Name:
MCB 421
EXAM #1
October 1, 2009
There are 9 Questions.
Be sure your name is on all pages.
Name:
1). (10 points) A Luria-Delbruck fluctuation test was done to determine the rate of
mutation to Dehydroproline resistance (a toxic proline analog) in E. coli. Twenty tubes of
rich medium were each inoculated with a few wild-type cells and the cultures grown to 5
x 109 cells /ml. A 0.1 ml sample of cultures 1-20 were then plated on minimal medium
supplemented with DHP to detect DHPR mutants. Ten 0.1ml aliquots from tube #21
were also plated. The results are shown in the following table.
Culture #
1
2
3
4
5
6
7
8
9
10
# DHPR mutants
32
25
35
17
42
53
32
22
29
23
Culture #
11
12
13
14
15
16
17
18
19
20
# DHPR mutants
34
32
39
31
47
32
30
40
42
46
Culture
21 (Multiple
Sampes)
30
36
40
15
43
51
57
43
75
72
a. (6 points). From inspection of the data shown in the table is the resistance to
dehydroproline due to induced or spontaneous mutation? Why?
Answer: Induced mutation. Even a superficial (non-statistical) analysis of the data
shows that there is very little variance between the cultures.
b. (4 points). If you picked a colony from the plate supplemented with dehydroproline
and grew it in LB broth for several generations, would the cells be resistant or sensitive to
dehydroproline?
Answer: The cells would all (or almost all) be resistant to dehydroproline because
the mutation is inherited by all the progeny.
Name:
2. (20 points)
a. (6 points) How would you isolate a temperature sensitive (TS) bacteriophage T4
mutant starting with a WT stock of the phage?
Answer: Mutagenize a stock of T4 and plate on an E.coli indicator for single plaques
at 30o. Pick individual plaques and inoculate duplicate plates seeded with the E.coli
indicator with several plaques [be sure to keep track of which is which]. Incubate
one plate at 30o and the other at 42o. A TS mutant will form plaques at 30o but not
at 42o.
b. (2 point)How would you isolate a revertant of your TS mutant?
Answer: Mutagenize a stock of your TS mutant and plate at 42o. Any plaques that
form are from revertants. [This would work for unmutagenized phage since there is
a selection for revertants].
c. (12 points) Suppose you have isolated three independent revertants of the TS mutant.
You further characterize the revertants and find the following:
Revertant #1- This revertant forms normal plaques at 30oC as well as at 42o.
Revertant #2- This revertant cannot make plaques at 30o.
Revertant #3- This revertant makes smaller than normal plaques at 30o.
What can you propose for each mutant with respect to the reversion event being
intragenic or intergenic? Assuming all the reversion events are intragenic. Propose a
molecular explanation for each answer.
Answer: Revertant #1- It could be a true revertant or pseudorevertant. The pseudorevertant would
encode an amino acid substitution that allows function of the protein. The pseudorevertant could
also encode a second site amino acid substitution that suppresses the defect of the original TS
substitution, possibly by interacting with the residue causing the TS defect.
Revertant #2- Is likely amino acid substitution causes a cold sensitive defect. This could occur by
changing the mutated residue to another residue that causes a cold sensitive defect or by changing a
different residue in the protein that causes a cold sensitive defect even though it repairs the TS
defect.
Revertant #3- Is likely the amino acid substitution allows function of the protein at high temperature
but is slightly impaired at low temperature. This could be due to change of the mutated amino acid
to another residue or an intragenic event that creates a second site substitution that allows the
protein to function at high temperature but impairs function at low temperature.
Name:
3. (5 points) An amber mutation in phage T4 can grow on strains carrying sup-1 but not
on strains containing sup-2, even though both sup-1 and sup-2 are amber suppressors.
Suggest an explanation for this result.
ANSWER: sup-1 and sup-2 are both amber suppressors -- due to a mutation in the
gene encoding a tRNA which allows recognition of the UAG codon. However these
two mutations affect two different tRNA genes such that, although both mutant
tRNAs recognise amber codons, they insert different amino acids (because they are
charged with the amino acid that charges each of the two different wild-type
tRNAs). For example, sup-1 might be a mutated tRNA-leu gene whereas sup-2 might
be a mutated tRNA-tyr gene. Thus, the supression in sup-1 would insert either
leucine and the suppression in sup-2 would insert a tyrosine at the position in the
protein corresponding to the amber codon. If the inserted amino acid is not similar
in size and/or charge to the amino acid at that position in the wild type protein, the
resulting amino acid substitution may interfere with the structure and function of
the resulting protein. If suppression is not efficient enough an insufficient amount of
active protein might be made.
4. (5 points) You have two strains of E. coli. One strain contains a temperature
sensitive trpA mutation and the other contains a temperature sensitive mutation in
the gene encoding the alpha subunit of RNA polymerase (RNAP). How could you
distinguish between the two strains using a simple test? Be sure to include the
composition of the media you would use and what growth temperature(s) you
would use.
Answer: The strains can only be distinguished at 42°, in media containing
tryptophan. In these conditions only the TS trpA strain will grow. Replica-plating
schemes are possible but not necessary.
5. (13 points).
a. (4 points) (5 points) If E coli only has only trp tRNA gene (trpT) and only one codon
(UGG) which codes for tryptophan, how is it possible to get amber (UAG) suppressors
that are mutants of the trpT gene? Why isn’t that lethal?
Answer: As discussed in class, mutated tRNA can decode UGG and UGA codons. It
can read both amber and tryptophan codons.
b. (4 points). Would you expect a bypass suppressor to be dominant or recessive to the
wild-type gene? Explain your answer with regard to the molecular mechanism involved.
