1. The recognition sequence is CPyCGPuG. For positions 1,3,4 and 6 in this sequence
only 1 base out of four will lead to cutting. For positions 2 and 5 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)
2. 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)
3. 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.
4. The restriction patterns seen for the first four individuals on the Southern blot
represent the locations of 4 EcoRI restriction sites, two of which are variable
(polymorphic) in the human genome. For individuals 2 and 3, a single band is visible
using the radiolabeled probe, 5 kb in size for individual #2; 4 kb in size for individual #3.
These data indicate that a restriction site is present in some individuals (vertical line with
asterisk) located 1 kb internal to the sites defined by the 5 kb fragment.
In individuals 1 and 4, a 1.9 kb fragment is observed
in both individuals; they differ in the size of the
larger fragment, one of which is 3.1 kb (individual
#1) the other 2.1 kb (individual #4). These
fragments differ by 1 kb, and this indicates that a
second variable restriction site (present in these
individuals, but absent in individuals 2 and 3) is
located 1.9 kb internal from the other restriction site
defined by the 5 kb fragment. Note that the 4 kb
fragment is broken into two pieces, 2.1 and 1.9 kb in
size; the second variable site must be 2.1 kb away from
the first site, since we see a 3.1, not a 2.9 kb fragment
on blots.
5.0
*
1.0
4.0
*
1.0
*
2.1
1.9
The probe used in this experiment must span the DNA on either side of the second
variable site, but can not be complementary to the 1 kb region. If this were so, a 1 kb
band would be visible on the Southern blot.
Ans: (d)
5.
A
5.0
*
1.0
B
4.0
*
1.0
*
2.1
*
1.9
*
3.1
C
D
1.9
There are 4 possible genotypes based on the restriction fragments in this region of the
genome. These can be indicated patterns A-D (above). Individual #1 carries two
chromosomes with the restriction site pattern in D; individual #4 carries two
chromosomes with the restriction site pattern in C. The cross between the two can
therefore be represented as:
DD x CC  CD
This heterozygote would on Southern blot analysis give the pattern seen in individual #7
above; there would be 1.9 kb, 2.1 kb and 3.1 kb which would hybridize with the probe.
Ans: (a) only 1.
6. If oligo dT is used as a primer, it will hybridize to the 3’ end of the mRNA…
A synthetic oligo for amino terminal of protein would hybridize to the 5’ end of the
mRNA. Therefore, the order of transcription on the diagram represented would be right
to left, towards the 1 kb fragment. Ans: (a) true.
7. Although cells fated to become red blood cells lose their nuclei as they mature, they
must contain a nucleus at some time in their developmental history, which serves as the
template for making mRNA. This mRNA can be made into cDNA and was the source for
cloning the human  and  globin coding sequences.
Ans: False (b)
8. 5’-ATCTCTATCAGA-3’ strand being synthesized
3’-TAGAGATAGTCT-5’ template strand
Ans: (d) 5’-TCTGATAGAGAT-3’
9. Ans: (d) dUTP. dUTP is not normally included in sequencing reactions; uridine,
remember, is usually a component of RNA not DNA.
10. The last compound in the pathway will support the growth of the greatest number of
mutants. A mutant will be blocked if it can not metabolize precursors upstream in the
pathway. Following this logic:
F E  D  A  B  C  Z
3 2
1
6
4
5
Ans: (b) 2.