GLYCOLYSIS

Video

The Video gives a visual explanation of glycolysis which I think provides useful in learning this may seem like a long complicated process. Guys, it is not difficult. It is just a little long and may be a little tedious but what do we have to lose. This is my interpretation of the video as well as some of my own knowledge in point form.
(1) Glycolysis has two main steps broken down into 5 enzyme catalyzed steps. In the first step glucose is phosphorylated and is converted to two glyceraldehydes-3-phosphate molecules. It is referred to as the investment stage. Whereas the second stage of glycolysis is referred to as the payoff phase where there is oxidative conversions of glyceraldehyde-3-phosphate to pyruvate and coupled formation of ATP and NADH molecules.
(2) Investment Stage Broken Down:
STEP1: Glucose is converted to glucose-6phosphate, the first irreversible reaction using ATP which is converted to ADP. It is catalysed by hexokinase and it cofactor Mg2+.
STEP 2: Glocuse-6-phosphate is converted to its isomer fructiose-6-phosphate which is a reversible reaction catalysed by phosphohexose isomerase. Mg2+ is again a cofactor.
Step 3: Another reaction where ATP is used and converted to ADP. This is an irreversible reaction catalysed by phosphofructose kinase 2 (pfk2). This enzyme is the most important regulatory enzyme in glycolysis. Fructose-6-phosphate is converted to fructose-1,6-phosphate.
STEP 4: Fructose-1,6-phosphate is converted into two, three Carbon compounds; glyceraldehydes-3-phosphate and dihydroxyacetone phosphate. The enzyme that catalyses this reaction is aldolase which is a reversible reaction.
STEP 5: Dihydroxyacetone phosphate is converted to its isomer, glyceraldehydes-3-phosphate catalysed by triose phosphate isomerase. This is a reversible reaction.
NOW WE ENTER THE 2ND STAGE (PAY OFF)
STEP6: Each glyceraldehydes-3phosphate is converted to 1,3-bisphosphateglycerates. In this reaction an inorganic phosphorous source converts two NAD+ to NADH + H. This is a reversible reaction catalysed by glyceraldehyde-3phosphate isomerase.
STEP 7: This is a dephosphorylation reaction where two molecules of ADP is converted to two molecules of ADP. The 1,3 bisphosphoglycerate molecules are converted to 3-phosphoglycerate by phosphoglycerate kinase and it cofactor Mg2+. It is a reversible reaction.
STEP 8: the two molecules of 3-phosphoglycerate is converted to 2-phosphoglycerate which is a reversible reaction catalysed by phosphoglucerate mutase.
STEP 9: The two 2-phosphoglycerate is converted to phosphoenolpyruvate catalysed by enolase. This is a reversible reaction in which two molecules of H2O is released.
STEP 10: the two molecules of Phosphoenolpyruvate is converted to two molecules of pyruvate. In this reaction two molecules of ADP is converted to ATP. Pyruvate kinase catalyses this reaction which is irreversible.

ENZYMES AND INHIBITION

Aside

Enzymes are biological molecules which catalyze metabolic reactions necessary to drive all life cycles. The do this lowering the activation energy of the reaction. However they do not change equilibrium, only the rate at which equilibrium is reached. Thus they do not change the free energies of the reactant or the products. Activation energy is the minimum amount of energy for a reaction to occur which is lowered by enzymes. This is the point of the reaction where the arrangement of intermediate structures between reactant and product is at its highest.

All enzymes are specific, some more than others. When an enzyme only reacts with a one certain substrate, it is said to have an absolute specificity. On the other hand some enzymes act on substrates of a particular functional group or side chains, having relative/group specificity and others, the least specific act on certain bonds. That is linkage specificity. Also there are stereochemical specific enzymes which act on a specific isomer.

Enzymes speed up the rate of reaction and inhibitors deter them from doing this. There are four types of irreversible inhibitors; competitive, non competitive, uncompetitive and mixed competitive inhibition. Inhibitors are important so to ensure that a surplus of product is regulated depending on the requirements of the body. 

