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1 Genetics Lab practices EXERCISES Exercise 1 The figure on the right shows a photograph of an agarose gel. The 2nd lane shows the result of the digestion of 500 ng of phage genomic DNA (total size: 48,502 bp) with PstI. Sizes of resulting fragments are shown in bp. The 1st lane shows the result of digestion of plasmid X (size: 20 kb) with BstYI. Keeping in mind that the mobility of a DNA fragment during electrophoresis is inversely proportional to the logarithm of its molecular size, calculate the length (bp) of the six fragments produced by BstYI digestion of plasmid X. To do this, first represent log of standard fragment size (on the y-axis, logarithmic scale) vs. the distance migrated (in mm, on the x-axis). Use the standard curve obtained to calculate the size of six fragments present in the problem lane. Knowing that the smallest band from plasmid X digestion and the 805 bp PstI band have the same intensity, calculate the total amount of plasmid X DNA used in the digestion. Exercise 2 We want to cleave 500 ng of a DNA molecule with HindIII restriction enzyme in a 20 μl final volume reaction. Design the reaction and fill in the table with missing volumes for each reagent, taking into account the stock concentrations and that we intend to use 3 U enzyme/μg of DNA. Molecular Techniques Third-year Biology 2015-16 λ PstI 10 2 10 3 10 4 10 5

cuestionario: análisis de fragmentos de restricción: determinación

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Page 1: cuestionario: análisis de fragmentos de restricción: determinación

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Genetics Lab practices

EXERCISES

Exercise 1 The figure on the right shows a photograph of an agarose gel. The 2nd lane shows the result of the digestion of 500 ng of phage genomic DNA (total size: 48,502 bp) with PstI. Sizes of resulting fragments are shown in bp. The 1st lane shows the result of digestion of plasmid X (size: 20 kb) with BstYI. Keeping in mind that the mobility of a DNA fragment during electrophoresis is inversely proportional to the logarithm of its molecular size, calculate the length (bp) of the six fragments produced by BstYI digestion of plasmid X. To do this, first represent log of standard fragment size (on the y-axis, logarithmic scale) vs. the distance migrated (in mm, on the x-axis). Use the standard curve obtained to calculate the size of six fragments present in the problem lane.

Knowing that the smallest band from plasmid X digestion and the 805 bp PstI band have the same intensity, calculate the total amount of plasmid X DNA used in the digestion.

Exercise 2 We want to cleave 500 ng of a DNA molecule with HindIII restriction enzyme in a 20 µl final volume reaction. Design the reaction and fill in the table with missing volumes for each reagent, taking into account the stock concentrations and that we intend to use 3 U enzyme/µg of DNA.

Molecular Techniques

Third-year Biology

2015-16

λ PstI

102

103

104

105

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PstI

pUAG211 pUAG222 pBGT9

Reagent (concn.) Vol. (µl)

DNA (100 ng/µl)

Restriction buffer (10)

BSA (5)

HindIII (5U/μl)

mQ H2O

Final volume 20 μl

Exercise 3 The gel on the right shows the BamHI digestion products from plasmids pUAG211 (5 kb and 1 kb bands) and pUAG222 (5 kb and 700 bp bands). We want to obtain a recombinant DNA molecule by cloning the small product of pUAG211 digestion (insert) into the big product of pUAG222 digestion (vector). To accomplish this, we excise the two gel bands, purify the DNA they contain and dissolve it in 10 l of TE. Molecular weight size standards used are PstI-digested λ genomic DNA (see exercise 1). Assuming we have used 100 ng for each plasmid digestion and an efficiency of 100% of DNA recovery after purification (even though it is not the case in the real world!), calculate the amount and concentration of the final vector and insert preparations.

To favour insert-vector ligation, we will use a vector: insert ratio of 1:5. Fill in the table with the concentrations and missing volume values (in µl), taking into account that we want to use 100 U of T4 DNA ligase in a final reaction volume of 20 µl.

Reagent (concn.) Vol. (µl)

Vector (__ ng/ μl) 1

Insert (__ ng/μl)

Ligation buffer (5)

T4 DNA ligase (50U/μl)

mQ H2O

Final volume 20 μl

Exercise 4 To separate nucleic acid molecules (DNA or RNA) by electrophoresis, we normally use agarose gels (hundreds of bp or kb fragments), in a 0.6-2% concentration range. Nevertheless, for very small molecules (tRNAs, oligonucleotides, etc.) we use polyacrylamide gels, in a 5-20% concentration range. The electrophoresis buffer in both cases is TBE (Tris/Borate/EDTA). The main difference between both types of gel is that agarose is melted in a microwave oven and forms a gel when it cools, whereas acrylamide polymerizes (to form polyacrylamide) when ammonium persulfate (APS) and TEMED are added. In both cases, we can add ethidium bromide (EtBr) to the gel during its preparation, which avoids subsequent gel staining. Fill in the following tables with volume or weight values for the different components in the indicated units:

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Exercise 5 We want to construct a recombinant DNA molecule by joining the insert present in the “I” lane (1.2 kb) and the vector present in the “V” lane (5.5 kb). The photograph on the right was obtained after loading 1 l of both insert and vector preparations for their quantification, together with 400 ng of a λ DNA EcoRI+HindIII digest as a molecular-weight standard set. Fill in the table with missing values:

Note: Sizes of λ DNA EcoRI+HindIII digestion products are: 21226, 5148-4973-4268-3530, 2027-1904, 1584, 1375, 947, 831 and 564 bp.

