Lecture 2 Flashcards
What four general characteristics must the genetic material possess?
(1) The genetic material must contain complex information.
(2) The genetic material must replicate or be replicated faithfully.
(3) The genetic material must have the capacity to vary or mutate to generate
diversity.
(4) The genetic material must encode the phenotype or have the ability to code for
traits.
What experiments demonstrated that DNA is the genetic material?
- Hershey and Chase 1950s:
- —-T2 bacteriophage experiment (tagging sulfur or phosphorus of bacteriophage in two separate trials. radiation showed up in bacteria when P tagged)
-Avery, Macleod, McCarty (“transforming material” of Griffiths’ S bacteria was DNA; expose material to Dnase, Rnase, or protease, then grow R cells in it. Only Dnase one failed to kill mouse).
What is transformation? How did Avery and his colleagues demonstrate that the transforming principle is DNA?
Transformation occurs when a transforming material (DNA) genetically alters the bacterium that absorbs the material. Avery et al demonstrated that DNA is the transforming material by using protease, DNase, and RNase. Only DNase had an effect on the material’s biological activity. They were also able to isolate the transforming material and demonstrate that it had chemical properties similar to DNA.
Predict what would happen if Griffith had mixed some heat-killed type IIIS bacteria and some heat-killed type IIIR bac and injected these into a mouse. Would the mouse have contracted pneumonia and died? Explain.
No. Although the mouse would have received IIIS DNA, which codes for the virulent strain, there are no live bacteria for this DNA to transform.
Imagine you are a student in Alfred Hershey and Martha Chase’s lab in the late 1940s. You are given five test tubes containing E. coli bacteria that were infected with T2 bacteriophage that have been labeled with either 32P or 35S. Unfortunately, you forgot to mark the tubes and are now uncertain which were labeled with 32P and which with 35S. You place the contents of the each tube in a blender and turn it on for a few seconds to shear off the protein coats. You then centrifuge the contents to separate the protein coats and the cells. You check for the presence of radioactivity and obtain the following results. Which tubes contained E. coli infected with 32P-labeled phage? Explain your answer.
For each tube, Presence of radioactivity in :
Tube 1 (T1): cells T2: protein coats T3: protein coats T4: cells T5: cells
Tubes 1,4,5. The DNA of the bacteriophage contains phosphorous and the protein contains sulfur. When the bacteriophages infect the cell, they inject their DNA into the cell, but the protein coats stay on the surface of the cell. The protein coats are sheared off in the blender, while the cells with the DNA pellet at the bottom of the tube. Thus, cells infected with 35S -labeled bacteriophage will have radioactivity associated with the protein coats, whereas those cells infected with 32P-labeled bacteriophage will have radioactivity associated with the cells.
Draw and identify the three parts of a DNA nucleotide
The three parts of a DNA nucleotide are phosphate, deoxyribose sugar, and a nitrogenous base. (I can’t paste a picture here)
How dies an RNA nucleotide differ from a DNA nucleotide?
DNA nucleotides have a deoxyribose sugar (lacks an OH on the 2’ C). RNA nucleotides have a ribose sugar (with the OH on the 2’ C).
Also, RNA uses uracil instead of thiamine, while DNA uses thiamine instead of uracil
How does a purine differ from a pyrimidine?
What purines and pyrimidines are found in nucleic acid?
Purine: six-sided ring attached to a five-sided ring.
-A and G
Pyrimidine: six-sided ring.
-C,T,U
An epidemic disease affecting sheep grazing near the hot springs of Thermopolis, Wyoming, was found to be due to a new virus. This virus could infect sheep kidney cells cultured in vitro. The virus appeared to contain four chemically defined biomolecules, which we will call W, X, Y, and Z. Investigators wished to determine which of these components carried the genetic information of the virus.
An experiment was conducted in which one of the four components was radioactively labeled in different batches of kidney cells cultured and infected with the virus. Thus, four batches of labeled virus were generated (labeled W, X, Y, or Z).
Radioactive virus from each of these batches was allowed to attach to nonradioactive kidney cells. The cells were centrifuged down to remove the unattached viruses. The cells were then briefly exposed to a vigorous agitation to release viral particles on the cell surface, and again centrifuged. The supernatant (containing shaved-off viral parts) and the pelleted infected cells were examined for radioactivity, with the following results:
For labled supernatant:
W = 100% radioactivity; X=Y=Z=0%
For labeled cells:
W = 0%; X=Y=Z=100%.
