Lecture 2

Question 1

Lipids are more related by their physical properties than their chemical properties true or false

Answer 1

True

Question 2

What are lipids?
They are related more by their chemical than by their physical properties.
True or false

Answer 2

False

Fat and oils structures
The lipids are a heterogeneous group of naturally occurring compounds; including fats, oils, steroids, waxes, and related compounds, that are related more by their physical than by their chemical properties.

Question 3

State the different categories of lipids

Note that sugars are hydrophilic

Answer 3

1.Fatty acids- saturated and unsaturated

2.Glycerides-neutral glycerides and phosphoglycerides: Neutral glycerides, also known as neutral fats or triglycerides, are a type of lipid composed of glycerol and three fatty acids. These molecules are called “neutral” because they are nonpolar and do not carry a charge. They are the most common form of fat found in the human body and in many types of food.

Structure

•	Glycerol Backbone: A three-carbon molecule with each carbon bearing a hydroxyl group (-OH).
•	Fatty Acids: Long hydrocarbon chains with a carboxyl group (-COOH) at one end. Each fatty acid forms an ester bond with one of the hydroxyl groups on glycerol.

Formation

The formation of a triglyceride involves an esterification reaction, where the hydroxyl groups of glycerol react with the carboxyl groups of three fatty acids, releasing three molecules of water. Glycerides, also known as acylglycerols, are esters formed from glycerol and fatty acids. Glycerides are different from glyco- which is related to sugar or carbs.
They are a major component of lipids in the body and in food. Glycerides are classified into three types based on the number of fatty acids attached to the glycerol molecule:

1.	Monoglycerides: One fatty acid attached to glycerol.
2.	Diglycerides: Two fatty acids attached to glycerol.
3.	Triglycerides: Three fatty acids attached to glycerol.

Triglycerides are the most common form of glycerides and are the main constituents of body fat in humans and animals, as well as vegetable fat. They serve as an important energy source and are stored in adipose tissue. Glycerol is hydrophilic due to its three hydroxyl (–OH) groups. These hydroxyl groups are highly polar and can form hydrogen bonds with water molecules, making glycerol soluble in water.however, while glycerol is part of a molecule that can be hydrophobic overall, glycerol itself is hydrophilic. When a fatty acid combines with glycerol, the result is a glyceride. The process involves esterification, where the hydroxyl groups of glycerol react with the carboxyl groups of fatty acids, forming ester bonds. Glycerides are generally hydrophobic due to their structure.

Here’s why:

•	Fatty Acids: The fatty acid chains in glycerides are long hydrocarbon chains, which are nonpolar and hydrophobic. This hydrophobic nature dominates the molecule’s overall properties.
•	Glycerol Backbone: While glycerol itself is hydrophilic due to its hydroxyl groups, the hydrophobic fatty acid chains reduce the overall hydrophilicity when attached.

Thus, glycerides are mostly hydrophobic, which makes them insoluble in water but soluble in nonpolar solvents. This property is important for their role in energy storage and forming lipid bilayers in cell membranes.

3.Non glyceride lipids-waxes,steroids,sphingolipids(sphingomyelins and Glycolipids): These lipids do not contain glycerol in their structure. Some key types of non-glycerine lipids include:

1.	Fatty Acids: Simple lipid molecules consisting of a carboxyl group attached to a hydrocarbon chain.
2.	Steroids: Characterized by a four-ring carbon structure, including cholesterol and steroid hormones.
3.	Waxes: Esters of long-chain fatty acids with long-chain alcohols, providing protective coatings.
4.	Terpenes: Composed of isoprene units, including essential oils and carotenoids.
5.	Sphingolipids: Contain a sphingosine backbone and are important in cell membranes, particularly in the nervous system. 6. Glycolipids: Glycolipids are molecules that consist of a lipid tail and a sugar (or carbohydrate) group. Glycolipids can have either a glycerol backbone or a sphingosine backbone. glycolipids can indeed have glycerol if they are glycoglycerolipids. If they are glycosphingolipids(especially in the nervous system)I, they do not contain glycerol but instead have a sphingosine backbone. 7. Phospholipids: Contain a phosphate group and are essential components of cell membranes. The most common type is phosphatidylcholine. Categorization Clarification

•	Non-Glycerine Lipids: Phospholipids with a sphingosine backbone (sphingophospholipids) fall into this category since they do not contain glycerol.
•	Complex Lipids: Both glycerophospholipids and sphingophospholipids fall into this category because they contain additional groups like phosphate and nitrogenous bases, making them more complex than simple lipids.

Summary

•	Glycerophospholipids: Complex lipids with a glycerol backbone.
•	Sphingophospholipids: Complex lipids with a sphingosine backbone, also considered non-glycerine lipids.

So, phospholipids can be classified based on their complexity (as complex lipids) and their backbone structure (as either glycerophospholipids or sphingophospholipids).

4.Complex lipids- lipoproteins

Key Differences

1.	Structure:
•	Non-Glycerine Lipids: Do not contain glycerol.
•	Complex Lipids: Contain glycerol and additional components like phosphate groups, carbohydrates, or proteins.
2.	Functions:
•	Non-Glycerine Lipids: Serve various roles including hormone production (steroids), energy storage (fatty acids), and protective barriers (waxes).
•	Complex Lipids: Primarily structural components of cell membranes (phospholipids), involved in cell recognition and signaling (glycolipids), and lipid transport (lipoproteins).
3.	Presence in Biological Systems:
•	Non-Glycerine Lipids: Found in various tissues and perform a wide range of functions.
•	Complex Lipids: Predominantly found in cell membranes and involved in cellular interactions and metabolism.

Question 4

State and define four physical and chemical characteristics each of lipids (this is biochem so we are more interested in the chemical aspect)

Answer 4

Physical
➢Lipids are either liquid or non-crystalline solid at room temp-example of liquid is oil and example of solids are waxes
➢Insoluble in water-example is waxes
➢pure fats and oils are colorless, odorless and tasteless
➢No ionic charges- only is sphingophospholipids have ionic charges . Most lipids are neutral and do not carry ionic charges. However, sphingophospholipids, such as sphingomyelins, can have ionic charges due to their phosphate groups.
➢Pure fats and oils are energy rich molecules. Pure fats and oils are energy-dense molecules, providing a high amount of energy when metabolized.
➢Soluble in organic solvents like chloroform, ether and acetone. Organic Solvents:
- Contain carbon.
- Dissolve organic compounds.
- Examples: hydrocarbons(such as Benzene, toluene,xylene),alcohols(ethanol,isopropanol),ethers such as(diethyl ether and tetrahydrofuran),esters(such as ethyl acetate and butyl acetate) ,ketones(such as acetone and methyl ethyl ketone)

Inorganic Solvents:
- Do not contain carbon.
- Dissolve inorganic compounds.
- Examples: Water, acids(hydrochloric acid),bases(sodium hydroxide),salts(ammonium chloride in solution)

•Chemical
➢hydrolysis:- hydrolysis of lipid such as triacylglycerol is done through lipases enzyme during the digestion of fat which results in fatty acids and glycerol. Hydrolysis is basically using water to break something down. So the lipase enzyme adds a hydroxyl group and hydrogen atom to the triacylglycerol to break it down or apart into glycerol and three fatty acids
Types are acid catalyzed hydrolysis and base catalyzed hydrolysis. Saponification is a base catalyzed hydrolysis

In saponification, hydrogen ions (H⁺) are not involved. The reaction specifically relies on hydroxide ions (OH⁻) provided by the strong base.

Here’s a quick comparison:

•	Saponification: Uses a strong base (like sodium hydroxide, NaOH) to provide hydroxide ions. These ions hydrolyze triglycerides into glycerol and fatty acid salts. There is no direct role for hydrogen ions in this process.
•	Acid-Catalyzed Hydrolysis: In contrast, acid-catalyzed hydrolysis (also known as acid hydrolysis) uses hydrogen ions (H⁺) from an acid to hydrolyze ester bonds in triglycerides, producing glycerol and free fatty acids.

➢Saponification:- hydrolysis of fat by an alkali (any base such as sodium hydroxide or potassium hydroxide) is known as saponification. the products formed are glycerol and alkali salts of fatty acids (plus the metal. So if sodium hydroxide,sodium metal is also released) which are known as soaps. The metal salts are also known as the salts of the fatty acids released during the reaction. For instance, if sodium hydroxide is used, you get sodium stearate or sodium palmitate (depending on the fatty acids involved), which are common types of soap.

Summary: The “metal salts” in saponification are indeed the soap, resulting from the reaction between fatty acids and the metal hydroxide base.

Hydrolysis generally involves breaking bonds in a molecule using water. It typically involves the addition of a hydrogen ion (H⁺) and a hydroxyl ion (OH⁻) from water to split the molecule.

Saponification is a specific form of hydrolysis where a fat or oil (triglyceride) reacts with a strong base (like sodium hydroxide or potassium hydroxide) rather than with water alone. Here’s how it works:

•	In saponification, the base (OH⁻) reacts with the ester bonds in triglycerides. This reaction breaks the ester bonds, resulting in the formation of glycerol and fatty acid salts (soap).

Yes, in saponification, the base facilitates the hydrolysis of triglycerides. Here’s how it works:

1.	Base-Catalyzed Hydrolysis: The base (such as sodium hydroxide or potassium hydroxide) provides hydroxide ions (OH⁻). These hydroxide ions attack the ester bonds in triglycerides, breaking them apart.
2.	Reaction: The triglyceride (a type of fat or oil) reacts with the hydroxide ions in a base-catalyzed reaction, leading to the formation of glycerol and fatty acid salts (soap)

➢Hydrogenation:- hydrogenation of unsaturated fat in the presence of a catalyst is known as hardening. It is mainly used for converting the liquid fats and unsaturated fats(have double bonds). So you add hydrogen to remove the double bonds. Double bonds aren’t so good because they disrupt structures and you can’t stack more structures on top of the other if they have double bonds.double bonds (unsaturated fatty acids) are not stable too. Liquid fats, also known as unsaturated fats, have double bonds in their fatty acid chains because these bonds prevent the molecules from packing tightly together. Here’s how it works:

1.	Double Bonds in Unsaturated Fats:
•	The presence of one or more double bonds between carbon atoms in the fatty acid chains causes a kink or bend in the structure. This makes it difficult for the fatty acids to stack closely or align neatly, which reduces the strength of the interactions between the molecules.
•	Unsaturated fats (liquid) have one or more double bonds in their fatty acid chains.
•	Hydrogenation adds hydrogen atoms to these double bonds.
•	Saturated fats (solid) have no double bonds in their fatty acid chains, making them more stable and solid at room temperature. Hydrogenation also increases the shelf life and stability of the fats. Catalysts used include Nickel,platinum and palladium. Also, clinically, hydrogenated fats(Produce trans fats which are also bad)  are bad because they raise LDL. Which can cause heart problems. So hydrogenated fats are seen in margarine and baked goods. For hydrolysis, if a person has a problem with production of lipase from the pancreas, there won’t be any hydrolysis of fats. This will make the fat go into the stool (steatorrhea) cuz the body can’t absorb the triglycerides. It can only absorb it as fatty acids and glycerol ➢Peroxidation:- peroxidation of lipids exposed to oxygen is responsible not only for breaking of foods but is also used for damage to tissues where it may cause cancer.its dangerous if it occurs in biological systems because it’s a form of oxidative stress where ROS like hydroxyl and superoxide radicals will bind to lipids and make them unstable or radicals too. This sets of a chain reactions causing everything in the cell to be unstable . Process: peroxidation is Oxidative damage of lipids due to exposure to oxygen, forming reactive oxygen species (ROS) like hydroxyl and superoxide radicals. This leads to cell instability and potential tissue damage.
•	Clinical Impact: Increases the risk of cancer and accelerates aging processes. Clinically relevant in:
•	Oxidative Stress: Associated with various diseases including cancer and neurodegenerative disorders. A patient with atherosclerosis is found to have elevated levels of lipid peroxides in their blood, indicating oxidative stress. A patient with early signs of neurodegenerative disease, such as Alzheimer’s disease, exhibits increased oxidative damage to lipids in brain tissue. A patient with acute inflammatory conditions, such as rheumatoid arthritis, may have increased levels of lipid peroxidation products in their serum. •	Antioxidant Therapy: May help mitigate oxidative damage. ➢Rancidity:- When any oil or fat produces unpleasant odour, they are termed as rancidity. Rancidity is caused after oxidation and hydrolysis. Oxidative rancidity occurs in triacylglycerol containing unsaturated fatty acids.rancid oils don’t have essential fatty acids in them cuz those essential acids have been broken down   So someone who takes in rancid oils usually have nutritional deficiencies and they also have GIt discomfort

Question 5

Change at the R group of the structure of fatty acids increases by a factor Of twos. Ranges from 2-24 carbons. True or false

Question 6

What are the classifications of fatty acids
Omega fatty acids are always unsaturated
Oils have what type of fatty acids?
Fats have what type of fatty acids?
Where does the methyl group point and where is it located in isobranched chain fatty acids ?
What about in anteisobranched chain fatty acids

Answer 6

Saturated-straight chain
Unsaturated(due to presence of double bonds)- cis and trans unsaturated. Oils have more cis unsaturated fatty acids while fats have more saturated fatty acids or trans unsaturated fatty acids cuz it’s more easy to pack them on top of each other. The cis one has a bent shape so it’ll be difficult to do so. Trans Unsaturated Fatty Acids: Formed during hydrogenation.

Branched chain fatty acids: isobranched and anteiso-branched chain. The isobranched is closer to the end of the fatty acid chain and the methyl branch points down while the anteisobranched is a bit away from the end of the fatty acid chain as compared to the iso and it’s methyl group points up. Isobranched: Methyl branch close to the end, pointing down.
• Anteiso-branched: Methyl branch farther from the end, pointing up.

Isobranched vs. Anteiso Branched: Both have branching, but the location of the branch (methyl group) differs. “Iso-” refers to the branch on the penultimate carbon, while “anteiso-” refers to the branch on the third-to-last carbon.

Complex fatty acids(possess other functional groups)- Beta hydroxyl(generating hydroxyl group on beta carbon to get energy from fatty acids), gamma cyclohexyl, cyclopropene

Question 7

What is the difference between cis and trans fatty acids

Answer 7

Note: cis fatty acids are present in cell membranes so that they won’t be packed on top of each other and will be more flexible.

Cis and trans fatty acids are types of unsaturated fatty acids, distinguished by the configuration of hydrogen atoms around their double bonds.

• Structure: In cis fatty acids, hydrogen atoms are on the same side of the double bond. This causes a bend or kink in the fatty acid chain. Cis: Think of “Cis = Same”. In cis isomers, the substituent groups on either side of the double bond are on the same side. You can visualize it as if they are “sitting together”. Cis: Picture two friends “Sitting Side by Side”. The letter “S” in “Cis” can help you remember that the groups are on the same side of the double bond.
  example of cis; • Cis-2-butene: Both methyl groups are on the same side of the double bond.
• Natural Occurrence: Most naturally occurring unsaturated fats in plant oils and fish are in the cis configuration.
• Health Impact: Generally considered healthier, cis fatty acids can help reduce bad cholesterol (LDL) and increase good cholesterol (HDL).

• Structure: In trans fatty acids, hydrogen atoms are on opposite sides of the double bond. This results in a straighter chain, similar to saturated fats. • Trans: Think of “Trans = Transverse” (meaning across). In trans isomers, the substituent groups are on opposite sides of the double bond. You can visualize it as if they are “across from each other”. Trans: Picture a “Transverse Line” crossing the double bond, where the groups are across from each other. The “T” in “Trans” can remind you that the groups are on opposite sides.

Example of trans:
• Trans-2-butene: Methyl groups are on opposite sides of the double bond.
- Formation: Trans fats can be found in small amounts naturally in some animal products, but are primarily formed during industrial hydrogenation of vegetable oils (e.g., in margarine, snack foods, and fried foods).
- Health Impact: Trans fats are associated with increased risk of heart disease, as they raise bad cholesterol (LDL) and lower good cholesterol (HDL). They can also increase inflammation and contribute to insulin resistance.

Regulations have been implemented in many countries to reduce or eliminate the use of trans fats in food products due to their adverse health effects.

Question 8

Whag fatty acids are the most common in the body?
How many. At one does stearic acid have?
How many carbons does palimitic acid have?
Difference between oleic acid, stearic acid and linoleic acid

How many carbons does butyric acid have?
Another name for butyric acid is ?

How many carbons does ethanoic acid have?

Answer 8

Palimitic acid - 16 carbons. Systemic name: hexadecanoic acid
Stearic acid -18 carbons (oleic acid has 18 carbons and is unsaturated but Stearic acid is saturated, Structure of linoleic acid is 18 carbons with two double bonds (polyunsaturated))
Systemic name: octadecanoic acid
Acetic-2 carbons Systemic name:ethanoic acid
Butyric acid -4 carbons systemic name:butanoic acid

oSaturated fatty acids may be envisaged as based on acetic acid (CH3 —COOH) as the first member of the series in which —CH2 — is progressively added between the terminal CH3 — and —COOH groups

oFatty acids in biological systems usually contain an even number of carbon atoms, typically between 14 and 24. The 16- and 18-carbon fatty acids are most common

oThe hydrocarbon chain is almost invariably unbranched in animal fatt

Question 9

State the types of unsaturated fatty acids
Most naturally occurring unsaturated fatty acids are cis
So the bad one is the saturated and trans fatty acids cuz they stick together and can stick on blood vessels to cause atherosclerosis that’s why they say take more cis unsaturated and omega fatty acids

Answer 9

Monounsaturated Monounsaturated (monoethenoid, monoenoic) acids, containing one double bond

•Polyunsaturated (polyethenoid, polyenoic) acids, containing two or more double bonds.

The configuration of the double bonds in most unsaturated fatty acids is cis geometric isomer.

The double bonds in polyunsaturated fatty acids are separated by at least one methylene group.

Question 10

State the biological importance of fatty acids. 4

Answer 10

1)-Fatty acids are the main components of dietary lipids. The human body stores energy such as fats in the form of triglycerides.

2)- Fatty acids are also required for the formation of membrane lipids such as phospholipids and glycolipids.

3) -They are required for the esterification of cholesterol to form cholesteryl esters. Cholesterol rarely occurs in its free form. fatty acids are combined with cholesterol so that the cholesterol can easily be managed and shuffled across the body . Cholesteryl esters are formed through the esterification of cholesterol with fatty acids. This process is important for cholesterol transport and storage in the body. Biological significance includes Cholesteryl Esters in Lipoproteins: Cholesteryl esters are found in lipoproteins like LDL (low-density lipoprotein) and HDL (high-density lipoprotein). They help in the transport and storage of cholesterol in the bloodstream.

Esterification is a chemical reaction in which a carboxylic acid(example is fatty acid ) reacts with an alcohol(example is cholesterol which is a sterol and is an alcohol because of the hydroxyl group in it) to form an ester(example is cholesteryl ester) and water(which is a by product of the reaction).

Another Example of Esterification Reaction

Formation of Ethyl Acetate:
• Reactants: Acetic acid (a carboxylic acid) and ethanol (an alcohol).
• Products: Ethyl acetate (an ester) and water.

Formation of Triacylglycerols (Triglycerides)
Another example:
• Reactants: Glycerol (an alcohol. A three-carbon alcohol with three hydroxyl (-OH) groups. ) and three fatty acids (carboxylic acids).
• Products: Triacylglycerol (a triglyceride) and three molecules of water.

4) They act as fuel molecules and are oxidized to produce energy.
5. Used to cushion vital organs in body

Question 11

What are triglycerides or triacylglycerides (tri-3carbon atoms,acyl-fatty acids,glycerol. Glycerol by nature has three carbon atoms so it can only bind to three fatty acid chains)
What is the end product of the formation of triglycerides?
What is dehydration synthesis
State two reasons dehydration synthesis is important for triglycerides
What enzyme is responsible for undertaking the triglyceride formation reaction?

Answer 11

Three fatty acid chains are bound to glycerol by dehydration synthesis.

Esterification Process:
• Reaction: Fatty acyl transferase enzymes facilitate the esterification of glycerol with fatty acids to form triglycerides (triacylglycerols).
• Dehydration Synthesis: This process involves the removal of a water molecule each time a fatty acid is added to the glycerol molecule.
1. Formation of Strong Covalent Bonds

•	Creation of Esters: In the synthesis of triglycerides, dehydration synthesis forms ester bonds between glycerol and fatty acids. These covalent bonds are essential for the stability and function of the triglyceride molecule. Dehydration synthesis results in the removal of water and the formation of a compact, hydrophobic molecule. This is essential for the storage of fats in adipose tissue and for forming lipid bilayers in cell membranes.

Triglycerides or neutral fat and 3 water molecules are the end products of the combination of glycerol and three fatty acids
This reaction is undertaken by fatty acyl transferase

Question 12

State five characteristics of triglyceride
Between Mono- and Diacylglycerol, which is an important signaling molecule that regulates calcium release inside cells?
What is the specific gravity or density of triglycerides?
oils are liquid at what temperature ?
Fats are solid at what temperature?

Answer 12

Characteristics
ØThe triacylglycerols are esters of the trihydric alcohol(alcohol molecule with three hydroxyl groups in it. example of this alcohol is glycerol), glycerol and fatty acids
ØMono- and Diacylglycerol, wherein one or two fatty acids are esterified or bound with glycerol, are also found in the tissues. Diacylglycerol is an important signaling molecule that regulates calcium release inside cells
ØNaturally occurring fats and oils are mixtures of triglycerides.
ØSpecific gravity or density is less than 1.0, that’s why all fats and oils float in water
ØOils are liquids at 20C, they contain higher proportion of unsaturated fatty acids
ØFats are solid at room temperature and contain saturated long chain fatty acid
ØTriglycerides are the storage form of energy in adipose tissue

Question 13

State five functions of triglycerides
What percentage of the body’s resting energy needs does compact stored fat provide?

