Practicals Flashcards

1
Q

What is the role of BSA?

A

Large globular protein that acts as as blocking buffer
- Preventing non-specific binding
= Reduces background noise and increases accuracy

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2
Q

Describe an epifluorescence microscope set up

A
  1. Light is passed through filters for a specific wavelength
  2. Reflected, passed through a lense and hits the object
  3. Is detected by a detector
    Images can be stacked up to generate a 3D image
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3
Q

What are the technical differences between an epifluorescence
microscope and CSLM that allow the higher resolution imaging?

A
  1. Laser beam scans the specimen in CSLM to create an image, contributing to greater spatial resolution
  2. Light is absorbed through a very small pin hole, so out of focus light is not detected. This also reduces background noise in CSLM
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4
Q

What are the disadvantages of CSLM compared to epifluorescence microscopy?

A
  • More expensive to carry out
  • Takes longer to analyse a specimen
  • Laser illumination could lead to photobleaching, so limit ability to capture long-term lapse images
  • Due to scanning technique and time, field of view may be limited
  • Less portable and bulkier than normal microscopes
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5
Q

Explain, briefly, how the Rubisco content has been reduced in this line.

A
  • Antisense construct is used to target the ssu in Rubisco
  • Introduces a genetic construct, complementary to mRNA
  • Forming a double stranded RNA
  • This is then targeted for degradation, by DICER
  • Leading to a reduced expression of the small subunit of Rubisco
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6
Q

Describe the measurements and assumptions that underly the estimation of the CO2 concentration in the intercellular airspaces (Ci) from leaf gas exchange.

A

Measurements:
- Ambient carbon dioxide
- NET photosynthesis, measured as uptake of CO2
- Temperature
Assumptions:
- Steady state conditions- assumed that it reaches steady state gas exchange
- Concentration of carbon is uniformly spread through a leaf
- CO2 diffuses 1.6x slower than water
- No issue with experimental system

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7
Q

What is the purpose of each component in an ELISA assay?
- Sodium bicarbonate
- Ab coupled to alkaline phosphate
- Bovine serum albumin (BSA)
- Positive control solution

A
  • Sodium bicarbonate
    = aids solubility and stability of proteins, acts as a buffer
  • Ab coupled to alkaline phosphate
    = Ab for specific recognition, enzyme catalyses pNPP to a yellow product
  • Bovine serum albumin (BSA)
    = Blocks non-specific binding
  • Positive control solution
    = Checks assay works
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8
Q

What are the advantages of analysing CO2 assimilation as a function of Ci rather than CO2 surrounding the leaf?

A
  • Stomatal conductance taken into account
  • May vary in response to the environment
  • More accurate
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9
Q

What are the role of the following in DNA extraction?
- Lithium chloride
- w/v SDS
- isopropanol
- 70% ethanol

A
  • Lithium chloride
    = salt to neutralise the charge of the DNA, more likely that DNA will separate and form a pellet
  • w/v SDS
    = Detergent, breaks down membranes and releases DNA from nucleus
  • isopropanol
    = Precipitates DNA
  • 70% ethanol
    = Remove the salt from earlier as this would affect PCR
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10
Q

Why should the student include a protein ladder/standard when running the gel?

A
  • To use as an indicator/calibrator for protein size
  • Can be used to determine the molecular weight of proteins
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11
Q

Why should samples be kept on ice?

A
  • Prevention of enzymatic activity - slowing down reactions
  • Minimize protein degradation
  • Reduction in protease activity
  • Prevent sample oxidation and oxidative damage
  • Helps preserve native protein structure
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12
Q

What could they be doing incorrectly when pipetting to not get the right value? Provide a suggestion for how they could improve their technique

A
  • in initial uptake be pushing down hard to the stop and then drawing up liquid
    (because all the values are greater)
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13
Q

A disc removed from the leaf is imperfect. What should the demonstrator advise?

A
  • Try again and get a new disc
  • As it may effect accuracy and amount of DNA/ protein content extracted
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14
Q

What are the key components of a figure legend?

A

Title
Techniques (PCR/Gel/ etc.)
Groups/ Treatments (WT, mut)
Result (line, increase, significant difference)
Other details on experiment (e.g. Temp)

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15
Q

Explain briefly how mechanical homogenisation and incubation temperature can impact recovery of sucrose

A

Mechanical homogenisation
- physical breakdown of cell wall and membrane to release sucrose
- increase recovery
Incubation temperature
- more extreme temperatures will denature the membrane
- higher will lead to more movement and gaps, increasing sucrose recovery
- snap-freezing, can render the wall and membrane brittle, also increasing recovery

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16
Q

To understand the function of CYCLOPS, it is necessary to generate loss-of-function mutants via CRISPR-Cas9. Describe the main steps and key considerations in the process of generating homozygous cyclops mutants in rice

A

CRISPR/ Cas9
1. Generate a guide RNA, which will guide it to the CYCLOPS gene
2. Transform the rice
- Achieved by A.tumefaciens
3. Select for mutants (positive transformants, by applying a selection pressure
4. Cross mutant heterozygous plants
5. Check by screening for homozygous e.g. PCR or NGS
RNAi
- introduce an RNA transcript that will post-transcriptionally silence the CYCLOPS gene
- Can be self crossed to generate homozygous mutants

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17
Q

Why does determination of maximal photosystem II (PSII) efficiency require dark-adaptation?

