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Summary BMS75 Advanced Tools in Molecular Biology - notes €6,99
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Summary BMS75 Advanced Tools in Molecular Biology - notes

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All my notes from the course BMS75; including lectures and self study assignments

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  • 15 september 2021
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BMS75 Advanced tool in molecular biology


Self study assignment 1

1. Plasmids and phages are the two most commonly used vectors. What are the
characteristics of both vectors in terms of size, possible size of insert,
structure, replication and transformation?


Plasmids Phages

size 1,2-3 kb (in the radboud cell up to 53 kb
department mostly 5-7kb)

possible size of insert few base pairs up to 20 kb up to 25 kb

structure circular, double stranded Virion consists of a head,
DNA. Separate from a cell’s which contains the phage
chromosomal DNA DNA genome; and a tail,
which functions in infecting
E. coli host cells.

replication extrachromosomal extrachromosomal and
packed into phages


transformation chemical (heat shock) or phage DNA packaged into
physical (electroporation) bacteriophage particles in
vitro. These can then be
used to infect the bacteria.
*kb = kilobase, unit of measurement of DNA/RNA size (=1000 base pairs)

2. Name three important elements in a plasmid to function as a vector for cloning
purposes and give the function of those elements.
- a replication origin; to replicate in a bacterial or eukaryotic host cell
- a marker that permits selection, usually a drug-resistance gene (resistance gene;
used for selection of positive bacteria)
- Promoter; to drive the expression of open reading frames in bacteria or eukaryotic
cells

3. Recombinant DNA technology makes use of many different enzymes that
modify DNA in one way or another. Name the four classes of enzymes and
shortly specify their function. Give two enzyme names for each class.
The basis of recombinant DNA technology is the ability to manipulate DNA molecules in the
test tube.
1. DNA polymerases; enzymes that synthesize new polynucleotides complementary to an
existing DNA or RNA template.
- DNA polymerase I; unmodified E. coli enzyme
- Reverse transcriptase; RNA-dependent DNA polymerase
- * lecture

, - Taq doesn’t do a proper job since it has no proofreading → used for
analysis
- Pfu however, has proofreading and is a better option for production
purposes, clone purposes etc.
- Q5
2. Nucleases; degrade DNA molecules by breaking the phosphodiester bonds that link one
nucleotide to the next.
- Restriction endonucleases; binds to DNA molecules at a specific seq and
makes a double-stranded cut at or near that seq. Because of the specificity
this can be predicted
- 3 types exist of which type II is most used cause it always cuts at the
same place (unlike the others). for example Type II enzyme EcoRI
- Type I; enzymes recognize DNA seq. but cut the DNA in
random sites that can be as far as 1000 base pairs away from
the recognition site
- Type III; enzymes recognize seq. but cut at a different location
close to (within 25 bp) of the recognition site.
- Most bacteria use restriction enzymes as defense against
bacteriophages
- restriction enzymes prevent the replication of the phage by cleaving its
DNA at specific sites.
- Host DNA is protected by methylases which add methyl groups to
adenine or cytosine bases within the recognition site thereby
modifying the site and protecting the DNA.
- other examples; BamHI, Hindlll
- SI nuclease; endonuclease specific for single-stranded DNA and RNA
3. Ligases; join DNA molecules together by synthesizing phosphodiester bonds between
nucleotides at the ends of two different molecules, or at the two ends of a single molecule.
Only joins ends when they contain 5’phosphate groups and if they ‘fit’

, - DNA ligase; obtained from E. coli cells infected with T4 bacteriophage.
- examples: T4, Taq, T7
4. End-modification enzymes; make changes to the ends of DNA molecules, adding an
important dimension to the design of ligation experiments, and providing one means of
labeling DNA molecules with radioactive and other markers.
- Klenow (DNA polymerase fragment); makes 5’ overhangs blunt by filling in
the missing nucleotides
- Shrimp Alkaline Phosphatase; removes phosphate group at the 5’ end of
DNA strands, preventing ligation of these fragments.
- T4 Polynucleotide Kinase; adds (labelled) phosphate group to the 5’ end of
DNA strands, enabling detection or ligation.

4. What is the main difference between endonucleases and exonucleases?
- exonucleases; removing nucleotides from the ends of DNA and/or RNA
- endonucleases; making cuts at internal phosphodiester bonds

5. What are the three possible ends that endonucleases may create after cutting
DNA. Pick one enzyme for each possibility and draw the ends.
- blunt




- AluI;
- Sticky 5’ overhang
- Eco RI;




- sticky 3’ overhang
- Kpn1;




6. What are isoschizomers and neoschizomers. Browse through the New England
Biolabs (NEB) website and give and draw an example of each.
Isoschizomers and neoschizomers is a way to classify restriction enzymes.
- Isoschizomers; restriction enzymes that have the same recognition seq. and the
same specificity. However, they may differ in site preferences, reaction conditions,
methylation sensitivity, and star activity.
- i.e. Xmal - TspMI

, - Neoschizomers; recognize the same nucleotide seq. but cleave DNA at different
positions.
- i.e. SmaI and Xmal, which both recognize the same nucleotides but cleave
them differently and thus generate different types of ends

7. The enzyme BamH-I recognizes 5'-GGATCC-3' and creates a 5'-GATC end. Bgl-
II recognizes 5'-AGATCT-3' and also creates a 5'-GATC end. Can a BamH-I
fragment be ligated to a Bgl-II fragment? If yes, can BamH-I or Bgl-II still cut the
newly formed fusion DNA?
BamH-I 5’-GGATCC-3’ → 5’-GATCC-3’
Bgl-II 5’-AGATCT-3’ → 5’-GATCT-3’

If you only want to fuse the fragments that are made which are similar, then yes, they can be
ligated:

5’ G A T C 3’
3’ C T A G 5’

Can BamH-I or BgI-II still cut the newly formed fusion DNA? → No, both BamHi and BgIII
cannot recognize the newly formed DNA fragment.

8. The X-Gal selection or blue-white screen can be used to select for bacteria that
contain the plasmid with the insert of interest. Give a short explanation of this
selection system.
X-Gal (= 5-bromo-4-chloro-3-indolyl-β-D-galactopyranoside), is a compound used in a
histochemical test to see functional presence of β-galactosidase molecules in the cells. This
compound is converted into a blue product by β-galactosidase. The gene lacZ is part of the
E.coli gene that forms the enzyme. The lacZ gene contains several restriction sites,
therefore DNA can be inserted here, and when this happens, functional β-galactosidase is
lost.

Therefore a distinction can be made between recombinants and non-recombinants by
plating transformed cells onto agar containing the antibiotic for which there are resistance
(for instance ampicillin) and X-gal.

→ only the antibiotic resistant cells will survive and the white colonies indicate cells without
β-galactosidase function and therefore you can tell that these are the recombinants.

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