BCCB2000 - Practical Exam

Question 1

Water’s Importance

Answer 1

As most of the chemical reactions that occur in organisms occur in water. As many organic chemicals and ions in biochemistry dissolve in water. As it is what makes molecules mobile and allows for interactions between them. In transporting nutrients, metabolites, ions, drugs and waste products in vivo. • As it dissociates to form acidic and basic ions.

Question 2

Water’s Dissociation

Answer 2

H2O + H2O ⇋ H3O+ + OH- simplified to H2O ⇋ H+ + OH-

Question 3

What does adding/detaching a H+ change?

Answer 3

Can change a molecule’s physical/chemical properties.

Question 4

H+ Effects

Answer 4

Has effects on: The chemical reactions that occur inside and outside of cells. The physicochemical properties of many small and large molecules. The structure, function and stability of macromolecules such as proteins. Biochemical laboratory techniques. • Growth and viability of all cells.

Question 5

How do we control the effects of H+?

Answer 5

We can control these effects by using buffers. Buffers interact with H+ to maintain constant pH.

Question 6

pH (potential of hydrogen)

Answer 6

H + concentration defines the pH of a solution. H + concentration is written as mol/L (M).

Question 7

pH equation

Answer 7

pH7 = basic. A water-based solution that is neutral (pH = 7) will have 1 x 10-7 M of H+ and 1 x 10-7 M of OH- , that is there will be 1 x 10-14 M ions in solution. H + concentration can be calculated if OH- concentration is known and vice versa.

Question 8

Equilibrium of Acids and bases

Answer 8

HA + H2O ↔ HB + H3O + (CH3COOH + H2O ↔ CH3COO- + H3O + )

Question 9

Acid Dissociation

Answer 9

[A- ] and [H+ ] = products and [HA] = reactants

Question 10

What does a small and a large Ka mean?

Answer 10

A large Ka = more products, small Ka = more reactants.

Question 11

Base Dissociation

Answer 11

[B+] and [OH- ] = products and [BOH] = reactants

Question 12

pKa and it’s relation with Ka

Answer 12

Ka is usually a small number so instead we work with pKa (pKa = -logKa).

Question 13

What does a small and a large pKa mean?

Answer 13

This means that a large pKa = more reactants (weak acid), small pKa = more products (strong acid).

Question 14

Buffering Region of Buffers

Answer 14

pKa +/- 1 Horizontal region represents buffering range for this system.

Question 15

Henderson-Hasselbalch Equation

Answer 15

Change in pKa, [acid] or [base] can change pH. A change in pH can change the ratio of [acid] and [base].

Question 16

pH and charges of amino acids

Answer 16

At low pH all amino acids have a net positive charge. At high pH, all amino acids have a net negative charge

Question 17

Using Henderson-Hasselbalch Equation for amino acids

Answer 17

You calculate the a-amino group and the a-carboxyl group to find the predominant groups and determine if deprotonated (basic) or protonated (acidic), or even a zwitterion.

Question 18

Isoelectric Point (pl)

Answer 18

pI is the point at which a molecule has no charge, that is, it is in it’s zwitterion form.

Question 19

Isoelectric Point (pl) Formula

Answer 19

pI=pKa1 + pKa2/2

Question 20

Equivalence Point

Answer 20

The point at which all of a titratable side chain has been converted from its basic form to acidic form (or vice versa depending on the experiment). Usually indicated by a sharp drop in pH.

Question 21

Equivalence Point of a-amino group

Answer 21

First equivalence point will be the point at which all of the NH2 has been converted to NH3 +

Question 22

Equivalence Point of a-carboxyl group

Answer 22

Second equivalence point will be the point at which all of the COO- has been converted to COOH.

