Biochemistry Exam 2 Material PDF

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Biochemistry Exam 2 Material study...

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122In addition, you are expected to have a working knowledge of the materials covered in Exam I. In particular, you are responsible for the following items and topics:

● Molecular structure of the 20 natural amino acids

Good website to review some of this concepts http://leah4sci.com/amino-acid-charge-in-zwitterions-and-isoelectric-point-mcat-tutorial/

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Drawing molecular structures of polypeptides, the dehydration synthesis and hydrolysis of the peptide bonds, the formation of disulfide bonds

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The concept of pH and pKa’s of weak acids and bases (including amino acids) Freddy was here pH and pKa

★ The lower the pH, the higher the concentration of hydrogen ions [H+]. ★ The lower the pKa, the stronger the acid and the greater its ability to donate protons. ★ pH depends on the concentration of the solution. This is important because it means a weak acid could actually have a lower pH than a diluted strong acid. For example, concentrated vinegar (acetic acid, which is a weak acid) could have a lower pH than a dilute solution of hydrochloric acid (a strong acid). ★ On the other hand, the pKa value is constant for each type of molecule. It is unaffected by concentration. ★ Even a chemical ordinarily considered a base can have a pKa value because the terms "acids" and "bases" simply refer to whether a species will give up protons (acid) or remove them (base). For example, if you have a base Y with a pKa of 13, it will accept protons and form YH, but when the pH exceeds 13, YH will be deprotonated and become Y. Because Y removes protons at a pH greater than the pH of neutral water (7), it is considered a base. Relating pH and pKa With the Henderson-Hasselbalch Equation If we know either pH or pKa, you can solve for the other value using an approximation called the Henderson-Hasselbalch equation: pH = pKa + log ([conjugate base]/[weak acid]) pH = pka+log ([A-]/[HA])

pH is the sum of the pKa value and the log of the concentration of the conjugate base divided by the concentration of the weak acid.

At half the equivalence point:

pH = pKa

It's worth noting sometimes this equation is written for the Ka value rather than pKa, so you should know the relationship:

pKa = -logKa

● Computing net charges of amino acids and polypeptides given any pH ● Computing isoelectric points of amino acids and polypeptides

● Primary, secondary, tertiary, and quaternary structures

http://biostat.jhsph.edu/~iruczins/teaching/260.841/notes/c2.6.pdf https://www.chemguide.co.uk/organicprops/aminoacids/proteinstruct.html https://dasher.wustl.edu/bio5477/lectures/lecture-03-sep09.pdf

● The concept of chirality Chirality is an important concept for stereochemistry and biochemistry. Most substances relevant to biology are chiral, such as carbohydrates (sugars, starch, and cellulose), the amino acids that are the building blocks of proteins, and the nucleic acids . The main cause of chirality in a molecule is that it has an atom (often a carbon atom) that is connected to four different groups in such a way that it is possible to have a non-superimposable image of the molecule. Such an atom is called a chiral center. Although the carbon atom is the most common chiral center, Si, N, and P atoms are also known to act as chiral centers.

Enantiomers

This type of stereoisomer is the essential mirror-image, non-superimposable type of stereoisomer introduced in the beginning of the article. Figure 3 provides a perfect example; note that the gray plane in the middle demotes the mirror plane.

Compounds with Multiple Chiral Centers We turn our attention next to molecules which have more than one stereocenter. We will start with a common four-carbon sugar called D-erythrose.

. We will use the D/L designations here to refer to different sugars, but we won't worry about learning the system. As you can see, D- erythrose is a chiral molecule: C2 and C3 are stereocenters, both of which have the R configuration. In addition, you should make a model to convince yourself that it is impossible to find a plane of symmetry through the molecule, regardless of the conformation. Does D-erythrose have an enantiomer? Of course it does – if it is a chiral molecule, it must. The

enantiomer of erythrose is its mirror image, and is named L-erythrose (once again, you should use models to convince yourself that these mirror images of erythrose are not superimposable).

