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H3 Molecular Biology > Chapter 4-Nucleic Acids And Protein Methods > Flashcards

Chapter 4-Nucleic Acids And Protein Methods Flashcards

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1
Q

Function of the Y2H system

A

To test the interaction between 2 proteins through the downstream activation of a reporter gene

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2
Q

Advantages of Y2H system

A
  1. Relatively fast and easy method to screen for protein-protein interactions
  2. Requires little hands-on time and technical skill
  3. Can be scaled up by screening yeast libraries of tagged prey proteins against a single bait, allowing thousands of potential interactions to be screened rapidly
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3
Q

Limitations of Y2H system

A
  1. High rates of false positives and false negatives
  2. Tagged proteins may not fold correctly and may not bind to their targets
  3. Interaction must occur in the nucleus for the reporter gene to be activated
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4
Q

Function of Y1H system

A

Tests for protein-DNA interaction

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5
Q

In the Y1H system, which transcriptional domain is the protein bound to?

A

Activating domain

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6
Q

Advantages of Y1H system

A
  1. Able to detect protein-DNA interactions that are not actually activating transcription e.g. DNA repair proteins and repressor proteins
  2. Compatible with existing libraries
  3. Can detect isoform specific interactions, detection of highly specific interactions
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7
Q

Disadvantages of Y1H system

A
  1. No information on the functional consequences of DNA-protein interaction
  2. Rates of false-positive results are high
  3. Improper protein folding may occur resulting in loss of interactions
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8
Q

Function of EMSA

A

Tests for DNA-protein interaction (need to know DNA sequence)

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9
Q

Advantages of EMSA

A
  1. Highly sensitive, assay can be performed with small protein and nucleic acid concentrations and small sample volumes
  2. Assay works well with both highly purified and crude cell extracts
  3. A wide range of nucleic acid sizes are compatible with the assay
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10
Q

Function of DNA footprinting

A
  • to study DNA-protein interactions (need to know general protein-binding region)
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11
Q

Function of ChIP

A

To study protein-DNA interactions (when you don’t know which part of the DNA the protein of interest binds to)

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12
Q

3 types of protein purification chromatography

A

Affinity, ion-exchange, gel-filtration

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13
Q

How does affinity chromatography work?

A
  • separation method based on specific binding interaction between an immobilised ligand and its binding partner on the protein
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14
Q

Advantages of affinity chromatography

A
  1. Cheap
  2. High specificity
  3. High degree of purity of protein
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15
Q

How does ion-exchange chromatography work?

A
  • separates proteins based on total charge
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16
Q

How does gel filtration chromatography work?

A
  • separates proteins based on size through filtration through a gel
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17
Q

What is the principle behind isolation of nucleic acids using spin columns?

A
  • nucleic acids bind to silica under certain conditions but not others
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18
Q

Conditions for the use of mini prep kits

A
  • small scale preparations of plasmids with yields of 2-5ug of DNA/RNA
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19
Q

Conditions for use of maxi prep kits

A
  • large scale preparation of plasmids with yields of 500ug-2mg DNA/RNA
  • more time consuming and expensive
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20
Q

How to obtain plasmids and not genomic DNA?

A
  • perform alkaline lysis: shorter stretches of DNA like plasmids renature following denaturation much faster than longer stretches of DNA like gDNA
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21
Q

What are the components in a PCR mix?

A
  • buffer
  • MgCl2
  • forward and reverse primers
  • template DNA
  • dNTPs
  • taq polymerase
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22
Q

Purpose of qPCR

A
  • relative quantification of the increase in amt of cDNA as it is amplified
  • measures amt of amplified product at each stage during PCR cycle
23
Q

What is the dye used in qPCR and what does it bind to?

A
  • SYBR green; dsDNA
24
Q

What is the Ct in qPCR?

A
  • fractional PCR cycle no. at which fluorescence is greater than the threshold
25
Q

How to calculate relative amt of DNA in qPCR using Ct?

