatically to form uracil. If this happens in DNA, it constitutes a mutation. Cells have a proofreading system that recognizes the error and replaces the U by C. 5-methyl cytosine occurs in DNA at a low but significant frequency (1/100-1/1000). When it is deaminated, it forms thymine, a normal constituent of DNA. This is a more serious alteration because there is now a T:G “base-pair”. O NH2 N O C C N CH H2O CH cytosine HN O C C N CH CH uracil Deamination of cytosine is of less consequence in RNA, because RNA is not the permanent repository of genetic information. The phosphate groups of DNA and RNA are negatively charged 5’ HO-CH2 N O O M+ OPO CH 2 O O N O A phosphodiester group has a pKa of about 1, and so will always be ionized and negatively charged under physiological conditions (pH ~7). Nucleic acids require counterions such as Mg2+, polyamines, histones or other proteins to balance this charge. OPO M+ CH 2 O N O O M+ OPO CH 2 O O N 3’ O-PO32 M+ OH HO-CH2 H-C=O O 5’ HO-CH2 H-C-OH The sugars are always in the β -furanose (cyclic) form N O H-C-OH OH H-C-OH OH β -furanose (cyclic) form CH2OH ribose in its aldehyde form O OPO OH CH2 O 5’ O O 4’ N OH OPO N endo O 1’ 3’ CH2 C-2’ exo N O O OPO OH CH2 O The ring can adopt various puckered conformations in which C-2’ and C-3’are in either exo or endo positions relative to the base and C-5’. O N 3’ O-PO32 OH NH2 The nucleotide base can rotate with respect to the sugar C N C HC C N HOCH2 N N CH HC N N HOCH2 O OH The bases can adopt either syn or anti conformations, but anti conformations are preferred. NH2 syn-Adenosine anti-Adenosine NH2 N O HOCH2 OH CH HC C CH HC N HOCH2 O OH syn-Cytosine N OH NH2 C C O OH OH C C OH C N N O OH anti-Cytosine C O N CH The pattern of X-ray diffraction by DNA fibers reveals a helical structure with steps of 3.4 and 34 Å This This x-ray diffraction by calf thymus DNA was measured by Franklin & Gosling in 1952. The X pattern is indicative of a helix with a pitch of 34 Å per turn. The strong spots at the top and bottom reveal internal steps of 3.4 Å. http://osulibrary.oregonstate.edu/specialcollections/coll/pauling/dna/pictures/franklin-typeBphoto.html H-bonds between Watson-Crick base pairs provide specificity in the double helix but most of the stability comes from base stacking interactions H O N C C N NH N NH N A C O N N CH3 CC C C C N G and C form three H-bonds O H C C C-H C NH N H N C H H H C C C G HN C C H O A forms two Hbonds with T or U C-H HN N T Base-pairing explains Chargaff’s rules for the base composition of DNA: A = T; G = C “It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.” -- J.D. Watson & F.H.C. Crick, Nature 171: 737 (1953) The B-form DNA helix has a diameter of about 20 Å ~20 Å Base Base pairs fill the center of the helix; the phosphates are on the outside. A base pair is more exposed to the solvent on one side (the “major groove”, at the top in these views) than the other (the “minor groove”, bottom). B-form DNA consists of a right-handed double helix with antiparallel strands 3.4 Å per base-pair 5’ 3’ minor major groove 34 Å (10 bp) per turn major groove minor groove 3’ 5’ These dimensions are for DNA fibers. In solution, there are ~10.5 base-pairs per turn. To find whether the two strands are parallel or antiparallel, Kornberg synthesized DNA enzymatically from labeled nucleoside triphosphates 1. Synthesize DNA from a mixture of ATP, GTP, CTP & TTP in which one of the four nucleotides is labeled with 32P in the α position (e.g., pppA). 2. Hydrolyze the DNA with a DNase that releases the 3’-mononucleotides. 3. Measure the amount of 32P transferred to various nucleotides. pppA + pppT + pppG + pppC DNA polymerase 5’ ... pGpApTpCpApGp ... 3’ 3’ ... pCpTpApGpTpCp ... 5’ ... Gp + Ap + Tp + Cp + Ap + Gp ... H2O DNase ... Cp + Tp + Ap + Gp + Tp + Cp ... 4. Repeat the experiment starting with the label in other nucleotides. 4. If the strands are antiparallel: A C = G T and A G = C where the arrows indicate the transfer pf the radioactive phospate/ What if the chains are parallel? High-resolution structures of small DNA molecules have been obtained by x-ray crystallography and NMR NMR NMR structure of a duplex DNA dodecamer in the DNA-binding site of an interferon promotor. J. R. Huth et al., Nature Struct. Biol. 4: 657 (1997). 