analytical techniques

Question 1

UV-VIS

Answer 1

detection and quantification of compounds

based on the ability of molecules to absorb light at wavelengths in UV and visible light region 180nm - 800nm

absorption of these wavelengths is characteristics of bonds and functional groups in molecules

Question 2

wavelengths and energy

Answer 2

longer wavelength = lower energy

shorter wavelength = higher energy

Question 3

UV VIS benefits

Answer 3

non destructive of compounds

simple and easy to operate

relatively low cost

used for both qualitative and quantitative analysis

extensive/complex sample prep not needed

used to test wide range compounds

example applications - chromatographic separations, quantification of protein/DNA, monitoring enzymatic reactions

Question 4

principles of UV VIS

Answer 4

Absorption of light causes electron excitation from lower to higher energy orbitals.

Larger energy gaps between orbitals require more energy to excite electrons.

Greater energy jumps correspond to shorter wavelengths absorbed, while smaller gaps result in longer wavelengths absorbed.

Question 5

UV VIS jumps between states

Answer 5

usually from pi to pi* (bonding pair of electrons pi bond to anti bonding pi*)

or from lone pairs electron (non bonding) = n to pi*

the energy difference between these jumps determines wavelength absorbed

after excitation the electron returns to ground state and releases the energy in from of light or heat

Question 6

effect of conjugation - alternating pi and sigma bonds

Answer 6

Alternating pi and sigma bonds (electron delocalization) decrease the energy gap, requiring less energy for electron excitation, leading to absorption at longer wavelengths.

More pi bonds in a molecule make it easier for electron transitions from pi to pi* due to reduced energy required for the jump.

Delocalized electrons are more stabilized as they spread over atoms, reducing electron-electron repulsion.

Question 7

homo and lumo electrons

Answer 7

homo = highest occupied orbital - contains highest energy electron

lumo - unoccupied energy level above homo electron

difference between these is called the band gap = transition energy. this is what determines wavelengths that are absorbed

Question 8

more than one absorption peaks

Answer 8

Multiple Electron Transitions: Different types of transitions (e.g., pi to pi, sigma to sigma) require different amounts of energy, leading to multiple peaks.

Conjugation & Delocalization: Conjugated systems with multiple double bonds cause various transitions at different energy levels, creating several peaks.

Question 9

beer Lambert law

Answer 9

intensity of light absorbed is quantified by beer Lambert law
transmittance = intensity of light out / intensity of light in

use this for measuring the concentration as it is difficult to weight out biological molecules accurately due to presence of water and salts and small quantities used

Question 10

schematics of UV VIS

Answer 10

single beam instrument - monochromator determines wavelength

double beam instrument - uses reference and sample cell

Question 11

UV VIS of proteins

Answer 11

chromophores are amide backbone with various side chains
dominant chromophore is amide group
- weak n->pi* transition at 220nm
- intense pi->pi* transition at 195nm

2nd absorption band around 280nm due to aromatic amino acids

Question 12

pi to pi* peak

Answer 12

200-400nm
double bonds or benzene
e- from pi to pi*
as number of c=c increases, peak shift to longer wavelengths

Question 13

n to pi* peak

Answer 13

300-600nm
molecules with lone pair e- adjacent to double bonds c=o n=o
energy required generally lower than pi - pi* so longer wavelength
peaks are less intense

Question 14

sigma to sigma * peaks

Answer 14

below 200nm
molecules with single bonds only
requires high energy in UV region out of range

Question 15

peak intensity

Answer 15

high intensity peak = high absorptivity
lower intensity = lower absorptivity
this relates to beer Lambert rule

broad peaks = overlapping transitions
narrow = more discrete specific electron transitions

Question 16

uv vis of nucleic acids

Answer 16

chromophores
- nucleic bases
intense pi - pi* transitions

Question 17

absorption spectrum of DNA

Answer 17

polynucleotides and nucleic acids absorb less per nucleotides than free ones

so depending on solvent, graph may look different

Question 18

hypochromism

Answer 18

decrease in absorbance intensity
- due to molecular interactions and pi stacking
- structural changes
- electron delocalisation

Question 19

example of hypochromism with heating and cooling

Answer 19

as heating, DNA becomes unstacked, interactions between bases decrease and absorbance increases

Question 20

uv vis to monitor self assembly

Answer 20

by tracking changes in the absorption spectra as the molecules interact and form complexes or aggregates.