Notice that both chiral centers in L-erythrose both have the S configuration. In a pair of enantiomers, a ll of the chiral centers are of the opposite configuration. What happens if we draw a stereoisomer of erythrose in which the configuration is S at C2 and R at C3? This stereoisomer, which is a sugar called D-threose, is not a mirror image of erythrose.  both D-erythrose and L-erythrose. D-threose is a diastereomer of

The definition of diastereomers is simple: if two molecules are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers, then they are diastereomers by default. In practical terms, this means that at least one - but not

 y definition, two molecules all - of the chiral centers are opposite in a pair of diastereomers. B that are diastereomers are not  mirror images of each other. L-threose, the enantiomer of D-threose, has the R configuration at C2 and the S configuration at C3. L-threose is a diastereomer of both erythrose enantiomers. In general, a structure with n stereocenters will have 2n different stereoisomers. (We are not considering, for the time being, the stereochemistry of double bonds – that will come later). For example, let's consider the glucose molecule in its open-chain form (recall that many sugar molecules can exist in either an open-chain or a cyclic form). There are two enantiomers of glucose, called D-glucose and L-glucose. The D-enantiomer is the common sugar that our bodies use for energy. It has n = 4 stereocenters, so therefore there are 2n = 24 = 16 possible stereoisomers (including D-glucose itself).

In L-glucose, all of the stereocenters are inverted relative to D - glucose. That leaves 14 diastereomers of D-glucose: these are molecules in which at least one, but not all, of the stereocenters are inverted relative to D-glucose. One of these 14 diastereomers, a sugar called D- galactose, is shown above: in D-galactose, one of four stereocenters is inverted relative to D-glucose. Diastereomers which differ in only one stereocenter (out of two or more) are called epimers. D-glucose and D-galactose can therefore be refered to as epimers as well as diastereomers.

PROTEIN INTERACTIONS & FUNCTION 1. What are cofactors? What is the difference between essential ions and coenzymes? What are vitamins, and what roles do they play in the biochemistry of life? Cofactors carry out essential chemistry that is not possible with the 20 amino acids. Cofactors may be ions, or small organic molecules. Also known as “helper molecules.” Vitamins are organic cofactors therefore are called coenzymes that are non-essential means they have to be consumed. The role Vitamins play in biochemistry life is the absence or shortage of a vitamin may result in a vitamin-deficiency disease. For example, Scurvy, Beriber and Pellagra (nicotinic) acid (niacin). 2. What is functional channeling? What are the benefits of channeling? In channeling, the three enzymes are associated together in a very specific way. There is a channel between the active sites o f the first, second enzyme, and third enzyme. This is important for two reasons. The product of the first enzyme is the substrate for the second. Rather than have the first product leave its enzyme and then wander around randomly inside of the cell until it finds the second enzyme, channeling allows the molecule to pass directly from enzyme to enzyme. This is efficient – random diffusion would take longer, and thus be less efficient. The second reason is also significant. Water can react with many biological molecules, generating unwanted side-products. By passing the molecule from enzyme to enzyme, interactions with water are impossible – and so there is no formation of side products.

3. Given a general binding reaction (for instance, a ligand binding to a protein),

write the equilibrium equation, and define K eq  , and in particular Ka , and K d . Reaction: P1+ L ---> P1:L Ka= Kd=

Ke =

[P 1:L] units: m-1 and s-1 [P ][L] [P 1] [L] [P : L] units: s-1 [P 1:L] [P ][L]

4. What are the challenges of transporting non-polar gases in animals, in particular

for humans? F  reddy was here ( thank You) A.F The molecular oxygen (and carbon dioxide) are nonpolar and have a low solubility in water. This presents a problem since all eukaryotes and many prokaryotic organisms require oxygen to survive. This necessitates the use of specialized oxygen transport tissues to take up oxygen and deliver it to the appropriate cells. Carbon dioxide provides essentially the same challenge. It must be effectively removed from tissues but is, in general, poorly soluble in water.

Non polar gases such as carbon Dioxide are transported through the blood but they can affect the affinity of other ligands such as oxygen. If there is an increase of carbon dioxide then the affinity of oxygen is limited and increases the acidity of blood by increasing concentration of hydrogen. As a result pH decreases and vice versa.

This slide illustrates the mechanism – and the challenges – of transporting oxygen throughout the body and CO2 back to the lungs for expiration. Do not worry about the details – this isn’t an Anatomy and Physiology class. Rather, it's worthwhile to perceive this as a problem of chemistry– how we manage, in practical terms, to maintain the necessary concentrations of oxygen and CO2 at the appropriate points in the organism. The red blood cells shown here are called erythrocytes. Oxygen is carried from the gills/lungs and distributed to the tissues. 5. What are the functions of myoglobin and hemoglobin? Where would you find

these molecules? Are they structurally and functionally homologous? FUNCTIONS OF THESE PROTEINS:

Hemoglobin, or Hb, is a protein molecule found in red blood cells (erythrocytes) made of four subunits: two alpha subunits and two beta subunits. Each subunit surrounds a central heme group that contains iron and binds one oxygen molecule, allowing each hemoglobin molecule to bind four oxygen molecules. Molecules with more oxygen bound to the heme groups are brighter red. As a result, oxygenated arterial blood where the Hb is carrying four oxygen molecules is bright red, while venous blood that is deoxygenated is darker red.

