Download GENETICS 603 Exam 1, September 27, 2013 1. Which of the

GENETICS 603 Exam 1, September 27, 2013 1. Which of the experiments that led to acceptance that DNA is the genetic information or that DNA replication is semi-­‐conservative did you find to be either the most convincing or the least convincing? Justify your answer. Lots of good answers here; I generally subtracted 1 point if you didn’t mention why your choice was more/less convincing than at least one other experiment. 2. A catalyst used in organic chemistry can convert cysteine to serine by replacing the side chain -­‐SH with -­‐OH. Predict the outcome of an experiment conducted in 1963 by Chappeville et al. where the following tests were made. (Assume 100% conversion). a) Rainey nickel was added to a mix of amino acids before they were added to a cell free translation system with the tRNAs, activating enzymes, ATP, GTP, Mg++, ribosomes and an artificial mRNA made of alternating U and G. A bit confusing since I only inferred the catalyst in question was Rainey nickel, but I think all reached that conclusion. Here there would be no cysteine available so no charged tRNA-­‐
cys could be made to read the UGU codons so no protein would form. b) Rainey nickel was added to a translation system that included all the activated tRNAs and ribosomes, GTP, Mg++ along with the -­‐UGUGUGUGUGUGUGUG-­‐-­‐ mRNA. Here the cys already on its tRNA was converted to ser causing the protein made to be alternating ser•val (note; ATP is not needed once the tRNAs are activated) c) What was the significance, at the time, of their results? It proved that the tRNA, without regard to the amino acid it carried, deciphered the codons. 3. In 1969 Brusick found that ICR-­‐170 induced ad-­3 mutants of Neurosopora could not be reverted by nitrous acid but could be reverted by a second treatment with ICR-­‐170. He also noted that many of the revertants grew very slowly on minimal medium and had lower enzymatic activity than wild type. In one example where the mutation was traced to a change in only one tryptic fragment of the ad-­‐3 protein that normally has the sequence trp-­‐ala-­‐gln-­‐met-­‐thr, the enzyme in the leaky revertant had the sequence trp-­‐ser-­‐ser-­‐val-­‐thr. A) What is the significance of the use of ICR-­‐170 and the failure to revert with NA? They cause a different type of mutation; ICR-­‐170 is a frameshift mutagen while NA mostly causes oxidative deamination and therefore transitions. B) Show potential changes that occurred in the gene (or just the mRNA) to create the non-­‐
functional mutant and then the leaky revertant. WT: TRP ALA GLN MET THR Possible codons UGG GCU CAA AUG ACU GCC CAG ACC GCA ACA GCG ACG Mutant: Frameshift so do not detect the protein Revertant TRP SER SER VAL THR (the +/-­‐changes occur between TRP and THR) UGG UCU UCU GUU ACX UCC UCC GUA UCA UCA GUA UCG UCG GUG AGU AGU AGC AGC Possibility: Adding an A after the UGG in the WT and deleting the A from AUG using the codons indicated will produce the observed result; events could occur +/-­‐ or-­‐/+. 4. Although we have not covered gene-­‐cloning technology so far, we have made comparisons of pro-­‐ and eukaryotes with regard to gene expression. What differences would be critical factors in the ability to express a eukaryotic gene in E. coli? 1) introns must not be present 2) there must be a leader with a Shine Delgarno sequence to allow translation 3) the coding sequence of the gene should be fused to a bacterial promoter or it will not be transcribed 4) the preferred codons may have to be substituted 5) the protein made may have to be modified in order to function 5. There are multiple mechanisms for repairing DNA adducts such as 8-­‐oxopurine and TT dimers. List at least 4 different mechanisms and mention, by function, the enzymes that would be required for each. 1) Remove TT dimers using photolyase (and blue light) 2) Create an AP site using a glycosylase, an endonuclease like AP lyase to nick the strand by the AP site, an exonuclease (or PolI) to remove some bases, a polymerase to fill the gap and ligase to seal 3) Dark repair, short patch and long patch repairs also require a ‘nicking’ enzyme (endonuclease), an exonuclease to open the gap and polymerase and ligase to complete the mend. 4) Post replication or SOS repair requires Rec A in addition to the other enzymes. 5) Some adducts will pair poorly and be removed by the 3’-­‐5’ exonuclease activity of DNA polymerases. 6) Pre-­‐lesion repair can be via glycosylases to repair modified bases or superoxide dismutase and catalase to eliminate free radicals. 7. 5-­‐Fluoruracil and 6-­‐azaguanine are often part of the mix of compounds used in multidrug chemotherapy to treat cancer. Present a hypothesis as to why these compounds are included. These base analogs readily form tautomers if incorporated into DNA (like 5BU they can beconverted to nucleotides), which is most likely to happen in replicating cells, a feature of cancer. Some somatic mutations they cause will be lethal to that cell. There is also some chance they will interfere with normal nucleotide biosynthesis and slow growth by starving the cancer cells for the nucleotides essential for continued growth. 