Exam 1 Questions

Question 0

B. Existence as a hybrid of resonance structures

Answer 0

The peptide bond exists as a hybrid of resonance structures giving it partial double bond character that causes it to be rigid and stronger and shorter than a typical amide bond.

Question 1

Describe the following characteristics of the peptide bond. Be brief but specific.
A. Functional group reactants and mechanism of bond formation?

Answer 1

A peptide bond results from a dehydration synthesis reaction between the alpha amino group of one amino acid and the alpha carboxyl group of another amino acid.

Question 2

C. Stereochemistry (ie. energetically favorable configurations for achieving minimal free energy.

Answer 2

The trans configuration is more energetically favorable than cis for achieving minimal free energy.

Question 3

D. Bond rotation within the C1 alpha-C-N-C2 alpha- peptide group in which it is located.

Answer 3

There is no rotation about the peptide bond but there is limited rotation about the phi and psi bonds in a peptide group.

Question 4

Consider the tripeptide group X- cis proline-Y In a section of a protein. Briefly describe steric, electrical, and geometrical characteristics of the X-cis proline dipeptide group that result in
A. Disruption of alpha helical secondary structure and?

Answer 4

Cis-proline disrupts alpha helical structure because of steric hindrance by it’s pyrrole ring and the inability of it’s imino-N to function as a hydrogen bond donor to a C double bond O functional group several residues away in the helix.

Question 5

Formation and stabilization of a reverse turn in the section of the protein where it is located.

Answer 5

The cis configuration provides the proper geometry for introducing a turn a peptide chain. The peptide bond between the downstream residue and proline. And the phi bond between the imino-N and the alpha-C of proline are both rigid which stabilizes the reverse turn and exposes the pyrrole ring to the solvent

Question 6

C. Steric implications to the cis proline-Y dipeptide group that follows

Answer 6

The residue upstream from cis proline in most instances is glycine which has a H for it’s side chain that can easily fit into the reverse turn.

Question 7

What types of noncovalent electrical forces can and cannot exist between molecules of methane CH4 and why or why not is each type of force able to exist?

Answer 7

Methane is nonpolar and methane molecules are thus unable to participate in permanent dipole-dipole interactions (eg. Hydrogen bonds) with each other. However weak van der waals attractions are possible when a temporary dipolar molecule induces a dipole moment in a neighboring nonpolar molecule by distorting the spatial arrangements of it’s bond electrons. Because methane does not ionize, methane molecules do not form salt (ionic) bonds.

Question 8

Consider the mixing of oil with water. Briefly describe the principles of molecular association and thermodynamics that explain why this system equilibrates by aggregation of small oil droplets into a single large glob or layer.

Answer 8

Because oil is nonpolar, the polar water molecules rearrange their hydrogen bonds and surround the small oil droplets in cages know as clathrates, which introduces order to the system because the water molecules at the oil interface are now restricted in their movement and freedom to hydrogen bond. The system equilibrates by aggregation of the small oil droplets into a glob or layer stabilized by hydrophobic interactions such that the oil/water interface is decreased. Water molecules released from the interface are now less restricted in their movement and freedom to hydrogen bond, which represents an increase In randomness (entropy) in accordance with the second law of thermodynamics.

Question 9

If you reacted .05 M NaOH with an equal volume of .1 M NaH2PO4 (pKa =7.20) would the resulting solution be a buffer system? Briefly explain why or why not. If it is a buffer system within what range of pH values does it effectively resist change in pH following addition of H+ or OH?

Answer 9

Upon mixing equal volumes the concentrations of NaOH and NaH2PO4 are halved. Upon reaction .025 M NaOH will convert half of .05 M NaH2PO4 (acid) to .025 M NaH2PO4 (conjugate base). There are now equal concentrations of acid and conjugate base a perfect buffer for which pKa =7.2 according to the Henderson hasselbach equation. The solution buffers effectively between pH 6.2 and 8.2 according to the +1/-1 rule.

