Biochemistry (Exam 1)

Question 1

Name the major bonding forces in biological systems and discriminate one from another (i.e. strength).

Answer 1

• Covalent bonds: shared electrons between atoms. Very strong (50-110 kcal/mol)!
• Non-covalent bonds: weaker than covalent bonds and, thus, more dynamic. Includes (4):
  
  • Ionic bonds: between ions of opposite charge. ~100-fold weaker than covalent (1-6 kcal/mol)
  • Hydrogen bonds: permanent dipoles created by unequal sharing of electrons in a bond. Atoms must have different electronegativity. (1-7 kcal/mol)
  • Van der Waals: induced dipoles where electron clouds around atoms of similar electronegativity repel one another. Very weak (<1 kcal/mol)
  • Hydrophobic interactions: result from inability of molecules to interact effectively with an aqeous environment. Not dependent on attractive forces.

Question 2

Which type of non-covalent bond relies on differences in electronegativity of atoms? Which is a result of similar electronegativites?

Answer 2

Hydrogen bonds involve permanent dipoles from unequal sharing of electrons in a bond. These rely on differences in electronegativity of atoms.

Van der Waals are induced dipoles by bringing molecules of similar electronegativities in close proximity.

Question 3

What is the Henderson-Hasselbalch equation? What does it measure?

Answer 3

pH = pK + log[A-] / [HA]

where pK = -logK(eq) and pH = -log[H+]

HH equation allows us to determine whether a species will be found predominantly in its conjugate base or conjugate acid form at a given pH.

Question 4

What is the pK of:

CH3COOH <–> CH3COO- + H+

Answer 4

pK = ~4

Question 5

In terms of the Henderson-Hasselbalch equation, at what pH range is a weak acid most effective as a buffer?

Answer 5

A weak acid is most effective as a buffer when its pK = pH. This is it’s buffer zone. From HH equation: if pH = pK then [A-] = [HA]

ex. CH3COOH pK = 4. It’s buffer zone is around pH of 4.

Question 6

Interpret the following result of a Henderson-Hasselbalch eqn calculation:

1000 = [CH₃COO^-]/[CH₃COOH]

Answer 6

At the pH (originally plugged in), virtually all of the COOH groups are in the conjugate base state. The net charge = -1

Question 7

What is the relationship between hydrocarbon tail length and melting point?

Answer 7

The longer the hydrocarbon tail, the higher the melting point (i.e. the longer it will stay in its solid state before melting)

Question 8

What is the relationship between the number of double bonds in a molecule and its melting point? For biological fatty acids, what does 18:0, 18:1, 18:2, etc mean?

Answer 8

The more double bonds, the lower the melting point.

__:__ = # of C’s : # of double bonds. Thus, 18:0 = 18 Carbons with 0 double bonds. Compare:

• Stearic acid (18:0), MP = 75
• Oleic acid (18:1), MP = 13.4
• Linoleic acid (18:2), MP = -9

Question 9

Select which ionization of phosphoric acid, H₃PO₄ (a polyprotic acid), would be best for creating a buffer at physiological pH.

Answer 9

H₃PO_{4 ,<–>} H₂PO₄^- <–> HPO₄^-2 <–> PO₄^-3

• H₃PO₄ ,<–> H₂PO₄^- has pK = 2
• H₂PO₄- <–> HPO₄^-2 has pK = 7.2
  
  • **biologically relevant b/c near physiological pH of 7**
• HPO₄^-2 <–> PO₄^-3 has pK = 12.7

Question 10

Define the term amphipathic and give an example of an amphipathic molecule.

Answer 10

amphipathic: a molecule that has two different affinities, as a polar end that is attracted to water and a nonpolar end that is repelled by it.

Fatty acids are amphipathic containing both:

• polar end = carboxylic acid functional group that ionizes at physiological pH
• non-polar end = long hydrocarbon tails

Question 11

You receive a genetics report which states that protein X has a P47G change. This report means that the primary structure of protein X has changed by having:

Answer 11

Glycine replaces Proline at position 47.

[The amino acid replacing the normal amino acid is listed second]

Question 12

Between what constituents does hydrogen bonding occur in an alpha helix v. beta sheets?

Answer 12

Both are between the functional groups of the polypeptide backbone. However, alpha helix has H-bonds between its functional groups within a segment of helix. While, beta sheets are stabilized by H-bonds forming between the functional groups of different segments of sheet structure

Question 13

What is required to fold a polypeptide chain into its native the 3-dimensional structure?

