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BCHM-3050 Lecture Notes - Lecture 3: Erwin Chargaff, Hydrogen Bond, Guanine

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scarletmouse537
19 May 2018
School
Clemson University
Department
Biochemistry
Course
BCHM-3050
Professor
Geoffrey Ford

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Primary Structure
Primary structure: nucleotide sequence
•
Main importance: genetic information is stored in the primary structure of
DNA
Gene is a particular DNA sequence
○
•
Early evidence: Avery, MacLeod and McCarty
DNA from pathogenic strains of pneumococcus can be transferred
into non pathogenic strains
Through transfer of DNA
§
○
•
Secondary and Tertiary Structure
Secondary structure: 3D arrangement of nucleotide residue w respect to
one another
•
Tertiary structure: longer range interactions in 3D
•
Watson-Crick Double Helix Model and other Data
Propose double helical model for DNA structure
•
Erwin Chargaff: determined that bases pair with each other
A=T
○
G=C
○
•
Watson and Crick: two strands in the double helix are stabilized by
hydrogen bonding
•
X-Ray Analysis of DNA fibers
Rosalind Franklin: Characteristics of DNA molecules
Characteristics:
The structure repeats itself at intervals of 10 nucleotides
§
Distance parallel to helix axis at which the helix makes a turn
is 3.4 nm
§
Distance parallel to the helix axis b/w two successive
nucleotide residues is 0.34 nm
§
○
•
Base-Stacking
stacking interaction: strong interaction between bases via van der Waals
•
Bases are closely packed within the helix
•
Major grooves: give direct access to bases
•
Minor grooves: face the sugar backbone
•
Grooves play a role in replication, transcription and other protein
interactions
•
DNA replication: semiconservative
Semiconservative: replication of DNA molecule yields two daughter
duplexes which consist of one half parental DNA
Half of the original material is conserved in each copy
○
•
Conservative: one pair is the same as parental pair while the other is
synthesized and new
•
Dispersive: parental material is scattered through the structures of the
daughter pairs
•
Meselson-Stahl experiment
Bacteria with high density DNA (1.724 g/mL) was grown on media with
low density (1.710 g/mL) bacteria.
Subsequent generations had intermediate density DNA (1.717 g/mL)
○
•
A and B form Helices
A form: DNA under low humidity conditions
dsRNA and DNA-RNA hybrid molecules
○
Bases lie farther to the outside
○
Grooves are equally width
○
•
B form: seen in DNA under high humidity conditions
dsDNA
○
Predominant form
○
Grooves are distinguishable
○
10.5 base pairs per turn
○
•
A and B forms are right hand helices
•
DNA and RNA molecules in Vivo
DNA molecules: Double stranded in most organisms
Two strands are complementary
○
•
DNA shapes
Circular DNA: can be small, involve single strand or two strands
intertwined in B form double helix
•
Supercoiled: strained circular molecules
Involved in tertiary structure
○
More compact
○
Essential for compaction
○
•
DNA ligase: seal DNA together covalently
•
Twist: complete helical turn
•
Writhe: recreating a helix
•
Hairpins and Cruciform
Cruciform: double hairpin
Can be formed in some DNA sequences
○
•
Palindromic: sequence is the same read forward to backward or backward
to forward'
•
G-Quadruplexes
G- quadruplex: structure consisting of several G-quartets
G- quartet: sequence consisting of 4 guanine molecules
○
May exist in:
telomeres
§
Transcriptional control sites
§
Oncogenes
§
○
•
Denaturation!
Denaturation: loss of secondary structure
•
DNA denaturation follows laws of thermodynamics
•
Biological function of nucleic acid
Replication : DNA to DNA1.
Transcription: DNA to RNA 2.
Translation : RNA to protein 3.
Chapter 4: Helix and structure
Saturday, May 19, 2018
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Document Summary

Main importance: genetic information is stored in the primary structure of. Dna from pathogenic strains of pneumococcus can be transferred into non pathogenic strains. Secondary structure: 3d arrangement of nucleotide residue w respect to one another. Erwin chargaff: determined that bases pair with each other. Watson and crick: two strands in the double helix are stabilized by hydrogen bonding. The structure repeats itself at intervals of 10 nucleotides. Distance parallel to helix axis at which the helix makes a turn is 3. 4 nm. Distance parallel to the helix axis b/w two successive nucleotide residues is 0. 34 nm. Base-stacking stacking interaction: strong interaction between bases via van der waals. Grooves play a role in replication, transcription and other protein interactions. Semiconservative: replication of dna molecule yields two daughter duplexes which consist of one half parental dna. Half of the original material is conserved in each copy.

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Related Questions

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.

magine that the year is 1928 and the scientific world is in an uproar trying to determine what component of the cell isresponsible for genetic inheritance. Is it DNA, protein, perhaps even RNA? You are busy in your lab, trying to help solve this crucial mystery that still plagues (then) modern day biology. Your work focuses on a microorganism, a bacterium named Streptococcus pneumoniae, and its effects on mice. There are two strains of S. pneumoniae (which is responsible for pneumonia), the “Rough” strain (R) and “Smooth” strain (S). The strains are very similar except the S strain has a smooth polysaccharide coat that enables the microorganism to avoid the organism’s immune system and cause illness (and death in mice). The R strain has a rough coat which is detected and dealt with by the immune system (and as a result causes no death in mice). You decide to perform a series of experiments using the R and S strain of S. pneumoniae involving injections of these strains in mice. Your results are documented in the figure below: living S strain of S. pneumoniae mouse dies of infection living R strain of S. pneumoniae mouse lives mouse lives S strain heat-killed living R strain mouse dies of infection living, pathogenic S strain recovered S strain heat-killed From these results, you figure out that something from the heat killed S strain is transforming the R strain into the virulent strain. What is this “Transforming principle”? Is it related to the cellular component that was passed from one generation to the next? As you know, this was the initial experiment performed by Dr. Frederick Griffith that eventually led to the discovery of the true cellular component that is the genetic inheritance molecule as well as the “transforming principle” (Hint: It’s DNA!).