A
  • All the PSII reactions centres are open, ready to accept electrons
  • Represents baseline/ ground state (F0)
  • Minimal NPQ, due to the lack of light
  • Steady state because it has been in darkness for enough time to be uniformly adapted
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18
Q

How is it possible that photosystem II subunit S (PsbS) overexpression mutants have much higher NPQ values than wild-type plants, yet very similar PSII operating efficiency?

A
  • The role of PsbS is to activate NPQ by dissipating excess energy
  • Overexpression generates a greater capacity to initiate NPQ
    = More rapid response to high light is signalled
  • So the NPQ values are higher
  • Operating PSII is a signal of plant health, and overexpression doesn’t effect this
  • PsbS is only involved in photoprotection when levels of light are too high, however doesn’t effect the normal function of PSII
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19
Q

What is the role of ICS1 and ICS2 gene?

A

Isochorismate synthase 1 & 2
- Isoforms of each other
- Important in the biosynthesis of salicylic acid (SA)
- Exhibit a degree of functional redundancy

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20
Q

What are the requirements for a primer?

A
  • 3’ CG clamp
  • 20bp minimum length
  • Equal distribution of A,T,C,G
  • No palindromes (same backwards + forwards)
  • No homopolymeric sequences
  • No self complementary regions
  • Tm of 50-60°C
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21
Q

Why might an adult leaf, have a greater sucrose content than younger leaves?

A

Storage location
Mature- enhanced chloroplast structure
- Higher expression of enzymes

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22
Q

What gene fusions will you use to measure the timing of PNE1 expression over a vircadian time course in wild type and blackpool null you use? How would you visualise this in the plant?

A
  • To measure over a time course, fuse the PNE1 promoter to a luciferase gene
  • Measure luminescence with a luminometer
  • Luminescence is only generated when the promoter is being activated
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23
Q

Describe the main differences in the transformation methods used for Arabidopsis thaliana and Nicotiana tabacum (tobacco).

A
  • Both can be transformed using A.tumefaciens, which involves the insertion of a plasmid
  • Different promoters and selection markers
  • Biolistics may be used instead for plants that are recalcitrant to A.tumefacien transformation
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24
Q

Why would an adult leaf have less sucrose than a young leaf?

A
  • Lower metabolic activity
  • Sink- source dynamic, sucrose is moved away
  • More used for storage of starch instead of shorter term energy (sucrose)
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25
Q

What is the Beer’s Lambert law?

A

A= c x ε x b
A = absorbance
c = conc
ε = extinction coefficient
b = length of path

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26
Q

What kind of microbe do you think is growing out from the rotten apple and explain your reasoning?

A

Oomycete
- Have filamentous hyphae, that resemble fungal but aren’t
- Formation of zoospores, round
- Can have flagella

27
Q

What is the delay between being infected and being able to infect known as?

A

Latent period
- Where pathogen replicates, before spreading

28
Q

What are the components of a phytomer?

A

Nodes
Internodes
Axillary bud (w/SAM)
Subtending leaf
Terminal bud (w/SAM)

29
Q

What is the role of Koch postulates? How is it carried out?

A
  • Identity whether a pathogen is causing the phenotype seen
    1. Record the phenotype
    2. Isolate the pathogen
    3. Cultivate the pathogen, or just
    4. Reinfect a healthy organism
    5. Record the phenotype
30
Q

What are key principles and rules when carrying out Koch postulates?

A
  • Degree of aseptic technique to avoid contamination and ensure that pathology is caused by the isolated organism
  • Aseptic technique is important when growing the pathogen (oomycete) on the agar
31
Q

What are Charles Bessey’s 5 DICTA?

A
  1. Reduction in number of parts
  2. Ancestral condition is bisexual
  3. Parts evolve from separate to fuse
  4. Radial evolved before bilateral
  5. Ovary moves from superior to inferior positions
32
Q

What is the reason behind Bessey’s 5 dicta?

A
  1. Reduce number due to energy requirement, more specialised so fewer required
  2. Selfing important for survival
  3. Fusing allows to control what insects can reach pollen
  4. Bilateral symmetry controls what insects and from where from
  5. Protect ovaries, by moving to an inferior position
33
Q

What do K, C, A and G stand for in the flower morphology equation?

A

K = Kalyx (Sepals), small green leaves
C = Corolla (petals)
A = Androecium (Stamens)
G = Gynoecium (Carpels)

34
Q

What are carpels and stamens? Name their structures

A

Carpels - female reproductive part contains
Stigma for pollen
Style as the stalk
Ovaries - ovules
Stamens - male reproductive part
Anthers- pollen bearing
Filament - stalk

35
Q

What is the first leaf that is found before a flower?