Question 10

Describe how hydrophobic interactions and the second law of thermodynamics contribute to proper as we’ll as improper (eg. As in certain neurodegenerative diseases such as Alzheimer’s, Parkinson’s, and huntingtons diseases) folding of proteins. Be specific.

Answer 10

In proper protein folding spontaneous hydrophobic interactions explain why side chains of nonpolar amino acids cluster together In the interior of the protein, although the driving force is actually the increase in entropy of the solvent (water) that results from increased freedom of movement, rotation, H bonding etc in Alzheimer’s, Parkinson’s, and huntingtons diseases, protein aggregates (ie. plaques) form when there are exposed nonpolar areas on protein surfaces that interact with hydrophobic regions of other similarly misfolded proteins. As with proper folding these hydrophobic interactions are spontaneous and increase entropy of solvent in accordance with the second law of thermodynamics.

Question 11

Briefly explain why dietary trans unsaturated fatty acids might be observed to elicit physiological effects similar to those of dietary saturated fats.

Answer 11

Trans configured unsaturated fatty acids would be expected to elicit physiological effects that are similar to those of saturated fatty acids because both are straight extended hydrocarbon chains ie. ( no kink as in cis-configured unsaturated fatty acids).

Question 12

Briefly describe how specific physical, stereochemical and compositional characteristics of membrane lipids contribute to membrane fluidity and asymmetry.

Answer 12

1) membrane fluidity is deceased as proportion of saturated fatty acids and trans unsaturated fatty acids increases because these are extended, have higher melting points and enable tighter packing of membrane lipids compared with kinked cis unsaturated fatty acids.
2) membrane fluidity is decreased as proportion of long chain fatty acids increases because they have higher melting points than fatty acids with shorter chain lengths.
3) the fused ring structure of cholesterol contributes to rigidity of lipid membranes which decreases fluidity.
4) polar head groups of lipids in the outer leaflet (sphingomyelins, cerebrosides, gangliosides) are larger and bulkier than those in the inner leaflet (primarily phosphoglycerides) requiring more room to accommodate them.

Question 13

Briefly distinguish between peripheral and integral membrane proteins in terms of a) their respective locations with respect to surfaces and the interior of the lipid bilayer. (B) stabilizing molecular forces between their respective amino acid side Chains and lipid components of the bilayer.
(C) an example of a membrane related function for each.
(D) an appropriate physical or chemical method for removing each from the membrane

Answer 13

A) peripheral proteins are located on the membrane surface whereas integral proteins are embedded within the membrane.
B) peripheral proteins are usually bound to polar/charged head groups by dipole dipole interactions (eg. Hydrogen bonds), ionic (salt) bonds or both. Nonpolar amino acid side chains of integral proteins are stabilized largely by van der waals forces with nonpolar lipids in the interior of the membrane, and may be covalently attached to lipid anchors.
C) examples of membrane related functions of both peripheral and integral proteins include catalysis, hormone receptor, signal transduction and transport.
D) peripheral proteins can usually be readily removed by raising ionic strength. Integral proteins on the other hand typically require detergents and or sonication for removal from the membrane.

Question 14

Which amino acids can have a net molecular charge of +2 at pH 0?

Answer 14

Lysine, histidine, arginine

Look at the molecules charges

Question 15

Which amino acids can have a net molecular charge of -2 at pH 13?

Answer 15

Cysteine, tyrosine, Aspartate, glutamate

Question 16

Consider the mixing of acetic acid (CH3COOH) with water. Briefly describe the principles of molecular association and thermodynamics that explain why the acetic acid molecules are readily dispersed into solution and not aggregated into a single glob or layer separate from the water.

Answer 16

Being a polar molecule acetic acid will dissociate forming the acid CH3COOH and it’s conjugate base CH3COO- in polar water. Conforming with the 2nd law of thermodynamics this dissociation increases the randomness or entropy of the system making it thermodynamically favorable. The molecules do not aggregate or layer separate because doing so would actually hinder the possible number of binding sites available to water.

Question 17

Provide a critique of the following statement: among all the noncovalent electrical forces that drive protein folding and contribute to stabilizing conformation, dipolar and ionic interactions between side chains of amino acids such as leucine, isoleucine, and valine are the most important for achieving minimal free energy?

