Name_______________________________
Microbial Genetics
BIO 410/510
Fall Quarter 2007
Exam I
StudentpID#__________________________
1.) Draw the structure of guanine. Indicate where the hydrogen bonds form with cytosine. (4pts)
hydrogen
bonds
form
at these
sites
O
HN
N
N
R
HN
H
2.) Thermosensitive mutations exist for several of the E. coli proteins associated with DNA
replication. Describe the role that each of the following proteins play during semiconservative
replication AND predict what would happen to replication if that protein were inactivated.
EXPLAIN your rationale. (10pts)
DnaB helicase: separates and unwinds the two strands DNA as replication progresses. Following
DnaB inactivation, the replication fork would rapidly be arrested.
DnaG primase: Repeatedly synthesizes a short RNA primer that allows the lagging strand
polymerase to reinitiate on the lagging strand template. Following DnaG inactivation, the
replication fork would either arrest or, if it continued, the lagging strand synthesis would not
occur.
DNA Ligase: Joins the Okazaki fragments on the nascent lagging strand following synthesis.
Following inactivation of DNA Ligase, the synthesis on the nascent lagging strand would remain
discontinuous (contain nicks).
Pol I: Removes the RNA primer on each Okazaki fragment and resynthesizes it with DNA on the
lagging strand. Following inactivation of Pol I, the synthesis on the nascent lagging strand would
remain discontinuous (contain nicks).
DNA Gyrase topoisomerase: Relieves the positive supercoiling induced by replication forks
progressing along the DNA. Following inactivation of DNA Gyrase, replication would arrest as
the supercoiling stress would prevent further DNA winding (replication fork progression) from
occurring.
Name_____________________________
page 2
3.) Escherichia coli replication machinery makes a mistake about once every 1010 times it
incorporates a nucleotide. E. coli O157 is a pathogenic strain that was isolated from hamburgers
at Jack in the Box restaurants in the late 1990’s. It caused severe fevers and death in several
customers. Its genome contains approximately 5 * 106 basepairs. How many changes
(mutations) would be expected to occur in its genome each time it replicates? Show your work.
(3pts)
(5 * 106 bases / genome) * (1 error/ 1 * 1010 bases) =
(5 * 106 bases / genome) * (1 error/ 1 * 1010 bases) =
5 * 10-4 errors / genome
4.) The actual error (mutation) rate per cell division in E. coli strain O157 was measured and it
was found to make a mistake about once every 106 times it incorporates a nucleotide. How many
changes (mutations) occur in its genome each time it replicates? Show your work. (2pts)
(5 * 106 bases / genome) * (1 error/ 1 * 106 bases) =
(5 * 106 bases / genome) * (1 error/ 1 * 106 bases) =
5 * errors / genome
5.) The mutS gene of E. coli O157 is inactivated by a mutation. MutS is part of the methyl
directed mismatch repair system. Describe how DNA methylation allows replication to correct
errors that were made during replication. (4pts)
Methylation occurs at GATC sites on the DNA. Since the methylation process takes
some time to occur, the methyl-directed mismatch repair system is able to identify which strand
is the newly replicated (unmethylated) strand of the DNA. Mismatched base pairs are then
excised from the unmethylated, daughter strand and the region is then re-replicated to correct the
error.
6.) The E. coli origin or replication contains multiple DnaA boxes and AT-rich 13-mers.
Describe what role each of these plays in the initiation of replication? (5pts)
Multiple DnaA proteins bind in a sequence specific manner to the DnaA boxes and
associated with each other. The DnaA binding induces stress on the surrounding helical DNA
regions and promote strand separation at the AT-rich regions. These single stranded regions then
serve as the binding sites for the delivery and loading of the DnaB helicase by DnaC to initiate
the subsequent steps of establishing the replication forks.
Name_____________________________
page 3
A soil bacteria that was isolated replicates in every 10 hours when grown in lab cultures.
To examine whether replication occurs conservatively or semiconservatively in this bacteria, you
decide to utilize a variation of the approach that Meselson-Stahl originally used to examine this
question in E. coli.
For your controls, you grow the bacteria in two separate culture media for several generations.
C1.) One culture is grown in normal media
C2.) The other culture is grown in media that contains 5-bromouracil, an analog of thymine that
has a much higher buoyant density.
For your experimental analysis, you inoculate a third culture of the bacteria in normal media and
allow it to grow for 2 days. At this time, you then transfer the cells into media containing 5bromouracil. To examine the mode of replication, you collect a portion of the cells
E1) immediately after changing the media,
E2) 10 hours after you changed the media, and
E3) 20 hours after you changed the media.
