Problem Set 5 Answers

1.a. The frequency of cutting in a random DNA sequence for a given restriction enzyme is once per every 4ⁿ, where n is the number of bases in the restriction enzymes recognition sequence.  The “4” derives from the fact that there are four different possible nucleotides that may be inserted at any one position (G, A, T, or C).

Eco RI and Hin dIII have a six-base recognition site, so they will cut once per every 4⁶, or 4096 bases.

Hin dII has a five-base recognition site, so it will cut once per every 4⁵, or 1024 bases.

Hpa II has a four-base recognition site, so it will cut once per every  4⁴, or 256 bases.

b. To determine the number of sites, divide the number of bases in the DNA by the average length of the restriction fragments resulting from the given enzyme.   The size of l is 48,502 base pairs, so the number of expected sites are:

 2a. Cutting with Hin dIII will yield fragments of 1, 2, and 2.5 kilobases.  Cutting with Eco RI will yield fragments of 1.5, 2, and 3.5 kilobases.  Cutting with both Hin dIII and Pvu II will yield fragments of 1, 1.5, and 2 kilobases.

b. The cDNA in pBR322 will only consist of the two exons, which are flanked at the ends by EcoRI.  When cut by EcoRI, the cDNA will yield 0.5, 1.5, and 2 kilobase fragments.  Digesting with both EcoRI and HindIII, will yield 0.5, 1, and 1.5 kilobase fragments.

c. The 1 kb radiolabelled probe is from the second exon in the cDNA.  If hybridized to genomic DNA that has been digested to EcoRI, it will hybridize to the 3.5 kilobase fragment.

d. The radiolabelled probe derived from the wild-type cDNA will only be able to hybridize to the section of genomic DNA that contains the first exon.  When the genomic DNA is cut with HindIII, this fragment is 2.5 kilobases long.

e. Because the length of the PvuII restriction fragment in the mutant is the same as that of the wild-type, there does not appear to be any deletion between the two PvuII sites.  Therefore, the mutation is not the same as that in part d.  The mutation here is likely to be a point mutation.

3. a.

E                  H         M  H                                  E            

|_________|_____|__|________________|

         2.5          1.5   0.5              4.5    

b. There is at least one intron, because the radiolabelled cDNA did not hybridize to the 1.5 and 0.5 fragments, but did hybridize to the flanking fragments. Other introns may reside in the 2.5 and 4.5 fragments.

c. Two methods that can be used to labed cDNAs are primer extension and nick translation.  Primer extension uses an oligonucleotide primer and Klenow fragment to synthesize a labeled DNA strand.  In nick translation, a small amount of DNase is used to create nicks in the dDNA.  Labeled oligonucleotides are then added with DNA Polymerase I in order to fill in at the nicks.

4.a. The "sticky ends" resulting from digesting with the enzymes are the same, but the actual restriction sites are different.

b. After isolating the plasmid, digest it with both enzymes, purify the 4.5 kb fragment, self-ligate the fragment, transform the new plasmid into bacteria. 

c. There will be no BamHI or BglII sites.

d. You can cut with both enzymes again after ligation – this would ensure that only a plasmid lacking both sites would transform the bacteria.

5. Walk to a clone that overlaps the end of a translocation or deletion (this will show difference in Southern blots from organisms with and without the genetic abnormality). Use a clone from wild-type to pick up a junction fragment in the mutant strain.  Use this fragment to pick up a clone at the other end of the chromosomal abnormality from a wild library.

6.a. A restriction map of the fragment is shown below:

                      P                  H                             P         E            

|__________|////////////////////||_______________|_____|////////////////////|

         2 kb             2 kb                   3 kb              1 kb          2 kb      

b. Isolate mRNA and genomic DNA from an animal.  Run the mRNA out on a gel and transfer it to a blot (this is called Northern blotting).  Using primer extension or nick translation, label isolated genomic DNA and use it as a probe to hybridize to the blot.  The probe should hybridize to its corresponding mRNA, and the location of the probe on the blot should tell you the size of the mRNA.  

c. See above; the cDNA will map to areas designated by |/////////|

7.a. To isolate only DNA, use RNase (or base).  To isolate RNA, use Dnase.

b. Use a restriction enzyme, rerun the gel, and add up the band sizes; if more than one DNA, they should add up to greater than the original band.

c. One method to isolate a D. melanogaster homologue of the gene is to create a labeled probe from the cloned C.elegans gene and hybridize it to a library of D. melanogaster DNA.  Another method is to create primers from sequences in the C. elegans gene and use it to PCR the homologue out of D. melanogaster genomic DNA.

d. Run the mRNA out on a gel, transfer the RNA to a blot, and then hybridize the Northern blot with a labeled probe generated from the gene.

e. Let the wild-type progeny self-mate.  If the mutation exhibits incomplete penetrance, then the wild type progeny should yield some mutant progeny.

8. Reverse transcription is used to make cDNA from mRNA, and PCR is used to amplify it. 

9. a. i) Bam HI    ii) Hin dIII    iii) Eco RI    (iv) Hin dIII and Eco Ri?

       6               12                 9.5                  6

                   5                6                    4                   4

                   4                                    2.5                 3.5

                   3                                      2                  2.5

                                                                                 2

b. Only the 4 and 5 kb Bam HI fragments (which contain the exons) will hybridize on the Northern blots (to a 6 kb mRNA) and this will only be found in the liver lane.

c. Only a 2 kb and 3 kb band will be found on the blot; these result from the DNA/RNA hybrid regions that were not digested by the S1 nuclease, which degrades single-stranded nucleic acids. 

d. By eliminating the initial 6 kb HindIII region, regulatory elements needed for transcription have been omitted.

10. False, the opposite is true.

11. YACs (yeast artifical chromosomes) contain yeast ARS elements (for replication), telomere segments, and CEN elements (centromere segments); they can contain several hundred Kb of DNA.  Cosmids are modified plasmids that have the cos (cohesive ends) of phage lambda and plasmid sequences (for replication and drug selection); they can contain 20-40 Kb of DNA.