Organic Lecture 5: Epimers, Anomers, and Meso Compounds

These topics are skimmed over in most of our organic classes but, nevertheless, it's important to have a cursory understanding for the MCAT.

Epimers: differ in their absolute configuration at a single chiral center; they are a subclass of diastereomers (see picture).

The prefix D on the name of molecules (i.e. D-Galactose) refers to the orientation of the hydroxyl group (-OH) on the highest-numbered chiral center in a Fischer projection. When the hydroxyl group is on the right of this carbon in the Fischer projection, the molecule is a D sugar. When it's on the left, the molecule is an L sugar.

**Note: D and L are entirely unrelated to optical activity. Distinctions between D and L (or between R and S) can be made just by looking at a drawing of the molecule, but distinctions between (+) and (-) can be made only by running experiments in a polarimeter

Once again:
R or S = absolute configuration (structure)
D or L = relative configuration (structure)
(+) or (-) = observed optical rotation (property)

Epimeric carbon: place at which stereochemistry is different between two molecules

Anomers: epimers that form as a result of ring closure; for MCAT it pertains to only sugar chemistry (i.e. glucose)

Anomeric carbon in glucose: can assume two forms--with hydroxyl group down, it is alpha...with the hydroxyl group up, it is Beta.

Meso compounds: internal plane of symmetry in a molecule that contains chiral centers; are not optically active because one side of the molecule is a mirror image of the other (this means optical activity imparted by one side is canceled out by its other side)

Organic Lecture 4: Enantionmers, Diasteriomers, Optical Activity

We are going to continue with isomerism.

Enantiomers: non-superimposable mirror images. They can only occur when chiral centers are present. NOTE: two molecules that are enantiomers will always have opposite absolute configurations (one has R, the other will have S, and vice versa).

Optical Activity: a compound that rotates the plane of polarized light is said to be optically active.

**A pair of enantiomers will rotate plane-polarized light with equal magnitude, but in opposite directions

If it rotates clockwise it is said to be dextrorotarory (d), also denoted (+)

If it rotates counterclockwise it is said to be levorotatory (l), also denoted (-)

Specific rotation: the magnitude of rotation of plane-polarized light for any compound

**Racemic mixtures are not optically active

One last EXTREMELY IMPORTANT NOTE: (+) and (-) say nothing about whether the absolute configuration is R or S. Therefore, there is no correlation between the sign of rotation and the absolute configuration.

Diastereomers: stereoisomers that are not enantiomers; they are non-superimposable, non-mirror images. Take a look at the picture for a better understanding.

Organic Lecture 3: Chirality and Determining R/S Configurations

I'll do my best to put these concepts into words but keep in mind that this is generally taught with a molecule kit so students can visually see how this works.

Any molecule that cannot be superimposed on its mirror image is said to be chiral. For the MCAT, it's important that you be able to identify chiral centers. For carbon, a chiral center will have four different groups bonded to it--therefore, it must also be sp3 hybridized with 109 degree bond angles. A chiral carbon can also be referred to as a stereocenter, a stereogenic center, or an asymmetric center.

Chiral centers can be assigned an absolute configuration. This is an arbitrary set of rules known as the Cahn-Ingold-Prelog rules.

Priority is assigned to the four different substituents according to increasing atomic number of the atoms directly attached to the chiral center.

Note on isotopes: MCAT likes to test on these. For example, the isotopes of hydrogen are 1H, 2H=D (deuterium), and 3H=T (tritium). Just assigned on the basis of atomic weight if given isotopes.

If two identical atoms are attached to a stereocenter, then the next atoms in both chains are examined until a difference is found.

A multiple bond is counted as two single bonds for both of the atoms involved. For example, if you see a carbon double bonded to an oxygen, it is treated as a carbon bonded to two oxygens.

Once priorities have been assigned, rotate the molecule so that the lowest priority groups points directly away from the viewer. Then simply trace the path from the highest priority to the lowest priority. If the path is clockwise, the absolute configuration is R. If the path is counterclockwise, the absolute configuration is S.

Note: if the lowest priority is sticking out of the page (pointed toward the viewer), trace your path and then flip it. So if you traced a clockwise path, the configuration will be an S because the lowest priority is not pointed away from the viewer.

