Flashcards in Chapter 10: DNA: Structure, Replication, and Variation Deck (64):
In 1928, Frederick Griffith established that _______.
heat-killed bacteria harbor the constituent(s) necessary to convey genetic properties to living bacteria
Because some of the nonvirulent bacteria acquired properties of the virulent bacteria, instructions for this transformation must be carried by the virulent bacteria.
To be certain that the extract prepared from virulent cells still contained the transforming principle that was present prior to lysis, Avery _______.
incubated nonvirulent cells with the complete extract
The complete extract possessed the same ability to induce transformation in IIR bacteria as whole heat-killed IIS bacteria.
If Avery had observed transformation using only the extracts containing degraded DNA, degraded RNA, and degraded protein, but NOT the extract containing degraded polysaccharides, he would have concluded that _______.
polysaccharides are the genetic material
Failure to transform suggests that the chemical degraded in that preparation is the one responsible for transformation, in this case polysaccharides.
The Hershey and Chase experiments involved the preparation of two different types of radioactively labeled phage. Which of the following best explains why two preparations were required?
It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment.
Because it was concluded that the component associated with bacteria at the end of the experiment must be the genetic material, it was critical that the component be identifiable as either DNA or protein.
Which of the following statements best represents the central conclusion of the Hershey-Chase experiments?
DNA is the identity of the hereditary material in phage T2.
Because phage DNA and not protein was associated with bacteria at the end of the experiment, it could be concluded that DNA - not protein - must be the genetic material.
Which of the following outcomes would be most likely if the Hershey-Chase experiments were repeated without the step involving the blender?
Both preparations of infected bacteria would exhibit radioactivity.
Instead of being removed from the preparation, the "ghosts" would be retained. Because both bacterial preparations would include ghosts as well as viral DNA, both would be radioactive, one with P32, one with S35.
What results did Avery, McLeod, and McCarty obtain in their experiments with virulent bacteria?
DNase destroyed the transforming activity.
Treatment of the transforming principle with DNase destroyed the DNA and thus its ability to transform bacteria.
What observation did Griffith make in his experiments with Streptococcus pneumoniae?
The mouse did not survive when injected with a mixture of live, avirulent (rough) Streptococcus pneumoniae and heat-killed, virulent Streptococcus pneumoniae.
Something in the heat-killed preparation was able to transform the avirulent strain to a virulent form.
Guanine and adenine are ____ found in DNA.
Guanine and adenine are indeed purines found in DNA; thymine and cytosine are the pyrimidines found in DNA.
The following statements about DNA structure are true?
The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions.
The 5′–3′ orientation of each chain runs in opposite directions.
What is the complementary DNA sequence to 5′ ATGCTTGACTG 3′?
5′ CAGTCAAGCAT 3′
Regarding the structure of DNA, the covalently arranged combination of a deoxyribose and a nitrogenous base would be called a(n) ________.
If 15% of the nitrogenous bases in a sample of DNA from a particular organism is thymine, what percentage should be cytosine?
Identify three possible components of a DNA nucleotide.
deoxyribose, phosphate group, thymine
DNA and RNA have similar structures: a pentose sugar with a nitrogenous base and a phosphate group. DNA and RNA differ in the type of pentose sugar each possesses (DNA has deoxyribose; RNA has ribose) and in one base (DNA has thymine; RNA has uracil).
List 3 main differences between DNA and RNA.
RNA often occurs as both single- and partially double-stranded forms, whereas DNA most often occurs in a double- stranded form.
Ribose in RNA replaces deoxyribose in DNA.
Uracil in RNA replaces thymine in DNA.
One of the most common spontaneous lesions that occurs in DNA under physiological conditions is the hydrolysis of the amino group of cytosine, converting the cytosine to uracil. What would be the effect on DNA structure of a uracil group replacing cytosine?
A base substitution of G:C to A:T after two rounds of replication.
In 1953, Watson and Crick published a paper that described the ____.
