Lecture test 2 Q2d: How pH changes affect enzymatic reaction

I have finished marking the lecture tests and I wish to highlight one question that was very poorly attempted in the test here. Although most of you are already aware that a change in pH from the optimum (more than or less than optimum) decreases enzyme activity, alot of you were unable to explain the mechanism behind this effect. Actually, the mechanism is already spelt out very clearly in your lecture notes so please go over it again to make sure you are clear about what I am saying here.

I also want to clarify some terms here as some of you seem to be confused over these terms. The enzyme is a globular protein that has undergone FOLDING to achieve its respective secondary/tertiary/quaternary (if more than one polypeptide chain) structures. Hence it has a specific CONFORMATION/configuration/3D structure as a result of the folding. And it is due to this conformation that the enzyme has a specific SHAPE in its ACTIVE SITE, which is complementary to the shape of its substrate. To summarise, "conformation" is not the same as "shape" so please don't use them interchangeably. The enzyme has a specific "conformation" while the active site has a specific "shape". Hope that this is clear.

Now let's move on to the mechanism. As I have already mentioned in class,  a change in pH changes the concentration of H+ and OH- ions, which causes the neutralisation of polar/charged R groups in the enzyme.

But what you should also know is that this effect can occur at the follwoing 2 areas of the enzyme. 

(i) At the active site -

At the active site, the catalytic and contact residues have charged R groups.

Hence, NEUTRALISATION of these CHARGED R groups disrupts the formation of INTERmolecular IONIC AND HYDROGEN bonds between the active site on the ENZYME and the SUBSTRATE. 

(ii) At other regions of the enzyme -

The bulk of the globular structure of the enzyme contain polar R groups (that form hydrogen bonds) and charged R groups (that form ionic bonds), which stabilise/maintain the secondary, tertiary and/or quaternary structures of the enzyme.

Hence, NEUTRALISATION of these POLAR/CHARGED R groups disrupts the INTRAmolecular IONIC AND HYDROGEN bonds maintaining the SECONDARY, TERTIARY and/or QUATERNARY structure of the enzyme. This causes the enzyme to unfold and lose its specific conformation (ie. denaturation) of the enzyme, hence causing the active site to lose its specific shape.

With both disruptions, binding of the substrate to the active site of the enzyme to form enzyme-substrate complexes, and hence products are prevented. And that is why the enzymatic activity decreases!

Importance of time management

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DNA and Genomics - Have you wondered...

Q1) DNA is a polymer of deoxyribonucleotides (that contain either A/C/G/T as its nitrogenous bases). For the genetic code contained in DNA to be able to code for all 20 amino acids, the genetic code has to be a triplet code.

Given the above information, calculate (means workings must be shown) how many codons there are altogether. How many of these codons code for amino acids? Explain your answer. 

Q2) Given that there are XXX number of codons coding for amino acids, there should be the same XXX number of tRNA molecules available, as the anticodon on tRNA complementary base pairs to the codon on the mRNA.


However, this is not true. Explain why. 

Enzymes tutorial Q4b - Although Km is not in your syllabus, you still have to draw the curves accurately!

Starch, amylase, and iodine test

1) Starch (amylose + amylopectin):

"Starch is generally insoluble in water at room temperature. Because of this, starch in nature is stored in cells as small granules which can be seen under a microscope. Starch granules are quite resistant to penetration by both water and hydrolytic enzymes due to the formation of hydrogen bonds within the same molecule and with other neighboring molecules.

However, these inter- and intra-HYDROGEN BONDS can become weak as the temperature of the suspension is raised. When an aqueous suspension of starch is heated, the hydrogen bonds break. "

(Source: http://www.eng.umd.edu/~nsw/ench485/lab5.htm)

Hence for boiled starch --> H bonds broken --> exposed amylose and amylopectin structure --> allow amylase to hydrolyse starch. 


In contrast, for unboiled starch --> amylase unable to hydrolyse starch as the starch granules are intact. 

2) Amylase:

Acts on α-1,4 glycosidic bonds (present in amylose and amylopectin). 

However, amylase is unable to break down amylopectin completely due to the α-1,6 glycosidic bonds present at the branch points. 

(Further reading: http://books.google.com.sg/books?id=xIj3YPKSpJkC&pg=PA453&lpg=PA453&dq=amylopectin+red+brown&source=bl&ots=uokPwYlqHY&sig=v_zrQ-kF7hCYj7So3lKARSbhBCo&hl=en&sa=X&ei=NM2dT6LCH8TJrAe8kcBm&ved=0CCcQ6AEwAQ#v=onepage&q=amylopectin%20red%20brown&f=false)

3) Iodine test 

The amylose (unbranched/linear portion of starch) forms helices, which allow iodine molecules to assemble, forming a dark blue/black color (refer to diagram below). 

The amylopectin (branched portion of starch) forms much shorter helices due to the branching present, and iodine molecules are unable to assemble, leading the color to be of red brown or red violet. 

As starch is broken down or hydrolyzed into smaller carbohydrate units, the blue-black color is not produced. Therefore, this test can determine completion of hydrolysis when a color change does not occur.

(Further reading: http://www.webexhibits.org/causesofcolor/6AC.html)

***Please refer to the SPA answer scheme (when it becomes available at IVLE) to see how the answers should be phrased for Q18,19 and 22.

Enzymes tutorial Q2ciii

Q) Suggest why the lysosomal membrane remains intact? [3]

Possible Answers:

[1st mark] 
- Lysosomal enzymes hydrolyse polysaccharides, lipids, proteins, and nucleic acids.

[2nd mark] 
- These differ from the constituents of the lysosomal membrane which comprise of phospholipids, membrane-bound proteins, glycolipids, and glycoproteins. [1]
OR 
- The inner surface of the lysosomal membrane is coated with an extensive glycocalyx (refer to pictures below).
OR
- The integral and inner peripheral membrane proteins and membrane phospholipids on the inner surface of the lysosomal membrane are highly glycosylated.
OR 
- There are large numbers of glycoproteins and glycolipids on the inner surface of the lysosomal membrane.

[3rd mark]
- As the constituents of the lysosomal membrane are not complementary to the active sites of the lysosomal enzymes, they cannot fit into the active sites and thus, remain unhydrolysed. [1]
OR
- The glycosylation on the inner lysosomal membrane prevents the lysosomal enzymes from accessing the membrane phospholipids and proteins.