Assignment 2 due 11 March 2011

What is insulin and why is it important in carbohydrate metabolism? (include the structure of insulin in your work.

Everything should fit in one page.

BIOC 208 gmail

Someone hacked into our account and locked me out. So I have decided to open another account and just e-mail you the notes from there. Send me an e-mail and direct it to:
biochemistry208@gmail.com
And I will send you the notes
In the subject field write: Notes

e-mail to: biochemistry208@gmail.com
Moss

Assignment 1 for 2011 Due 11 February 2011

The importance of biochemistry in Pharmacy (Agriculture ---if registered for this)

Best Student Award and Best Poster Award in BIOC 208

Nucleic Acid Videos

The Central Dogma: The flow of genetic information from DNA to RNA to proteins.It is divided into replication, transcription and translation. Here are the videos.

DNA Replication
The process of duplicating or synthesizing DNA.
DNA Transciption
The process of synthesizing RNA from DNA.

RNA Translation
The process of synthesizing proteins from RNA

Nucleic Acids

Central Dogma
Nucleic Acids

The central dogma of molecular biology was first articulated by Francis Crick in 1958 and re-stated in a Nature paper published in 1970:

The central dogma of molecular biology deals with the detailed residue-by-residue transfer of sequential information. It states that information cannot be transferred back from protein to either protein or nucleic acid. In other words, 'once information gets into protein, it can't flow back to nucleic acid.' The dogma is a framework for understanding the transfer of sequence information between sequential information-carrying biopolymers, in the most common or general case, in living organisms. There are 3 major classes of such biopolymers: DNA and RNA (both nucleic acids), and protein.

DNA and RNA
Living organisms are complex systems. Hundreds of thousands of proteins exist inside each one of us to help carry out our daily functions. These proteins are produced locally, assembled piece-by-piece to exact specifications. An enormous amount of information is required to manage this complex system correctly. This information, detailing the specific structure of the proteins inside of our bodies, is stored in a set of molecules called nucleic acids.

The nucleic acids are very large molecules that have two main parts. The backbone of a nucleic acid is made of alternating sugar and phosphate molecules bonded together in a long chain.

Each of the sugar groups in the backbone is attached (via the bond shown in red) to a third type of molecule called a nucleotide base. Though only four different nucleotide bases can occur in a nucleic acid, each nucleic acid contains millions of bases bonded to it. The order in which these nucleotide bases appear in the nucleic acid is the coding for the information carried in the molecule. In other words, the nucleotide bases serve as a sort of genetic alphabet on which the structure of each protein in our bodies is encoded.

DNA
In most living organisms (except for viruses), genetic information is stored in the molecule deoxyribonucleic acid, or DNA. DNA is made and resides in the nucleus of living cells. DNA gets its name from the sugar molecule contained in its backbone(deoxyribose); however, it gets its significance from its unique structure. Four different nucleotide bases occur in DNA: adenine (A), cytosine (C), guanine (G), and thymine (T).

Chemical Structure of the DNA Nucleotides


These nucleotides bind to the sugar backbone of the molecule as follows:

A-->T G-->C
sugar phosphate sugar phosphate sugar phosphate sugar ...

The versatility of DNA comes from the fact that the molecule is actually double-stranded. The nucleotide bases of the DNA molecule form complementary pairs: The nucleotides hydrogen bond to another nucleotide base in a strand of DNA opposite to the original. This bonding is specific, and adenine always bonds to thymine (and vice versa) and guanine always bonds to cytosine (and vice versa). This bonding occurs across the molecule, leading to a double-stranded system as pictured below:

sugar phosphate sugar phosphate sugar phosphate sugar ...
T A C G
¦ ¦ ¦ ¦
A T G C
sugar phosphate sugar phosphate sugar phosphate sugar ...


In the early 1950s, four scientists, James Watson and Francis Crick at Cambridge University and Maurice Wilkins and Rosalind Franklin at King's College, determined the true structure of DNA from data and X-ray pictures of the molecule that Franklin had taken. In 1953, Watson and Crick published a paper in the scientific journal Nature describing this research. Watson, Crick, Wilkins and Franklin had shown that not only is the DNA molecule double-stranded, but the two strands wrap around each other forming a coil, or helix. The true structure of the DNA molecule is a double helix.