Exercise 6 Most diseases are originated by point mutations in specific genes, some of which can also result in the appearance with Mendelian inheritance of a site for a particular restriction enzyme. This makes genetic analysis easier, since the mutation can be detected by PCR followed by digestion with such enzyme.

Cystic fibrosis (CF) or mucoviscidosis is caused by the presence of mutations in the gene coding for the protein cystic fibrosis transmembrane conductance regulator (CFTR), residing in chromosome 7, that encodes the chloride ion (Cl) channel located in the plasma membrane. CF is an autosomal recessive disorder: disease appears when both copies of the gene are mutated, in which case no active protein is produced. Individuals with a single working allele are carriers and do not exhibit CF symptoms.

Here, we describe the procedure to detect two different mutations in CFTR gene: Arg1162Ter (R1162X), which originates a premature stop codon (nonsense mutation), and Asn1303Lys (N1303K), which originates a single amino acid change:

aas: S V S R V F K N L D WT sequence: TCT GTG AGC CGA GTG TTT AAG AAC TTA GAC R1162X: TCT GTG AGC TGA GTG TTT AAG AAC TTA GAC N1303K: TCT GTG AGC CGA GTG TTT AAG AAG TTA GAC

Reagent (concn.) Vol. (µl)

Vector (___ ng/ μl) 1

Insert (___ ng/μl)

Ligation buffer (5)

T4 DNA ligase (50U/μl) 2

mQ H2O

Final volume 20 μl

To prepare an agarose gel:

Reagent Final Concn

TBE 50 ml 1

Agarose g 0.7%

H2O ml ----

EtBr 10 mg/ml μl 0.5 μg/ml

Final volume 50 ml

To prepare a polyacrylamide gel:

Reagent Final Concn

TBE 50 ml 1

Acrylamide/Bisacryl. 40:1 ml 12%

H2O ml ----

APS 10% ml 0.1%

TEMED μl 0.1%

Final volume 100 ml

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Both mutations are located in different regions of the CFTR gene, which therefore need to be analysed independently by PCR using two pairs of primers. The 1st PCR (for the R1162X mutation) yields a 377 bp product, whereas the 2nd PCR (for the N1303K mutation) yields a 188 bp product. Both PCR products are purified and digested with restriction enzyme DdeI, which recognises the sequence 5’-CTNAG-3’. The resulting products are analysed by electrophoresis in polyacrylamide gels. Positions of DdeI sites in wild-type and mutant alleles are indicated with asterisks in the following figure:

Draw the expected band pattern for genetic analysis made on:

A healthy non-carrier individual (+/+ genotype).

An R1162X carrier (R/+ genotype).

A N1303K carrier (N/+ genotype).

An R1162X homozygous patient (R/R genotype).

A N1303K homozygous patient (N/N genotype).

The following pedigree represents a family with an affected individual. The gel shows the results of the molecular analysis made for detecting the R1162X and N1303K mutations in all the members of the family. Figure out the genotype for each member of the family and indicate it on the pedigree.

I-2 I-1

II-1 II-2 II:3

Mutation R1162X

gcccgacaaataacaagtga agccgagtg ctcag tcaataagtcctggccagag

gcccgacaaataacaagtga ctcag tcaataagtcctggccagagagctgagtg

275 pb 102 pb

102 pb130 pb145 pb

*

WT allele

Mutant allele (R)

Mutation N1303K

ggtaagtacatgggtgtttc aagaacttagaccctatgaacagtgga

ggtaagtacatgggtgtttc aagaagttagaccctatgaacagtgga*

167 pb 21 pb

178 pb

WT allele

Mutant allele (N)

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Exercise 7 The human YFG gene contains, at both ends, restriction sites for the enzymes BamHI, EcoRI, SalI, PstI and HindIII. We want to clone it in vector pJMN2 (see map), which harbours a replication origin (ori) that functions in E. coli and resistance genes for ampicillin (Ap) and tetracycline (Tc). Select the more appropriate restriction enzyme for this cloning and give a short explanation.

Fill in the table indicating if E. coli clones with pJMN2 or its pJMN2-YFG derivative will grow (+) or not (–) in media containing ampicillin or tetracycline.

If pJMN2 has a 60% A+T content and we cleave it with TaqI, whose restriction site is 5’-TCGA-3’, how many fragments would be produced approximately? How many fragments would you expect for a lineal DNA molecule with characteristics similar to pJMN2? Growth +Ap +Tc

Without plasmid

pJMN2

pJMN2-TGF

(5.0 kb)

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ANSWER SHEET. TESTS

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