Homework Set # 1

Question 1

Q # 1 (Pt.1) How can just a few elements give rise to all biological diversity? (Pt. 2) At what level, if any, are all biological organisms similar? (Pt.3) Given this biochemical similarity, how is the structural and functional diversity of living things possible?

Answer 1

Living things are composed primarily of macromolecules, polymers of simple compounds of just a few different types. The properties of these polymers are determined by their sequence of monomers and these can be combined in many different ways. Diversity is thus achieved through the nearly limitless variety of sequences that can exist when amino acids are linked to form proteins, nucleotides are linked to form nucleic acids, and monosaccharides are linked to form polysaccharides. Branching in the latter can contribute additional heterogeneity. Each type of organism constructs a unique set of macromolecules from these monomeric units, resulting in the structural and functional diversity among species

Question 2

Q # 2 What are five periodic elements most frequently seen incorporated into compounds of biological organisms? Name three occasionally seen elements too.

Answer 2

A) Oxygen, Nitrogen, hydrogen, carbon and Phosphorous. B) Trace elements: Fe, Cu, Zn

Question 3

Q # 3 Draw the structures of the following functional groups in their un-ionized forms: (a) hydroxyl, (b) carboxyl, (c) amino, (d) phosphoryl.

Question 4

Q # 4 (a) List the types of noncovalent interactions that are important in providing stability to the three-dimensional structures of macromolecules. (b) Why is it important that these interactions be noncovalent, rather than covalent, bonds?

Answer 4

(a) hydrophobic, van der waals, ionic, hydrogen; (b) It is important that these interactions be non-covalent, rather than covalent bonds because they need to be readily able to break and form. If macromolecules interacted via covalent bonds, catabolism and the disuption of macromolecular interactions would require more energy to break.

Question 5

Q #5 Why is an asymmetric carbon atom called a chiral center?

Answer 5

An asymmetric carbon atom is called a chiral center to due the different conformations asymmetry confers in 3-D space. A carbon atom with four different substituents that can create two chiral molecules (two molecules that are mirror images, but are non superimposable)

Question 6

Q # 6 A chemist working in a pharmaceutical lab synthesized a new drug as a racemic mixture. Why is it important that she separate the two enantiomers and test each for its biological activity?

Answer 6

Biomolecules such as receptors for drugs are stereospecific, so each of the two enantiomers of the drug may have very different effects on an organism. One may be beneficial, the other toxic; or one enantiomer may be ineffective and its presence could reduce the efficacy of the other enantiomer.

Question 7

Q # 7 How is the genetic information encoded in DNA and how is a new copy of DNA synthesized?

Answer 7

The genetic information is encoded in DNA by the organization of nucleotide bases in its double helix. Central dogma states that DNA-> RNA -> Protein. DNA leads to the production of proteins and other biologically important compounds of organisms. A new DNA copy is synthesized when DNA helicase unzips the DNA double helix (Specifically the hydrogen bonds), DNA synthesizes from 5’ to 3’ end via DNA polymerase with the help of an RNA primer. DNA ligase links okazaki fragments.

Question 8

8. Name two functions of (a) proteins, (b) nucleic acids, (c) polysaccharides, (d) lipids.

Answer 8

(a) Proteins: Structure & chemical reactions (b) Nucleic Acids: Genetic information storage & production of proteins (c) Polysaccharides: Structure and Energy Storage (d) Lipids: Structure, Cell signaling, and energy storage

Question 9

Q # 9.How does the electronegativity of each atom affect the polarity of a bond? Use electronegativity to explain why water is a good solvent.

Answer 9

Electronegativity is the tendancy for molecules to draw electron density toward themselves. This creates a dipole which gives partial ionic character to a molecule. Electronegativity is a property that makes water a solvent for mulitple reasons. Oxygen is a more electronegative atom than hydrogren. It has negative character while both H atoms have positive character. H is attracted to the more electronegative oxygen molecule in adjacent water molecules, and vice versa. Each water molecule has the potential to interact with four other oxygen molecules

Question 10

Q # 10 Explain the fact that ethanol (CH3CH2OH) is more soluble in water than is ethane (CH3CH3).

Answer 10

Ethanol is more soluble than ethane, because ethanol has an OH group, which is capable of participating in hydrogen bonds.

Question 11

Q# 11 Describe how van der Waal’s interactions work. What types of molecules can participate in van der Waal’s interactions?

Answer 11

Van der Waal’s interactions work because of mutually attractive induced dipoles. They are highly dependent on the distance between two participating atoms. They are very weak, but any atoms can participate in them

Question 12

12. Phosphoric acid (H3PO4) has three dissociable protons, with the pKa’s shown below. Which form of phosphoric acid predominates in a solution at pH 4? Explain your answer.