You then lyse the cells and load the cell lysates (and DNA) from each sample in neutral CsCl
gradients before centrifuging them to equilibrium. The results of your controls, C1 and C2, are
shown below.
7.) On the tubes below,
clearly indicate where the
DNA would be expected to
appear if the bacteria replicate
conservatively? (5pts)
5-bromoNormal media
uracil
media
C1
C2
0 hrs
E1
10 hrs
E2
8.) On the tubes below,
clearly indicate where the
DNA would be expected to
appear if the bacteria replicate
semi-conservatively? (5pts)
20 hrs
E3
0 hrs
E1
10 hrs
E2
20 hrs
E3
Name_____________________________
page 4
9.) What DNA sequences are important for factor independent transcriptional termination? How
are these thought to promote transcription termination? (3pts) Termination by this mechanism
relies upon an inverted repeat sequence that is followed by a stretch of UUUUs in the RNA
transcript. Transcription of the inverted repeats produces a hairpin in the RNA that destabilizes
the RNA polymerase enough to dissociate it when followed by a string of UA base pairs in the
transcription bubble. The UA base pairs are less stable than GC base pairs due to the lower
number of hydrogen bonds formed between these base pairs.
10.) What DNA sequences are important for factor dependent transcriptional termination? How
are these thought to promote transcription termination? (3pts) Factor dependant termination
occurs when a protein factor, such a Rho, binds to a specific sequence on the RNA transcript. In
the case of Rho, the rut sequence is bound. The Rho factor is an RNA helicase that, in effect,
chases or follows the RNA polymerase and dislodges it at specific pause sites downstream on the
transcript. Since Rho can only bind to rut when translation is not "hiding" the rut sequence, this
also provides a mechanism to regulate the transcription of polycistronic messages.
11.) Briefly describe (or draw) the events involved during translation elongation (do not include
events that are associated specifically with initiation or termination).
Include the following terms, if appropriate (not all terms are appropriate). TATA Box, 30S
Subunit, 50S Subunit, 70S Subunit, RNA Polymerase, Shine Delgarno, rho, ATP, GTP, formylmethionine, P site, A site, RF1, IF3 and IF1, IF2, EF-Tu, EF-G, t-RNA, sigma factor. (6pts)
A. EF-Tu loads a charged tRNA into the empty A site of the 30S ribosome subunit using GTP
as an energy source. (EF-Ts then exchanges the GDP for GTP on Ef-Tu to "recharge" the
enzyme for the next round.
B. The polypeptide attached to the tRNA in the P site is transferred (bonded) to the amino
acid-tRNA in the A site using a peptidyltransferase activity found on the 50S subunit.
C. GTP charged EF-G powers the ribosome forward moving the tRNA containing the
polypeptide into the P site. (The cycle then repeats.)
You’ve identified a small gene
from a strain of Streptomyces whose
expression prevents other bacteria from
growing when Streptomyces is present.
You are interested in this peptide because
of its potential as a new antimicrobial
agent. The entire open reading frame (ie. the
TRANSLATED portion of the mRNA
transcript has following the sequence:
5'… GUG AUC AUU AUA ACU ACC ACA ACG UAA… 3'
12.) Using the table for the genetic code and your knowledge of how translation occurs, translate
the given DNA sequence into its appropriate amino acids. (4pts)
fMet-Ile-Ile-Ile-Thr-Thr-Thr-Thr
13.) What’s the absolute minimum number of different tRNA molecules that would be needed to
translate this peptide? Why? (2pts)
At least three, an initiation tRNA, an Ile-tRNA and a Thr-tRNA. The wobble position of an anticodon on a given tRNA may recognize multiple bases, meaning that the same tRNA may deliver
a given amino acid to codon sequences that differ at the wobble position.
Name_____________________________
page 5
The gene below encodes Homo sapiens coagulation factor III, a cell surface glycoprotein that
enables cells to initiate the blood coagulation cascade. This protein is the only one in the
coagulation pathway for which a congenital deficiency has not been described. Your working
for a biotech company that is interested in producing drugs to treat specific forms of hemophelia.
Your task is to clone this gene into an expression vector as a potential mode of treatment.