Answer 17

Leucine and isoleucine and valine all have nonpolar hydrophobic side chains. To conform with the 2nd law of thermodynamics and achieve minimal free energy these side chains from hydrophobic interactions making pockets to avoid contact with the hydrophilic environment. Without these interactions the molecule would have to great an entropy requirement due to water methyl group contact.

Question 18

Explain how stereochemical and electrical properties of cis-proline are inconsistent with alpha helix conformation in proteins.

Answer 18

Cis-proline does a 180 degree turn. It’s bulky planar side chain limits rotation which is why it is often followed by glycine, it is small enough to fit after it. H+ bonds do not occur no -NH

Question 19

Briefly describe the mechanisms by which detergents disrupt molecular forces that stabilize tertiary structure in proteins.

Answer 19

Detergents disrupt hydrophobic interactions on the inside nonpolar portion of a proteins tertiary structure which leads to unraveling and denaturation. Detergents with a charge can also electrostatically interact with proteins giving the protein a charge to salt (ionic) bond with besides itself which leads to denaturation.

Question 20

Briefly describe how chemical composition of the cytosolic (inner) and extra cellular (outer) leaflets of a lipid bilayer membrane contributes to it’s asymmetry.

Answer 20

In order to have equal pairs of phospholipids in the lipid bilayer membrane, outer leaflet polar heads must be larger to accommodate the extra space created geometrically. The larger outer polar heads must be of different chemical composition than the inner (smaller) polar heads that are needed to fit more tightly together.

Question 21

The term homeoviscous adaptation refers to compositional adaptation of membrane lipids that enable maintenance of more or less constant membrane fluidity under different temperatures. How would you expect chain length and degree of saturation of membrane fatty acids to change with decreasing temp? Why?

Answer 21

To maintain fluidity the chain length and the degree of saturation should decrease because the more saturated a fatty acid becomes the less flexible it becomes.

Question 22

The plasma membrane of cells contains sphingolipids and protein receptors organized in microdomains called lipid rafts that are more ordered and tightly packed than the surrounding bilayer, but float freely in the membrane bilayer. How would you expect degree of lipid saturation and cholesterol content in lipid rafts to compare with those of the surrounding bilayer? Briefly explain why. What are the primary molecular forces that stabilize the lipid components of rafts.

Answer 22

The lipid rafts would have a higher degree of saturated fatty acids and cholesterol than the surrounding lipid bilayer. Bother saturated fats and cholesterol increase order and decease fluidity. Van der waals and hydrophobic interactions stabilize the lipid components of rafts. The area facing the interior of rafts would be hydrophobic.

Question 23

Ribonuclease denatured by urea and B-mercaptoethanol is observed to spontaneously convert to fully active, native ribonuclease following removal of both denaturing agents and air oxidation of sulfhydryl groups. However when denatured ribonuclease is air oxidized following removal of B-mercaptoethanol but not urea very little (~1%) of the enzymatic activity of the native protein is restored why?

Answer 23

When denatured ribonuclease is air oxidized following removal of B-mercaptoethanol but not urea, disulfide bridges (cysteine) can be re-formed from thiol groups of cysteine residues, but enzymatic activity is not restored because the protein is still denatured. Unless the polar molecule urea is removed it continues to disrupt dipolar interactions such as hydrogen bonds.

Question 24

Phospholipids synthesized in the cytoplasm are first incorporated into the inner leaflet of a lipid bilayer membrane. Enzymes called flippases can then transfer them within the plane of the membrane to the outer leaflet via transverse diffusion or flip flop. Briefly explain why phospholipids synthesized in the cytoplasm can’t simply diffuse laterally through the membrane from the inner to the outer leaflet.

Answer 24

Because they have polar and nonpolar parts. Outside the membrane is polar, inner is nonpolar. Two distinct areas and while polar part of phospholipid can come into contact with polar heads of the membrane, once it hits the nonpolar section will not be able to cross. Opposites repel, will need a carrier molecule or pump to cross.