Final Exam Flashcards
(135 cards)
1
Q
Explain the Griffith experiment.
Define the independent variable, dependent variable and
A
Two strains of bacteria were chosen: rough and smooth strains. Rough bacteria were given such a name because they had a mutation that prevented them from forming a glycocalyx, a slimy protective layer. Because of this, the rough strain was non-virulent. Subsequent injections of the two strains into rat populations showed this. Next, the smooth, viral strain was boiled and when injected, it had no effect. Finally, the boiled smooth strain was mixed with a normal rough strain, and when injected, it killed mice.
Dependent variable: state of mice
Independent variable: smooth, rough, or mixed strain.
2
Q
How did Avery work upon what Griffith had found?
A
Avery knew boiled smooth bacteria mixed with the alive rough strain would have all the macromolecules, and it seemed that something had moved from the dead smooth to the alive rough strain - he wanted to figure out which macromolecule it was.
To do this, he mixed the boiled smooth bacteria with the alive rough under a few circumstances - different enzymes. Using proteases, DNAses, and RNAses, he saw that only when the DNAses were mixed in did the glycocalyx NOT form.
3
Q
Explain the Hershey/Chase experiment.
A
Bacteriophages were used to infect an E Coli host. These viruses contain, generally, have DNA to give instructions to their host as well as a protein coat. In two separate groups, an isotope for either phosphate or sulfur was given to the phages. Marking was to see whether the phages passed DNA or protein. When the phages were put into a medium containing E Coli that did not have any isotopes, and that had been scraped, it was clear that the original phages progeny only contained the isotopes in those that had their DNA marked.
4
Q
How are chromosomes numbered?
Why would we never see a trisomy 1?
A
They are numbered, for the most part, by size.
If there was an extra first chromosome, it would be much too big for the organism to survive.
5
Q
What was the impact of Rosalind Franklin on DNA elucidation?
A
She was an expert in x-ray diffraction, which shoots an ray through a pure DNA extract, and the results shows the general structure.
She deduced that it was symmetric and that it had major and minor grooves, meaning it must be helical.
6
Q
How did Watson and Crick expand on her work?
A
Using the information from Franklin, they new that for the structure to be possible, it must have upside down strands.
7
Q
Explain the structure of Histones and what they function to do.
A
They complex proteins that have a core of eight subunits that have positively charged groups that interact with the phosphates of DNA, and another subunit that has non-polar groups that act to aggregate multiple histone molecules.
2x H2A
2x H2B
2x H3
2x H4
H1
They act to pack up DNA and organize it.
8
Q
What is the difference between a nucleosome and chromatin?
A
A nucleosome is the bead of wrapped DNA and associated proteins that forms when histones aggregate.
Chromatin is the more broad structure of the DNA, where nucleosomes are the units.
9
Q
Explain the role of scaffolding proteins in coordination with histones for DNA packing.
A
Scaffolding proteins are acidic by nature and they act to anchor the compacting chromatin creating loops that eventually become compact enough to become a chromatid.
10
Q
What stage of the cell cycle are we in once scaffolding (not maximal compaction though) has taken place?
Why does it make sense for it to be this phase?
A
G1
A newly duplicated cell, directly after cytokinesis, has DNA that is decondensed but not completely loose. To effectively read the DNA during G1 the DNA must be continue to be scaffolded.
11
Q
Take for example the 11 genes that work together to encode eye color.
Are these genes localized/organized?
A
No, the genes that encode eye color are spread out throughout the genome.
12
Q
What are telomeric sequences?
A
They are repeated sequences at the ends of chromosomes that are used to protect genes or other information from being deleted.
13
Q
What are SINES?
A
SINES are short for “short interspersed nuclear elements” and they are about 400 base pairs long
14
Q
What are LINES?
A
LINES are short for “long interspersed nuclear elements” and they are about 10,000 base pairs long.
15
Q
What is junk DNA / ambiguous DNA?
A
It is the DNA that doesn’t code for anything, including SINES and LINES.
16
Q
How did Chargaff influence what we know about DNA?
A
He took DNA from many different organisms and noticed that there were always the same amount of purines to pyrimidines.
17
Q
For double stranded DNA, which of the following base ratios always equals 1?
A. (A+T)/(G+C)
B. (A+G)/ (C+T)
C. C/G
D. A/G
A
B and C
18
Q
Explain the Meselson Stahl experiment and what was figured out.
A
They had three models for how DNA replicated: conservative, semiconservative, and dispersive. To find out which was correct, they took E Coli and grew it in a medium that contained an isotope of Nitrogen (N-15).
Once the E Coli was grown, it was transferred to another medium that only contained normal Nitrogen. If the conservative model was correct, after one replication, they should have saw two distinct bands. Instead they saw a single clump, meaning it was either the semiconservative model or the dispersive model.
