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BCH 5425 Molecular Biology and Biotechnology

Spring 1998

Exam I (100 Points)

1. (10 points) A bacteriophage is discovered on the planet Mars. It is observed to contain both DNA and protein. An experiment similar to that of Hershey and Chase is performed: progeny phage are isolated from bacterial cultures containing either ³²P labeled ATP or ³⁵S labeled methionine. These progeny phage were then used to infect bacteria. The following results were observed:

^{32P labeled phage:

Adsorbed phage shells 85%
Infected cells (prior to lysis) 15%
Lysed cell debris 15%
Progeny phage <1%

^{35S labeled phage:

Adsorbed phage shells 30%
Infected cells (prior to lysis) 70%
Lysed cell debris 40%
Progeny phage 30%

How would you interpret these results?
1. It would appear that for this life form protein is the genetic material, or
2. In this organism proteins contain phosphorous instead of sulfur, and nucleotides contain sulfur in their ribose backbones instead of phosphorous
2. (15 points) Draw the deoxy-dinucleotide 5' GT 3'. The 5' end has a monophosphate group and the 3' end has a hydroxyl group. Include the numbering for the bases and ribose groups (labeling of one of the ribose groups is sufficient). ALSO: draw the appropriate Watson and Crick base pair for the T base. Indicate what this base is, and show the hydrogen bonds (it is not necessary to provide numbering for this base).

3. (14 points) Match up the restriction enzymes on the left with characteristics on the right. More than one characteristic may apply for each enzyme. DON'T DRAW LINES FROM ENZYME TO CHARACTERISTIC. JUST LIST THE LETTERS AFTER THE ENZYME.
1. Alu I A,D A. Cuts leaving blunt ends
2. EcoR I B, E B. Cuts leaving 5' overhangs
3. Aat II C, E C. Cuts leaving 3' overhangs
4. Not I B, F D. "four cutter"
5. Nci I B, G, I E. "six cutter"
6. EcoO 109I B, H, I F. "eight cutter"
G. Cuts every 1024 basepairs on average
H. Cuts every 16.4 Kb on average
I. May have non-complementary ends
4. (15 points) Given the following DNA duplexes, diagram the products after treatment with the listed DNA modifying enzymes. Assume the overhangs are quite long. INDICATE CLEARLY WHAT IS GOING ON. ADD A SENTENCE IF NECESSARY.

E. coli DNA Pol I, dNTP's:

T4 DNA Polymerase, dNTP's:

Nuclease Bal 31:

Exonuclease III:

T4 DNA Polymerase, T4 polynucleotide kinase, T4 DNA ligase:

5. (15 points) Draw a diagram of one of the replication forks of the E. coli genome during replication. Assume that replication has already begun and that leading and lagging strand synthesis is well underway. Diagram how the system accomplishes leading and lagging strand synthesis. Show the unique topology which allows lagging strand synthesis. Show also all proteins which are required. Show the oligonucleotides which are present.

6. (5 points) Regarding DNA replication: What would be the consequence if PolI was absent? Diagram the results. Would the result be lethal to the organism?
UNRESOLVED OKASAKI FRAGMENTS ON LAGGING STRAND. LETHAL.
7. (10 points) A virus with a circular genome 6,940 base pairs in length finishes replication with a Linking number of 600. The host of the virus is a bacteria which lives in hot springs under high temperature and salt conditions. Under such conditions, the DNA prefers to adopt a conformation of 10.1 base pairs per turn of the helix. What supercoiling topology will the viral genome adopt to achieve this preferred conformation? (i.e. determine writhe under these conditions)
L = T + W
600 = 6940 bp(1 turn/10.1 bp) + W
W = -87.1
8. (5 points) Is the DNA "overwound" or "underwound"?
UNDERWOUND
9. (5 points) Given the above situation, if we wanted to treat the DNA with topoisomerases so as to eliminate writhe, which topoisomearase would we use?
TOPOISOMERASE I
10. (5 points) Examine the rubber duplex in class. What is the linking number?
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