Chemistry 242--Biological Chemistry

Spring 2000 Midterm #1

Friday-Tuesday Feb. 18-22, 2000

Part I of this midterm is to be completed using only two 8.5 x 11 inch pages of your hand-written notes. Closed book, closed internet etc.

Time Limit 1 hour

Part II

Open notebooks, textbook chapters 1-6, lab reports, handouts for laboratory and lecture. Closed library, internet.

Time Limit 2 hours.

You will need to use a calculator and you may word process your answers. Time limits do not apply to time needed for printing answers. No consultations are permitted; if a question is unclear, please call me in my office (610-526-5107) or at home (610-527-8054) or restate the question in a clearer manner.

 

 

Good Skill!!

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

1 calorie = 4.18 Joule

 

 

 

 

 

A brief answer key is posted outside of Room 202.

 

Problem 1 _________________18

Problem 2 _________________12

Problem 3 _________________10

Problem 4 _________________30

Problem 5__________________30

TOTAL_________________100

Part I--Closed book.

Problem 1--Multiple Choice

Please circle the best answer.

1. A microgram of DNA dissolved in 1 mL of water would have a higher absorbance in the ultraviolet region than a microgram of lysozyme dissolved in 1 mL of water because

a. DNA has a higher proportion of aromatic bases than does lysozyme.

b. The average DNA molecule has a larger molecular weight or molar mass.

c. When DNA is dissolved in water it hydrolyzes into individual bases.

d. Lysozyme binds the Bradford reagent more strongly.

 

2. Watson and Crick’s model of DNA structure can easily be used to explain

a. how a retrovirus such as the HIV virus integrates itself into the host genome.

b. how DNA replicates.

c. the nature of the genetic code.

d. how DNA is damaged by ultraviolet light.

3. Glycine is frequently found in the turns of proteins because

a. it is the only amino acid which is not chiral.

b. its side chain can form hydrogen bonds.

c. the small size of its side chain reduces steric hindrance.

d. its peptide bond is not planar.

4. The incorporation of 2’3’ dideoxy adenosine into a growing (or replicating) DNA chain would result in

a. an A which would pair with a C instead of a T.

b. normal DNA.

c. a DNA with a branch point.

d. a truncated DNA chain.

5. A and B form double helices have different sized grooves because

a. G:C-rich sequences have a higher melting temperature.

b. Their base pairs have different numbers of hydrogen bonds.

c. The A-form helix has base pairs which are displaced away from the helical axis towards the minor groove.

    1. The B-form helix has sugars which are flat instead of puckered.
  1. All transfer RNAs have the same L-shaped three-dimensional shape because
    1. different amino acids are attached to their 3' ends.
    2. The anticodon must base pair with the messenger RNA.
    3. Their primary and secondary structures are different
    4. They share the common function of bringing an amino acid into the ribosome.

 

 

 

Problem 2 (closed book)

a) The normal pKa of the side chain of free histidine amino acid is 6.04. At pH = 4.00 what fraction of the histidines would be protonated?

 

 

 

 

 

 

 

 

 

 

 

b) An NMR spectroscopist conducts an experiment on a protein containing one histidine residue. She assigns several peaks to histidine protons and follows their chemical shifts as a function of pH and discovers that this histidine residue has a pKa of 7.5.

At pH = 7.5 in what fraction of protein molecules is this histidine protonated?

 

Problem 3 (closed book)

Linus Pauling was among the first chemists to point out the existence of two resonance structures for the peptide bond. Draw both resonance forms and discuss their implications for protein structure.

Problem 4 (Open book) (based on an article discussed in last year's lab)

Questions based on the RNA article "Rpp14 and Rpp 29, two protein subunits of human ribonuclease P".

  1. Examine the sequences of the two tryptic peptides. Is there evidence that they were indeed the result of hydrolysis by the protease, trypsin?

    2.  

     

     

     

  2. Based on the DNA sequences shown in Figure 1A, the authors predict a molecular weight of 13,821 g/mol for the smaller protein. Explain how this prediction is made and what assumptions must hold.

    3.  

     

     

     

     

  3. One of the proteins has a pI (isoelectric point) of 10.1. At neutral pH is this protein electrically positive, negative, or neutral? Given that the protein is a part of a protein-RNA complex does this make sense?

    4. More problem 4 (last year had more molecular biology emphasis)
  4. Restriction enzymes bind to and hydrolyze specific DNA sequences and were used in cloning. The Eco RI endonuclease recognizes the sequence GAATTC by using one GLU and two ARG sidechains to form 6 hydrogen bonds in the major groove of the GAA (and the GAA remains base paired). Knowing that ARG forms 2 H-bonds with G and that ARG and GLU can each form 2 H bonds with adjacent A's, draw a reasonable hydrogen bonding scheme.

    5.  

     

     

     

     

     

     

     

     

     

     

     

     

     

     

  5. Using your drawing in d) explain why E. coli bacteria are resistant to hydrolysis by Eco RI because the second adenine has a methyl group instead of an amino group.

 

 

Problem 5 (open book)

In the United States, glaucoma is one of the leading causes of blindness. Degeneration of the optic nerve is often accompanied by an increase in pressure inside the eye. The gene responsible for a type of glaucoma was recently identified by examining the sequence of the GLC1A gene in patients having familial glaucoma. The first step was to amplify the DNA sequence of interest according to the procedure below.

The polymerase chain reaction (PCR) was discussed in class and two primers complementary to the ends of the gene of interest are used. Thus, after each reaction two new strands are made from two self-complementary template strands.

"A 12.5 nanogram sample of each patient’s DNA was used for the template in an 8.35 ml PCR mixture containing 1.25 ml of 10X buffer (100 mM tris-HCl, pH 8.3, 500 mM KCl, 15 mM MgCl2); the deoxynucleotides dCTP, dATP, dGTP, and dTTP (300 mM for each); 1 pmol of each primer; and 0.25 units of Taq polymerase. Samples were denatured for 5 min. at 94°C and incubated for 35 cycles under the following conditions: 94°C for 30 seconds, 55 °C for 30 seconds, and 72 °C for 30 seconds in a DNA thermocycler."

Briefly explain the importance of each ingredient or step in the Polymerase Chain Reaction (PCR) procedure. Use drawings where appropriate.

a. dCTP, dATP, dGTP, and dTTP

 

 

 

b. template

 

 

 

c. primer

 

 

 

d. Taq polymerase

 

 

 

 

e. 94°C for 30 seconds

 

 

 

 

 

(still more Problem 5)

f. 55 °C for 30 seconds

 

 

 

 

 

 

g. Assuming that the molecular weight (molar mass) of one DNA base pair is 660 g/mol, how many moles of DNA were present in the initial patient samples? Each human has 2.9 x 109 base pairs of chromosomal DNA.

 

 

 

 

h. In principle, what mass of DNA could have been produced by 35 cycles of the PCR? Is this a realistic value? Why or why not?

 

 

 

 

 

 

The amplified DNA was sequenced and the amino acid sequences of the mutant proteins were deduced. The normal and mutant DNA (coding strand) sequences are shown below.

i. Fill in the blanks with the correct amino acids.

Normal Normal Amino Acid Mutant Mutant Amino Acid

1 TAC CAC

2 GGC GTC

3 CAG TAG

j. A conservative mutation is one where an amino acid is replaced with a chemically similar amino acid and many conservative mutations have only a small effect on the function of the protein. Are any of the changes above conservative changes?

 

 

 

 

k. Which of the changes above would be the most harmful? Why?