Absorption Changes: Self-assembly causes shifts or new peaks in absorption spectra as biomolecules interact.

Red/Blue Shifts: Assembly can cause red (longer wavelength) or blue (shorter wavelength) shifts in absorption peaks.

Intensity Changes: Peak intensity may decrease or increase due to changes in the molecular environment, like aggregation or complex formation.

Question 21

fluorescent spectro

Answer 21

Fluorescence Emission: Measures light emitted after absorption by the sample.

Excitation: Sample is illuminated with light to excite electrons from the ground state to an excited state.

Excitation Wavelength: Based on the absorption characteristics of the fluorophore.

Vibrational Relaxation: After excitation, the molecule loses energy via vibrational relaxation before reaching a stable vibrational state.

Photon Emission: Once stable, the molecule returns to the ground state and emits a photon.

Stokes Shift: Emitted light has a longer wavelength than the excitation light.

Emission Detection: Detects emitted light and measures intensity across different wavelengths.

Question 22

applications of fluorescence

Answer 22

detect presence of proteins
measure enzymatic activity
cell imaging
DNA concentration
biological events

Question 23

advantages of fluorescence spectroscopy

Answer 23

high detected sensitivity - single molecule detection possible

high selectivity - 3 characteristic parameters, excitation wavelength, emission wavelength, lifetime

sensitive to molecular environment

sensitive to intermolecular interactions
both organic and inorganic species

Question 24

fluorescence comes from …

Answer 24

arises from pi to pi* transitions - lots of aromatic rings and double bonds
- requires compounds with pi bonds and conjugated electrons

Question 25

fluorescence quantum yield

Answer 25

number photons emitted/number photons absorbed

a higher quantum yield, a better compound for fluorescence - easier to detect

this is different from UV VIS as for that, interested in absorption of product

Question 26

fluorescence spectrum - jablonski diagram

Answer 26

Excitation (Absorption): Molecule absorbs light, exciting electrons to vibrational levels of the first excited singlet state.

Vibrational Relaxation: In solution, the molecule rapidly loses vibrational energy to the solvent through collisions, converting energy into heat.

Fluorescence Emission: Molecule emits fluorescence from the lowest vibrational level of the singlet state.

Thermal Population: Higher vibrational levels can be thermally populated.

Transition to Ground State: The molecule then transitions from the excited singlet state to vibrational levels of the ground state, emitting light in the process.

Question 27

stokes shift

Answer 27

Definition: The difference in wavelength between the excitation light (absorbed) and the emitted fluorescence (after energy loss).

Cause: After excitation, a molecule loses some energy through vibrational relaxation before emitting light, resulting in longer wavelength (lower energy) fluorescence compared to the excitation light.

Key Point: Stokes shift is observed because the emitted light has less energy than the absorbed light.

will lie more to the right of graph than absorption, often mirror images

Question 28

emission spectrum - The emission spectrum tells you which wavelengths are emitted after excitation.

Answer 28

Fixed Excitation Wavelength: The sample is excited with a specific wavelength of light.

Emission Monochromator: A monochromator scans the emitted light at different wavelengths.

Fluorescence Intensity: The intensity of fluorescence is recorded as a function of the emission wavelength.

Spectrum Overview: The emission spectrum shows where fluorescence occurs and the peak intensity at the wavelength with the highest emission.

Question 29

excitation spectrum - The excitation spectrum tells you which wavelengths can excite the fluorophore.

Answer 29

Fixed Emission Wavelength: The fluorescence emission is detected at a fixed wavelength.

Scan Excitation Monochromator: The excitation light is varied by scanning the excitation monochromator.