Answer 13

Does NOT require elaborate cellular machinery. All that is required to fold a protein into its 3D structure is its interaction with an aqueous environment.

Question 14

What is the major thermodynamic driving force for protein folding?

Answer 14

the hydrophobic effect

[In an unfolded state there are hydrophobic residues exposed to the aqueous environment, which is a thermodynamically unfavorable state. Thus, the hydrophobic residues will seek refuge from water by sequestering themselves on the interior of the evolving structure = hydrophobic collapse]

Question 15

What is hydrophobic collapse?

Answer 15

It is the hydrophobic effect during protein folding, where the hydrophobic residues of the polypeptide chain seek refuge from water by hiding themselves on the interior of the structure.

*The hydrophobic effect is the major thermodynamic driving force for protein folding*

Question 16

Diagram a peptide bond, including permanent dipoles, and show which bonds associated with C(alpha) atoms are free to rotate.

Answer 16

Red X = NON-rotatable

Yellow highlight = rotatable

Question 17

Provide examples of the interaction of water with polar and non-polar molecules.

Answer 17

Charged Polar- Water has a high dielectric constant and insulates oppositely charged molecules from mutual electrostatic attraction. That is why table salt (NaCl) dissolves in water.

Polar- water interacts with other polar molecules via hydrogen bonding.

Non-Polar-Water causes aggregation of non-polar molecules. For example how oil floats on top of water instead of being dispersed throughout. This is a result of thermodynamic favorability. In this case, entropy is favored.

Question 18

Give an example of how collections of molecules differ in their properties and how substitutions of one class for another can lead to human disease.

Answer 18

In Sickle Cell Anemia, a valine is substituted with a glutamic acid. The glutamic acid is a polar, charged molecule meaning that is hydrophillic. It is replaced in hemoglobin beta chains by valine, which is a non-polar, hydrophobic molecule. The presence of a hydrophobic patch on the surface of the hemoglobin causes large numbers of hemoglobins to aggregate which lead to the characteristic sickling seen in Sickle Cell Anemia.

Question 19

Describe how buffers work.

Answer 19

Buffering is the ability of a solution to resist changes in pH. Solutions containing conjugate acid/base pairs can act as buffers. Any added base or acid will react with the conjugate base or acid in the solution, which will only change the pH very slightly. This works best if the pH of the solution is near the pK of the weak acid.

Question 20

Classify fatty acids according to their length, number of double bonds, and boiling points.

Answer 20

Saturated fatty acids lack double bonds.

Unsaturated fats have double bonds.

Fewer double bonds means a higher boiling point.

Longer hydrocarbon tail means higher boiling point.

Question 21

Explain how changing the fatty acid composition of a polyunsaturated tricylglycerol to a more saturated state changes its physical properties.

Answer 21

Changing the fatty acid composition to a more saturated state would cause the tricylglycerol to be more solid at room temperature.

Question 22

List some of the common modifications to amino acid residues in proteins.

Answer 22

• peptide bond created between alpha-carbonyl and alpha-amino group
• amino and carboxyl groups are lost upon peptide bond formation. The only charge from the peptide backbone is at the terminus, which cancels at neutral pH

Question 23

Design a peptide of 15 amino acids (only 4 of which are non-polar) that would segregate the non-polar residues to a single surface of an alpha helix.

Answer 23

L - Y - S - L - C - S - N - L - S - K - L - Y - T - C

In order to create a total non-polar surface in an alpha helix, the non-polar amino acids have to be in the 3/4 sequence with a nonpolar amino acid every 3rd, then every 4th spot

Question 24

List the four primary elements of proteint structure.

Answer 24

Primary structure: linear sequence of amino acids in polypeptide chain.

Secondary Structure: short segments of polypeptide backbone that assume certain structures: alpha helixes and beta sheets.

Tertiary Structures: How segments of secondary structure fold back on one another to form a complete three dimensional structure of a protein.

Quarternary Structure: non-covalent assembly of more than one polypeptide chain.

Question 25

Describe the fluid mosaic model of biological membranes. How can changes in membrane composition affect protein function?

Answer 25

B/c of the large polar heads on the glycerophospholipids, they do not readily flip-flop from one leaf to another. However, they do diffuse laterally within their bilayer = the "fluid"-like consistency. Fluidity of the membrane is dependent on temperature and lipid composition of the membrane. The more unsaturated the fatty acids are, the lower the fluid/gel transition temperatures. Protein function can be influenced by changes in fluidity.

Question 26

Which (7) amino acids have ionizable side chains? What are each of their side chains and respective pKs?

