Exam 1 Flashcards

(44 cards)

1
Q

Characteristics of hereditary material

A
  1. Localized to the nucleus and a component of chromosomes.
  2. Present in stable form in cells.
  3. Sufficiently complex to contain information needed for structure, function,
    development, and reproduction of an organism.
  4. Able to accurately replicate itself so that daughter cells contain the same
    information as parent cells.
  5. Mutable; undergoing a low rate of mutations that introduces genetic variation
    and serves as a foundation for evolutionary change.
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2
Q

Avery, McLeod, and McCarty Experiment

A

Original observation: Heat killed S type (smooth, nonvirulent) was injected into mice along with live R type (rough, nonvirulent). The R type was transformed into S type and killed the mouse.

In order to test what the transforming factor (and therefore likely hereditary material), Avery et al set up assays that eliminated (through enzymes) for RNA, proteins/polysaccharides, DNA, and lipids. Only the assay that removed DNA from the heat killed S type prevented the transformation of R type, therefore indicating that DNA is the hereditary material

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3
Q

Hershey-Chase Experiment

A

Bacteriophages were grown in mediums that radioactively labelled the DNA and proteins and were then set to infect bacteria. In the final bacterial culture, only the radioactive DNA was found whereas the radioactive proteins remained in the phage ghosts

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4
Q

The 4 nucleotides

A

A, G (purine/2 rings) and T,C (pyrimidine/ 1 ring)

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5
Q

Phosphodiester bonds

A

The 3’ OH group will react with the 5’ phosphate group

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6
Q

How did the Meselson-Stahl experiment prove the semiconservative replication model? How would the results have change for either conservative or dispersive?

A

Heavy parent strand was then 50/50 light heavy, which then separated into two distinct bands

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7
Q

Huberman and Rigg’s Experiment

A

Pulse-chase experiment in which pulse cells would be darkly visible in regions of active replication. In their results, there were dark/pulses regions on either side of each of the origins of replication, indicating that replication was happening bidirectionally

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8
Q

Replication Initiation in Bacteria

A

DnaA binds to 9-mer seq. which forces the AT rich 12-mer seq. to unwind. DnaC will help recruit DnaB(helicase) which opens the bubble for replication

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9
Q

Characteristics of DNA polymerase

A

Only adds nucleotides to the 3’ end
Can not initiate replication de novo (needs RNA primer)
Contains proofreading 3’->5’ exonuclease domain

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10
Q

Replication in bacteria

A

1) DnaB(helicase) and topoisomerase unwind the DNA
2) Protect the ssDNA
- single-stranded binding proteins
3)DnaG(RNA primase/polymerase) begins synthesis of a new strand utilizing RNA nucleotides
4) DNA polymerase III extends the daughter strands from the RNA primers
5) RNA primers must be removed
-DNA polymerase I or RNase H
6) Ligation of Okazaki fragments

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11
Q

Replisome

A

topoisomerase, Helicase, SSB, primase, DNA pol III, DNA pol I, DNA ligase

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12
Q

End problem of linear chromosomes

A

Circular chromosomes can just be nicked and ligated together at the end of replication.
In linear chromosomes, the lagging strand will have a primer on the 5’ end of the new strand that leaves and 3’ overhang after the primer is removed

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13
Q

Telomerase mechanism

A

1) telomerase uses its associated RNA template to synthesize a complement that extends the 3’ overhang. It shifts and repeats this before primase and DNA pol synthesize the complementary strand and fill in the gap

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14
Q

Why is telomerase activity so tightly regulated

A

Telomerase being dysfunctional would result in the eventual shortening of coding portions of chromosomes. Additionally, reactivation of telomerase can lead to ageing cells continually replicating (cancer)

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15
Q

PCR

A
  1. Denature the double-stranded template DNA into two
    single-stranded template strands at 95-100°C
    * This is why we need thermal stable DNA
    polymerases.
  2. Anneal the primers to the denatured template strands
    at 45-68°C (this can vary depending on your primer
    sequence).
  3. Extension (72°C): The Taq DNA polymerase extends
    from the 3’-OH of the primers generating the fragment
    of DNA of interest.
  4. Repeat 30-35 times usually
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16
Q

VNTRs

A

Variable Number Tandem Repeats
-inherited, detected by PCR
-sequencing flanking the repeats are identical in different people

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17
Q

Sanger DNA sequencing

A

DNA pol is ised to replicate new DNA from a ss template. Replication is terminated by a ddNTP which lacks a 3’ OH group which prevents the addition of more phosphodiester bonds

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18
Q

Bacterial Promoters

A

pribnow box, -35 and -10 consensus sequences (the sigma subunits in holoenzymes are specific to these sequences)

19
Q

Why are eukaryotic promoters more diverse and complex?