Fluorescence Intensity: Fluorescence intensity is recorded as a function of excitation wavelength.

Absorption Equivalent: The excitation spectrum is equivalent to the absorption spectrum of the fluorophore at the emission wavelength.

Information Provided: It gives information about the absorbance properties of the fluorophore.

Question 30

intrinsic fluorescence of proteins

Answer 30

intrinsic fluorphores are the aromatic amino acids
- tryptophan
- tyrosine
- phenylalanine

Question 31

collisional quenching

Answer 31

Definition:
Collisional quenching occurs when a molecule in an excited state loses energy due to a collision with another molecule, instead of emitting light (fluorescence or phosphorescence).

What happens? Excited molecule + quencher molecule → Energy lost as heat, no light emitted.

Cause? Physical collision between molecules.

Common quenchers: Oxygen, iodide, or other molecules that can absorb the energy.

Where? Happens in solutions or gases where molecules move freely and collide often.

Question 32

use fluorescence to monitor protein changes

Answer 32

if protein is buried in a protein and less exposed to solvent - blue shift and peak/graph moves to the left

if protein becomes more exposed to solvent then red shift and peak/graph moves to the right

Question 33

quenching

Answer 33

causes the glow to decrease

Question 34

effect of calcium binding on calmodulin structure

Answer 34

ca2+ binding protein

activates proteins by binding, causing conformational changes - opens it up the protein

Question 35

solvent effect

Answer 35

Definition:
how the polarity or properties of the solvent influence the fluorescence intensity, emission wavelength, and lifetime of a fluorophore (fluorescent molecule).

What changes?
Emission wavelength: Polar solvents stabilize the excited state, causing red-shifting (longer wavelengths).

Fluorescence intensity: Can increase or decrease depending on solvent interactions.

Lifetime: Solvent polarity and quenching affect how long the molecule remains fluorescent.

Why does this happen?
Polarity: A polar solvent stabilizes the fluorophore’s excited state differently than the ground state, altering energy levels.

Hydrogen bonding or dipole-dipole interactions: These interactions can either enhance or quench fluorescence.

Example:
A fluorophore like anthracene emits at different wavelengths in ethanol (polar) vs. hexane (non-polar).

Question 36

foster resonance energy transfer FRET

Answer 36

energy is transferred non-radiatively (without emitting light) from an excited donor fluorophore to an acceptor fluorophore through dipole-dipole coupling.

What happens?
A donor fluorophore absorbs light and enters an excited state.
Instead of emitting light, the donor transfers energy to a nearby acceptor molecule.
The acceptor can then fluoresce or dissipate the energy in other ways.

Applications:
Measuring molecular distances (e.g., protein folding or interactions).
Biosensors to monitor cellular processes.

Question 37

FRET

Answer 37

observe interactions in cells transferring energy without light emission

donor and acceptor molecule need to be close

see emission from acceptor

Question 38

unpolarised light

Answer 38

polarisation potential

random oscillation- all directions

common sources

pass through polarising filter to become polarised

Question 39

polarised linear light

Answer 39

single oscillation plane

through polarising filter
scattering

used in LCD screens and sunglasses

Question 40

circular polarised light

Answer 40

direction of oscillation plane is circular, helical

rightly circular polarised = clockwise rotation

left = anticlockwise rotation

Question 41

circular dichroism

Answer 41

asymmetric chiral molecules interact differently with right and left circular polarised light

used to measure the difference in absorption of left and right circular light by a chiral molecule - as optically active

Chiral molecules absorb left- and right-handed circularly polarized light differently.
The difference in absorption is recorded as the ellipticity of light.
The CD spectrum is a plot of ellipticity vs. wavelength.

Question 42

circular dichroism use

Answer 42

Applications:
Determining protein secondary structure (alpha-helices, beta-sheets).
Monitoring structural changes in response to temperature, pH, or ligand binding.
Characterizing chiral pharmaceuticals.

