• The enzyme that converts pyruvate to lactate is lactate dehydrogenase
  • The conversion of 1 molecule fructose 1,6-bisphosphate to 2 molecules of pyruvate by the glycolytic pathway results in a net formation of 2 NADH and 4 ATP.
  • In the alcoholic fermentation of glucose by yeast, thiamine pyrophosphate is a coenzyme required by pyruvate decarboxylase.
  • In eukaryotes, glycolysis typically occurs in the cytosol
  • Hexokinase is the enzyme that catalyzes the transfer of a phosphoryl group from ATP to glucose
  • During glycolysis, the steps between glucose and formation of the triose phosphates consume 2 ATP molecules
  • Pyruvate dehydrogenase is the enzyme that catalyzes the conversion of pyruvate to acetyl CoA (Link Rxn).
  • In skeletal muscle cells, the NADH that is produced by glycolysis under anaerobic conditions (vigorous exercise) is regenerated to NAD+by the conversion of pyruvate to lactate
  • The anaerobic conversion of 1 molecule of glucose to 2 molecules of lactate by fermentation is accompanied by a net gain of 2ATP
  • Pyruvate decarboxylase and alcohol dehydrogenase is the enzymes responsible for catalyzing the conversion of pyruvate to ethanol
  • Phosphoglycerate kinase and pyruvate kinase are the enzymes that are involved in substrate level phosphorylation in glycolysis
  •  In 1,3-Bisphosphoglycerate → 3-phosphoglycerate ATP “consumed” during glycolysis
  • Glycolysis in the erythrocyte produces pyruvate that is further metabolized to lactate
  • The process by which ATP is formed from ADP in glycolysis is referred to as oxidative phosphorylation
  • Fructose-6-phosphate → fructose-1,6-bisphosphate is considered to be the major control point of glycolysis
  • NAD+/NADH participates directly in most of the oxidation-reduction reactions in the fermentation of glucose to lactate
  • The steps of glycolysis between glyceraldehyde 3-phosphate and 3-phosphoglycerate involve all of the following :
  1. ATP synthesis
  2. the formation of 1,3-bisphosphoglycerate
  3. catalysis by phosphoglycerate kinase
  4. utilization of Pi


  • There are 238 amino acids are present in GFP. Fluorescence is caused by thee of these amino acids (Ser-65-Tyr-66-Gly-67). Coelenterazine binds to apoaequorin when oxygen is present to produce active aequorin. Coelenterazine is activated to colenteramide due to the bond of Ca++ aequorin that causes the conversion. The emission of blue light occurs when the colenteramide that was activated goes back to ground state. The GFP is activated due to the blue light and emits green light.
  • Federic Sanger acquired the 1st protein sequencing of bovine insulin. This was achieved in the year 1953. (
  • Disulfide bonds make the cross-links in α-keratin.
  • Gelatin is almost entirely composed of protein but yet contains a very insignificant amount of nutrition. This is because it contains various amino acids but has a high proportion of non-essential amino acids such as proline and glycine and fail to contain some essential amino acids (tryptophan, isoleucine, threonine and methionine).
  • GLYCINEIn collagen glycine occurs third in the chain of amino acids. It is known as the smallest amino acid which provides for its flexibility. Glycine modifies itself to give allowance for the chain to form a helix.
  • PROLINEThe backbone of proline consists of a 5-membered ring. Proline appears approximately as every 5 amino acids. This ring structure of proline aids in the bending of the chain where this amino acid displays its presence.
  • Myoglobin is an oxygen binding molecule which is in high concentration in whales, seals and porpoise to aid in providing sufficient oxygen for deep diving. Myoglobin contains a high amount of iron which is brown in nature, thus its high concentration in the muscles of these animals results in its brown appearance.
  • Osteogenesis imperfecta results in fragile bones or breakage of bones. It is caused by a genetic disorder. The modification of collagen occurs due to the switch of amino acids glycine and a larger one. Due to this larger appearance of this amino acid the side chains are expanded much more than glycine causing steric hindrance. This poorly formed collagen structure is hydrolyzed. Due to the inability to get rid if this poor structure, proper formation of the bone is compromised due to the denied interaction between collagen fibrils and hydroxyapatite.
  • A deficiency of vitamin C may result in scurvy. The formation of collagen requires the presence if ascorbic acid. Proline, an amino acid that is present in collagen is converted to hydroxyproline via post-translational modification. The enzyme prolyl hydroxylase is responsible for catalyzing this reaction. Fe2+ ion is necessary for the functioning of the enzyme. Due to the unstable nature of Fe2+ it is readily oxidized. Ascorbic acid plays an important role in aiding the Fe to remain unstable in the 2+ state reather than a stable 3+ state.  