Chylomicrons are a specific type of lipoprotein particle that primarily transports dietary lipids from the intestines to other tissues in the body.

Note: No, triglycerides do not contribute directly to cell membrane formation. Instead, phospholipids are the primary lipids involved in forming cell membranes. Phospholipids have a hydrophilic (water-attracting) head and two hydrophobic (water-repelling) tails, which arrange themselves into a bilayer, creating the structural foundation of cell membranes. This bilayer forms a stable barrier between the interior of the cell and its external environment, allowing selective permeability and communication.

Triglycerides, on the other hand, primarily serve as energy storage molecules and are stored in adipose tissue. They do not have the amphipathic properties necessary for forming the bilayer structure of cell membranes.

Answer 13

Functions

üMajor lipid in the body and diet
üStored fat provides about 60% of the body’s resting energy needs – compactly!
üInsulation and protection-especially in babies. They have a lot of brown fat which is easily metabolized for them to keep warm. Animal bear also has huge triglycerides stores so that during the winter season, it breaks it down to keep warm. Brown fat, also called brown adipose tissue, helps maintain your body temperature when you get too cold. It’s the same fat that bears use to stay warm when they hibernate. Babies are born with a lot of brown fat behind their shoulder blades. White fat stores excess energy while brown fat burns it to generate heat, which is why hibernating bears and newborns need it.

üCarrier of fat-soluble compounds:example is fat soluble vitamins,formation of lipoproteins(phospholipids also help in this), Also Within cells, triglycerides can be stored in lipid droplets and provide a source of fatty acids for membrane synthesis and signaling molecules.
üSensory qualities – flavor and texture: Triglycerides enhance food’s sensory qualities by acting as carriers for fat-soluble flavor compounds and enhancing aromas, particularly when cooked. They provide a creamy, smooth mouthfeel in products like butter and cream, and contribute to the moist, tender texture of baked goods. In pastries, solid fats from triglycerides create layers that result in a flaky texture. Overall, triglycerides enrich the flavor and texture, making foods more enjoyable and satisfying.

Question 14

What is the importance of diacylglycerol

Answer 14

Diacylglycerol is an important signaling molecule that regulates calcium release inside cells

Question 15

Why are phospholipid called amphipathic molecules

Answer 15

Phospholipids have a polar or hydrophilic head which is formed by the phosphate backbone and glycerol group of the fatty acid. It also has hydrophobic tails which interact with themselves. They are amphipathic cuz they can exist in both aqeous solutions and other solutions due to the polar head and hydrophobic tails

Contain in addition to fatty acids and glycerol/or other alcohol, a phosphoric acid residue, nitrogen containing base and other substituents.

Most phospholipids may be regarded as derivatives of phosphatidic acid , in which the phosphate is esterified with the —OH of a suitable alcohol.

They are amphipathic molecules containing a polar head and a hydrophobic portion

Question 16

When phospholipids R is replaced with the following molecules, what are their functions and locations?
1. an H molecule
2.ethanolamine
3.choline
4.serine
5.glycerol
6. myo-Inositol 4,5-bisphosphate
7. phosphatidylglycerin

Answer 16

When replaced with an H, it is phosphatidic acid
When replaced with ethanolamine,it is phosphatidylethanolamine- (this is distributed across the cell membranes all over the body but is more present in the brain)
When replaced with choline, it is phosphatidylcholine-(present in the brain)
When replaced with serine, it is phosphatidylserine-(signal molecule that warns body that this molecule that the cell is found on is in danger and needs to be destroyed to protect the neighboring cell.
When replaced with glycerol, it is phosphatidylglycerol
When replaced with myo-Inositol 4,5-bisphosphate it becomes Phosphatidylinositol
4,5-bisphosphate(this serves as a signaling molecule that regulates calcium balancing)
When replaced with phosphatidylglycerin, it becomes cardiolipin-(predominates cell membranes in the heart)

Question 17

State ten functions of phospholipids

Answer 17

Components of cell membrane, mitochondrial membrane and lipoproteins(triglycerides won’t help in this cuz they are not ampiphatic)

●Participate in lipid absorption and transportation from intestine

●Play important role in blood coagulation- allow for clotting factors and enzymes involved in coagulation cascade to embed on blood vessel membrane and be fixed to it. In bleeding, the integrity of the cell membrane in the vessels have been destroyed. Phospholipids are crucial in blood coagulation due to their involvement in several key processes:

1. Platelet Activation: Phosphatidylserine flips to the outer leaflet of platelet membranes, promoting clotting factor activation and aggregation.
2. Coagulation Cascade: Phosphatidylinositol and its metabolites are involved in cellular signaling and calcium ion release, which are essential for clotting factor activation.
3. Formation of Coagulation Complexes: Phospholipid membranes provide a surface for the assembly of enzyme-substrate complexes needed for the coagulation cascade.
4. Intrinsic and Extrinsic Pathways: Phospholipids support the assembly of tenase and prothrombinase complexes in the intrinsic pathway and facilitate tissue factor interaction in the extrinsic pathway.

Overall, phospholipids play a vital role in ensuring efficient and regulated blood clot formation.

●Required for enzyme action- especially in mitochondrial electron transport chain

●It plays a major role in the transportation and removal of cholesterol from the cells.

●It forms the structural components of the cell membrane with the association of proteins.

●They act as surfactants (these make the lungs able to expand freely without difficulty and contract freely without collapsing completely. Surfactants in the lungs are produced in the last four weeks of gestation. So if baby is premature, the surfactant won’t be fully developed so their given dexamethasone which will help with production of surfactant) in the respiratory system and are also involved in the coagulation of blood cells.

●It helps in the synthesis of different lipoproteins, prostacyclins(blood vessel dilation), prostaglandins(blood vessel constriction and inflammation ) and thromboxanes

F Certainly! Here’s a detailed explanation of each function of phospholipids:

• Components: Phospholipids are fundamental components of cell membranes. They form a bilayer with hydrophilic (water-attracting) heads facing outward and hydrophobic (water-repelling) tails facing inward.
• Function: This bilayer creates a semi-permeable membrane that maintains the integrity of the cell, controls the movement of substances in and out, and supports membrane proteins involved in various cellular functions. In mitochondria and lipoproteins, phospholipids also contribute to membrane structure and function.

• Role in Digestion: Phospholipids are crucial for the digestion and absorption of dietary fats. They form micelles, which are tiny lipid droplets that facilitate the absorption of lipids and fat-soluble vitamins in the intestine.
• Transport: Once absorbed, lipids are incorporated into chylomicrons (a type of lipoprotein) and transported through the lymphatic and circulatory systems to various tissues.

• Platelet Activation: Phosphatidylserine, a type of phospholipid, flips from the inner to the outer leaflet of the platelet membrane upon activation. This exposure promotes the activation of clotting factors and aids in the formation of blood clots.
• Coagulation Cascade: Phosphatidylinositol and its derivatives play a role in signaling pathways that lead to clotting. They help release calcium ions, which are crucial for the activation of clotting factors.
• Complex Formation: Phospholipid membranes provide a surface for the assembly of enzyme-substrate complexes necessary for the coagulation cascade. This includes the intrinsic and extrinsic pathways of coagulation, where phospholipids support the assembly of tenase and prothrombinase complexes and facilitate tissue factor interaction.

• Mitochondrial Function: In mitochondria, phospholipids are essential for the function of the electron transport chain, which is critical for ATP production. They help anchor and stabilize the enzymes involved in oxidative phosphorylation.

• Lipoprotein Formation: Phospholipids are involved in the formation of lipoproteins, which transport cholesterol and other lipids through the bloodstream. This process is crucial for maintaining lipid homeostasis and preventing cholesterol accumulation in cells.

• Membrane Structure: Phospholipids, along with proteins, form the structural backbone of cell membranes. They help maintain membrane fluidity and flexibility, which is essential for various cellular processes including cell signaling and movement.

• Respiratory Function: In the lungs, phospholipids form surfactants, which reduce surface tension in the alveoli, preventing collapse and aiding in lung expansion. This is especially important for premature infants, whose surfactant production may be insufficient, leading to respiratory distress.
• Clinical Relevance: Surfactant therapy with compounds like dexamethasone can help stimulate surfactant production in premature infants to improve lung function.

• Lipoproteins: Phospholipids are integral to lipoproteins, which transport lipids throughout the body and play a role in lipid metabolism.
• Prostacyclins and Prostaglandins: Phospholipids are precursors for the synthesis of prostacyclins (which cause vasodilation) and prostaglandins (which have various roles including vasoconstriction and inflammation).
• Thromboxanes: These are involved in blood clotting and vascular contraction. Phospholipids provide the substrate for their synthesis.

These detailed functions highlight the multifaceted roles of phospholipids in maintaining cellular function, supporting metabolic processes, and contributing to overall health.

Question 18

What is the structure or cholesterol
Where can cholesterol be gotten from
Cholesterol has two forms, name them
How is cholesteryl ester formed?
In plasma, the two forms of cholesterol are transported in what?
What is serum total cholesterol
Why is ldl bad cholesterol?
Why is hdl good cholesterol?
When you hear “ol “ in a molecule’s name, know there is a hydroxyl group located in the molecule

Answer 18

•Four linked carbon rings with –OH functional group

•From diet or synthesised de novo (hepatocytes) De novo cholesterol synthesis is the process by which cholesterol is synthesized from simpler molecules within the body, particularly in the liver
LDL is bad cholesterol because the lipoprotein contains lots of cholesterol more than protein content. The more cholesterol a lipoprotein has, the more the likelihood that lipoprotein will be distributing the cholesterol as it travels through the blood to the various parts of the body and if it deposits a lot of the cholesterol in the coronary arteries, you’ll get a heart attack or a stroke
•Cholesterol occurs both as free form or in ester form

•In cholesteryl ester, the hydroxyl group on position 3 is esterified with a long-chain fatty acid. When fatty acids bind to hydroxyl group, cholesterol ester is formed and this helps packaging of cholesterol into lipoproteins for the transport.

•In plasma, both forms are transported in lipoproteins

• sum total of free and ester cholesterol in serum is called serum total cholesterol

HDL is good cholesterol because it gathers cholesterol that has been distributed all over the body and sending it to the liver

Question 19

Which part of the body is the cholesterol particularly distributed in?
Cholesterol is a major constituent of what?
What is the chief role of cholesterol in pathogenic processes ?

Answer 19

➢Cholesterol is widely distributed in all cells of the body but particularly in nervous tissue.

➢It is a major constituent of the plasma membrane and of plasma lipoproteins.

➢Its chief role in pathologic processes is as a factor in the genesis of atherosclerosis of vital arteries, causing cerebrovascular, coronary, and peripheral vascular disease.

