1. (100 x 106) base pairs/haploid genome x 3.4 Ao/base pair x 10-8 cm/Ao x 2 haploid
genomes/somatic cell = 6.8 cm/somatic cell
Ans: (b)
2. When cleaved by BglI, the plasmid is linearized within the tet gene. Insertion of
foreign DNA into the site will disrupt the tet gene coding sequence. Transformed cells
with therefore be tetracycline sensitive and kanamycin resistant.
Ans: (c)
3. If the DNA sequence is included in the clone, it can be amplified by specific primers
using PCR. Only those clones that contain the sequence of interest (i.e. capable of being
amplified by a primer set specific for the sequence) will have a ‘positive’ sign in the
chart. Therefore, clones that represent overlapping sequences in the genome can be
ordered:
PCR product
Clones containing
specific DNA
sequences
No PCR product
Specific
oligonucleotide
primers for PCR
Clone C: (1,5)
Clone D: (1,6)
Clone E: (4,6)
Clone F: (2,5)
Clone A: (3,4)
Clone B: (2,7)
Ans: (d)
(4,6,1,5)
(3,4,6,1,5,2,7)
4. The PCR reactions from sperm comes from a DNA source that is going to be
haploid…i.e. it can only contain an X or a Y. If the gene sequence is X-specific you
would therefore expect it to show up in only ½ of the sperm sampled. Gene Q shows up
in 11/20 sperm sampled.
Ans: (a)
5. Using the logic described above, an autosomal gene would be expected to show up in
every sperm. If an individual was homozygous, we would expect to see only a single
allele represented. Gene T fits these criteria.
Ans: (d)
6. The R and S genes are linked. If you examine the
pattern of inheritance, it is non-random- the R1 gene is
most likely to segregate with the S2 gene, and the R2 gene
is most likely to segregate with the S1 gene:
R1
S2
R2
S1
When examining the 20 individual sperm, this inheritance pattern is rearranged as a result
of a crossover in 4 out of 20 sperm. This is 20% recombinant sperm or 20 map units.
Ans: (d)
7. The DNA fingerprint pattern on the cigarette matches that under the victim’s
fingernails. It does not match the DNA of any of the suspects. Another person was in
the room and could have been the murderer- indeed, the DNA under the victim’s
fingernails makes this person suspect.
Ans: (e)
8. Regan has exactly the same fingerprint pattern as her sister. This indicates that they
must be monozygotic twins. They would therefore have the same birthday.
Ans: (e)
9. EDFACBZ
The last compound in the pathways supports the growth of the greatest number of
mutants- following that logic for the rest of the pathyway, B is the penultimate precursor
in the pathway…this is not given as a choice.
Ans: (e)
4
5
6
3 1
2
10. Mutant 5 can not grow on E, but, since it
is wild type for all other enzymes, can grow
EDFACBZ
on any of the other compounds in the
pathway. Following the same logic, a mutation in gene 2 would block the D to F
conversion. We would expect compound D to accumulate.
Ans: (c)