2eze.pdb The two strands of the double helix separate reversibly at high temperatures If the temperature is lowered, the strands recombine. The rate of recombination is inversely proportional to the complexity of the DNA. 100 80 % Denatured The temperature at which this “denaturation” or “melting” occurs (tm) depends on the pH and salt concentration, and increases approximately linearly with the GC content of the DNA. 60 40 20 0 70 40 40 50 60% GC 30 50 60 70% GC 80 90 100 o Temperature / C 110 Double-stranded and single-stranded DNA differ in their optical absorption at 260 nm dA dG dU dC The conjugated π-electron systems of the purine & pyrimidine bases absorb strongly in the UV. nucleotides ssDNA dsDNA The The absorbance of double-stranded DNA (dsDNA) at 260 nm is less than that of either single-stranded DNA (ssDNA) or the free bases. This is called “hypochromism.” Hypochromism results from dipole-dipole interactions between neighboring bases The excited states of an interacting pair of molecules can be described as linear The combinations of the excited states of the individual molecules. In certain geometries, some of the absorption strength in the near-UV moves to bands at higher energies. increasing energy empty orbitals strong absorption band light filled orbitals individual base strong band near 260 nm interacting bases weaker band near 260 nm individual base strong band near 260 nm Summary of the main structural features of B-form DNA •Right-handed helix •Two antiparallel strands held together by Watson-Crick hydrogen bonds •Pitch (repeat length) = 34 Å (3.4 nm) •36o rotation between residues •Helix diameter of 20 Å (2.0 nm) •Wide major groove, narrow minor groove •Chargaff’s Rules: A = T; G = C •Charged phosphates •Bases in anti configuration •The strands separate at high temperatures •The solution structure is dynamic Tautomeric forms of G & T can cause mutations due to mis-pairing during DNA replication O OH C HN C H2N-C C N C N CH N Guanine (keto form) N C H2N-C C N N CH N Guanine (enol form) xx O C HN O C N OH C C CH3 CH Thymine (keto form) N O C C CH3 CH N Thymine (enol form) The ring NH atoms of G and T have pKa values of about 9. At physiological pH, about 99% of the base is in the keto form and 1% in the enol form. Palindromic* sequences (inverted repeats) in DNA or RNA can form hairpin or cruciform structures 5’ TGCGATACTCATCGCA 3’ 3’ ACGCTATGAGTAGCGT 5’ inverted repeats in an antiparallel double helix C T A 5’ A T C G C A T A C T A G C G T A T C G C A 5’ hairpin 3’ Mirror repeats cannot form such structures. 3’ 3’ C T A G C G T C 5’ cruciform structure A C G C T A T G C G A T T G G A *A palindrome reads the same in either direction (“Radar,” “Madam, I’m Adam”). DNA base sequences can be determined by using DNA polymerase and dideoxynucleotides Synthesize DNA enzymatically in the presence of a small amount of a labeled dideoxynucleotide (e.g., ddTTP) and larger amounts of ordinary ATP, GTP, CTP and TTP, using a primer and one strand of the DNA to be sequenced as a template. DNA polymerase synthesizes the new strand in the 5’ → 3’ direction. The dideoxynucleotide interrupts DNA synthesis wherever it is incorporated. This happens at random points where the corresponding base occurs in the sequence. In this illustration, each DNA fragment ends in ddT, which is labeled with a red dye. O O O O P O P O P O-CH2 5’ primer 3’ synthesis GCCGCATGTCGTAACTT CGGCGTACAGCATTGAA 3’ template 5’ O NH O O O ddTTP (continued …) N O O DNA sequencing can be automated Repeat the procedure with ddC, ddA and ddG, using a different fluorescent dye to tag the dideoxynucleotide in each case. Then mix all the DNA fragments and separate them according to size by electrophoresis. Electrophoresis of the DNA fragments gives a “ladder” of bands, each ending in a particular dideoxynucleotide at its 3’ end and tagged with the corresponding dye. This procedure has been automated in machines that can sequence more than 3x106 bases in a month. GC C G C A T G T C GT A A C TT 5’ primer 3’ synthesis GCCGCAT GT C GT AACT T CGGCGT ACAGCAT T GAA 5’ template 3’ Alternative Sequencing Paradigms Massively parallel sequencing Nanopore sequencing Sequencing by hybridization Real-time sequencingby-synthesis What can a sequencer do today? In 8 days… (Illumina HiSeq 2000) “Paired end” 100 bp reads >2 billion read-pairs >400 gigabases (Gb) of total output 1 in 1,000 error rate for most base-calls Initial draft of human genome based on 23 Gb Next Generation Biology Erez Lieberman-Aiden, Lauren Solomon The human (and other) genomeswhat have we learned? Number of human genes is fewer than expected (but alternative splicing increases that number). Presesnt estimate is ~30,000. Gene arrangement is often conserved between species (synteny). Polymorphisms, differences between individuals in the same species, are very frequent (~1 change per 1000 nucleotides). The number of gene families is similar in flies and humans. Most human genes have orthologs in other species-even down to yeast. This fact is important for studying gene function. There are many paralogs: related genes in the same genome due to gene duplication. (Homologs are genes related by descent from a common ancestor, an example of divergent evolution). 27 www.wellcome.ac.uk/en/genome/genesandbody ...
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