Question 20

What is a major constituent of gallstones?
How does high cholesterol cause gallstones

Answer 20

.
➢Cholesterol is a major constituent of gallstones(liver is where cholesterol is metabolized to produce bile acids.the bile acids are conjugated in the liver to produce bile salts.the bile salts are sent to the gallbladder for storage.
Bile salt is supposed to emulsify fats.
If there is too much cholesterol in the bile salts such that it can’t emulsify the fat, the cholesterol will form salts with potassium and sodium electrolytes and form the stones. The stones may crystallize together as gallstones

Question 21

Cholesterol is the precursor of what?
What is the starting molecule for the biosynthesis of cholesterol

Answer 21

It is synthesized in many tissues from acetyl-CoA. Cholesterol Synthesis: Acetyl-CoA is the starting material for the biosynthesis of cholesterol.
It is the precursor of all other steroids in the body, including corticosteroids, sex hormones, bile acids, and vitamin D

Question 22

State five biological importance of cholesterol

Answer 22

Lipid raft formation
Precursor of vitamin D
Plasma lipoprotein constituent
Bile acid synthesis
Insulates nerve fiber-especially sphingomyelin and phosphatidylcholine
Membrane fluidity- cholesterol gives the membrane rigidity
Membrane homeostasis
Steroid hormone precursor (aldosterone,cortisol,progesterone,estrogen and testosterone)

Cholesterol and Membrane Homeostasis: Cholesterol regulates membrane fluidity and stability, contributing to the barrier function and dynamic adaptation of the membrane.
• Cholesterol and Lipid Rafts: Cholesterol is critical for the formation and maintenance of lipid rafts, which are important for organizing cellular signaling and membrane processes.

Question 23

What are nucleus acids?
What are nucleotides ?
What two forms can a nucleic acids take on?
What are the types of RNA?
What are the two types of 5 carbon sugars?
A nucleotide is made of three things. Name them

Nucleic acids are monomers of nucleotides true or false

Answer 23

most important macromolecules for the continuity of life

•false. Nucleic acids are polymers of nucleotides.

Yes, it would be incorrect to say that nucleic acids are monomers composed of nucleotides. The correct statement is:

“Nucleic acids are polymers composed of nucleotides.”

• Nucleotides: These are the monomers, the individual building blocks that join together to form nucleic acids.
• Nucleic Acids: These are the polymers made up of long chains of nucleotides.

So, the accurate way to describe the relationship is that nucleotides are the monomers, and nucleic acids (like DNA and RNA) are the polymers composed of these nucleotide monomers.

•Nucleotides are building blocks of Nucleic acid. They are carbon ring structures containing nitrogenous base linked to a 5-carbon sugar (a ribose).So nucleotide is made up of phosphate,nitrogenous base and 5 carbon sugar

Nucleic acids are either RNA or DNA. The RNA is divided into ribosomal RNA(produced by nucleolus),Transfer RNA,Messenger RNA(read by ribosomes for protein synthesis),MicroRNA

Deoxyribose meaning one of the carbons lack oxygen or a hydroxyl group
•5-carbon sugar is either a ribose or a deoxy-ribose making the nucleotide either a ribonucleotide or a deoxyribonucleotide

Question 24

Which macromolecule is most important for the continuity of life

Answer 24

Nucleic acid
If there’s a problem with your Nucleic acid, you can’t procreate cuz you don’t have genetic material to procreate

Question 25

what is a nucleoside
Why is DNA said to have a deoxyribose Nucleic acid ?
Why is DNA less susceptible to hydrolysis and more suitable for long-term storage of genetic information.
What is a deoxynucleoside?
What is a deoxynucleotide?

Answer 25

If you have a nucleotide that has only 5 carbon sugar and the nitrogenous base without the phosphate group

Structure: Deoxyribose is a pentose sugar with the chemical formula C₅H₁₀O₄. It is a cyclic molecule with five carbon atoms, forming a ring structure.
2. Absence of an Oxygen Atom: Unlike ribose (the sugar in RNA), deoxyribose lacks an oxygen atom on the 2’ carbon. This means that instead of a hydroxyl group (-OH), there is just a hydrogen atom at the 2’ position. This difference is crucial and gives DNA its name (deoxy means “lacking oxygen”).
3. Nucleotide Formation: Deoxyribose bonds with a nitrogenous base (adenine, thymine, cytosine, or guanine) to form a deoxynucleoside. When one or more phosphate groups are attached to the 5’ carbon of the deoxyribose, it becomes a deoxynucleotide (e.g., deoxyadenosine triphosphate, dATP).
4. Role in DNA Structure: In DNA, deoxyribose forms part of the sugar-phosphate backbone. Each deoxyribose molecule is connected to a phosphate group at the 3’ carbon of one nucleotide and the 5’ carbon of the next nucleotide. This linkage forms the phosphodiester bonds that hold the DNA strand together.
5. Stability: The absence of the hydroxyl group at the 2’ position in deoxyribose contributes to the chemical stability of DNA compared to RNA, making DNA less susceptible to hydrolysis and more suitable for long-term storage of genetic information.

Question 26

What is the modified central dogma of molecular biology and how is it different from the original
Which part of the cell cycle does DNA replication occur ?

Answer 26

Original:
Starts with DNA AND ITLL BE TRANSCRIBED TO RNA AND THE RNA WILL BE TRANSLATED TO PROTEINS USING RIBOSOMES

MODIFIED:
. You can also go back from RNA to DNA. Example is viruses such as HIV AND HEP B. These viruses have RNA by nature but they convert their Nucleic acid into DNA so they’ll insert new “DNA” into human DNA so that it becomes part of it so whenever the human DNA is replicating itself, the virus too is replicating or multiplying. This is why they destroy cd4 cells cuz they make more copies of themselves at the expense of cd4 cells and keep destroying the cd4 cells until you are defenseless

DNA replication occurs inside cells cycle specifically the interphase phase

The central dogma of molecular biology, both classical and modified, outlines how genetic information is used to produce proteins, but they differ in how they account for additional complexities and exceptions. Here’s a breakdown of the differences:

1. DNA → RNA → Protein:
   
   • Transcription: DNA is transcribed into messenger RNA (mRNA).
   • Translation: mRNA is translated into protein at the ribosome.

1. DNA → RNA → Protein (same as classical):
   
   • Transcription: DNA is transcribed into mRNA.
   • Translation: mRNA is translated into protein.
2. RNA → DNA (Reverse Transcription):
   
   • Reverse Transcription: Some viruses (e.g., HIV, Hepatitis B) use reverse transcriptase to convert their RNA genome into DNA, which can then integrate into the host’s genome. This process does not involve directly making proteins from DNA without RNA.

1. Reverse Transcription: The modified central dogma includes the process of reverse transcription, where RNA can be converted into DNA, which was not part of the classical theory.
2. No Direct Protein Synthesis from DNA: The idea that proteins can be directly synthesized from DNA without going through RNA is not supported by either the classical or modified central dogma. Proteins still require an RNA intermediary for their synthesis.

In summary, while the modified central dogma recognizes additional processes like reverse transcription, it does not propose that proteins can be synthesized directly from DNA without passing through RNA.

RNA has the enzyme reverse transcriptase, which is not present in DNA. Reverse transcriptase is responsible for converting RNA into DNA, a process used by retroviruses (like HIV) and in certain laboratory techniques.

Question 27

What link forms the 3prime five prime phosphodiester bonds
How can nucleosides form phosphodiester bonds since they lack phosphate groups ?
What molecular group on the five prime binds to what molecular group on the three prime?

Answer 27

When combining nucleotides together, the phosphate group on the five prime carbon binds to the hydroxyl group on the three prime carbon of another molecule.
This is what formed the 3 prime five prime phosphodiester links. It is a very strong link which needs special enzymes to break them

Nucleosides themselves cannot directly form phosphodiester bonds because they lack the phosphate group. For nucleosides to participate in forming the phosphodiester bonds that link nucleotides into a nucleic acid strand, they first need to be converted into nucleotides. This conversion involves the addition of one or more phosphate groups to the nucleoside, typically on the 5’ carbon of the sugar.

Here’s a brief overview of the process:

1. Nucleoside Conversion to Nucleotide: A nucleoside (which consists of a nitrogenous base and a sugar) is phosphorylated (or undergoes phosphorylation reaction) by specific kinases to form a nucleotide (which consists of a nitrogenous base, a sugar, and one or more phosphate groups).
2. Formation of Nucleic Acid Strand: Once the nucleosides are converted into nucleotides, they can participate in the formation of a nucleic acid strand. During nucleic acid synthesis (e.g., DNA or RNA synthesis), the 3’ hydroxyl group of the sugar of one nucleotide forms a phosphodiester bond with the phosphate group attached to the 5’ carbon of the sugar of another nucleotide.
3. Phosphodiester Bond Formation: This reaction involves a condensation reaction where a molecule of water is released, linking the nucleotides together in a 5’ to 3’ direction along the sugar-phosphate backbone of the nucleic acid.

Thus, the conversion from nucleosides to nucleotides is a crucial step that enables the formation of phosphodiester bonds, allowing nucleotides to link together to form the long chains of DNA and RNA.

Question 28

State some differences between RNA and DNA

In a DNA double helix, the two strands are parallel to each other
True or false

Answer 28

DNA strands move in 5 prime to three prime direction while the opposing DNA strand moves in
The opposing DNA strand moves in the 3’ to 5’ direction. In a DNA double helix, the two strands are antiparallel, meaning they run in opposite directions.

• One strand runs from the 5’ end to the 3’ end.
• The complementary strand runs from the 3’ end to the 5’ end.

This antiparallel orientation is essential for the base pairing interactions and the overall structure of the DNA double helix.

Function DNA RNA
DNA-Carries genetic information
rNA-Involved in protein synthesis

Location:
dNA-Remains in the nucleus
rNa-Leaves the nucleus
Structure:
dNa-Double helix
RNA-Usually single-stranded(so it can be double stranded but not usually. Example is in rotavirus)
Sugar:
DNA-Deoxyribose
RNA-Ribose

Pyrimidines:
DNA-Cytosine, thymine
RNA-Cytosine, uracil
Purines:
DNA-Adenine, guanine
RNA- Adenine, guanine

Purines bind to pyrimidines
Adenine will always bind to thiamine
Guanine will always bind to cytosine
In RNA,thiamine is replaced with Uracil so the Adenine binds to uracil

Question 29

What is the direction that DNA synthesis occurs in?
What enzyme is responsible for adding nucleotides to a growing DNA strand?
Which direction is the DNA strand being copied read in?
What are Okazaki fragments
What enzyme joins these fragments together?
The enzyme responsible for adding nucleotides to a growing DNA strand, can only add new nucleotides to which part of the DNA strand. 3’ or 5’?

Answer 29

DNA synthesis and replication proceed in a specific direction due to the chemical structure of nucleotides. Here’s a brief explanation:

1. 5’ to 3’ Direction:
   
   • Synthesis: DNA polymerase, the enzyme responsible for adding nucleotides to a growing DNA strand, can only add new nucleotides to the 3’ end of the DNA strand. Therefore, DNA synthesis occurs in the 5’ to 3’ direction.
   • Template Strand: The DNA strand being copied is read in the 3’ to 5’ direction, ensuring that the new strand is synthesized in the 5’ to 3’ direction.(so it starts reading from 3prime so by the time it gets to 5prime, it can start from 5 prime and 5 prime will take the lead as more nucleotides are being added to 3prime.)

1. Leading Strand:
   
   • Continuous Synthesis: During DNA replication, the leading strand is synthesized continuously in the 5’ to 3’ direction as the replication fork progresses.
2. Lagging Strand:
   
   • Discontinuous Synthesis: The lagging strand is synthesized discontinuously in short segments known as Okazaki fragments, which are also synthesized in the 5’ to 3’ direction. These fragments are later joined together by DNA ligase to form a continuous strand.

Understanding this directional synthesis is crucial for comprehending the mechanisms of DNA replication, repair, and transcription.

Question 30

Adenine thiamine bond is weaker than the guanine cytosine bond. Why?

Answer 30

The GC bond have three hydrogen bonds linking them together (remember Minister GUC is strong in God and has three hydrogen bonds. GUC) while the AT bonds have two hydrogen bonds linking them together.

Question 31

State five functions of nucleotide

Answer 31

Nucleotide Functions:
•Forms the constituents of DNA and RNA –
•They serve as building blocks of nucleic acids (dNTP)
•Carriers of activated metabolites for the process of biosynthesis (UDP-Glucose for synthesis of glycogen. Functions of UDP:

1.	Glycogen Synthesis:
•	UDP-glucose is an activated form of glucose that serves as a substrate for the enzyme glycogen synthase, facilitating the synthesis of glycogen from glucose.
2.	Biosynthetic Pathways:
•	UDP is involved in the synthesis of complex carbohydrates and glycoproteins. For example, UDP-glucose is a precursor in the biosynthesis of lactose and the glycosylation of proteins and lipids) •Involved in storing chemical energy (ATP, GTP. Between the two, ATP is mostly used) •Provides cellular energy sources and other metabolic functions (ATP) •Required for chemical associations in the response of cells to the hormones and other extracellular stimuli (cAMP(secondary messenger), GTP, ATP, ADP-for coagulation cascade. Certain hormones are hydrophilic and polar so they can’t cross and have access to the inner part of the cell so These stimuli  are on the inside of the cell (soeciallically Camp And Cyclic GMP) and these stimuli will convey the message to the interior part of the cell. •Serve as structural components of enzyme cofactors and other metabolic intermediates (NAD, FAD, NADP) these are important for proper functions of enzymes and energy metabolism. NADPH is crucial for the synthesis of fatty acids.  •Drugs against(HIV, Hepatitis, Gout, Antivirals) gout from too much Nucleic acids in the meat especially in older men. Break down of Nucleic acids can form uric acid. The uric acid can form salts and these salts will form crystals that can deposit in your joints and soft tissues especially in your big toe and this triggers inflammation and pain.  •

Question 32

Nucleic Acid Functions

Answer 32

Nucleic Acid Functions:
•Storage of genetic info (DNA)
•Transmission of genetic info (mRNA)
•Processing of genetic information (ribozymes)
•Protein synthesis (tRNA and rRNA)

Question 33

State six uses of carbohydrates

Answer 33

Are most abundant organic molecules in nature

• Provide a significant fraction of the energy in the diet of most organisms

• Important source of energy for cells-best storage forms for energy is lipids but if you want more readily available sources, it’s glycogen which is stored in liver and muscles

• Can act as a storage form of energy

• Can be structural components of many organisms

• Can be cell-membrane components mediating intercellular communication

• Can be cell-surface antigens
-example is blood groupings
• Can be part of the body’s extracellular ground substance

• Can be associated with proteins and lipids

• Part of RNA, DNA, and several coenzymes (NAD+, NADP+, FAD, CoA)

Here’s a summary of the functions of carbohydrates:

1. Structural Components: Carbohydrates, such as cellulose in plants and chitin in arthropods, provide structural support to various organisms.
2. Cell-Membrane Components: Carbohydrates are integral to cell membranes, where they mediate intercellular communication and recognition through glycoproteins and glycolipids.
3. Cell-Surface Antigens: Carbohydrates on cell surfaces serve as antigens, with blood groupings (e.g., A, B, AB, O) being a key example. Glycoproteins or Glycolipids as antigens in j cell surface
4. Extracellular Ground Substance: Carbohydrates, like glycosaminoglycans, are components of the extracellular matrix, contributing to tissue structure and function.

Question 34

Explain the classifications of carbohydrates based on complexity

Answer 34

1.Complexity
Simple Carbohydrates- monosaccharides(examples are glucose,galactose,fructose,mannose,ribose)
Complex carbs: disaccharides,oligosaccharides (
1. Raffinose
2. Maltotriose
3. Fructooligosaccharides (FOS) ) and polysaccharides (starch,amylose,glycogen)

Question 35

What comprises these disaccharides?
sucrose
lactose?
maltose
trehalose

Answer 35

1. Sucrose (Glucose + Fructose):
   
   • Sources: Table sugar, sugar cane, sugar beets, processed foods.
   • Role: Common sweetener.
2. Lactose (Glucose + Galactose):
   
   • Sources: Milk and dairy products.
   • Role: Main sugar in milk.
3. Maltose (Glucose + Glucose):
   
   • Sources: Malted foods and beverages (e.g., malted milkshakes, beer).
   • Role: Formed during starch digestion.
4. Trehalose (Glucose + Glucose):
   
   • Sources: Mushrooms, honey, seafood, and certain insects.
   • Role: Used by organisms to survive extreme conditions.

Trehalose: Composed of two glucose molecules linked by an α,α-1,1-glycosidic bond.
• Maltose: Composed of two glucose molecules linked by an α-1,4-glycosidic bond.
2. Occurrence:
• Trehalose: Found in organisms such as fungi, bacteria, plants, and some invertebrates. It serves various biological roles, including stress protection.
• Maltose: Commonly found in malted grains, such as barley, and is produced during the digestion of starches. It is involved in the breakdown of starches in humans and other animals.

Question 36

State the Carbs classification based on size and C=0 function

Answer 36

Size: Tetrose C4 sugar, pentose C5,hexose C6,heptose C7,octose C8, nonose C9,decaose C10

Diose: A sugar with two carbon atoms. It is the simplest form of sugar, but it is not commonly encountered in nature. Examples include erythrose and threose, which are used in some biochemical pathways.
2. Triose: A sugar with three carbon atoms. Triose sugars include:
• D-glyceraldehyde (an aldose)
• Dihydroxyacetone (a ketose)
Triose sugars play crucial roles in metabolic pathways, such as glycolysis and the Calvin cycle.
3. Ose: This is a general suffix used to denote sugars. The suffix “-ose” indicates that the molecule is a carbohydrate, with examples including:
• Glucose (a hexose)

С=0 Function: Aldose
sugars having an aldehyde function or an acetal equivalent.

1. Aldose Carbohydrates:
   
   • Aldose carbohydrates are characterized by having an aldehyde functional group (-CHO) at one end of their carbon chain.
   • In the case of aldoses, the carbon at the end of the chain is part of an aldehyde group.
   • The most common example of an aldose is glucose, which is a six-carbon sugar with an aldehyde group at one end of its carbon chain.
2. Ketose Carbohydrates:
   
   • Ketose carbohydrates are characterized by having a ketone functional group (C=O) within their carbon chain.
   • In ketoses, the carbonyl group (C=O) is located within the carbon chain, not at the end.
   • Fructose is a common example of a ketose carbohydrate. It’s a six-carbon sugar with a ketone group located within the carbon chain, typically at the second carbon position.

Ketones and aldehydes are both types of carbonyl compounds, which contain a carbonyl group (a carbon-oxygen double bond, C=O) but differ in their structures and properties:

1.	Aldehydes:
•	Structure: The carbonyl group is bonded to at least one hydrogen atom and an alkyl or aryl group (R-CHO).
•	Example: Formaldehyde (HCHO) and acetaldehyde (CH3CHO).
•	Properties: Aldehydes are typically more reactive than ketones because the carbonyl carbon in aldehydes is bonded to at least one hydrogen, making it more electrophilic. They are often used in organic synthesis and as intermediates in various biochemical processes.
2.	Ketones:
•	Structure: The carbonyl group is bonded to two alkyl or aryl groups (R-CO-R’).
•	Example: Acetone (CH3COCH3) and butanone (CH3COC2H5).
•	Properties: Ketones are generally less reactive than aldehydes because the carbonyl carbon is surrounded by two alkyl or aryl groups, which provide steric and electronic stabilization. Steric stabilization is a phenomenon in colloid and surface science that prevents particles from aggregating due to the presence of steric (physical) hindrances

Classification Based on the C=O Function (Aldehyde or Ketone Group)

1.	Aldoses:
•	These are monosaccharides that contain an aldehyde group (-CHO).
•	The carbonyl group (C=O) is located at the end of the carbon chain.
•	Examples: Glucose, Ribose, Glyceraldehyde, Galactose.
2.	Ketoses:
•	These are monosaccharides that contain a ketone group (C=O).
•	The carbonyl group (C=O) is located at the second carbon of the carbon chain (not at the end).
•	Examples: Fructose, Ribulose, Dihydroxyacetone.

In summary, the key distinction is the location of the functional group within the carbon chain: at the end for aldoses (aldehyde group) and within for ketoses (ketone group).

Ketose
sugars having a ketone function or an acetal equivalent.

You’re correct in that aldehydes and ketones are typically found in specific positions in carbon compounds:

• Aldehydes: These functional groups are found at the end of a carbon chain. In sugars, this means the carbonyl group (C=O) is located at the first carbon (C-1). This is why glucose, which has an aldehyde group, is classified as an aldose.
• Ketones: These functional groups are located within the carbon chain, meaning the carbonyl group is attached to a carbon that is not at the end of the chain. In the case of fructose, the carbonyl group is located at the second carbon (C-2), making it a ketose.

So, your understanding is correct: aldehydes are at the end of the chain, while ketones are in the middle.

Question 37

What is isomerism?
What is epimerism?
What is stereoisomerism
What are enantiomers?
What is an asymmetric carbon?
At carbon 2, where is the hydroxyl group of mannose located ? Left or right?
What about glucose?
On which carbon are glucose and galactose termed as epimers?
Why are glucose and galactose epimers?

Answer 37

An isomer refers to a compound that has the same molecular formula as another compound but a different structural arrangement of atoms.

There are several types of isomers:

1.	Structural (constitutional) isomers – Compounds with the same molecular formula but different connectivity of atoms.
2.	Stereoisomers – Compounds with the same connectivity but different spatial arrangements of atoms. They include:
•	Geometric isomers (cis-trans isomers) – Due to restricted rotation, typically around a double bond.
•	Enantiomers – Non-superimposable mirror images, important in biochemistry because they can have different biological activities.
•	Diastereomers – Stereoisomers that are not mirror images of each other.

