MCB421 HOMEWORK #3 Answers
Due: Tue Sept 20
Page 1 of 4
FALL 2011
1. Strain FC40 is unable to grow on lactose as a sole carbon source (Lac -). A Luria-Delbruck fluctuation
test was done by plating 108 cells from 10 independent cultures (#1-10) onto minimal lactose plates, and at
the same time plating 108 cells from a single culture (#11) onto 10 minimal lactose plates. After a long
incubation, the following results were obtained.
Independent Cultures
(Culture #)
1
2
3
4
5
6
7
8
9
10
Number of
Colonies
22
18
19
24
20
23
21
22
21
17
Culture 11
(Plate #)
1
2
3
4
5
6
7
8
9
10
Number of
Colonies
17
24
23
26
20
22
21
24
20
19
Based upon these results, would you conclude that the mutation to Lac + is random or adaptive? (Briefly
explain your answer.)
[The variance between the number of Lac+ colonies from individual cultures equals that seen with
multiple samples of a single culture. These results suggest that the phenotype is not likely to be due
to a random, spontaneous mutation. (5 points for this) (A random mutation which would show much
greater variance between individual cultures.) The simplest interpretation of these results is that the
mutation to Lac+ is an example of “adaptive or induced mutagenesis” (5 points for this). Note – your
answer must describe the comparison of the variance between cultures #1-10 vs the multiple platings
of culture #11. When given a problem like this, you need to evaluate the data given, do not simply
say that it must be random mutagenesis because I said in lecture that mutagenesis is random!]
2.
In the following table, briefly diagram or indicate the common properties of each type of mutation.
Mutation
Missense
Nonsense
Frameshift
Deletion
Insertion
effect on
DNA
base substitution
base substitution
resulting in a stop codon
insertion or
deletion of 1 or 2
base pairs
loss of
multiple base
pairs
addition of multiple
base pairs
effect on
Protein
substituted amino
acid
truncated polypeptide;
inactive protein
altered amino acid
sequence
downstream of
mutation; usually
truncated
polypeptide;
inactive protein
usually
absent;
usually
inactive
protein
may insert extra amino
acid (if in frame) or
cause premature
truncation; usually
inactive protein
effect on
Phenotype
may result in loss
of function or
conditional
phenotype
usually null
null
null
usually null; if insertion
is in frame and doesn’t
cause premature
termination and is in a
permissive site,
sometimes protein
remains functional
MCB421 HOMEWORK #3 Answers
Due: Tue Sept 20
Page 2 of 4
FALL 2011
3. An elegant genetic experiment to determine the number of bases required to code for each amino acid
took advantage of a large collection of frameshift mutations in the rII gene of phage T4 [Crick, F., L.
Barnett, S. Brenner, and R. Watts-Tobin. 1961. Nature 192: 1227-1232]. The wild-type DNA and amino
acid sequence corresponding to the first portion of the rII gene are shown below.
bp 1
31
61
ATG TAC AAT ATT AAA TGC CTG ACC AAA AAC
Met tyr asn ile lys cys leu thr lys asn

 CTG TAT TCA
GAA CAA GCT GAA ATT GTT AAA
glu gln ala glu ile val lys leu tyr ser
30
 60
 CAA CAG GAA TTG GCT
AGT GGT AAT TAC ACC
ser gly asn tyr thr gln gln glu leu ala
Four frameshift mutations isolated were FCO (an insertion of an A at base 50), FC1 (a deletion of the A at
base 32), FC40 (an insertion of an A at base 60), and FC88 (a deletion of the C at base 75).
a.)
Write out the amino acid sequence for the double mutant FCO FC1.
ANSWER:
AAA AAC GAc aag ctg aaa ttg tta aAA CTG
Lys Asn Asp Lys Leu Lys Leu Leu Lys Leu
b.)
Write out the amino acid sequence for the double mutant FC1 FC40.
ANSWER:
AAA AAC GAc aag ctg aaa ttg tta aac tgt att caa AGT
Lys Asn Asp Lys Leu Lys Leu Leu Asn Cys Ile Gln Ser
c.)
Write out the amino acid sequence for the double mutant FCO FC88.
ANSWER:
GTT Aaa act gta ttc aag tgg taa tta cac CAA
Val Lys Thr Val Phe Lys Trp STOP
d.)
Write out the amino acid sequence for the double mutant FC40 FC88.
ANSWER:
TCA aag tgg taa tta cac CAA
Ser Lys Trp STOP
e.)
The double mutants FCO FC1 and FC1 FC40 both produce functional rII protein, but FCO FC88
is inactive. What does this indicate about the structure and function of the wild-type rII protein?
ANSWER: The N-terminus and the C-terminus of the wild-type rII protein are essential for function.