1
61
121
181
241
301
361
421
481
541
601
661
721
781
841
901
961
1021
1081
1141
1201
1261
1321
1381
1441
1501
1561
1621
1681
1741
1801
1861
1921
1981
2041
2101
gggggggggg
cccaacctcc
gacatggaga
ctcctgctcg
gcagcatata
aaacccgtca
aaatgctttt
aagcagacgt
tctgctgggg
ctcggacagc
gaagatgaac
ggcaaggact
gccaaaacaa
agtgttcaag
gagtgtatgg
gtatttgtgg
gcaggagtgg
tgttggagct
tacatggaaa
gatatgacct
atgtaacgaa
tgcacataac
caaaaacaaa
tttttttttt
tctcggctca
gagtagctgg
gagatggggt
ccaccttggc
agcttttgag
atttctagga
ggaatacatt
gagacattgg
taagtggcat
agagtttatg
aaaaaggttt
atattgagat
cccccgcgca
ccagccccac
cccctgcctg
gctgggtctt
atttaacttg
atcaagtcta
acacaacaga
acttggcacg
agcctctgta
caacaattca
ggactttagt
taatttatac
acactaatga
cagtgattcc
gccaggagaa
tcatcatcct
ggcagagctg
actgcaaatg
cgcaaatgag
gttattacca
tggtactaca
atgctttaga
tgggaaaatg
tttgagacgg
cttgcaccct
gattacaggt
ttcaccatct
ctcccaaaga
gggctgactt
cttttctaac
tggaaattca
tattctgggc
taaacatttg
atttaaggta
ttctatatgg
aatttattta
ccccctcgca
gggcgccacg
gccccgggtc
cgcccaggtg
gaaatcaact
cactgttcaa
cacagagtgt
ggtcttctcc
tgagaactcc
gagttttgaa
cagaaggaac
actttattat
gtttttgatt
ctcccgaaca
aggggaattc
tgtcatcatc
gaaggagaac
ctatattgca
tatttcggag
ttagcattct
accaattcca
ttatatattc
tcttaaaaaa
agtcttgctc
ccgtctctcg
gcgcactacc
tggccaggct
tgctagtatt
caatccatgt
atatgtctat
aaacaattgg
agcttcctaa
agagctaact
cttaaagctt
ggattttcta
atatacttta
ctccctctgg
gaacccgctc
ccgcgccccg
gccggcgctt
aatttcaaga
ataagcacta
gacctcaccg
tacccggcag
ccagagttca
caggtgggaa
aacactttcc
tggaaatctt
gatgtggata
gttaaccgga
agagaaatat
ctggctatat
tccccactga
ctgtgaccga
catgaagacc
ggttttgaca
agttttaatt
cgcacttaag
tcctgggtgg
tgttgcccag
ggttcaagca
acgccaagct
ggtcttgaat
atgggcgtga
aggaaagtaa
aatatagtgt
gcaaactttg
tatgctttac
atatttttat
ctatggttga
tttatgtagg
aataaaggtg
ccggcccagg
gatctcgccg
agaccgccgt
caggcactac
caattttgga
agtcaggaga
acgagattgt
ggaatgtgga
caccttacct
caaaagtgaa
taagcctccg
caagttcagg
aaggagaaaa
agagtacaga
tctacatcat
ctctacacaa
atgtttcata
gaacttttaa
ctggagttca
tcagcattag
tttaacacca
gattaaccag
acttttgaaa
gctggagtgc
attgtctgcc
aatttttgta
tcctgacctc
accaccatgc
aatggaagga
ttaggttctt
tattaatgtg
aatctgcact
aagactacta
cattgtatat
taatattgtt
actggaaaaa
gcgccttcag
ccaactggta
cgctcggacg
aaatactgtg
gtgggaaccc
ttggaaaagc
gaaggatgtg
gagcaccggt
ggagacaaac
tgtgaccgta
ggatgttttt
aaagaaaaca
ctactgtttc
cagcccggta
tggagctgtg
gtgtagaaag
aaggaagcac
gaggatagaa
aaaaactctt
tcactttgaa
tggcaccttt
gtcgtccaag
agcttttttt
agtagcacga
tcagcctccc
ttttttagta
agtgatccac
ccagccgaaa
aattgggtgc
ttttttttca
ttaagtgcag
ttaactgact
tacaaactac
ataatttttt
ctatttgtat
tttttttttt
14.) Design two 15 base primers that you could use in a PCR reaction to amplify this gene from
human cells? (4pts)
5’ GGG GGG GGG GCC CCC 3’
5’ AAA AAA AAA ATT TTT 3’
15.) You know that you will need to clone your PCR product into a BamHI restriction site,
GGATCC, found on the expression vector. Modify your primers below such that you will be
able to easily clone the PCR product into a BamHI site on an expression vector. (3pts)
5’ NNN NNN GGA TCC- GGG GGG GGG GCC CCC 3’
5’ NNN NNN GGA TCC- AAA AAA AAA ATT TTT 3’
16.) The human genome contains 3.3*109 basepairs. Is the 15 base sequence of your primer
likely to be unique in the human genome? On average, how many times would your primer
sequence be found on the human genome? Show your work (4pts)
4 possible bases at every position… so a sequence 15 bases long would be found every…
4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 * 4 bases
415 bases = a specific15 base sequence would be found once in every 1,073,741,824 bases.