Answer 26

* D (Asp) - pK = 4 (carboxyl) * E (Glu) - pK = 4 (carboxyl) * C (Cys) - pK = 8 (sulfhydryl) * K (Lys) - pK = 10 (amino) * H (His) - pK = 6 (imidazole) * Y (Tyr) - pK = 10 (phenol) * R (Arg) - **pK = 12** (guanidium = C bonded to 3 Ns!) Rest have non-ionizable side chains

Question 27

Name (5) polar, CHARGED amino acids. What are each of their pKs?

Answer 27

* Asp (D) * Glu (E) * Both Asp and Glu have pK of 4, which is pK of COOH side group * Lys (K) pK = 10 (amino group) * His (H) pK = 6 (near phys pH! Thus, 50/50 for being in acid or base conj) * Arg (R) pK = 12 \*most polar side chain

Question 28

Define leucine zipper. How is it used in forming elements of quarternary structure in certain transcription factors?

Answer 28

A leucine zipper is a protein sequence where leucine is every 7th position. This forms an alpha helix which aids in *dimerization* of complexes such as DNA, i.e. it brings two strands of proteins together so they can bond.

Question 29

Design a sequence of 4 amino acids that would have a relatively high probability of forming a reverse turn.

Answer 29

**P-G-N-S** ## Footnote Reverse turns typically are initiated by Proline (P, Pro) and are generally formed with small side chain residues, such as ***Gly, Ser, Asn, Asp***

Question 30

What is the only thing you need for a denaturated polypeptide chain to refold?

Answer 30

An aqueous environment. Denatured proteins will *spontaneously* refold into their active state in an aqueous environment.

Question 31

Calculate the net charge of the following peptide at pH of 10: ## Footnote **Asp-Lys-Leu-Phe-Asn**

Answer 31

Leu, Phe, and Asn are non-ionizable, so disregard them. * Asp pK = 4, which is more than 1 less than 10, so **charge = -1** * Lys pK = 10. Equal to pH and Lys is already (+), so **chage = +1/2** * Terminus end groups = (+1/2) and (-1) Thus, -1 + 1/2 + 1/2 - 1 = **_-1_**

Question 32

2 characteristics of Simple electrophoresis

Answer 32

* Separates by charge at *low resolution* * can be used to determine sickle cell trait v. normal

Question 33

3 features of Isoelectric focusing

Answer 33

* separates by charge at *high resolution* * pH gradient is established by the gel matrix * proteins migrate through the gel in an electric field until they reach the **isoelectric point**

Question 34

4 features of SDS-PAGE

Answer 34

* separates by size at *high resolution* * causes equal charge to mass ratio for ALL proteins * smaller proteins go through the pores and go further * larger proteins get stuck in the pores

Question 35

3 features of Two-dimensional gels

Answer 35

* combines isoelectric focusing with SDS-PAGE * separates by charge AND size * can separate over 1,000 proteins on a single gel → looks for abnormal cells for disease

Question 36

Compare the ligand binding properties of ferrous iron v. ferric iron.

Answer 36

**\*\*Ferrous iron** = **Fe²⁺.** It can form up to 6 bonds (4 N, 1 Histadine, and 1 O)\*\* **Ferric iron** = **Fe³⁺**. More oxidized, so it does not bind with O

Question 37

Why is ferrous iron not readily oxidized to ferric iron when oxygen binds to myoglobin or hemoglobin?

Answer 37

The binding site in myoglobin and hemoglobin is in a protective environment formed by hydrophobic molecules from valine and phenylalanine. This prevents water from accessing the binding site, thereby preventing the oxidation of Fe²⁺ to Fe³⁺

Question 38

Compare and contrast myoglobin and hemoglobin

Answer 38

Hemoglobin (Hb): * tetramer * _each of 4 globin chains binds a single heme_ in a hydrophobic pocket * secondary and tertiary structures of Hb globin chains are similar to myoglobin * has alpha AND beta chains Myoglobin: * single polypeptide chain of 153 amino acids * most in an alpha helix * 8 distinct helices labeled A-H * binds a _single heme_ in a hydrophobic pocket * ***Proximal His*** forms bond w/ heme iron * ***Distal His*** forms a H-bond with oxygen

Question 39

Diagram the developmental profile of globin gene expression in human fetuses and adults

Answer 39

Adult humans have (3) forms of Hb: * HbA: α₂β₂ (92.5%) * HbA2: α₂δ₂ (2.5%) * HbA_1C: α₂(β2-NH-Glc)₂ (5%) Fetusus have HbF = α₂γ₂

Question 40

How does the oxygen-binding curive of Hb diff from myoglobin's? What is the physiological relevance of this?