Acid pKa H3PO4 2.14; H2PO4 – 6.86; HPO4 2– 12.4

Answer 12

At pH 4, the first dissociable proton (pKa = 2.14) has been titrated completely, and the second (pKa = 6.86) has just started to be titrated. The dominant form at pH 4 is therefore H2PO4 – , the form with one dissociated proton (see Fig. 2-15).

Question 13

Q # 13 Define pKa for a weak acid in the following two ways: (1) in relation to its acid dissociation constant, Ka, and (2) by reference to a titration curve for the weak acid.

Answer 13

(1) pKa = log 1/Ka or -logKa. The smaller the value for pKa, the more acidic the proton is, while a larger value for Ka corresponds to more acidic proton. (2) Titration curves are used to find the pKa. This is the point (pH) at which the conjugate acid is equal to its conjugate base. At +/- 1 of the pKa is its buffering region.

Question 14

Q # 14 Give the general Henderson-Hasselbalch equation and sketch the plot it describes (pH against amount of NaOH added to a weak acid). On your curve, label the pKa for the weak acid and indicate the region in which the buffering capacity of the system is greatest.

Answer 14

1) pH = pka + log [A-]/[HA] 2) Draw titration curve on you own.

Question 15

Q # 15 You have just made a solution by combining 50 mL of a 0.1 M sodium acetate solution with 150 mL of 1 M acetic acid (pKa = 4.7). What your solution’s pH?

Answer 15

(1) pH = pKa + log [A-]/[HA]. (2) Work pH = 4.7 + Log [0.1]/[1]

Question 16

Q # 16 What are the structural characteristics common to all amino acids found in naturally occurring proteins?

Answer 16

(1) Carboxylic acid group, (2) free amino group, with the proline, pro, P as an exception (3) free H group. with the exemption of proline (4) Chirality. with glycine, G, Gly as an exception. All amino acids found in naturally occurring proteins have an alpha carbon to which are attached a carboxylic acid, an amine, a hydrogen, and a variable side chain. All the amino acids are also in the L configuration.

Question 17

Q # 17 Only one of the common amino acids has no free alpha-amino group. Name this amino acid and draw its structure.

Answer 17

Proline.

Question 18

Q # 18. Draw the structures of the amino acids phenylalanine and aspartate in the ionization state you would expect at pH 7.0. Why is aspartate very soluble in water, whereas phenylalanine is much less soluble?

Answer 18

(1) Phenylalanine, Phe, F (2) Aspartate, Asp, D. I would expect phenylalanine to have no charge, while I would expect aspartate to have a negative charge. Aspartate is very soluble in water, because of the presence a carboxylic acid group which is capable of relatively easy dissociation and hydrogen bonding. Phenylalanine is much less soluble because of the presence of no polar groups on its aromatic group.

Question 19

Q # 19 The amino acid histidine has three ionizable groups, with pKa values of 1.8, 6.0, and 9.2. (a) Which pKa corresponds to the histidine side chain? (b) In a solution at pH 5.4, what percentage of the histidine side chains will carry a positive charge?

Answer 19

(a) 6.0; (b) 80%. 4 = [acid]/[conjugate base], or 4[conjugate base] = [acid] Therefore, at pH 5.4, 4/5 (80%) of the histidine will be in the protonated form

Question 20

Q # 20. Define the primary structure of a protein.

Answer 20

The primary structure of a protein is its unique sequence of amino acids and any disulfide bridges present in the native structure, that is, its covalent bond structure.

Question 21

Q # 21 Which one of the above tripeptides: (a) is most negatively charged at pH 7? (b) will not form an alpha helix or beta sheet? (c) contains the largest number of nonpolar R groups? (d) contains sulfur? (e) will have the greatest light absorbance at 280 nm?

Answer 21

(a) = D; (b) = B; (c) = E; (d)= A; (e) = C

Question 22

Q # 22. Any given protein is characterized by a unique amino acid sequence (primary structure) and three-dimensional (tertiary) structure. How are these related?

Answer 22

The primary sequence of amino acids determines its secondary, tertiary and quaternary structure for a variety of reasons. The R groups of each amino acid in a sequence,H-bonding between the n and n+4 amino acids, and non-covalent interactions that arise from the initial sequence of amino acids influences tertiary structure.

Question 23

Q # 23. When a polypeptide is in its native conformation, there are weak interactions between its R groups. However, when it is denatured there are similar interactions between the protein groups and water. What then accounts for the greater stability of the native conformation?

