Name:
MCB 421
EXAM #1
September 29, 2011
There are 8 Questions.
There is a genetic code on the last page.
Be sure your name is on all pages.
Name:
1. (12 points) The antibiotic streptomycin inhibits protein synthesis by binding to
ribosomes and inhibiting translation. It is possible to isolate spontaneous mutants that
are resistant to streptomycin. The mutations map in the rpsL gene which encodes a
protein found in the 30S subunit of ribosomes.
a. (2 points) How would you isolate streptomycin resistant mutants?
Answer: Plate a large population of cells, 109, on plates supplemented with
streptomycin. Mutants will form colonies.
b. (2 points) What class of mutation is likely present in the mutant?
Answer: It is almost certainly a base substitution mutation that makes an amino
acid substitution in the RpsL protein so that the protein functions in translation but
does not bind to streptomycin. Frameshift, deletions and insertions in rpsL would
be inviable.
c. (4 points) When you construct a strain that contains a copy of the wild-type rpsL gene
and a copy of the allele that makes cells resistant to streptomycin, the resulting strain is
sensitive to streptomycin. How can this be explained at the molecular level?
Answer: Half of the ribosomes that contain the wild type protein will bind
streptomycin and protein synthesis will be inhibited. The other half will be resistant
and still synthesize proteins. However, the amount of protein synthesis is
apparently insufficient to make the cells viable.
d. (4 points) When several independent streptomycin resistant mutants were analyzed, it
was found that only three different substitutions in the rpsL gene were found. What is the
likely explanation for this finding?
Answer: The RpsL protein is an essential protein because it is a component of
ribosomes. Thus, most substitution mutations in the rpsL gene are lethal because
they make inactive ribosomes. Thus, it is very rare to isolate a mutant that encodes
a protein that is both active in protein synthesis and does not bind streptomycin.
Name:
2. (5 points) You have a culture of E. coli that contains 99% wild type cells (prototrophs)
and 1% tryptophan auxotrophs. How would you isolate the tryptophan auxotrophs? Is
this a selection or a screen?
Answer: Plate on minimal media with tryptophan. Aim to obtain 100-200 colonies a
plate. Then replica plate onto minimal media without tryptophan because both the
mutant and wild type cells grow on minimal media with tryptophan. Look for
colonies that only grow on minimal media with tryptophan. This is a screen.
3. (8 points) Mutants of E. coli that lack superoxide dismutase (SOD) cannot grow on
minimal medium aerobically. Supplementation with amino acids partially restores growth
under these conditions, suggesting that the defect may be due to the sensitivity of gene
products in several amino acid biosynthetic pathways to superoxide. In addition, a
revertant was obtained that partially restored growth on minimal medium aerobically
without amino acid supplementation [Imlay J and I Fridovich. 1991. MoL. Gen. Genet.
228: 410-416].
a. (4 points) Why did the authors conclude that the reversion was due to a
pseudorevertant rather than a true revertant?
Answer: If the revertant restored the wild-type sequence, it would behave exactly
like the wild-type strain and grow well aerobically on minimal medium.
b. (4 points) Describe how a potential intergenic and intragenic suppressor could cause
the pseudoreversion phenotype. For each case, briefly explain how the second mutation
would repair the mutant phenotype. (Only one suppressor for each type).
Answer: There are several possible explanations for this question.
Intergenic- Affect a protein that interacts with mutant SOD to partially restore
function. If the mutation was a nonsense mutation, a suppressor tRNA could arise.
Other possible answers.
Intragenic- Second site substitution of an amino acid that interacts with the original
mutant residue to restore function. Substitution mutant in the SOD protein that
improved the function of the mutant protein so it functions well enough to let the
cell survive.
Name:
4. (16 points)
a. (5 points) Would you expect a bypass suppressor to be dominant or recessive to the
wild-type gene?
Answer: Dominant because it is expressed independently of the wild type gene.
b. (5 points) What does allele-specific mean? What does it tell you if a suppressor is
allele-specific?
An allele-specific suppressor is a second-site mutation that repairs the mutant
phenotype but only in strains with certain, specific mutations at the first-site.