The double-stranded DNA molecule has the unique ability that it can make exact copies of itself, or self-replicate. When more DNA is required by an organism (such as during reproduction or cell growth) the hydrogen bonds between the nucleotide bases break and the two single strands of DNA separate. New complementary bases are brought in by the cell and paired up with each of the two separate strands, thus forming two new, identical, double-stranded DNA molecules.


RNA

Ribonucleic acid, or RNA, gets its name from the sugar group in the molecule's backbone - ribose. Several important similarities and differences exist between RNA and DNA. Like DNA, RNA has a sugar-phosphate backbone with nucleotide bases attached to it. Like DNA, RNA contains the bases adenine (A), cytosine (C), and guanine (G); however, RNA does not contain thymine, instead, RNA's fourth nucleotide is the base uracil (U). Unlike the double-stranded DNA molecule, RNA is a single-stranded molecule. RNA is the main genetic material used in the organisms called viruses, and RNA is also important in the production of proteins in other living organisms. RNA can move around the cells of living organisms and thus serves as a sort of genetic messenger, relaying the information stored in the cell's DNA out from the nucleus to other parts of the cell where it is used to help make proteins.

Fatty acid synthesis and beta-oxidation

Beta-oxidation


Fatty acid synthesis

Test 1 Memorandum for BIOC 208

Test 1 BIOC 208

Date: 22 February 2010
Time: 18h00-19h00 (16h00-17h00 for those having STATS Class)
Venue: Life Science Building(Q-Block Extension)
Ground Floor Labs
Total: 60 points
Covers: Chapters 1-4 (Proteins)

GOOD LUCK

Chapter 7: Glycolysis

Metabolism is the set of chemical reactions that happen in living organisms to maintain life.

CATABOLISM AND ANABOLISM
1. Catablosim breaks down large molecules into smaller molecules whereas Anabolism joins small molecules to produces larger molecules.
2. Catabolism produces energy whereas Anabolism requires energy.

For an interactive animation of glycolysis click HERE
Glycolysis is the anaerobic catabolism of glucose.
•It occurs in virtually all cells.
•In eukaryotes, it occurs in the cytosol.
•C6H12O6 + 2NAD+ -> 2C3H4O3 + 2NADH + 2H+
•The free energy stored in 2 molecules of pyruvic acid is somewhat less than that in the original glucose molecule.
•Some of this difference is captured in 2 molecules of ATP.

The Fates of Pyruvic Acid
In YEAST
•Pyruvic acid is decarboxylated and reduced by NADH to form a molecule of carbon dioxide and one of ethanol.
•C3H4O3 + NADH + H+ → CO2 + C2H5OH + NAD+
•This accounts for the bubbles and alcohol in, for examples, beer and champagne.
•The process is called alcoholic fermentation.
•The process is energetically wasteful because so much of the free energy of glucose (some 95%) remains in the alcohol (a good fuel!).

In active MUSCLES•Pyruvic acid is reduced by NADH forming a molecule of lactic acid. •C3H4O3 + NADH + H+ → C3H6O3 + NAD+
•The process is called lactic acid fermentation.
•The process is energetically wasteful because so much free energy remains in the lactic acid molecule. (It can also be debilitating because of the drop in pH as the lactic acid produced in overworked muscles is transported out into the blood.)

In MITOCHONDRIA•Pyruvic acid is oxidized completely to form carbon dioxide and water.
•The process is called cellular respiration. Link to a discussion of cellular respiration.

•Approximately 40% of the energy in the original glucose molecule is trapped in molecules of ATP.

Test yourself:
1. The end product of Glycolysis is:
A) Pyruvate Kinase
B) Phosphoenol pyruvate
C) Glucose
D) Pyruvate

2. Glycolysis occurs in the:
A) Mitochondrion
B) Nucleus
C) Cytoplasm
D) Cell membrane

3. Catabolic processes
A) make complex molecules from simpler ones
B) break complex molecules into simpler ones
C) occur only in autotrophs
D) occur only in heterotrophs
E) none of the above

4. The proper sequence of stages in glycolysis is
A) glucose priming, cleavage and rearrangement, oxidation, ATP generation
B) cleavage and rearrangement, glucose priming, ATP generation, oxidation
C) glucose priming, oxidation, cleavage and rearrangement, ATP generation
D) ATP generation, oxidation, glucose priming, cleavage and rearrangement
E) oxidation, cleavage and rearrangement, ATP generation, glucose priming