1. Chain isomers: Isomers that differ in the arrangement of the carbon skeleton.
2. Position or point isomers: Isomers that differ in the position of a functional group on the same carbon chain.
3. Functional group isomers: Isomers that differ in the type of functional group present.
4. Tautomers: Isomers that exist in dynamic equilibrium, usually involving the movement of atoms within the molecule.

Isomers of Aldoses

Ribose and xylose are isomers of each other
Glucose,galactose and mannose are isomers of each other
This is because each group of isomers have the same number of carbon and oxygen atoms. The distinction is, at particular carbon atoms, the arrangement varies.
Mannose is an epimer of glucose. At carbon 2, mannose has hydroxyl group on left while glucose has its hydroxyl group on the right

The difference between epimers and structural isomers lies in the type and extent of their structural variations:

• Epimers: These are a specific type of stereoisomers where two compounds differ only in the configuration around a single carbon atom (usually at one of the asymmetric carbons) in a molecule with multiple stereocenters. For example, mannose and glucose are epimers because they differ only at the C-2 carbon.
• Structural isomers: These are compounds that have the same molecular formula but differ in how their atoms are connected (bonded). For example, glucose and fructose are structural isomers because they have the same molecular formula (C₆H₁₂O₆) but differ in their bonding: glucose is an aldose (contains an aldehyde group), while fructose is a ketose (contains a ketone group).

In short, epimers differ by the spatial arrangement at one carbon, while structural isomers differ in the actual connectivity of atoms.

isomerism (structural and epimers)

Epimers are compounds which have more than one asymmetric carbon and differ only in the
configuration around one carbon. They are not mirror images of each other (which distinguishes them from enantiomers).

An asymmetric carbon (also called a chiral carbon) is a carbon atom that is attached to four different groups or atoms. This uniqueness in its attachments gives the carbon a chiral center, meaning it can exist in two different forms (called enantiomers) that are mirror images of each other but cannot be superimposed on one another, just like your left and right hands

Example of an Asymmetric Carbon:
• Consider a carbon atom with the following four groups attached to it:
• —H (Hydrogen)
• —OH (Hydroxyl group)
• —CH₃ (Methyl group)
• —NH₂ (Amino group)
• Since all four groups are different, this carbon is an asymmetric (chiral) carbon.
4. Visualizing Asymmetric Carbon:
• Imagine holding a tetrahedral carbon model in your hand, with four different colored balls representing different groups attached to each arm of the tetrahedron. If you create a mirror image of this model, you will notice that you cannot rotate or flip it to make the original and the mirror image look exactly the same.

Structural
All have the formula C6H12O6

Epimers
-OH in space differ at asymmetric Carbon
Glucose and Mannose at C2
Glucose and Galactose at C4

Table sugars have a cyclic structure cus in a straight chain, they’ll be unstable
Galactose at carbon 2 has the same arrangement as glucose but at carbon 4,galactose pulls hydroxyl group to left while glucose own is on the right

Here’s a clearer breakdown of the different types of isomerism:

• Definition: Isomers are molecules with the same molecular formula but different structures or spatial arrangements.
• Types: Includes all forms of isomerism such as structural isomerism and stereoisomerism.

• Definition: Structural isomers have the same molecular formula but differ in the connectivity of their atoms.
• Types:
  
  • Chain Isomerism: Differences in the carbon chain structure. For example, butane and isobutane (2-methylpropane) are chain isomers.
  • Functional Group Isomerism: Differences in the type of functional group. For example, ethanol (an alcohol) and ethanal (an aldehyde) are functional group isomers.
  • Position Isomerism: Differences in the position of the functional group within the molecule. For example, in glucose and fructose, the placement of the carbonyl group varies.

• Definition: Stereoisomers have the same molecular formula and connectivity but differ in the spatial arrangement of atoms.
• Types:
  
  • Geometric Isomerism (cis-trans): Differences in the spatial arrangement around a double bond or ring structure. This is more common in alkenes and cyclic compounds.
  • Optical Isomerism (Enantiomerism): Molecules are non-superimposable mirror images of each other. For example, D-glucose and L-glucose.
  • Epimerism: A type of diastereoisomerism where two sugars differ in configuration at exactly one chiral center. For example, glucose and galactose are epimers because they differ only in the configuration at the fourth carbon.

No, the fourth carbon in glucose and galactose is considered a chiral center because it is bonded to four different groups, not because it is near the side group. A chiral center (or asymmetric carbon) is a carbon atom that has four distinct substituents attached to it.

In the case of glucose and galactose, they are epimers because they differ in the configuration at the C-4 carbon (the fourth carbon), which is a chiral center. This means the orientation of the hydroxyl group (-OH) at that specific carbon differs between the two sugars:
- In glucose, the hydroxyl group on the C-4 carbon is on the right.
- In galactose, the hydroxyl group on the C-4 carbon is on the left.
This difference in configuration at a single chiral center is what makes glucose and galactose epimers.

Yes, amino acids also contain chiral centers. In the context of amino acids:

• Most amino acids (except glycine) have a central carbon atom known as the alpha carbon (Cα) that is attached to four different groups:
  
  1. An amino group (-NH₂)
  2. A carboxyl group (-COOH)
  3. A hydrogen atom (-H)
  4. A variable side chain (R group) that differs among amino acids.
• This arrangement makes the alpha carbon a chiral center, leading to two possible stereoisomers, which are often referred to as the L (levo) and D (dextro) forms.

In summary, like sugars, amino acids can also exhibit chirality due to the presence of chiral centers. However, in amino acids, chirality is primarily associated with the alpha carbon, while in sugars, it can involve multiple chiral centers.

• Epimerism: A subtype of diastereoisomerism. Epimers differ in configuration at only one chiral center.
• Chain Isomerism: A form of structural isomerism where different carbon chain arrangements create isomers.
• Functional Group Isomerism: A form of structural isomerism where different functional groups lead to isomers.

In summary, isomerism encompasses all types of structural and stereoisomerism, while specific forms like epimerism, chain isomerism, and functional group isomerism are subcategories that describe different ways in which molecules can differ in structure or spatial arrangement.

Question 38

Classification of isomers based on dextro rotatory and levo rotatory
Between D and L, which is more common in sugars and which is more common in amino acids?
If the hydroxyl group on the penultimate carbon in a Fischer projection is on the right, is it L or D?
If it’s on the left, is it L or D?
What is a penultimate carbon?

Answer 38

D and L isomerism
The orientation of the —H and —OH groups around the carbon atom adjacent to the terminal alcohol carbon

D – form is found most in biological systems including humans.
Most amino acids are L amino acids

•Most enzymes are specific for either the D or the L form, but enzymes known as isomerases are able to interconvert D- and L-isomers”

Position of last but one carbon group(penultimate carbon group) tells you whether the sugar molecule is D or L. If on the right, it’s D and if left, it’s L

Here’s a clearer summary of the concepts:

1. D and L Forms of Sugars:
   
   • D-Form (Dextrorotatory): In biological systems, including humans, most sugars are in the D-form. The designation “D” refers to the configuration of the hydroxyl group (-OH) on the penultimate carbon (the carbon farthest from the aldehyde or ketone group). If this hydroxyl group is on the right in a Fischer projection, the sugar is in the D-form.
   • L-Form (Levorotatory): This is the mirror image of the D-form. If the hydroxyl group on the penultimate carbon is on the left in a Fischer projection, the sugar is in the L-form.
2. Amino Acids:
   
   • Most naturally occurring amino acids are in the L-form. This is important for protein structure and function.

So, while D-forms are prevalent in sugars in biological systems, L-forms are the standard for amino acids.

Question 39

Explain the Class of carbs based on isomers(anomers)
What is mutarotation?
What is cyclization

Answer 39

Yes, that’s correct! Here’s a more detailed explanation:

1. Anomeric Carbon:
   
   • The anomeric carbon is indeed the carbon that is “sacrificed” or transformed from a carbonyl group (C=O) to a new stereocenter (chiral center) during the cyclization of a sugar molecule. In the open-chain form of a sugar:
     
     • For aldoses (e.g., glucose), the anomeric carbon is carbon-1 (C1) because it is the carbon in the aldehyde group (—CHO).
     • For ketoses (e.g., fructose), the anomeric carbon is carbon-2 (C2) because it is the carbon in the ketone group (C=O).
   • Upon cyclization, the hydroxyl group (—OH) on another carbon (the penultimate carbon) reacts with the carbonyl carbon (anomeric carbon), resulting in the formation of a ring (hemiacetal or hemiketal). This reaction converts the carbonyl carbon into a new chiral center, known as the anomeric carbon.
2. Penultimate Carbon:
   
   • The penultimate carbon is the second-to-last carbon in the open-chain form of a sugar. It is called “penultimate” because it is one position before the last carbon in the chain:
     
     • For example, in glucose, the penultimate carbon is carbon-5 (C5) in the six-carbon chain.
   • The hydroxyl group (—OH) on the penultimate carbon is the one that attacks the carbonyl carbon (the anomeric carbon) during cyclization, leading to ring formation. The orientation of this hydroxyl group determines the D or L configuration of the sugar (D-sugars have the hydroxyl group on the right in the Fischer projection, while L-sugars have it on the left).

• Anomeric Carbon: The carbon that was part of the carbonyl group and becomes a new stereocenter upon ring formation (it can form α or β anomers).
• Penultimate Carbon: The second-to-last carbon in the sugar chain that participates in the ring-closing reaction by attacking the carbonyl carbon. It helps determine whether the sugar is a D or L isomer.

So, when you say the “carbon that is sacrificed,” it refers to the anomeric carbon, which undergoes a change in bonding during the cyclization process. The penultimate carbon is the last but one carbon that helps form the ring structure by reacting with the anomeric carbon.

When a sugar cyclizes, it can form two possible configurations at the anomeric carbon: α (alpha) or β (beta).

Mutarotation

•	Mutarotation is the change in the optical rotation of a solution due to the interconversion between the α and β anomers of a sugar when it is dissolved in water.
•	Mutarotation occurs because the cyclic form of the sugar can temporarily open back up to its linear form and then reclose, potentially forming the opposite anomer (e.g., from α to β or vice versa).
•	The process of mutarotation involves the sugar going through cyclization and ring-opening multiple times. As it opens and recloses, it can switch between the α and β forms, resulting in an equilibrium mixture of the two.

The predominant forms of ribose, glucose, fructose, and many other sugars in solution are not open chains. They are cyclic. For them to become cyclic, an internal reaction occurs. The penultimate group attacks the carbonyl or aldehyde group (if it’s a ketose too it attacks the ketosegroup) or it attacks the group attached to the oxygen. When this happens, the carbon has to sacrifice one of its bonds cuz it can only bond to four atoms or have four bonds at a time. It forms a type of structure. The structure it forms looks bulky with atoms all over the molecule and so if another molecule wants to bind to the carbon atom in the first molecule, it’ll be difficult so they now form the cyclical structure to allow easy access for binding to prevent stearic hindrance. Mutarotation is a process that involves cyclization and switching back and forth between the cyclic structure and the open structure which also means switching between the alpha and beta cyclic forms.
The molecule can switch from cyclical to the straight chain and vice versa when they undergo mutarotation. Carbs prefer the ring or cyclical structure cuz it helps them polymerize easily.