Myoglobin i s found in muscles tissues only while hemoglobin is found all over the body. The function of myoglobin, an iron containing protein in muscle, receives oxygen from the red blood cells and transports it to the mitochondria of muscle cells, where the oxygen is used in cellular respiration to produce energy. Each myoglobin molecule has one heme prosthetic group located in the hydrophobic cleft in the protein.

STRUCTURE AND FUNCTIONALLY HOMOLOGOUS:

No, they are not structurally and functionally homologous. Hemoglobin is called tetrameric hemoprotein, while myoglobin is called monomeric protein.

6. Be familiar with the structure of the heme group. How is ferrous coordinated with the heme group and with the protein itself? How does oxygen bind to the heme group? STRUCTURE OF HEME GROUP:

OXYGEN BINDS TO HEME: Each  subunit surrounds a central heme  group that contains iron and binds one oxygen molecule, allowing each hemoglobin molecule to bind four oxygen molecules. Hemoglobin is made up of four symmetrical subunits and four heme groups. Iron associated with the heme binds oxygen. FERROUS COORDINATED WITH HEME AND PROTEIN ITSELF: In the body, the iron in the heme is coordinated to the four nitrogen atoms of a porphyrin and also to a nitrogen atom of a histidine amino-acid residue in the hemoglobin protein. The sixth position (coordination site) around the iron of the heme is occupied by O2 when the hemoglobin protein is oxygenated. 7. In the binding site of myoglobin and hemoglobin, there are two histidines—the proximal histidine and the distal histidine. Describe the functions of these two conserved residues. Heme consists of two components, consists of an organic component called protoporphyrin which contains carbon, hydrogen, nitrogen, oxygen atoms, and the entire region in black is the protoporphyrin. Epicenter of the protoporphryin is an inorganic atom, a metal atom, the iron atom, which makes the inorganic component of that heme group. That FE atom is responsbile for binding not only to the portein but also oxygen, Fe atom is bound to form nitrogen atoms, 4 N atoms. Fe can have an oxidation state of +6 and in this particular case, since we have 4 bonds, this Fe is in its Ferrous state which has an oxidation state of +2. The Fe in the cetner of heme group can form 2 other bonds. One of the bond is formed between 1 of the amino acids of that polypeptide chain. If we take heme gorup and we flip in, and the bottom portion of the heme group we have an amino acid, more specifically a histidine that is part of that protein either

myoglobin or hemoglobin. On one side of the protoporphyrin plane, the iron atom is bound to the histidine residue on that polypeptide chain which can be part of the myoglobin molecule or hemoglobin molecule. Each poly chain contains a single heme group, Myoglobin contains a single heme group, but hemoglobin contains 4 diff heme groups, so hemoglobin can bind 4 diff oxygen molecules, and myoglobin only forms 1 oxygen molecule. 8. Discuss the binding of oxygen vs. carbon monoxide. How do myoglobin and hemoglobin limit the binding of carbon monoxide despite its very low K d ? 9. Be able to interpret a binding curve, for instance, the binding curve of oxygen to

myoglobin and hemoglobin. What are the axes of a binding curve? How do you  reddy was here read K d from a binding curve? F

10. Distinguish between the T state and the S state of hemoglobin. Structurally, what

actually happens to the alignment around the heme group when an oxygen binds to a ferrous ion? Describe the effect of this slight structural shift in the conformation of the rest of the hemoglobin complex. F  reddy was here

T - Sate (tight)

S - State (relaxed)

Lower affinity for O2

Higher affinity for O2

Releases O2

Binds O2

If [H] increases favors T state (low If [H] decreases favors S state (high pH) → releases O2 pH)→ binds O2

When oxygen binds to the heme group, the ferrous group is coming at an angle to be able to bind to the oxygen. This is because it is trying to prevent the binding of carbon dioxide which is perpendicular to the heme group. One of the factors of limiting or allowing certain ligands to be attached to the heme group is 2,3 Bpg it alters the conformation of the T - state of hemoglobin causing the releases of oxygen. 11. Describe allostery. How is the binding of oxygen to hemoglobin an example of

allostery? binding of oxygen to one h  emoglobin subunit induces conformational changes. making them more able to b ind oxygen by raising their affinity for this molecule. An example is fetal hemoglobin (hbf) has a higher affinity to oxygen because it is less inhibited than maternal or adult hemoglobin.