8. ‘Amber suppressors’ are very specific for the UAG codon. A) What aminoacyl-­‐tRNAs could become amber suppressors by a 1) transition, 2) transversion? 1) GLN, TRP 2) LEU, SER, TYR, LYS, GLU b) Ochre (UAA) suppressors are not as specific and the suppressor strains grow very poorly. Why are they ‘less specific’ and speculate on possible reasons they are ‘sicker’ than amber suppressor cells. 1) There may be more UAA stops where inserting an amino acid is detrimental 2) The UAA suppressors would wobble to also suppress UAG whereas the reverse is not true. 3) The RF/tRNA competition could differ for the two stop codons 9. Aromatic amino acids (phe, tyr and trp) all are derived from a common precursor, chorismate. Isolation of Neurospora ‘arom’ mutants that could grow if provided chorismate allowed the initial steps of the pathway to be elucidated, as shown in the feeding studies illustrated below (modified from actual for test purposes): (+ = growth; -­‐ = no growth) Mutant Minimal + S Minimal + Minimal + Minimal + Min + strain DHQ 3ESA 5PS chorismate arom1 -­‐ -­‐ -­‐ -­‐ + arom2 + -­‐ + + + arom3 -­‐ -­‐ + -­‐ + arom4 -­‐ -­‐ + + + arom5 -­‐ -­‐ + + + A) Create a pathway assuming all the compounds are in fact components and show the site where each arom mutant creates a block. Arom2 arom4 & 5 arom3 arom1 DHQ S 5PS ESA Chorismate B) Explain how strains 4 and 5 that are mutations in different genes can show the same response in the growth tests. They may code for different subunits required by one enzyme There may be a missing intermediate 10. The two amino acids shown below are both found in natural proteins and their presence is essential for function. How do they get there? Hydroxy proline Selenocysteine Hydroxy proline is formed by enzymatic addition of a hydroxy group to proline after translation. Selenocysteine, is created by enzymatic addition of Se to a serine on a special tRNA that recognizes a UGA stop codon, in context, with the aid of another factor that competes with the release factor. (It is the ‘21st’ amino acid.) 11. Place the letter or letters of the enzymes or proteins in the blank before each function for which it applies. A. Polymerase I E Unwinds a DNA double helix B. Polymerase III A, B Has 3’ to 5’ exonuclease activity C. SSB G Relieves supercoiling D. Primase A, B 5’ dNTPs added to 3’OH of a polynucleotide E. Helicase A 5’ to 3’ exonuclease activity F. ligase C Prevents reassociation of complementary bases G. Gyrase F Binds a 3’ OH of one polynucleotide to the 5’ end of another (topoisomerase) polynucleotide A, B, D Is a 5’ to 3’ polymerase E Unwinds DNA double helix D connects NTPs 5’ to 3’ on a DNA template 12. A series of different HFR strains of E. coli that could all grow on minimal medium with either maltose or mannose were allowed to mate for 40 minutes with an F-­‐ strains with mutations that was auxotrophic at 8 loci and unable to use either sugar. After disrupting the mating at 40 minutes, and killing the males, the exconjugants were plated on glucose complete medium. Replica plating revealed the following markers had been transferred. (Markers are listed in alphabetic order, since no attempt was made to time the entry.) HFR-­‐1 arg+, cys+, mal+, tyr+ HFR-­‐2 arg+, ilv+, leu+, thr+ HFR-­‐3 aro+, leu+, pur+, thr+ HFR-­‐4 aro+, pur+, trp+ HFR-­‐5 man+, try+, tyr+ A) Map, as far as possible the order of the genes and possible HFR insertion sites. ilv arg ( thr leu) (mal (aro cys) pur) tyr trp man pairs in brackets could be either way, HFRs could transfer either direction B. Suggest a method that could be used to kill the males that should be effective for all 5 HFR strains. A male specific lytic phage would be best since it would work for all HFRs and not kill any recipients after only 30 minutes of mating C. If the specific order of genes was not revealed in the data, propose another mechanism that could be used to make that determination. Include the markers that you would use. Test for co-­‐transformation or co-­‐transduction: select for a flanking marker such as tyr+ and determine if cys+ or mal+ were most often co-­‐transferred D. List components of the media used in the replica plating to differentiate arg+ from arg-­‐ colonies. Glucose complete would allow all combinations to grow: colonies missing replicate plates containing glucose + low levels of the other compounds except for arg would be arg-­‐ E. What media would be used to differentiate mal+ from mal-­‐ colonies? Plate on glucose or another sugar with all potential supplements (low levels) needed for
auxothophs to allow mal+ and mal- colonies to grow.