To distinguish between the two models, a second replication in the clean medium was allowed to happen.
If the semiconservative model was correct, they knew they would see two distinct bands: a lighter band from the full N-14 strands that are now forming and a denser band that still contained the isotope. If the dispersive model was correct, they knew they would see a single band that got progressively more and more light.
It was determined that the semiconservative model was correct.
19
Q
What are ARSs?
What do they do?
A
ARS stands for Autonomously Replicating Sequences.
They are specific sequences on the DNA that act as binding sites for ORCs
20
Q
What are ORCs and what do they do?
A
ORC stands for Origin of Replication Complexes.
They are enzymes that are made en masse during G1 and they run around the nucleus looking for binding sites that fit their active site.
21
Q
DNA A is the first enzyme involved in replication in prokaryotes.
What does it do?
A
DNA A binds directly to the prokaryotic DNA and causes a sequence near to it to become kinked.
22
Q
What is DNA B and C called?
What do they do?
A
They are separate parts of what is commonly called DNA helicase.
They break H bonds between the kinked region.
23
Q
What proteins are responsible for keeping the replication bubbles open?
A
single-stranded binding proteins.
24
Q
After the replication bubbles have formed, how does Primase (modified RNA Polymerase) bind to start making RNA primers?
A
DNA helicase (DNA-C/B) recruit Primase to bind.
Primase then makes short RNA primers that provide a 3’ OH group that is needed for DNA Poly to work.
25
What are the three things that DNA Polymerase can do?
1. Adds nucleotides in the 5' to 3' direction (5' end of incoming nucleotide is attached to 3' end of strand)
5' to 3' Polymerase activity
2. Proofreads as it goes along. (3' to 5' exonuclease activity)
3. Removes primers in the 5' to 3' direction as it goes along.
26
Why is it sort of misleading when we learned about there being a leading and lagging strand?
Because there is always more than just one replication fork given that we form replication bubbles.
Because of this, we know that there are both leading strands and lagging strands on a single strand of DNA.
27
What specifically allows DNA Polymerase to remove primers with relative ease?
On the second carbon of RNA nucleotides in a primer, there is a hydroxyl that is not present in DNA nucleotides.
This makes it very prone to nucleophilic attack and unstable.
28
On a template strand that is 3' to 5' direction DNA Polymerase removes RNA primers and runs in to the back of the DNA.
What is left?
There are small nicks between the 3' end of the segment that has been added in place of the primer and the 5' end of the leading strand that came from the original primer.
29
Describe the End Replication Problem.
At 3' end of a template strand, a primer will exist at the very end to provide a 3' OH group, whereas at the 5' end of the same template strand, DNA nucleotides are added all the way to the end.
This primer at the 3' end of the template strand will eventually need to be removed leaving an overhang, risking loss of genetic information.
3'__________________________5'
5'|||3' 5'|||||||||||
(primer for
leading strand)
30
How do some cells solve the end replication problem?
Using an enzyme called telomerase, which contains an RNA sequence AAUCCC within the enzyme will bind to the 3' end of the template strand, extending it.
Then, primase is recruited to add a primer to this extended template strand, providing a 3' OH group to grow the smaller strand.
Eventually, the last primer is removed, and the overhang is cut, preserving the DNA near the ends.
31
What does degradation of the ends of chromosomes trigger?
Semiscence - state of a cell, in which it is not completely dead, but not doing a whole lot.
32
What is a common hallmark of cancer in regards to telomerase activity?
Increased telomerase activity is seen.
33
How many replication cycles can one cell typically go through?
60
34
What causes aging?
A lack of telomerase activity.
35
What would the DNA sequence be at the end of a template strand?
What is this called?
Because the telomerase enzyme contains an RNA sequence AAUCCC, the corresponding DNA sequence would have to be TTAGGG.
These are called telomeric regions.
36
What are mRNAs?
How big are they?
Messenger RNA - the transcript made of RNA responsible for carrying the code for the order of amino acids in translation.
The size of mRNA is between 1500-8000 bases, but usually 3000
37
What are tRNAs?
How big are they?
What do they look like?
Transfer RNA - escorts amino acids to the ribosome complex to convert mRNA to a polypeptide.
The size is 91 bases, and they fold on themselves forming clover looking things.
38
What are rRNAs?
How big are they?
Ribosomal RNA - structurally important for the ribosome, connecting all the protein complexes within.
No size was given as there are many subunits of rRNA, but we do know from the first exam that it is generally smaller than mRNA.
39
What are snRNAs?
Small nuclear RNAs.
They interact with proteins to form snURPs (small nUclear Rna and Proteins).
These complexes play a role in RNA processing and are a few hundred bases long.
40
What are miRNAs?
How big are they?