With 3.3 * 109 bases in the human genome a specific 15 base sequence would appear, on
average,
3.3 * 109 bases/ 1.07 * 109 bases = 3.08 times… so its not long enough to be unique.
Name_____________________________
page 6
Your biotech company is using 4 different experimental
cloning vectors called pEXPRESS1, pEXPRESS2,
pEXPRESS3, and pEXPRESS4. All the vectors have similar
constructions (shown to the right). However, the cloning region
on each vector is slightly different. Each of the vector cloning
regions is shown below.
18.) For each vector, determine if the Coagulation factor III
product is likely to be expressed when your PCR product
is cloned into the Bam HI site. Explain Why or Why not in
each case. (8 pts)
Cloning Region
pEXPRESS Vectors
ampicillin
resistance
origin of
replication
Promoter
Shine
Stop
Transcription
BamHI site
-35-10
Delgarno ATG codons
termination
GGA TCC consensus sequence
UAA UAG
sequences
Cloning Region of pEXPRESS 1
pEXPRESS1:
Coagulation factor III product will not be expressed when the PCR product is cloned into the
Bam HI site because the gene will be upstream of the promoter and ribosome binding site that
are needed for transcription and translation to occur.
Shine
Promoter
Stop
Transcription
ATG BamHI site codons
Delgarno
-35-10
termination
GGA TCC UAA UAG
sequence consensus
sequences
Cloning Region of pEXPRESS 2
pEXPRESS2:
Coagulation factor III product will not be expressed when the PCR product is cloned into the
Bam HI site because the ribosome binding site that is needed for transcription is upstream of the
transcriptional promoter. Thus the mRNA containing the gene will not be translated because it
does not contain the Shine Delgarno sequence.
Promoter
Shine
Stop
Transcription
-35-10
Delgarno ATG BamHI site codons
termination
GGA TCC UAA UAG
consensus sequence
sequences
Cloning Region of pEXPRESS 3
pEXPRESS3:
Coagulation factor III product will be expressed when the PCR product is cloned into the Bam
HI site. The open reading frame, when cloned into the BamHI site, has a transcriptional
promoter, the mRNA have a ribosome binding site before the first start codon as well as
additional stop codons prior to the end of the transcript.
Promoter
Shine
-35-10
Delgarno ATG BamHI site
GGA TCC
consensus sequence
Transcription
Stop
termination
codons
sequences UAA UAG
Cloning Region of pEXPRESS 4
pEXPRESS4:
Coagulation factor III product will probably be expressed when the PCR product is cloned into
the Bam HI site. If the open reading frame includes its own stop codon, the gene should be
expressed normally.
Name_____________________________
page 7
The Luria-Delbruck experiment demonstrated that bacteria can become resistant to bacteriophage T1 infection
through random mutations, rather than through a directed change. You decide to try and see if the random mutation
hypothesis also applies to how bacteria become resistant to the antibiotic, rifampicin.
You do the following experiment. You inoculate 200 mls of media with a dilute bacterial culture. Then,
you split the culture A) growing 100 mls in a single flask, and B) growing the other half in 100 separate, 1 ml test
tubes.
After allowing the cultures to grow for several hours, you spread A) 1ml aliquots from the single flask on 100 plates
that contain rifampicin, and B) the 100 1ml cultures onto 100 plates that contain rifampicin. You allow each set to
incubate overnight and count the number of rifampicin resistant colonies on each plate in the morning.
Five representative plates from A) the single 100 ml flask are shown below:
19.) If mutations arose primarily through a process of directed change, CLEARLY diagram how
you would expect to five representative plates to look from B) the 100 individually grown 1ml
cultures. Explain why. (4pts)
The experiment would predict that each cell has an equal probability of generating resistance to the antibiotic
WHEN exposed to it. Since all cells were exposed to the antibiotic at the same time, each plate should have roughly
equal numbers of mutations.
20.) If mutations primarily arose randomly within growing populations, CLEARLY diagram
how you would expect to five representative plates to look from B) the 100 individually grown
1ml cultures. Explain why. (4pts)
The experiment would predict that mutations occur randomly within the population prior to any exposure to the
antibiotic...some of these mutations confer resistance to the antibiotic and depending on when they first occur in
each culture, a given culture may have many (if the mutation occurred early) or few to no (if the mutation occurred
late to never) resistant mutants. The number on each plate should vary significantly.
