Answers to Quiz 4
Spring 2013
1. see http://news.sciencemag.org/scienceinsider/2013/04/can-human-genes-bepatented.html?ref=em. The correct answer is a): Myriad Genetics, of Salt Lake City,
currently holds patent rights to the BRCA1 and BRCA2 genes, which is being
contested. The Court of Appeals ruled in favor of Myriad’s claims. Eric Lander
argued against these claims in a ‘friend of the court’ brief.
2. see http://news.sciencemag.org/sciencenow/2013/04/living-fossil-gets-itsgenome-se.html?ref=em. The correct answer is b): the coelacanth genome does
appear to be evolving slower within the protein-coding regions than ray-finned
fishes. This may be a consequence of decreased selective pressure in its
environment. There are two species of coelacanths with different geographical
distributions, discovered over the last 75 years. Although it is closely related to
tetrapods, genomic analysis indicates that extant lungfish are closer.
3. see http://www.fasebj.org/content/8/2/201.long and
http://www.nytimes.com/2013/04/18/us/politics/officials-intercept-suspiciousletter-sent-to-obama.html. The correct answer is c): the compound leads to a
depurination of an adenine nucleotide in ribosomal RNA and blocks translation. It is
produced from the castor bean, and since the anthrax attacks in the early 1990s,
Capitol and White House mail is screened at a remote facility outside Washington
before delivery.
4. Ans. (c). DNA gyrase is a topoisomerase; it is responsible for introducing negative
supercoils in DNA and relieving the torsional stress incurred when positive
supercoils are introduced into the DNA downstream of helicase action during DNA
replication.
5. Ans: (a). The pattern that is observed
here shows no 1.6 kb fragment
identified by the probe- therefore the
region hybridizing with the DNA does
NOT overlap into the 1.6 kb region of the
genome, and (c), (d), and (e) are not
possible answers. We do, however, see a
2,3, 4.1, 3.9 and 8 kb fragments in these
individuals, indicating a variable
restriction sites at both internal
locations. Answer (b) can’t be correct,
because we see a 3.9 kb fragment in
some individuals. Thus, only (a) would produce the pattern seen.
6. Ans: (e). None of the above; these individuals are homozygous- both maternal and
paternally derived chromosomes contain the same restriction sites for this enzyme
in for this region- it is the same RFLP pattern for both.
7. 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…
STS marker
BAC Clone
1
2
3
4
5
6
7
A
-
-
-
+
-
+
-
Clone A: contains 4,6
B
-
+
-
-
+
-
-
Clone B: contains 2,5
C
+
-
-
+
-
-
-
Clone C: contains 1,4
D
-
+
-
-
+
+
-
Clone D: contains 2,5,6
E
-
+
-
+
-
+
-
Clone E: contains 2,4,6
F
-
+
+
-
+
-
-
Clone F: contains 2,3,5
G
-
-
+
-
+
-
+
Clone G: contains 3,5,7
Clones A and C: indicates: (146) order
Above with clone E indicates: (1462) order
Above with clone D indicates: (14625) order
Above with clone F indicates: (146253) order
Above with clone G indicates: (1462537) order. Ans:(a).
8. The strand being synthesized from the primer site is:
5’-AGACTATCTCTA-3’
The template strand would run antiparallel and is:
5’-AGACTATCTCTA-3’
3’-TCTGATAGAGAT-5’
Ans: (c) 5’-TAGAGATAGTCT-3’
9. Ans: (e). All of the normal deoxynucleotide
triphosphates would be necessary to elongate the newly
synthesized DNA strand; in the ‘A’ reaction, dideoxy-ATP would also be present as a
chain terminator to accumulate fragments when T is being read off of the template
strand.
10.
A
precursors added to minimal medium
B
C
D
E
F
Z
1
+
+
+
-
-
-
+
2
+
+
+
+
-
-
+
3
+
+
+
+
+
-
+
4
-
-
+
-
-
-
+
5
-
-
-
-
-
-
+
6
-
+
+
-
-
-
+
mutant number
The compound that supports the growth of the greatest number of mutants occurs
latest in the pathway, therefore
FEDABCZ
A mutant that blocks the F to E conversion should survive on all nutrients
downstream of F; following this logic:
FEDABCZ
3 2 1 6 4 5
Ans: (d): the enzyme that catalyzes the production of compound C is mutant in
strain 4.