Micro interference RNAs play a role in gene regulation.
They act to cut up mRNA and degrade it by forming RISC complexes.
They are 21 bases long.
41
What is the core promoter?
The core promoter is a region of DNA sequence near to the gene that includes the TSS (transcription start site) the lnr (initiator) and the TATAAA box in order of origin from the gene.
42
What is the TATAA box?
Region of DNA sequence that is upstream from the TSS by around +30.
Acts directly to promote transcription (you need TATAA).
43
What is the CATT box?
Region of DNA outside the core promoter that is around -75 bases upstream of the TSS.
It will also help to promote transcription, but not as much.
44
What is the GC box?
Region of DNA outside the core promoter that is around +90 base pairs upstream of the TSS.
It will also help to promote transcription, but even less than the CATT box.
45
Explain initiation of transcription.
TFIID bounces around the nucleus until it binds to TATA box, flexing the protein. This opens a binding place for other proteins involved in initiation to bind to TFIID - TFIIA and TFIIB.
TFIIA and TFIIB act to create a binding place for TFIIF which carries along with it RNA Polymerase II.
Then, other proteins TFIIE and TFIIH bind tightly to the whole complex to stabilize RNA Polymerase II, completing the initiation complex.
46
Explain elongation in transcription.
RNA Polymerase II has its own helicase activity, meaning it can unwind the DNA as it chugs along. RNA strands are created and drag behind. The reason the RNA does not stick to the DNA is because it is slightly bigger and can't hold on with the turns.
This continues until termination is initiated.
47
What are the three things that RNA Polymerase II can do?
1. Helicase activity
2. 5' to 3' Polymerase activity
3. Like Primase, it does not require a 3' OH group to start adding.
48
What are the two things that we talked about that RNA Polymerase II cannot do?
1. No proofreading (3' to 5' exonuclease)
2. No primer removal (5' to 3' exonuclease)
49
Can initiation and elongation happen simultaneously?
Yes.
50
Explain how a gene that is Rho dependent terminates.
As RNA Poly II approaches the termination site on the DNA, Rho protein binds to a site on the growing RNA strand called the Rut site past the stop codon. Rho then travels towards the RNA Polymerase (backwards) and when it reaches it, there is no room and RNA Polymerase breaks off.
51
Explain how Rho independent termination works.
At the end of transcribed RNA sequences that terminate using Rho independent methods, there exists special sequences that aid in the process. First, repeated GC bases, followed by TA sequences.
When the RNA Polymerase makes the repeat GC sequences, it dissociates forming a hairpin structure due to hydrogen bonding.
After the GC repeats, there are only TA repeats that easily break away from the strand, terminating transcription.
52
What must happen to the mRNA before it may leave the nucleus?
If any of the three processing steps fail, the RNA does not undergo the necessary structural changes.
Enzymes may remain attached and the mRNA will not be able to pass through the small nuclear pores.
53
What is 5' capping and what does it accomplish?
5' capping is the addition of an upside down guanosine triphosphate to the 5' end of the transcript.
This GTP can then be methylated, protecting the mRNA from degradation as well as helping to align the mRNA on a ribosome for translation.
Kind of acts like a shoestring cover.
54
What is intron removal (exon splicing)?
How does it work?
Intron removal, generally speaking, is the removal of non-coding sequences (introns) from the mRNA sequence.
As mentioned previously, snRNAs bound to proteins are vital to this process. Each intron has a GU sequence at the 5' splice site. Within the intron there is also what is called the A branch site. Additionally, each intron also has a AG sequence at the 3' splice site.
During splicing, U1 snURP will bind to the GU sequence at the 5' end of the intron. Then, U2 snURP will bind to the A branch site, bending the mRNA so that the splice sites are closer together. After this, more proteins apart of the spliceosome, U4, U5, U6 come in and stabilize the entire complex, allowing for the catalyzation of the two exons.
Ribozymes that are within the spliceosome catalyze the process.
55
What is alternative splicing?
When an unprocessed mRNA is spliced in different ways, it allows for multiple proteins to be made from the same gene.
56
What is polyadenylation?
How does it work?
What does it accomplish?
Polyadenylation is the process of adding Adenosine (200 or so) to the mRNA transcript.
A specific sequence near the 3' end of the mRNA allows PolyA Protein to cut the strand and add the As.
57
RNA has a pretty short half-life.
Why is this adventageous?
1. It controls how much protein we get
2. The As protect the gene within the transcript (buys us time)
58
Which side of a tRNA molecule acts as the "acceptor stem"?
The 3' end of the clover looking thing.
59
What is the difference between a codon and an anticodon?
Codons are present within the mRNA strand while anticodons are present on the tRNA.
60
Explain the wobble hypothesis.