Question 23

Dilutions (Volume : Volume)

Answer 23

e.g. 1:4 represents 1 part added to 4 parts Would be the same as 1/5

Question 24

Dilutions (Volume / Volume)

Answer 24

e.g. 1/4 represents 1 part out of a total of four parts Would be the same as 1:3

Question 25

Fold dilution

Answer 25

## Footnote refers to the number of parts

Question 26

Dilution volumes (E.g., 1:4 dilution of 100 ml solution)

Answer 26

## Footnote Divide final volume by the number of parts 100/5 = 20 ml Volume of concentrated solution = 1 x 20ml Volume of solvent = 4 x 20ml

Question 27

Types of dilutions

Answer 27

## Footnote Dilution by addition/Stepwise Dilution Serial Dilution

Question 28

Dilution by addition/Stepwise Dilution

Answer 28

## Footnote Requires calculations (C1V1 = C2V2 ) Multiple steps Used when dilutions required are not sequential in concentration.

Question 29

Serial Dilution

Answer 29

## Footnote Used when dilutions are sequential in concentration. Less steps

Question 30

Constant Volume Formula

Answer 30

## Footnote Final volume/dilution factor (DF)-1 DF= total # of parts

Question 31

E.g., Prepare three 1:1 (1/2) dilutions from a 200g/L stock solution. Final volume in each tube must be no less than 2ml. Find the concentration of each tube if the stock is 20 mM

Answer 31

## Footnote We first add a volume of solvent to each tube that is equal to the final volume required. So we would add 2ml of dH2O to tubes 2-4. To work out the volume of stock : 2ml/2-1 = 2/1 = 2 ml of stock To calculate concentration in each tube divide by the dilution factor (DF = total # of parts (1:4 = 5 parts)) First dilution = 20 ÷ 5 = 4mM Second dilution = 4 ÷ 5 = 0.8mM Third dilution = 0.8 ÷ 5 = 0.16mM

Question 32

Factors affecting Absorbance

Answer 32

## Footnote Pathlength (l) Concentration (c) Molar absorptivity (ε)

Question 33

Beers Law

Answer 33

## Footnote where A = the absorbance (no units) ε = molar absorptivity for a molecule at a particular wavelength (M-1cm-1 ) l = the path length (cm) c = the concentration (M)

Question 34

y = mx + c Vs A = εcl

Answer 34

## Footnote y = A m = εl (l is usually 1cm, so can say m = ε) x = c c = y-intercept (this will be zero as there will be zero absorbance when there is zero concentration)

Question 35

Molar absorptivity (ε)

Answer 35

## Footnote The probability that a photon of a particular wavelength will be absorbed by the specific material in the sample. The wavelength that a molecule absorbs maximally at. Indicates how strongly the molecule absorbs light at that wavelength. Large number = strong absorbance Value is equal to the gradient of a straight line.

Question 36

Estimating molar absorptivity

Answer 36

## Footnote Step 1 First we have to rearrange the formula for Beers Law (A= εcl) Step 2 Calculate the gradient of the line. Step 3 Convert the gradient in mass to moles.

Question 37

If the Gradient of the graph is in (mg l^-1)^-1 cm^-1, how do we convert it to molar absorptivity (M^-1 cm^-1)

Answer 37

## Footnote The units for 'ε' include the inverse of the units of concentration so rather than divide the gradient (in mg) by the Mwt, we would have to multiply the gradient by the Mwt. Our gradient is in mg, therefore when we convert to moles, our answer is expressed in mmoles. 'ε' needs to be expressed in moles so we must also multiply by 1000. Our answer would be reported with the units M-1 cm-1 .

Question 38

Calculating concentration in Beers Law

Answer 38

## Footnote Use Beers Law A = εcl Use a standard curve (and its equation)

Question 39

Standard Curve of Beers Law

Question 40

Protein structure and Absorption

Answer 40

## Footnote Amino acids join via peptide bonds to form primary structure of proteins. All amino acids absorb infrared. Some amino acids absorb UV light - particularly aromatic amino acids. Properties of amino acids within a protein allow us to quantitate protein levels using spectrophotometry. Proteins absorb at 280nm - UV. Often easier to react the protein with another compound to produce a coloured compound that can be measured at a wavelength within the visible range (380-750nm).