The region in the middle of the protein, between base 30 and 75, does not appear to be important, as
frameshift mutations do not alter protein function. In the double mutant FCO FC88, there is a
premature stop codon that results in a truncated (non-functional) protein. For both double mutants
FCO FC1 and FC1 FC40, the correct reading frame is restored at base 50 and base 61, respectively,
resulting in a mutant rII protein with a functional C-terminus.
f.) Would the double mutant FCO, FC40 produce a functional rII protein?
MCB421 HOMEWORK #3 Answers
Due: Tue Sept 20
Page 3 of 4
FALL 2011
ANSWER:
GTT Aaa act gta ttc aaa gtg gta att aca ccc aac agg
Val Lys Thr Val Phe Lys Val Val Ile Thr Pro Asn Arg
aat tgg ct
Asn Trp
The protein would not be active because of the frameshift caused by the two base insertions.
4). Your laboratory has been working on tryptophan biosynthesis in Pseudomonas illini, a strain isolated
from the South Farms. They have isolated mutants that require tryptophan at 42 o C to grow (require
tryptophan supplementation when grown at 42o C) but they have not been tested for growth in the absence
of tryptophan at 30o C. (Don’t worry about how they were isolated for this question). They grow the
mutant strains in minimal medium at 42o C supplemented with a small amount of tryptophan so the cells
are able to grow. Next they make cell-free extracts and assay for enzymatic activities known to be present
in E. coli from previous studies. For example, they assay for the equivalents of the E. coli TrpC, TrpB.
And TrpA proteins encoded by the trpC, trpB and trpA genes, respectively.
They find the results for the P. ilini strains presented in the table below:
Enzyme Activity
Strain
TrpC
TrpB
TrpA
Wild-type
100
100
100
Mutant #1
0
2
Mutant #2
0
0
2
Mutant #3
2
100
100
Mutant #4
100
5
100
2
They also find that Mutants #1, #3, and #4 revert either spontaneously or with mutagens whereas Mutant #2
never reverts.
They also have antibodies that react with the TrpB protein of E. coli and find that the Wild type and
Mutants #3 and #4 make material that cross reacts with the antibodies (CRM) while Mutants #1 and #2 do
not.
Assume that each mutant strain has only 1 mutation.
1). For each mutant, explain what type of mutation is likely involved (ie base substitution, frameshift,
deletion or insertion) and how it affects the expression of trp genes. Explain your reasoning. It might not
be possible to determine the exact type of mutation for each mutant with the data given.
[Mutant #1 lacks TrpC activity and is polar on TrpB and TrpC so it could be 1). A substitution
mutation in trpC that creates a nonsense codon in trpC, leading to polarity on expression of the trpB
and trpA genes leading to reduced expression of the TrpB and TrpA proteins, 2). A frameshift
mutation in trpC that leads to truncated and inactive TrpC protein due to translation of the incorrect
reading frame downstream of the frameshift site and premature translation termination. This leads
to polarity because premature transcription termination leads to lower levels of TrpB and TrpC. 3).
(One could consider an internal deletion in trpC that creates a frameshift mutation that causes
MCB421 HOMEWORK #3 Answers
Due: Tue Sept 20
Page 4 of 4
FALL 2011
polarity on TrpB and TrpA as in (B). However, this mutant reverts (regains TrpC activity) so this
explanation is very unlikely). 4). One could consider an insertion mutation in trpC that also causes
polarity. The fact that it does revert indicates that it could be an insertion. Since mutant #1 does not
make CRM for antibodies to TrpB, it indicates that the level of TrpB is lower in the mutant
consistent with the polarity hypothesis.
Mutant #2 is missing both TrpC and TrpB activity so it is likely a deletion that deletes parts or most
of trpC and trpB. This conclusion is supported by the fact that it doesn’t revert, has lost the activities
encoded by 2 genes, is polar on expression of trpA, and doesn’t make detectible CRM to TrpB. It also
suggests that trpC and trpB are adjacent so a deletion mutation could affect both genes.
Mutant #3 is likely a missense mutation in trpC that makes a defective protein. Note that expression
of trpB and trpA are unaffected ruling out polarity and CRM detects TrpB as expected. It is possible
that the mutation makes the protein TS but we don’t have the data yet.
Mutant #4 is likely a missense mutation for reasons in (3).]
2). What is the probable order of trpC, trpB, and trpA relative to the transcritionnal startpoint? Why?
[Likely order is: Promoter....trpC....trpB....trpA. Mutants #1 and #2 indicate that trpC and trpB and
adjacent to each other. Since mutant #1 is polar on TrpB and TrpA expression, it indicates trpB and
trpA are downstream of trpC. Therefore the order is likely trpC, trpB, trpA.]
3). Which mutant(s) would likely encode a temperature sensitive TrpB protein?
[Mutant #4 because TrpB is wild type in Mutant #3, deleted in Mutant #2, and likely wild type in
Mutant #1 because of the polar mutation in trpC.]
4). How would you demonstrate if any of the mutants were temperature sensitive?
[Streak on 2 minimal plates supplemented with only glucose. A TS mutant will grow at 30 o C but not
at 42o C. ]