The hypothesis explains how it could be that multiple codons with varying 3rd bases can code for the same amino acid.
The idea is that a single charged tRNA molecule can bind to a few different codons because less precise base pairs form between the 3rd base of the codon and the 1st base of the anticodon.
61
What are the various stop codons?
UAA UAG UGA
62
What is the ORF?
The open reading frame is the area between the start codon and the stop codon.
63
How does a tRNA get charged?
How is it ensured the right amino acid gets placed on the right tRNA?
An enzyme called amino acetyl tRNA synthetase carries out this job.
The enzyme, which is specific to the different tRNAs and amino acids, will bind to a free amino acid. Using the process of ATP hydrolysis, the enzyme attaches the amino acid to the 3' acceptor stem.
Specificity and accuracy are ensured by the fact that there are different tRNAs for different amino acids, and there are even different amino acetyl tRNA synthetases.
64
Explain initiation of translation.
Various proteins are involved in setting up translation on a free ribosome.
First, eIF-2 binds to a specific charged tRNA (methionine) that contains the anticodon complementary to the start codon (AUG). eIF-3 then binds to the 40S ribosomal subunit, and eIF-4 binds to the 5' cap of the mRNA.
These complexes converge, with the tRNA anticodon (UAC) pairing with the start codon (AUG) on the mRNA. The binding of eIF-4 to the 5' cap ensures the proper alignment of the mRNA, facilitating its correct positioning. eIF-2 also helps slide the tRNA along the mRNA until the start anticodon matches the start codon.
Once this alignment occurs, the 60S ribosomal subunit, with eIF-5 attached, binds to the complex. The binding of 60S displaces eIF-2 and eIF-3, allowing the 60S and 40S subunits to join together, forming a functional ribosome. eIF-5 then dissociates, leaving the mature ribosome with the mRNA in the proper orientation for translation to begin.
A final protein, eIF1, helps to dislodge all the eIFs except eIF4.
65
Explain elongation of translation for eukaryotic mRNA.
During elongation, the poly(A) tail of the mRNA, bound by poly(A)-binding protein (PABP), interacts with the ribosome to stabilize the complex. Then an enzyme called EF-Tu binds to a charged tRNA; the tRNA complex then moves through the cytoplasm until it enters the A-site of the ribosome. If the anticodon of the tRNA matches the codon on the mRNA the tRNA remains in the A-site, delivering the appropriate amino acid.
Next, the ribosome's peptidyl transferase activity facilitates the formation of a peptide bond between the amino acid in the A-site and the growing peptide chain in the P-site. This breaks the bond between Met and its original tRNA.
Following this, EF-G promotes the movement of the ribosome along the mRNA. The tRNA in the P-site, now carrying the growing peptide chain, shifts to the E-site, where it is released. The tRNA in the A-site, carrying the Met-Pro chain, moves to the P-site, leaving the A-site open for the next tRNA. The ribosome slides over the mRNA, exposing the next codon in the A-site.
This cycle repeats until a stop codon is reached.
66
What site of the ribosome will accept the very first tRNA containing Met?
The P site
67
Is it possible for there to be many different ribosomal complexes to be working on the same strand at a time?
Yes, but they all start at the same place.
If this were happening, we are trying to make a good amount of protein from this mRNA
68
Explain termination of translation of eukaryotic mRNA.
Once the stop codon enters the A-site of the ribosome, termination of translation is triggered. A release factor (RF1) binds to the ribosome complex at the A-site, recognizing the stop codon.
This allows RF3 to bind, facilitating the disassembly process. The large (60S) and small (40S) ribosomal subunits dissociate from the mRNA strand, marking the end of translation. The final tRNA, which remains bound to the newly synthesized polypeptide chain, is released as well.
69
Explain protein trafficking.
All proteins made and released into the cytoplasm. Each proteins may contain a signal peptide - 15 to 30 AAs starting with Met that tell the polypeptide where to go.
If a protein has no signal peptide, it will float in the cytoplasm and stay there. If a protein does have a signal peptide, Signal Recognition particle proteins (SRP proteins) can bind to the signal peptides of the newly made proteins and take them to docking proteins on the endoplasmic reticulum.
When the SRP bind on docking proteins, the protein is fed into the ER and the signal peptide is cut off. A cis vesicle forms around the protein and buds off from the ER. Then, walking proteins take the cis vesicle to the golgi, fusing it to the golgi, and releases the inside proteins inside.
Here, protein modification takes place. From the golgi, modified proteins bud off into trans vesicles. Trans vesicles can be taken to the cell membrane to do 2 things: if protein is hydrophilic, will be released from the membrane into the bloodstream or if protein is hydrophobic w hydrophilic arms, will remain in the membrane.
70
What effects can mutations have on the protein product?
