midterm Flashcards
HW 3 and 4 (42 cards)
(3.1) Franklin’s experimental data (for A form DNA) indicated that the DNA backbone has dyad symmetry with regard to an axis perpendicular to the long axis of the DNA fiber.
A) Why is this a surprising finding?
B)Explain how this finding refuted a proposal for the DNA by Corey and Pauling.
C)How did Watson and Crick explain dyad symmetry?
Single polynucleotide strands lack dyad symmetry. The dyad symmetry,
therefore, rules out any structure with parallel polynucleotide strands, including a
single strand, and also any structure with an uneven number of strands (Corey and
Pauling’s structure), since at least two strands must have the same orientation.
However, two antiparallel strands - Watson and Crick’s solution - exhibit the
required symmetry.
(3.2)
A)How did Watson and Crick explain the association of two polynucleotide strands?
B)What was the logical implication of their solution that exposed it to experimental testing by other means than solving the atomic structure?
Watson and Crick explained the association of two strands by virtue of
complementary base pairing: adenine with thymine and guanine with cytosine. This
implied that DNA contains equal amounts of adenine and thymine and equal amounts
of guanine and cytosine, as was shown earlier by E. Chargaff. (Any results with ratios
significantly different from the expected 1:1 ratio would have refuted the Watson-Crick
structure.)
(3.3) Which observation in Franklin’s data was explained by offset between the two helices in the Watson-crick double helix 3/8th of the a pitch?
The lack of reflections on 4th layer line
(3.4) Explain the finding that DNA in solution with higher salt concentration have a higher melting temperature
The cations of the salt shield the negative charges of negative phosphate
groups, which destabilize the double helix.
(3.5) The Watson Crick structure suggested a mechanistic solution to a fundamental biological problem. What was this problem? How is this mechanistic solution called?
The structure suggested a mechanism for the reduplication of genetic information for cell division. The solution is called semiconservative replication.
(3.6) Why is pyrophosphate a leaving group in phosphoester bond formation during DNA polymerization and not deoxyribose? (Hint: Invoke pKa values of potential leaving groups.)
The deoxyribose would leave the alpha-phosphorus atom as an alkoxide. The pKa of its conjugated acid, an alcohol, is about 16, which makes for a poor leaving group, i.e., the energy of the transition state is very high. The pKa of 𝐻𝑃(𝑂*+, is 9.4, a weak acid, but more than six orders of magnitude more acidic than an alcohol,
suggesting that pyrophosphate is a much better leaving group. The enzyme furthermore lowers the transition state for phosphoanhydride bond cleavage and not phosphoester cleavage, e.g., by stabilizing the pyrophosphate anion (i.e., lowering the activation energy) via ionic bond formation with lysine and arginine residues, but not the alkoxide sugar or hydroxyl anion.
(3.7) Explain why DNA polymerases require deoxynucleoside triphosphates, and not monophosphates, as substrates, although the building blocks of DNA are deoxynucleoside monophosphates.
Phosphoester bond formation is strongly endergonic, i.e., the equilibrium of the
hydrolysis reaction lies far on the side of the deoxynucleoside mono-phosphates and not
DNA. Efficient polymerization, therefore, requires the coupling of phosphoester bond formation to exergonic reactions: the cleavage of two phosphoanhydride bonds by DNA polymerase and pyrophosphatase, respectively. The polymerase activity of DNA polymerases, thus, does not catalyze a hydrolysis reaction (and its reverse) but a pyrophosphorolysis reaction (and its reverse)
(3.8) The use of deoxynucleoside diphosphates may be sufficient to drive phosphoester bond formation. Why do DNA polymerases prefer deoxynucleoside triphosphates as substrates instead?
With the cleavage two phosphoanhydride bonds rather than one more free energy is expended per phosphoester bond formed, i.e., the reverse reaction becomes further unlikely, decreasing the variation of length of synthesized DNA strands within a given time interval. (This evidently is important for DNA replication during S phase.
(3.10) What limits the read length of Illumina sequencing?
Bridge amplification requires template molecules of relatively short length to limit the size of clone clusters. Small cluster sizes increase the number of molecules that can be sequenced per slide
(3.9) What are the three mechanisms that increase the specificity of DNA polymerases?
Helical recognition, induced fit, and kinetic proofreading.
(4.1) DNA polymerases, even those that lack proofreading activity, select correct nucleotides (i.e., nucleotides that are Watson-Crick complementary to the corresponding base pair in the template strand) with greater fidelity than primer-template junctions in solution (i.e., not bound to the enzyme). How is this increase in selectivity explained?
While the primer template junction in free solution selects correct nucleotide via Watson-Crick base pairing alone, the enzyme also recognizes the helical geometry of
the primer template junction, which is perturbed upon insertion of incorrect nucleotides.
(4.2) Which step of the reactions cycle that is catalyzed by DNA polymerases transfers free energy from a chemical work reservoir to the reaction cycle, and which step dissipates this free energy for kinetic proofreading?
The free energy that is dissipated in the proofreading step -hydrolysis of the
phosphoester bond- is provided by the polymerization reaction, which energetically
couples phosphoester bond formation to the cleavage of a phosphoanhydride bond. The
latter provides the free energy to drive the hydrolysis reaction away from equilibrium
by virtue of the fact that the reaction
𝑑𝑁𝑇𝑃 ⇌ 𝑑𝑁𝑀𝑃 + 𝑃𝑃( (the chemical work
reservoir) is maintained far from equilibrium by metabolic reactions.