NADH and FADHbring electrons (high transfer potential) that pass along the cytochrome complex. There are four complexes located on the inner membrane of the mitocondria, they use energy to pump hydrogen ions out of the matrix into the inter membrane space against a gradient. This process is known as oxidative phosphorylation. Complex II does not pump hydrogen ions. Hydrogen ions would want to return to the matrix but are unable to because the inner membrane is impermeable to hydrogen ions. A proton motive force (PMF) is applies where hydrogen ions goes through ATP syntase. Protons pass through ATP syntase to give conformational change to give energy to convert ADP to ATP.


  • Glycolysis is an example of fermentation
  • The anaerobic conversion of 1mol of glucose to 2mol of lactate by fermentation is accompanied by a net gain of 2mol ATP
  • NAD+ is a cofactor in the reaction catalysed by glyceraldehydes 3-phosphate dehydrogenase.
  • Three irreversible enzymes in glyciolysis are
  1. Hexokinase
  2. Phosphate fructokinase 1 (PK1)
  3. Pyruvate kinase
  • Two enzymes involved in substrate level phosphorylation
  1. Phosphoglycerate kinase
  2. Pyruvate kinase
  • Cells perform glycolysis to generate ATP
  • Fermentation reactions are important
  1. To generate NAD+ for glycolysis to continue
  2. There is a limited amount of NAD+ in the cell
  3. Prokaryotes rely solely on glycolysis to obtain energy 


The Effects of Enzyme Inhibitors

Enzymes can be inhibited

  • competitively, when the substrate and inhibitor compete for binding to the same active site or
  • noncompetitively, when the inhibitor binds somewhere else on the enzyme molecule reducing its efficiency.

The distinction can be determined by plotting enzyme activity with and without the inhibitor present.

Competitive Inhibition

In the presence of a competitive inhibitor, it takes a higher substrate concentration to achieve the same velocities that were reached in its absence. So while Vmax can still be reached if sufficient substrate is available, one-half Vmax requires a higher [S] than before and thus Km is larger.

Noncompetitive Inhibition

With noncompetitive inhibition, enzyme molecules that have been bound by the inhibitor are taken out of the game so

  • enzyme rate (velocity) is reduced for all values of [S], including
  • Vmax and one-half Vmax but
  • Km remains unchanged because the active site of those enzyme molecules that have not been inhibited is unchanged.


  • Four internationally accepted classes of enzymes include
  1. Hydrolases
  2. Ligases
  3. Oxidoreductases
  4. Transferases

ü  Polymerases are NOT included

  • Enzymes are potent catalysts because they lower the activation energy for the reactions they catalyse
  • Enzyme catalysts can increase the reaction rate for a given reaction by a thousand fold or more
  • The Michaelis-Menten Curve
  • A plot of V0 vs [S] for an enzyme that follows Michaelis-Menten kenetics shows

ü  As [S] increases, the initial velocity of reaction V0 also increases

ü  Km is the [S] at which V0 = ½ Vmax

ü  The shape of the curve is a hyperbola

ü  The y-axis is a rate term with units of amount or concentration/unit time


  • Several simplifying assumptions allow for the derivation of the Michaelis-Menten equation:

ü  (1) The binding step (E + S ES ) is fast, allowing the reaction to quickly reach equilibrium ratios of [E], [S], and [ES].