1. There is no effect
2. There is an abnormal function or partial function
3. No function
4. No protein product, period
71
Explain the Luria/Delbruck Experiment.
What was the purpose?
What was the conclusion?
The purpose of the experiment was to discern whether variation results from random mutations that happen to be adaptive OR mutations and variation result from an adaptation to an environment.
The idea of the experiment was to grow multiple small colonies of E Coli (wild type) and introduce a T1 bacteriophage to each. If mutations happen as a result to the introduction of the phage, then it would be expected that the number of survivors would have little variation.
Because they observed a large fluctuation in the number of resistant colonies, they knew that mutation must be random.
72
What is a transition mutation?
A type of point mutation that results in a "transition" from one purine-pyrimidine base pair to another. (A-T to G-C)
The key here is that a purine is replaced by a another purine and a pyrimidine is replaced with another pyrimidine.
73
When is the only time that an actual mutation (changing one or more base pair) takes place?
In the S phase of the cell cycle.
74
What is a transversion mutation?
A type of point mutation that results in a "transversion" from one purine-pyrimidine base pair to a pyrimidine-purine base pair.
(C-G) to (G-C)
The key here is that a pyrimidine is replaced with purine and a purine is replaced by a pyrimidine.
75
What is a missense mutation?
A gene mutation that happens when a base-pair change causes a change in an mRNA codon so that a different amino acid is inserted.
Key is that this mutation causes an amino acid to change.
76
What type of mutation causes sickle-cell anemia?
A point mutation that is further specified as a missense mutation.
77
What is a nonsense mutation?
A gene mutation that results when a base pair change alters a codon of the mRNA to a stop codon.
Key is that a codon that once coded for an amino acid is now a stop codon.
UGA UAG UAA
78
Why are certain people tolerant to lactose?
There is a mutation in the regulatory region of the gene for lactase that keeps the protein product producing into adulthood.
79
What causes cystic fibrosis?
A single gene trait change in the gene for CFTR, which normally keeps the mucous liquidy. Can be caused by a variety of mutations but the two mentioned were nonsense mutations and frameshift mutations.
Must be homozygous!
80
What is a neutral mutation?
A type of missense mutation, that results when a codon in the mRNA is changed to a codon that codes for a very similar amino acid.
81
What is a silent mutation?
A type of mutation that results when a change in a codon specifies the same amino acid as prior.
*This is the exact opposite of what a missense mutation is*
82
When do silent mutations most often occur?
They usually occur when there is a base change in the third base of a codon - the wobble hypothesis.
83
What is a frameshift mutation?
A mutation that results when either an addition or deletion of a base pair causes a shift in the reading frame of the mRNA.
Causes incorrect amino acids to be added after the mutation site or a stop codon to pop up earlier.
84
What is Taysach's disease?
What does it result from?
A disease that occurs when there is a non-functional HEX-A enzyme, resulting in death of neurons etc.
It is caused by a deletion in the coding region of the HEX-A gene causing a frameshift mutation, leading to a truncated/nonfunctional enzyme.
85
One frameshift mutation confers resistance to HIV.
How does this work?
There is a deletion or insertion into the coding region for CCR5 gene, making the protein product non-functional - HIV cannot infect cells.
86
What is a suppressor mutation?
A mutation that occurs a different base pair than an original mutation which acts to mask the original effects of the initial mutation.
87
Explain an example of two suppressor mutations.
1. A mutation occurs that changes a codon in the middle of the reading frame from UAC to UAA stop codon, preventing the whole mRNA from being read.
Another mutation occurs on the gene for a tRNA that gives it the anti codon AUU (normally tRNAs that match stop codons do not exist)
2. A mutation occurs changing one amino acid important for correct folding from hydrophobic to hydrophilic. Now the protein cannot fold correctly. Another mutation occurs changing a second amino acid important into folding from hydrophobic to hydrophilic. Now the protein can fold how it did before.
88
What is the difference between a mutagen and a carcinogen?
A mutagen increases the mutation rate within our cells, while a carcinogen is something that has been shown to contribute to cancer.
Not all mutagens are carcinogens and not all carcinogens are mutagens.
89
If your DNA is to suffer a base modification, which stage of the cell cycle makes it least likely to become a permanent mutation?
G2
90
What type of mutagen causes the mutations we see from UV radiation?
Non-ionizing radiation.
91
What types of mutations result from UV (non-ionizing) radiation?
A variety of mutation can happen when thymine dimers form.
Could skip addition of Adenine to next replication cycle resulting in a frameshift if not fixed.
Could result in a transversion mutation from a (T-A) to a (G-C)
Pyrimidine went to a purine and purine went to pyrimidine.
92
How is ionizing radiation different than non-ionizing radiation?
Ionizing radiation sends particles that displace electrons, leading to the formation of free radicals.
