Download Problem Set 4B

Problem Set 4B
Biology 210
1) How is the expression of the trp operon controlled in bacteria?
The trp operon in bacteria produces enzymes that convert chorismate to the amino
acid tryptophan. There are two methods of expression control. In the first,
transcription of the operon is repressed by a regulatory protein when tryptophan is
present. Thus, the trp operon is a negative repressible operon. The second mode of
expression control is through attenuation. When tryptophan concentrations are
low, there are few charged tRNAs present that encode tryptophan. When
translation of the 5’UTR of the trp operon starts, the ribosome stalls on the trp
codons. This allows the formation of a secondary structure in the mRNA that
allows further transcription. When tryptophan concentrations are high, the
ribosome does not pause, and a second secondary structure is formed by the mRNA.
This secondary structure resembles a termination hairpin. Bond between nearby
Uracil (mRNA) and Adenines (DNA) are broken, and transcription is terminated
prematurely.
2) For the following lac operon partial diploid genotypes, indicate whether enzymatic
activity would be detected for Beta Galactosidase and Permease when lactose is absent
and when lactose is present. In each case, can the bacteria survive with lactose as the
only food source? What happens to the expression of the lac operon when both lactose
and glucose are present? Why?
A.
B.
C.
D.
lacI+ lacP- lacO+ lacZ+ lacY- / lacI- lacP+ lacO+ lacZ+ lacY+
lacI+ lacP- lacO+ lacZ+ lacY- / lacI- lacP+ lacOC lacZ- lacY+
lacIS lacP+ lacOC lacZ+ lacY- / lacI+ lacP+ lacO+ lacZ+ lacY+
lacI- lacP+ lacOC lacZ+ lacY- / lacI+ lacP+ lacO+ lacZ- lacY+
A
B
C
D
Lactose Absent
Beta-Gal
Permease
N
N
N
Y
Y
N
Y
N
Lactose Present
Beta-Gal
Permease
Y
Y
N
Y
Y
N
Y
Y
Survive?
Y
N
N
Y
When glucose and lactose are present, the lac operon is repressed via catabolite
repression. In this case, low concentrations of cAMP result in infrequent formation
of a cAMP complex with catabolite activator protein (CAP). This cAMP/CAP
complex is necessary for transcription. This occurs because it is energetically
wasteful to transcribe genes for the lac operon (which helps convert lactose to
glucose and galactose) when an efficient source of energy (glucose) is already
present.
3) At what levels can gene expression be controlled in eukaryotes? For each level,
provide one example mechanism.
Eukaryotic genes can be regulated at the following levels:
Gene Structure: CpG islands can repress transcription of adjacent genes
Transcription: Regulators, silencers, enhancers. Insulators, response elements
mRNA processing: Alternative splicing
Regulation of mRNA stability: stability affected by 5’Cap, Poly A tail, 5’ UTR, 3’
UTR
Translation: availability of translational apparatus
Post translational protein modification: chemical modification of proteins
4) For the following, indicate the genetic (what happens to the DNA sequence) and
phenotypic effect, if any. If the phenotypic effect cannot be determined, indicate that it is
“Unknown” in your answer.
A. Nonsense mutation in the lacY gene.
Nonfunctional permease. A stop codon is introduced in the protein coding
sequence. Lactose is no longer transported into the cell.
B. Neutral mutation in the DNA Glycosylase gene.
The amino acid sequence of the enzyme is changed, but there is no effect on the
DNA repair pathway.
C. Deletion (4 bases) mutation in the DNA polymerase gene.
Shift of the reading frame will cause a non functional protein to be translated. If
this is the only DNA polymerase gene, the cell will likely die.
D. Loss of function mutation in a Mismatch Repair protein gene.
Mismatch repair is disabled because one of the proteins is no longer functional.
This could be the result of many different kinds of changes at the nucleotide level.
E. Missense mutation in lacP. Note, this was changed to “Large deletion of LacP”.
The promoter is deleted, RNA polymerase cannot bind, and transcription of the lac
Operon is halted. The cell cannot use lactose as an energy source.
F. Suppressor mutation of lacZThe function of the lacZ gene is restored, functional Beta Galactosidase is produced.
5) Describe the function and mechanism of the 4 major DNA repair pathways.
For functions (type of damage repaired) see table 17.5
A. Mismatch Repair:
Detects mismatches through distortions in 3
dimensional config.
Exonucleases remove newly synthesized strand and it is
replaced by DNA polymerase
Direct Repair:
Damaged nucleotides are replaced with their correct structures
Base Excision:
Modified bases are excised by DNA glycosylases and replaced
Nucleotide excision: detects distortions of 3 dimensional configuration
DNA strands separated and stabilized with single stranded
binding proteins
Damaged strand is removed and polymerase fills the gaps.
6) For each of the following mutagens (or mutagenic processes), describe the kind of
mutation that can be induced, and identify the DNA repair pathways that can be
employed to repair the mutation.
i.
ii.
iii.
iv.
Strand slippage during replication
Chemical modification (deamination)
Exposure to UV light
Exposure to mustard gas
i. strand slippage induces small insertions and deletions and is repaired by
mismatch repair.
ii. deamination changes cytosine to uracil, mutating a CG pair to an AT pair
(repaired by direct repair, or base excision repair).
iii. produces pyrimidine dimers that can be repaired with base excision
repair
iv. mustard gas is an alkylating agent, and can modify bases by adding alkyl
groups (e.g. to guanine). This can cause a mismatch in pairing. This can be
repaired by base excision or nucleotide excision