(4.3) The product of a Sanger (dideoxy) sequencing reaction is a set of DNA strands with lengths ranging from “short” to “long.” Explain the underlying chemical mechanism for this length distribution. What would the product of such a reaction be if polymerization proceeded from 3’→5’ and not 5’→3’? Explain your answer
Random incorporation of the dideoxynucleotide leads to chain termination, because the dideoxynucleotide lacks the 3’-hydroxyl group necessary of linkage of the next nucleotide, and a distribution from short to long molecules. Only long DNA molecules would be produced because the dideoxynucleotide would never be incorporated into a strand that polymerizes from 3′ → 5′, and thus no chain termination would occur.
(4.4) Sanger and colleagues (1977) used the Klenow fragment of DNA polymerase I of E. coli for DNA sequencing. This enzyme has a 3’→5’ exonuclease. Provide an explanation for why this activity does not interfere with Sanger sequencing!
The dideoxynucleotide does not perturb helix geometry (so long as it’s correctly
base paired to the corresponding nucleotide on the template strand). The enzyme,
therefore, does not recognize dideoxynucleotides as incorrect.
(4.5) You wish to sequence the following DNA molecule using the dideox-method according to Sanger: 5’-GCTGTGGCATGCAGGACACGTGTCG-3’. The sequence of your primer is 5’-CGACAC-3’. In the sequence above, indicate the binding site of the primer to the template DNA. In the table below provide the expected band pattern in a sequencing gel by drawing horizontal bars into table boxes. Assume that the products of the four Sanger reaction were run on four adjacent lanes in the sequence A, C, G, T (from left to right). The cathode (-) is at the top, the anode (+) at the bottom of the gel (table).
(4.6) An extensive analysis of protein sequencing data shows that any amino acid (total of 20) can be linked to any other amino acid in biological proteins. Show that that this finding refutes the possibility of an overlapping triplet code
Every amino acid (20) may have 20 neighbors. In an overlapping triplet code, a
single codon allows for no more than 4 neighbors. Hence, for every amino acid there
must be at least 5 codons to account for 20 neighbors each. The total number of codons
required, therefore, is 5 x 20 = 100. However, only 64 distinct triplets can be formed with
four bases. This contradiction rules out the possibility of an overlapping triplet code.
(4.7) A nonoverlapping (triplet) code poses a formidable problem that is avoided by an overlapping triplet code. Which is that problem?
A non-overlapping code requires additional information for how to choose the
correct reading frame. Since an overlapping code reads off code words with a steps size
of one nucleotide, the reading frame problem does not arise.
(4.8) Explain why the enzymatic error fraction decreases with decreasing catalytic speed (k^cat).
The error fraction is minimal when the incorrect enzymatic substrate is rejected
(i.e., dissociates from the active center prior to catalysis) as often as it binds to the active
center. This is the case only when the binding reaction is in equilibrium, which requires
that 𝑘./0 «_space;𝑘2-
(4.9) If the maximal discrimination factor for correct versus incorrect substrate binding is k_2^-/k_1^-=0.3×10^4 for DNA polymerase, then what is the minimal error fraction?
The minimal error fraction is
(𝑘1-/𝑘2-)2 = (3.3 × 10-4)2 = 10-7
what three major finding made by Waston and Crick?
1.double helix
2.base pairing
3.anti-parallelism
Rosalind Franklin
Experiment:
-Used X-ray diffraction to study the structure of DNA. Prepared DNA fibers and exposed them to X-ray beams. When the X-rays hit the DNA, they scattered (diffracted) in a specific pattern.
Known for:
1.contributed in finding A & B form DNA structure
2.bases are inside, and the sugar-phosphate backbone is outside
3.DNA has dyad symmetry
4.double helix model of DNA
Arthur Kornberg
Experiment:
-Used biochemical assays to study the enzymatic properties of this polymerase, demonstrating its role in DNA synthesis. Purifying the first DNA polymerase (bacterial DNA polymerase I).
Conclusion: DNA polymerase I catalyzes DNA replication by adding nucleotides to a growing DNA strand using a primer-template junction.
Known for:
1.Discovery of DNA polymerase I, which is essential for DNA replication.
2.Understanding of the enzymatic properties of DNA synthesis,
3.foundation for techniques like PCR (Polymerase Chain Reaction)
Erwin Chargaff
Experiment:
-Used paper chromatography to analyze the ratios of nitrogenous bases in DNA from different organisms. Hydrolyzed DNA into its nucleotide components and quantified the amounts of (A), (T), (G) and (C).
Conclusion:
Base paring ratio was 1:1 because amount of adenine (A) is always approximately equal to the amount of thymine (T), and the amount of guanine (G) is always approximately equal to the amount of cytosine (C).
Known for:
Chargaff’s Rule (A = T, G = C), helped Watson and Crick deduce the base-pairing mechanism in DNA.
Robert Corey
Experiment:
worked with Linus Pauling on studying the structure of proteins and DNA using X-ray crystallography and model building techniques.
proposed a triple-helix model for DNA, in which three intertwined polynucleotide strands were positioned with the sugar-phosphate backbones in the center and bases facing outward.
Conclusion:
Corey-Pauling triple-helix model was incorrect
Known for:
Contributions to understanding protein structures, particularly in defining the alpha-helix and beta-sheet
Attempted, but incorrectly proposed, a triple-helix model of DNA.