More than just thymine dimers can now form: guanine to 8-oxoguanine, strand breaks (double or single), and cross linking between different parts of the DNA or proteins.
93
What are base analogues?
They are compounds that resemble normal nitrogenous bases, but they are non-shapeshifting, meaning that they are unable to structurally change to fit their desired base pair (T-A), so they are often prone to pairing with the wrong one.
An example is when a base analogue for thymine often makes pairs with guanine.
94
What is 5-bromo-uracil a base analogue to?
What process allows it to behave more like cytosine?
Thymine
The process of tautomerization causes the analogue to transition from its keto form (where it really acts like thymine) to its enol form where it resembles cytosine.
95
5-bromo-uracil is in place of Thymine forming the base pair (5-bromo-uracil - A).
During the S-phase, it transitions from its keto form to its enol form, causing the analogue to pair with guanine. In the next replication the guanine will pair with cytosine, causing a (T-A) to (C-G) mutation.
What type of mutation is this?
This is a transition mutation.
We see that the base analogue for thymine was replaced with another pyrimidine and the adenine was replaced with another purine.
96
What is a base modifying agent?
A chemical that acts as a mutagen by modifying the chemical structure/properties of bases.
97
Nitrous acid is a base modifying agent that deaminates cytosine, making it behave like uracil.
What type of mutation would occur given the base pair C-G (modified)?
The agent would cause the cytosine to behave like uracil causing the first replication to result in the modified base pairing with adenine.
A second replication would result in the adenine pairing with thymine, causing an overall (C-G) to (T-A) transition mutation.
98
What is an intercalating agent?
They are molecules that insert themselves between adjacent bases on one or more strands of the DNA.
99
What type of mutations are most commonly caused by intercalating agents?
The most common are insertions or deletions that cause frameshift mutations.
100
What does the Ames test aim to accomplish?
Determine if something is mutagenic.
101
Explain the basis for how the Ames test is performed.
Wild-type Salmonella has the ability to produce Histidine (His+). Using a lab technique called site directed mutagenesis, the gene responsible for Histidine production in wild-type Salmonella, Bill, is mutated. These modified Salmonella are unable to make Histidine (His-).
Modified (His-) Salmonella are grown in a test tube by adding Histidine and are transferred to three different agar plates containing no Histidine: a plate that contains the suspected mutagen (independent variable), a plate that contains a known non-mutagenic species (negative control), and a plate with a known mutagen (positive control).
The plates are then incubated and growth is observed. The presence of many colonies tells us that there were so many mutations that the mutated Bill gene was turned back on in a single colony, which thrived in the presence of no Histidine. We expect that the positive control will have growth and the negative control with show little to no growth.
Relevant amounts of growth in the plate with the suspected mutagen, would tell us that it is mutagenic.
102
How could you perform an Ames test keeping in mind that some compounds need to become metabolized to be ultimate mutagens.
You would perform the same exact steps as before, but you would also introduce S9 extract to the plates.
S9 extract is ground up rat liver that has the ability to break down the suspected mutagen into intermediates.
103
What is base excision repair?
How does it work?
A repair mechanism that is responsible for recognizing and removing damaged bases like oxo-guanine or a methyl on a guanine.
Process starts with DNA glycosylase, which scans the DNA looking for damaged bases. It then flips out the damaged base, cleaving the glycosidic bond leaving an AP site.
AP endonuclease then comes along and cuts the DNA backbone at the 5' of the AP site, leaving a 3' OH group of the adjacent sugar.
Another protein Phosphodiesterase then cuts the 3' position of the AP site's backbone leaving a 5' phosphate group of the other sugar.
DNA Polymerase then fixes the spot, using the undamaged as a template.
DNA Ligase seals up the bonds.
104
What is nucleotide excision repair?
How does it work?
A repair mechanism that is responsible for removing bulky lesions like thymine dimers or large adducts.
Process starts with UVr AB complex that scans the DNA and finds damage.
Once damage is detected, the UVr A dimer is released, and UVr C binds, cutting out a few dozen nucleotides that are next to the damage. UVr C cuts the 5', UVr B cuts the 3'.
Another protein, UVr D then binds to the DNA and unwinds the region, dislodging the messed up strand using its helicase activity.
DNA Polymerase fixes the damage, DNA ligase seals it up.
105
Xeroderma Pigmentosum is a genetic disease in which one or more of the genes involved in the NER system are mutated.
Propose a mutation that might happen to a particular gene in this disease.
There could be a nonsense mutation in the gene that codes for UVr A.
106
What is methyl-directed mismatch repair?
How does it work?
A repair mechanism that is used to fix errors made by DNA Polymerase (no other issues with the bases, they are just wrong) by reading the methylation state of a nearby sequence.