Organic Lecture 2: Structure, Bonding, and Saturation

A quick note on hybridization: it is essentially the mixing of atomic orbitals to rationalize observed chemical and structural trends. Simply, if add an S orbital to a P orbital we get a SP hybrid. To determine the hybridization for an atom in a molecule, add the number of attached atoms to the number of non-bonding electron pairs (non-delocalized). Then use the table in the picture.

Sigma Bonds: consists of two electrons that are localized between two nuclei. It is a very strong bond bond is always the first type of bond to be formed between any two atoms; a single bond must be a sigma bond.

Pi bonds: composed of two electrons that are localized to the region that lies on opposite sides of the plane formed by the two bonded nuclei and immediately adjacent atoms (not directly between like sigma bonds). Is formed by the side-to-side aligment of two unhybridized p orbitals. The electrons in a pi bond are further from the nuclei than the electrons of a sigma bond, and therefore at a higher energy level, less stable, and form a weaker bond. Note that pi bonds prevent rotation.

**In any multiple bond, there is only one sigma bond and the remainder are pi bonds.

Saturation: a molecule is said to be saturated if it contains no pi bonds and no rings; therefore, it is unsaturated if it has at least one pi bond or a ring.

To determine the degree of unsaturation use this formula: {(2n+2)-x}/2
n=number of carbons
x=number of hydrogens (plus any monovalent atoms such as halogens like F, Cl, Br, or I.

One degree of unsaturation indicates the presence of one pi bond or one ring, two degrees indicates two pi bonds (2 separate double bonds or one triple bond), or one pi bond and one ring, two rings, etc.

**Each oxygen (or other divalent atom) "replaces" one carbon and 2 hydrogen atoms
**Each nitrogen (or other trivalent atom) "replaces" one carbon and 1 hydrogen atom

Let's touch on bond dissociation energy (BDE). By definition, it is the energy required to break a bond homolytically. In homolytic bond cleavage, one electron of the bond being broken goes to each fragment of the molecule. Two radicals form in the process. This is different from heterolytic bond cleavage where both electrons of the electron pair that make up the bond go to the same atom; this forms a cation and an anion

**The higher the bond order, the shorter and strong the bond.
**When comparing the same types of bonds, the greater the S character in the component orbital, the shorter the bond (because s-orbitals are closer to the nucleus than p-orbitals).
**The longer the bond, the weaker it is

ERROR: on my picture it should say "degree of unsaturation", not "saturation" at the bottom. Sorry

Organic Lecture 1: The Effects of Hydrocarbon Branching on MP and BP

Let me preface our first organic lecture with a brief excerpt on how to succeed in studying for this portion of the test. My advice is to approach studying organic like you would approach studying a foreign language. It's a mistake to believe that you can simply memorize this material and perform well on any organic chemistry test, especially the MCAT. There are an infinite number of possible reactions out there so you first need to understand the vocabulary (mechanisms) and grammar of organic. How does one accomplish this? Work lots of problems and seek help when needed. With this said, I understand that by reading this post most of you have already taken at least one semester of the course. If you have and did well, congratulations. It's a class that has ruined more medical careers than any other. Studying for MCAT organic will luckily be a little more mundane than the preparation you put forth for the actual class. Flash cards won't be necessary and there are no synthesis problems on the MCAT--although they do find a way to test these concepts indirectly as we will soon discover. So without further adieu, let's begin our first organic lesson with melting/boiling points and how hydrocarbon branching affects each.

For starters, melting point (mp) and boiling point (bp) are indicators of how well identical molecules interact with (attract) each other. Nonpolar molecules interact principally due to the London dispersion force, one of the intermolecular forces. Such forces can be overcome to melt a nonpolar compound or to boil a nonpolar compound. The greater the attractive force between molecules, the more energy will be required to get the compound to melt or boil. Branching is the most significant factor in determining the degree to which molecules will interact. Branching tends to inhibit van der Waals forces by reducing the surface area available for intermolecular interaction. Thus, branching tends to reduce attractive forces between molecules and to lower both melting points and boiling points.

Another influencing melting point and boiling point for hydrocarbons is molecular weight. The greater the molecular weight, the more surface area there is to interact, the greater number of van der Waals interactions, and the higher the melting point and boiling point.

So, to get to the meat of the lesson, what is the effect of hydrocarbon branching on mp and bp?

Here's the short answer: Branching will decrease mp and bp. Condensing and freezing happen for alkanes because of dispersion forces caused by temporary, induced dipoles, which are the only intermolecular forces holding nonpolar molecules like alkanes (hydrocarbons) together in the liquid and solid states. Branched alkanes have smaller dispersion forces compared to straight-chain alkanes of the same MW, so they will be harder to liquify or freeze.