21.) Does this experiment involve positive or negative selection? (2 pts)
positive selection
22.) If 20 out of your 100 plates of 1ml cultures had no colonies on them, and the cultures had
108 cells/ml, calculate the mutation rate (a) for generating resistence to rifampicin using the
Poisson expression, where the probability of having i mutations per culture is represented by
P(i) = ( mi e-m )/ i! and a = m/n. (5pts).
20 out of 100 cultures had 0 mutations.
So the probability of having zero mutations (Pi ) is 20/100,
and i= 0 mutational events per culture in this situation.
Name_____________________________
Pi
(20/100)
0.2
0.2
-ln(0.2)
.916
=
=
=
=
=
=
( mi e-m )/ i!
( m0 e-m )/ 0!
( 1 e-m )/ 1
e-m
m
m
1.6 mutational events per culture
and each culture has 1 x 108 cells
a
=
m/N
a
a
=
=
1.6 mutational events/1 x 108 cells
1.6 x 10-8 rifR mutational events per cell division
page 8
Name_____________________________
page 9
Primer design (10pts):
Design primers for recombineering that would delete the recF gene of E.coli and replace it with
the catRsacB cassette from plasmid pEL4.
>E. coli
1
31
61
91
121
151
181
211
241
271
301
331
361
391
421
451
481
511
541
571
601
631
661
691
721
751
781
811
841
871
901
931
961
991
1021
1051
-
|EG10828|recF: 1074 bp - Recombination and repair
g
ccagagcgcggcttatgttgtcatgccaatgagactgta
atg tcc ctc acc cgc ttg ttg atc cgc gat
ttc cgc aac att gaa acc gcg gat ctc gcc
tta tct ccc ggc ttt aac ttt ctg gta ggt
gcc aac ggc agt ggc aaa acc agc gtg ctg
gaa gcc atc tat acg ctc ggc cat ggt cgg
gcg ttt cgc agt ttg cag att ggt cgc gtc
att cgc cat gag cag gag gcg ttt gtt ctc
cac ggg cga tta cag ggc gaa gag cgc gag
aca gcg att ggc tta acc aaa gac aaa cag
ggc gac agc aaa gtc cgc atc gac ggt aca
gac ggg cat aag gtc gcg gaa ctg gcg cac
ctg atg cca atg cag ttg ata acg cca gaa
ggg ttt act tta ctc aac ggc ggc ccc aaa
tac aga aga gca ttc ctc gac tgg gga tgc
ttt cac aac gaa ccc gga ttt ttc acc gcc
tgg agc aat ctc aag cga ttg ctc aag cag
cgc aat gcg gcg ctg cgc cag gtg aca cgt
tac gaa cag cta cgc ccg tgg gat aaa gag
ctg atc ccg ctg gcg gag caa atc agc acc
tgg cgc gcg gag tat agc gcc ggt atc gcg
gcc gat atg gct gat acc tgt aag caa ttt
ctc cct gag ttt tct ctg act ttc tct ttc
cag cgc ggc tgg gag aaa gag aca gaa tat
gct gag gtg ctg gaa cgt aat ttt gaa cgc
gat cgc cag cta acc tac acc gcg cac ggc
ccg cac aaa gcg gac tta cgc att cgc gcc
gac ggt gcg ccg gtg gaa gat acc tta tcg
cgt ggg cag ctt aag ctg ttg atg tgc gcc
tta cgt ctg gcg caa gga gag ttc ctc acc
cgt gaa agc ggg cgg cgg tgt ctc tac ctg
ata gat gat ttt gcc tct gag ctt gat gat
gag cgt cgc ggg ctg ctt gcc agc cgc tta
aaa gcg acg caa tca cag gtc ttt gtc agc
gcg atc agt gct gaa cac gtt ata gac atg
tcg gac gaa aat tcg aag atg ttt acc gtg
gaa aag ggt aaa ata acg gat taa
cccaagtataaatgagcgagaaacgttgatgtcgaattc
t
Primers used to PCR amplify the cat-sacB cassette:
Primer-L: CCTGTGACGGAAGATCACTTCG
Primer-R: CTGAGGTTCTTATGGCTCTTG
Forward primer:
5’gccagagcgcggcttatgttgtcatgccaatgagactgtaCCTGTGACGGAAGATCACTTCG
Reverse primer:
5’agaattcgacatcaacgtttctcgctcatttatacttgggCTGAGGTTCTTATGGCTCTTG