Answer 23

The greater stability in the native conformation arises for a variety of reasons. In the naive conformation, hydrophobic regions of the polypetide face inward away from polar water, while the polar R groups associate with water face the outside. This allows polar and non-polar regions to associate in thermodynamically favorable ways. When a protein denatures, hydrophobic regions of the amino acid structure associate with water which is polar. Hydrophobic regions cause water to order around those regions resulting in a decrease in entropy. This is less thermodynamically favorable.

Question 24

Q #24. Draw the resonance structure of a peptide bond, and explain why there is no rotation around the C—N bond

Answer 24

(B) There is no rotation around the C-N bond due to the overlapping of pi orbitals that occur from resonance structures between the double bonded oxygen and the nitrogen with the free lone pair in the C-N bond.

Question 25

Q # 25. Draw the hydrogen bonding typically found between two residues in an alpha helix.

Answer 25

Hydrogen bonds occur between every carbonyl oxygen in the polypeptide backbone and the peptide —NH of the fourth amino acid residue toward the amino terminus of the chain. (See Fig. 4-2, p. 116.)

Question 26

26. Describe three important features of an alpha-helix structure. Provide one or two sentences describing why each feature is important.

Answer 26

(1) Internal hydrogen bonding. H-bonding is important to stabilize the overall helical structure especially the interaction between the hydrogen atom attached to the electronegative nitrogen atom of the fourth amino acid on the amino terminal side of that peptide bond. Each succesive turn is stabilized by 3-4 hydrogen bonds on each turn. (2) Hydrophobic interactions in the interior of the protein are responsible for strength as well. (3) psi and phi angles are also important to the stability of alpha-helix structure. Angles of -57˚ of Phi and -47˚ of psi are characteristic of the most stable conformation. (minimzes repulsive forces and steric hindrance)

Question 27

Q # 27. Describe three of the important features of a beta sheet polypeptide structure. Provide one or two sentences for each feature.

Answer 27

(1) Hydrogen bonds. Adjacent segements hydrogen bond to stabilize the structure. (2) The beta conformation. Beta sheets form a zig-zag backbone deriving from stability of their dihedral angles. (3) Antiparallel vs parallel conformations. The alignment of two segments of a beta sheet give rise to differences in length of the sheet. Parallel (6å) antiparallel (7å).

Question 28

Q # 28 Why are glycine and proline often found within a beta turn?

Answer 28

(a) Proline favors B- turns because it readily forms the cis configuration via the imino nitrogen of proline, which makes a tight turn. (b) Glycine is flexible

Question 29

Q # 29. Explain how circular dichroism spectroscopy could be used to measure the denaturation (unfolding) of a protein.

Answer 29

Circular dichroism spectroscopy measures the amount of -helix in a given protein. As the protein denatures, the amount of -helix should decrease as the protein chain becomes disordered; this change would be detectable using CD spectrography.

Question 30

Q # 30. How can changes in pH alter the conformation of a protein?

Answer 30

Changes in pH alter the net charge on the protein, causing electrostatic repulsion and the disruption of hydrogen bonding.

Question 31

Q # 31. Name four factors (bonds or other forces) that contribute to stabilizing the native structure of a protein, and describe one condition or reagent that interferes with each type of stabilizing force.

Answer 31

Any of the following forces stabilize native protein structures and are disrupted by the listed conditions or reagents: (a) disulfide bonds by reducing conditions or mercaptoethanol or dithiothreitol, (b) hydrogen bonds by pH extremes, high salt or heat, (c) hydrophobic interactions of non-polar groups in aqueous solvent by detergents, urea or guanidine hydrochloride, or organic solvent, (d) ionic interactions by changes in pH or ionic strength, and (e) van der Waals interactions by any unfolding condition.

Question 32

Q # 32. What important concepts regarding protein thermal denaturation can be inferred from the egg white of a boiled egg?

Answer 32

1) Denatured proteins often precipitate and/or aggregate. 2) Denaturation is often not a reversible process.

Question 33

What is a chaotrophic agent?

Answer 33

A chaotrophic agent is a compound (i.e urea, guanidinium, and hcl) cause disruptions in the maintenance of hydrogen bonds.

Question 34

Q # 33. Once a protein has been denatured, how can it be renatured? If renaturation does not occur, what might be the explanation?

Answer 34

Once a protein has been denatured, it can be renatured by being returned to its optimal environment that promoted its folding or if chaperones aid in the process of returning proteins to its structure. If renaturation does not occur, metabolites or compounds may still be in the environment that do not promote the reformation of the native structure. Chaperones may not be available to reform the protein. Lastly, denaturation could result in a protein that has been so badly damaged that it will not refold.