ü  The catalytic step (ES E + P) is slower, and thus rate-limiting.

ü  (2) At early time points, where initial velocity (Vo) is measured, [P] ≈ 0.

ü  (3) ES immediately comes to steady state, so [ES] is constant (throughout the measured portion of the reaction).

ü  (4) [S] >> [ET], so the fraction of S that binds to E (to form ES) is negligible, and [S] is constant at early time points.

ü  (5) The enzyme exists in only two forms: free (E), and substrate-bound (ES). Thus, the total enzyme concentration (ET) is the

ü  sum of the free and substrate-bound concentrations: [ET] = [E] + [ES]

  • The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the turnover number
  • A small molecule that decreases the activity of an enzyme by binding to a site other than the catalytic site is termed an allosteric inhibitor
  • The Line-Weaver-Berk Plot
  • Plotting the reciprocals of the same data points yields a “double-reciprocal” or Lineweaver-Burk plot. This provides a more precise way to determine Vmax and Km.

ü  Vmax is determined by the point where the line crosses the 1/Vi = 0 axis (so the [S] is infinite).

ü  Note that the magnitude represented by the data points in this plot decrease from lower left to upper right.

ü  Km equals Vmax times the slope of line. This is easily determined from the intercept on the X axis.



  • Phospholipids, carbohydrates, proteins and cholesterol are all components of biological membranes. Triacylglycerol is a neutral fat and is not present in biological membranes.
  • Beeswax, prostaglandins, sphingolipids and triacylglycerols contain or are derived from fatty acids.
  • Cholesterol is a sterol that is commonly found in mammals. Sterols are more common in plasma membranes than the intracellular membranes (mitochondria, lysosomes etc.) and are precursors of steroid hormones. Sterols have a structure that includes four fused rings.
  • Tay-sachs disease is the result of a genetic defect in the metabolism of gangliosides.
  • Prostaglandins mediates pain and inflammation
  • Sphingolipids are an important component of myelin membranes
  • Thromboxanes aid in blood clotting
  • Vitamin A is necessary for sight
  • Vitamin K is necessary for blood clotting
  • Vitamin D aids in Ca2+ and phosphate metabolism
  • Vitamin E is necessary for the prevention of oxidative damage


  • The amino acids found in naturally occurring proteins are all the L isomers.
  • The ionizable side chains of amino acids all have pKas that are higher than that of the α carbonyl group
  • All amino acids have an α amino and α carboxyl group
  • Hydrophobic amino acids prefer non polar solutions. They are found in the interior of a protein
  • Protein molecules only contain quaternary structure if multimeric
  • Multimeric proteins must contain more than one polypeptide chain
  • Myoglobin is used for storageand is a monomeric polypeptide
  • What bonds are responsible for primary structure?

ü  Peptide bonds

ü  Linear form

  • The bonds responsible for secondary structure are hydrogen bonds between the peptide bonds.
  • Α helix is a stable structure due to its hydrogen bonds
  • Proline is not in the α helix form since

ü  α-C-N bond can’t rotate freely

ü  a kink is formed

ü  hydrogen bonds keeps α helix together

  • Glycine destabilizes the helix since

ü  It has a high conformational flexibility

ü  It locks a hydrogenon nitrogen, therefore, it can’t form a proper network of hydrogen bonds

  • The α helix is stabilized by hydrogen bonds in parallel. Each hydrogen bond gives a dipole to give a macrodipole.
  • The ninhydrin test is used to test for amino acids. A positive test gives a purple colour change, a yellow colour change may be seen for some amino acids.
  • The biuret test is used to test for proteins. Positive test gives a purple colour change.



This video review is based on an advertisement for a product – IC-5. IC-5 is a synergistic blend of five unique hard to combine ingredients that have been clinically proven to increase insulin sensitivity and bold sugar management. It is a product designed for weight loss.