Here's the long story: Let's imagine that we are cooling down a gaseous alkane like hexane that is straight-chained, versus another of the same MW that is branched, like 2,3-dimethylbutane. Remember that gases don't have any intermolecular interactions, at least not if they're ideal. As we lower the temperature, the molecules stop moving as much, and they begin to have intermolecular interactions that are due to induced, temporary dipoles called London dispersion forces.

Ok, so now we need to consider what kinds of molecules will have the strongest induced dipoles. The strength of the induced dipoles is directly proportional to the surface areas of the molecules that are coming into contact. This is intuitive, because if contact can be made over a greater area, there is a greater chance that electrons will distribute unequally at some point over that surface, causing the temporary dipole and inducing dipoles in the neighboring molecule. You may know that the shape with the smallest surface area-to-volume ratio is a sphere. So molecules that are more spherical (ie, highly branched) do not have very much surface area relative to molecules that are long and extended (straight chains). That is why branched molecules have weaker dispersion forces versus straight-chains. Since they have weaker dispersion forces, the branched molecules will tend to want to stay in the gaseous phase longer, and you'll have to cool them further to force them to condense into a liquid. This means that branched compounds have a lower bp (condense at a lower temperature) versus straight chains.

As we continue to cool, the molecules continue to move closer and closer together, and their interactions continue to increase. Eventually, we reach a point where they begin to crystallize, or at least form an amorphous solid. So we need to consider how well the molecules pack together at this point. Branched compounds are like little spheres, or like porcupines. It's hard to get them to pack well, and this means that you will have to cool them to a lower temperature to freeze them (lower mp) compared with straight chains, which can stack up nicely, more like a cord of firewood. So the mp of a branched compound will be lower than that of a straight chain, assuming that they have the same MW.

Sorry for the shoddy image. You can click on it and it will get bigger if it appears too small.

BIO Lecture 8: Glycolysis (All you need to know)

Lucky for you and me, the MCAT doesn't require the we memorize all the enzyme names and structures of the intermediates in glycolysis. Heck, we don't even really need to memorize the order of the pathway save the inputs and outputs.

Glycolysis is an extremely old pathway and is the universal first step in glucose metabolism--all cells from all domains possess the enzymes of this pathway. In short, a glucose molecule is oxidized and split into two pyruvate molecules producing a net surplus of 2 ATP and 2 NADH. It also takes place in the cytoplasm.

Hexokinase catalyzes the first step in glycolysis, the phosphorylation of glucose to G6P. As a side note, anytime you see the word "kinase" think transfer of a phosphate group.

Phosphofructokinase (PFK) catalyzes the third step, the transfer of a phosphate group from ATP to F6P to form F16bP. This is an important step in the pathway because the reaction is very favorable (a.k.a. irreversible). In addition, the PFK reaction is the primary control point for glycolysis and is known in many MCAT books as the committed step. As a general rule, very favorable steps in enzymatic pathways are the ones that are usually subject to allosteric regulation--PFK is no exception and is regulated by ATP. High concentrations of ATP inhibit PFK.

BIO Lecture 7: Outline of Respiration

Here is a big picture summary of how all of the different stages of metabolism fit together. We'll start looking at the specifics in the next lecture. Keep in mind that when glucose is oxidized there is very little ATP generated directly. Instead, the oxidation of glucose is accompanied by the reduction of high-energy electron carriers, primarily the reduction of NAD+ to NADH (remember from last lecture that when something is oxidized something else must be reduced). The energy in reduced NADH is then used to pump protons out of the interior of the mitochondria and create a proton gradient--this is what finally drives the production of ATP. You should know that oxygen is the final electron acceptor of the electron transport chain, and that anaerobic respiration is insufficient to sustain human life. In addition, fermentation produces lactic acid as a byproduct in humans, and ethanol in yeast. Finally, you should know where in the cell each stage of respiration occurs.

Here are the stages that oxidize glucose to produce CO2 and ATP:

1. Glycolysis: "glucose splitting"; anaerobic (no oxygen required) and occurs in cytoplasm; glucose is partially oxidized while it is split in half into two pyruvate molecules
--2 net ATP (4 total made, but 2 needed in investment phase)
--2 NADH produced (3 ATP in ETC for eukaryotes and 5 ATP for prokaryotes)

2. Fermentation: In short, in anaerobic conditions (without oxygen), electron transport cannot function and the limited supply of NAD+ becomes entirely converted to NADH. Therefore, fermentation has evolved to regenerate NAD+ in anaerobic conditions thereby allowing glycolysis to continue in the absence of oxygen. This process also occurs in the cytoplasm.
--0 ATP; main purpose is to reoxidize the NADH produced in glycolysis (pyruvate is the electron acceptor)

3. Pyruvate Dehydrogenase complex (PDC): the pyruvate produced in glycolysis is decarboxylated to form an acetyl group which is then attached to coenzyme A (a carrier that transfers the acetyl group into the Krebs cycle). Only occurs when oxygen is available but doesn't use oxygen. Occurs in mitochondria matrix for eukaryotes and in the cytoplasm for prokaryotes. It is an aerobic process.
--0 ATP produced
--2 NADH produced (makes 5 ATP in ETC).

4. Krebs cycle (aerobic process): the acetyl group from PDC is added to oxaloacetate to form citric acid--the citric acid is then decarboxylated and isomerized to regenerate the original oxaloacetate. Only occurs when oxygen is available but doesn't use oxygen. Occurs in the mitochondria matrix of eukaryotes and in the cytoplasm for prokaryotes. To fully oxidize the equivalent of one glucose molecule, two acetyl-CoA must be metabolized by the cycle.
Per glucose molecule:
--2 GTP produced (equivalent of 2 ATP)
--6 NADH produced (making 15 ATP in ETC)
--2 FADH2 produced (making 3 ATP in ETC)

4. Electron transport/oxidative phosphorylation: NADH and FADH2 are oxidized by ETC in the inner mitochondrial membrane for eukaryotes and across the inner cell membrane for prokaryotes. In other words, oxidation of metabolic fuels such as glucose and the oxidation of acetyl carbons to CO2 via the citric acid cycle yields the reduced cofactors NADH and FADH2. These compounds are forms of energy currency because their reoxidation--ultimately by molecular oxygen--is an exergonic reaction. The free energy thereby released is harvested to synthesize ATP. It's an aerobic process as well.
--NADH oxidized back to NAD+ and FADH2 oxidized back to FAD occur along with ATP production allowing earlier stages to continue

Summary: 30 ATP (eukaryotes) and 32 ATP (prokaryotes)

I know this is a lot to comprehend right now but the key is to understand how all of the processes function together.

BIO Lecture 6: Intro to Metabolism and Redox

First off, those of you that have taken biochemistry or have studied cellular respiration extensively can breathe easy--relative to its true complexity, MCAT metabolism is a joke. Forget the seemingly endless array of names in each pathway and throw away all of your flash cards, it's just the bare bones here. Let us begin.

Metabolism is the essentially the set of chemical reactions that occur in a cell to maintain life. Metabolism is generally divided into 2 categories, catabolism and anabolism.

Definitions you should know:
Catabolism: the process of breaking down molecules; includes breaking down and oxidizing food molecules (as we oxidize food, we release the stored energy that plants got from the sun). The purpose of catabolic reactions is to provide the energy and components necessary for anabolic reactions. Example: we extract glucose through oxidative catabolism (more on glycolysis later)
Anabolism: is "building-up" and the opposite of catabolism; think of this as constructive
Photosynthesis: the process by which plants store energy from the sun
Photoautotrophs: organisms (plants) that use energy from light (photo) to make their own (auto) food
Chemoheterotrophs: use the energy of chemicals (chemo) produced by other (hetero) living things; this is what humans are

The production and utilization of energy boil down to a series of oxidation/reduction reactions. Let's define what this means.

The three Meanings of Oxidize:
1. Attach oxygen or increase number of bonds to oxygen
2. Remove hydrogen
3. Remove electrons

The three Meanings of Reduce:
1. remove oxygen or decrease number of bonds to oxygen
2. add hydrogen
3. add electrons

When you reduce something, it's like compressing a spring--you store potential energy. The reduced substance wants to be oxidized back to where it started. NOTE: When one atom gets reduced, another must be oxidized. This is what is meant by the term redox pair.

BIO Special Lecture: Cancer and p53 Gene

This is a post from one of my other blogs that I made last year. While the MCAT does not directly test what causes cancer at the cellular level, I figured that this could be helpful if the topic did pop-up in a verbal passage or was maybe even mentioned in a biology passage testing your experimental knowledge (i.e. DNA microarrays and cDNA via reverse transcriptase). Please note that this writing was not intended for us future doctors; therefore it is slightly oversimplified at times but the underlying concepts are what's important. Enjoy

Killing more than 7.6 million people in 2007, cancer is the second leading cause of death in the world. In fact, half of all men and one third of all women will develop some type of cancer during their lifetimes. I recently took a Genetic Analysis class in college where there was a heavy emphasis on understanding what causes cancer at the cellular level and how treatments vary upon what we discover about the cancer (i.e. chemotherapy vs. surgery, etc.). I found this information to be overwhelmingly useful as an increasing number of my relatives have been diagnosed with some type of cancer in recent years. It's also shocking how little they know about the inner workings of this disease and how their doctors have arrived at their prognoses.

To preface what I am about to share with you, I think it is important to first define what cancer is. The American Cancer Society says that cancer results when cells in a part of the body begin to grow out of control. This definition is entirely true but I would like to clarify a little bit before we go any farther. First, most human cells are frequently replaced and reproduced during our life in order to protect us from cells that become too old and carry mutations. Mutations are changes in the base pair sequence of our DNA and are almost always deleterious to humans. So how do we get mutations? Mutations can result from copying errors during cell division or from carcinogens like tobacco smoke or radiation--hence what all smokers and frequent sun bathers have in common. Because our cells are frequently dividing, the older a cell gets the more prone it is to carry harmful mutations--this is because it has been constantly dividing and creating new cells for the body. Lucky for us, the body has the ability to distinguish between old and new cells, much like how it is able to differentiate between a healthy cell and a virus. WARNING: it's about to get more scientific. Our body is able to distinguish between old and new cells by measuring the telomeres of each cell which are located at the ends of the DNA. Telomeres are buffers that keep the DNA from being degraded every time it is replicated. For this part of our discussion, picture DNA as a long strand that contains a telomere at each end. Now every time our cell divides to make a new cell, the telomeres get shorter (the reason for this is complicated and beyond the scope of this post). Thus, as a cell keeps dividing and becomes older, the telomeres are shortened and will eventually be detected by the body. Once a cell is deemed hazardous (meaning the telomeres are too short and the cell is thus, too old), the targeted cell will actually undergo a programmed cell death called apoptosis. This is how our body protects us, by killing off the potentially harmful cells. HOWEVER, if a cell is cancerous (meaning the cell's DNA has multiple mutations), there is an interference with apoptosis so the cell continues to divide unregulated and uncontrollably--this is what the ACS meant when they said that cancer cells grow out of control. Because there is no regulation of the cell's life span via apoptosis, the cancer cells continue to divide and possibly spread around the body to vital organs (this will be later discussed in my benign vs. malignant tumor post).

So here are some takeaways:
-Cancer is cell division gone terribly wrong
-Cancer is fundamentally a disease of the genes
-Cancer is caused by mutations which can result from copying errors during cell division or from environmental carcinogenic agents like radiation and tobacco smoke--for our MCATers, multiple defects contribute to cancer. They are: cell division and differentiation, apoptosis, telomere erosion, contact inhibition, angiogenesis.
-The shortening of telomeres will trigger apoptosis (cell death) unless the cell is cancerous

One last note: for those of you who have had either biochemistry or a heavy dose of genetics are probably wondering why I failed to mention p53 and its role. Well here it is, compliments of my biochem text: The tumor suppressor gene p53 is found to be mutated in at least half of all human tumors. The level of p53 in the cell is controlled by its rate of degradation--this is because it is ubiquinated and targeted for destruction, like cyclins. Therefore, the concentration of p53 increases when its degradation is slowed. Activated p53 stimulates the synthesis of a protein that inhibits cyclin-dependent kinases, thereby block progression in the cell cycle. This regulatory mechanism would conceivably buy time for the cell to repair DNA using enzymes whose synthesis is also stimulated by p53. Some of p53's other target genes encode proteins that carry out apoptosis. Lastly, the position of p53 at the interface of pathways related to DNA repair, cell cycle control, and apoptosis indicate why the loss of the gene is so strongly associated with the development of cancer.

BIO Lecture 5: Enzyme Kinetics and More Inhibition

This lecture is definitely my bread and butter. Therefore I'll try to make this as painless as possible considering the nature of the material ahead.