A

There are multiple transcription factors in eukaryotes, which are polymerase specific and therefore bind to different promoters

20
Q

Bacterial Transcription (intiation)

A

Due to the intrinsic helicase activity of the holoenzyme, it begins to unwind the DNA after promoter recognition to form the open promoter complex.

21
Q

Bacterial Transcription (elongation)

A

RNA polymerase begins transcription and shortly after the sigma subunit disassociates. DNA is unwound ahead of the enzyme to maintain ~18 bp of open DNA

22
Q

Bacterial Transcription (termination)

A

Transcription continues until the RNA pol reaches the termination sequence in which it then dissociates from the DNA. Termination can either be rho dependent or intrinsic

23
Q

Eukaryotic promoters

A

TATA box, CAAT box, GT rich box

24
Q

Eukaryotic termination and how it compares to bacterial

A

The DNA’s CTD has an associated RNase that will cleave ~30 bp downstream from the Poly-A signal. The RNase then goes and eats the rest of the RNA around RNA pol II< facilitating its release

25
Promoter recognition and initiation
In prokaryotes, there is a -35 and a -10 (pribnow box) consensus sequences that the holoenzyme binds to
26
role of CTD to RNA Pol II
Gets phosphorylated for initiation and has an associated RNase that enacts termination
27
3 types of eukaryotic mRNA processing (why does it undergo processing?)
1) intron splicing 2) 5' capping 3) 3' polyadenylation mRNA undergoes processing as it is transcribed in order to increase stability
28
Mechanisms of intron splicing
Group I, Group II, Spliceosome assisted
29
Tricks a cell can use to get more proteins than genes
1. Alternative splicing. 2. Alternative promoter usage. 3. Alternative polyadenylation signal usage.
30
4 levels of protein structure
Primary, Secondary, Tertiary, Quaternary
31
Structure of ribosomes and functional sites
Made up of proteins and rRNA. Consists of two subunits, small and large (60S/40S eukaryotic, 30S/50S prokaryotic). E,P, and A sites + channels for mRNA and emerging polypeptide
32
tRNA structure and function
synthetase, anticodon
33
Prokaryotic translation intitiation
At the 5' end, IF3+mRNA, small subunit binds to form preinitiation complex that scans the mrNA for the Shine-Delgarno sequence which places the AUG (start codon) at the P site The initiator tRNA will base pair with start codon in the P site. IF2-GTP – helps bring in the initiator tRNA. IF1 – binds at the A site and prevents the large subunit from binding. mRNA + small subunit + IF3 + IF2 + IF1 + initiator tRNA = 30S initiation complex. IF2 hydrolyzes GTP – this stimulates the release of IF1, IF2, and IF3. The A site is open and waiting for the next tRNA. This is called the 70S initiation complex. Translation will commence!
34
Eukaryotic translation intiation
1. eIF1 + eIF3 + eIF1a + small subunit = preinitiation complex 2. eIF5 + initiator tRNA (normal methionine) + preinitiation complex = initiation complex 3. Secondary structure is removed from mRNA by eIF4 complex. 4. mRNA binds to initiation complex and it scans mRNA until AUG base pairs with UAC anticodon of initiator tRNA 5. Large subunit binds, IF’s are released, and translation begins
35
Overall translation process (initiation, elongation, termination)
Recruit tRNA to A site. Formation of peptide bond between two AA. Translocate the ribosome down the mRNA in the 3’ direction.
36
Polyribosomes (and how they differ between eukayotes and prokaryotes)
Multiple ribosomes translating an mRNA at once
37
polycistronic mRNAs
Some bacterial genes use one promoter to transcribe for one mRNA that can be translated into multiple genes
38
What does saying "the genetic code is degenerate" mean? What is its benefit?
wobble position, helps to decrease required number of tRNAs
39
Properties of inosine and why it is used in tRNA molecules
Can be substituted for any of the NTPs, used in tRNA molecules as a way to decrease specificity and therfore number of required tRNAs
40
Draw a gene and its mRNA
41
Draw a replication fork
42
Role of enhancer and silencer sequences
Regulates transcription rate Enhancers increase transcription rate by recruiting proteins that promote transcription Silencers decrease transcription levels by recruiting proteins that's inhibit transcription
43
Holoenzyme
RNA polymerase + sigma subunit
44
Eukaryotic Transcription (Initiation)
TFIID binds to the TATA box forming the initial committed complex TFIIA,TFIIB, TFIIF, and RNA pol II join to form the minimal initiation complex TFIIE and TFIIH are added to form the preinitiation complex (PIC) The PIC directs the RNA polymerase to the +1 site where it begins generating mRNA