Question 43

CD equation

Answer 43

ΔA=A
L
​
−A
R
​

where:

Δ
A
ΔA = Difference in absorbance
A
L
A
L
​
= Absorbance of left-handed circularly polarized light
A
R
A
R
​
= Absorbance of right-handed circularly polarized light

Question 44

molar ellipticity

Answer 44

[θ]=
10⋅c⋅l
θ
​

Where:

[
θ
]
[θ] = Molar ellipticity (in deg·cm²·dmol⁻¹).
θ
θ = Observed ellipticity in millidegrees (mdeg).
c
c = Concentration of the sample in mol/L.
l
l = Path length of the cuvette in cm.

Molar ellipticity is a normalized measure of how much circularly polarized light is absorbed by a chiral molecule, taking into account the concentration of the molecule and the path length of the light through the sample.

[θ]=3298⋅
c⋅l
ΔA
​

Question 45

example question CD and molar ellipticity

Answer 45

Problem:
You perform a CD experiment on a chiral protein solution. The following data is provided:

Observed ellipticity (
θ
θ) = 2.5 mdeg at 222 nm.
Concentration of the protein solution (
c
c) = 0.002 M (mol/L).
Path length of the cuvette (
l
l) = 1 cm.
Calculate the molar ellipticity (
[
θ
]
[θ]) of the protein at 222 nm.

Solution:

Using the formula:

[
θ
]
=
θ
10
⋅
c
⋅
l
[θ]=
10⋅c⋅l
θ
​

Convert the observed ellipticity to degrees:
θ
=
2.5
 
mdeg
=
0.0025
 
degrees
θ=2.5mdeg=0.0025degrees
Substitute the values into the formula:
[
θ
]
=
0.0025
10
⋅
0.002
⋅
1
[θ]=
10⋅0.002⋅1
0.0025
​

Simplify:
[
θ
]
=
0.0025
0.02
=
0.125
 
deg
\cdotp
cm
2
\cdotp
dmol
−
1
[θ]=
0.02
0.0025
​
=The molar ellipticity (
[
θ
]
[θ]) of the protein is 0.125 deg·cm²·dmol⁻¹.

Question 47

another CD example bit different

Answer 47

Solution:

To calculate the molar ellipticity using the 3298 constant, use the formula:

[
θ
]
=
3298
⋅
Δ
A
c
⋅
l
[θ]=3298⋅
c⋅l
ΔA
​

Substitute the given values:
Δ
A
=
0.015
ΔA=0.015
c
=
0.001
 
M
c=0.001M
l
=
1
 
cm
l=1cm
Plug the values into the formula:
[
θ
]
=
3298
⋅
0.015
0.001
⋅
1
[θ]=3298⋅
0.001⋅1
0.015
​

Simplify the calculation:
[
θ
]
=
3298
⋅
0.015
0.001
=
3298
⋅
15
=
49470
 
deg
\cdotp
cm
2
\cdotp
dmol
−
1
[θ]=3298⋅
0.001
0.015
​
=3298⋅15=49470deg\cdotpcm
2
\cdotpdmol
−

Question 48

enzymes

Answer 48

highly efficient catalysts
high substrate specificity
high stereospecificty

Question 49

binding cleft

Answer 49

also known as active site

Question 50

binding interactions

Answer 50

involves range of weak invectives between substrate and enzyme sidechains

• hydrogen bonds
• ionic interactions/salt bridges
• van der Waals
• hydrophobic effect

Question 51

substrate binding

Answer 51

E+S –><– ES
THIS IS AN EQUILIBRIUM EXPRESSED AS

Ka = [ES]/ [E] x [S]

Question 52

Ka meaning

Answer 52

high Ka = high association constant so enzyme binds tightly with substrate - very stable complex and high affinity

low ka means the opposite

Question 53

dissociation constant Kd

Answer 53

1/Ka

low Kd = stable ES complex

Question 54

lock and key mechanism

Answer 54

active site of protein has complementary shape to substrate that fits substrate

binding cleft and substrate are viewed as rigid
- both groups are sterically and chemically complementary

Question 55

induced fit mechanism

Answer 55

enzymes are flexible - conformational changes occur when substrate binds during reaction to get maximal complimentary in transition state

