1. a) Using any isotope(s) and tools you like that were available before DNA sequencing became routine, roughly outline a test to determine if the dsDNA genome from a virus can integrate directly into the genome of its host.
A good option would be to grow the phage in E. coli with 3H-thymidine in the growth medium; The viruses produced with tritium lableld DNA can then be allowed to infect unlablelled or lightly labeled bacteria. Those that survive after a short time should have the phage DNA integrated into the chromosome; detectable by autoradiography as in the case of the replicating DNA, but in this case, only a smaa portion of the circle of DNA should be labeleled.
If 32P was used, auoradiography would not be realistic since the whole film would be exposed. In that case, it may be possible to collect surviving cells after several numbers of divisions and demonstrate the fall-off of radioactivity (adjusted for the short half life) fit the expected values for the number of chromosome replications that occurred.
2. DNA replication turned out to require more than just a polymerase. What was the ‘key’ observation (in your opinion) that led to that realization and list at least five additional proteins are critical to DNA replication. Describe the role of the other proteins.
Lots of possibilities: the facts that PolI mutants were not lethal, that all only worked 5’ to 3’, that fragments with 5’ RNA bases were found all work well.
Other proteins:
SSB single stranded binding proteins protect from endonucleases
Gyrase- emovesupercoilsdy double stranbreakage nd rejoining
Topoisomerase- removesupercoils by single strand break and swaps
Helicase – unwind DNA double helix
Primase –create RNA primers for laggindstran Okazaki fragment syntheis
DNA polymerase A to remove RNA from fragments
Ligase- to seal the fragments
Telomerase- for maintaining ends of linear chromosomes
Rtc.
3. Compare initiation of translation of mRNA in a eukaryote to that in a prokaryote. What significance does this have to expressing a eukaryotic gene in a prokaryote?
Prokaryotes recognize a Shine-Delgarnosequnnce in the 5’ leader by homology to a region on the small ribosomal RNA in a context that aligns the start codon so that when the large subunit joins with the aid of transcriptions factors an F-met tRNAi is in position. This can happen as soon as transcription clears the area;
For eukaryoties that have a 5’CAP and polyA-tail with introns removed andthat have exited the nucleus, binding tlarger ribosomes requires more eIFs, including a CAP- binding protein to orient the mRNA to align the first AUG in the proper context and a factor that brings the A tail to the complex. In this case the tRNAi does not carry f-met.
To translate a eukaryotic mRNA in a prokaryote, the gene must be attached to a prokaryotic leader at the correct spacing.
4. CGA is a codon for Arginine. Each base could be altered as either a transition (TI) or transversion (TV). Show the potential outcomes for changes in terms of type and its effect on translation for
a) All possible first base changes
CGA to IUGA (TI) creates a stop codon (nonsense mutation)
CGA to AGA (TV) creates an ARG codon so is a samesense or synonymous mutation
CGA to GGA (TV) replaces ARG with GLY so is a missense (and changes the protein charge)
b) All possible 3rd base changes.
CGA to CGC (TI) samesense
CGA to CGU and to CGC (TV) both samesense
c) Of all the mutations, which do you think would have the most deleterious effect on the protein that should be produced?
The nonsense mutation since it would terminate the chain of amino acids prematurely.
5. 8-oxo-guanine is one of the most common effects of ROS on DNA. If unrepaired, it can pair with adenine. List a series of steps (enzymes, preferably) that could be involved in effective repair.
If not yet incorporated, superoxide dismutase and peroxidase could help;
Once incorporated, the ‘first’ mechanism would use a
glycosylase to remove the 8-oxoG from the backbone to make an AP site
alyase would nick the phosphate-sugar backbone
DNApolymerase I (most likely) would start at the nick and replace bases
A DNA ligase would seal any nicks remaining.
(other possibilities exist)
6. In his work with identifying mutants, reversions and suppressors using the 392 amino acid enzyme tryptophan synthetase (TS) of E. coli, Yanofski identified a mutant that produced no enzyme activity and had no response to anti-TS. When slow growing revertants were found, a specific protease fragment showed the following changes which were the only changes in the gene product.
Normal amino acid sequence: ala•pro•pro•leu•gln•asp•phe•
Recovered amino acid sequenceala•pro•pro•ile•ala•gly•phe
Propose a model, as detailed as possible to account for the twomutations that resulted in this protein.
Make a table showing all the possible codons in order for the original andrecovered over the region of change (leucine is critical as it has UUA/G and CUX codons. Careful inspection shows that inserting an A before a UUG would create an ileu and with correct choices of original 3rd base of the codons, provide the als and gly. To get back to the original reading frame the third base of codon preceeding the phe must be deleted. While these two changes could occur in either order, the only way to make the change requires one single base insertion and a single base deletion.
