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MCB 421 Exam #1
Fall 2007
There are 7 questions and the genetic code dictionary on
last page.
Be sure your name is on each page
Average 67
Low 44
High 92
S.D. 14
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
69
32
72
31
47
32
62
40
42
46
Culture
21
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.
2. (5 points) A frameshift mutation in the arc gene results in premature termination of
the encoded protein at an amber codon. However, the mutant is not suppressed by an
amber suppressor.
A. Explain why.
Answer: Downstream of the frameshift mutation translation occurs in the wrong
reading frame. Thus even if premature termination is prevented by suppression of
the “out of frame” nonsense codon, translation in the wrong reading frame will
produce multiple amino acid changes, resulting in an inactive protein.
Note: This question specifically asked about the effect of an amber suppressor on a
frameshift mutation, not how amber suppressors work or how frameshift
suppressors work.
3. (10 points) It is not possible to obtain true revertants of a deletion mutation in a
haploid organism. However, it is sometimes possible to obtain second site revertants of a
deletion mutation.
a. (5 points) What type of second site reversion event is most likely? [Explain].
Answer: The most likely type of reversion event is a bypass suppressor because it
completely avoids the need for the gene product.
Note: failure to obtain true revertants indicates that the deletion mutation described
is not a simple frameshift mutation. Thus, your answer should explain how a large
deletion can revert. It is true that in some very rare cases an insertion mutation
could compensate for the deletion, but this is less likely than a bypass suppressor.
Furthermore, if you suggest that an insertion mutation could compensate, you
would need to describe i. how it would compensate and ii where the insertion of
DNA might come from.
b. (5 points) Would this second site mutation be dominant or recessive? [Explain].
Answer: A bypass suppressor is an altered/gain of function mutation and thus
would probably be dominant to all types of loss of function mutations in the gene
and the wild-type gene. (In the case of the wild-type gene, both pathways would
probably function simultaneously. Sometimes such mutations are said to be “codominant”.
4. (20 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). (5 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. (5 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. (5 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. (5 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.
5. (30 points) The rII region of phage T4 has two genes called rIIA and rIIB. The
phenotypes of rII mutants on E. coli B and on E. coli K are shown below. They give
plaques of the type indicated when plated with an excess of the E. coli B or K. (DNA
sequencing is not an acceptable answer to any of the questions below).
rII+
rII-
E. coli B
wild type plaques
rough (“r”) plaques
E. coli K
wild type plaques
no plaques
a). (5 points) S. Benzer isolated several thousand mutagen induced and spontaneous rIImutants. Describe how he was able to isolate the mutants. Is this a selection or screen?
Answer: He plated T4+ lysate or mutagenized T4+ lysate on plates on E. coli B and
SCREENED for rough plaques.
b). (5 points) Crick, Barnett, Brenner and Watts-Tobin (1961, Nature 192:1227-1232)
Isolated revertants of an rIIB- mutant (FC0) made by mutagenesis with proflavin which is
an acridine. They found that almost all of the mutants were second site mutations within
the rII gene. How did they isolate the mutants? Is it a selection or screen?
Answer: They plated the rIIB mutant FCO on E. coli K and looked for plaques.
This is a selection because only revertants can grow on the K strain.
c). (5 points) They found that the frequency of revertants was very high when the FC0
mutant was treated with proflavin and virtually undetectable when treated with 5BU,
What does this suggest about the type of mutation in FC0?
Answer: It is extremely likely that the mutation is a frameshift. Frameshift
mutations are often reverted by second site mutations. Base substitutions like those
induced by 5BU are usually reverted by 5BU. Since 5BU doesn’t revert FC0 well,
the mutation is likely a frameshift.
d). (5 points) How did they show that the revertants were second site mutations within
the rII gene? Be sure to describe the type of E. coli strain used at each step.
Answer: They did mixed infections of cells where most cells were infected with both
T4+ and the revertant. Recombination during T4 infections is very high so
recombinants containing only FC0 or only the “suppressor” occur frequently. This
cross is done in E. coli B, which lets all three types of phages grow. Recombinants
will have the “r” phenotype when plated on E. coli B so they can be distinguished
from T4+ parents. (If the mechanism of suppression proposed is correct, the “r”
plaques should contain pahge with the original FCO mutation or the suppressor)
When this experiment was done they got “r” plaques on E. coli B. To actually show
that the “r” type plaques contained either the FCO or suppressor mutations, they
did a “backcross” (in E. coli B) where they crossed several of the “r” type phages
with FCO. They found that some crosses produced WT plaques on E. coli while
others did not.
Based upon proflavin-induced mutants in the rII locus of phage T4 and pseudo-revertants
(suppressors) of these original mutants, Crick et al. deduced that the genetic code must be
a multiple of three nucleotides. DNA sequence analysis of several rII mutants showed
that they had the indicated mutation shown in the sequence below.
5’ C C A A A A A C G A A G A A G C T G A A A T T G T T A A A C 3’
+A
-G
-A
+A
FC41
FC1
FC9
FC0
The phenotype of phage with combinations of mutants is shown in the following table.
"r" means that the phage makes rough plaques. "w" indicates that the phage makes wildtype plaques.