A
  • 2^n where n is the difference in Ct number
26
Q

Purpose of Sanger sequencing

A
  • to determine the nucleotide sequence in DNA
27
Q

Difference between dideoxynucleotides (ddNTPs),and deoxyribonucleotide triphosphates (dNTPs)

A

ddNTPs contain a hydrogen group on the 3’ carbon instead of a hydroxyl group

28
Q

Components in a Sanger sequencing master mix

A
  • template
  • forward and reverse primers
  • excess dNTPs
  • DNA polymerase
  • small amounts of ddATP/ddTTP/ddCTP/ddGTP
29
Q

Advantages of Next-Gen Sequencing compared to Sanger sequencing

A
  1. Ability to sequence many copies of genes simultaneously
  2. Lower cost
  3. Able to sequence samples with low input amounts of starting material
30
Q

Purpose of microarrays

A
  • to determine which genes are expressed in different cells
31
Q

Purpose of RNA-seq

A
  • measures gene expression across the entire transcriptome using NGS
32
Q

Purpose of BCA assay

A
  • for quantitation of proteins
33
Q

What’s the wavelength used for BCA assay?

A

562 nm

34
Q

Purpose of Bradford assay

A
  • for quantification of proteins
35
Q

Wavelength for absorbance for Bradford assay

A

595 nm

36
Q

Why is polyacrylamide suitable for SDS-PAGE?

A
  • chemically inert, electrically neutral, hydrophilic and transparent for optical detection
37
Q

Substrate of HRP

A

Luminol

38
Q

Why is a loading control needed?

A
  • to demonstrate that there was equal protein loaded in the sample
39
Q

Function of X-ray crystallography

A
  • determine protein structure
  • to produce well-ordered crystals that is lacking in contaminants
  • proteins of any size can be precipitated
40
Q

When does crystallisation occur?

A
  • protein concentration is greater than its solubility limit and the protein is in a supersaturated state
41
Q

Conditions for protein crystallization

A
  • at least 97% protein purity
  • pH should correspond to pI since solubility of protein is lowest here
  • suitable conc of protein
  • temperature
  • precipitants (increase ionic strength and lowers solubility of protein to enable ppt)
42
Q

Bragg’s law

A
  • when x-rays are scattered from a crystal lattice, peaks of scattered intensity are observed which correspond to the following conditions: angle of i=angle of scattering, path length diff is equal to an integer number of wavelengths
  • calculate details about protein structure
43
Q

What is Nuclear Magnetic Resonance?

A
  • physical phenomenon which occurs when the nuclei of certain atoms are immersed in a static magnetic field and exposed to a radio frequency energy source
  • conducted in solution
44
Q

What is NMR suitable for?

A
  • small proteins (25kDa max, around 220aa)
45
Q

What is nuclear spin?

A
  • fundamental nuclear property
  • comes in multiples of 1/2
  • individual unpaired protons, electrons and neutrons possess nuclear spins of 1/2
  • In 1H atom, proton has a nuclear spin of 1/2
  • 1H, 13C most commonly used in NMR
46
Q

What is resonance in NMR?

A
  • nuclei absorb energy of radio waves that enable a change in the angle of nuclear spin
47
Q

Range of magnetic field strength of NMR

A
  • 200-900 MHz
48
Q

Define chemical shift

A
  • frequency of absorption for a nucleus of interest relative to that of the molecular standard
  • serves as a guide to analyse organic chemical structures
  • molecular standard for 1H and 13C is TMS bc of molecular symmetry: all 12 H absorb at the same frequency and all 4 C absorb at the same frequency
49
Q

How to understand the number of signals in NMR spectra?

A
  • corresponds to the number of diff protons in molecule

- a set of protons in the same magnetic environment are equivalent and will give a single signal

50
Q

How to understand position of signals? (Chemical shift)

A
  • reveals what kind of proton gave that signal
  • dependent on nuclear shielding
  • with high nuclear shielding, less resonant energy is required to flip nucleus hence low chem shift
51
Q

How to understand splitting in NMR?

A
  • due to nuclear spin interactions of one nucleus with one or more nearby nuclei through spin spin coupling
  • provides info on how many H neighbours (split into N+1 peaks where N=no. of H on adjacent atom)
52
Q

Which group has no signal splitting?

A
  • OH bc of H20 interference
53
Q

Cryo-EM

A
  • generally >200 kDa, sometimes 80 kDa

H3 Molecular Biology (10 decks)

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