Image

Competitive inhibitors compete for the active site with substrates where they bind. Competitive inhibitors can do this since they resemble the substrates. This will decrease the affinity of the active site to the substrate to an extent depending on how much inhibitor to substrate concentration in present. The higher the substrate concentration the inhibitor effect will decrease. Thus the Km increases and Vmax remains the same as show on graph (a)

Noncompetitive inhibitors bind on any site that is not the active site. It makes the enzyme undergo a conformational change. Competitive inhibitor can bind to either the free enzyme or the enzyme substrate complex. It does not resemble the substrate. The km remains the same and the Vmax decreases as shown in graph (b). Thus increasing enzyme substrate will not prevent the effect of the inhibitor. Mixed inhibitors have similar charactoistics however they differ by their kinetics in that they either decrease or increase km values and

Uncompetitive inhibitors bind only to the enzyme-substrate complex, not on the active site. Therefore there is no need for this inhibitor to resemble the substrate. Look at graph (c) and you will notice that the lines representing enzyme catalyzed reaction and inhibited enzyme catalyzed reactions you will notice the lines are parallel and you can see that the km is reduced by the amount that Vmax is being reduced.

Phenolase, peroxidase and xanthine dehydrogenase are enzymes, all of which catalyse oxidation reactions and therefore belong to the first class of enzymes, oxidoreductases. Phenolase also call polyphenol oxidase has the enzyme commission number 1.10.3.1.  Copper (II) ions is its inorganic cofactor.

AMINO ACIDS AND PROTEINS

Image

 

 

All amino acids have this basic structure. They differ by their R groups. The smallest R group is a hydrogen atom. The R groups can be of varying lengths of C chains whether they are polar, uncharged R groups, non polar aliphatic R groups, aromatic R groups or positively charged R and negatively charged R groups. Hence the R group is used to classify R groups.

Amino acids undergo oxidation reactions with each other to form peptide bonds between each other. Many amino acids can be linked by peptide bonds to form proteins. Proteins make most of the enzymes that allow the catalysis of metabolic reactions necessary for life to continue.

 

MCQS

Aside

HOW ABOUT A NICE SHORT AND SWEET QUESTIONAIRE FOR YOU

Select the correct multiple answer using one of the keys A, B, C, D OR E as follows:

  1. A. 1,2 and 3 are correct
  2. B. 1 and 3 are correct
  3. C. 2 and 4 are correct
  4. D. Only 4 is correct
  5. E. All are correct

I. In what type of cell would mitochondria be found in relatively large quantities?

1) Liver cells

2) Nerve cells

3) Muscle cells

4) Euthrocytes

II. In what type of cell would lysosomes be found in the greatest quantities?

1) Eukaryotes

2) Muscle cells

3) Nerve cells

4) Liver cells

III. Which statement/s best describes the golgi-apparatus

1) The golgi-apparatus receives mRNA which it uses to transcribe for proteins and then modifying the proteins and or adding signal sequences befoor it shipped off.

2) The golgi-apparatus receives vesicles containing proteins on it trans side which will then be modified, signal sequence may be added and packaged to be shipped off in vesicles on the cis side.

3) The golgi-apparatus is the site of lipid synthesis.

4) The golgi-apparatus receives vesicles containing proteins on it cis side which will then be modified, signal sequence may be added and packaged to be shipped off in vesicles on the trans side.

IV. Mannose and glucose are epimers. This means that the orientation of a H atom and an OH group differ only on a certain C atom. Which C atom does this occur?

1) 1

2) 2

3) 3

4) 4

V. Some importance functions of carbohydrates include:

1) Structure

2) Energy source

3) Precursor molecule

4) growth

Need help studying carbohydrates

http://ull.chemistry.uakron.edu/genobc/Chapter_17.pdf
Hey guys so this pdf was really helpfull.. Maybe it will be for you to. It is all abou the carbohydrates
It includes a few things i didnt tlk about in my posts including some starch and cellulose. Those are important to know. Also a small bit on lactose.
Also another important thing to remember some functions of carbohydrates:
1) energy source: we need to break down carbohydrates to obtain its stored energy to drive everyday metabolic reactions.
2) storage: listen we cant let sucrose build up in our blood. Yes it gives us energy but guys remember too much of a good thing is a bad thing. We animals use glycogen and plants use starch for storage. In animals hormones such as insulin and glucagon regulate concentrations of sucrose in our blood. Insulin convert glucose to glycogen. Glucagon convert glycogen to glucose.
3) structure: the more complex carbohydrates provide good, strong and rigid cell walls; cellulose for plant and murein for prokaryotes and insects have exoskelotons made of chitin.