(Interaction suppressors are usually allele specific). Allele specificity argues that the
mutant and suppressor residues interact.
d. (3 points) Often, investigators grow multiple independent cultures inoculated from
independent colonies of mutator strains for mutagenesis. Why?
ANSWER: This reduces the chances of isolating sibling mutations.
e. (3 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.
Name:
5. (10 points) You have four different leucine auxotrophs of Salmonella
typhimurium that require leucine at 30o C. 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, "NQNO."
Mutant
leu-1
leu-2
leu-3
leu-4
Mutant Number of Revertants /108 Cells
ICR-191
EMS
Spontaneous
0
0
0
1926
18
20
5
859
18
8
9
9
NQNO
0
22
367
6
a. (4 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. (4 points) Is compound NQNO a mutagen? If so, what type of mutations does it
cause? How do you know?
Answer: NQNO 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,
NQNO probably causes base substitution mutations. Note that in some
versions, NQNO is an ICR-191-like mutagen and not an EMS-like mutagen.
c. (2 points) Which mutant(s) is the most likely one to be a CS mutant? Why?
How would you show it is a CS mutant?
Mutant leu-3 would most likely be a candidate for a CS mutant. It’s a base
substitution while the others are a deletion, insertion or frameshift that would
not make an intact protein. If his-3 is a CS mutant it will grow on minimal
plates at 42o C.
Name:
6. (9 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. (9 points) 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:
7. (20 points) Miller and Roth [(1971) J. Mol. Biol. 59:63 – 75] isolated what they called
a “recessive lethal nonsense suppressor” in Salmonella typhimurium. They did the
following series of genetic experiments to isolate a strain containing a UGA suppressor
(mutant t-RNA gene that decodes the UGA codon).
They used a strain with a known hisUGA mutation. It also contained missense mutations
in the met and ilv genes. This strain would normally require histidine (his), methionine
(met), isoleucine and valine (ilv) to grow on minimal medium. However the strain also
contained a low copy plasmid that carried the region of the chromosome that carries wildtype copies of met+ and ilv+ genes. Thus the strain containing the plasmid only requires
histidine to grow on minimal medium. The duplicated region of the chromosome (about
10 % of the chromosome) contains the met+ and ilv+ genes but not the his+ gene.
A typical cell containing the plasmid has 2 to 3 copies. However segregants that
spontaneously lose the plasmid and require his, met, ile, and val occur at a frequency of
about 5%.
+
-
+
-
UGA
a. (5 points) How would you demonstrate that the above segregants occur?
Answer: Plate the cells on minimal plus his + met + ile + val. All the cells should
grow to colonies. Colonies that arise from cells that lost the plasmid will not grow
on Min + his medium.
Name:
Starting with the strain containing the plasmid, they selected for His+ revertants. They
were hoping to isolate mutants that had an UGA suppressor linked to the met – ilv region
of the chromosome.
b. (5 points) How would you select for His+ revertants?
Answer: Plate cells on minimal lacking histidine. Only His+ revertants will grow.
When they analyzed the His+ revertants they found that some of them (Class A)
segregated cells that were met- ilv - and his-. Other revertants (Class B) did not
segregate met- ilv- cells. (i.e no colonies grow on medium lacking met, ile, leu).
c. (5 points) What is the explanation for each class of revertants?
Answer: The Class A cells could be true revertants or contain a UGA suppressor on
the plasmid that suppresses the his UGA mutation. The latter type could lose the
plasmid and the suppressor and become met-, ilv- and his-.
(The Class B mutants are more tricky). If the suppressor mutation is in the
chromosome and mutates a tRNA (to a UGA suppressor) that is required for
translating cellular mRNA and thus cell viability, the cell would need a second copy
of the wild-type version of the tRNA gene on the plasmid to decode the cognate
codon(s). However, if the cell loses the plasmid (and the wild-type version of the
tRNA gene), it is a lethal event. This is because loss of the plasmid removes the
wild-type copy of the tRNA gene so that only the UGA suppressor copy is present.
This tRNA does not decode normal mRNA and the cell dies.