5. What substance is regenerated by fermentation?
A) O2
B) NAD+
C) acetyl-CoA
D) ATP
E) glucose

6. The oxidation of glucose to two molecules each of pyruvate, ATP, and NADH is called ________ and occurs in the ________.
A) glycolysis; cytoplasm
B) fermentation; cytoplasm
C) the Krebs cycle; matrix of the mitochondrion
D) anaerobic respiration; cytoplasm
E) the respiratory electron transport chain; cristae of the mitochondrion

7. A cell culture was supplied with radioactively labeled O2. The cells were monitored. In a few minutes the radioactive oxygen atoms were present in which of the following compounds:
A) carbon dioxide
B) NADH and FADH2
C) water
D) ATP
E) lactic acid

8. The final electron acceptor in lactic acid fermentation is:
A) NAD+
B) pyruvate
C) O2
D) lactic acid
E) ATP

9. During the oxidation of glucose, a net gain of ATP only occurs under aerobic conditions.
A) True
B) False

10. ATP formation by glycolysis
A) occurs through aerobic respiration
B) is an extremely efficient method of acquiring energy by the cell
C) requires oxygen
D) involves substrate-level phosphorylation
E) both a and c

Chapter 6: Carbohydrates

Carbohydrates (also called saccharides) are molecular compounds made from just three elements: carbon, hydrogen and oxygen. Monosaccharides (e.g. glucose) and disaccharides (e.g. sucrose) are relatively small molecules. They are often called sugars. Other carbohydrate molecules are very large (polysaccharides such as starch and cellulose).

Carbohydrates are:• a source of energy for the body e.g. glucose and a store of energy, e.g. starch in plants
• building blocks for polysaccharides (giant carbohydrates), e.g. cellulose in plants and glycogen in the human body
• components of other molecules eg DNA, RNA, glycolipids, glycoproteins, ATP

Monosaccharides
Monosaccharides are the simplest carbohydrates and are often called single sugars. They are the building blocks from which all bigger carbohydrates are made.
Monosaccharides have the general molecular formula (CH2O)n, where n can be 3, 5 or 6.
They can be classified according to the number of carbon atoms in a molecule:
n = 3 trioses, e.g. glyceraldehyde
n = 5 pentoses, e.g. ribose and deoxyribose ('pent' indicates 5)
n = 6 hexoses, e.g. fructose, glucose and galactose ('hex' indicates 6)

There is more than one molecule with the molecular formula C5H10O5 and more than one with the molecular formula C6H12O6. Molecules that have the same molecular formula but different structural formulae are called structural isomers.
Glyceraldehyde's molecular formula is C3H6O3. Its structural formula shows it contains an aldehyde group (-CHO) and two hydroxyl groups (-OH). The presence of an aldehyde group means that glyceraldehyde can also be classified as an aldose. It is a reducing sugar and gives a positive test with Benedict's reagent.
CH2OHCH(OH)CHO is oxidised by Benedict's reagent to CH2OHCH(OH)COOH; the aldehyde group is oxidised to a carboxylic acid and Benedict's reagent is reduced (Cu2+ to Cu+).

Pentoses and hexoses can exist in two forms: cyclic and non-cyclic. In the non-cyclic form their structural formulae show they contain either an aldehyde group or a ketone group.
Monosaccharides containing the aldehyde group are classified as aldoses, and those with a ketone group are classified as ketoses. Aldoses are reducing sugars; ketoses are non-reducing sugars. This is important in understanding the reaction of sugars with Benedict's reagent.
However, in water pentoses and hexoses exist mainly in the cyclic form, and it is in this form that they combine to form larger saccharide molecules.

Glucose
Glucose is the most important carbohydrate fuel in human cells. Its concentration in the blood is about 1 gdm-3. The small size and solubility in water of glucose molecules allows them to pass through the cell membrane into the cell. Energy is released when the molecules are metabolised (C6H12O6 + 6O2 6CO2 + 6H2O). This is part of the process of respiration.
There are two forms of the cyclic glucose molecule: α-glucose and β-glucose.
Two glucose molecules react to form the dissacharide maltose. Starch and cellulose are polysaccharides made up of glucose units.

Galactose
Galactose molecules look very similar to glucose molecules. They can also exist in α and β forms. Galactose reacts with glucose to make the dissacharide lactose.
However, glucose and galactose cannot be easily converted into one another. Galactose cannot play the same part in respiration as glucose.
This comparison of glucose and galactose shows why the precise arrangement of atoms in a molecule (shown by the displayed formula) is so important.

Fructose
Fructose, glucose and galactose are all hexoses. However, whereas glucose and galactose are aldoses (reducing sugars), fructose is a ketose (a non-reducing sugar). It also has a five-atom ring rather than a six-atom ring. Fructose reacts with glucose to make the dissacharide sucrose.

Ribose and deoxyribose
Ribose and deoxyribose are pentoses. The ribose unit forms part of a nucleotide of RNA. The deoxyribose unit forms part of the nucleotide of DNA.

Disaccharides
Monosaccharides are rare in nature. Most sugars found in nature are disaccharides. These form when two monosaccharides react.
A condensation reaction takes place releasing water. This process requires energy. A glycosidic bond forms and holds the two monosaccharide units together.
The three most important disaccharides are sucrose, lactose and maltose. They are formed from the a forms of the appropriate monosaccharides. Sucrose is a non-reducing sugar. Lactose and maltose are reducing sugars.

Disaccharide Monosaccharides
sucrose from α-glucose + α-fructose
maltose from α-glucose + α-glucose
α-lactose * from α-glucose + β-galactose

* Lactose also exists in a beta form, which is made from β-galactose and β-glucose
Disaccharides are soluble in water, but they are too big to pass through the cell membrane by diffusion. They are broken down in the small intestine during digestion to give the smaller monosaccharides that pass into the blood and through cell membranes into cells.
C12H22O11 + H2O C6H12O6 + C6H12O6
This is a hydrolysis reaction and is the reverse of a condensation reaction. It releases energy.

Monosaccharides are used very quickly by cells. However, a cell may not need all the energy immediately and it may need to store it. Monosaccharides are converted into disaccharides in the cell by condensation reactions. Further condensation reactions result in the formation of polysaccharides. These are giant molecules which, importantly, are too big to escape from the cell. These are broken down by hydrolysis into monosaccharides when energy is needed by the cell.

Polysaccharides
Monosaccharides can undergo a series of condensation reactions, adding one unit after another to the chain until very large molecules (polysaccharides) are formed. This is called condensation polymerisation, and the building blocks are called monomers. The properties of a polysaccharide molecule depend on:
• its length (though they are usually very long)
• the extent of any branching (addition of units to the side of the chain rather than one of its ends)
• any folding which results in a more compact molecule
• whether the chain is 'straight' or 'coiled'

Starch
Starch is often produced in plants as a way of storing energy. It exists in two forms: amylose and amylopectin. Both are made from α-glucose. Amylose is an unbranched polymer of α-glucose. The molecules coil into a helical structure. It forms a colloidal suspension in hot water. Amylopectin is a branched polymer of α-glucose. It is completely insoluble in water.

Glycogen
Glycogen is amylopectin with very short distances between the branching side-chains. Starch from plants is hydrolysed in the body to produce glucose. Glucose passes into the cell and is used in metabolism. Inside the cell, glucose can be polymerised to make glycogen which acts as a carbohydrate energy store.

Cellulose
Cellulose is a third polymer made from glucose. But this time it's made from β-glucose molecules and the polymer molecules are 'straight'.
Section of a cellulose molecule

Cellulose serves a very different purpose in nature to starch and glycogen. It makes up the cell walls in plant cells. These are much tougher than cell membranes. This toughness is due to the arrangement of glucose units in the polymer chain and the hydrogen-bonding between neighbouring chains.
Cellulose is not hydrolysed easily and, therefore, cannot be digested so it is not a source of energy for humans. The stomachs of Herbivores contain a specific enzyme called cellulase which enables them to digest cellulose.

REDUCING SUGARSA reducing sugar is any sugar that, in an alkaline solution, forms some aldehyde or ketone. This allows the sugar to act as a reducing agent, for example in the Maillard reaction and Benedict's reaction.
Examples
Reducing sugars include glucose, fructose, glyceraldehyde, lactose, arabinose and maltose. All monosaccharides which contain ketone groups are known as ketoses, and those which contain aldehyde groups are known as aldoses.
Also in glucose polymers as glucose syrup, maltodextrin and dextrin the macromolecule, begins with a reducing sugar, a free aldehyde. The more hydrolysed starch, the shorter molecule, the more reducing sugars are present. The percentage reducing sugars present in these starch derivatives is called dextrose equivalent (DE).
Significantly, sucrose and trehalose are not reducing sugars.

Chemistry of reducing sugars
A reducing sugar occurs when its anomeric carbon (the carbon which is linked to two oxygen atoms) is in the free form. Since sugars occur in a chain as well as a ring structure, it is possible to have an equilibrium between these two forms. When the hemi-acetal or hemi-ketal hydroxyl group is free, i.e. it is not locked in an acetal or ketal linkage, the aldehyde or ketone (i.e. the chain-form) is present. The aldehyde can be oxidized to a carboxyl group via a redox reaction. The chemical that causes this oxidation becomes reduced. Thus, a reducing sugar is one that reduces certain chemicals. Even though a ketone cannot be oxidized directly, a keto sugar can sometimes be converted to an aldehyde via a series of tautomeric shifts to migrate the carbonyl to the end of the chain. Therefore, keto sugars are also sometimes reducing.



Test your knowledge
1. Saccharides contain the following combination of elements:
A. carbon, nitrogen and hydrogen
B. carbon, hydrogen and phosphorus
C. carbon and hydrogen
D. carbon, oxygen and hydrogen

2. Aldoses are reducing sugars because in their non-cyclic form they contain:
A. an ester group
B. a ketone group
C. an aldehyde group
D. an hydroxyl group

3. Which is the most important carbohydrate fuel in human cells?
A. ribose
B. glucose
C. fructose
D. galactose

4. Which two monosaccharides combine to form sucrose?
A. α-galactose and α-fructose
B. α-fructose and α-ribose
C. α-glucose and β-glucose
D. α-glucose and α-fructose

5. The type of reaction that occurs when a disaccharide is formed from two monosaccharides is
A. reduction
B. addition
C. hydrolysis
D. condensation

6. The type of bond that forms when a disaccharide is formed from two monosaccharides is called:
A. a peptide bond
B. a carbohydrate bond
C. an ester bond
D. a glycosidic bond

7. The products of hydrolysis of lactose are:
A. α-glucose and α-fructose
B. α-galactose and α-ribose
C. α-fructose and α-galactose
D. α-glucose and α-galactose

8. Starch is a polymer made from the following monomer:
A. α-glucose
B. β-glucose
C. α-fructose
D. α-galactose

How TB challenges the host (Try answering)

I was attending a Colloquim on Innate Immunity in Johannesburg on Monday and Tueday 15-16 Feb and here is a scenario that you will find interesting. The figure above shows how TB is attacked by the cells of the immune system in an attempt to kill it. TLR4 is a protein on cells and it binds TB bacteria using the extracellular component and once this is achieved, Protein A binds the intracellular domain of TLR-4, leading to internalization and killing of TB.

Try these Questions:
1. Which of the two strains of TB is likely to be killed in this way. Study the amino acid composition of the structures to make your argument.
2. TLR4 has 3 components: the extracellular domain, transmembrane domain, and intracellular domain. Each of the domains has different amino acid composition. Can you comment on the relevance of these differences in structure.
3. TB is ultimately supposed to be killed by an enzyme called Lysozyme at pH2. If lysozyme is rich in Tryptophan, why would a pH of 2 be required?
4. What type of amino acids would you expect to find dominating Protein A.
5. If you were to design a drug that could directly kill TB, what kind of drug would you design (look at the amino acid composition of TB to guide you).

BIOC 208 Lecture Schedule for 2010

UNIT 1: PROTEIN STRUCTURE AND FUNCTION (Chapters 1-4)

ASSIGNMENT NO. 1: WHAT IS THE IMPORTANCE OF BIOCHEMISTRY IN PHARMACY?

Minimum: 500 words, Maximum: 1200 words. Submission Date: 01 February 2010 in class

Typed using Times New Romans, Font size 12pt and Line spacing of 1.

On top of first page mention the Surname & Initials, Student Number, Course Code, Assignment One: What is the importance of Biochemistry in Pharmacy? … and then start with your assignment(don’t make a traditional cover page with your personal details, just go on and write)

Proteins are very important molecules in our cells. They are involved in virtually all cell functions. Each protein within the body has a specific function. Some proteins are involved in structural support, while others are involved in bodily movement, or in defense against germs. Proteins vary in structure as well as function. They are constructed from a set of 20 amino acids and have distinct three-dimensional shapes. Below is a list of several types of proteins and their functions.

Protein Functions

1. Antibodies - are specialized proteins involved in defending the body from antigens (foreign invaders). One way antibodies destroy antigens is by immobilizing them so that they can be destroyed by white blood cells.

2. Contractile Proteins - are responsible for movement. Examples include actin and myosin. These proteins are involved in muscle contraction and movement.

3. Enzymes - are proteins that facilitate biochemical reactions. They are often referred to as catalysts because they speed up chemical reactions. Examples include the enzymes lactase and pepsin. Lactase breaks down the sugar lactose found in milk. Pepsin is a digestive enzyme that works in the stomach to break down proteins in food.

4. Hormonal Proteins - are messenger proteins which help to coordinate certain bodily activities. Examples include insulin, oxytocin, and somatotropin. Insulin regulates glucose metabolism by controlling the blood-sugar concentration. Oxytocin stimulates contractions in females during childbirth. Somatotropin is a growth hormone that stimulates protein production in muscle cells.

5. Structural Proteins - are fibrous and stringy and provide support. Examples include keratin, collagen, and elastin. Keratins strengthen protective coverings such as hair, quills, feathers, horns, and beaks. Collagens and elastin provide support for connective tissues such as tendons and ligaments.

6. Storage Proteins - store amino acids. Examples include ovalbumin and casein. Ovalbumin is found in egg whites and casein is a milk-based protein.

7. Transport Proteins - are carrier proteins which move molecules from one place to another around the body. Examples include hemoglobin and cytochromes. Hemoglobin transports oxygen through the blood. Cytochromes operate in the electron transport chain as electron carrier proteins.

Summary: Proteins serve various functions in the body. The structure of a protein determines its function. For example, collagen has a super-coiled helical shape. It is long, stringy, strong, and resembles a rope. This structure is great for providing support. Hemoglobin on the other hand, is a globular protein that is folded and compact. Its spherical shape is useful for maneuvering through blood vessels.

Chapter 1: Amino Acids

Titration of Aspartate with Hydroxide

On the left in both the chemical reaction and the titration curve, you should imagine that aspartic acid is in a very acidic solution at a pH of about 0. As you move from left to right accross the page, you are adding hydroxide to the solution. This increases the hydroxide concentration and decreases the hydrogen ion concentration. Note that aspartate loses protons as you move from left to right. At the first pKa, the alpha-carboxyl dissociates. At the second pKa, the R-group carboxyl dissociates, At the third pKa, the alpha-amino dissociates.

Note that at each pKa, the solution is buffered. That is, it resists changes in pH as hydroxide is added. Also note, that the pI occurs where aspartate has no net charge.

The main objective of this section is "Given any amino acid you must be able to predict the isoelectric point." How do I start? How do I figure this out without a graph to look at?

For this course, there are only four categories of isoelectric points that you need to learn:

• amino acids without any charged R-group (alanine, glycine, ...)
• lysine and arginine
• aspartate and glutamate
• histidine

The isoelectric point (pI) is the pH at which an amino acid or protein has no net charge and will not migrate towards the anode or cathode in an electric field. The charges on any amino acid at a given pH are a function of their pKas for dissociation of a proton from the alpha-carboxyl groups, the alpha-amino groups, and the side chains (R-group). The pKa for the alpha-amino groups and the alpha-carboxyl groups are about 2 and 10.

You start by having a very rough idea of the structure of the amino acid. What are the acidic groups and what are their pKas. Next, you try to visualize the amino acid fully associated with hydrogen and what the charge on the molecule would be. Next, you visualize removing hydrogen ions by titrating with hydroxide ions. You will remove hydrogen ions from the group with the lowest pKa first and, then from the next higher pKa. Eventually you reach the pI.

Example: Calculate the pI for Aspartate. Aspartic acid has an alpha-carboxyl, and alpha-amino and an R-Group that is a carboxyl group.