Oxygen inside a carb molecule will not be counted as a carb. So if ideally the molecule is to be called a hexose cuz it has six sides, the presence of the oxygen at one side will make it five carbons not six
Mutarotation is the exchange eod the position of the hydroxyl group around the anomeric carbon so that the one that was formerly pointing downwards will now point upwards and vice versa. If the hydroxyl group is now pointing downwards after the mutarotation, it’s an alpha something something and if upwards, it’s a beta something something
When the ring is being formed, if the hydroxyl group or hydrogen atom is on the left, it’ll point upwards and if right, it’ll point downwards.

Do alpha and beta are in hawthorn projection and left and right is Fischer projection.

Your description of mutarotation and the positioning of hydroxyl groups is almost correct, but it can be refined for clarity. Here’s a more precise summary:

• Definition: Mutarotation is the process by which the configuration of the anomeric carbon in a cyclic sugar changes, leading to the interconversion between alpha (α) and beta (β) forms. This occurs due to the exchange of the position of the hydroxyl group around the anomeric carbon (C-1).

1. During Mutarotation:
   
   • If the hydroxyl group (-OH) on the anomeric carbon (C-1) is downward after mutarotation, it indicates the sugar is in the alpha (α) form.
   • If the hydroxyl group (-OH) on the anomeric carbon (C-1) is upward, it indicates the sugar is in the beta (β) form.
2. Formation of the Ring:
   
   • When the cyclic form is being formed:
     
     • If the hydroxyl group (-OH) or hydrogen atom on the anomeric carbon is on the left in the Fischer projection, it will point upwards in the Haworth projection (resulting in the beta form).
     • If it is on the right, it will point downwards (resulting in the alpha form).

• Mutarotation:
  
  • Down (after mutarotation) = Alpha (α).
  • Up (after mutarotation) = Beta (β).
• During Ring Formation:
  
  • Left in Fischer = Up in Haworth = Beta (β).
  • Right in Fischer = Down in Haworth = Alpha (α).

This understanding is essential in carbohydrate chemistry for identifying the structure of sugars and their interconversions.

Not exactly. Cyclization and mutarotation are related concepts in carbohydrate chemistry, but they are not the same thing. Let’s clarify the difference between them:

• Cyclization refers to the process by which a linear (open-chain) form of a sugar molecule converts into a ring (cyclic) form.
• In solution, monosaccharides such as glucose exist in equilibrium between their open-chain and cyclic forms. The cyclic forms are more stable and are predominantly present.
• During cyclization, the hydroxyl group (—OH) on the penultimate carbon (the second-to-last carbon) attacks the carbonyl carbon (the aldehyde or ketone group), resulting in the formation of a hemiacetal (for aldoses) or hemiketal (for ketoses).
• This process creates a new chiral center at the anomeric carbon, leading to the formation of α (alpha) and β (beta) isomers (anomers).

• Mutarotation is the process by which a sugar molecule interconverts between its α and β anomeric forms in solution.
• When a sugar dissolves in water, it can switch between its open-chain form and its cyclic forms. As it does so, the hydroxyl group (—OH) on the anomeric carbon can change orientation from downward (α) to upward (β), or vice versa.
• This interconversion happens because the sugar can open up to its linear form and then reclose into either the α or β form. The equilibrium mixture of α and β forms, along with the open-chain form, is what gives rise to mutarotation.
• Mutarotation is observed as a change in the optical rotation of the solution over time until equilibrium is reached.

• Cyclization is the process of forming a ring from an open-chain structure.
• Mutarotation involves the equilibrium between the α and β forms of a cyclic sugar due to the opening and reclosing of the ring structure.

Cyclization is the initial step that allows a sugar to form a ring structure, while mutarotation is the ongoing process that allows the sugar to switch between its α and β forms in solution. Mutarotation results from the dynamic equilibrium between the open-chain and cyclic forms, whereas cyclization is the actual formation of the ring itself.

You’ve got a good grasp of the basics of sugar chemistry and mutarotation! Here’s a more refined breakdown of the concepts you mentioned:

1. Cyclic Forms of Sugars: Most sugars, like ribose (a five-carbon sugar), glucose, and fructose (six-carbon sugars), predominantly exist in cyclic forms rather than as open chains when they are in solution. This is because the cyclic form is more stable.
2. Cyclization Reaction: For a sugar molecule to form a ring (cyclic) structure, an internal reaction occurs. In this reaction, the hydroxyl group (-OH) on the penultimate carbon (the second-to-last carbon in the chain) attacks the carbonyl group (a carbon double-bonded to oxygen, either an aldehyde or ketone group) of the same molecule. This attack forms a new bond, converting the linear (open-chain) form of the sugar into a ring.
3. Limitation of Carbon Bonds: In this reaction, because carbon can only form four bonds at a time, one of the existing bonds of the carbonyl carbon is “sacrificed,” or broken, to allow the new bond to form, leading to a stable cyclic structure.
4. Steric Hindrance: When sugars form a ring, the bulky atoms in the structure (like various -OH groups) are better positioned to avoid “steric hindrance” — which is a fancy way of saying that they avoid bumping into each other and blocking other molecules from binding. This makes it easier for other chemical groups or molecules to interact with the sugar, especially if it needs to form polymers (long chains of sugars) like starch or cellulose.
5. Mutarotation: Mutarotation specifically refers to the change in the optical rotation that happens when a sugar in a cyclic form interconverts between its two different forms: **alpha (α) anomer ** and **beta (β) anomer. These forms differ in the position of the hydroxyl group attached to the anomeric carbon (the carbon derived from the carbonyl group).
   • If the hydroxyl group on the anomeric carbon is downwards, it is called the alpha (α) form.
   • If the hydroxyl group is upwards, it is the beta (β) form.
6. Determination of Upward or Downward Position: When the ring forms, the direction in which the groups point (up or down) depends on their positions in the Fischer projection (a 2D representation of the molecule). Generally:
   
   • Groups on the left in the Fischer projection will point upwards in the ring (Haworth) projection.
   • Groups on the right will point downwards.
7. Carbohydrates Prefer Cyclic Forms: Carbohydrates often prefer their ring forms because these structures allow for better stability and easier polymerization. Polymers like cellulose or glycogen are long chains of glucose units connected together, and this is more efficient when the glucose is in its cyclic form.

Mutarotation describes the dynamic process where the sugar changes back and forth between its α and β forms through the open-chain form, allowing an equilibrium mixture in solution. This dynamic equilibrium is crucial for the reactivity and functionality of sugars in biological systems.

Imagine you have a sugar that can change its shape, like a magic toy that can flip between two forms. Let’s call these forms “Form A” and “Form B.” When you put the sugar into water, it starts switching back and forth between Form A and Form B until there’s a mix of both forms.

At first, each form might bend light differently. Form A might bend the light one way, and Form B might bend it another way. But when they keep switching and find a balance, the way they bend light settles in between.

This back-and-forth switching that changes how the sugar bends light is called mutarotation. It’s like the sugar is dancing between two moves until it finds its favorite dance mix!

Most of the time, hydroxyl groups point upwards
they become anomers
Fluorine or fluoride is the most electronegative atom
•• For an aldohexose such as glucose, Formation of a hemiacetal by reaction of the C-1 aldehyde group with the C-5 hydroxyl group to form an intramolecular hemiacetal. The resulting cyclic hemiacetal, a six-membered ring, is called pyranose

Question 40

How are disaccharides formed?
What type of bonds exist between disaccharides?
These bonds can be of what types?

Answer 40

Two monosaccharides undergo a dehydration reaction(this reaction involves sacrificing one molecule of water (a hemiacetly means the oxygen is bonded to one carbon and one hydrogen. when the oxygen is bound to two carbons then it’s an acetal. Two hemiacetals make one acetal)
Adding water across breaks the bonds in disaccharides )
•Forms glycosidic bonds
•Glycosidic bonds (or glycosidic linkages) can be an alpha or beta type. An alpha bond is formed when the OH group on the carbon-1 of the first glucose is below the ring plane, and a beta bond is formed when the OH group on the carbon-1 is above the ring plane.
•Example: formation of maltose from D-glucose

It is produced in germinating grain (such as barley) as starch is broken down during malting

Formed during the hydrolysis of starch to glucose during digestion
So glycosidic for glucose,peptide bonds for amino acids and esters for fatty acids

Question 41

What is the difference between cellulose and starch
Why is bacteria needed in the gut

Answer 41

Cellulose is beta 1,4 glycosidic linkage and starch are alpha 1,4 glycosidic linkages
Lactose also has beta 1,4

Human body doesn’t have enzymes to break down beta 1,4 glycosidic linkages. This is why bacteria is needed in our gut for them to break the beta thing down through hydrolysis using cellulase enzymes

Starch is made up of amylose and amylopectin. These are polysaccharides that primarily contain α-1,4-glycosidic bonds.
Amylose contains this in a linear form and amylopectin contains this in a branched form

You’re right to point out the distinction between amylopectin’s branched structure and the glycosidic bonds involved. Here’s a clearer explanation:

Amylose is completely alpha 1,4
### Glycosidic Bonds in Amylopectin

• Amylopectin: This is a branched polysaccharide found in starch. It has the following characteristics:
  
  • Straight Chains: The linear portions of amylopectin are linked by α-1,4 glycosidic bonds. This means that glucose units in the straight chains are connected through these bonds.
  • Branches: The branching points occur where glucose units are attached via α-1,6 glycosidic bonds. These bonds create branches off the main chain.

• α-1,4 Glycosidic Bonds: Present in the straight (unbranched) portions of amylopectin and also in amylose, which is entirely unbranched.
• α-1,6 Glycosidic Bonds: Present at the branching points of amylopectin.

In summary, amylopectin contains both α-1,4 glycosidic bonds in its linear segments and α-1,6 bonds at its branch points, allowing it to be a branched structure. This is why the α-1,4 glycosidic bonds are associated with the straight chains in amylopectin despite it being a branched polymer.

Yes, the concept of glycosidic bonds applies similarly to other polysaccharides. Here’s a summary for some common polysaccharides regarding their structure and the types of glycosidic bonds they contain:

• Structure: Unbranched polymer.
• Glycosidic Bonds: Composed entirely of α-1,4 glycosidic bonds linking glucose units in a straight chain.

• Structure: Branched polymer.
• Glycosidic Bonds: Contains α-1,4 glycosidic bonds in the linear segments and α-1,6 glycosidic bonds at the branch points.

• Structure: Highly branched polymer (more branching than amylopectin).
• Glycosidic Bonds: Composed of α-1,4 glycosidic bonds in the straight chains and α-1,6 glycosidic bonds at the branching points.