Yup yup yup

20130412-052300.jpg

CARBOHYDRATES

Aside

CHARACTORISING CARBOHYDRATES
MONOSACCHARIDES DISACCHARIDES OLISACCHARIDES POLISACCHARIDES
One carbohydrate sub-unit. The simplest of the carbohydrateCan be further classified based on the length of the base carbon chain:

Triose – 3 C atoms

Tetros – 4 C atoms

Pentose – 5 C atoms

Hexose – 6 C atoms

Heptose – 7 C atoms

Can also be further classified based on the presence of either an aldyhyde or ketose fubctional group.  

Carbohydrate consists of two monosaccharides attached by a glycosidic bond. Examples:Sucrose – glucose and fructose

Lactose – glucose and galactose

Maltose – glucose and glucose

 

Carbohydrate consists of 5 to 10 monosaccharides attached by a glycosidic bond. Long chain carbohydrates consisting of hundreds or thousands of monosaccharides joined by glycosidic bonds.Examples: Amylose

Amylopectin

Cellulose

 

ALDOSE FUNCTIONAL GROUP vs KETOSE FUNCTIONAL GROUP

05_03Monosaccharides

ANOMER ISOMERISM

A monosaccharide can carry an alpha configuration or a beta comfiguration

Let us look at glucose: What is the difference between alpha (α) and beta (β) glucose?

image-3

Both pictures represent an isomer of glucose but take a closer look and tell me if you find a difference between the two… and guys it’s not a trick question. Yes they are both glucose and yes there is a difference.

Well look at the highlighted OH groups on both molecules; one is above the plane of the anomeric C atom and the other is below. So on the α glucose molecule the OH group is below the anomeric carbon and on the β glucose molecule the OH group is above the anomeric C atom. BUY OOOOOOO! What is that new word I’m using ‘anomeric’ what is that? Well that simply means the carbon atom that now the stereocenter as a result of cylization. It allows for this anomer isomerism.

CHIRALITY: simply means that there are four different bond attached to the central carbon atom. Molecules that express chirality are superimposable and have its mirror image.

D vs L ISOMER

glucos

The image on the right shows two isomers of glucose in the fischer projection; D-isomer and L-isomer.

NB: the functional group in glucose is the aldehyde group.

The D isomer shows that the OH group attached to the chiral carbon furthest from the functional group is on the right.

The L-isomer shows the OH group attacheched to the chiral carbon furthest from the functional group is on the left. 

images (32)

EPIMERS

Yet another type of isomerism. Epimers only differ in atomic configuration around one C atom. Take the example seen on the right ; mannose and glucose are epimers of each other since they differ only by the orientation of the OH group and H atom on C2. Likewise, galactose differs from glucose similarly only on C4.

GLYCOSIDIC BOND

glycosidic-linkage

Looking at the diagram on the right we can see a glycosidic bond being formed in a process called condensation.

2 α glucose molecules are bonding to form the maltose. The O atom on C1 of the 1st molecule is lost and the OH group. on C4 of the 2nd molecule is lost i.e. a water molecule is lost. The oxygen on C1 of the1st molecule can then bond with the C4 on the 2nd molecule. The reverse reaction would be hydrolysis.

Now do you think you can name some carbohydrates based on what we have learnt today?

maltose

If these two α glucose molecules under condensation what will we call the new product.

Well, the bond will take place between a C1 and a C4 and we know they are both α molecules and two glucose molecules make maltose

α1-4 maltose.

How about this one?

images (33)Since the answer is already there I’ll let you tackle this one on your own.

Last one. Remember guys practice makes perfect!! 

And this time I’m not giving you the answer

images (34)