Name Student ID# Bacterial Genetics, BIO 4443/6443 Spring Semester 2001 Final Exam 1.) Different bacteriophage utilize different mechanisms to ensure that their own genes (and not host s genes) are transcribed and regulated. Describe a mechanism that each phage uses to transcribe and regulate their own gene expression rather than those of the host. (6pts) phage T7: encodes its own RNA polymerase phage T4: encodes a sigma factor to modify the recognition of the E.coli RNA polymerase phage λ (lambda): uses the host RNA polymerase and encodes antitermination proteins that regulate which genes are tanscribed. (Several other answers would also be acceptable for each phage) 2.) T7 has a typical lytic life cycle. Describe two gene products that you expect to be expressed early after infection? (9pts) RNA polymerase Antirestriction Protein Kinase that inactivates the E. coli RNA polymerase RNA polymerase Host DNA degradation genes DNA ligase Describe two gene products that you expect to be expressed late after infection? DNA Polymerase Primase/Helicase Phage Particle Parts Lysis genes Hok lysozyme How does the early viral DNA replication differ from late viral DNA replication? Early: Controlled bidirectional replication initiated from a unique origin Late: Replication become highly recombinagenic utilizing a rolling circle form of replication to amplify the genetic material 1
Bacteriophage T4 normally produce large, fuzzy edged plaques when plated on a lawn of E.coli. However, rii mutants (named for rapid lysis, classii) produce smaller, clear edged plaques on E.coli. Additionally, it is known that rii mutants cannot grow at all in E.coli which carry a lysogen of the phage lambda, E.coli λ. By contrast, wild type phage, r+, grow normally in E.coli λ. The type of plaque formed in each case is sumarized below. Use Wildtype T4 (r+) to infect E.coli Use Wildtype T4 (r+) to infect E.coli λ Use rii mutants to infect E.coli Use rii mutants to infect E.coli λ No Growth 3.) Seymour Benzer made use of these properties to determine several fundamental genetic principles. He first isolated and collected several different rii mutants based upon their plaque morphology. Suppose you were trying to isolate rii mutants. After you mutagenize your phage, would you infect E.coli or E.coli λ to screen for your mutants? Why? (4pts) E.coli, because the mutants you are hoping to isolate would not grow in E.coli λ. After isolating several mutants, Benzer then tried to determine how many genes were represented by the rii mutations. He mixed two different rii mutations together and infected E.coliλ cells at a high MOI. Using a sample of just four of his mutants, the type of plaques produced by each combination is shown in the table below rii #1 rii #2 rii #3 rii #4 rii #1 No Growth rii #2 No Growth No Growth rii #3 No Growth rii #4 No Growth No Growth 4.) What type of analysis is this? (3pts) Complementation analysis 5.) How many genes are represented by the four mutants shown above? (6pts) Two: Gene A (rii #1 and rii #2) and Gene B (rii #3 and rii #4) 6.) Why is it necessary to infect at a high MOI for this experiment to work? What would have occurred had he used a very low MOI instead? (2pts) You need to make sure most cells are infected by both mutant phages. At a very low MOI, cells would only get infected by one of the mutations and would have no chance to assay for complementation. 2
riib is a gene that causes an rii phenotype when it is inactivated. Francis Crick and collegues utilized the riib gene product to demonstrate that the genetic code was most likely based on three letters (i.e three bases code for one amino acid). As part of this experiment, they mutagenized T4 phage by growing them in the presence of acridine dyes and isolated an riib mutant (remember that acridine dyes generally induce either a one base pair insertion or a one base pair deletion). 7.) Then they isolated several supressors of riib that restored wild type function (an r+ phenotype). Using E.coli λ, what would be an easy way to detect riib phage that have suppressor mutations? (5pts) Infect and plate on E.coli λ, phage that are able to grow and form plaques are the suppressors. Once they had obtained several riib suppressors, they needed to know which ones were true revertants and which ones were second site suppressors. (The second site suppressors were going to be central to their experiment). To do this, they decided to cross the riib suppressors with wild type phage and examine the type of recombinants that are produced. Shown to the right, are what each type of phage plaque looked like before the crosses were performed. If he used wild type T4 to infect E. coli, he observed only large, fuzzy edged plaques. If he used his riib mutant to infect E. coli, he observed only small, clear edged plaques. If he used any of his riib suppressors to infect E. coli, he observed only large, fuzzy edged plaques. To cross the phage, they infected E. coli with a mixture of a wild type T4 phage and an riib suppressor. Then, they allowed the co-infected E.coli to lyse and collected the progeny phage. To examine the recombinants that were formed, they infected the E.coli again, this time using only the progeny phage from the cross, and plated the infected cells out on a lawn of E.coli. Shown below are the types of plaques produced on each plate. The progeny phage from some crosses produced a mixture of large, fuzzy edged plaques and small clear edged plaques on E.coli. However, the progeny phage from other crosses produced only large, fuzzy edged plaques on E. coli. 8.) Based upon the above results, which suppressors are likely to be a true revertants and which are most likely to be a second site suppressors? Why? (9pts) Second site suppressors produced the mixed plate since a cross between a double mutant and wild type phage could produce single mutant recombinants. True revertants produced the plate with only wild type phage since a cross between two wild type phage can produce only wild type recombinants. 3
Phage λ is able to regulate its gene expression such that it can either enter a lytic or lysogenic life cycle upon infecting E. coli. The principle region that regulates this decision is shown below. pr attp tl2 ciii tl1 OL1 OL2 OL3 t RnaseIII cleavage site int xis N ci OR1 OR2 OR3 cro tr1 cii O P tr2 Q pi pl prm 9.) For each bacteriophage below, determine whether the bacteriophage will enter a lytic or lysogenic cycle upon infection and explain why. (6pts) A phage λ with a mutation inactivating the ci gene Lytic. CI is needed for repression of pr and pl (the lytic promoters) A phage λ with a mutation inactivating the cii gene Lytic. CII is needed to activate pre and pi which transcribe the ci and int genes, both of which are required for lysogeny. A phage λ with a mutation inactivating the cro gene Lysogenic. Cro competes with CI for the Or1 Or2 and Or3. Without Cro, CI binds and promotes lysogeny 10.) Phage integration into the chromosome requires that only the int gene product be present, whereas phage excision from the chromosome requires the function of both the int and xis gene products. How does phage λ regulate the expression of these genes such that int and xis are only expressed at the proper time? (4pts) Before integratation, the transcripts from pl containing int and xis are degraded because they also contain the RNaseIII degradation signal. Only transcripts from pi, which contain only the int gene can be expressed in a stable form. Following integration using the attp site, the circular lambda genome is linearized removing the RNaseIII degradation sequence from the region. This stabilizes the pl transcript which contains both the int and xis genes. You need to determine if two mutant phage belong to the same complementation group. In your experiment you decide to infect at an MOI of 2. 11.) Using the poisson expression, P i = ( m i e -m )/ i!, what fraction of the bacteria in your experiment will actually be infected by exactly two bacteriophage? Show your work. (6pts) probability of getting infected by exactly 2 phage= = ( 2 2 e -2 )/ 2! = (4)(.135)/ 2 =.27 or 27% or 27/100 bacteria What fraction of the bacteria will be infected by two or more phage? probability of getting infected by more than 2 phage= 1-( +P 1 +P 0 ) + = 1-(( 2 2 e -2 )/ 2! + ( 2 1. e -2 )/ 1! + ( 2 0 e -2 )/ 0! + = 1-((.27) + (.27) + (.14) + = 1-(.68) =.32 or 32% or 32/100 bacteria pre 4
Shown below are two operons that control genes for the metabolism of different carbon sources. The lactose operon and the L-arabinose operon. plac pbad laci O lacz lacy laca arac O2 O1 I arab araa arad placi parac The lac operon is regulated by the laci gene product, while the L-arabinose operon is regulated by the arac gene product. Although these operons are similar both laczlacylaca and arabaraaarad are only expressed when either lactose or L-arabinose is present in the media, the are regulated through very different mechanisms. 12.) How does LacI regulate the lac operon? What type of regulation is this called? (2pts) In the absence of lactose (allolactose), LacI binds to the Operator (O) and represses transcription. In the presence of lactose (allolactose), allolactose binds to LacI preventing it from binding to the Operator, allowing the lac operon to be transcribed. Negative regulation. 13.) How does AraC regulate the L-arabinose operon? What type of regulation is this called? (2pts) In the presence of arabinose, arabinose binds to AraC allowing it to bind to the Operator (I) and activate transcription. In the absence of arabinose, AraC cannot bind to the Operator and the operon cannot be transcribed. Positive regulation. Suppose that you were trying to isolate mutants that constitutively express either the lactose operon or the L-arabinose operon. 14.) What is the most likely type of mutation would you expect to get that would produce a lac constitutive phenotype? (2pts) Any mutation that inactivates the laci gene 15.) What what is the most likely type of mutation would you expect to get that would produce an ara constitutive phenotype? (2pts) A specific (subtle) mutation that changes the conformation of AraC so it will bind the operator in the presence or absence of arabinose. 16.) If you were to screen equal numbers of total cells in each case, would you expect to obtain more lac constitutive mutants or more ara constitutive mutants? Why? (2pts) You would expect to get many more lac constitutive mutations since a mutation anywhere within the laci gene or promoter that inactivate it would produce the desireed phenotype, whereas there are probably only one or two bases in the arac gene that would produce the subtle change that you are looking for. 17.) If a wild type lac operon was introduced on an F into the lac constitutive mutant, would the constitutive phenotype be dominant or recessive to the wild type operon? Why? (2pts) Recessive. The wild type copy of LacI could still bund and repress both repressors. 18.) If a wild type ara operon was introduced on an F into the ara constitutive mutant, would the constitutive phenotype be dominant or recessive to the wild type operon? Why? (2pts) Dominant. *If we assume AraC is simply an activator. Then, the constitutive mutant would always bind both copies of the operon and activate them. (In reality, AraC also functions as an antiactivator protein, so Recessive would also be correct as long as you explained why.) 5
You have an auxotrophic bacterial strain that is resistant to naladixic acid, his-, trp-, nal R. Because all three markers appear to map closely, you decide to map these genes by transduction using a three factor cross. You use a wild type strain, his+, trp+, nal S, as your donor and the his-, trp-, nal R as your recipient You select 100 his+ transductant colonies for further analysis and score them as having the following genotypes: his+trp+nal R his+trp+nal S his+trp-nal R his+trp-nal S 5 colonies 19 colonies 49 colonies 27 colonies 19.) What is the cotransduction frequency of nal with his? (2pts) How often does his+ also bring in nal S with it... (19+27)/(5+19+49+27)= 47/100= 47% cotransducable 20.) What is the cotransduction frequency of nal with trp? (2pts) Typo. Everybody gets this one for free. 21.) What is the map order of the three genes? Show your reasoning. (6pts) Rarest phenotype is likely to involve the most exchanges. his+trp+nal R 5 colonies his+ nal S trp+ his- nal R trp- This gene order requires that 4 exchanges must occur. The other two possible gene orders would only require 2 exchanges A pathogenic strain has been isolated from a local hospital. They suspect that the strain may be a mutator (ie. it has an abnormally high mutation rate) and they have asked you to determine its 6
mutation rate. Since this bacteria is sensitive to rifampicin, you decide to measure the mutation rate for rif R (risistance to rifampicin) a commonly used and widely accepted measurement of an organism s overall mutation rate. 22.) How would you screen for rif R mutants? Does this screen utilize a positive or negative selection? (6pts) Plate cells on plates containing rifampicin. Cells that grow are rif R mutants. This is a positive selection. The number of rifampicin resistant colonies that you obtained from 30 individual, 1 ml cultures of bacteria are shown in the table. By counting the bacteria from a separate culture, you determine that there were 1 * 10 6 bacteria/ml in each culture. PLATE # of rif R colonies PLATE # of rif R colonies PLATE # of rif R colonies #1 12 #11 22 #21 2 #2 4 #12 16 #22 4 #3 34 #13 33 #23 136 #4 126 #14 14 #24 25 #5 0 #15 0 #25 0 #6 0 #16 0 #26 0 #7 6 #17 1 #27 9 #8 15 #18 22 #28 71 #9 1 #19 1 #29 11 #10 42 #20 11 #30 4 23.) Assuming that the mutation rate, a = m/n, estimate the mutation rate for rif R mutations by using the Poisson expression where the probability of having i mutations per culture is represented by P i = ( m i e -m )/ i!. Is the isolate a mutator? (10pts) 6 out of 30 cultures had 0 mutations, so the probability of having zero mutations (P 0 ) is 6/30, and i=0 mutational events per culture in this situation. P i (6/30) = = ( m i e -m )/ i! ( m 0 e -m )/ 0! 0.2 = ( 1 e -m )/ 1 0.2 = e -m -ln(0.2) = m 1.61 = m 1.61 mutational events per culture and each culture has 1 x 10 8 cells a = m/n a = 1.61 mutational events/1 x 10 6 cells a = 1.61 x 10-6 rif R mutational events per cell division This is a high mutation rate for a bacteria so it probably is a mutator. (No points off if you got this part of the question wrong though) 7