Replicate plate to maltose plates with same supplements; any that cannot grow should be mal-.
13. Which of the following mutations that alter the anticodon GUU (written 3’ to 5’ as it would pair with its codon) would you predict to have the most damaging effect on cell survival, and why? a) GUU to GGU or b) GUU to GUC? GUU to GGU would cause proline to be inserted where glutamine should which would likely cause protein misfolding. GUU to GUC would still be inserting a glutamine. Another thing to consider is the consequences of loss of the GUU antidodon-­‐containing DNA; if it is the only one that can pair with CAA then this could be a lethal defect and if it is one of very few, the general slowdown of translation could be a major problem. (This assumes that the anticodon is not the entire basis for the specificity of the activating enzyme, which for certain is true for the amber and ochre suppressor mutations. In some cases the anticodon has a role, but is not the sole bases for selectivity in charging.) 14. The three genes required by E. coli to utilize arabinose are transcribed on one mRNA, in the order, araB, araA araD. AraC is a distinct gene whose protein shows positive regulation of the Ara-­‐BAD operon and of itself. A) Make a drawing showing how regulation could be managed in the presence of glucose or arabinose as the sole carbon source. B) Predict the effects of mutations that disabled araC. Since the araC product is required to make the araBAD mRNA, the enzymes will not be made no matter the sugar source. C) Would regulation of the Ara-­‐BAD operon be subject to cAMP levels? Justify your answer. Most likely, since glucose is still the optimal carbon source, it would be useful to have the araBAD operon activated only when cAMP levels increase indicating a deficiency in ATP. (This is in fact the case.) 15. Predict the effect of the following mutations as they occur in different gene in the tryptophan operon of E. coli. Tell whether the enzymes made by the indicated genes will be present at high or low levels in the haploid and partial diploid strains shown below when grown with or without tryptophan. Anthranilate Synthase (AS) is made of subunits from trp E and trp D, Indole glycerolphosphate synthase (IGPS) is from trp C and Tryptophan synthase (TS) has subunits made by trp B and trpA. The Regulator gene ‘R’ is separated from the structural components that are shown in correct map order in the strains below. Use H to indicate high levels of an enzyme L for low levels and N for no enzyme activity. A superscript – after an allele indicates that component is inactive while FS indicates a frameshift. tryptophan added no tryptophan Strain genotype(s) AS IGPS TS AS IGPS TS pR p O L E D C B A L L L H H H H H N H H N pR-­‐ p O L E D C B-­‐ A pR p O-­‐ L E D CFS B A H N N h N pR p O L E-­‐ D C B A/ p O L E D C B-­‐ A L L L H H pR p O L E D C B-­‐ A/ p O-­‐ L E D C B A-­‐ H H L/N H H pR-­‐ p O L EFS D C B A/ p-­‐ O L E D C B A N N N N N N H H N B) Suppose the two successive UGG codons were deleted in the Leader (L). What effect would this have on transcription? This would lead to the loss of attenuation and increased mRNA even when there is plenty of tryptophan since there would be no slowing of translation in the leader due to low levels of try-­‐tRNAtry