In 1953, Watson and Crick published a paper that described the structure of DNA.
RNA differs from DNA EXCEPT what way?
the 5'-3' orientation of the polynucleotide strand
Both RNA and DNA have the same 5' amino group and 3' hydroxyl group chemical orientation.
All EXCEPT which of the following statements are evidence that DNA, and not protein, is the genetic material in eukaryotes as well as bacteria?
DNA has four nucleotides.
DNA is located only where the primary genetic function is known to occur.
UV light is most mutagenic at a wavelength at which DNA and RNA strongly absorb.
Introduction of a cloned DNA into another organism results in the production of the corresponding protein product.
DNA has four nucleotides, and proteins have 20 amino acids. Protein was thought to have the chemical diversity and complexity necessary for the genetic material and DNA was not.
Which technique can be used to identify the location of genes on a chromosome?
Fluorescent in situ hybridization is a technique used to identify the location of a gene or a specific DNA sequence on a chromosome. The technique involves hybridizing single-stranded DNA probes, labeled with a fluorescent dye, to their complementary DNA strand in the genome. Their location on the chromosome can be viewed under a fluorescent microscope.
___ has no sulfur, and ____ have no phosphorus.
DNA has no sulfur, and proteins have no phosphorus.
In a DNA sequence, the purine ____ always pairs with the pyrimidine ____ , and the purine ____ always pairs with the pyrimidine ____.
In a DNA sequence, the purine adenine always pairs with the pyrimidine thymine, and the purine guanine always pairs with the pyrimidine cytosine.
Write the complementary sequence for the following DNA sequence, in order from 3' to 5':
The base pair adenine-cytosine occurs very rarely in nature. It only happens during a mutation event. When the DNA is replicated, one of the two daughters will contain a guanine-cytosine base pair in the location of the mutation, and the other daughter will contain an adenine-thymine base pair.
How does the number of hydrogen bonds between the two bases affect the stability of a base pair?
All other factors being equal, the renaturation of the three classes of complementary nucleic acid sequences occurs in what order, from fastest to slowest?
A) unique sequences
B) moderately repetitive sequences
C) highly repetitive sequences
DNA is used for storage of genetic information. The presence of deoxyribose as the sugar in DNA makes the molecule more stable and less susceptible to hydrolysis.
The 2'-oxygen on the ribose found in RNA makes RNA much more susceptible to breakdown. It is important that mRNA be easily broken down, to ensure that the correct levels of protein are maintained in the cell.
Genetic material qualities
Variation by mutations
DNA to Protien
DNA >>> TRANSCRIPTION >>> RNA
mRNA, tRNA, rRNA >>> Ribosome
Ribosome >>> TRANSLATION >>> Protien
What is the genetic material?
Proteins and nucleic acids were considered possibilities.
Pre 1940’s, protein was leading contender:
-Protein is abundant in cell
-There is a wide variety of proteins in cells
-DNA was known to be made up of 4 nucleotides, nucleic acids were “too simple” to be genetic material
-Proteins contained 20 different amino acids, thus more potential variation
Model systems and differential labeling
Choice of the correct organism to perform an experiment
Methodology to deferentially label or separate molecules of interest
Evidence of DNA as Genetic Material:
Early work (1927) by Frederick Griffith
He analyzed virulence of various strains of Diplococcus pneumoniae
Some strains caused pneumonia in humans and mice (VIRULENT)
Some did NOT cause pneumonia (AVIRULENT)
Virulence of Diplococcus pneumoniae caused by polysaccharide capsule of bacterium
Those with capsule virulent & grow as smooth, shiny (S) colonies
Those without capsule are avirulent & grow as rough (R) colonies
Therefore, you can distinguish virulent from avirulent by growth characteristics
Griffith’s “Transforming principle” experiments
Model system: D. pneumoniae
Many sereotypes of rough and smooth colonies, eg. type I, type II, type III
Differential identification: Rough colonies= aviurlent; Smooth colonies= virulent
Griffith used seroytpes II and III of rough and smooth strains in his experiments (IIR and IIIS)
Griffith’s critical experiment Results:
Lastly also known, if you inject heat killed virulent bacteria:
Inject heat-killed virulent IIIS bacteria & living avirulent IIR bacteria into mouse:
DEAD MOUSE, Also recover living virulent IIIS bacteria from dead mouse tissue!