Give an example of a mutagen that would be expected to cause the changes. An acridine dye
7. Would the following partial diploid and mutant strains of E. coli make high (H), or low including zero (L) levels of -galactosidease andpermease ? A minus superscript indicates a nonfunctional allele, and fs a frameshift, while is indicates a repressor that can’t interact withallolactose.
Carbon Source in Medium
Glucose onlyZ Y / Lactose only
Z Y / Gluc + Lactose
Z Y
p I p O Z Y- A / p I p O- Z- Y A / L / H / H / H / L / L
p I- P O Zfs Y A / p I P O Z- Y / L / L / L / H / L / L
p- I P O Z Y A / P Is P O Z Y A / L / L / L / L / L / L
p I p O Z Y A but CAP protein absent / L / L / L / L / L / L
p I p O Z Y A but CAP protein can bind DNA without cAMP / L / L / H / H / H / H
8. Compound “E” is a derivative of histidine that is not required for survival of Neurospora crassa, but it is essential for conidia germination in the presence of H202. A set of “e-“ mutants unable to make E were used to establish the biosynthetic pathway. Results of the tests with prospective precursors S, P, and C are shown in the table below, with + indicating functional synthesis of compound E.
Mutant / Histidine / S / P / C / Ee1 / - / + / - / + / +
e2 / - / - / - / + / +
e3 / - / - / - / - / +
e4 / - / + / + / + / +
e5 / - / - / - / + / +
a) Propose a protocol for isolation/identification of mutants that can’t make compound E
Mutagenizeconidia, grow individually in tubes, (complete would be OK)
Take conidia from each tube and identify those that will not grow in the presence of H2O2
These should be blocked ion some step of E synthesis.
b) Use the data in the table to show the E biosynthetic pathway, indicating where each mutant blocks the pathway.
Histidine -- P - S ------ C - E
E4 E1 E2/E5 E3
c) Mutants e2 and e5 were shown to be in different genes. Give at least 2 possible explanations for this observation.
1) They code for different polypeptides of the enzyme
2. There is another intermediate that isn’t listed
9. An Hfr strain of E. coli with the genotype Met+, pur+, thi+ ,StrS was mated to an F- strain with the opposite markers. A test for met+ transfer among the excongugants after 10 minutes gave no colonies, but after 12 minutes, there were. When 200 of these colonies were replica plated to test for pur+ and thi+, 100 grew on minimal medium, 70 were met+ only and 30 were met+pur+ but none were just mt+thi+.
a) make a ‘map’ to show the most likely order of the genes.
Thi-----pur------met ------
< 2min
b) Describe the components of the media used to identify:
1. met+ colonies after the mating was interrupted
minimal + Thiamin + purines + Strep (otherwise only all +++ would grow)
2. pur+ and thi+ mutants via replica plating.
For pur + (after met+ , strep + selection) min + Thi
For Thi+ min + thiamin
c) Outline another experiment that would verify the gene order of these markers.
Since the genes are all within 12 minutes and likely coser than that, either transpuction with P1 or transformation would work. Use the same concept as before: select for one marker and see how often each of the others were transferred in the same event. (To get the middle marker left out would mean that the two outside genes were transferred but then two additional crossovers would have to occur in a short distance.
10. As part of a study of the His operon in E. coli, Ames examined the level of expression of the enzymes present in a set of polar mutants. In the table below, P, A, D, PP and T are shorthand for 5 of the enzymes in the pathway. Activity levels, as compared to levels in wild type growing on minimal medium, are listed simply as N for normal or L for low (less than 20 % of normal).
Mutant / P / A / D / PP / T1 / N / L / N / N / N
2 / L / L / N / N / N
3 / L / L / N / N / L
4 / L / L / L / N / L
5 / L / L / L / L / L
a) What is the order of these enzymes in the histidine pathway.
------------ ---His
PP DTP A
(Or opposite direction) All enzymes after the frameshift mutation should be missing since it is an operon.
b) How can the fact that many the mutants often showed some enyme activity rather than 0 activity be explained?
There must be some cryptic restart sites in the mRNA
11.94% of mice homozygous for an asp to ala mutation at position 400 of a protein called DNA polymerase delta developed cancer and died by age 10 months. Only 3-4% of heterozygotes or homozygous normal animals developed cancer in 18 months. Humans who are heterozygous for a ser to asn mutation at position 478 of the equivalent human gene have a high risk of colorectal cancer, but no individuals homozygous for the mutation have been reported. Propose a function for DNA polymerase delta and an explanation of the apparent difference in man and mouse regarding copy number of the defective allele.
PolD is most likely involved in DNA repair
In mice one defective copy does not lead to increased mutation and cancer while in humans heterozygotes are affected. While homozygosityin humans could be lethal, t could also just mean it is a very rare defect. The two mutation may have a different effect on enyme function, which is easiest to explained if the enzyme is at least a homodimer. Then having a change that affects the ability of the protein to assemble correctly could explain why even heterozygotes can’t function as well in repair.