Question 41

Colourimetric assays

Answer 41

## Footnote Use reagents that react with the protein to produce a coloured compound which can be measured using a spectrophotometer. Requires a standard curve. Albumin (BSA) often used as a standard for protein assays although not ideal. If samples (unknowns) have an absorbance that does not fall within the range of the standard curve we must dilute the original sample and use our diluted sample to go through the protein assay again.

Question 42

Biuret assay

Answer 42

## Footnote Involves the formation of a complex between amides involved in peptide bonds between amino acids of your protein and copper ions. Uses a reagent called Biuret, which is made up of sodium hydroxide, hydrated copper sulphate, sodium potassium tartrate and potassium iodide. If proteins are present in your solution, the amides will form a bond with the Cu2+ ions. In the presence of strong alkaline conditions, the Cu2+ ions will be reduced to Cu+ ions. This results in a colour change from blue (no bond formation ⸫ no protein present) to purple (bond formation ⸫ protein present). The deeper the colour, the higher the protein concentration. The amount of coloured product is then measured on a spectrophotometer at 555nm. The absorbance is plotted on a protein standard curve to determine protein concentration.

Question 43

Factors that Affect Reaction Rate

Answer 43

## Footnote Concentration Temperature Physical Nature Presence of Catalyst

Question 44

Reaction rate

Answer 44

## Footnote Rate = reactants consumed = products produced This can be quite slow and as such catalysts are used

Question 45

Catalysts Characteristics

Answer 45

## Footnote Are not consumed during a reaction Do not alter equilibrium of a reaction Provide an alternate pathway for a reaction to occur with a lower activation energy Increase the rate of reaction Can be enzymatic or non-enzymatic

Question 46

Enzymes - Nomenclature

Answer 46

## Footnote Proteins. Globular. End in -ase, -me or -in. Named for their function.

Question 47

Types of Enzymes

Answer 47

## Footnote Oxidoreductases Transferases Hydrolases Lyases Isomerases Ligases

Question 48

Enzymes Characteristics

Answer 48

## Footnote Highly specific Can be regulated Catalyse reactions in mild conditions (25-37°C) Stereospecific Enzymes catalyse reactions at their active site (a highly specific 3D arrangements of residues arising from the different amino acids comprising the protein structure)

Question 49

Enzyme structure

Answer 49

## Footnote Enzymes that work alone = unconjugated Enzymes that need help = conjugated

Question 50

Conjugated Enzymes

Answer 50

## Footnote Conjugated enzymes require cofactors. Cofactors are a nonprotein component - organic molecules (coenzymes) - inorganic ions (Fe2+, Ca2+) Cofactors may be - permanently attached (prosthetic groups) - temporarily attached (coenzymes)

Question 51

Factors of Enzyme catalysed reactions

Answer 51

## Footnote Temperature, pH, ionic strength, substrate/product/enzyme concentration and presence of inhibitors, activators or cofactors By varying the conditions above we can determine how an enzyme works (enzyme kinetics).

Question 52

Catalase

Answer 52

## Footnote Catalyses the breakdown of hydrogen peroxide (H2O2 ) to water (H2O) and oxygen (O2 ) High concentrations of catalase are found in the liver, kidney and RBCs (also potatoes) Optimum pH for catalase is pH 7.0

Question 53

Enzyme kinetics

Answer 53

## Footnote Describe an enzyme catalysed reaction.

Question 54

Enzyme Kinetics Parameters

Answer 54

## Footnote Km Vmax kcat kcat/Vmax kcat/Km We can calculate these parameters by looking at the enzyme activity at different substrate concentrations We use the Michaelis-Menten and Lineweaver-Burk plots to achieve this

Question 55

Michaelis-Menten

Answer 55

## Footnote Describes the relationship between initial velocity (Vo) and substrate concentration [S]. Gives us values for: Km (Michaelis constant) = substrate concentration point at which enzyme catalysed reaction is at half of its maximal velocity (½ Vmax). Vmax (Maximal velocity) = the fastest rate at which an enzyme can catalyse a reaction