Somehow the heat-killed IIIS bacteria were transforming the live avirulent IIR bacteria into virulent IIIS bacteria!
Griffith suggested the transforming principle might be something in the capsule, however the capsule alone did not cause pneumonia
Further transformation studies
Work extended in the 1930’s showed that transformation could happen in vitro
Soluble filtrate of heat-killed S cells was sufficient to cause transformation
Experimental evidence suggested that a chemical substance was responsible for the observed transformation phenomena
What is the “transforming” molecule?
Experiments of Avery, McLeod, and McCarty
Classic paper, published in 1944 after 10 years of work!
Conclusion: DNA was the transforming molecule
Avery, McLeod, and McCarty experiment, which demonstrated that DNA is the transforming principle.
First, they transformed with crude extract, but still got transformation
Next: treat remaining extract with protease to remove any remaining protein. Still got transformation
Next: treat remaining extract with RNase, degrades RNA: Transformation still occurs
Next: treat remaining extract with deoxyribonuclease, degrades DNA: Transformation does NOT OCCUR, therefore DNA must be transforming material!!!
Summary of Avery, MacLeod, and McCarty’s experiment
Differential separation of cell molecules from virulent strain
Test these separate components for ability to transform avirulent into virulent
Conclusion: The only molecule capable of transformation is DNA
Provided convincing evidence of DNA as genetic material
Key was the model system used (E. coli and its viruses) and differential labeling of proteins and nucleic acids
-Provides method to “differentiate one thing from another thing”
Hershey-Chase Experiments: Model system
E. coli: bacteria easily grown and analyzed
T2 bacteriophage: a virus of E. coli that replicates within E. coli following infection
Proteins labeled with radioactive 35S, DNA labeled with radioactive 32P. This is Differential Labeling!
State of knowledge of bacteriophage T2 in early 1950’s
T2 phages ~50% protein, 50% DNA
Infection of E. coli is initiated by adsorption of phage to bacterial cell
The production of new viruses occurs within the E. coli cell
Summary of Hershey–Chase experiment. demonstrating that DNA, and not protein, is responsible for directing the reproduction of phage T2 during the infection of E. coli.
Demonstrated that DNA, and not protein, is responsible for directing the reproduction of phage T2 during the infection of E. coli.
Differentially labeled phages,
DNA with 32P; proteins labeled with 35S
Infect non-radioactive E. coli with labeled phage:
32P DNA, 35S proteins
Add phage to E. coli 32P medium.
Phage progeny labeled.
32P labeled phages infect unlabeled bacteria.
Add phage to E. coli 35S medium.
Phage progeny labeled.
35S labeled phages infect unlabeled bacteria.
KEY RESULTS of the Hershey–Chase experiment
Removed phage ghosts, only radioactive phage ghosts were those labeled with 35S
Viable phage were radioactive only from those that were labeled with 32P, therefore it was DNA that served as the genetic material!
DNA was acting as the genetic material in T2 bacteriophage
Additional evidence followed in subsequent years that supported this conclusion
DNA as genetic material in eukaroytes
Indirect evidence: DNA found only where primary genetic function occurs, correlation of ploidy and the quantity of genetic material molecule
Indirect evidence: UV light is most mutagenic at wavelength that RNA and DNA absorb UV light the most
determines the most effective mutagenic UV wavelength
shows the range of wavelength where nucleic acids and proteins absorb UV light
The structure of DNA?
Nucleic acid chemistry helped to work out the structure of DNA
DNA is a nucleic acid, nucleotides are building blocks of nucleic acids
3 components: nitrogenous base, pentose (5-carbon) sugar, and a phosphate group
Purines: 9 member double ring.