The consumption of carbohydrates makes weight loss very difficult. On consumption of carbohydrates a great amount of sugar becomes present in the bold stream. A hormone insulin is responsible for maintaining blood sugar levels in our body. Due to consumption of processed insulin and blood sugar spiking carbohydrates most people suffer from some level of insulin resistance; insulin is no longer able to efficiently remove blood sugar from the blood stream causing dramatically reduced fat burning and increased fat storage. Insulin resistance leads to type II diabetes and other health problems. This occurs due to insulin sensitivity.

Upon consumption of carbohydrates, the body needs to:

  1. Minimize insulin release – When the body is highly sensitive to insulin, only a small amount of insulin is needed to clear glucose from the blood and its storage site. The body has a difficult time in burning fat in the presence of insulin.
  2. Have quick and efficient blood sugar clearance – Again, the body is highly sensitive to insulin.
  3. Have maximum glycogen uptake – Glycogen is used to restore carbohydrates in muscles. When muscle tissue is highly sensitive to insulin the majority of blood glucose is stored in the muscles and not as fat.
  4. Minimize fat storage – there is high insulin sensitivity especially in skeletal muscles. The body chooses to store carbohydrate intake as energy in muscle tissue instead of body fat.

The body’s ability to process carbohydrates is dependant on insulin sensitivity and the body’s ability to quickly and efficiently shuttle carbohydrates to muscle tissue and not fat.

Insulin sensitivity can be increased by going on a low carbohydrate diet for an extended period of time. This attempts to repair insulin receptors. However, carbohydrates need to be slowly reintroduced to the body to prevent massive fat gain.

The video seeks to show that there is no need to go on low carbohydrate diet since there is a product IC-5 that is a super charger insulin sensitivity product that could be taken before each carbohydrate meal. As previously stated IC-5 is made up of five ingredients:

  1. Cinnamomum Burmannii – extract harvest from Indonesian cinnamon bark that assist in managing blood sugar after a carbohydrate rich meal while increasing glucose metabolism by up to 10 fold.
  2. Berberine – rare plant alkaloid. Improves the signaling of insulin between insulin and its associated receptors while also increasing glucose uptake by skeletal muscle and not fat.
  3. Pterocarpus  Marsupium – used for the treatment of diabetics. It reduces blood sugar levels 2hours after a meal by 21%
  4. 4-hydroxyisoleucine – natural photochemical extracted from the Fenugreek herb that increases glycogen storage while decreasing fat storage through the sensitizing of insulin receptors in muscle tissue. It also interferes with gluconeogenesis, thus sparing muscle protein while encouraging fat loss.
  5. R-α-Lipoic-Acid. Consists of two isomers, one of which is shown to be more effective and biologically available. Hence, it is included in the IC-5 formula.

As for all products there is a convincing advertisement along-side it. In my opinion, the best way to loose fat/weight is to exercise and maintain a balanced diet. Most of these formulas consist of chemicals that does more harmful to the body and causes major side effects. In this video however, the five active ingredients seem to be all natural but as an important aspect of advertising, the harmful substances are left out and the beneficial ones are highlighted.



So far we have studied cells, their organelles and carbohydrates.

Sometimes we are so focused on individual topics that we forget to look at the ‘big picture’ or the application of this knowledge.

Scientists have used, what we know now of cells and carbohydrates, to make advancements in medicine, nutrition and agriculture.

Cell Size

As cell size increases the surface area to volume ratio decreases. This restricts the

maximum size of cells. Some organisms have adapted to overcome these restrictions.

We shall look at some of them now:

Question 1

The nucleus is responsible for controlling metabolism in the cytosol. For efficient

control, it is important to maintain a critical nucleus to cytoplasm ratio. As cells

increase in size, this ratio diminishes. Suggest how some cells have adapted to this

challenge giving an example of such cells. [1mk]


Particular cells such as muscle cells in animals and haphae of fungi contain more than one  nucleus. This allows for the movement via osmosis to occur efficiently since the volume ratio of the nucleus to cytoplasm would be in proportion. Also, in liver cells, due to cell volume and proliferative activity the cells swell which occurs due to osmotic stress.


Question 2

Explain what is meant by cytoplasmic streaming. [2.5 mks]


Cytoplasmic streaming deals with the movement of nutrients, proteins and organelles within plant and animal cells. It involves a molecule of 2 proteins that moves 1 protein wrt the other. To move organelles, 1 protein must be fixed to a substrate and the motor protein moves organelles and molecules through the cytoplasm.

As the cell size increases the process won’t be as efficient since the movement of organelles and molecules would have to move further and it would therefore take longer.


Organelles and Human Disease

Organelles can contribute to a disease state in several ways. First, the organelle itself may be dysfunctional either because it contains one or more defective biomolecules that impair function, or because it has been damaged by exposure to harmful substances such as chemicals, heavy metals, or oxygen radicals. Second, an organelle can, through its normal function, exacerbate damage occurring elsewhere in the cell.

Pathophysiology is the study of the biologic and physical manifestations of disease as they correlate with the underlying abnormalities and physiologic disturbances. Pathophysiology does not deal directly with the treatment of disease. Rather, it explains the processes within the body that result in the signs andsymptoms of a disease.

For example the underlying pathophysiologic defect in type 1 diabetes is an autoimmune destruction of pancreatic beta cells. Following this destruction, the individual has an absolute insulin deficiency and no longer produces insulin. Autoimmune beta cell destruction is thought to be triggered by an environmental 2event, such as a viral infection. Genetically determined susceptibility factors increase the risk of such autoimmune phenomena. Since the pancreas no longer produces insulin, a type 1 diabetes patient is absolutely dependent on exogenously administered insulin for survival. People with type 1 diabetes are highly susceptible to diabetic ketoacidosis. Because the pancreas produces no insulin, glucose cannot enter cells and remains in the bloodstream. To meet cellular energy needs, fat is broken down through lipolysis, releasing glycerol and free fatty acids. Glycerol is converted to glucose for cellular use. Fatty acids are converted to ketones, resulting in increased ketone levels in body fluids and decreased hydrogen ion concentration (pH). Ketones are excreted in the urine, accompanied by large amounts of water. The accumulation of ketones in body fluids, decreased pH, electrolyte loss and dehydration from excessive urination, and alterations in the bicarbonate buffer system result in diabetic ketoacidosis (DKA). Untreated DKA can result in coma or death.

Question 3

Describe the pathophysiology of the following diseases.

(i) Tay-Sachs disease [2.5mks]

(ii) Leigh’s disease [2.5mks]

(iii) Pompe’s disease [2.5 mks]


1)      Tay Scah’s disease

  • Rare inherited disorder
  • Destroys nerve cells in the brain and spinal cord
  • HEXA gene mutation
  • HEXA gene allows the enzyme β-hexosaminidase A to be made
  • The enzyme breaks down GM2 ganglioside
  • Too much GM2 ganglioside causes the destruction of neurons

Leigh’s disease

  • Severe neurological disorder
  • Mutation in 1 of over 30 genes in nuclear DNA
  • These genes correspond with energy production via mitochondria
  • In oxidative phosphorylation 5 protein complexes are involved which drives the production of ATP. The 5 complex are disrupted by the mutation

Pompe’s disease

  • Inherited disorder
  • Build up of glycogen causes malfunctioning of organs, tissues and muscles
  • GAA gene is mutated
  • GAA gene is associated with the formation of the enzyme acidic α-glucosidase
  • This enzyme functions simultaneously with lysosomes. It breaks glycogen to glucose



Honey, the first sweetener known to humankind, is the only sweetening agent that can be stored and used exactly as produced in nature. Bees process the nectar of flowers so that their final product is able to survive long-term storage at ambient temperature. Used as a ceremonial material and medicinal agent in earliest times, honey was not regarded as a food until the Greeks and Romans. Only in modern times have cane and beet sugar surpassed honey as the most frequently used sweetener. The bees’ processing of honey consists of (1) reducing the water content of the nectar (30% to 60%) to the self-preserving range of 15% to 19%, (2) converting the significant amount ofsucrose in nectar to glucose and fructose by enzyme actionand (3) producing small amounts of gluconic acid from glucose by the action of the enzyme. Most of the sugar in the final product is glucose and fructose, and the finalproduct is supersaturated with respect to these monosaccharides.

Question 4

State the type of reaction occurring and the enzyme that the bees use to convert:

(i) sucrose to glucose and fructose [1mk]

(ii) glucose to gluconic acid (draw the structure of gluconic acid. [2mks]

The fructose in honey is mainly in the β-D-pyranose form. This is one of the sweetest carbohydrates known, about twice as sweet as glucose; the β-D-furanose form of fructose is much less sweet.

(iii) Draw β-D-Fructopyranose and β-D-Fructofuranose. [1mk]

The sweetness of honey gradually decreases at a high temperature. Also, highfructose corn syrup (a commercial product in which much of the glucose in corn

syrup is converted to fructose) is used for sweetening cold but not hot drinks.

(iv) Account for these observations? [1mk]


i) Bees use the enzyme invertase to convert surose to glucose and fructose via hydrolysis.

ii) The enzyme glucose oxidase converts glucose to gluconolactone, which is hydrolysed to gluconic acid.


 β-D-Fructopryanose (

 β-D-Frutofuranose (

iv) The cyclization of straight chain fructose results in pyranose or furanose. These are not as sweet as fructose ;when temperature increases the equilibrium shifts.


Side note:

Although most textbooks show fructose exclusively in its furanose form, the predominant form of fructose (67% of total fructose) is β-D-fructopyranose, with the β- and α-fructofuranose forms accounting for 27% and 6% of the fructose, respectively.

Question 5

Lactose exists in two anomeric forms, but no anomeric forms of sucrose have been reported. Why? [1mk]


Sucrose has 2 monosaccharides that contain glycosidic bonds where the anomeric carbons are associated. No mutarotation ours since there are no free anomeric carbons in the disaccharide.

Question 6

Explain why pectin is sometimes added to fruit extracts to make jams and jellies. [1 mk]


Pectin holds cell walls together. It forms a barrier that aids in trapping liquids. It works well with aid and sugar to form a jelly like substance.

Question 7

Insects use an open circulatory system to circulate hemolymph (insect blood). The ‘blood sugar” is not glucose but rather trehalose, an unusual, nonreducing disaccharide. Explain why trehalose is a nonreducing sugar. [2mks]


2 glucose molecules bonded by 1-1 α bond gives the nonreducing sugar trehalose. This bond allows trehalose to be resistant to acid hydrolysis. The enzyme trehalase is responsible for breaking these 2 glucose molecules for absorption into the gut.

Question 8

The following words were matched with their descriptions. [3mks]

Chitin – Give rigidity and strength to exoskeletons.

Chondroitin sulphate – Contribute to the tensile strength of cartilage, tendons, ligaments, and the walls of the aorta.

Dermatan sulphate – Is a component of the extracellular matrix of skin also present in blood vessels and heart valves.

Hyaluronate – Is an important component of the vitreous humor in the eye and of synovial fluid, the lubricant fluid of joints in the body.

Lectin – Found in all organisms, are proteins that bind carbohydrates with high specificity and with moderate to high affinity

Peptidoglycan – Give rigidity and strength to cell envelopes.


Did You Know

‘In 1869, concerned over the precipitous decline (from hunting) of the elephant population in Africa, the billiard ball manufacturers Phelan and Collander offered a prize of $10,000 for production of a substitute for ivory. Brothers Isaiah and John Hyatt in Albany, New York, produced a substitute for ivory by mixing guncotton with camphor, then heating and squeezing it to produce celluloid. This product found immediate uses well beyond billiard balls. It was easy to shape, strong, and resilient, and it exhibited a high tensile strength. Celluloid was eventually used to make dolls, combs, musical instruments, fountain pens, piano keys, and a variety of other products. The Hyatt brotherseventually formed the Albany Dental Company to make false teeth from celluloid. Because camphor was used in their production, the company advertised that their teeth smelled “clean,” but as reported in the New York Times in 1875, the teeth also occasionally exploded!’