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BSCI-1510 Lecture Notes - Lecture 23: Glycosidic Bond, Hydrogen Bond, Transfer Rna

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skytoad282
10 Feb 2016
School
Vanderbilt University
Department
Biological Sciences
Course
BSCI-1510
Professor
Patton Jim

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Document Summary

Patton lecture 3 nucleic acids & replication. Nitrogen has 3 bonds in it, so you don"t have to put hydrogen in it and you don"t have to memorize the double bonds in it. *** draw nucleotide with glycosidic bond and the sugar and the phosphate. Proteins that decode dna look at the edge of the bases (bases are flat, planar structures) Side chains of amino acids interact with dna along the major groove and they read atoms that aren"t involved in hydrogen bonding. Proteins see atoms that aren"t involved in hydrogen bonding. As a protein follows a major groove and reads the bases, those are the items that it looks at and will have a lock and key fit. Protein that define where dna replication starts read a specific sequence in the major groove. We always pair a with t and c with g. a = purine (2 rings) t is a pyrimidine (1 ring)

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Related Questions

The recognition of specific DNA sequences by proteins is thoughtto depend on two types of mechanisms: one that involves theformation of hydrogen bonds with specific bases, primarily in themajor groove, and one involving sequence-dependent deformations ofthe DNA helix. By comprehensively analyzing the three-dimensionalstructures of protein-DNA complexes, we show that the binding ofarginines to narrow minor grooves is a widely used mode forprotein-DNA recognition. This readout mechanism exploits thephenomenon that narrow minor grooves strongly enhance the negativeelectrostatic potential of the DNA. The nucleosome core particleoffers a striking example of this effect. Minor groove narrowing isoften associated with the presence of A-tracts, AT-rich sequencesthat exclude the flexible TpA step. These findings suggest that theability to detect local variations in DNA shape and electrostaticpotential is a general mechanism that enables proteins to useinformation in the minor groove, which otherwise offers fewopportunities for the formation of base-specific hydrogen bonds, toachieve DNA binding specificity

1. What three properties of the major groove are advantageous toalpha helix insertion?

2. Which nucleotides are prevalent in narrow minor grooves?

3. Which amino acid tends to bind to the minor groove and howare minor groove width and electrostatic potential correlated?

1. In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that
DNA is a double helix.
DNA may be taken up by bacterial cells and be active.
DNA is present in bacteriophage T2.
eukaryotic cells may be genetically transformed.
DNA binds to the cell wall of the bacterium.
2. In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.
temperatures high enough to kill cells
enzymes that hydrolyze proteins
enzymes that hydrolyze DNA
enzymes that hydrolyze polysaccharides
enzymes that hydrolyze RNA
3. Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?
Pure RNA from S cells
Cell-membrane extracts from S cells
A mixture of pure RNA and protein from S cells
Cell-free extracts from S cells
Pure protein from S cells
4. A-T base pairs in a DNA double helix
form three hydrogen bonds with each other.
are more heat-stable than G-C base pairs.
are of a substantially different length than G-C base pairs.
are not accessible to DNA binding proteins.
are chemically distinct from G-C base pairs.
5. If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?
27
50
0
73
23
6. The uniform diameter of the DNA structure provides evidence for
antiparallel DNA strands.
a right-handed helix.
a double-stranded helix.
3′ end phosphorylation.
purine–pyrimidine pairing.
7. If a sequence of one strand of DNA is 5′-TGACTATC-3′, what is the complementary strand?
5′-ACTGATAG-3′
3′-TGACTATC-5′
3′-ACTGATAG-5′
3′-ACTGATAG-3′
5′-TGACTATC-3′
8. What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?
3′ end phosphorylation
Purine–pyrimidine pairing
Antiparallel DNA strands
Covalent bonds between sugars and phosphate groups
Hydrogen bonding between paired bases
9. If the Meselson–Stahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only 15N and the second only 14N, this result would have indicated that replication was
unidirectional.
dispersive.
conservative.
bidirectional.

semiconservative.
10. The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3′ hydroxyl group (–OH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.
The replication complex will no longer bind to origins of replication.
The DNA polymerase will only incorporate acyclovir molecules.
Acyclovir will bind single-strand binding proteins.
DNA polymerase will not be able to link a successive nucleotide.
Acyclovir will block the activity of the DNA helicase.
11. Which of the following does not demonstrate the stability of the DNA double helix?
Single-strand binding proteins are required to keep the strands apart.
Paired bases form hydrogen bonds.
High temperatures are required to denature it.
DNA synthesis requires energy.
DNA helicases require ATP.
12. What effect would a primase inhibitor have on DNA replication?
DNA polymerase would not be able to add bases to the DNA.
Primase would be blocked from joining the DNA replication complex.
Primase would not be able to provide primers for DNA polymerases.
DNA polymerase would not be able to use DNA as a template.
There would be no effect on DNA replication.
13. To replicate their DNA in a timely manner, most eukaryotic chromosomes
normally have uncoiled DNA.
have multiple origins of replication.
contain linear molecules of single-stranded DNA.
have two replication forks that move in the same direction.
are composed entirely of DNA.
14. Which statement about DNA replication is false?
Okazaki fragments are synthesized as parts of the leading strand.
Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
The sliding clamp increases the rate of DNA synthesis.
Replication forks represent areas of active DNA synthesis on the chromosomes.
Ligases and polymerases function in the vicinity of replication forks.
15. In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.
leading; telomerase
lagging; telomerase
lagging; DNA polymerase I
leading; DNA polymerase I
leading; replicase
16. A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?
DNA polymerase is constantly active.
Primase is able to extend the telomeres.
The mismatch repair system is extra efficient.
The telomerase gene is being expressed.
The cells do not need the ends of their chromosomes.
17. If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?
The replication complex excises the incorrect nucleotide.
The excision repair proteins remove the incorrect nucleotide.
The mismatch repair system scans for base-pairing mismatches.
There is no correction, because nucleotide errors by DNA polymerases are not repaired.
DNA ligase connects the correct nucleotide.
18. Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.
1, 2, 3, 4
2, 1, 3, 4
3, 4, 2, 1
2, 4, 1, 3
4, 2, 3, 1
19. Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.
64
32
12
128
6
20. When double-stranded DNA is heated to temperatures above 90°C, it denatures. Denaturation is a process that
breaks covalent bonds.
hydrolyzes DNA.
separates double-stranded DNA into two single strands.
causes new hydrogen bonds to form.
irreversibly destroys the double helical structure.

Identify amino acids which when being part of a protein structure have side chains that can participate in the hydrogen bond formation. Draw the structure of each amino acid and indicate how it can form the hydrogen bonding.

2. Identify amino acids which when being part of a protein structure have side chains that can participate in chemical reactions. Explain why and how. For instance which residues are used for post-translational modifications, for regulation of protein activity, how are glycans and lipids attached to proteins?

3. Identify amino acids which when being part of a protein structure have side chains that can ionize or are charged at pH 7. Based on the pKR values of the free amino acids estimate/calculate what percentage of the amino acid is in the charged form.

Explain why some amino acids, like serine in serine proteases, can ionize/lose proton at neutral pH while their pKR values suggest that this should not be possible.