Enzyme kinetics is simply the study of the rate of formation of products from substrates in the presence of an enzyme. The reaction rate (V denoting velocity) is the amount of product formed per unit time. This depends on the concentration of the substrate [S] and the enzyme. (Before we go any further I want to point out the importance of understanding graphs for this lecture. While I am clearly new at blogging, I have not mastered the art of including images with my lectures. Therefore, please go find a figure in either a biochemistry textbook or online that shows "saturation kinetics" so that you will have a visual to follow along with. In the meantime, I will try my best to find a way to post a drawing or some other type of visual aid.) If there is a small amount of substrate added to the enzyme preparation, enzyme activity appears to increase on the graph almost linearly. However, the enzyme's activity increases less dramatically as more substrate is added. At very high substrate concentrations, enzyme activity appears to level off as it approaches a maximum value (denoted Vmax). At this point the enzyme is said to be saturated. We say that this graph is hyperbolic.

Cooperativity is a different animal. Many multi-subunit enzymes do not behave in the simple kinetic manner described above. In such cases, the binding of substrate to one subunit allosterically increases the affinity of other subunits for the substrate. This is a cooperative case and occurs when the enzyme subunits are structurally linked to each other so that a substrate induced conformational change in one subunit elicits conformational changes in the remaining subunits. EXAMPLE: hemoglobin, when oxygen binding to the heme group in one subunit alters the oxygen affinity of the other subunits. The result is a SIGMOIDAL curve. NOTE: cooperative enzymes must have more than one active site. They are usually multisubunit complexes composed of more than one protein chain held together in a quaternary structure.

INHIBITION: inhibitors reduce enzyme activity through the following mechanisms--competitive inhibition and noncompetitive inhibition.
Competitive: inhibitors compete with substrate for active site (remember that the most effective competitive inhibitors resemble the transition state). NOTE that competitive inhibition can be overcome by adding more substrate; this is because the most substrate we have the more it can out compete the inhibitor. Vmax is not affected but it increases the Km value
Noncompetitive: they bind to an allosteric site, not the active site. Therefore, no matter how much substrate you add, the inhibitor will not be displaced from its site of action. Vmax is diminished therefore.

Now let's touch on the Michaelis constant, Km. Essentially, Km is related to the affinity of the enzyme for the substrate and is the substrate concentration at 1/2 of Vmax. An enzyme with a low Km reaches 1/2 Vmax at very low concentrations because the enzyme has a high affinity for the substrate. An enzyme with a high Km, though, doesn't have a strong affinity for the substrate so it takes a lot more of the substrate to get the enzyme up to 1/2 Vmax. If you haven't done so already, please look at a Lineweaver-Burke plot and a velocity vs. concentration graph for this to make total sense.

One last thing: notice that there is no mention of the Michaelis-Menton equation or the Lineweaver-Burke equation. Questions involving those are usually memorization with plug & chug calculation. Understanding what happens on the graphs is much more intuitive.

BIO Lecture 4: Regulation of Enzymes

Inside a cell, an enzyme is subject to a variety of factors that influence its behavior. Metabolic pathways in the cell are not always on and must be tightly regulated to maintain health. Therefore, the activity of key enzymes in metabolic pathways is usually regulated in one or more of the following ways:
1. Covalent modification: addition of a phosphate group by a protein kinase--this can activate or deactivate an enzyme. Note that this is always a reversible modification
2. Proteolytic cleavage: many enzymes are synthesized in inactive forms (zymogens) that are activated by cleavage by a protease
3. Association with other polypeptides: some enzymes have catalytic activity in one polypeptide subunit that is regulated by association with a separate regulatory subunit. If you remove the regulatory subunit then continuous rapid catalysis results from the catalytic subunit (known as constitutive activity)
4. Allosteric Regulation: modification of active site activity through interactions of molecules with other specific sites on the enzyme known as allosteric sites

ALLOSTERIC REGULATION: means at another place. The binding of the allosteric regulator to the allosteric site is generally noncovalent and reversible. In other words, small molecules bind to particular sites on an enzyme that are different from the active site.

Feedback inhibition: negative feedback is the most common form of feedback regulation. Essentially, an end product will shut off an enzyme early in the pathway.

BIO Lecture 3: Enzyme Structure and Function

Enzymes are biomolecules that catalyze chemical reactions.  Most enzymes are proteins (some are RNA too) that fold into specific 3-D structures.  The reason for the importance of folding in enzyme function is the proper formation of the active site (region that is directly involved in catalysis).  