At very acidic pHs, the R-group is in the COOH form, the alpha-amino group is in the –NH₃⁺ form and the alpha-carboxyl group is in the COOH form so aspartate has a net charge of +1

As we titrate with hydroxide ion, we remove hydrogen ions. They combine with hydroxide ions and become water. When we reach pH 2, the protons on half the alpha-carboxyl groups are removed. This is not the pI because half of the alpha-carboxyl have a negative charge but all of the alpha-amino groups –NH₃⁺ have a positive charge and all of the R-Group (COOH ) have no charge. So, the net charge on the aspartate molecules is a positive 0.5.

As we titrate with more hydroxide ions, we reach a point half way between pK_a1 and pK_a2. At this pH, half of the protons have been removed from the two carboxyl groups and half of the carboxyl groups are not dissociated so the net charge on the carboxyl groups is a -1. The alpha-amino group is fully charged so it has a net charge of +1. The net charge on the aspartate molecules is 0. This is the pI

As we titrate with more hydroxide ions, we reach the pK_a2, At this pH, all of the protons have been removed from the alpha-carboxyl group and half of the half of the protons have been removed from the R-group carboxyl groups. The net charge on the carboxyl groups is a -1.5. The alpha-amino group is fully charged so it has a net charge of +1. The net charge on the aspartate molecules is -0.5.

As we titrate with more hydroxide ions, we reach the pK_a3, a point where all the carboxyl groups are dissociated and only half of the alpha-amino groups still have a positive charge. The net charge is a negative 1.5. We did not have to go this far to determine the pI but I thought it might be useful.

To review, you started with knowing that aspartate had three dissociable groups and the pKas for those groups. You know that as you titrate, the molecule will change as follows:

COOH, COOH, –NH3+	at a pH below pKa1. The net charge is about +1.
COO-, COOH, –NH3+	at a pH between pKa1 and pKa2. Net charge is 0.
COO-, COO-, –NH3+	at pH above pKa2. The net charge is about -1.
COO-, COO-, –NH2	at a pH above pKa3, the net charge is about-2.

The only neutral solution of aspartate must be when the pH is between pK_a1 and pK_a2. The pI is half way in between.pK_a1 and pK_a2

ACID-BASE PROBLEMS AND THE HENDERSON-HASSELBALCH EQUATION

A. A pharmaceutical molecule with antifungal properties is only active when deprotonated and negatively charged (A^-). The protonated state (HA) is inactive. If the pK_a of this drug is 10.0, (a) calculate the ratio of protonated to deprotonated compound at physiological pH (7.4). (b) Is this drug likely to be a useful pharmaceutical agent?

(a) calculate the ratio of protonated to deprotonated compound at physiological pH (7.4).

Since we are given both the pH and pK_a of the compound, we can use the Henderson-Hasselbalch equation to solve for the ratio of [HA] to [A^-].

pH = pK_a - log([HA] / [A^-])

log([HA] / [A^-]) = pK_a – pH

log([HA] / [A^-]) = 10.0 – 7.4

log([HA] / [A^-]) = 2.6

([HA] / [A^-]) = 398.11

([HA] / [A^-]) = 400 (correct sig figs)

The ratio of protonated (inactive) compound to deprotonated (active) compound is 400 to 1 at physiological pH.

(b) Is this drug likely to be a useful pharmaceutical agent?

Since the vast majority of the compound is in the inactive form at physiological pH, it is unlikely to be a useful pharmaceutical agent.* Ideally, most of the compound would be active in the body.

*However, if the active compound is highly potent, it is possible that a small fraction of active compound is sufficient for useful antifungal activity.

B.

Absorption of aspirin (acetylsalicylic acid, C₉H₈O₄,) into the bloodstream occurs only when the molecule is in its conjugate base form.

(a) If a patient takes two tablets of aspirin (325 mg each), how many grams of aspirin are available for immediate absorption in the stomach? The pH of the stomach is 1.6, and the pK_a of aspirin is 3.5.

Since we are given both the pH of the stomach and the pK_a of aspirin, we can use the Henderson-Hasselbalch equation to solve for the ratio of [HA] to [A^-].

pH = pK_a - log([HA] / [A^-])

log([HA] / [A^-]) = pK_a – pH

log([HA] / [A^-]) = 3.5 – 1.6

log([HA] / [A^-]) = 1.9

([HA] / [A^-]) = 79

The ratio of protonated aspirin to its conjugate base is 79 to 1.