Question 35

Q # 34. Each of the following reagents or conditions will denature a protein. For each, describe in one or two sentences what the reagent/condition does to destroy native protein structure. (a) urea / guanidine hydrochloride (b) high temperature (c) detergent (d) low pH

Answer 35

(a) Urea or guanidine hydrochloride acts primarily by disrupting hydrophobic interactions. (b) High temperature provides thermal energy greater than the strength of the weak interactions (hydrogen bonds, electrostatic interactions, hydrophobic interactions, and van der Waals forces, breaking these interactions. (c) Detergents bind to hydrophobic regions of the protein, preventing hydrophobic interactions among several hydrophobic patches on the native protein. (d) Low pH causes protonation of the side chains of Asp, Glu, and His, preventing electrostatic interactions.

Question 36

Q # 35 What is the pI, and how is it determined for a protein?

Answer 36

pI -the isoelectric point is the pH at which the net electric charge of a protein is zero, and it is determined by titrating a protein

Question 37

Q # 36 A biochemist is attempting to separate a DNA-binding protein (protein X) from other proteins in a solution. Only three other proteins (A, B, and C) are present. The proteins have the following properties:

What type of protein separation techniques might she use to separate: (a) protein X from protein A? (b) protein X from protein B? (c) protein X from protein C?

Answer 37

(a) Size Exclusion Chromatography; (b) Ion-exchange chromatography (c) Affinity chromatography

Question 38

Q # 37. A biochemist has accidentally isolated a second protein with the protein they were trying to isolate. The reason for its co-purification is unknown, as is its identity. How might the biochemist have identified its presence and what should they do to determine its identity?

Answer 38

The biochemist most likely identified the second protein in a PAGE gel, either SDS or isoelectric focusing or both. Since the biochemist presumably knows what species they are working with, they only need a small amount of sequence to correctly identify the protein. In this circumstance, either Edman degradation or mass spectrometry may be performed. However, mass spectrometry is the most likely to be used, since it is more readily available.

Question 39

Q # 38. What factors would make it difficult to interpret the results of a gel electrophoresis of proteins in the absence of sodium dodecyl sulfate (SDS)?

Answer 39

Without SDS, protein migration through a gel would be influenced by the protein’s intrinsic net charge—which could be positive or negative—and its unique three-dimensional shape, in addition to its molecular weight. Thus, it would be difficult to ascertain the difference between proteins based upon a comparison of their mobilities in gel electrophoresis.

Question 40

Q # 39 Describe a reservation about the use of x-ray crystallography in determining the three dimensional structures of biological molecules.

Answer 40

In order to crystallize proteins, they need need to be highly ordered. Proteins have proven to be difficult to crystallize. X-ray analysis also take several steps making it both long and labor intensive. X-ray analysis does not allow for understanding proteins in vivo.

Question 41

Q #40 Name two ways to determine the precise (high-resolution) three-dimensional structure of a protein complex. Describe why you think these methods are appropriate.

Answer 41

The protein complex could be crystallized, and its structure determined by x-ray crystallography. The pattern of diffracted x-rays yields, by Fourier transformation (black magic), the three-dimensional distribution of electron density. By matching electron density with the known sequence of amino acids in the protein, each region of electron density is identified as a single atom. This would be an appropriate choice if the complex can be crystallized, but inappropriate if not. Sometimes, the three-dimensional structure of a small protein or peptide can be determined in solution by sophisticated analysis of the NMR spectrum of the polypeptide. However, this technique is rarely appropriate for complexes, which usually exceed the complexitiy limit. A three dimensional structure can also be determined by cryo-electron tomography, which is an excellent method to use with complexes that don’t crystallize. If an advanced machine is available, molecular details may be sufficient to determine high resolution structure. This is most appropriate for larger complexes, or those with individually crystallized components, but may also be used as the first approach. Homology modeling is only an appropriate answer if the above three are likely to fail, and a good structure of a highly similar protein is available.

Question 42

Q# 41. For the binding of a ligand to a protein, what is the relationship between the Ka (association constant), the Kd (dissociation constant), and the affinity of the protein for the ligand?