12. How is the cooperativity of oxygen binding to hemoglobin an advantage as far as

the delivery of oxygen from oxygen-rich lungs to oxygen-poor tissues? (freddy was here) When we breathe In oxygen it is transferred to the hemoglobin (hb) which then takes oxygen and drops it onto myoglobin. Thus myoglobin has to have the higher affinity of oxygen levels. ( s ee graph on question 9)

13. Identify and describe the different ways carbon dioxide is transported from the

tissues to the lungs. Carbon dioxide must also be transported but in the opposite direction—from tissues to the lungs. Because the solubility of carbon dioxide in water is low, it is more easily transported in the blood as bicarbonate. CO2 + H2O H+ + HCO3The bicarbonate product formed is more soluble as water and the pH decreases due to more H+ ions being released as the reaction moves in the forward direction.

14. What is the Bohr effect? How does the Bohr effect facilitate the increased

deoxygenation of hemoglobin?  t the relatively low pH and high CO2 concentration of the peripheral tissues, the A affinity of hemoglobin for oxygen decreases as H+ and CO2 are bound and O2 is released to the tissues. The effect of the H+ and CO2 concentration on the binding and release of oxygen by hemoglobin is the Bohr effect. Carbon Dioxide binds as a carbonate group to the alpha amino group at the terminal end of each globin chain forming the carbaminohemoglobin. This reaction produces H+ contributing to the Bohr effect.

15. Describe the role of 2,3-bisphosphoglycerate in hemoglobin regulation.

W  hen hemoglobin is isolated it contains a substantial amount of bound BPG which binds at a site distant from the oxygen-binding site and regulates the O2 binding affinity and regulates the O2 affinity of hemoglobin in relation to the partial pressure of Oxygen pO2 in the lungs. 16. Fetal hemoglobin is made up of a2g2 quaternary structure. Describe the binding

of oxygen to fetal hemoglobin compared to adult hemoglobin. How is this an advantage during fetal development?

2,3-BPG also contributes to the obligatory transfer of oxygen from the mother to the fetus. Without this, well, you would probably not be here. Unlike adult hemoglobin, which is composed of a2b2 subunits, fetal hemoglobin is formed as a tetramer with an a2g2 subunit composition. In the g (g is for gamma) subunits, there is a HisàSer mutation. As a result of this mutation, 2-BPG does not bind as well to fetal hemoglobin as adult hemoglobin, and the affinity for oxygen is increased. As a result, fetal hemoglobin binds oxygen better than adult hemoglobin, and oxygen is transported (net) from the maternal to the fetal tissues. This is necessary for the developing fetus to adequately extract oxygen from the mother’s blood. Notice how, at the same partial pressure of oxygen, fetal hemoglobin binds ~1.5 X more oxygen than maternal hemoglobin. Fetal  red  blood  cells  have  a  higher  affinity  for  oxygen  than  maternal red  blood  cells  because  fetal  hemoglobin doesn't bind 2,3-BPG as well as maternal hemoglobin does.  The result of this difference in oxygen affinity allows oxygen to be transferred effectively from maternal to fetal red blood cells.

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17. Discuss the molecular mechanism associated with the development of sickle cell

anemia.

Normal and sickle-cell hemoglobin. Notice the subtle differences between the conformations of hemoglobin A and hemoglobin S result from a single amino acid change in the β  chains. Gene variants are called alleles. One can be homozygous or heterozygous for an allele. To develop sickle cell anemia, the individual must be homozygous for the HbS allele. Hbs has a Glu à Val substitution, resulting in two fewer negative charges The substitution creates a hydrophobic patch on the surface of the hemoglobin molecule.

ENZYME FUNDAMENTALS & MECHANISMS 1. Describe the key scientific developments that occurred, along with the scientists who made them happen, in the study of biochemistry and enzymology. Louis Pasteur- He realised that “ferments” caused the fermentation of sugar into alcohol by yeast...