So one-eightieth (1/80) of the total aspirin taken will be in the conjugate base form and available for immediate absorption in the stomach:

2 x 325 mg x (1/80) = 8.75 mg

9 mg

(b) Would you expect more or less aspirin to be absorbed in the small intestine (pH ≈ 7.5) compared to the stomach? Briefly explain your answer (no calculation is required).

More aspirin will be absorbed in the small intestine. The higher pH in the intestine means that more aspirin will be in the conjugate base form and therefore available for absorption.

TEST YOURSELF

1.1 Which of the following is the smallest of all amino acids?

A).Glycine, B).Valine, C).Alanine, D).Serine

1.2 Which of the following amino acids can form a covalent bond called a disulfide bond?

A). C, B). K, C). M, D). Y

1.3 Which of the following amino acids is actually an imino acid?

A). Glutamate, B). Alanine, C). Proline D). Glycine

1.4 Which of the following amino acids is the largest of all amino acids?

A). E, B). Y, C). W, D). H

1.5 Which of the following amino acids is not positively charged?

A). H, B). K, C). D D). R

1.6 The three letter “asn” is for which amino acid?

A). Alanine, B). Aspartic acid, C). Asparagine, D). Arginine

1.7 Which of the following is a basic amino acid with a positive charge?

A). Lys, B). Glu C). Trp, D). Gln

1.8 Which of the following amino acids’ side chain is a single methyl group?

A). I, B). V, C). A, D). L

1.9 Which of the following amino acids can form an ionic bond?

A). E & V, B). D & K, C). R & Q, D). P & G

1.10Which of the following amino acids() are polar?

A). All of these, B). Tyrosine, C). Serine, D). Arginine

1.11Titration of Valine by a strong base reveals two pKa’s. The titration reaction occurring at pKa2(pKa2=9.62) is:

A. –COOH + OH^- à -COO^- + H₂O C. –COO^- + -NH₂^+- à -COOH + -NH₂

B. –COOH + -NH₂ à -COO^- + -NH₂⁺ D. –-NH₃⁺ + OH^- à -NH₂ + H₂O

1.12For amino acids with a neutral R group, at any pH below the pI of the amino acid, these amino acids in solution will have:

A). a net negative charge, B). a net positive charge, C). no charged groups, D). no net charge

1.13The peptide AEGAL has:

A). a S-S bridge, B). 5 peptide bonds, C). 4 peptide bonds, D). no free carboxyl groups.

1.14Charmaine was upset that her boyfriend dumped her over her best friend; so she decided to commit suicide by overdosing with Panado (pKa=3) and drinking it with lots of fresh milk (pH 6.7). She was admitted to the hospital but was later discharged without any serious complication. Drug tests revealed that most of the Panado remained in the stomach rather than being absorbed, can you explain these? (include calculations).

1.15You are working for a MERCK Pharmaceutics and the manager gives you a task of designing a drug to treat breast cancer and produce it in the form of a pill. The drug must have an absorption success of atleast 60% in the stomach (pH=2), which is where it must be in unionized form. What will the pKa of your drug be?

1.16A student titrates acidified Lysine using NaOH to completed all of its titrable groups. Draw the titration curve of Lysine showing the state of these Ionizable groups at each point and also calculate its pI.

1.17Hemoglobin is a protein responsible for transporting oxygen from the lungs to the tissue via the blood circulatory system. For it to be able to perform its function which of the following groups of amino acids will dominate the surface of hemoglobin?

A). A, V, F, L B). H, R, L, E C). Y, G, M, I D). W, D, A, V

1.18Insulin binds to insulin receptor on cell surfaces to be able to facilitate uptake of glucose by the cells. Which types of amino acids will dominate the surface of the Insulin receptor?

A). AVFL B). HRLE C). YGMN D). WDAV

1.19All the 19 amino acids have a chiral carbon center except for A). F B). G C). H D). W.

1.20Suppose you were a bacterium and had a choice of living in the small intestines (pH 8) or the stomach (pH 1). A new antibacterial drug (pKa = 2.5) has just been released into the market to be taken by humans. Which environment would choose to live in? The stomach or small intestine?