A. On the basis of these results, which of these components does NOT carry the viral genetic info?
Other experiments revealed that one of these purified biomolecules was contaminating sheep DNA, and the two remaining components did appear to be derived from a virus. The nitrogenous base compositions were as follows: %(A,G,C,T,U): X = (21,21,32,0,26) Y = (26,32,21,21,0) Z = (28,22,22,28,0)
B. What can you say about the type of nucleic acid found in each of these three components?
C. Can you determine which of these purified biomolecules is most likely sheep DNA and which are more likely to belong to the virus? Explain your response.
a) As biomolecule W was found in the supernatant and not within the infected cells, it is most likely not carrying genetic information within its structure. The data are consistent with W being some type of structural component of the virus outer coat.
b) Component X is RNA, as it contains uracil rather than thymine as a nitrogenous base, while Y and Z contain nitrogenous bases consistent with DNA.
c) Sheep DNA is double stranded, and thus should be consistent with Chargaff’s rules (A=T, G=C). The biomolecule Z appears to be double-stranded DNA, while Y is most likely single-stranded DNA and potentially a viral genome.
Note to self: What about X?
One nucleotide strand of DNA molecule has the base sequence illustrated below.
5’—ATTGCTACGG—3’
Give the base sequence and label the 5’ and 3’ ends of the complementary DNA nucleotide strand.
3’—TAACGATGCC—5’
If a double-stranded DNA molecule is 15% thymine, what are the percentages of all the other bases?
A ≈ 15%
C ≈ G ≈ 0.5(100%-2•15%) = 35%
Draw out/describe the models for DNA replication.
Conservative: The parent is conserved, and offspring are completely new
Semiconservative: each offspring has one strand of its parent and one new strand.
Dispersive: Each daughter has a random mix of parent and new DNA segments.
Explain the results of Meselson and Stahl experiment
They grew initial cells in heavy N15. Then they grew cells (Generation 1) in light N14 for a generation. They found the DNA was all of an intermediate density (one intermediate band).
- This ruled out Conservative model (would expect two bands; one heavy, for parent and one light, for daughters)
Note bands separated by density; light bands at top of tube, heavy at bottom.
They then continued the experiment through a second generation after which both intermediate and low density DNA was observed, supporting semi-conservative over dispersive (dispersive would predict still one band that increases in lightness as the generation number increases because the heavy bases are being dilluted equally among all offspring. In semiconservative, the heavy bases stay together in one strand, so two daughters out of all, no matter how many generations later, will always make that intermediate band).
Imagine it is the year 2050. Because the Martian landers discovered that liquid water was present on Mars, another probe was sent to look for signs of life. Cells containing double-stranded DNA were found, and using an automated variant of the Meselson-Stahl experiment, the original DNA was labeled with 14N and then transferred to 15N (note that this is the reverse of the original M-S experiment). In the low temperatures of Mars, the replication machinery works slower, so each cell division takes 60 hours.
Assume you do not know if replication machinery uses a dispersive, conservative, or semiconservative approach. Predict what the centrifuge tube patterns should be at 60, 120, 180, and 240 hours
Note bands separated by density; lighter at top of tube, heavier at bottom of tube
Dispersive: hours (60,120,180,240) = (thin light band, thicker intermediate band, thicker heavier band, thicker heavier band)
Conservative: hours (60,120,180,240) = (thin light band, thin light band and thin heavy band, thin light band and thicker heavy band, thin light band and thicker heavy band)
Semiconservative: hours (60,120,180,240) = (thin light band, medium-thickness intermediate band, medium-thickness intermediate band and medium-thickness heavy band, medium-thickness intermediate band and thick heavy band)
List the stages of interphase and the major events that take place in each stage.
Three predominant stages are found in interphase of cells active in the cell cycle.
(1) G1 (Gap 1): In this phase, the cell grows and synthesizes proteins necessary for cell division. During G1, the G1/S checkpoint takes place. Once the cell has passed this checkpoint, it is committed to divide.
(2) S phase: During S phase, DNA replication takes place.
(3) G2 (Gap 2): In G2, additional biochemical reactions take place that prepare the cell for mitosis. A major checkpoint in G2 is the G2/M checkpoint. Once the cell has passed this checkpoint, it enters into mitosis.
(4) A fourth stage is frequently found in cells prior to the G1/S checkpoint. Cells may exit the active cell cycle and enter into a nondividing stage called G0.
What are checkpoints? List some of the important checkpoints in the cell cycle.
Checkpoints function to ensure that all the cellular components, such as important proteins and chromosomes, are present and functioning before the cell moves to the next stage of the cell cycle. If components are missing or not functioning, the checkpoint will prevent the cell from moving to the next stage. The checkpoints prevent defective cells from replicating and malfunctioning.
These checkpoints occur throughout the various stages of the cell cycle. Important checkpoints include the G1/S checkpoint, which occurs during G1 prior to the S phase; the G2/M checkpoint, which occurs in G2 prior to mitosis; and the spindle-assembly checkpoint, which occurs during mitosis.