Question 56

lowering activation energy

Answer 56

enzymes favour the transition state of reaction - uses weak interactions to stabilise the high energy stable so less energy needed to reach transition state

• changing the reaction pathway

Question 57

rate determining step

Answer 57

the slowest step, the step with the highest activation energy

Question 58

Michaelis menten terms

Answer 58

determine key enzyme kinetic constants

V max = maximum velocity - max rate of an enzymatic reaction

Km = substrate concentration required to reach half V max

K cat = turnover number - V max x K cat = V max / [E]o

K cat/ Km - enzyme efficiency

Question 59

MM equation

Answer 59

v = V max x initial sub conc / Km + initial sub conc

v = reaction velocity = rate of product formation
​

Question 60

km

Answer 60

Kₘ: The substrate concentration at which the reaction rate is half of Vₘₐₓ.
A low Kₘ means the enzyme has a high affinity for the substrate, while a high Kₘ indicates lower affinity.

Question 61

steady state assumption for MM

Answer 61

need concentration of enzyme to be constant

total conc of active sites is constant

create limiting step is the catalytic step

initial conc of product is zero

Question 62

MM plot

Answer 62

velocity (v) y axis
substrate concentration x axis
Shape of the Graph:
The plot shows a hyperbolic curve.
At low substrate concentrations ([S]), the reaction velocity (v) increases almost linearly with increasing [S] (first-order kinetics).
As [S] increases, the reaction velocity approaches a maximum value Vₘₐₓ, where the enzyme is saturated with substrate, and no further increase in reaction velocity occurs, despite increasing substrate concentration.
Key Points on the Graph:

At low [S]: The reaction velocity is roughly proportional to substrate concentration. This corresponds to the first-order reaction rate.
At high [S]: The velocity reaches its Vₘₐₓ value, where the enzyme is saturated, and adding more substrate doesn’t increase the rate.
Half-maximal velocity occurs when [S] = Kₘ, the Michaelis constant.
Graph Interpretation:

Vₘₐₓ is the asymptote of the curve, representing the maximum reaction velocity.
The Kₘ value can be estimated from the substrate concentration at which the reaction velocity is half of Vₘₐₓ.

Question 63

Lineweaver-Burk Plot (1/v vs. 1/[S])

Answer 63

Formula for the Lineweaver-Burk Plot:

1
v
=
K
m
V
max
⁡
[
S
]
+
1
V
max
⁡
v
1
​
=
V
max
​
[S]
K
m
​

​
+
V
max
​

1
​

This equation is in the form of a straight line
y
=
m
x
+
b
y=mx+b, where:

y = 1/v (reciprocal of the reaction velocity),
x = 1/[S] (reciprocal of the substrate concentration),
m = Kₘ / Vₘₐₓ (slope of the line),
b = 1/Vₘₐₓ (y-intercept).
Graph Interpretation:

The y-intercept of the Lineweaver-Burk plot gives the value of 1/Vₘₐₓ.
The slope of the Lineweaver-Burk plot gives Kₘ / Vₘₐₓ.
The x-intercept corresponds to -1/Kₘ.
Advantages:

It makes it easier to estimate the values of Vₘₐₓ and Kₘ from experimental data.
It can also be useful in identifying enzyme inhibition mechanisms by analyzing deviations from a straight line.

Question 64

K cat = turnover number

Answer 64

number substrate molecules converted into product by one molecule of enzyme active site per unit of term when enzyme is fully saturated

factors effecting
- enzyme active site and structure
- environmental conditions like temp or ph

Question 65

enzyme kinetics in lab setting

Answer 65

Km and V max must be determined by measuring the velocity at a variety of substrate concs

measure V at very early times in action before [S] decreases significantly

Question 66

chromatography basic principle

Answer 66

1. mixture to be separated is dissolved in the mobile phase
2. mobile phase is added throughout the process
3. components start to separate and move through the stationary phase at different rates
4. each component is collected as it reaches the bottom of the column
   - components have different affinities from the SP causing them to move through the chromatography different rates

