1. False. Bacterial plasmids used for propagation of recombinant DNA in E. coli do not
integrate into the genome. They replicate from their own origins of replication and are
often found at much higher numbers of copies than the bacterial chromosome.
2. The recognition sequence is PuAATTPy. For positions 2,3,4 and 5 in this sequence
only 1 base out of four will lead to cutting. For positions 1 and 6 in the sequence, two
bases out of 4 will lead to cutting. Therefore, the odds of having this exact sequence in a
random DNA molecule will be:
1/4 x1/2 x 1/4 x 1/4 x 1/2 x 1/4 = 1/1024; or it will cut once every 1024 base pairs.
Ans: 1024 bp (c)
3. The cut sites for the three enzymes are:
5’...GGATC C...3’
3’...C CTAGG...5’
5’...TCTAG A...3’
3’...A GATCT...5’
5’...AGATC T...3’
3’...T CTAGA...5’
The results of separating the two strands of DNA will leave a 4 base overhang…this 4
base overhang happens to be THE SAME for the enzymes BamHI and BglII:
5’...G
GATCC...3’
3’...CCTAG
G...5’
5’...T
CTAGA...3’
3’...AGATC
T...5’
5’...A
GATCT...3’
3’...TCTAG
A...5’
Therefore, the single strand overhangs for these two enzymes would allow them to base
pair with each other. Please note, however, that if we used these two enzymes for
cloning, we would not reconstitute either a BamHI or BglII site after ligation: e.g.
5’…AGATCC…3’
3’…TCTAGG…5’
Ans: (c)
4. The restriction patterns seen for the first six individuals on the Southern blot represent
the locations of 5 HindIII restriction sites, three of which are variable (polymorphic) in
the human genome. For individuals 3, 5 and 6, a single band is visible using the
radiolabeled probe, 6 kb in size for individual #3; 9 kb in size for individual #5 and 15 kb
in size for individual 6. These data indicate that the size of the largest fragment bound by
the probe is 15 kb. In addition, individuals 1, 2 and 4 contain a 2 kb fragment that
hybridizes, and the size of the larger fragments in these individuals would add to 6, 9 and
15 kb respectively. These data indicate that a restriction site is present in some
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
individuals (vertical line with
asterisk) located 2 kb internal to a
fixed site. In addition, two variable
sites located 4 kb, and 3 kb
15 kb
13 kb
from the variable 2 kb site
would explain the 6, 9 and 15 kb 9 kb
7 kb
patterns. If all three variable
6 kb
sites were missing (2 + 4 + 3 +
4 kb
6) the 15 kb fragment would be
expected to hybridize. Since
2 kb
individuals 1, 2 and 4 show two
bands, the probe must extend into
the 4 kb fragment. So the probe
used in this experiment must span the DNA on either side of the first variable site, but
can not be complementary to the 3 kb fragment. If this were so, a 3 kb band would be
visible on the Southern blot.
Ans: (c)
5. Individuals 7-15 represent individuals who are carrying two homolog chromosomes
which differ with regard to the variable sites present- i.e. they are carry homologs with
different polymorphisms. For example, individual 1 would be homozygous for two
homologs that contained the first two polymorphic sites (site three can’t be determined):
*
1)
2
3
4
6
Individual 5 would be homozygous for two homologs that were missing the first two sites
and contained the third site:
5)
2
4
3
6
The cross between the two can therefore be represented as:
1,1 x 5,5  1,5
This heterozygote would on Southern blot analysis give the pattern seen in individual #10
above; there would be a 2 kb, 4 kb and 9 kb fragment which would hybridize with the
probe.
Ans: (a) only 1.
6. The 720 nt probe hybridizes to the 2 kb fragment, and to other fragments containing
this sequence (2, 6 and 9 kb). If the 13 kb and 15 kb fragments hybridized, the carboxy
terminal end of the coding sequence would lie to the right of the 2 kb fragment, but these
fragments do not hybridize. The 720 nt probe must therefore lie within the 2 kb
fragment. Ans: (true)
7. The size of the coding sequence needed to code for 1600 amino acids would need to
be 1600 x 3 = 4800 nucleotides. However, the amino terminal coding sequence did NOT
hybridize to the 6 kb fragment; this suggests that the coding region is not contiguous, and
that there is an intron interrupting the coding sequence. Ans: True.
8. The TEMPLATE strand is the strand that is complementary and runs antiparallel to the
strand being synthesized. If the strand being synthesized is:
5’-AGCTGCACGCAT-3’
Then the template strand would be:
5’-ATGCGTGCAGCT-3’.
Ans: (d)
9. The A reaction must contain ALL of the dNTPs and ddATP. Ans: (e)
10. F→E→D→A→B→C→Z
3 2 1 6 4 5
The E to D conversion would be performed by enzyme 2. Ans: (b)