1. Several experiments were required to demonstrate how traits are inherited. Which scientist or team of scientists first demonstrated that cells contain some component that can be transferred to a new population of cells and permanently cause changes in the new cells?

a. Griffith b. Watson and Crick c. Avery, MacLeod, and McCarty d. Hershey and Chase

2. From your results, which one of the cellular components can be determined NOT to be the component responsible for genetic inheritance? Why?

a. Proteins because these macromolecules become denatured with heat. b. DNA because these macromolecules form a single helix in solution. c. RNA because it's double stranded 100% of the time d. Sugars because glycolysis happens.

3. Which one of the following experimental trials below are considered to be controls for this experiment? a. S Strain grown in 10% ethanol medium. b. Living S strain; Living R strain; Heat Killed S Strain c. Living R Strain with heat killed S strain; Living S Strain d. Heat Killed S strain; Living R strain; R strain grown in 10% ethanol medium

4. Which scientist or team of scientists obtained definitive results demonstrating that DNA is the genetic molecule? a. Griffith b. Watson c. Crick d. Hershey and Chase

5. Hershey and Chase used radiolabeled macromolecules to identify the material that contains heritable information. What radioactive material was used to track DNA during this experiment?

a. 3H b. 14C c. 35S d. 32P

6. DNA molecules, like proteins, consist of a single, long polymeric chain that is assembled from small monomeric subunits.

a. True b. False

7. The polarity of a DNA strand results from the polarity of the nucleotide subunits.

a. True b. False

8. There are five different nucleotides that become incorporates into a DNA strand.

a. True b. False

9. Hydrogen bonds between each nucleotide hold individual DNA strands together.

a. True b. False

1. Characters that show a continuous range of variation, such as height and eye color, usually are controlled:

a. by a single gene with two alleles that are codominant.
b. by many genes with an additive effect.
c. by epistatic interactions between two genes.
d. mainly by the environment, with only a small genetic component.

2. In humans, red-green colorblindness is inherited as a sex-linked recessive trait. In order for a woman to be red-green colorblind, which of the following statements must be true.

a. Her mother must be red-green colorblind.
b. All of her brothers must be red-green colorblind.
c. Her father must be red-green colorblind.
d. All of the above statements must be true if a woman is red-green colorblind.

3. The x-ray crystallography data collected by Rosalind Franklin suggested to Watson and Crick that the:

a. structure of DNA is a double helix.
b. two strands of the DNA molecule are joined by hydrogen bonds between the bases.
c. four bases within DNA pair in a specific way.
d. two strands of the DNA molecule are joined by covalent bonds between the bases.

4. In the genetic code, _________ one amino acid.

a. one nucleotide specifies
b. two nucleotides specify
c. three nucleotides specify
d. four nucleotides specify

5. During Meiosis I, a homologous pair of chromosomes may not separate, resulting in daughter cells that have extra chromosomes or are missing chromosomes. This can lead to genetic disorders, including Down Syndrome. This phenomenon is called:

a. independent assortment.
b. nondisjunction.
c. segregation.
d. crossing over.

6. You are a human geneticist studying the incidence of retinitis pigmentosa in the residents of Tristan de Cunha, a group of small islands in the middle of the southern Atlantic Ocean. The allele for retinitis pigmentosa, which causes a form of blindness, is inherited as an autosomal recessive. You have determined that the frequency of this allele (r) in the population is 0.4 (40%). Using the principles of the Hardy-Weinberg rule, you would estimate the frequency of individuals who are heterozygous for this allele (Rr) in the population to be:

a. 0.16 (16%)
b. 0.24 (24%)
c. 0.36 (36%)
d. 0.48 (48%)

7. Natural selection acts at the level of the:

a. phenotype.
b. gene.
c. population.
d. nucleotide.

8. You are working with pea plants, trying to recreate the experiments that Mendel performed. You are doing a dihybrid cross with a plant that is heterozygous for both seed shape and seed color, with the genotype RrYy. Which allelic combinations would you expect to find in the gametes produced by this plant?

a. This plant would produce only RY and ry gametes.
b. This plant would produce only RrYy gametes.
c. This plant would produce RY, Ry, rY, and ry gametes.
d. You cannot determine which gametes this plant can produce without knowing the genotypes of its parents.

9. Biochemist Erwin Chargaff found that in DNA there is a special relationship between the four bases that we now call Chargaff's rule. His observation was that, in an organism's genome the:

a. percentage of A nucleotides = the percentage of T nucleotides, and the percentage of C nucleotides = the percentage of G nucleotides.
b. four bases all occur in an equal frequency (25%) within each organism.
c. percentage of A nucleotides = the percentage of G nucleotides, and the percentage of C nucleotides = the percentage of T nucleotides.
d. genetic material is composed of proteins, not DNA.

10. During DNA replication:

a. each strand of the double helix acts as a template for the synthesis of a new strand.
b. the enzyme DNA polymerase adds nucleotides to the strand being synthesized.
c. the bases A,C,G and T are required.
d. All of the above are true of DNA replication.

11. During translation, amino acids are joined by peptide bonds to make polypeptides. The formation of these peptide bonds is catalyzed by:

a. DNA.
b. mRNA.
c. tRNA.
d. rRNA.

12. If an allele (R) at a gene with two alleles shows complete dominance, individuals with the genotypes ______ will have the same phenotype.

a. RR and rr.
b. RR and Rr
c. Rr and rr
d. Each of the three possible genotypes will have a different phenotype.