To start the process, Mut S binds to the mismatch (A-C)
Nearby, at a sequence GATC (palindrome) a sequence of proteins Mut H and Mut L bind to the sequence that is not yet methylated (it is newly made). Then, a new complex is formed between Mut H Mut L and Mut S, bringing the mismatch closer to the new strand.
Then, Mut H nicks the unmethylated strand and endonucleases start cleaving nucleotides all the way to the mismatch.
DNA Polymerase then comes in and fills in the gap and DNA Ligase seals it.
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What are class I transposable elements?
They are retrotransposons (LINES) that contain a gene called transposase. They also contain terminal inverted repeats that flank the gene.
The gene is first transcribed into mRNA, and then a reverse transcriptase can reverse transcribe the RNA into DNA that can be inserted into the genome.
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What are class II transposable elements?
These are transposable elements, often called "jumping genes" that also have a gene called transposase and sequences that flank the gene called inverted repeats.
The gene for transposase becomes its protein product, allowing the enzyme to cut at the inverted repeats, freeing the gene itself.
This element can then insert itself into the genome at a different place and interrupt important genes.
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What is SCID?
How is it related to class I transposable elements?
Severe Combined Immunodeficiency
Class I transposable elements may contribute to the malfunction of the genes required in T-cell production.
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What genes have evolved with the help of transposons?
Genes for antibodies.
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What is the difference between an enhancer and an activator?
What do they do?
Enhancer is a DNA sequence, whereas activators are proteins that bind to enhancer regions.
They both act to upregulate the transcription of a gene.
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What is the difference between a silencer and a repressor?
What do they do?
A silencer is a DNA sequence, whereas a repressor is a protein that binds to silencer regions.
They both act to downregulate the transcription of a gene.
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What is a mediator?
A protein that bind to either activators or repressors to mediate the action of the proteins on the GTFs.
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Which of these proteins is not needed for transcription to occur?
A. TFII D
B. RNA Polymerase
C. Mediator protein
D. They are all required
C.
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What is an example of how transcription is regulated by something that enters the cell. (Galactose metabolism)
In the absence of Galactose, a protein called Gal 80p binds to an activation domain of the activator of the genes required for the metabolism of galactose, called Gal 4p. Binding covers up the activation domain blocking transcription.
Secondary active transport allows galactose uptake in the cell; hydrogen ions are pumped out of the cell and galactose moves against its gradient. An enzyme called Gal 3p then converts galactose into an inducer, which can then bind Gal 80p, causing a conformational change, revealing the activation domain on Gal-4p.
Transcription can now be activated.
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What is another example of how transcription can be regulated from something entering into the cell?
(steroid hormone)
A steroid hormone (glucocorticoid) can bind to its receptor within the cell, changing its conformation by displacing HSP 90 on the receptor, allowing the receptor to enter the nucleus.
In the nucleus, the receptor-hormone complex can directly bind to DNA and act as a activator or repressor.
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How can transcription be regulated by something that does not enter the cell?
(Signal Transduction Pathway)
Something may bind to a receptor that triggers a cascade of events that ends with the activation of a Protein Kinase A via cAMP.
This can then go phosphorylate activators or repressors to either turn them on or turn them off.
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How can transcriptional control via signal transduction pathways play a role in sex determination?
The Y chromosome carries the SRY gene that encodes the TDF. The TDF along with another protein called SF1 act as activators for the SOX gene, producing the SOX protein.
The SOX protein can then act as an activator for the sex coordination gene. The sex coor. protein product has a signal sequence that directs them to the ER, then the Golgi, eventually into the bloodstream.
They can then travel to gonad cells where they bind to receptors initiating signal transduction.
They then bind to androgen receptor and migrate to the nucleus to upregulate genes associated with bone mass/density and hair growth. They also act to downregulate genes associated with estrogen.
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How does a DNAse I sensitivity assay work?
What does it aim to accomplish?
Aims to see which genes are open and which ones are condensed.
You isolate intact DNA from two types of cells of the same DNA. You then expose the DNA to DNAse I which will only chew up DNA that is in the euchromatin form (loose).
Now we run the fragments in a gel, and denature the fragments after. Then we will complete a southern blot by transferring the DNA to a nylon membrane and hitting it with UV.
Then, because we know the some sequence of the gene, we create probes with a fluorescent tag.
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How do mediators actually allow for transcription by making TATA box available?
They recruit HATs which add acetyl groups to histones (H3).
Acetyl groups are negatively charged and so they compete with DNA resulting in a more open chromatin.
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What do HDACs do?
What do HATs do?
HDACs take the acetyl groups away making the localized region heterochromatin.
HATs add acetyl groups, making the localized region euchromatin.
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What gets methylated?
What does methylation do?
The actual DNA gets methylated.
Methylated areas attract HDACs
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Why is methylation inherited but HATs are not?