Mutation(s)
FC41
FC0
FC1
FC9
FC41
FC1
Plaques on E. coli B
r
r
r
r
w
FCO
FC1
w
FCO
FC41
FC9
FC0
w
r
FC41
FC9
r
FC1
FC9
r
Reasons for plaque phenotype
The + and – restores the correct reading frame.
The + and – restore the correct reading frame. Note: the
mRNA sequence in region between the frameshifts would
have at least six amino acid residues that are wrong. This
indicates that the amino acid residues in this region of the
protein are not essential for function.
The + and – restore the correct reading frame.
Two + frameshifts do not restore the proper reading
frame.
This is a + and a – that restores the proper reading frame
but not function. This means that there is a nonsense
codon in the correct reading frame (UGA) or that an
amino acid residue that does not allow protein function in
present. The latter is unlikely because the experiment
with the FC0 / FC1 combination showed that this region
of the protein is not essential for function.
Two – frameshifts do not restore the correct reading
frame.
e. (6 points). In the last column of the table above, explain the reason why each
combination produces “w” or “r” plaques. Include an explanation that explains the
phenotype at the level of messenger RNA translation.
f. (4 points). Brenner isolated a mutant of E. coli that allowed the double mutation FC 41
and FC 9 combination to make wild-type plaques on E. coli K-12. What is the most likely
explanation for this result?
ANSWER: The most likely explanation is that Brenners’s strain had a UGA
suppressor that suppressed the UGA codon in the mRNA between the FC 41 and
FC 9 mutations. It is very unlikely that an amino acid that disrupts function occurs
in the region between the mutations because this region is not necessary for protein
function.
6). (10 points) Four independent mutants were obtained that affect the synthesis of
leucine. The properties of the mutations are described in the table below (where +
indicates growth on minimal media and - indicates no growth).
Mutation
leu+
leu (null)
leu1
leu2
leu3
leu4
leu1 leu2
leu1 leu3
leu1 leu4
leu2 leu4
leu3 leu4
30°C
+
+
+
-
42°C
+
+
+
-
30  42°C
+
-
42  30°C
+
-
+
+
+
-
a. (4 points) Strains were constructed that contained combinations of double mutations
(don’t worry about how this was done). The mutants were grown at 30°C for several
hours and then incubated at 42 for several hours (30 → 42). They were also grown at 42
before growth at 30 (42 → 30). Based upon the above results, indicate whether each of
the single mutations is a temperature sensitive (Ts) or cold sensitive (Cs) mutation.
ANSWER (4 pts): 1 and 3 are CS, 2 and 4 are TS.
b. (6 points) Based upon the above results, order the mutant gene products in the pathway
of leucine synthesis as best as possible. Explain your answer.
1. (1,2) leu2 must act before leu1 since the double mutant must first be incubated at
leu2's permissive temperature and then shifted to leu1's in order to grow.
2. (1,3) leu3 cannot grow because it is only shown in combination with another CS
mutant.
3. (1,4)By the same reasoning, leu1 must act before leu4.
4. (2,4). leu2 cannot grow because it is only shown in combination with another CS
mutant.
5. (3,4) leu4 before leu3.
So the order is 2-1-4 -3.
7. (15 points) You have obtained five phage T4 plaque morphology mutants that form
tiny plaques on E. coli whereas the wild-type phage forms large plaques. The mutants are
designated r1, r2, r3, r4, and r5. In order to try to determine the type of mutation present
in the r gene of each phage, you perform a reversion analysis of the mutants and measure
the frequency that you get wild-type plaques (Don’t worry about how this was done). The
results are shown in the table below.
Mutagen X is an unknown mutagenic compound.
(+) means that revertants are frequent
(-) means revertants are very rare
(+/-) means that the frequency is intermediate
Mutant 5BU EMS NG ICR191 X HA
r1
+
+/- +/+
r2
+
r3
+
+/- +/+
r4
+
+
+
+
r5
a). (5 points) What is the likely type of mutation (ie. missense, nonsense, frameshift,
insertion, or deletion) in each mutant phage? Why?
ANSWER: r1, r3, r4 are missense or nonsense because they revert with mutagens
that make base substitutions. r2 is a fs because it reverts with ICR191, a mutagen
that favors frameshifts. r5 is likely a deletion or insertion. (Possibly a fs that can’t be
reverted by a second site mutation.)
b). (5 points) Can any specific predictions be made about the base changes made by any
of the mutagens?
ANSWER: Mutagen X must revert AT to GC because it doesn’t revert HA induced
mutations and HA doesn’t revert any X induced mutations. A mutagen that is
bidirectional (5BU, reverts both HA and X as expected. 4 is reverted by HA and
EMS and NG that prefer GC to AT changes,1 and 3 could be reverted by AT to GC
because EMS and NG prefer GC to AT.
r1, r3- reverted by mutagens that favor transitions.
r2- fs because only reverted by ICR
r4- reverted by HA so GC- AT. r4 is reverted by 5BU, EMS, NG, and HA. These
are likely reverted by a GC to AT transition because HA is specific for GC to AT.
r5- deletion
c). (5 points) Which mutagens would most likely produce revertants that are TS or CS?
How could such revertants occur?
All but ICR191 make base substitutions so revertants that are TS or CS must not be
true revertants. Thus secondary site substitutions, either in the original mutant
codon or elsewhere in the gene would be able to make TS or CS mutants. ICR most
likely will not make TS or CS because they usually make inactive proteins.