Answer: Dominant because it is expressed independently of the wild type gene.
c. (5 points) What does allele-specific mean? What does it tell you if a suppressor is
allele-specific?
Name:
ANSWER: An allele-specific suppressor is a second-site mutation that repairs the
mutant phenotype but only in strains with certain, specific mutations at the firstsite. (Interaction suppressors are usually allele specific).
6.
(12 points). You have four different leucine auxotrophs of Salmonella
typhimurium. To characterize the mutants, you compare their spontaneous
reversion rates to their reversion rates in the presence of ICR-191, EMS, and a
novel compound, "Carbox."
Mutant
leu-1
leu-2
leu-3
leu-4
Mutant Number of Revertants /10 Cells
ICR-191
EMS
Spontaneous
0
0
0
1926
18
20
5
859
18
8
9
9
8
Carbox
0
22
367
6
a). (8 points) Based on the above reversion data, what type of mutation is present
in each mutant strain? How do you know this?
Answer:
leu-1 is probably a deletion because it doesn't revert.
leu-2 is probably a frameshift mutation because reversion is increased by the
intercalating agent ICR-191.
leu-3 is probably a base substitution mutation because reversion is increased
by the alkylating agent EMS
leu-4 is probably an insertion mutation because reverts at a low frequency,
but the frequency of reversion is not stimulated by mutagens.
b). (2 points) Is compound Carbox a mutagen? If so, what type of mutations does
it cause? How do you know?
Answer: Carbox seems to be a mutagen because it causes a dramatic increase
in the reversion frequency of leu-3. (Although not as powerful of a mutagen
as EMS.) Because it increases the reversion frequency of leu-3 which is likely
to be a base substitution mutation (as described above) but does not increase
the reversion frequency of leu-2 which is likely to be a frameshift mutation,
carbox probably causes base substitution mutations. Note that in some
versions, carbox is an ICR-191-like mutagen and not an EMS-like mutagen.
c). (2 point) What type of medium would you use to select the His+ revertants?
Answer: Minimal medium, no His
Name:
7. (10 points) Four independent mutations were obtained that affect the synthesis of
tryptophan. The properties of the mutations are described in the table below (where +
indicates growth on minimal media and - indicates no growth).
a. (4 points) Based upon the above results, indicate whether each of the single mutations
is a temperature sensitive (Ts) or cold sensitive (Cs) mutation.
Answer: trp1 and trp3 are CS, trp2 and trp4 are TS.
b. (6 points) Based upon the above results, order the mutant gene products in the pathway
of tryptophan synthesis as best as possible. Explain your answer.
Answer: Trp2 must act before Trp1 since the trp1trp2 double mutant must first be
incubated at Trp2's permissive temperature and then shifted to Trp1's in order to
grow. By the same reasoning, Trp1 must act before Trp4. Trp3 cannot be ordered
because it is only shown in combination with another CS mutant. So the order is 21-4. Note that in test versions B and D, the data are changed and the order is 4-1-2.
Name:
8. (10 points) A group of genes, yfgABCDEFG, are involved in protein export. You are
curious as to whether these genes work together or by separate pathways. Therefore, you
perform epistatic analysis by constructing mutants of each gene and double mutants of
each combination of genes. The results are shown below, where the numbers are
deficiency factors (a deficiency factor of 1000 means the mutant has 1/1000th the activity
of a wildtype strain).
a. Identify multiple pathways within this group of genes. What genes are parts of which
pathway? Do any of the genes appear to be required for all activity, regardless of
pathway? If so, what does this tell you about its role in the phenomenon? Be sure to
explain your answers.
Answer: – Two separate pathways may be identified, one containing yfgACF and
the other yfgBEG. These are identified by comparing the phenotypes of double
mutants. If two genes are part of the same pathway, a double mutant will show no
greater defect than each mutant in isolation. If they are part of a different pathway,
then a double mutant will be more defective than either mutant alone. yfgD appears
to be essential for the export, since a yfgD single mutant and all combinations of
double mutants are severely defiecient. Thus, yfgD must be required for both the
yfgACF and the yfgBEG pathway.
Name:
9. (15 points) Phage T4 is a bacterial virus that lyses E. coli, producing a clearing in a
lawn of sensitive bacteria. This clearing is called a “plaque”. Benzer isolated a class of
conditional mutations in phage T4 that depended upon the strain of E. coli that the phage
infected. T4 rII mutants grow on E. coli B strains but not E. coli K-12 strains.
Phage
T4
T4 rII
Strain infected
E. coli K-12
E. coli B
small plaques small plaques
no plaques
large plaques
a. (5 points) How could you select for revertants of T4 rII mutants?
Answer: Revertants could be isolated by plating a pool of rII phage on K12; only
revertants will be able to form plaques. This is a selection. Looking for wild type
plaques on B would require a lot more work since most of the plaques would be r
type and wild type plaques would be exceedingly rare)]
b. (4 points) How could you distinguish deletion mutations from point mutations?
Answer: Deletion mutants will not revert by true reversion while point mutations
will revert either spontaneously or with mutagens.
c. (3 points) If a mutation were stimulated to revert by proflavin (an intercalating agent),
what would you conclude about the mutation?
Answer: Proflavin induces frameshift mutations when phage are grown in cells
treated with proflavin.
d. (3 points) If a mutation were stimulated to revert by hydroxylamine, what would you
conclude about the mutation?
Answer: HA is specific for G:C to A:T transitions. Thus the mutant had a G:C base
pair that reverted to A:T.