Genetics 603, Exam 1, Oct. 1, 2010

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1 Genetics 603, Exam 1, Oct. 1, 2010 Name 1. Avery et al showed that DNA caused transformation of rii to SIII bacteria before isotopes were available. Describe (in principle, not in experimental detail) how they might they have used one or more of the isotopes 3 H, 32 P, 15 N or 35 S to diminish skepticism of their results. Since bacteria have more than just protein and DNA, the best bet might have been to prove that the purified DNA that caused transformation did not contain protein as shown by the lack of any 35 S in DNA extracted from bacteria grown in that isotope. It would also help to show that the recovered transformed bacteria did have some 32 P in the DNA extracted from the transformant SIII bacteria. Lots of other possibilities could apply, but N and H aren t too useful since they would end up in both DNA and protein. 2. ADAR enzymes (adenosine deaminase acting on RNA) act on double stranded RNA through oxidative deamination of adenines to create the base inosine. a) Propose an explanation as to why ADARs are assumed to have evolved to protect against RNA viruses that have a double strand as part of their life cycle. Changing an A to an I would allow pairing with A, T or U, and C (as discussed in trnas). Thus many mutations would be incorporated during replication, many aberrant proteins formed during translation, etc.. all of which would be predicted to disrupt viral reproduction. b) Hepatitis delta virus, the smallest virus known to infect humans at 1700 bases, 70% of which are self complementary, has one open reading frame but in the presence of ADAR1 can make 2 proteins, one of which is 20 amino acids longer than the other. Only if both proteins are made can the virus reproduce. How has the virus thwarted the protection presumed to be provided by ADARs? It can only make both proteins needed to replicate f an A in a stop codon is changed to an I so that the longer protein can be made. 3) The universal genetic code has codons for 20 amino acids, but two others are found (individually) in a few rare proteins. How do they get there? Both selenocystine and pyrolysine are made by enzymatic addition of the seleno- or pyro- groups to special trnas charged with cystine and lysine, respectively. The trnas anticodons allow them to bind to a nonsense codon within a coding context recognized by a specific factor that displaces the normal release factors. 4) At ph near 7, the pyrimidine 5- OH, 2 deoxycytidine is 99% in the amino form (pairs with G) and 1% in the imino form (pair with A). Explain why it is likely to be a mutagen and the type of changes it would be predicted to cause. Also tell why it is possible that it would not be mutagenic to a detectable level. The analog could be incorporated as either a C or T analog, leading to a high frequency of potential transitions in subsequent replications. Like 5 BU, this rate of tautomer formation is much greater than that of the normal bases. It would only have this effect if it can be converted to a nucleoside tri- phosphate that is incorporated into DNA during replication or repair.

2 5) The base analog 6- mercaptopurine along with compounds that inhibit de novo synthesis of natural purines have been combined for use in chemotherapy against cancer. What is the rationale for this combined drug treatment? Preventing the de novo synthesis of purines will increase the chance that 6- SH- purine is Incorporated into DNA and/or RNA where it will lead to hopefully lethal mutations and aberrant proteins. Since cancer cells are continuously going through DNA replication and cell division, the base analog mutagens will affect them to a much greater extent than non- dividing somatic cells. (side effects such as loss of hair and cells lining the stomach are also a consequence) 6) A single A to G transition within the human β- globin gene has been found to cause partial anemia in homozygous individuals. Ninety percent of the β- globin in these individuals has a block of 19 extra amino acids within the normal chain of 146 amino acids, while 10 % of the β- globin made is normal. Propose an explanation for these observations. Most likely (and confirmed) is that the change alters a splice site normally used to remove an intron which happens to code, in phase, for 19 amino acids in the β globin transcript. 7) Srb and Horiowitz found evidence that 4 different genes in Neurospora could mutate to cause auxotrophy for the amino acid methionine (met - mutants). When tested for the ability to grow on potential precursors, the following results were obtained: Mutants/Compounds minimal cystine Homocystine methionine met met met met When grown on limiting amounts of methionine, the met3 mutants accumulated a compound X that allowed met2 and met4 mutants to grow. A) Make a pathway for methionine biosynthesis that accounts for these data, showing the compounds and sites of each of the blocks. met4 met2 met3 met1 cys X homo methionine B) Why was it essential to grow the met3 mutants on limiting methionine in order to observe compound X? Excess methionine would repress expression of gene in the pathway and feedback inhibition of the first enzyme would shut down the synthesis of X. C) How did they know that met2 and met3 were not mutations in the same gene even before compound X was discovered? Crosses between the two mutants gave some normal progeny (or complementation tests in heterokaryons) 2

3 8) The original segment of a gene that codes for a gene in the lysine biosynthetic pathway of yeast has the amino acid sequence: - - Gln- Asp- Leu- Glu- Arg- Thr- A lys2 - mutant with no activity was screened for revertants on minimal medium. Among several revertants detected, one had the sequence: - Gln- Val- Asp- Leu- Glu- Arg- Thr- and another had the sequence: - Gln- Asp- Arg- Ser- Gly- Glu- Ser- Thr- All amino acids preceeding Gln and following Thr were the same as in the wild type yeast. A) Noting that these two revertants (and many others that are not shown) increase the number of amino acids propose a mechanism that might account for the original mutation and revertant 1. The original mutation was most likely a frameshift involving addition of one or two bases, and the reversion added 2 or 1, respectively, to add a codon and restore the reading frame with an extra amino acid. For example, adding a U between the Gln and Asp codons would create an in- frame stop and then adding another GU before that U could account for reversion1. B) Now, realizing that both revertants had the same original change, can you come up with a model for revertant 2? This means the hypothesis for part A isn t sufficient. There are likely several possible explanations, one being the original mutation added 4 bases (GATC for example) that included the VAL codon and correcting revertants took one base away (A if GATC) or added two more bases downstream to get back to the correct reading frame with one and two more extra amino acids, respectively. C) What kind of mutagen would be most likely to cause these mutations? An acridine dye (intercolating agent) D All of the base changes leading to lys2 + reversions occurred within a 30 base upstream or 110 bases downstream of the original mutation site, ie., the region of altered amino acids was limited. Propose an explanation fro this observation, Reading frame corrections that are too far away would mean a longer out of normal sequence of amino acids, or even an in- frame stop codon occurring within the affected area. If this part of the protein is critical in any way, too many changes would almost certainly be detrimental. 3