Pyrimidines: 6 member single ring.
DNA has A, C, G, T
RNA has A, C, G, U
Trick to remember:
Short name (purine), BIG (double ring) structure.
Long name (pyrimidine): SMALL (single ring) structure.
join Ribose to purine or pyrimidine in specific manner; result is ribonucleosides.
2 -OH groups in ribonucloeside
Structure resembles DNA, with some exceptions
Ribose sugar replaces deoxyribose
Uracil replaces Thymine
Most RNA is single stranded, but can base pair with itself
join 2-Deoxyribose to purine or pyrimidine in specific manner; result is deoxyribonucleosides
1 -OH groups
Deoxy= the difference in deoxyribonucleotide vs. ribonucleotide
DNA & RNA
Nucleotides are the building blocks of each
NMP: nucleoside monophosphate
NDP: nucloeside diphosphate
NTP: nucleoside triphosphate
formed between mononucleotides with phosphodiester bond, linkage between C-3’ to C-5’
Linkage of two nucleotides by the formation of a 5' phosphodiester bond, produce a dinucleotide.
Watson-Crick DNA model
Previous studies were integrated into their model
Built models based on the known constraints experimentally determined
The ribbonlike strands constitute the sugar-phosphate backbones, and the horizontal rungs constitute the nitrogenous base pairs, of which there are 10 per complete turn. Major and minor grooves are apparent.
The antiparallel nature of the helix and the horizontal stacking of the bases.
The Watson-Crick DNA model features:
2 polynucleotide chains coiled around a central axis, forming right-handed helix
The 2 chains are anti-parallel: 5’-->3’ runs in opposite directions on each strand
The bases lay flat, perpendicular to the axis, stacked on one another 3.4 Angstroms apart
Each complete turn of the helix is 34 Angstroms
In the chain, there are major grooves alternating with minor grooves
The double helix is 20 Angstroms in diameter
Alternative DNA Forms
Using different isolation conditions, different forms of DNA have been obtained
2 forms known at the time of Watson & Crick model: A & B
B-DNA is the “normal” form of DNA in vivo
Other forms now found: C, D, E, P, & Z DNA (roles in vivo still not understood)
RNA & DNA analysis
Both RNA and DNA absorb UV light most strongly at 260nm
The amount of UV light absorbed is related to the concentration of RNA or DNA in a sample
This can be used to quantify RNA or DNA
A260 = 1 = 50ug/ml
Nucleic Acid Centrifugation
RNA’s differentiated by their sedimentation behavior
Measure is called Svedberg coefficient
Higher S, in general, larger molecule
Ribosomal RNAs the largest, eg. eukaryotic 18S rRNA
Gradient centrifugation of nucleic acids
Collection of fractions
Separation of a mixture of 2 types of nucleic acid by gradient centrifugation.
To fractionate the gradient, successive samples are eluted from the bottom of the tube.
Each is measured for absorbance of ultraviolet light at 260 nm, producing a profile of the sample in graphic form.
Nucleic Acid Separation
Another method: gel electrophoresis
Hydrogen bonds of DNA can be broken with thermal energy (heat)
Lower temperature, DNA reanneals
Basis of molecular hybridiztion techniques (like FISH)
DNA reassociation/ Cot curves
Slow initial association
Once association begins, then goes faster
Last association takes the longest, unique sequences finding one another
Cot1/2 = point where 50% of the DNA is reassociated
In general, smaller genomes reassociate faster than larger genomes
Linear relationship between Cot1/2 and genome size
The ideal time course for reassociation of DNA (C/C0) when, at time zero, all DNA consists of unique fragments of single-stranded complements.
C0: initial single stranded DNA in moles per liter of nucleotides
C: single stranded DNA concentration after time t has elapsed
The abscissa (C0t) is plotted logarithmically.
We can use comparisons of unknown DNA reassociation to known curves to characterize the unknown