Laboratory: Unit 3: agarose gels & sequencing template preparation (page 56) Lecture: review & agarose gel electrophoresis In-Class Writing: practice exam (pages ) Read: pages 96, & xlii, section E Next Class: exam (class 13) Due Class 14: flow chart 4 & draft of editorial on GMO crops (pages 96 & xlii, section E)

DNA Preparation for Sequencing DNA must be free of contaminants. Submit samples in dH 2 O or Tris, not Tris/EDTA.

DNA Preparation for Sequencing Remove unincorporated dNTPs & primers (Qiagen kit). 10 µL of 50 ng/uL = 500 ng of amplicon in 10  L

DNA Preparation for Sequencing Estimate PCR product concentration by agarose gel electrophoresis. Compare to DNA Mass Ladder.

DNA Preparation for Sequencing Difficult to estimate PCR product concentration with conventional spectrophotometer. Use nanodrop spectrophotometer to estimate DNA concentration.

Sequencing Primers Primers (10 10 µM = 10 pmol/µL = 61.6 ng/  L for 8-27F primer (MW = 6161)

Low Mass Ladder (Invitrogen) 2% 3:1 agarose gel

ng: (10) bp: (100) 2% 3:1 agarose gel

5 kb 0.5 kb BamHI HindIII 2 kb 5.5 kb 3 kb 4.5 kb 3 kb 1.5 kb BamHIHindIII 5 kb

> 1 solution EcoRI coordinate 6/0

6/ pMB311 = 6 kb EcoRI

EcoRI cuts 3.5-kb PstI fragment  kb PstI sites 1 & 2.5 kb from coordinates 2.5 & 5.0 EcoRI PstI 1 kb2.5 kb PstI

6/ pMB311 = 6 kb EcoRI 2.5 PstI

3rd PstI site 0.5 kb from another PstI site (coordinate 4.5)

6/ pMB311 = 6 kb EcoRI 2.5 PstI 4.5

PstI cuts 3.8-kb SalI fragment  1.8, 0.5, 1.5 kb PstI cuts 2.2-kb SalI fragment  kb SalI 1.5 kb 0.5 kb PstI 1.8 kb SalI PstI SalI 0.5 kb 1.7 kb SalI PstI

coordinates 4.5 & 5.0  0.5-kb PstI fragment in 3.8 kb SalI fragment Place 3.8-kb SalI fragment on map in both possible orientations. Only one  kb PstI-SalI fragments  SalI coordinates 0.8 & 3.0

SalI 1.5 kb 0.5 kb PstI 1.8 kb SalI PstI EcoRI PstI 6/ PstI EcoRI 6/054 3 SalI PstI EcoRI 6/ kb SalI 1.8 kb 0.5 kb 1.7 kb0.5 kb kb

6/ pMB311 = 6 kb EcoRI SalI SalI PstI 4.5

20-nucleotide primer (MW = 6600) in 1 ml 5  l  495  l of water 260 nm = OD 260 unit = 33  g/ml for ssDNA What is concentration of undiluted primer stock?

MW = 6600  1M stock = 6600 g/l 1  M stock = 6600  g/l = 6.6  g/ml 5  l  495  l = 1/100 dilution OD 260 of diluted stock = 0.61 concentrated stock = 0.61 OD 260 x 100 x 33  g/ml/ OD 260 = 2013  g/ml 2013  g/ml x 1  M/6.6  g/ml = 305  M

How much must you dilute concentrated primer stock to make 10  M solution?

Have 305  M Want 10  M Divide what you want by what you have: 10  M/305  M = = 3.28/100

29 nmol/290  l = 0.1 nmol/  l = 100  mol/l = 100  M

25-nucleotide primer = 50% G+C & 100% complementary to template PCR reaction = 100 mM NaCl What is melting temperature (T m ) of duplex DNA between primer & template?

T m = 16.6 log [Na] (% G+C) /bp [Na] = molar salt concentration % G+C = whole number (50% = 50) bp = length of DNA:DNA hybrid

Tm = 16.6 log[0.1] x – 500/25 = 16.6 (-1) – 20 = 65.5 o C

From the results of this experiment, it was indicated that blue dextran (blue band) had a greater molecular weight than DNP-glycine (yellow band), as it was eluted first. Blue dextran (blue band) eluted first, indicating that it had a greater molecular weight than DNP-glycine (yellow band).

The procedures on pages of the lab book were generally followed with the following three deviations as listed below: The procedures on pages of the lab book were followed with three exceptions:

Sentence #1: “From the results …” “From this experiment…” “This experiment indicated that…” Sentence #2: do not end the sentence with “below.” (Day, page 193) “A preposition is a poor word to end a sentence with.”

Both the crude and purified PCR products were determined to be 560 base pairs in length. The crude and purified amplicons were 560 base pairs long.

It was estimated that the crude product was two times brighter than the ladder. The crude product produced a band twice as bright as the ladder.

Identification of the unknown environmental organism was identified as being Escherichia coli. The unknown organism isolated from the environment was Escherichia coli.

DNA was extracted as described for Agrobacterium above. DNA was extracted as described (reference).

Ends of restriction fragments produced by PstI cleavage contain four unpaired bases that can hydrogen bond to the complimentary bases on the end of another PstI fragment. Ends of PstI restriction fragments contain four unpaired bases that can form hydrogen bonds with the complementary bases of another PstI fragment.

Colonies that were yellow in color and white in color were seen. The colonies were either white or yellow.

A possible explanation for this result could be due to the fact that molecules larger then the largest pore size of the gel cannot diffuse into the gel pores. Molecules larger than the pores are excluded from the gel.

Proper uses of “that” and “which.” (Day, page 196) “Which”  “nonrestrictive” Introduce a nonessential clause. “That” introduces an essential clause.

“CetB mutants, which are tolerant to colicin, also have…” All are tolerant. “CetB mutants that are tolerant to colicin also have…” Some are tolerant.

Active versus passive voice (Day, 198-9) Active voice is usually better. S. aureus produced lactate. (active) Lactate was produced by S. aureus. (passive)

In methods, passive voice is often valid. (Day, page 67) Mice were injected with… (passive) We injected mice with … (active)

In unit 3, you: cultured unidentified bacteria, isolated genomic DNA, used this DNA as template for PCR.

Authors of the paper on bacterial diversity in Amazon isolated bacterial DNA directly from soil without culturing bacteria. Why did they use this approach?

They did not attempt to grow bacteria from soil in culture because ~99% of these species do not grow in culture using current methods.

List 3 parameters that affect Tm of primer-template duplex DNA. salt concentration length GC content number of mismatched bases