Draw a molecule of DNA undergoing eukaryotic linear replication. On your drawing, identify (a) origin of replication, (b) polarity (5’ and 3’ ends) of all template strands and newly synthesized strands, (c) leading and lagging strands, (d) Okazaki fragments, and (e) location of primers
I can’t insert a pic onto these cards :(
The number of proteins in the prokaryotic replisome is 13, while the number in a eukaryotic cell is 27. What are important protein components of the replisome in general? Why might a eukaryotic cell have more components?
Important components of the replisome are:
polymerase III holoenzyme: responsible for the addition of nucleotides
primase: enzyme that synthesizes the RNA primer
helicase: disrupts the hydrogen bonds that hold the two strands of the double helix together
topoiomerase: relaxes supercoiled DNA by breaking a single DNA strand or both strands, allowing DNA to rotate into a relaxed molecule
single-stranded binding protein: binds to single-stranded DNA and prevents the duplex from reforming.
One reason for the added complexity of the eukaryotic replisome is the higher complexity of the eukaryotic template. Unlike the bacterial chromosome, eukaryotic chromosomes exist in the nucleus as chromatin; thus, the replisome has to not only copy the parental strands but also disassemble the nucleosomes in the parental strands and reassemble them in daughter molecules.
How would DNA replication be affected in a bacterial cell that is lacking DNA
gyrase?
DNA gyrase or topoisomerase II reduces the positive supercoiling or torsional strain that develops ahead of the replication fork due to the unwinding of the double helix. If the topoisomerase activity was lacking, then the torsional strain would continue to increase, making it more difficult to unwind the double helix. Ultimately, the increasing strain would lead to an inhibition of the replication fork movement.
What three mechanisms ensure the accuracy of replication in bacteria?
1) Highly accurate nucleotide selection by the DNA polymerases when pairing bases
(2) The proofreading function of DNA polymerase, which removes incorrectly inserted bases
(3) A mismatch-repair apparatus that repairs mistakes after replication is complete
Consider the following segment of DNA, which is part of a much longer molecule consisting a chromosome:
5’-ATTCGTACGATCGACCTGACTGACAGTC-3’
3’-TAAGCATGCTAGCTGGACTGACTGTCAG-5’
a. Which is the template strand?
b. Draw the molecule when the DNA polymerase is halfway along the segment.
c. Draw the two complete daughter molecules.
d. Is your diagram in part b compatible with bidirectional replication from a single origin, the usual mode of replication?
A. If replication is proceeding such that the DNA on the right is replicated first, then the top strand is the templte for the leading strand
B,C,D. good luck
Do bacteria require telomerase? Why or why not?
No because the DNA is circular and therefore does not have linear ends that would require telomerase.
What is the end-of-chromosome problem for replication? Why, in the absence of telomerase, do the ends of chromosomes get progressively shorter each time the DNA is replicated?
For DNA polymerases to work, they need the presence of a 3’ OH group to which to add a nucleotide. At the ends of the chromosomes when the RNA primer is removed, there is no adjacent 3’ OH group to which to add a nucleotide. Thus, no nucleotides are added, leaving a gap at the end of the chromosome. Telomerase can extend the single-stranded protruding end by pairing with the overhanging 3’ end of the DNA and adding a repeated sequence of nucleotides. In the absence of telomerase, DNA polymerase will be unable to add nucleotides to the end of the strand. After multiple rounds of replication without a functional telomerase, the chromosome will become progressively shorter. Additional research has indicated that in some single-cell eukaryotic species, the RNA primer is directly placed at the 3’ template end of the DNA. However, in human cells it has been demonstrated that the RNA primer is not placed directly at the end of the 3’ end of the chromosome. Typically, it is placed 70 to 100 nucleotides from the end, which will result in these nucleotides not being replicated and the chromosomes decreasing in length even faster because the end is not being replicated.
A number of scientists who study cancer treatment have become interested in telomerase. Why? How might cancer drug therapies that target telomerase work?
Telomerase is an enzyme that functions in cells that undergo continuous cell division and may play a role in the lack of cellular aging in these cells. Cells that lack telomerase exhibit progressive shortening of the chromosomal ends or telomeres. This shortening leads to unstable chromosomes and ultimately to programmed cell death. Many tumor cells also express telomerase, which may assist these cells in becoming immortal. If the telomerase activity in these cancer cells could be inhibited, then cell division might be halted in the cancer cells, thus controlling cancer cell growth. Chemically modified antisense RNAs or DNA oligonucleotides complementary to the telomerase RNA sequence might block the telomerase activity by base pairing with the telomerase RNA, making it unavailable as a template. A second strategy would be to target the DNA synthesis activity of the telomerase protein preventing the telomere DNA from being synthesized.