• Structure: Linear polymer.
• Glycosidic Bonds: Contains β-1,4 glycosidic bonds linking glucose units. This type of bond leads to a straight-chain structure that forms strong fibers.

• Structure: Linear polymer, similar to cellulose but with an N-acetyl group.
• Glycosidic Bonds: Composed of β-1,4 glycosidic bonds, linking N-acetylglucosamine units in a linear fashion.

• Linear Polymers: Amylose, cellulose, and chitin are linear, with specific types of glycosidic bonds (α-1,4 for amylose and β-1,4 for cellulose and chitin).
• Branched Polymers: Amylopectin and glycogen have both linear and branched structures, characterized by both α-1,4 and α-1,6 glycosidic bonds.

In summary, the same principles regarding the types of glycosidic bonds apply across different polysaccharides. Linear structures will typically have one type of glycosidic bond, while branched structures will have both types, corresponding to their straight and branched sections.

Question 42

Why is glucose and fructose a reducing sugar but not sucrose

Answer 42

Sucrose is table sugar .sucrose is not a reducing sugar cuz it has a very strong glycosidic bond and so it can’t easily go back to the way it was and I t’s not easily broken hence it can easily open up for aldehyde to be exposed. . Reducing sugars: a sugar whose reaction that made it a cyclic group can be reversed. This reversal will cause the cyclic structure to open up and expose the aldehyde group leading to oxidation of aldehyde group leading to coloration(Benedict’s solution,Felix solution). Glucose and fructose are reducing sugars

Anomeric carbons have two features, bound to hydroxyl group and are closer to intracyclic oxygen group (CH2OH)

Whether a sugar is a reducing or non-reducing sugar does not depend on whether it is a monosaccharide, disaccharide, or polysaccharide but rather on the presence of a free anomeric carbon that can open up to form an aldehyde or ketone group

Sucrose as a Non-Reducing Sugar

Sucrose, commonly known as table sugar, is a non-reducing sugar. This is because it has a strong glycosidic bond that links glucose and fructose through their anomeric carbons (C1 of glucose and C2 of fructose). In sucrose, the glycosidic bond is a 1,2-α,β linkage. This bond locks both of the anomeric carbons and prevents them from opening up to form an aldehyde or keto group. Because of this, sucrose cannot participate in redox reactions, such as those with Benedict’s or Fehling’s solutions, which would lead to a color change due to the reduction of copper(II) ions to copper(I) oxide.

Yes, exactly! The fact that copper(II) ions (Cu²⁺) gain electrons is what makes them reduced.Key Points:
• Reduction is defined as the gain of electrons.
• In Benedict’s and Fehling’s tests, Cu²⁺ (copper(II) ions) in the solution gain electrons from a reducing sugar. This gain of electrons converts them to Cu⁺ (copper(I) ions). Not CU3+ cuz it’s unstable in the Benedict and fehling solution. So the copper I combines with something bit of rom copper( I )oxide

Reducing Sugars

A reducing sugar is one that contains a free aldehyde group or a free ketone group that can be oxidized, leading to a reduction reaction (like with Benedict’s or Fehling’s solutions). Sugars like glucose and fructose are examples of reducing sugars. In their cyclic forms, they have an anomeric carbon that is not involved in a glycosidic bond, allowing them to revert to their open-chain forms. In the open-chain form, the aldehyde group of glucose or the keto group of fructose can react with oxidizing agents, leading to oxidation of the sugar and a color change.

A reducing sugar can donate electrons (and thus reduce other compounds) because it has a free anomeric carbon that can revert to an open-chain form.
• Reason for Non-Reducing Nature: In sucrose, both anomeric carbons are involved in the bond formation, so neither is free to revert to an open-chain form. Therefore, sucrose cannot act as a reducing agent.

Conclusion:

•	Sucrose is classified as a non-reducing sugar because the glycosidic bond involves both anomeric carbons, preventing it from being oxidized or participating in redox reactions.

Anomeric Carbon

The anomeric carbon is a specific carbon in the carbohydrate ring structure that is bound to two oxygen atoms: one is part of a hydroxyl group (-OH) and the other is part of an intracyclic oxygen that forms part of the ring. It is also the carbon at which the cyclic form of a sugar can open up to revert to the linear form. In glucose, the anomeric carbon is at position C1, while in fructose, it is at C2. Anomeric carbons are key to determining whether a sugar can act as a reducing sugar or not, depending on whether this carbon is free (not involved in a glycosidic bond) or locked in a bond

Question 43

Lactose constitutes 5% of cow’s milk and 7% of human milk. It is digested by the enzyme lactase. Pure lactose is found in whey, the watery byproduct of cheese production.
D- Galactose and D-glucose
What type of glycosidic bond is this?

Answer 43

Lactose is composed of two monosaccharides: D-galactose and D-glucose. The glycosidic bond between these two sugars is a β-1,4-glycosidic bond.

• β-1,4-Glycosidic Bond: In lactose, the β-1,4-glycosidic bond is formed between the anomeric carbon (carbon 1) of D-galactose and carbon 4 of D-glucose. The β notation indicates that the hydroxyl group on the anomeric carbon of galactose is in the equatorial position (up) relative to the plane of the ring, while the 1,4 specifies the carbons involved in the linkage.

This bond is crucial for the formation and function of lactose, which is hydrolyzed by the enzyme lactase into its constituent sugars, D-galactose and D-glucose, during digestion.

The designation of β or α in glycosidic bonds refers to the orientation of the hydroxyl group (-OH) attached to the anomeric carbon of a sugar in relation to the plane of the sugar ring:

• β (Beta) Glycosidic Bond: The hydroxyl group on the anomeric carbon of the sugar is positioned above the plane of the sugar ring. In lactose, the β-1,4-glycosidic bond means that the hydroxyl group on the anomeric carbon of D-galactose is above the plane of the ring, while the bond is formed with the carbon 4 of D-glucose.
• α (Alpha) Glycosidic Bond: The hydroxyl group on the anomeric carbon is positioned below the plane of the sugar ring.

In lactose:
- D-Galactose: The hydroxyl group on the anomeric carbon (carbon 1) is in the β position (above the ring).
- D-Glucose: The hydroxyl group on carbon 4 is linked to this β-positioned anomeric carbon of galactose.

The formation of the β-1,4-glycosidic bond between D-galactose and D-glucose determines the structural and functional properties of lactose, including how it is recognized and processed by the lactase enzyme in the digestive system.

Question 44

Why aren’t polysaccharides reducing sugars?
What two components are starch made up of?
The chain of polysaccharides may be branched or unbranched, and it may contain different types of monosaccharides true or false
State three examples of polysaccharides

Answer 44

Starch is made up of amylose and amylopectin
Polysaccharides comprise of repeat glucose molecules.

Polysaccharides contain hundreds or thousands of carbohydrate units.

The chain may be branched or unbranched, and it may contain different types of monosaccharides.

The molecular weight may be 100,000 daltons or more

Polysaccharides are not reducing sugars, since the anomeric carbons are connected through glycosidic linkages.

Examples: Starch, glycogen, cellulose, and chitin

Polysaccharides are not considered reducing sugars because their anomeric carbons are involved in glycosidic bonds, which makes them unable to participate in redox reactions that characterize reducing sugars.

Reducing sugars, such as glucose, have a free anomeric carbon (the carbon linked to the oxygen in the carbonyl group) that can either open up to form a free aldehyde group (in aldoses) or be converted into a free ketone group (in ketoses). This free aldehyde or ketone group can then act as a reducing agent in redox reactions, such as with Benedict’s or Fehling’s solution.

However, in polysaccharides:

1. Glycosidic Linkages Block Anomeric Carbons: The anomeric carbon of each monosaccharide unit is involved in a glycosidic bond with another monosaccharide. This bonding “locks” the anomeric carbon and prevents it from opening to form the free aldehyde or ketone group needed for reduction.
2. No Free Aldehyde or Ketone Group: Because both or all anomeric carbons in polysaccharides are occupied by glycosidic linkages, they lack the free aldehyde or ketone group necessary for reducing properties.
3. Example with Sucrose: Sucrose, a disaccharide, is a common example where both anomeric carbons (glucose and fructose) are involved in the glycosidic bond. This makes sucrose a non-reducing sugar, similar to how polysaccharides behave.

Therefore, the absence of a free anomeric carbon due to glycosidic linkages is the key reason why polysaccharides are not reducing sugars.

Question 45

What is starch
What is its importance?
Why is glycogen more readily mobilized for energy in animals than starch
Why is it insoluble in water
What are the types of starch

Answer 45

•Starch is a polymer consisting of D-glucose units
1,6 branching point occurs more rapidly in glycogen than starch . Your statements are correct and highlight key differences between starch and glycogen:

1. Starch is a polymer consisting of D-glucose units:
   
   • Starch is composed of two types of glucose polymers: amylose and amylopectin.
   • Amylose is a linear polymer of α(1→4) linked D-glucose units.
   • Amylopectin is a branched polymer with α(1→4) linked D-glucose units and α(1→6) branching points.
2. 1,6 branching points occur more frequently in glycogen than in starch:
   
   • Glycogen, the storage form of glucose in animals, is similar to amylopectin but has more frequent α(1→6) branching points, approximately every 8-12 glucose units.
   • Starch, specifically amylopectin, has α(1→6) branching points less frequently, approximately every 24-30 glucose units.

• Structure:
  
  • Starch: Contains both linear (amylose) and branched (amylopectin) components.
  • Glycogen: Highly branched, similar to amylopectin but with more frequent branches.
• Branching:
  
  • Starch (Amylopectin): α(1→6) branching points occur approximately every 24-30 glucose units.
  • Glycogen: α(1→6) branching points occur approximately every 8-12 glucose units, making it more highly branched than amylopectin.
    These differences in structure affect their physical properties and biological roles, with glycogen being more readily mobilized for energy in animals due to its highly branched structure.

•Storage form of glucose synthesised by plants

•Starches (and other glucose polymers) are usually insoluble in water because of the high molecular weight.

•There are two forms of starch: amylose and amylopectin.

Question 46

What is the most abundant bio polymer?

Answer 46

Cellulose is a beta 1,4 glycosidic linkage

The most abundant natural biopolymer.

•Cellulose mostly comprises a plant’s cell wall. This provides the cell structural support

•monomers are packed tightly as extended long chains. This gives cellulose its rigidity and high tensile strength—which is so important to plant cells.

The statement can be clarified as follows:

• Cellulose is a polymer composed of monomers. Specifically, cellulose is made up of repeating monomers of β-D-glucose, linked together by β-1,4-glycosidic bonds. This polymer forms the structural component of plant cell walls.
• Nucleic Acids are also polymers, but they are composed of monomers known as nucleotides. Nucleic acids, such as DNA and RNA, consist of long chains of nucleotides linked together by phosphodiester bonds.

In summary:
- Cellulose: Polymer of glucose monomers.
- Nucleic Acids: Polymer of nucleotide monomers.