Answer 40

Myoglobin's curive is a rectangular hyperbola. Myoglobin has a single binding site for oxygen. Hb's curve is sigmoidal due to the fact that its oxygen binding is cooperative. Means binding of one ligard (O2) positively increases the affinity for the next one. **The last O2 binds with a 100x greater affinity that the first.** This curve works well, because if there were *only* high affinity, tissues would be starved for oxygen. And if there was only low affinity, the lungs couldn't load efficiently.

Question 41

How do H+ affect the K_d of Hb for oxygen and how this influences the binding curive of O2 to Hb? What is the physiological relevance of this?

Answer 41

An increase in H+ causes an increase in the dissociation constant (K_d) for Hb. This is b/c Hb will release the oxygen in order to take up the extra H+ floating around. This decreases affinity and shifts the _curive to the right._ This process occurs in exercising muscles. Active muscles produce H+ as a result of their metabolism. The increase in acid causes the Hb to release more oxygen to the muscles and, therefore, take up the excess protons.

Question 42

How does BPG play an important role in the difference in oxygen binding properties of fetal and adult hemoglobin?

Answer 42

The strongly negative BPG binds to the central cavity of deoxyHb, stabilizing it (**T-conformation**) and lowering its affinity for oxygen. Fetuses are not able to bind BPG and therefore have a higher affinity for oxygen than their mothers.

Question 43

Identify regions of Hb structure where substitutions would be a problem.

Answer 43

Changes in residues on the *outer surface* of Hb molecule rarely affect function or have pathological effects (_except for sickle cell!)_ Changes in residues critical for 3D structure can lead to protein instability and hemolytic anemias. These unstable proteins precipitate, forming **Heinz bodies** at the surface of the plasma membrane. Substitutions include: * polar for non-polar residues in the hydophobic core * proline for residues in an alpha helix

Question 44

Identify regions of Hb structure where amino acid substitutions would increase or decrease the oxygen-binding affinity of a variant form of Hb v. normal Hb.

Answer 44

Changes that destabilize the **T (deoxy) state** create Hb variants with *increased* affinity for O2. An increase in # of RBCs offsets the reduction in O2 affinity and decreases delivery to the tissue. People with this will have ruddy appearance. Caused by _substitutions at the α2β2 interface that destabilize the T state._ Changes that *decrease* the affinity of Hb for O2 are associated w/ *increased amounts of deoxyHb*, which can create pronounced cyanosis. Caused by substitutions that: * promote oxidation in heme binding pocket from Fe²⁺ Fe³⁺ * _destabilize the R (oxy) state_

Question 45

How can a defect in the GTPase activity of the Ras protein promote cancer?

Answer 45

GTPase hydroyzes GTP to GDP, which turns the switch "off." A defect in GTPase of the Ras protein means the output from the protein is never turned off and the protein no longer needs a signal to keep the switch on (GDP to GTP). So the cells keeps dividing with nothing to stop them.

Question 46

2 major fates of mis-folded proteins within cells. How is one of these fates responsible for CF?

Answer 46

Mis-folded proteins within cells are either degraded or aggregated. If the proteins that form the Cl- channels in the epithelial cells of the lungs are mis-folded, they are degraded and the Cl- channels will not exist. This causes mucous in th eloungs to be less viscous and causes Cystic Fibrosis

Question 47

Transform one element of secondary structure into another in forming the pathogenic form of PrP

Answer 47

In CJD, the normal proteins form an alpha helix structure. But the infectious prions form beta pleated sheets to make *larger aggregates*. This decreaes the [] of beta pleated sheets that were in an equilibrium with the alpha helix structures, causing a shift to make more beta sheets. The aggregates build up, causing necrosis and holes in the brain.

Question 48

(3) major forms of transmission of prion diseases and the molecular basis for each transmission

Answer 48

* **_Infectious_** - this is how Kurur, Scrapie (sheep), Mad Cow are transmitted. For Mad Cow, the **homozygous 120 M/M polymorphism** in the PrP protein is needed in humans for the prion to transmit from cows to humans (why UK outbreak wasn't as bad as should have been). * **_Iatrogenic_** - prions spread infectiously by med intervention (ex. bone marrow transplant) * **_Inherited_** - disease can be inherited trait that usually shows in people in late 60-70s. Takes a long time for the aggregates to build up and necrosis to occur * **_spontaneous_** - proteins in body randomly mis-fold into the beta pleated sheets and are encouraged by toxic aggregates

Question 49

What enzyme is responsible for the first committed step in heme synthesis - i.e. major regulator in this pathway?