Question 67

chromatography components

Answer 67

inlet - sample mixture is inserted

column - sample is separated into its individual components

detector

collector

transport always occurs in the mobile phase

Question 68

elution development

Answer 68

gradually changing the condition of the chromatography like the solvent composition or temperature, to separate compounds to achieve optimal separation

• changing polarity or flow rate of mobile phase

Question 69

retention time and graph

Answer 69

area under peak is proportional to the amount of sample present

retention time is time spent in phases

Question 70

different in peak widths

Answer 70

sharp peak = thin layer in one column

wide peak = spreads in chromatography

Question 71

adjusted retention time

Answer 71

tR’ = observed retention time tR - void time t0 (time mobile phase to pass through)

Question 72

retention factor (k)

Answer 72

more important than retention time

ratio of the time a substance spends in the stationary phase to the time it spends in the mobile phase

k = time in stationary phase / time in mobile phase

time = mass

Question 73

partition coefficient

Answer 73

K = molar conc in stationary phase / molar conc in mobile

Question 74

phase ratio

Answer 74

B = volume of mobile phase/ volume stationary phase

Question 75

retention partition and phase ratio

Answer 75

k = K/B

Question 76

separation factor

Answer 76

measure of how well two compounds are separated in chromatography

alpha = t’R2/t’R1

IF ALPHA = 1 - NO SEPARATION
higher the value, better separation

Question 77

resolution

Answer 77

separation of 2 peaks is defined at their resolution

ration between separation of 2 peak maxima and their average base width

Rs = R
s
​
=
W1+W2
2(t
′
R2−t
′
R1)
​

Question 78

resolution and separation

Answer 78

Rs>1.5 is good

1.5 is baseline separation

if less than 1.5, overlap

Question 79

normal phase

Answer 79

polar stationary phase and non polar mobile phase

the least polar compounds elute first and most polar are last

as SP is polar, polar compounds have greater affinity

Question 80

reverse phase

Answer 80

non polar stationary phase and polar mobile phase

solute reaction in both cases depends on polar/non polar interactions between solute and column

Question 81

choice of mobile phase

Answer 81

elution rate can vary depending on polarity of solvent

solutes elute from normal phase faster than polar solvent

solutes elute from reverse phase faster in non polar

Question 82

LIQUID CHROMO - ION EXCHANGE

Answer 82

separation based on charger interactions

cation or anion exchange

Question 83

size exclusion LC

Answer 83

separates based on molecular size

retention differences are controlled by the extent to which molecules can diffuse through pores of stationary phase

no separation occur if either none or all of the sample can access the pores

Question 84

affinity chromatography LC

Answer 84

Binding: Target molecules in the sample bind to ligands on the stationary phase, while others are washed away.

Washing: Unbound impurities are removed using a buffer.

Elution: The target molecule is released using an elution buffer (e.g., by changing pH, salt concentration, or adding a competitor).

Applications:
Purification of proteins (e.g., antibodies, enzymes)
Isolation of nucleic acids
Drug discovery

Advantages:
High specificity
High purity of the eluted product

Example Ligand-Target Pairings:
Antibody-antigen
Enzyme-substrate
Receptor-ligand

Question 85

gas chromatography

Answer 85

used to separate and analyze volatile compounds in a mixture.

Principle:
The mixture is vaporized and carried by an inert mobile phase (a gas like helium or nitrogen) through a stationary phase (a liquid or solid lining the column). Components are separated based on their volatility and interactions with the stationary phase.

Question 86

GC

Answer 86

separation is possible in GC if solutes differ in vapour pressure or intensity of solute stationary phase interactions

minimum requirement - solute or derivative thereof must be stable at temp

Question 87

resolution in microscopy

Answer 87

is the shortest distance between two points on a specimen that can still be distinguished as separate

it is dependent on the wavelength of the beam

the smaller the resolution, the better

Question 88

TEM

Answer 88

Direct Beam: Electrons pass through the specimen without interacting; forms the bright background in the image.