4 9) The bacterium Acinetobacter calcoaceticus has a natural transformation system that takes in segments up to 9 kb. Gene pca+ is required for growth using protocatechuate as a carbon source which can be made from p- hydroxybenzoate if the gene pob+ also is present. DNA from a pob+, pca+ donor was used to transform a pob-, pca- recipient and colonies that grew on protocatechuate were replica plated to p- hydroxybenzoate medium. 16% of the colonies grew. Based on this information is it likely that the pob and pca genes are close together? Explain and tell any assumptions you are making. Likely fairly closely linked since transformation is a rare event and the odds of 2 different fragments entering and incorporating at the same time would be very low. This assumes the genome is not exceptionally small (9 kb would be less than 1% of most bacterial chromosomes so even with a high rate of transformation, co- transformation would be expected to be less than I in 10 4 for genes that are not close together. 10. A gene (ace + ) that allows acetate to be used as a carbon source is normally the last gene transferred in a particular HFR strain of E. coli. In some cases of crosses with an ace - recipient, ace + recipients can be recovered just a few minutes after a mating mixture was made. This change can occur even if the recipient is also reca -, which is not true when the transfer occurs after 100 minutes of mating. How are these observations explained? Hfr ace+ If the F forms an F taking along the ace+ allele it would transfer quickly. Congugation would still occur and the ace+ gene would be expressed in the recipient without being incorporated (ie without the aid of the rec protein 11. How does translation of mrna in prokaryotes differ from that in eukaryotes? What makes these differences important? Initiation in prokaryotes uses base pairing of the Shane- Delgarno sequence to the small subunit rrna to position the start codon which inserts an F- met. Eukaryotes use different IF factors, one of which recognizes the 5 CAP at the beginning of the message and the initiation trna does not carry F- met. Translation is concurrent with transcription in prokaryotes, but in eukaryotes, mrnas must move to the cytoplasm to be translated on larger ribosomes than are found in bacteria. Stop codons in prokaryotes require 2 release factors, only one is needed in eukaryotes. Importance: Leaders must be changed to read eukaryotic messages in prokaryotes; the translation systems are sensitive to different antibiotics. 12. Conjugation using an Hfr strain that had the genotype lac +, arg +, his + and Ton S was mated to a female with the opposite markers. (lac = lactose, arg = arginine, his = histidine and Ton = phage T1) After 30 minutes, the matings were disrupted, phage T1 was added and the cells plated on glucose complete medium. a) What was the rationale for adding the T1 phage? Kill donor cells b) Since the insertion site of the F- factor was unknown initially it would possible that the Ton S alleles was also transferred in the 30 minutes. Suggest an experiment that would 4

5 allow this possibility to be tested. All cells could be plated on lactose complete medium (or medium deficient in arg or his to eliminate unmated females. Replica plating could be used to determine if any recipients with a combination of male and female markers also contained the Ton sensitive marker from the male. c) What media should be prepared to distinguish the 4 possible genotypic combinations for the lac, and arg loci? If recipients are plated on glucose + his + arg, all combinations (lac+/-, arg+/- ) will grow. Replica plating on Lac + his + arg will identify which colonies are lac+ (and lac- ); plating onto glucose + his but no arginine will identify those that are arg+ (and arg- ) 13. Shown below are a series of strains of E. coli that have either one or two copies of the lac operon. Defective components are indicated with a superscript except for frameshift mutations which are shown as FS, and S indicates a superrepressor. A) Indicate in the appropriate boxes whether synthesis of β- galactosidase, permease and transacetylase will be constitutive (C), regulated (R) or always off (A). Strain β- gal permease TA ase 1. P I P O - Z Y A C C C 2. P I S P O Z Y A A A A 3. P I - P O Z Y - A C A C 4. P I P - O Z Y A A A A 5. P I S P O - Z Y A - // P - I P O Z - Y A C C A 6. P I P O - Z Y FS A - // P I P O Z - Y A C R R 7. P I - P O Z Y - A// P - I P O Z Y A - C C C B) Would any of the strains show different responses if lactose was used rather than IPTG to test for induction? If so, which one(s). None- the only ones that are either Z- or Y- that might require IPTG are already constitutive. 14. As described in class, β- thienylalanine is a toxic analog of phenylalanine used to screen newborns for PKU. It is possible that mutations in the bacteria used in the test could lead to either false negative or false positive results. Give examples of mutations that could do so. False negative (don t grow despite high phe in blood sample) these could result from mutations that prevent uptake of phe or have a new unsupplemented auxotrophic mutation. False positive (grow on samples with no excess of phe) These could be mutations that overproduce phe on their own or that have gained ability to degrade or incorporate the β- thienylalanine.. 5