BCH 5425 Molecular Biology and Biotechnology

Spring 1998

Exam III (100 Points)

 

1. The figure below represents an agarose gel. You are going to analyze a sample of pBR322 plasmid DNA on this gel. The sample contains DNA forms I, II and III.

Explain what the three forms are, and in lane ‘A’ diagram how the three forms will typically migrate (there is no ethidium bromide present when you run the gel)

In lane ‘B’ diagram how the sample will run if it has been digested with EcoR I restriction endonuclease prior to loading the sample (20 points)

 

 

2. A series of protein molecular weight standards are run on an acrylamide gel with the following results:

Molecular Mass (Kda)

Mr (cm)

92

1.53

68

1.64

43

1.84

31

2.01

22

2.23

14

2.62

An unknown protein is run on the same gel and has a Mr value of 1.6 cm.

What is the Molecular Mass of the unknown protein? (Note: plot this on graph paper) (20 points)

75 KDa

 

 

3. A 30 basepair DNA fragment is sequenced using dideoxy sequencing and the following sequencing primers:

The sequencing gels for each sequencing primer are as follows:

 

 What is the sequence of the original template? (20 points)

5’ GGTACGCCATTGAGCCCATAAGGATGCAAA 3’

 

4. In a PCR experiment 1 x 10-9 moles of a duplex DNA template is amplified 22 PCR cycles. At the end of the experiment how many moles of each of the following would you expect? (20 points)

A. Parental oligonucleotides

1 x 10-9 moles each

B. Oligonucleotides with undefined 3’ ends

22 x 10-9 moles each

C. Oligonucleotides with defined 3’ ends

4.19 x 10-3 moles each

D. Assume that parental oligonucleotides and oligonucleotides with undefined 3’ ends will form duplexes with oligonucleotides with defined 3’ ends. If the remaining oligonucleotides with defined 3’ ends form duplexes, what is the concentration of this duplex after the 22 cycles?

 Essentially 4.19 x 10-3 moles

 

5. For the following PCR primer determine values for Tm, Tms and Tmp values in 0.1M NaCl (20 points)

5’ GCGTGACGATAGGACACTGA 3’

 

Tm:

Tm = [(number of A+T residues) x 2 °C] + [(number of G+C residues) x 4 °C] = 62°

 

Tms:

Tms = 81.5 + 16.6(log10[J+]) + 0.41(%G+C) - (600/l) = 52.5°

 

Tmp:

Tmp = 22 + 1.46([2 x (G+C)] + (A+T)) = 63.1°