BCH5425 Molecular Biology and Biotechnology

Spring 1999

 

Exam I

Friday February 29

 

1. In the follow diagram is an experiment like that which was performed by O.T. Avery. Answer the following questions: (8 points total)

a) What is the fate each mouse?

Mouse A will most likely have a virulent infection and die.

Mouse B will most likely not become infected.

 

b) Will viable Pneumococcus be isolated from each mouse? If so, what type will they be, and what will be their physical appearance?

Viable bacteria will be isolated from mouse A and they will be type I and appear smooth.

No viable bacteria will be isolated from mouse B.

 

 

2. Draw the structure of dideoxy cytidine diphosphate. Include the numbering for all atoms of the base, ribose and phosphate(s). (8 points)

3. Draw the deoxydinucleotide: 5' GT 3' Include a 5' monophosphate group. (8 points)

 

 

 

4. If might be possible for the two purine bases, A and G to hydrogen bond in a similar fashion as an A T base pair. However, this arrangement would result in a DNA molecule with structural features or consequences that would disagree with two facts known about DNA. Draw the hydrogen bond arrangement for an A-G base pairing (5 points). List the two facts about the DNA molecule that would not agree with such a basepairing? (5 points)

 

 

1. This would predict a duplex with a width greater than a purine/pyrimidine base pairing, i.e. 20Å (x-ray data of Rosalind Franklin)

2. This would violate Chargaff's law of G=C and A=T from analysis of DNA from various species

 

5. Restriction Enzymes:

 

Alu I

Bfa I

Nci I

EcoR I

Hae II

EcoO109 I

Bgl I

BsaH I

Aat II

Bpm I

Not I

Bsm I

 

Answer the following by providing one example for each: (12 points)

 

a. This enzyme would be most useful for isolating large DNA fragments

Not I

 

b. This enzyme leaves blunt ends

Alu I

 

c. This enzyme cuts once about every 1000 base pairs of DNA

Nci I

 

d. The fragments from this enzyme digest cannot be made blunt by "filling in" with DNA polymerase

Hae II, Aat II, Bpm I, Bsm I

 

e. Some fragments from this enzyme digest may not be able to re-ligate with each other

Nci I, EcoO109 I, Bgl I, Bpm I, Bsm I

 

f. This enzyme is useful for Restriction Fragment Length Polymorphism (RFLP) analysis of DNA

Alu I, Bfa I

 

6. For the following restriction endonuclease recognition sequences indicate which ones could potentially be protected by dam methylase, dcm methylase, or neither (12 points)

 

Enzyme

Sequence

Protected by ?

Nar I

GGCGCC

dcm

Sca I

AGTACT

neither

Bcl I

TGATCA

dam

EcoR I

GAATTC

neither

Msc I

TGGCCA

dcm

Mbo II

GAAGA

dam

 

 

7. For the following oligonucleotides indicate the expected product after treatment with the following DNA modifying enzymes. Indicate clearly if an end is blunt or ragged after treatment. Note: all overhangs are considered to be of indeterminate length. (15 points)

 

 

a. E. coli DNA pol I, dNTP's

 

b. Nuclease Bal-31

 

 

c. Mung Bean Nuclease

 

8. For the following two DNA fragments: indicate how you would treat the DNA to be able to ligate fragment B to A, and at the same time minimize self-ligation of fragment A. Note: all overhangs are considered to be of indeterminate length (10 points)

 

 

Fragment A: Treat with mung bean nuclease to produce blunt ends 3' ends, treat with calf intestinal phosphatase to remove 5' phosphate groups (to minimize self-ligation)

Fragment B: Fill in with polymerase (T4 DNA, or Klenow, etc.), then phosphorylate with T4 polynucleotide kinase.

Finally, combine fragments A and B in the presence of T4 DNA ligase (will ligate blunt end DNA duplexes) and ATP.

 

 

9. Draw a replication fork during DNA replication showing the detail of lagging strand synthesis. Indicate and identify all enzymes and transient structures that are required to produce a continuous oligonucleotide on the lagging strand. Note: there is no need to diagram or describe how replication is initiated, or show the leading strand synthesis. (9 points)

 

10. A certain plasmid contains 6,324 basepairs. At physiological pH and NaCl it prefers to adopt 10.6 bp/turn. The plasmid is observed to have a writhe value of +19.4.

a) What is the value for the Twist and Linkage parameters? (4 points)

 

Twist = 6324/10.6 = 596.6

L = 596.6 + 19.4 = 616

 

b) The DNA is placed into a solution of higher NaCl where it prefers to adopt 11.1 base pairs per turn. What are the new values for the Twist and Writhe parameters? (4 points)

 

Twist = 6324/11.1 = 569.7

W = L - T = 616 - 569.7 = 46.3