Key for Exam One: Spring, 2007
1. Explain why non-polar molecules and regions of molecules form stable associations. (6 pts)
Water molecules are driven by enthalpy to form hydrogen bonds. When a nonpolar group or molecule enters the picture, the water must cage around it to make the hydrogen bonds. This increases the order of the system and is energetically unfavorable. Entropy moves the system towards maximum disorder. By having the nonpolar molecules associate with one another, water caging is minimized since many cages around individual nonpolar molecules creates more order than does one larger case around a group of nonpolars.
2. Draw the outline of two amino acids. Call the side chains R1 and R2. Then show how they interact chemically to form a peptide bond. (4 pts)
See Figure 2-15 in our textbook.
Now show the region of planarity of this peptide bond. (4 pts)
All atoms in the peptide bond are included.
Explain why the peptide bond must remain planar. (4 pts)
Resonance of the electrons in the peptide bond. The C-N and C-O each have partial double bond character and therefore the region must remain planar to accommodate this.
3. Answer the following questions about protein structure:
A. define the term structural motif. (4 pts)
A combination of secondary structures that are frequently seen together in proteins. They perform a similar function in these different proteins. For example, DNA binding.
B. Describe how the alpha-helix of a protein is stabilized into that structure. (4 pts)
There are hydrogen bonds between the carbonyl oxygen of each peptide bond with the amine hydrogen of a peptide bond 4 residues away. This forces the polypeptide chain into an alpha helix with 3.6 amino acids per turn.
C. Illustrate an antiparallel and a parallel beta pleated sheet, using arrows to indicate polarity and dotted lines to indicate hydrogen bond locations and orientation. (4 pts)
See the link to our website that describes this. Antiparallel sheets have parallel H bonds that are stronger. Parallel have angled H bonds not as strong.
4. Phosphofructokinase is an allosteric enzyme involved in the early reactions of glycolysis. It is regulated by either ATP or by ADP/AMP in opposite directions.
A. Knowing that glycolysis is an essential pathway for ATP production, predict the effect (activation or inhibition) upon the enzyme’s activity when ATP levels are high or when ADP/AMP levels are high. (4 pts)
If ATP levels are high, the enzyme is inhibited.
If ADP/AMP
levels are high, the enzyme is activated.
B. Explain the mechanism that allows such regulation of this enzyme. (6 pts)
The enzyme exists in equilibrium between an active and inactive form. If ATP is high, it binds to the allosteric site on the regulatory subunit of the inactive form, stabilizing it. This shifts the equilibrium towards the inactive form (more inactives since active can still shift to inactive but not vice-versa). If ADP/AMP is high, it binds to the active form, stabilizing it and shifting the equilibrium towards active.
5. Serine proteases all use a similar mechanism to catalyze peptide bond cleavage. Three amino acids are responsible for this. Briefly describe the initial “charge-relay” that activates the serine. (5 pts)
The negatively charged asp draws a proton from his towards itself. This leaves his somewhat negative and it draws a proton from ser to itself. This activates the ser which then interacts with the substrate to form the first transition state intermediate.
What is different in the manner that trypsin and chymotrypsin interact with their substrates. (5 pts)
Chymotrypsin cleaves peptide bonds after nonpolar, usually aromatic side chains. Thus its active site has a binding region rich in nonpolar side chains where the amino acid binds. Trypsin cleaves after basic amino acids lysine or arginine and therefore its binding site contains a negatively charged asp that can ionically interact with them.
6. Describe B-DNA structure as thoroughly as you can. You may illustrate it if you prefer. (8 pts)
A double stranded polymer of deoxynucleotides connected by 3’-5’ phosphodiester bonds. Each strand runs antiparallel to the other (3’-5’ opposite 5’-3’), each winding around the other in a helical configuration. The sugar-phosphate backbone is on the exterior of the double helix. The nucleotide bases extend into the interior of the double-helix. Bases from one strand hydrogen bond to bases from the other strand. A forms two hydrogen bonds with T and C forms three hydrogen bonds with G. The diameter is 2nm. The interior is hydrophobic because the bases are hydrogen bonded, eliminating the polarity. There are approximately 10 base pairs per turn of the helix (3.4 nm per turn). The structure has alternating major and minor grooves on the surface through which protein-DNA interactions can occur.
What do we mean by the term chromatin (describe its structure)? (4 pts)
DNA organization whereby approximately 146 base pairs of DNA wraps twice around a histone octamer consisting of 2 each of H2A, H2B, H3, and H4 forming a nucleosome. These are separated by variable lengths of linker DNA not organized into nucleosomes, averaging 200 base pairs in length. Histone H1 sits at the junction between nucleosomes and linker regions and is responsible for stacking the nucleosomes into 30 nm cylinders or solenoids.
7. When scientists want to introduce radioactivity into a DNA molecule, they use radioactive deoxynucleoside triphosphates (dNTPs) that contain P32 (a radioactive form of phosphorus) at one of the three phosphates. Which form of dNTP (alpha, beta, or gamma labeled phosphate) would they choose and why? (6 pts)
Labeled alpha dNTP is used because when a new nucleotide is incorporated into a growing DNA polymer, the dNTP is used as the energy source for formation of the phosphodiester bond. The bond between the alpha and beta phosphates is broken, releasing pyrophosphate which breaks down to 2P. The nucleotide monophosphate (containing alpha) is attached to the polymer.
8. Draw a replication fork and indicate on it the following: polarities of the template strands; the direction of synthesis of the leading strand; the direction of synthesis of the lagging strand. (4 pts)
See Figure 6-3 in our textbook.
9. Describe the role in DNA replication of each of the following: (3 pts each)
Topoisomerase
Works ahead of the
moving replication fork to remove supercoiling
induced by the separation of the strands. Enzyme binds to a
phosphodiester bond
in one strand, causing a nick. The other strand swivels through,
relieving the
supercoiling. The nick is repaired by the enzyme as it releases.
Primase
An RNA polymerase that synthesizes a short RNA nucleotide to serve as a primer, giving a 3’OH for DNA polymerase to use as it begins a new polymer of DNA.
DNA ligase
Creates the final
phosphodiester bond that connects
Helicase
Using the energy of
ATP, its melts the hydrogen bonds of the
double stranded DNA, making the single stranded template available for
replication.
3’-5’ exonuclease activity
Removes incorrectly
added new nucleotide from the 3’ end of
a growing DNA polymer by sensing a mismatch with the template strand.
It breaks
the phosphodiester bond to the 5’ side of the nucleotide, releasing it.
5’-3’ exonuclease activity
Removes the RNA primer from
10. Describe one of the following DNA repair systems: nucleotide excision repair, base excision repair, or mismatch repair. (5 pts)
Refer to either Figure 6-21 (base excision repair), Figure 6-23 (nucleotide excision repair), or Figure 6-24 (mismatch repair).
11. What have we learned about the initiation of DNA replication at origins of replication from studies of yeast? (5 pts)
The yeast origin of replication, called an ARS, contains an 11-base pair consensus sequence, ACS, rich in A and T residues. The large Origin Recognition Complex (ORC) binds there and recruits several other initiation-related proteins to the origin region, including the MCM helicases that make the first single-stranded bubble to allow the formation of the replication fork.