Key – HW3
BIOL 417 – Fall 2012
Guided Activity I
1.
1
1873
HhaI PstI
XbaI
2.
Lane
1
2
3
4
5
6
7
1
RE used
EcoRI
XbaI
PstI
XbaI & PstI
N/A
N/A
N/A
2
3
Length (bp)
1873
527 & 1346
478 & 1395
478, 527 & 865
N/A
N/A
N/A
4
5
6
7
2000
1700
1400
800
3.
400
1
2
3
4
5
6
7
2000
1700
1400
800
400
2b and 3b are shown in the gel diagram above.
Note, in the gel diagram above, the increased thickness of the ~800bp band is not meaningful.
The bottom band in Lane 5 would have run off the gel, since it is less than 50bp in length and
the 400 bp marker is already near the bottom of the gel.
3a. The alternative map would lack the HhaI site, because it is not part of the digest. Also, the XbaI
would be just to the right of the PstI site, and would not be present in the location shown for #1 above.
1
1873
PstI
XbaI
GUIDED ACTIVITY II: Using the computer to map and locate the exons in the chicken ovalbumin
gene.
II.1.
On a separate sheet of paper, draw a linear restriction endonuclease map to scale, using your
results from NEBcutter for the chicken ovalbumin gene (not the cDNA). Use a scale of 1 cm =
500 bp.
a.
For the chicken ovalbumin mRNA, the line was:
1873/500 = 3.746 cm long.
For the gene, the line will be:
9206/ 500 = 18.412cm long.
b.
Label the 5’ end of the map “1” for the beginning nucleotide number and label the 3’ end
with the appropriate ending nucleotide number.
c.
On the map, use your ruler to position the cut sites for each of the following enzymes, as
appropriate according to your printouts.
Bam HI, EcoRI, HhaI, Hind III, PstI, XbaI, NotI
d.
You should have five Eco RI sites on the genomic map, resulting in six fragments.
Number those six fragments 1-6, from left to right. Remember to number the fragments,
not the cut sites.
See attached.
II.2a and b.
III.1.
I hypothesize that the EcoRI and Hind III sites are within introns.
In Lane 1 of the diagram below, sketch the positions in the gel at which you would expect EcoRI
fragments 1-6 to be located if you were to digest complete chicken genomic DNA. (Remember
that if you were actually to run a gel of digested chicken genomic DNA, it would appear as a
continuous smear because there would be so many other fragments in addition to the ones
containing the sequence of the ovalbumin gene.) Remember that if you were to actually digest
chicken genomic DNA, you would not know the exact positions of Fragments 1 and 6 based on
the sequence information you have, because you do not know the location of the next EcoRI site
to either side of the mapped region. Put each of these two fragments in any logical place.
Reproduce the diagram you made in Lane 1 in Lanes 2-4. At this point, Lanes 1-4 should look
identical. In each lane, as specified below, circle the fragments that would be positive after
hybridization with the cDNA probe mentioned. Use the table you filled out to help you.
Lane 1 – the full-length cDNA
Lane 2 – the cDNA fragment 5’ of the Pst I site
Lane 3 – the cDNA fragment located between the PstI and Xba I sites
Lane 4 – the cDNA fragment located 3’ of the Xba I site.
1
10,000 bp
7,0001
4,000
1,000
250
Analyses of Southern Data
Imagine that you are one of the early investigators of the ovalbumin cDNA and gene. You have
restriction enzyme and gel electrophoretic analysis of ovalbumin cDNA, and you have Southern blotting
data of chicken genomic DNA probed with the ovalbumin cDNA and its fragments, but you have no
sequence yet. In other words, you have only the experimental information from the gel electrophoresis of
the cDNA from the third homework assignment and the Southern blotting and probing of the genomic
DNA as you did it on the previous page.
III.2.
Explain why the data shown in Lane 1 of the Southern blot and the data from restriction enzyme
digests and gel electrophoretic analysis of the ovalbumin cDNA (last time’s work) would lead you
to make the hypothesis that the ovalbumin gene must contain intervening sequences.
The results suggest intervening sequences since the gel electrophoresis for the cDNA only contained one
fragment after attempted digestion with EcoRI. The fragment was 1873bp, i.e., showed no
evidence of a cut. However, the Southern blot contained 4 hybridization-positive fragments
varying in lengths. With this discrepancy there may be more sequences in the gene than in the
cDNA.
III.3.
Explain how the results in Lanes 2-4 of the Southern blot further support the hypothesis.
Lanes 2-4 further support this by the hybridization of the cDNA.
Notice that each of the 3 probes from the cDNA (left, middle, right), hybridizes to a different subset of
genomic fragments on the blot. This can happen only if different parts of the cDNA sequence appear in
different EcoRI fragments. This can happen only if there are EcoRI cut sites located in sequences that
intervene between different parts of the sequence that appear in the cDNA.
III.4.
Explain in words and with a diagram how your cDNA RE map and your Southern results would
together allow you to partially determine the order of the genomic ovalbumin restriction digest
fragments in the gene without knowing sequence. Note that the answer to this question requires
absolutely no knowledge of sequence!
The 1731+ bp length should go first since it’s the only fragment that hybridized with sequence found
upstream of the PstI site in the cDNA and with no other cDNA fragment. Next would follow the
2362 bp fragment, since not only did it hybridize with the cDNA sequence upstream of PstI, but it
also hybridized with the cDNA fragment found between PstI and XbaI. The next sequence is the
1806bp which hybridized with the cDNA fragment from between the PstI and XbaI sites, but NOT
with the fragment from downstream of XbaI. And finally the 2363+ bp sequence, which
hybridized only with the cDNA sequence downstream of XbaI.