HATs are easier to remove.
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What is Fragile X Mental Retardation Syndrome?
Affects males because we only have one X chromosome.
Individuals have many more CpG island repeats (areas with high amounts of methylation) than normal folks.
These recruit HDACs which remove acetyl groups from the chromatin closing it up (heterochromatin)
This results in the inability to access the gene Fragile X mental retardation protein I, which is a very important protein in neuronal development.
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What type of control mechanism is imprinting?
Describe imprinting.
Imprinting is a transcriptional control mechanism where EZH2, stabilized by a long noncoding RNA (lncRNA), binds to hemimethylated DNA and methylates the non-methylated strand. This results in genomic imprinting, where methylated DNA attracts HDACs, causing the chromatin to become compact and less readable.
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What is Prader Willi Syndrome?
Prader-Willi syndrome is disease linked to DNA imprinting. It involves a lack of hypothalamus development because the maternal ICE gene is methylated and inactive, while the paternal ICE gene is active.
This means the ICE gene can only be inherited from the father. An lncRNA normally stabilizes an activator protein that binds to an enhancer to promote thalamus development.
In Prader-Willi syndrome, the chromosome containing the lncRNA is deleted on the maternal side, preventing activation and leading to the disorder.
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Explain X inactivation.
X-inactivation is a form of transcriptional control in females, where one of the two X chromosomes is randomly chosen to remain active in each cell. The X-inactivation process is regulated by the XCE sequence, which controls the genes XIST and TSIX. These genes produce lncRNAs that interact with the X-inactivation center (XIC).
If the XCE is unmethylated, both XIST and TSIX produce lncRNAs that bind together, preventing proper interaction with the XIC. This allows the X chromosome to stay active.
If the XCE is methylated, only XIST produces an lncRNA, which interacts with the XIC and triggers X chromosome inactivation.
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Explain Hutchinson Gilford.
In Hutchinson-Gilford Progeria Syndrome (HGPS), a mutation in the LMNA gene activates a cryptic splice site, leading to the inclusion of an abnormal exon that mimics an intron. This causes the production of progerin, a defective form of the lamin A protein, which disrupts the nuclear envelope and results in premature aging symptoms.
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Explain IPEX disease.
A mutation in DNA leads to an error in RNA, resulting in improper polyadenylation and the absence of poly-A tails. Without poly-A tails, the mRNA degrades quickly and cannot be translated into protein. This affects the FoxP3 gene, which is crucial for the development of T regulatory (Treg) cells.
Treg cells are responsible for regulating the Major Histocompatibility Complex (MHC), which plays a role in inflammation and killing foreign invaders. Without proper FoxP3 expression, MHC regulation fails, and the MHC remains active, leading to an uncontrolled immune response.
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What are the types of processing control?
Alternative splicing
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Describe alternative splicing.
When a gene is transcribed into RNA, it undergoes alternative splicing. During this process, introns are removed, and exons may either remain in the final mRNA strand or be excluded. The levels at which certain exons are retained or removed can lead to significant variation in the resulting protein.
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What are the types of post-transcriptional control?
eIF2 system
miRNA
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Explain the eIF2 system.
When a cell experiences stress from viral infection, fever, nutrient deprivation, etc., the eIF-α kinase is activated. This kinase phosphorylates eIF-α, which then binds to eIF-2β, inactivating it. As a result, translation cannot begin because eIF-2, which carries the initiator methionine to the ribosome, is disabled.
This response is beneficial during viral infections. When viruses inject their DNA into cells, the resulting stress activates the eIF-α pathway, inhibiting viral protein production.
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Explain how miRNA is a post-transcriptional regulator.
In our mRNA, there are short stretches of microRNA (miRNA) genes between protein-coding regions. These miRNAs are transcribed by RNA Polymerase II.
After transcription, miRNAs receive a poly-A tail and a G-cap. In the nucleus, a protein called Drosha modifies the miRNA by removing the poly-A tail and G-cap, allowing the miRNA to exit the nucleus.
Once in the cytoplasm, the Dicer protein trims the miRNA at both ends, leaving 3' overhangs. This structure then attracts an Argonaute protein, which processes the miRNA into a single strand. The resulting RNA-induced silencing complex (RISC) binds to complementary mRNA sequences and degrades them, preventing translation into protein.
In Alzheimer's disease, there is a problem with this mRNA degradation process. The enzyme BACE-1, which is stabilized by long non-coding RNAs (lncRNAs), generates beta-amyloid proteins. THe miRNA is non functional and cannot degrade the excess BACE-1.
The accumulation of beta-amyloid proteins increases the likelihood of misfolding. Misfolded beta-amyloid proteins accumulate in brain tissue, forming plaques that contribute to Alzheimer's pathology.
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