Question 56

Michaelis-Menten plot

Question 57

Lineweaver-Burk

Answer 57

## Footnote Uses the reciprocals of the velocity (1/Vo ) and substrate concentration (1/[S]) to reproduce the data in a straight line Km = -1/Km = x-axis intercept Vmax = 1/Vmax = y-axis intercept

Question 58

Turnover Number (kcat)

Answer 58

## Footnote Amount of substrate converted to product, per amount of enzyme in unit time when the enzyme is operating near Vmax. Reported in seconds (s). where [Et ] = total enzyme concentration

Question 59

Specificity constant

Answer 59

## Footnote indicates the specificity of an enzyme for its substrate(s). Reported in seconds/M (s-1M-1 )

Question 60

Carbohydrates

Answer 60

## Footnote Comprised of carbon, hydrogen and oxygen in a ratio of 1:2:1 (Cn (H2O)n ). Simple - monosaccharides or disaccharides Complex - polysaccharides or oligosaccharides

Question 61

Carbohydrate roles

Answer 61

## Footnote Cell recognition Structural components Communication Source of Energy

Question 62

Cellular respiration

Answer 62

## Footnote glucose + oxygen --> carbon dioxide + water + energy (C6H12O6) + (6O2 ) --> (6CO2 ) + (6H2O) ENERGY (2870 kJ/mol) = 36-38 ATP

Question 63

Types of carbohydrates

Answer 63

## Footnote All carbohydrates are either an aldose or a ketose. The difference between an aldose and a ketose is the position of the carbonyl. Aldehydes are terminal carbonyls while ketones are carbonyls surrounded on each side by a carbon.

Question 64

Aldose and Ketose Carbohydrates

Question 65

Benedict's solution

Answer 65

## Footnote Used to detect reducing sugars like glucose. In the presence of reducing sugars solution changes colour from blue to red. Contains CuSO4 .5H2O + Na2CO3 + Na3C6H5O7 CuSO4 .5H2O source of cupric ions (Cu2+) Na2CO3 dissociates to produce OH-, increasing pH Na3C6H5O7 Binds Cu2+ to prevent binding with OH- until reduced to Cu+ Detects carbohydrates that contain an aldehyde or ketone group.

Question 66

Glucose regulation Insulin

Answer 66

## Footnote small peptide hormone (5734 da) stimulates uptake of glucose into cells

Question 67

Glucose regulation Hyperglycemia

Answer 67

## Footnote glucose levels above normal reference range (>6.0mM) not enough insulin/insulin resistance

Question 68

Glucose regulation Hypoglycemia

Answer 68

## Footnote glucose levels below normal reference range (<3.4mM) too much insulin (treatment for diabetes)

Question 69

ATP

Answer 69

## Footnote Adenosine triphosphate Captures chemical energy from the breakdown of macromolecules and releases it to fuel other cellular processes

Question 70

Photosynthesis

Answer 70

## Footnote (ATP is also produced)

Question 71

Fermentation

Answer 71

## Footnote C6H12O6 --> 2C3H6O3 + ATP Glucose --> Lactic Acid + ATP C6H12O6 --> 2C2H6O + 2CO2 + ATP Glucose --> Ethanol + Carbon Dioxide + ATP

Question 72

Glycolysis

Answer 72

## Footnote Glucose + 2ADP + 2Pi --> 2Pyruvate + 2ATP

Question 73

Gibbs Free Energy (∆G)

Answer 73

## Footnote ∆G = ∆H - T∆S ∆G: Change in free energy (energy available to do work and make changes to the system) ∆H: Change in enthalpy (energy contained in bonds of the molecules in the system) T: Temperature in Kelvin ∆S: Change in entropy (energy contained in the motional energy of components in the system)

Question 74

Understanding Gibbs Free Energy (∆G) (Exothermic)

Answer 74

## Footnote Free energy (∆G'0): -ve Reaction direction: Towards products Spontaneous: Yes Enthalpy (∆H): 0

Question 75

Understanding Gibbs Free Energy (∆G) (Endothermic)

Answer 75

## Footnote Free energy (∆G'0): +ve Reaction direction: Towards reactants Spontaneous: No Enthalpy (∆H): >0 (endothermic) Entropy (∆S): <0

Question 76

Understanding Gibbs Free Energy (∆G) (Free energy of 0)

Answer 76

## Footnote A reaction will never spontaneously move away from equilibrium A reaction will always move spontaneously towards equilibrium

Question 77

Fate of pyruvate Lactate Fermentation

Answer 77

## Footnote End Product: Lactate Cell Type: Skeletal muscle and some bacteria Requires O2?: No NADH is converted to NAD+, which can be used by cells to conduct another cycle of glycolysis and produce energy in the form of ATP

Question 78

Fate of pyruvate Alcoholic Fermentation

Answer 78

## Footnote End Product: Ethanol Cell Type: Yeast and some plant cells Requires O2?: No NADH is converted to NAD+, which can be used by cells to conduct another cycle of glycolysis and produce energy in the form of ATP

Question 79

Fate of pyruvate Cellular Respiration

Answer 79

## Footnote End Product: Carbon Dioxide and Water Cell Type: All cells Requires O2?: Yes

Question 80

Electrophoresis

Answer 80

## Footnote Electrophoresis is the migration of electrically charged molecules under the influence of an electrical field.

Question 81

Types of Electrophoresis

Answer 81

## Footnote Native SDS, SDS-PAGE

Question 82

Native PAGE Characteristic

Answer 82

## Footnote Separates on charge Separates on size Separates on shape Does not denature protein Allows for visualisation of multiprotein complexes

Question 83

SDS-PAGE Characteristic

Answer 83

## Footnote Does not separates on charge Separates on size Does not separates on shape Denatures protein Does not allow for visualisation of multiprotein complexes

Question 84

SDS

Answer 84

## Footnote Sodium dodecyl sulfate. Is an organic anionic surfactant

Question 85

SDS Roles

Answer 85

## Footnote Cleaning Emulsifying Lysing

Question 86

SDS Protein Separation

Answer 86

## Footnote Disrupts bonds Coats protein

Question 87

How does SDS work?

Answer 87

## Footnote SDS molecule with anionic polar head group. 3D arrangement of amino acids forming protein. SDS disrupts hydrogen and ionic bonds and hydrophobic interactions between amino acids. β-mercaptoethanol disrupts di-sulfide bonds. SDS binds to protein in an approximate ratio of one molecule of SDS to every two amino acids. Treatment with SDS reduces folded protein to linear molecule with net negative charge.

Question 88

Sample loading buffer

Answer 88

## Footnote High heat - denatures the protein SDS - disrupts bonds and coats protein in negative charges β-mercaptoethanol - disrupts di-sulfide bonds Glycerol - adds density to the sample Bromophenol blue - dye to visualise movement of sample

Question 89

Electrophoresis procedure

Answer 89

## Footnote Samples are loaded onto a polyacrylamide gel and a voltage is applied allowing the negatively charged denatured proteins to move towards the positively charged anode Large proteins will find it more difficult to migrate through the acrylamide gel due to friction or resistance

Question 90

Running buffer Characteristics

Answer 90

## Footnote pH = 8.3 Tris-HCl - buffering solution SDS - keeps the proteins linear Glycine - important role in the stacking gel.

Question 91

How to Calculate Relative Mobility

Answer 91

## Footnote Distance molecular weight markers travelled (Rf) / Distance Dye Travelled aka. the dye front (Rm) Rf/Rm

Question 92

Plotting Electrophoresis Results

Answer 92

## Footnote Plot log of the molecular weight (x-axis) against relative mobility (Rf) on the y axis for each of the markers. You can use this to find the unknown molecular weight/relative mobility of an unknown by using gradient y = relative mobility, x = log molecular weight (log mwt) After finding log of the weight, you inverse it by doing: 10^log mwt