Elastically Scattered Electrons: Electrons interact with the specimen’s atoms without energy loss; create contrast by scattering differently based on the material’s density and thickness.

Inelastically Scattered Electrons: Electrons lose energy interacting with the specimen (e.g., exciting electrons or vibrations); reduce image quality by contributing to noise and signal blurring.

Contrast arises from electron scattering by different parts of the sample (density, thickness).

Question 89

SEM

Answer 89

uses a focused electron beam to scan the surface of a specimen, producing high-resolution images of its surface topology and composition.

Key features:

Electron beam interacts with the sample’s surface.
Signals: Secondary electrons (surface details), backscattered electrons (composition contrast), and X-rays (elemental analysis).
Applications: Examining surface structures, materials science, and biological specimens.

Question 90

contrast mechanisms in TEM

Answer 90

amplitude contrast - depends on thickness and Z (mass) of specimen

phase contrast - interference of the scattered and transmitted beams

Question 91

amplitude contrast

Answer 91

Amplitude Contrast in microscopy refers to image contrast generated by variations in the intensity of transmitted electrons caused by the specimen absorbing or scattering electrons out of the optical path.

Key Points:
How it Works: Some areas of the sample block or scatter electrons more strongly, reducing the electron intensity in those regions.
Result: Darker areas correspond to higher absorption or scattering, creating contrast.
Significance in TEM: Useful for dense or stained specimens, but not dominant in thin, unstained samples (phase contrast is more important there).

Question 92

objective aperture

Answer 92

used in TEM for amplitude contrast, it stops electrons with a scattering angle larger than alpha from passing down into the column and contributing to the image, so makes it a lot clearer

Question 93

phase contrast

Answer 93

Phase Contrast in microscopy is a technique that enhances contrast by converting phase shifts of transmitted electrons (caused by the specimen) into amplitude differences.

Key Points:
How it Works:
Electrons passing through the sample experience phase shifts due to variations in sample thickness or refractive index.
These shifts don’t directly affect intensity but can interfere constructively or destructively to create visible contrast.

Result: Subtle structural details, like membranes or protein complexes, become visible without staining.

Significance in TEM: Essential for imaging biological specimens or other thin, low-density materials in their native state.

Question 94

positive chemical staining

Answer 94

heavy salts that react with cellular structures are used

example - uranyl acetate, lead citrate

stained features appear dark - amplitude contrast, helps to see clearer

Why Choose It:
Highlights specific structures (e.g., organelles or proteins) by directly staining them.
Provides detailed information about the internal features of the specimen.
Best for studying fixed biological samples or when precise localization of features is needed.
Downside: Can require complex preparation, potentially altering the sample structure.

Question 95

negative chemical staining

Answer 95

contrast is applied to the environment using heavy metal salts instead of cells

the specimen appears white in dark background

Why Choose It:
Highlights the outline of the specimen by staining the background while leaving the sample unstained.
Suitable for visualizing overall shapes or surface structures of small, delicate samples (e.g., viruses, nanoparticles).
Easier and faster to prepare, preserving native structure.
Downside: Provides less internal detail compared to positive staining.

Question 96

TEM samples

Answer 96

thickness needs to be 200-300nm thin to allow electrons beam to pass through

thicker samples need to be sections

need a high vacuum, so samples are dehydrated - issue is that water is an important component of the cell structure

Question 97

cryoTEM

Answer 97

samples are frozen in native state instead of dehydrated and imaged at liquid nitrogen temperature

important water frozen amorphous ice which is transparent to electron beam

ice layer has to be very 250nm max

Question 98

considerations for cryotem

Answer 98

freezing rate 10,000 degress/second

lower cooling rate results in hexagonal ice

if sample warms after freezing, ice crystallised into cubic ice

Question 99

cryo tomography

Answer 99

collect images at different tilts angles and merge them to reconstruct the 3D object

images need to be acquired at every 1-2 degrees
65-130 images are generated = lots of data
samples can be sensitive to electron beam so electron dose has to be lowered
limited resolution in 3D