Answer 42

(a) Ka is an association constant that describes the equilibrium between the complex and the unbound components of the complex ;For P + L PL; Ka = [PL]/[P][L]; Where P=protein, L = ligand

;The larger the Ka, association constant, the higher the affinity there is of the ligand for a protein; Units M^-1

(b) the equilibrium constant for the release of a ligand ;A lower value of Kd results in a higher affinity of ligand for a protein;For PL -> P + L;
Kd = [P][L]/[PL]= kd/ka (b) Ka = 1/Kd. The larger the Ka (and hence the smaller the Kd), the higher the affinity of the protein for the ligand

Question 43

Q # 42 Describe how you would determine the Ka (association constant) for a ligand and a protein.

Answer 43

An experiment would be carried out in which a fixed amount of the protein is incubated with varying amounts of ligand (long enough to reach equilibrium). The fraction of protein molecules that have a molecule of ligand bound is then determined. A plot of this fraction (ø) vs. ligand concentration [L] should yield a hyperbola. The value of [L] when  = 0.5 is equal to 1/Ka.

Question 44

Q #43 What fraction of ligand binding sites are occupied (ø) when [ligand] = Kd? Show your work.

Answer 44

1) Where theta equals the fraction of binding sites occupied divided by total binding sites 2) When [L] equals Kd, half of the ligand binding sites are occupied. 3) A lower value of Kd results in a higher affinity of ligand for a protein. Kd = [P][L]/[PL]; [PL] = [P][L]/Kd; ø [L]/([L] + Kd)

Question 45

Q # 44 Explain why most multicellular organisms use an iron-containing protein for oxygen binding rather than free Fe2+. Your answer should include an explanation of (a) the role of heme and (b) the role of the protein itself.

Answer 45

Iron promotes the formation ROS’ (reactive oxygen species) such as hydroxyl molecules, that damage DNA and macromolecules. Free Iron would cause cellular damage. A heme group consists of a complex organic ring structure (Protoporphyrin) to which is bound a single iron atom (Fe2+). Fe2+ is buried deep within the heme molecule as well as being associated with four nitrogen molecules that donate electrons This prevents Fe2+ from going to Fe3+ which irreversibly binds O2. One of the two free groups is bound by His, the other allows binding of oxygen.

Question 46

Q # 45. Explain why the structure of myoglobin makes it function well as an oxygen-storage protein, whereas the structure of hemoglobin makes it function well as an oxygen transport protein.

Answer 46

(a) The heme group is at the center of the myoglobin molecule, and it is surrounded by 8 alpha helical segments. Myoglobin only has a single binding site for oxygen, where as hemoglobin has four subunits capable of transporting four oxygen molecules. Hemoglobin’s ability to transport more oxygen molecules makes it more suited to transport oxygen than myoglobin with its single binding site.

Question 47

Q # 46 How does 2,3 BPG binding to hemoglobin decrease its affinity for oxygen?

Answer 47

2,3 BGP Binds to the space between ß subunits in in a hemoglobin tetramer. Only one BGP molecule can bind to a single tetramer. BGP binds to the positively charged amino acid residues in hemoglobin. This binding stabilizes the T state of hemoglobin, leading to a decrease in the affinity for oxygen.

Question 48

Q # 47 Fetal hemoglobin binds BPG with lower affinity than adult hemoglobin. How does this property facilitate tranfers of O2 from mother to fetus?

Answer 48

Fetal hemoglobin has a greater affinity for oxygen, because the replacement of gamma subunit with the beta subunit does not bind BGP with the same affinity as beta subunits. This leads to a greater affinity of oxygen. This makes physiological sense, because the fetal environment is surrounded by fluid, and not readily available oxygen from the air.

Question 49

Q # 48 Why is carbon monoxide (CO) toxic to aerobic organisms?

Answer 49

Carbon monoxide is toxic to aerobic organisms, because CO binds to hemoglobin with greater affinity to hemoglobin than O2. When two molecules of CO bind to hemoglobin, its affinity for O2 in the other two subunits increases. Oxygen is less likely to be released when it is bound to a hemoglobin molecule which has also bound carbon monoxide, which leads to tissue death via oxygen deprivation.

Question 50

Q # 49 Describe briefly the structure of myosin.

Answer 50

Myosin has 6 subunits (2H 4L); at carboxyl terminus, they form tight left handed alpha helixes. At the amino terminus, each heavy chain occurs where ATP is hydrolyzed

Question 51

Q # 50 What is the role of ATP and ATP hydrolysis in the cycle of actin-myosin association and disassociation that leads to muscle contraction?

Answer 51

1) Myosin and actin dissociate when myosin binds to ATP 2) ATP is hydrolyzed to ADP + Pi, 3) Myosin then attaches to actin causing the release of Pi 4) Pi release causes a “power” stroke, which is a conformational change in the myosin head that moves actin and myosin filaments relative to one another.