A

Subtending bract

36
Q

What are sepals, what is their role?

A

Little leaves that protect flower when as a bud
- Group known as a calyx

37
Q

What is the name of the stem of the flower?

A

Pedicel

38
Q

What is the perianth?

A

Sepals and petals that are fused together
Undifferentiate = tepal, non distinguishable
Differentiated = distinguishable

39
Q

Compare and contrast rosids and asterids

A

Asterids - corolla are fused, rosids aren’t
Both eudicot, common number of 4 or 5

40
Q

Compare the characteristics of monocots and dicots

A

Seeds - one cotyledon (monocot), two cotyledons (dicot)
Leaves - Parallel vein (monocot), Branched (dicot)
Flower - x3 (monocot), x4/5 (dicot)
Stems - scattered vascular bundles (monocot), ring arranged vascular bundles (dicot)
Roots - fibrous (monocot), taproot (dicot)

41
Q

Describe 3 reporter genes
1. Detection method
2. Advantages
3. Disadvantages

A
  1. GFP
    - Fluorescence microscope
    - Accurate, alive, temporal resolution
    - Fluorescence may continue after expression, GFP is large
  2. Luciferase
    - Luminometer
    - Accurate, good for diurnal, temporal resolution
    - Requires substrate, detection is transient
  3. GUS
    - light microscope
    - requires substrate, cheap, easy
    - lack of temporal resolution
42
Q

How to distinguish between introns and exons in TAIR and general gene models?

A

Introns - thin lines
Exons - thicker regions

43
Q

By simply looking at the FASTA sequence, how can you tell whether it is the protein coding sequence or the genomic sequence?

A

Protein coding sequence (or the CDS) starts with the ATG
+ ends with a stop codon

44
Q

What are introns and exons?

A

Exons - coding
Introns - non-coding regions, removed in splicing

45
Q

If the length of the genomic sequence is 2390 bps, two introns are 151 and 227 bps long, and the mRNA is 2012 bps, why is the coding sequence 1575 bps? (Hint: what else do we need to take away?)

A

2390- 151- 227 = 2012 mRNA
1575 is coding
437 bps is for Untranslated regions at 5’ and 3’ end

46
Q

What is the condition of two flower morphs called? What’s the evolutionary value?

A

Dimorphic flowers
- Reproductive (pollination) strategies
- Increased fitness in diff environments
- Reduce inbreeding
- Speciation

47
Q

You wish to test the activity of a new promoter sequence in transformed plants. Describe the DNA features required for construction of a suitable Agrobacterium transformation vector.

A
  • Promoter gene
  • Reporter gene (GFP, GUS, Luciferase)
  • T-DNA border regions (L and R)
  • Stop codon/ Terminator region
  • Binary vector (Ti-plasmid)
48
Q
A
49
Q

Why is NADPH not a suitable option for the quantification of starch?

A
  • Absorbance peak clashes with an enzyme in the reaction
50
Q

What is the role of invertase?

A

Cleaves sucrose into glucose and fructose

51
Q

How is water use efficiency calculated?

A

Assimilation (CO2)/ Evaporation (water)

52
Q
A
53
Q

How is internal CO2 calculated?

A

[CO2out]- Assimilation/ gCO2

54
Q

How to calculate electron transport rate from fluorescence?

A

Electron transport rate (ETR) = Operating PSII x Light absorbed x Light to PSII x light intensity

55
Q

How is R0 calculated?

A

Overall rate of infection x infectious period
βN x 1/µ

56
Q

What is the role of glycerol in protein quantification?

A
  • Increases protein stability
  • Can increase resolution of bands and evenly distribute sample in well
57
Q

Why is a ubiquitin primer added into a PCR reaction?

A
  • As a control
  • Test for integrity of the DNA
58
Q
A
59
Q

What traits should a gRNA for CRISPR have?

A
  • PAM motif
  • Specificity to certain area, not multiple unless that’s the aim
  • Optimised GC content for stable hybridization
  • Avoid self complementation
60
Q

What is the Cas9 nuclease tagged to, to allow entry into the nucleus?

A
  • nuclear localization signal (NLS)
  • signal for host machinery to recognised and transport protein into the nucleus
61
Q

What methods can be done to check for successful CRISPR modification?

A
  • Sequence the genome
  • PCR - if remove section of DNA
  • RT-PCR to see modification to proteins
62
Q

What does Adnate, Connate, Congenital and Postgenital mean in flower development?

A

Adnate- fused to other organ type
Connate- fused to same organ
Congenital- fusion occurs as parts develop
Postgenital- fusion occurs after parts have developed

63
Q

Compare the advantages and disadvantages of PCR based assays versus immunodiagnostic assays for detecting viral pathogens in plants.

A

PCR adv
- High sensitivity
- Quantitative
- Versatile (many pathogens)
PCR disadv
- Cost
- Requires extraction and amplification
ID adv
- ELISA easier to perform
- Cost effective
ID disadv
- may not detect low levels
- specific Ab need to be developed

64
Q

Why are viruses difficult to do Koch’s postulates on?

A

Difficult to isolate