Miller and Roth speculated that the tRNA that decodes tryptophan could give rise to the
suppressor because they also isolated an amber (UAG) suppressor at the same region of
the chromosome.
d. (5 points) Why did they make that prediction? (A genetic code dictionary is attached
to the exam).
Answer: The only codon for tryptophan is UGG. The suppressors likely arise by a
single mutational step. Thus only a single mutation would be required to mutate
the trp-tRNA to decode amber (UAG; G to A)) or the UGA (G to A) codon.
FYI: making strains with partial chromosome duplications is a common type of
approach towards isolating mutants with complicated phenotypes as we will see in
future lectures. There are several ways to make duplications in bacteria.
Also, Charles Miller became a professor of Microbiology here at the U of I. Some of
you may have taken MCB 300 from him.
Name:
8. (20 points) Brenner, Stretton and Kaplan (Nature 208:994 – 999,1965) performed an
ingenious set of experiments using mutagens to determine the sequences of the amber
(UAG) and ochre (UAA) codons. They used the rII system of phage T4 to perform the
experiments.
It was known that wild-type T4 grows normally and forms plaques on both E. coli B and
K strains. However, rII mutants have a distinct phenotype. They grow normally on E.
coli B. However, rII mutants cannot grow on E. coli K strains because of a defect that
acts before phage DNA synthesis. Thus progeny phages are not formed in E. coli K cells
where the rII defect is expressed.
They used HA as a mutagen. A phage preparation was mutagenized with HA and divided
in half. One half (SetB) was grown in E. coli B to rescue all rII mutants. The other half
(Set K) was plated on E.coli K to eliminate all phage with an immediate rII phenotype.
Amber and ochre mutants were then isolated and mapped to the rII region. (Don’t worry
about how this was done).
They found the following results:
Number of Amber mutants found at each site
rII mutant site
SetB
HB118
27
N97
44
S116
31
S24
44
HB129
14
N19
15
SetK
15
1
2
3
12
9
They argued that there are two types of mutant. One class occurs many times in Set B
but rarely in SetK. In the other class mutants occur at equal frequencies in both Sets.
a. (10 points) Knowing the mechanism of action of HA, explain the difference in the
mechanisms that give rise to the 2 class of mutants.
Answer: Only one strand of the DNA is transcribed into mRNA. Thus, because the
rII function is required before DNA synthesis, the mutation will only register a
phenotypic effect in the first cycle of growth if it is in the template strand which is
transcribed into mRNA. Thus the modified C will specify an A in the mRNA rather
than the normal G.
The SetB first class fulfills the expectation for a C to U change in the template
strand. The mRNA contains an A rather than the normal G. This results in a
mutant rII protein which prevents growth in E. coli K and are not recovered. The
Name:
SetK class fulfills the expectation for a C to U change in the coding strand. In this
case, the mRNA contains the normal C and thus codes for a WT rII protein even
though the DNA contains a mutation. The phage can thus grow in E. coli K even
though it will ultimately produce an amber mutation (and thus grow in E. coli B).
(FYI: Sam Kaplan became a professor of microbiology here at the U of I where he
ran a research program for 30 years).
It was known that ochre suppressors can translate amber mutations but amber suppressors
cannot translate ochre codons. You have 3 types of strains available for the following
experiment: (1) a strain with no suppressors, (2) a strain with a known amber suppressor
and (3) a strain with a known ochre suppressor.
b. (10 points) They also showed that the amber codon could be mutated to an ochre
codon with HA. Starting with several different known rII amber mutants, how could you
isolate a mutant with an ochre mutation at the same codon? Would you expect to isolate
ochre mutants from all the amber mutants ochre mutants? Why?
Answer: If one selects for “revertants” of several different rII amber mutants on a
strain containing an ochre suppressor, some revertants will be wild type and some
will contain ochre mutations where the HA mutated the UAG codon to a UAA
codon However, only wild-type revertants will grow on a suppressor free strain
while ochre mutants will fail to grow on the suppressor free strain.
Not all amber mutants will yield ochre mutants because the amino acid inserted by
the ochre suppressing strain may not make an active rII protein.
Name: