Biology Laboratory Techniques

Question 1

Separations

Answer 1

a variety of lab techniques that use intermolecular forces to separate a mixture into its components parts

Question 2

Extraction

Answer 2

a separation technique based on solubility
“like dissolves like”
involves 2 immiscible (not mixing) phases = most commonly an aqueous layer and a less dense organic mixture

Question 3

Steps to perform an extraction

Answer 3

1/ Add WEAK ACID + shake. The acid protonates strong base in the organic layer => making them polar and thus causing them to move to the aqueous layer. The aqueous is then separated and washed with a base to neutralize the added acid, leaving only the strong organic bases.
2/ Add STRONG ACID + Shake. The acid protonates weak bases…..similar as above.
3/ Add WEAK BASE. The base deprotonates ONLY the strongest organic acids….. washed with acid to neutralize the added base
4/ Add STRONG BASE, deprotonates the weak acids…..

i.e. phenol-chloroform extraction is used to separate nucleic acids from cellular proteins

Question 4

Distillation

Answer 4

used to separate compounds with significantly different boiling points at least 20C
CAN NOT COMPLETELY separate two compounds

Question 5

Fractional distillation

Answer 5

a more precise method of distillation that can be used to separate liquids whose boiling points are fairly close together . The vapor is run through glass beads, allowing the compound with higher boiling point to repeatedly condense and fall back into the solution

Question 6

Crystallization

Answer 6

based on the principle that pure substances form crystals more easily than impure substances (i.e. pure substances have higher freezing/melting points)
This is a very difficult method of separation
For most salts, crystallization is an exothermic process

Question 7

Chromatography

Answer 7

to purify a compound from a mixture and/or to identify the ratio of compounds in a mixture.
By passing the mixture through or over a matrix that adsorbs (binds) different compounds more or less strongly according to their properties, altering the rate at which they lose contact with the matrix.
The mixture usually dissolved into a solution to serve as the mobile phase, while the matrix is often a solid surface.
Typically the stationary phase is polar, causing more polar compounds to elute more slowly

Question 8

Column chromatography

Answer 8

a solution containing the mixture is dripped down a column containing the solid phase (usually glass beads). The more polar compounds in the mixture the slowly it travel down.
Gravity drives the movement.

Question 9

High pressure liquid chromatography (HPLC)

Answer 9

column and solution use an apparatus that puts the system under high pressure

Question 10

Paper chromatography

Answer 10

a small portion of the sample to be separated is spotted onto paper
One end of the paper is then placed into a non-polar solvent
The solvent moves up the paper via capillary action and dissolves the sample as it passes over it
The more polar components of the sample the more slowly it moves b/c they are attracted to the polar paper
Rf factor = distance traveled by component/distance traveled by solvent

Question 11

Thin layer chromatography

Answer 11

similar to paper chromatography except that a coated glass or plastic plate is used instead of paper, and the result is visualized via an iodine vapor chamber

Question 12

Gas-liquid chromatography

Answer 12

the liquid phase is the stationary phase.
The mixture is dissolved into a heated carrier gas (usually helium or nitrogen) and passed over a liquid phase bound to a column.
Compounds in the mixture equilibrate with the liquid phase at different rates and pass through an exit port as individual components

Question 13

Size-exclusion chromatography

Answer 13

separated by size and sometimes molecular weight, often through gel filtration

Question 14

Ion exchange chromatography

Answer 14

molecules are separated based on their net surface charge

Utilizes cationic or anionic “exchanger” that slow down the movement of charged molecules

Question 15

Affinity chromatography

Answer 15

uses highly specific interaction

makes use of receptor-ligand, enzyme-substrate, and antigen antibody interactions

Question 16

Southern blotting

Answer 16

used to identify target fragments of a known DNA seq in a large population of DNA
DNA is first chopped up and run on gel electrophoresis, blot onto a membrane and incubated with a labeled probe

Question 17

Separation of enantiomers from a racemic mixture

Answer 17

challenging b/c enantiomers have almost exactly the same physical and chemical properties.
3 ways to resolve enantiomers:
1/ differences in crystallization
2/ use stereospecific enzymes
3/ converted the enantiomers into diastereomers, which have different physical and chemical properties

Question 18

Anode

Answer 18

positively charged

Question 19

Cathode

Answer 19

negatively charged

Question 20

Nuclear magnetic resonance spectroscopy

Answer 20

the study of the interaction between atomic nuclei and radio waves
most commonly used to study hydrogen nuclei, but can also be used for the nucleus of carbon-13 and other atoms as well.

Question 21

Principles of NMR

Answer 21

Nuclei with ODD atomic or mass numbers possess a mechanical property called NUCLEAR SPIN
A spinning proton, like any rotating sphere of charge, generates a magnetic field, in this case around the nucleus
When subject to an external magnetic field, the field of a nucleus aligns either with or against the external field
Nuclei ALIGN with the external field occupy LOWER energy alpha spin states
Nuclei AGAINST the external field occupy higher energy beta spin states.
Photons whose energy are equal to the delta E btw 2 spin states can cause nuclei whose spins are aligned with the external field to FLIP and align against it. Nuclei that flipped to the beta spin state are in resonance with the external field
When protons return to their original spin states, they release electrical impulses that are detected by an NMR spectrometer
In NMR, the frequency of the electromagnetic radiation is held constant while the magnetic field strength is varied

Question 22

NMR spectrum

Answer 22

_ a graph of the magnetic field strengths absorbed by the hydrogen atoms of a specific compound at a single frequency
_ The field strength increases from left to right
The leftward direction = downfield
The rightward direction = upfield
_ Each peak represents chemically equivalent hydrogens = whose position on the compound are indistinguishable by NMR, having the same chemical shift = the difference between the resonance frequency of the chemically shifted hydrogens and that on a reference compound.
_ Splitting of peaks is caused by neighboring hydrogens
_ The area under a peak is proportional to the number of hydrogens represented by that peak. The tallest peak does not necessarily correspond to the greatest area.
_ The position of each peak is dictated by electron shielding, so the positions of different groups of hydrogen can be predicted based on the presence of electron-withdrawing and electron-donating groups

Question 23

Electron withdrawing group (NMR)

Answer 23

lowers shielding, thus decreases the magnetic field strength at which resonance takes place
Shift toward the left

Question 24

Electron donating group (NMR)

Answer 24

increases shielding and increases the field strength required for resonance
Shift toward the right

Question 25

Splitting = spin-spin splitting

Answer 25

caused by neighboring hydrogens that are not chemically equivalent
a neighboring hydrogen is one that is bound to an atom adjacent to the atom to which the other hydrogen is connected.

Question 26

Aldehyde protons

Answer 26

have a very distinctive shift at 9.5 ppm

Question 27

IR spectroscopy

Answer 27

uses molecular dipoles to find info about functional groups.
When exposed to an electric field, these oppositely charged centers move in opposite directions.
In IR radiation, the direction of the electric field oscillates, causing the + and - centers within polar bonds to move toward each other and then away from each other. => the polar bonds within compound stretch and contract, causing intramolecular vibrations and rotations.
Atoms with greater mass resonate at lower frequencies.
Stiffer bonds, such as double and triple bonds, resonate at higher frequencies.
IR spectrum does not reveal the shape or length of the carbon skeleton.
Most predictable spectrum is from 1600-3500 cm-1

Question 28

IR spectrum -OH bond

Answer 28

a broad dip around 3200-3600 cm-1

Question 29

IR spectrum C = O

Answer 29

a sharp dip around 1700 cm-1

Question 30

IR spectrum saturated C-H

Answer 30

double peaks (almost triple) 2800-3000 cm-1

Question 31

IR spectrum N-H

Answer 31

small/ broad big dip at 3300 cm-1

Question 32

IR spectrum C-H

Answer 32

double dips at 2700 and 2800 cm-1

Question 33

IR spectrum C-N

Answer 33

sharp dip at 2200 cm-1

Question 34

IR fingerprint region

Answer 34

from 600-1400 cm-1 unique for nearly every compound but hard to interpret

Question 35

UV spectroscopy

Answer 35

_ detects conjugated systems (double bonds separated by one single bond) by comparing the intensities of two beams light from the same monochromatic light source.
_ One beam is shone through a sample cell and the other is shone through a reference cell.
_ The difference in energy absorbed by the sample and the reference is recorded as a UV spectrum of the sample compound.
_ Provides limited info about the length and structure of the conjugated portion of the molecule
_ Conjugated systems with lamda bonds have vacant orbitals at energy levels close to their highest occupied molecular orbital (HOMO) energy levels.

Question 36

Rule of thumb for UV absorption

Answer 36

30-40 nm increase for each additional conjugated double bond, and a 5 nm increase for each additional alkyl group
_ A = ecl
_ the longer the chain of conjugated bonds, the greater the wavelength of absorption
_ Isolated double bonds do not affect the absorption wavelength
_ Samples must be extremely pure
_ if a comp has > = 8 double bonds, its absorbance moves into the visible region

Question 37

Spectrometry

Answer 37

the study of interactions between matter and energy sources other than electromagnetic radiation

Question 38

Mass spectrometry

Answer 38

to determine a compound’s molecular weight, and, in the case of high resolution, its molecular formula
In mass spec, molecules are bombarded with electrons, causing them to break apart and to ionize. The largest ion is the size of the original molecule but has one less electron = molecular ion.
After being broken apart, the ions are accelerated through a magnetic field. The resulting force deflects the ions around a curved path. The radius of curvature of their path depends upon their mass to charge ratio.
The largest peak is called the base peak, represent the mass of the molecule being studied.
All the peaks are assigned abundances as percentages of the base peak.

Question 39

Grignard reagent

Answer 39

Grignard reagents are strong bases and will react with water

Question 40

Aromatic compounds

Answer 40

needs 2 criteria:
1/ containing a ring comprised of continuous, overlapping p orbitals.
2/ The number of electrons in that ring must be a Huckel number = the number of pi electrons must be equal to 4n + 2, where n is a whole number.