Answer 49

**delta-aminoleuvulinic acid synthase**. This enzyme has Vitamin B6 (pyridoxal phosphate) as a co-factor

Question 50

Lead and iron can inhibit steps in heme synthesis. Identify enzyme targets for each metal

Answer 50

* **Ferrochelatase** (final enzyme in the synthesis of heme) can be inhibited by LEAD. When this occurs, Zn replaces Fe and becomes fluorescent, which is a test for lead poisoning. * **Uroporphyrinogen Decarboxylase** is inhibited by Fe, preventing the formation of ***Coporphyrinogen III***

Question 52

What are porphyrias? How can regulation of heme synthesis factor into tx for porphyrias?

Answer 52

**Porphyrias** are diseases caused by abnormalities in the heme production process. Heme's function hinges on its ability to bind oxygen in Hb. For tx of porphyrias, need to stop the toxic substrate build up by inhibiting the enzymes at the *beginning of the pathway*. Many drugs stimulate the **delta-aminoleuvulinic acid synthase** enzyme, which produces **ALA**, which is a neurotoxin. If this enzyme is regulated, there will not be a backup of the toxins.

Question 53

How can defects at various points in heme synthesis pathway give rise to profoundly different disorders?

Answer 53

Remember that if an enzyme is blocked, the substrate it works on builds up, leading to the disorder. * If **porphobilinogen deaminase** is inhibited, get a back up of *ALA and porphobilinogen*, causing **Acute Intermittent Porphyria**. * If Fe is present, can inhibit **uroporphinogen decarboxylase** → **PCT (porphyria cutanea tarda)**, the most common porphyria. Get build up *uroporphyrinogen III.* Sx = skin lesions induced by light. * **Pb** can inhibit **ALA dehytratase**, also backing up *ALA* causing neuro issues. In this case, Zn will replace Fe in the Protoporphyrin IX, causing a fluorescent product that can be imaged

Question 54

What enzyme catalyzes the first step in heme breakdown? What is an unusual feature of the enzyme's rxn mechanism?

Answer 54

**Heme Oxidase**. This enzyme takes the ringed heme structure and makes it linear into *biliverdin*. \*_ONLY rxn in humans that produces CO.\*_ The biliverdin is then reduced (C=C to CH₂) to *bilirubin*.

Question 55

How are measurements of direct and indirect bilirubin used to identify causes of jaundice?

Answer 55

*Indirect bilirubin* = issue with the turning the unconjugated bilirubin into the conjugated form. Can be caused by: * **Gilbert Syndrome:** a deficiency in the conjugating enzyme that builds up indirect bilirubin; or * hemolysis. Increase in heme synthesis would mean an increase in heme degradation and more unconjugated bilirubin *Direct bilirubin* = issue with build up in the bile duct. Direct bilirubin is the conjugated form in the liver.

Question 56

What do defects in heme synthesis lead to? What causes sideroblastic anemia?

Answer 56

Defects in heme synthesis lead to _porphyrias_ Defects in delta-aminoleuvulinic synthase lead to _sideroblastic anemia_

Question 57

How can you distinguish lead poisoning from acute intermittent porphyria?

Answer 57

AIP → increased ALA *and* increased porphobilinogen Pb poisoning = only increased ALA \*Both affect nervous system

Question 58

What is the role of ransferrin in iron transport?

Answer 58

Transferrin takes the oxidized and *protected state* (Fe³⁺) and transfers it across the cell membrane. In healthy people, transferrin is in excess so that it can accept all the Fe3+ to keep it from reducing back to the Fe²⁺ state. The Fe2+ can react with a peroxide to form a radical that would mess with proteins and nucleic acids.

Question 59

What is hemochromatosis? How is it treated?

Answer 59

**Hemochromatosis** is a disease from iron overload. The iron sensing devices, such as ***HFE***, do not work properly and, thus, never signaled for ***Hepcidin*** to be made. (Hepcidin degrades iron when its levels are high in the body). This all eventually leads to increase in Fe3+ oxidized iron, which can damage organs it is stored in, such as liver, heart, pancreas, and pituitary. The ONLY way to tx hemochromatosis is **phlebotomy** (blood letting), primarily with leeches. This is b/c the only way to get rid of the excess iron is to bleed or get rid of the cells.