LAB 1

Question 1

Mol to mmol

Answer 1

X1000

Question 2

Mmol to mol

Answer 2

÷ 1000

Question 3

mol to µmol

Answer 3

x1000000

Question 4

µmol to mol

Answer 4

÷ 1000000

Question 5

mmol to µmol

Answer 5

X1000

Question 6

µmol to mmol

Answer 6

÷ 1000

Question 7

L to ml

Answer 7

X1000

Question 8

mL to L

Answer 8

÷ 1000

Question 9

ml to µL

Answer 9

X1000

Question 10

µL to ml

Answer 10

÷ 1000

Question 11

µL to L

Answer 11

÷ 1000000

Question 12

L to µL

Answer 12

X1000000

Question 13

100 ÷ 1000 pipette

Answer 13

P1000 pipette, the first/top digit is thousands of µL, the middle digit is hundred, while the third is tens

1000 µL would read 100
350 µL would read 035

100-1000 µL

Question 14

20 ÷ 200 pipette

Answer 14

P200 pipette, the first/top digit is hundreds of µL, the middle digit is tens, while the third is single units.

200 µL would read 200
95 µL would read as 095

20-200 µL

Question 15

2 ÷ 20 pipette

Answer 15

P20 pipette, the first/top digit is tens of µL, the middle digit is single units, while the third is tenths.

20 µL would read 200
2.5 µLwould read 025
2-20 µL

Question 16

Denaturation because of pH

Answer 16

At low pH ionisable side chains on amino acids become protonated, while high pH they become deprotonated. pH induced alteration of the ionisation status of a protein will result in the disruption of non covalent bonds that would normally stabilise the protein fold

Question 17

Denaturation because of heat

Answer 17

Heating a protein will increase the kinetic energy of the molecules in the solution. When you add sufficient heat energy to the molecules they start to vibrate so rapidly that the weaker non covalent chemical bonds will be disrupted . The breaking of the non-covalent bonds in a protein can destabilise the structure and cause the protein to denature

Question 18

Starch

Answer 18

Starch is a carbohydrate found in plants. It consists of two different types of polysaccharides that are made up of glucose monomer units which form two different structural arrangements. One is the linear but coiled amylose and the other is branched amylopectin.

Question 19

Salivary amylase protein structure and function

Answer 19

Salivary amylase is an enzyme that initiates carbohydrate breakdown in the mouth, digesting starch specifically, hydrolysing polysaccharides into oligo-, tri- and disaccharides. As the partially digested starch reaches the stomach, the acidic conditions inactivates the salivary amylase, and pancreatic amylase is secreted into the small intestine subsequently hydrolyses the partially digested starch into disaccharides

Salivary amylase is a monomeric protein whose single polypeptide chain comprises of 496 residues. Its secondary structure includes a total of 20 alpha helices and 42 beta strands. These form functional domains including a beta barrel made up of 8 antiparallel beta strands surrounded by 8 alpha helices i.e. an alpha/beta barrel - this makes up the catalytic domain and a greek key beta barrel. In addition to non-covalent bonds, the territory structure of the enzyme is stabilised by 5 disulphide bonds.

Salivary amylase relies on several key residues to perform catalysis, including active site residues that act as a nucleophile and proton donor. There are also several histidine (His) residues that are essential in the recognition and binding of the starch substrate, as well as being involved in catalysis. Charged and/or polar amino acids such as arginine (Arg), asparagine (Asn), aspartate (Asp) and His are critical in holding the catalytically important Cl- and Ca+ ions in place

Question 20

Iodine test

Answer 20

In a solution containing starch, iodine will bind to the amylose component of the carbohydrate, causing a colour change to dark blue/black. In the absence of starch, the iodine has nowhere to bind, so the colour of the solution will resemble to pale yellow/orange colour of the iodine itself.

Question 21

Results of amylase experiment

Answer 21

Colour change in positive control (no amylase), 80 degree treated salivary amylase, HCl toasted salivary amylase (the last two are a result of the denaturing of the amylase)

Following heating of the amylase protein to 80C, we see a deep blue/black colour following the iodine starch test. This indicates that starch is still present in the solution and has therefore not been broken down by the amylase protein, suggesting that the amylase protein has been denatured by the high heat. We do not see this effect at 40C or room temperature. With regards to HCl treatment, we also see a colour change to deep blue/black, again indicating that acidic conditions caused the amylase protein to denature and therefore unable to function in its role of carbohydrate digestion.

Question 22

What effect does heat have on chemical bonds?

Answer 22

High heat breaks the non-covalent bonds in the amylase causing it to denature

Question 23

What happens when it is heated to 80 degrees?

Answer 23

At this temperature, a significant number of the non covalent bonds break which impacts on the function of amylase as it will be denatured and the hydrophobic core will now be exposed

The secondary and tertiary structure will now be affected

Question 25

At low pH, explain how extensive protonation of ionisable side chains of salivary amylase affects the function of the protein

Answer 25

Low pH means lots of H+ - affects non covalent bonds - can cause structural change therefore change in function as well as exposing the hydrophobic core of the protein

Question 26

Dilution factor

Answer 26

Concentration of undiluted solution A/ concentration of dilute solution A = whole number

Concentration of diluted solution A/ concentration of undiluted solution A = fraction

Dilution factor = total volume / volume of solution A

Question 27

Dilution factor of 5 is also the same as …

Answer 27

1 in 5 dilution
1/5 dilution
5 fold dilution
1:4 dilution (1 part solution A to 4 parts solution B)

Question 28

Solving dilution calculations using a formula

Answer 28

Initial concentration x initial volum = final concentration x final volume

Final = from the diluted solution