NOTE: Enzymes are most likely to have globular shape

Enzymes work by lowering the activation energy for the reaction.  The active site is responsible for this by using its residues to stabilize the transition state of the reaction.  The active site is very specific for its substrates (the reactants), including stereospecificity (i.e. enzymes catalyzing reactions involving amino acids specific for D or L amino acids).  As an aside, the functional groups in the active site are so carefully arranged that the enzyme can select its substrates from among many others that are similar in size and shape.  

NOTE: In animals, L amino acids and D sugars are found.

Quick note on proteases: they are protein-cleaving enzymes that have an active site with a serine residue whose OH group can act as a nucleophile, attacking the carbonyl carbon of an amino acid residue in a polypeptide chain.  Examples include: trypsin, chymotrypsin, and elastase.  These enzymes are also known for having a recognition pocket (or specificity pocket) near the active site.  

BIO Lecture 2: Kinetics

In our last lecture we talked about how DeltaG says nothing about the rate of a reaction, only if it is spontaneous or not.  The study of reaction rates is called kinetics.  

All reactions proceed through a transient intermediate that is unstable and takes a great deal of energy to produce (called activation energy).  More specifically, activation energy is the barrier that prevents many reactions from taking place even though they may have a negative DeltaG.  Therefore, lowering the activation energy increases the rate of the reaction (lowering of the activation energy is done through catalysts--more on these below).  This is because we have essentially reduced the amount of energy that is required to reach the transition state.  Also, the lower the energy barrier, the more likely the reaction is to occur because more reactant molecules have enough energy to achieve the transition state.  

For clarification, a transition state exists for a very, very short time.  Because of this, we really don't know what they look like because we are unable to isolate them via analytical techniques.

A catalyst lowers the activation energy without changing the DeltaG.  It does this by stabilizing the transition state.  It is also not consumed in the reaction and is therefore regenerated with each cycle.  Note that enzymes are catalysts and they increase the rate of reactions that have a negative DeltaG.

Thermodynamically unfavorable reactions (positive DeltaG) are driven forward by reaction coupling.  This means that a very favorable reaction is used to drive an unfavorable one (i.e. ATP hydrolysis).  This is possible because free energy changes are additive.    

 

BIO Lecture 1: Thermodynamics

The MCAT tests our knowledge on thermodynamics in terms of how it relates to chemical reactions.  Therefore, there are 2 forms of energy in chemistry that we need to know: HEAT ENERGY and POTENTIAL ENERGY.  Heat energy refers to movement of molecules and potential energy refers to energy stored in chemical bonds (i.e. ATP).  

1st Law of Thermodynamics: energy in the universe is constant (conservation of energy - therefore if we lose energy in a "system" the energy of the rest of the universe must increase)

2nd Law of Thermodynamics: entropy (disorder or randomness) of the universe tends to increase over time (in terms of chemistry: a reaction will occur spontaneously--without input of energy--if it increases the disorder of the universe)

Gibbs Free Energy: we can't discuss thermodynamics without touching on this formula.  DeltaG = DeltaH - TDeltaS.  The change in Gibbs free energy determines whether the reaction is favorable (spontaneous) or unfavorable (nonspontaneous)

T = Temperature (in Kelvin) 

H = Enthalpy 

S = Entropy

-if DeltaG is negative the reaction is spontaneous and exergonic (energy exits the system)

-if DeltaG is positive the reaction is non spontaneous and endergonic (energy must be added)

NOTE: the value of DeltaG depends on the concentrations of the reactants and products

NOTE: spontaneous means that a reaction may proceed without additional energy input, but it says NOTHING about the rate of the reaction. Thus, DeltaG is only a measure of the difference in free energy between reactants and products 

-if DeltaH is negative then the reaction is exothermic (liberates heat)

-if DeltaH is positive then the reaction is endothermic (requires input of heat)

Preface

The purpose of this blog is simple--to make MCAT concepts more manageable and easier to understand.  I won't bore you with the basics of who I am or why I'm doing this.  Just be thankful that it's out there and email me with specific questions.  I intend to cover the most essential topics for the MCAT only, therefore this is by no means a comprehensive review.  However, I think you'll find that it will be easy to understand as I will be writing in simpler terms.  Good luck with your studies.